Einstein's theories of relativity

[Sam5]

.......that the twin who goes out and then comes back will be younger, and a simple understanding of what an "inertial frame" is will tell you that he can't do that if he stays in the same inertial frame.


There is NO “going out” and “coming back” in SR theory. There is only “relative motion” between the two frames, and in the example we’ve been discussing from the actual theory, they are moving toward each other.

That “Jack and Jill” stuff was made up only so that Jack could experience “acceleration” during his blast off and turn-around, so that the people who first used it could pretend they finally “solved” the paradox. Some of the more recent people claim not to use “acceleration” as an excuse, but they use other things, such as “Oh, well, my goodness, everybody knows that Jack changes frames and that’s what salvages the theory!”

I had at least one physics professor assure me that if Jack and Jill both move relatively in opposite directions and both turn around, and if we leave off any GR acceleration effects, then there will be “no” SR time dilation of either clock.

You still aren’t sure where the frequency shift takes place in the train-observer example, are you? I hope you sit around tonight thinking about it. Say, why don't you use some graph paper to figure it out?

[kilopi]

And if you leave out any consideraton of the “blast off” and “turn around” accelerations, you’ve still got the same SR clock paradox.
Not true (http://mentock.home.mindspring.com/twins.htm). There never has been a real paradox.

[Glom]

Okay Sam5, a few misunderstandings. First, you insist on treating shift as an event in space and time. We all agree that binary stars are detectable through Doppler shift, we simply have no need to assign an space-time event to the shift. Stop trying to argue the case over where the shift occurs because we're not going to agree it occurs at some location. Doppler shift has no position. Don't continue argue the case that it is.

Second, you are suggesting an ethereal medium, call it what you want. The Michelson-Morley experiment demonstrated conclusively that there is no such medium responsible for the propogation of light. Unless you provide us with experimental data to show this ether of yours, you're not going to convince anyone.

[kilopi]

I had at least one physics professor assure me that if Jack and Jill both move relatively in opposite directions and both turn around, and if we leave off any GR acceleration effects, then there will be “no” SR time dilation of either clock.
Is that example symmetric? Did he mean there is no relative time dilation?

[daver]

True, but there is an impression given in the 1905 theory, and this impression tends to be encouraged by certain physics professors, that the two spaceships are in two different ?worlds? at the same time. There is an impression given by the theory that there are two separate sets of ?existences? simultaneously, one that is ?contracted? and one that is not. And in fact, that is the very reason Einstein had to clear up the matter in his 1907 paper, because so many guys like me were complaining about his ambiguity in the 1905 paper.

SR gives a set of equations for transforming measurements in one reference frame into measurements in another. If you measure your spacecraft as 1 mile long and 1 mile wide I can apply the transforms to see which dimensions i'd measure your spaceship as. Your spaceship doesn't exist in two different worlds, but your measurements aren't necessarily equivalent to my measurements.


If in 1905 he had said, ?This length contraction isn?t real?, then that would have been fine. But if he had said that, then his clocks would not "really? time dilate. And that would have been even better, and we wouldn?t have wound up with the twins paradox.

But the clocks do time dilate, and we've known that for sure ever since particle accelerators started pumping things up to nearly the speed of light.


As I have mentioned, the biologists have known of it a long time, and in fact the latest type of time discussed in biology papers tends to be ?thermodynamic time?, which is not related to relative motion or acceleration.

That is also not related to what physicists mean by "time dilation". I don't know if biochemists refer to it as "time dilation", if they do then they're using it in a different sense than physicists.


What a lot of physics professors need to do today (and I hope I don?t offend anyone by saying this) is read a lot of biology ?thermodynamic time? papers.
No, they don't. The behavior of biological (or chemical) systems under low temperatures has nothing to do with relativistic time contraction.

[daver]

I'm lazy--i'm not going to look up your source. I'm assuming observer A watches observer B zoom by, each in an unaccelerated reference frame.. Observer A thinks Observer B's clock is running slowly. Observer B thinks Observer A's clock is running slowly.

Nobody in an unaccelerated reference frame ever thinks time in any other unaccelerated reference frame is moving quicker.

Right, but in Einstein?s 1905 thought experiment that started all of this ?twins paradox? debate, 98 years ago, he said that only one of the clocks winds up really time dilated. That is the paradox. Both ?see? the other time dilate, but when they unite, only one is ?really? time dilated, and that?s impossible, under the terms set forth in that theory.

No. This crops up again and again. The twin paradox is only an apparent paradox--there are three frames of reference involved, and if you're careful about maintaining a consistent frame of reference everything dovetails together quite nicely.


Uhh, do you know anything about how brains think and how the left side interacts with the right side?
Very little; i read some popularizations of the experiments with people whose corpus callosum had been severed for one reason or another, or where information was presented to one eye (or ear) and the like. I don't see how it pertains to SR or the perception of a paradox (I suppose you might be able to come up with something about how non-intuitive SR can be under certain circumstances, so right-brained people just can't glom onto what the left brain is trying to tell them, but that seems way too pop-psych).

[daver]

You also know that if Jack and Jill both ?blast off? in the opposite directions, go the same distance, then ?turn around?, their atomic clocks will both slow down exactly alike, as per GR, but during their unaccelerated ?relative motion? SR requires that they ?see? each other?s clocks ?slow down? the same amount, as per SR, and when they get back to earth SR also requires that only one of their clocks ?lag behind? due to SR. And in this experiment we have both of them doing exactly the same thing. And if you leave out any consideraton of the ?blast off? and ?turn around? accelerations, you?ve still got the same SR clock paradox.

Jack blasts off going east at 90%c, Jill blasts off at the same time going west at 90%c. Jane stays home. After 1 hour (ship time), Jack and Jill turn around and return (again at 90%c). Jack and Jill are the same age (two hours older than when they left), Jane is 4.6 hours older. I don't see the paradox.

[Sam5]

Okay Sam5, a few misunderstandings. First, you insist on treating shift as an event in space and time. We all agree that binary stars are detectable through Doppler shift, we simply have no need to assign an space-time event to the shift. Stop trying to argue the case over where the shift occurs because we're not going to agree it occurs at some location. Doppler shift has no position.

Yes it does. It’s a basic law of physics (and of nature). And just like most everything else, a Doppler shift has a “place” where it occurs, just as the revolving binaries have a “place” where they exist. But if you don’t want to think about where the shift occurs, then that’s your business.


Second, you are suggesting an ethereal medium, call it what you want. The Michelson-Morley experiment demonstrated conclusively that there is no such medium responsible for the propogation of light.

No it did not. Let me tell you something about the MM experiment that you might not know. In the 19th Century, physicists and astronomers didn’t realize that so many stars and galaxies were moving so fast. They didn’t even know that the spiral “nebula” were outside our own Milky Way. They thought that mainly only comets and planets moved, and at fairly slow speeds. So, they thought the “universal ether” was all one big thing, fixed and stationary with a “stationary” universe, and they thought the earth moved “through” that “ether” at 18.6 mps, during its revolution around the sun.

It didn’t occur to them that each and every large astronomical body might regulate the speed of light, locally, to approximately c, depending on the body’s mass, at each body. So, it didn’t occur to them that the earth might be traveling through space with its own local c-regulator, or “local aether”, out to some distance from the earth, where, gradually, the sun’s local c-regulator begins to regulate the speed of light inside our solar system.

A few of them did entertain the idea that maybe the MM results suggested that a portion of the big overall “fixed” universal aether was “entrained” by the earth as the earth moved through it and carried some of it along with the earth, sort of like the way a car sometimes pulls along some smoke when it drives through an area near a forest fire, but they gradually dropped that idea.

Using GR theory, in 1920 Einstein toyed with an idea of local ethers, based on the local “fields” concept, when he wrote:

”Recapitulating, we may say that according to the general theory of relativity space is endowed with physical qualities; in this sense, therefore, there exists an ether. According to the general theory of relativity space without ether is unthinkable; for in such space there not only would be no propagation of light, but also no possibility of existence for standards of space and time (measuring-rods and clocks), nor therefore any space-time intervals in the physical sense. But this ether may not be thought of as endowed with the quality characteristic of ponderable inertia, as consisting of parts which may be tracked through time. The idea of motion may not be applied to it.

Now, if he had gone just a little further and endowed this “ether” with “motion”, so that local areas of it travel through space with the large astronomical bodies that generate the gravitational fields, then he would have had the very same idea that I am suggesting here. Anyway, this is not my idea. I got it from a couple of physics professors.

It is possible that light is “self propagating”, but, something alters its speed when it passes near large masses, and this something appears to be the gravitational fields. The lack of a “concentration” of this something allows light to speed up again when it leaves the near vicinity of those masses. So, light might be “self propagating”, but its speed might be “regulated” by something such as the gravitational fields of massive bodies, and those fields extend everywhere into space, but they are concentrated around the massive bodies.

If the distant galaxies are indeed traveling at several times the speed of light, relative to the earth, and if we believe that light travels at approximately “c” within them, then we must recognize that what we have here at the earth (the local c-regulator) is moving with the earth through space relative to the c-regulators that are moving through space (or with “expanding space”) with those galaxies. In modern cosmology papers, this “local c-regulator” is presently being called “comoving space”, and even that is an “ether theory”, since the space that is “comoving” with a distant galaxy is moving relative to our own local “comoving” space.

I’m not inventing this stuff. I’m just reporting it in a way you’ve never heard it reported before.

[Sam5]

Jack blasts off going east at 90%c, Jill blasts off at the same time going west at 90%c. Jane stays home. After 1 hour (ship time), Jack and Jill turn around and return (again at 90%c). Jack and Jill are the same age (two hours older than when they left), Jane is 4.6 hours older. I don't see the paradox.

The paradox in SR theory is that Jack and Jill will “see” each other age more slowly, due only to relative motion, and they will “see” this while traveling in both directions. The problem is, the theory winds up with only one of them “really” aged. If you’ve got three observers and three frames, in SR theory, all of them are supposed to “see” each other age, while they don’t see themselves age. Jack will “see” Jane and Jill age. Jane will “see” Jack and Jill age. Jill will “see” Jane and Jack age. So when they unite, all three of them will disagree about who aged. The theory requires that all three of them will be “older” than each other.

[Sam5]

No. This crops up again and again. The twin paradox is only an apparent paradox--there are three frames of reference involved, and if you're careful about maintaining a consistent frame of reference everything dovetails together quite nicely.


In SR theory, you can’t be “constant” about maintaining a “frame of reference”, because all frames of reference “see” exactly the same thing happening inside the other frames. If you try to maintain your frame of reference always with “Jack”, you will think you’ve “solved” it, but you will only report what Jack “sees”, and you will disregard what Jill says she sees. You will “see” Jill aging, but Jill will “see” you aging the same amount. So who are we to believe when you and Jill get back together? This doesn’t happen in GR. Only one of you ages in GR, the one who experiences the acceleration, and the one who ages sees the other’s time “speed up”, not “slow down”. This is correct and it contains no paradox.

[daver]

Jack blasts off going east at 90%c, Jill blasts off at the same time going west at 90%c. Jane stays home. After 1 hour (ship time), Jack and Jill turn around and return (again at 90%c). Jack and Jill are the same age (two hours older than when they left), Jane is 4.6 hours older. I don't see the paradox.

The paradox in SR theory is that Jack and Jill will ?see? each other age more slowly, due only to relative motion, and they will ?see? this while traveling in both directions. The problem is, the theory winds up with only one of them ?really? aged. If you?ve got three observers and three frames, in SR theory, all of them are supposed to ?see? each other age, while they don?t see themselves age. Jack will ?see? Jane and Jill age. Jane will ?see? Jack and Jill age. Jill will ?see? Jane and Jack age. So when they unite, all three of them will disagree about who aged. The theory requires that all three of them will be ?older? than each other.
No. The theory says that Jane will have aged more than Jack or Jill; Jack or Jill (before turnaround) could calculate how much time will ellapse for them after turnaround and could conclude that they would be younger than Jane when they meet.

[Sam5]

But the clocks do time dilate, and we've known that for sure ever since particle accelerators started pumping things up to nearly the speed of light.


Particle accelerators accelerate particles. That’s GR.

[SeanF]

.......that the twin who goes out and then comes back will be younger, and a simple understanding of what an "inertial frame" is will tell you that he can't do that if he stays in the same inertial frame.

There is NO “going out” and “coming back” in SR theory. There is only “relative motion” between the two frames, and in the example we’ve been discussing from the actual theory, they are moving toward each other.

Well, if we're talking about a situation in which they are already moving relative to each other at the beginning of the experiment, then there is a simultaneity difference to deal with. As such, it would be meaningless to say they're both the same age at the beginning, or that the clocks are synchronized at the beginning.

If we're talking about a situation where they're not moving relative to each other at the beginning and then they start moving relative to each other, then there's a change in inertial frames to deal with.

In either case, the "paradox" is eliminated.



That “Jack and Jill” stuff was made up only so that Jack could experience “acceleration” during his blast off and turn-around, so that the people who first used it could pretend they finally “solved” the paradox. Some of the more recent people claim not to use “acceleration” as an excuse, but they use other things, such as “Oh, well, my goodness, everybody knows that Jack changes frames and that’s what salvages the theory!”

I had at least one physics professor assure me that if Jack and Jill both move relatively in opposite directions and both turn around, and if we leave off any GR acceleration effects, then there will be “no” SR time dilation of either clock.

He's right. If both of them move off in opposite directions, and then turn around and come back together, they'll both be the same age when they get back together. I'm not real up to speed on GR, but I would imagine that you wouldn't have to "leave off" the GR effects, either - they'd cancel out just like the SR effects.


You still aren’t sure where the frequency shift takes place in the train-observer example, are you? I hope you sit around tonight thinking about it. Say, why don't you use some graph paper to figure it out?
Nah, I think I got that one. Glom's right, it doesn't really happen at either "end", it happens overall because of the relative motion.

[Sam5]

No.

Sorry, I meant to say "age more slowly" rather than just "age".

[Sam5]

Well, if we're talking about a situation in which they are already moving relative to each other at the beginning of the experiment, then there is a simultaneity difference to deal with. As such, it would be meaningless to say they're both the same age at the beginning, or that the clocks are synchronized at the beginning.

Read section 4 again. Both clocks start off "stationary" relative to each other, they are synchronized and synchronous.

[Sam5]

If we're talking about a situation where they're not moving relative to each other at the beginning and then they start moving relative to each other, then there's a change in inertial frames to deal with.


That doesn’t matter. That’s got nothing to do with it. An inertial frame is an inertial frame. You can say that both “change inertial frames” at the same time. It’s the “relative motion” and the “light signals” that cause the observed dilation, and both observers observe the dilation in each other’s frame but not in their own. Paradox.

We’ve been over this several times before.

[Sam5]

He's right. If both of them move off in opposite directions, and then turn around and come back together, they'll both be the same age when they get back together.

No. Re: SR, the “relative motion” and distorted light signals cause the “time dilation” and it is seen by both observes when they look into each other’s frame but not when they look into their own frame. And it is seen in both direction of travel. It doesn’t matter whether they go out come back, etc., etc. Whatever they do, each one sees himself as stationary and the other as moving. Each one sees the other as aging more slowly, but not themselves as aging more slowly.

[Sam5]

Nah, I think I got that one. Glom's right, it doesn't really happen at either "end", it happens overall because of the relative motion.

So you mean “Redshift Happens”, and that’s it, huh? Do you sell bumper stickers that say that?

[Sam5]

Shift does not happen as a result of motion of the source.

Ok, let me ask you what you think about this idea:

You are on a planet in a distant star system. You aim a power laser at the earth and you turn it on. You are 4 light years away, so the light will take 4 years to reach the earth. The light originates “at the source”, your planet, 4 light years away from earth.

Then you start wiggling the laser from side to side. It will take earth people 4 years to notice your wiggle. The wiggle originates “at the source”.

Now you move the laser up and down, away from and toward the earth, generating a Doppler shift in the light. That Doppler shift will take 4 years to reach the earth. The Doppler shift originates “at the source”. Yes or no?

[SeanF]

In SR theory, you can’t be “constant” about maintaining a “frame of reference”, because all frames of reference “see” exactly the same thing happening inside the other frames.

An event that happens will happen "in" every reference frame. There is nothing that happens in one reference frame but doesn't happen in another. However, the question of "where" and "when" something happened is determined by the reference frame the observer is in.

For example, two events, A & B, occur. No matter what reference frame an observer is in, those two events both occur.

When viewed from one reference frame, the two events are simultaneous.

When viewed from a second reference frame, which is in relative motion to the first, event A occurs before event B.

When viewed from a third reference frame, which is in relative motion to the first but in the opposite direction as the second, event B occurs before event A.

Now, it is possible that events A and B directly concern the observers. For example, A might be the birth of a child who is in the third reference frame, while B is the birth of a child who is in the second reference frame.

However, when those two children eventually meet (quick quiz - is it possible, in the situation I have described, that they are moving apart and will never meet?) there will be absolutely no disagreement as to the ages of the children when they pass.

If the second reference frame and third reference frame are moving at the same velocities relative to the first, then an observer in the first reference frame will see the two children aging at the same rate. Since this observer also saw the births as simultaneous, when they pass each other, they will therefore be the same age.

This means that the observer who saw A born first (this would be someone in the second reference frame - perhaps B's mother) must see A aging more slowly than B, so that B can "catch up" to A by the time they pass.

Conversely, the observer who saw B born first (A's mother) would have to see B aging more slowly than A, so that A can "catch up" to B by the time they pass.

Therefore, although these two observers in relative motion both see the other aging more slowly, it is not really true that they see each other identically, and the disagreement over who was born first "cancels out" the disagreement over who is aging faster.

And that is true throughout SR. Disagreements cancel out, and so there are no paradoxes.

(by the way, the quick quiz answer is "nope, it's not.")

[SeanF]

If we're talking about a situation where they're not moving relative to each other at the beginning and then they start moving relative to each other, then there's a change in inertial frames to deal with.

That doesn’t matter. That’s got nothing to do with it. An inertial frame is an inertial frame. You can say that both “change inertial frames” at the same time. It’s the “relative motion” and the “light signals” that cause the observed dilation, and both observers observe the dilation in each other’s frame but not in their own. Paradox.

"At the same time" is questionable, though, ain't it? But you can't just "say" that they changed inertial frames. It's something that has to happen.

We’ve been over this several times before.
That we have, that we have.

[Sam5]

Glom, SeanF,

LINK TO UCLA (http://www.astro.ucla.edu/~wright/doppler.htm)

Ok, here is a link to a UCLA website. How do you guys account for: 1) the shifts originating at the moving light emitter, 2) the waves leaving the emitter in front at a slower speed than c relative to the emitter and in the rear at a higher speed than c, 3) what is this guy using for a “medium”, a “c-regulator” in space?

[Sam5]

Glom, SeanF,

U. of OREGON (http://zebu.uoregon.edu/~soper/Light/doppler.html)

University of Oregon. Same thing. The light shift originates at the moving emitter. The waves out front move slowly relative to the emitter, the waves in back move faster. The c-regulator is embedded into the white background.

[SeanF]

Glom, SeanF,

LINK TO UCLA (http://www.astro.ucla.edu/~wright/doppler.htm)

Ok, here is a link to a UCLA website. How do you guys account for: 1) the shifts originating at the moving light emitter, 2) the waves leaving the emitter in front at a slower speed than c relative to the emitter and in the rear at a higher speed than c, 3) what is this guy using for a “medium”, a “c-regulator” in space?

First of all, I need to know how you account for the fact that the light is black!

What's that you say? It's a "drawing"? A "representation"? You mean that's not what light leaving an emitter really looks like?

Well, I guess that answers my question and yours, doesn't it . . .

(Oh, except for your last one. Since I don't know what "an OmegaM=0.3 vacuum-dominated flat model" is, I have no idea how the speed of light is regulated in one)

[Sam5]

First of all, I need to know how you account for the fact that the light is black!

Hey! Now is not the time to get picky. If you don’t like it, reverse the polarity of your monitor screen. Take it over to PhotoShop and change it from a negative to a positive.

NASA DOT GOV (http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/961029a.html)

Ho ho, NASA dot Gov:

“Now imagine a source of light at a constant distance from us, emitting waves of light at a constant wavelength. Obviously, the wavelength of the waves we receive will be the same constant wavelength at which they are emitted by the source. Suppose now that the source starts moving directly toward us. When the source emits the next wave crest, it will be nearer to us, so the distance we will see between the two wave crests arriving will appear to be smaller than when the star was stationary. This means that the wavelength of the waves we receive will be shorter (or shifted toward the blue end of the spectrum) than when the source was not moving. Similarly, if the source is moving away from us, the wavelength of the waves will appear slightly longer, or shifted toward the red end of the spectrum. The relationship between wavelength and speed is called the Doppler effect. We experience it every day -- like the engine sounds of a car approaching us having a higher pitch...and a lower pitch when the car moving away from us. Light and sound are both waves, so they both exhibit the Doppler effect.”


”When the source emits the next wave crest, it will be nearer to us, so the distance we will see between the two wave crests arriving will appear to be smaller ....”

And this happens at the source, when next wave crest is emitted, after the star starts to move. We will not see that shifted crest until after the number of years it takes the crest to reach us.

[Musashi]

What if we moved at precisely the time the shifted crest was about to it us, making it appear to be no shift? That would mean it can happen at the destination too.

[Sam5]

What if we moved at precisely the time the shifted crest was about to it us, making it appear to be no shift? That would mean it can happen at the destination too.



If we move away, at just the right amount, just as a blueshifted crest reaches us, we would not see a shift at all. But our motion would not affect the waves, the wavelengths, or the frequency of the light that are traveling from the star. The star would never know that we moved.

However, there are some extenuating circumstances that I’ve been trying to figure out for years. I don’t think the earth’s local light speed regulator starts right exactly at the surface of the earth. I think it starts somewhere out in space, at some distance from the earth, sort of along the lines of how the earth’s gravitational field gets weaker the further from earth we go. So, I think we would probably have to gradually speed up as the blueshifted crest approaches near the earth, rather than make a sudden jump. I think that in the overall area of the solar system, the sun’s gravitational field probably regulates the light speed until the crests come within some fairly close distance to the earth. If I could figure out the exact distances and equations, I’d certainly try to publish this idea as a paper, but I don’t know the exact distances or equations.

[SeanF]

First of all, I need to know how you account for the fact that the light is black!

Hey! Now is not the time to get picky. If you don’t like it, reverse the polarity of your monitor screen. Take it over to PhotoShop and change it from a negative to a positive.

I suspect my point went over your head. Maybe I should try again...


NASA DOT GOV (http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/961029a.html)

Ho ho, NASA dot Gov:

...snip...
”When the source emits the next wave crest, it will be nearer to us, so the distance we will see between the two wave crests arriving will appear to be smaller ....”

And this happens at the source, when next wave crest is emitted, after the star starts to move. We will not see that shifted crest until after the number of years it takes the crest to reach us.

Yeah, but the first sentence of their answer says, "The fact that a star, or any light source, is moving toward or away from you does not affect what it emits - it affects what you (as the observer) perceive it emitting."

At any rate, that's nothing! Look at this NASA dot Gov (http://www-spof.gsfc.nasa.gov/stargaze/Sundial.htm) page and marvel in the knowledge that the sun actually moves around the Earth.

My goodness, one might think they were trying to keep things simple so as not to confuse people with information unneccessary to the task at hand, wouldn't one?

It's off to bed for me. I'll check in again in the morning.

Happy googling, Sam5!

[Sam5]

Yeah, but the first sentence of their answer says, "The fact that a star, or any light source, is moving toward or away from you does not affect what it emits - it affects what you (as the observer) perceive it emitting."

Right. That means the star will continue to emit light of the same frequencies that are normal for that kind of star. But what we see will be Doppler shifted.

You know, like the train whistle. Lol, did you think the train “really” emitted a higher and a lower pitch, at the whistle?

Do you not understand what a “Doppler shift” is?? Look, I could draw you some pictures on graph paper if you think that would help.

[Musashi]

If we move away, at just the right amount, just as a blueshifted crest reaches us, we would not see a shift at all. But our motion would not affect the waves, the wavelengths, or the frequency of the light that are traveling from the star. The star would never know that we moved.


You know, like the train whistle. Lol, did you think the train “really” emitted a higher and a lower pitch, at the whistle?

So.. is doppler shift at the source? At the destination? Both? Neither?

[kilopi]

You know, like the train whistle. Lol, did you think the train “really” emitted a higher and a lower pitch, at the whistle?
I did. Well, very near the whistle, anyway.

[Sam5]

So.. is doppler shift at the source? At the destination? Both? Neither?


In my personal opinion, when a source moves through a propagating medium, the shift occurs at the source, and the shift will require some time before the observer sees it, depending on how far away the observer is from the source and also depending on the speed of the wave.

And in my opinion, when an observer moves through a propagating medium, the shift occurs at the observer, and he sees it instantly.

[SeanF]

So.. is doppler shift at the source? At the destination? Both? Neither?


In my personal opinion, when a source moves through a propagating medium, the shift occurs at the source, and the shift will require some time before the observer sees it, depending on how far away the observer is from the source and also depending on the speed of the wave.

And in my opinion, when an observer moves through a propagating medium, the shift occurs at the observer, and he sees it instantly.

Okay, Sam5, bottom line re: Doppler shift.

What you have described in the quoted above is a pretty good description of "classical" physics predictions on when Doppler shifts will be observed.

SR's predictions are that if the emitter at the time of emission and the observer at the time of observation are in the same inertial frame, no Doppler. If they're in different inertial frames, Doppler.

So, if you can posit a situation in which "classical" predicts Doppler and SR doesn't, or vice-versa, then we've got something to talk about. The fact that SR and "classical" predict different causes for the shift(s) is meaningless, unless there's some way to detect those causes directly.

So, describe an experiment using Doppler which would give different results if SR is true than it would if SR is not true and we'll talk about it. Otherwise, Doppler's not going to lead us anywhere, so we might as well drop it.

[Sam5]

Okay, Sam5, bottom line re: Doppler shift.

What you have described in the quoted above is a pretty good description of "classical" physics predictions on when Doppler shifts will be observed.

SR's predictions are that if the emitter at the time of emission and the observer at the time of observation are in the same inertial frame, no Doppler. If they're in different inertial frames, Doppler.


Man, you don’t need SR for that! That’s the same thing Doppler discovered back in 1842! With the guy on the rear of the train in my sound example, he hears no Doppler shift from the moving train whistle, because he’s moving with the whistle, although both are moving through the propagating medium of the air. If the guy is on the ground or a different train, he hears the Doppler shift.

This is not Einstein stuff, it’s classical Doppler stuff.

[SeanF]

Okay, Sam5, bottom line re: Doppler shift.

What you have described in the quoted above is a pretty good description of "classical" physics predictions on when Doppler shifts will be observed.

SR's predictions are that if the emitter at the time of emission and the observer at the time of observation are in the same inertial frame, no Doppler. If they're in different inertial frames, Doppler.


Man, you don’t need SR for that! That’s the same thing Doppler discovered back in 1842! With the guy on the rear of the train in my sound example, he hears no Doppler shift from the moving train whistle, because he’s moving with the whistle, although both are moving through the propagating medium of the air. If the guy is on the ground or a different train, he hears the Doppler shift.

This is not Einstein stuff, it’s classical Doppler stuff.

Okay, I thought you were talking about Doppler shift because you thought that SR's predictions of it wouldn't match reality. I guess I was wrong.

So why were you talking about Doppler shift?

[Sam5]

Okay, I thought you were talking about Doppler shift because you thought that SR's predictions of it wouldn't match reality. I guess I was wrong.

So why were you talking about Doppler shift?


Because when you discussed Einstein’s moving train observer experiment, you claimed the Doppler shift he saw would take place “at the source”, and I wanted to educate you abut Doppler theory. After I did, you finally realized you were wrong about the shift taking place “at the source”.

[SeanF]

Okay, I thought you were talking about Doppler shift because you thought that SR's predictions of it wouldn't match reality. I guess I was wrong.

So why were you talking about Doppler shift?


Because when you discussed Einstein’s moving train observer experiment, you claimed the Doppler shift he saw would take place “at the source”, and I wanted to educate you abut Doppler theory. After I did, you finally realized you were wrong about the shift taking place “at the source”.

Well, before you get too full of yourself:

First, I didn't say the Doppler shift he saw would take place "at the source," I said the Doppler shift he saw would appear from within his reference frame to have taken place "at the source.'

Second, that is perhaps a tad simplistic, but it isn't actually "wrong."

Third, that realization came more from what Glom said than what you said.

Fourth, I haven't realized that you were right about anything, because what I did finally conclude about Doppler still isn't what you said Doppler shift is.

[Glom]

Third, that realization came more from what Glom said than what you said.

What did I say?

This is so cool! I'm actually contributing to a controversial thread.

[SeanF]

Glom, you said this:


Okay Sam5, a few misunderstandings. First, you insist on treating shift as an event in space and time. We all agree that binary stars are detectable through Doppler shift, we simply have no need to assign an space-time event to the shift. Stop trying to argue the case over where the shift occurs because we're not going to agree it occurs at some location. Doppler shift has no position. Don't continue argue the case that it is.

That was when I realized why I was having problems visualizing what would happen if the emitter/receiver relative motion changed while the light was already in transit.

(And I greatly appreciate it!) :D

[Musashi]

My take on doppler shift is that it you have to know what is going on with the sender and the reciever. It is an interaction between the two, and as such it does not happen at the source or the destination. This reminds me of the 'errata' thread about gravity...

Take the train whistle. What if I am on my bicycle and I am riding towards a stationary train that is blowing its whistle? I will observe doppler shift, won't I? Now, I keep riding and come to a railroad crossing. The arms are down so I stop. Now a train approaches, and I observe a shift again. In each of these cases, where was the shift? Now, I turn my bike onto a path the parallels the tracks and as I ride, another train approaches going the opposite direction. The whistle tone doppler shifts again, this time higher than the two previous times. Finally a train approaches me from behind, heading in the same direction I am. I pump my little legs and match speeds with the train (it's no going that fast). The engineer blows his whistle and there is no shift. Now I slow down and the whistle tone begins to shift. Where is the doppler?

[daver]

So when they unite, all three of them will disagree about who aged. The theory requires that all three of them will be ?older? than each other.

Again, no. Jack and Jill each will have been in two reference frames, measurements in one reference frame are not necessarily valid in another. Pretend that we have two Jacks--the outbound Jack and the inbound Jack. The outbound Jack thinks that Jane is moving slowly, and thinks that the inbound Jack is moving very slowly. The inbound Jack thinks the same about Jane and the outbound Jack. If the outbound Jack had passed a stopwatch to the inbound Jack, the inbound Jack would have thought the stopwatch had been running very slowly before it was passed, and would not be at all surprised when it showed less total elapsed time when he finally meets up with Jane and her stopwatch. The outbound Jack thinks that the stopwatch after the handoff is running very slowly and again would not be surprised when he hears that the watch is slow compared to Jane's stopwatch.

The situation changes only slightly when you get rid of the stopwatch and just have Jack turn around--measurements in one reference frame are not valid in another reference frame.

[daver]

In SR theory, you can?t be ?constant? about maintaining a ?frame of reference?, because all frames of reference ?see? exactly the same thing happening inside the other frames.

Roughly, yes. And all the differences cancel out, so that Jack thinks that Jill will be the same age, Jill thinks that Jack will be the same age, Jack and Jill think that Jane will be older, and Jane thinks that Jack and Jill will be younger.

You should try setting up a sample problem and cranking through the numbers. In the Jack, Jill, and Jane case you have five reference frames (really only three--outbound Jack and inbound Jill are in the same frame) to play with. You'd see that an observer in any of the five (three) frames would see exactly the same thing by the time everyone joins up (although they might disagree about the ages at any time before they joined--that's the simultaneity issue).

[Sam5]

Well, before you get too full of yourself:

First, I didn't say the Doppler shift he saw would take place "at the source," I said the Doppler shift he saw would appear from within his reference frame to have taken place "at the source.'


No. I asked you:

Page 8:

"And if I asked you where the “blueshift” and "redshift" originate for the moving observer, you will say “at the source”, rather than “at the observer”??"

And you answered:

“The observer on the train sees one light source as moving away from him and the other towards him, so the Doppler shift he measures would originate "at the sources."”

You didn’t say he would “think” it originated at the sources or that he would “see” it originate at the sources. You said it would originate “at the sources”.

Then I asked everyone else on:

Page 9:

“And let me ask you this, in the train thought experiment we have been discussing, do you think the shift that the moving observer sees occurs “at the source” or “at the moving observer”?”

And no one answered the question correctly.

I’ve warned you several times about what you are doing. You are not looking at the overall situation. You are trying to separate the two frames too much, and you are thinking that one thing happens in one frame, while something different happens in the other frame. But this isn’t true. Both frames are inter-related.

In the train example, both observers are sharing one light propagating medium. The embankment observer sees the two flashes at the same time, unshifted, because the medium is stationary relative to him. The moving observer sees the flashes at different time, shifted, because he is moving through the stationary medium. The blueshifts and redshifts he sees take place at him, not at the sources.

[daver]

But the clocks do time dilate, and we've known that for sure ever since particle accelerators started pumping things up to nearly the speed of light.


Particle accelerators accelerate particles. That?s GR.
Particles after collision aren't accelerated (appreciably). The decay times of the debris indicate that fast moving particles are time dilated. That's not GR, that's SR.

[daver]

So, describe an experiment using Doppler which would give different results if SR is true than it would if SR is not true and we'll talk about it. Otherwise, Doppler's not going to lead us anywhere, so we might as well drop it.

The amount of the Doppler effect would change depending on whether SR is true. I don't know if Sam5 would accept this as proof of time dilation in unaccelerated reference frames or not.

[Sam5]

My take on doppler shift is that it you have to know what is going on with the sender and the reciever. It is an interaction between the two, and as such it does not happen at the source or the destination.


No, it does originate either at the source or the destination. If the whistle moves through the air, the shifts take place at the whistle because the wavelengths are compressed in the air in one direction and stretched out in the air in the opposite direction. And this takes place at the whistle. If the observer moves through he air, the shifts take place at the observer. The wavelengths are normal, but the observer is receiving them at a faster or slower rate. And of course you can have both whistle and observer moving and have shifts taking place at both the whistle and the observer at the same time.

[Musashi]

Sounds good. So instead of neither, it is either, as the situation warrants.

[Sam5]

daver, give me a source for your particle decay time dilation information and I'll study it.

[Sam5]

Daver,

In the SR theory, the rates of the clocks depend not on what’s going on inside the clocks, but what’s going on outside them. Einstein adjusts the rates of the “moving” clocks by means of his distorted light signals. In SR theory, the propagating medium that he uses is always stationary with his “stationary” frame, while his “moving” frame moves through it. This is why the “stationary” observer sees normal clock rates in his own frame but “dilated” rates in the “moving” frame. He is adjusting the “moving” frame clocks based on the frame’s “motion” through the propagating medium that is stationary with the “stationary” observer.

So the internal clock rates don’t change in the “moving” frame. But Einstein adjusts their rates, based on what the “stationary” observer “sees” the light signals do in the “moving” frame. Since the “moving” frame is moving through the “stationary” frame’s stationary ether, he adjusts the “moving” clocks to run at a rate that is different than the rate of the “stationary” frame clocks. So, he’s using the distorted light signals to adjust the rates of the “moving” clocks. They don’t change just due to “relative motion”. Einstein mathematically changes their rates when he sees those distorted light signals while he sees himself inside the "stationary" frame.

I’m surprised high school physics classes don’t teach all this detailed stuff about Doppler shift. All they do is give equations to calculate the amount of shift, but they don’t explain how, why, and where all the shifts take place.

University physics classes avoid this kind of discussion when they discuss Doppler shifts of light, because if they discussed it in detail, they’d have to admit there is something that regulates the speed of light in space, some kind of “ether”.

[SeanF]

Well, before you get too full of yourself:

First, I didn't say the Doppler shift he saw would take place "at the source," I said the Doppler shift he saw would appear from within his reference frame to have taken place "at the source.'


No. I asked you:

Page 8:

"And if I asked you where the “blueshift” and "redshift" originate for the moving observer, you will say “at the source”, rather than “at the observer”??"

And you answered:

“The observer on the train sees one light source as moving away from him and the other towards him, so the Doppler shift he measures would originate "at the sources."”

Exactly. Your question set the context ("for the moving observer"), and my answer ("the Doppler shift he measures") was within that context.


I’ve warned you several times about what you are doing. You are not looking at the overall situation. You are trying to separate the two frames too much, and you are thinking that one thing happens in one frame, while something different happens in the other frame. But this isn’t true. Both frames are inter-related.

Again, nothing "happens" in one frame and doesn't "happen" in the other. It's a matter of where and when something happens, not if it happens.


In the train example, both observers are sharing one light propagating medium. The embankment observer sees the two flashes at the same time, unshifted, because the medium is stationary relative to him. The moving observer sees the flashes at different time, shifted, because he is moving through the stationary medium. The blueshifts and redshifts he sees take place at him, not at the sources.

There is no light propagating medium. Light doesn't require one - it's not a sound wave. It simply moves through space.


In the SR theory, the rates of the clocks depend not on what’s going on inside the clocks, but what’s going on outside them. Einstein adjusts the rates of the “moving” clocks by means of his distorted light signals. In SR theory, the propagating medium that he uses is always stationary with his “stationary” frame, while his “moving” frame moves through it. This is why the “stationary” observer sees normal clock rates in his own frame but “dilated” rates in the “moving” frame. He is adjusting the “moving” frame clocks based on the frame’s “motion” through the propagating medium that is stationary with the “stationary” observer.

You do not understand what is happening in Einstein's SR experiments. You're still under your misconception of the "theory" and the "examples." You still think there's some kind of light-propogating medium involved, and there isn't.

When Einstein says "We'll consider Clock A to be stationary and Clock B in motion," your brain immediately says "Okay, A is not moving through the 'medium' and B is moving through the 'medium.'" That's wrong (whether it's "wrong" in reality or not, it's "wrong" in Relativity), and that's why you keep ending up someplace else from where Einstein and the rest of us are.

[Wally]

Alright, campers. Let's start from scratch.

Fred resides on Earth and observes as his twin George hijacks the C-ship Lorentz and flies it away at v. We'll ignore the powerful and brief acceleration period. So Fred resides in inertial frame S while George resides in inertial frame S', which is travelling at speed v relative to S.

As Fred observes from frame S, he sees the Lorentz travel a distance x and then come to a stop relative to him. In S, a time t has elapsed. But what time, t', has elapsed for George? Use the Lorentz transformation.

t' = t sqrt(1-v²/c²)

Since 1-v²/c² must always be less than 1 for real v, this means that t' < t and hence George hasn't experienced as much time as Fred. Hence George is younger.

The Lorentz turns around, quickly gets up to speed v again, heading home and again travels x in time t relative to S and hence the duration of the return leg is the same as the outbound leg for the respective frames. You get the idea, George arrives home younger than Fred.

Now! What if we observe this from George's perspective? :-s

George observes Fred along with Earth, and the rest of the local area in fact, heading away from him at speed v. Because of this, everything starts squashing up according to the length contraction.

x' = x sqrt(1-v²/c²)

Clearly, x' < x.

Earth is observed to move away at speed v. Hence, the time of travel observed by George is

t' = x'/v = x sqrt(1-v²/c²)

Since x' < x, this means that t' < t, (and ditto for the return journey) which was also implied when observing from Fred's perspective.

Hence, the paradox is explained without the need for GR.

again, I haven't read ahead of the above quote, but want to comment before I lose my spot.

Glom, you left out the part where, when George "turns around" to head back to earth, he observes Fred's clock immediately leap ahead of his own. During the return trip, he still perceives it to run slow, but it'll always still be ahead of his. Hence when he gets back to earth, there's no disagreement that Fred has aged more than he.

[Wally]

All righty, I'm back! Anybody miss me? :)

Yes, we missed you very much.


First things first. The thought experiments used in Einstein's original papers and elsewhere are not the Theory of Relativity. They are merely demonstrations of it. SR itself can be summed up in less than 20 words:

"The speed of light in a vacuum (c) is the same for all observers in inertial frames."

Surely you don’t believe that. Light slows down when it passes near the sun. Light leaves a distant high speed galaxy traveling at about c relative to the galaxy but less than c relative to us. The light speeds up on its way to us and arrives at us traveling at about c relative to us. Light gets sucked into a “black hole” and can’t leave a “black hole”. There’s no evidence that light always has the same speed in a vacuum in deep space, and there is plenty of evidence that it does not.
.

Sam. You're tossing GR into the conversation where GR is not needed or warranted. Leave gravity wells out of the picture, and sure 'nuff, the speed of light in a vacuum is indeed the same in all inertial frames of reference. that is the essence of SR.

[Sam5]

Exactly. Your question set the context ("for the moving observer"), and my answer ("the Doppler shift he measures") was within that context.

But you were wrong and you are just refusing to admit it. The Doppler shifts “he measures” originate at him, not at the sources. He doesn’t necessairly know where they originate. He knows he sees them, and we should know – if we understand Doppler theory – that they originate at him, not at the sources. But you said they ”would originate at the sources” You didn’t say “he would see them as originating at the sources” or "as if they originated at the sources". You said they did originate at the sources. Then later you changed your mind and you said you weren't sure where they originated. You said maybe they took place all along the pathway of the light, or something like that.

Look, I've got to go out into the real world right now, so I'll get back to you later. In the mean time, I suggest that you study Doppler theory and learn how to understand it. Perhaps if you used some graph paper and drawings that would help. Let the graph background of the paper represent the propagating medium.

[Musashi]

What if there is no propigating medium? Do I use non-existent graph paper?

[Sam5]

Wally, in SR Einstein manipulates the “rate” of the “moving” clocks mathematically, by means of his method of adjusting the “normal” clock rate in the “stationary” system, when he sends out a beam of light from point A to point B and has it reflect back to point A. In the “stationary” system with a stationary light propagating medium, the light travel time from A to B and from B to A are the same.

But when the “moving” frame moves through that “stationary” medium, the travel times in the “moving” frame, “as seen by” the “stationary” frame are NOT the same. So Einstein uses an equation to ADJUST the rates of the “moving” frame clocks. They don’t “slow down” due to the “relative motion” or due to what’s taking place inside the clocks. He adjusts their rates mathematically, based on what he says about the bouncing light signals, so that mathematically they appear to be running slow. It is He, Einstein, who changes the rates of the “moving” clocks, it’s not anything physical that happens to or inside the clocks.

The paradox results when we switch frames and the light propagating medium is suddenly stationary with the other frame. That’s why no one can decide which clock “really” slows down, when we switch our point of view from one frame to the other.

I’ve got to go out for a while.

[Sam5]

What if there is no propigating medium? Do I use non-existent graph paper?



You can use plain white paper, but you’ll find that when you start moving observers and emitters around, you’ll need a background grid to measure the distances and the amount of Doppler shift. You can use just a ruler and plain white paper, but sooner or later you know that the white paper itself represents the light propagating medium, and I think it will help you to go ahead and put grid lines on it.

Notice that when the Doppler shift takes place at the emitter, the light waves will move away from the emitter slower in one direction and faster in the opposite directions, in a c – v and c + v manner, relative to the emitter.

[SeanF]

Exactly. Your question set the context ("for the moving observer"), and my answer ("the Doppler shift he measures") was within that context.

But you were wrong and you are just refusing to admit it. The Doppler shifts “he measures” originate at him, not at the sources. He doesn’t necessairly know where they originate. He knows he sees them, and we should know – if we understand Doppler theory – that they originate at him, not at the sources. But you said they ”would originate at the sources” You didn’t say “he would see them as originating at the sources” or "as if they originated at the sources". You said they did originate at the sources. Then later you changed your mind and you said you weren't sure where they originated. You said maybe they took place all along the pathway of the light, or something like that.

Sam5, you've still got it stuck in your head that the railcar is moving and the embankment is stationary. That is not absolutely true. It can be considered to be true in the classical sense, in which case the Doppler shift would occur at the observer.

However, it is equally valid and true to consider the embankment to be moving and the railcar to be stationary. From that consideration, in the classical sense, the Doppler shift occurs at the source, which is what I said.

When, in discussing SR, a phrase like "for the railcar observer" or "from the railcar observer's frame of reference" is used, that is understood to mean "from the frame of reference in which the railcar observer is motionless." If you understood SR, you would understand that, and you would understand what we're trying to tell you.

You expect us all to think about this experiment like you are, that there's some magical mystical aether controlling the light and you can determine that the railcar's moving through it and the embankment isn't. But nobody here who knows Relativity is going to do that. We're going to deal with the situation the way it was intended to be dealt with when it was presented - under the postulates of the Theory of Special Relativity.



Look, I've got to go out into the real world right now, so I'll get back to you later. In the mean time, I suggest that you study Doppler theory and learn how to understand it. Perhaps if you used some graph paper and drawings that would help. Let the graph background of the paper represent the propagating medium.

Look, I will state here and now, without concern, that in a universe with a "light-propogating medium" such as you have alluded to in several posts Doppler effect will be produced exactly how and where and when you say it will.

But that ain't this universe.

[Musashi]

You can use plain white paper, but you’ll find that when you start moving observers and emitters around, you’ll need a background grid to measure the distances and the amount of Doppler shift. You can use just a ruler and plain white paper, but sooner or later you know that the white paper itself represents the light propagating medium, and I think it will help you to go ahead and put grid lines on it.

But what if I do not believe there is any light-propigating medium? Do I use imaginary paper to draw on, or am I allowed to use regular graph paper? Sooner or later you know the paper represents the medium, but that is really just your opinion. I can use the graph paper all day long and still beileve (read know if you want) that it does not represent and light-propigating medium.

[Eroica]

Glom, you left out the part where, when George "turns around" to head back to earth, he observes Fred's clock immediately leap ahead of his own. During the return trip, he still perceives it to run slow, but it'll always still be ahead of his. Hence when he gets back to earth, there's no disagreement that Fred has aged more than he.
That's not right. When George turns around to head back to Earth, he observes Fred's clock immediately speed up. During the return trip Fred's clock will appear to George to be running faster than George's clock. Fred, however, will only see George's clock speed up when Fred "sees" George turn for home (which will be some time after George actually turns for home).

[Glom]

Glom, you left out the part where, when George "turns around" to head back to earth, he observes Fred's clock immediately leap ahead of his own. During the return trip, he still perceives it to run slow, but it'll always still be ahead of his. Hence when he gets back to earth, there's no disagreement that Fred has aged more than he.

Ah, the enigmatic element shows itself. I still have to grasp the reason for this clock jump.

[SeanF]

Glom, you left out the part where, when George "turns around" to head back to earth, he observes Fred's clock immediately leap ahead of his own. During the return trip, he still perceives it to run slow, but it'll always still be ahead of his. Hence when he gets back to earth, there's no disagreement that Fred has aged more than he.

Ah, the enigmatic element shows itself. I still have to grasp the reason for this clock jump.

Glom, it's the simultaneity issue of SR. Consider two clocks, A and B, motionless relative to each other. From that reference frame, they are located one light-minute apart and are synchronized.

Now, when viewed from a reference frame that is moving at 0.6c relative to those two clocks, they will appear:

A) to be only 4/5 of a light-minute apart.
B) to be running at 80% normal rate
C) to be un-synchronized by 36 seconds

Now, the important thing about C is that it is somewhat direction-dependent. If our moving reference frame is going in the direction A -> B, then B will appear to be 36 seconds ahead of A. If, however, the moving reference frame is going in the direction A <- B, then B will appear to be 36 seconds behind A.

If we consider our traveling twin's trip is A to B and back again, then he'll be in the A -> B frame on the way out and the A <- B frame on the way back. So when he switches at B, A switches from being 36 seconds behind B to being 36 seconds ahead of B (B can't change, because the observer is right there, so A has to).

I don't think I'm explaining it very well. It's a little easier on paper. Maybe we could borrow some of Sam5's graph paper . . . :)

[Glom]

That works. It's just that I don't quite grasp why they are un-synchronized.

[SeanF]

That works. It's just that I don't quite grasp why they are un-synchronized.

Hmm. If we bounce a light signal:

1. Leaves Clock A at A=12:00:00
2. Hits Clock B at B=12:01:00
3. Returns to Clock A at A=12:02:00 {I had to edit this post to fix this line. Stupid, stupid, stupid!}

What does A= when event 2 occurs? Figure this if the two clocks are stationary, then figure it if the two clocks are co-moving in one direction, then figure it if the two clocks are co-moving in the other direction.

[daver]

daver, give me a source for your particle decay time dilation information and I'll study it.
Do a google search of particle decay time dilation; you'll get a bunch of hits. I looked at the fourth one down, a report of a study of decay rates of mu mesons produced by cosmic rays striking the upper atmosphere.

[kilopi]

The "Twin Paradox" is this exact same thing with the "stay-at-home" twin as A and the "traveling" twin as B in the first half and C in the second. Since you don't need GR for this, why would you need it for the "Twin Paradox"?
I think I answered that here:

http://mentock.home.mindspring.com/twin2.htm

[SeanF]

The "Twin Paradox" is this exact same thing with the "stay-at-home" twin as A and the "traveling" twin as B in the first half and C in the second. Since you don't need GR for this, why would you need it for the "Twin Paradox"?
I think I answered that here:

http://mentock.home.mindspring.com/twin2.htm

So GR says that the 3.6 years is due to the time dilation caused by the gravity field-equivalent acceleration? SR says the 3.6 years is due to the simultaneity difference between the reference frames. So I don't know that you've told me you need GR to explain the 3.6 years, have you?

BTW, at the end of that page, discussing the gLT/c^2 equation, it says:


Dynamic calculations, which take account of the varying distance between Bob and Ann, would produce the same value, but would be much more complicated. Interestingly, the value 3.6 years does not depend upon the time T, as both factors cancel.

First, mathematically, the equation breaks down if T=0, doesn't it?

Secondly, in regards to the Dynamic calculations . . . in order to get the 3.6 years would you have the varying distance start at 3 light-years, end at 3 light-years, or average 3 light-years?

[daver]

Daver,

In the SR theory, the rates of the clocks depend not on what?s going on inside the clocks, but what?s going on outside them. Einstein adjusts the rates of the ?moving? clocks by means of his distorted light signals. In SR theory, the propagating medium that he uses is always stationary with his ?stationary? frame, while his ?moving? frame moves through it. This is why the ?stationary? observer sees normal clock rates in his own frame but ?dilated? rates in the ?moving? frame. He is adjusting the ?moving? frame clocks based on the frame?s ?motion? through the propagating medium that is stationary with the ?stationary? observer.

The speed of light is constant in all inertial reference frames. There is no medium, no preferred reference frame. If you are arguing differently, then you aren't arguing either SR or GR, and the discussion should be taken to Against the Mainstream.


So the internal clock rates don?t change in the ?moving? frame. But Einstein adjusts their rates, based on what the ?stationary? observer ?sees? the light signals do in the ?moving? frame. Since the ?moving? frame is moving through the ?stationary? frame?s stationary ether, he adjusts the ?moving? clocks to run at a rate that is different than the rate of the ?stationary? frame clocks. So, he?s using the distorted light signals to adjust the rates of the ?moving? clocks. They don?t change just due to ?relative motion?. Einstein mathematically changes their rates when he sees those distorted light signals while he sees himself inside the "stationary" frame.

I?m surprised high school physics classes don?t teach all this detailed stuff about Doppler shift. All they do is give equations to calculate the amount of shift, but they don?t explain how, why, and where all the shifts take place.

High school physics classes taught Doppler through a medium (air). This was pretty straightforward. My class didn't go into SR, though it could have--the pertinent equations are extremely simple (i think the SR equations fit on the bottom half of the first page of my second-year physics book, though they spent a few more pages on it just to make sure we understood).


University physics classes avoid this kind of discussion when they discuss Doppler shifts of light, because if they discussed it in detail, they?d have to admit there is something that regulates the speed of light in space, some kind of ?ether?.
No. SR doesn't require an ether, the relativistic doppler equation follows pretty easily from the classical doppler with relativistic time dilation.

[Sam5]

Sam5, you've still got it stuck in your head that the railcar is moving

Uhh, yes. Einstein said so right here in the book:

“Just when the flashes (as judged from the embankment) of lightning occur, this point M1 naturally coincides with the point M but it moves towards the right in the diagram with the velocity v of the train.”

Surely you aren’t going to dispute Einstein, are you?? He says the train “moves towards the right in the diagram” and I take his word for it. He also says:

“We suppose a very long train travelling along the rails with the constant velocity v and in the direction indicated in Fig 1. People travelling in this train will with a vantage view the train as a rigid reference-body (co-ordinate system); “

So, relative to the embankment, the train is “moving”. Relative to the embankment the two lightening strikes hit the embankment an equal distance from an observer at position M on the embankment. Relative to the train, the train observer is moving toward the origin point of flash B and he is moving away from the origin point of flash A.

Are you saying this is not true??

I don’t understand your confusion. It’s a simple thought experiment.

[Sam5]

However, it is equally valid and true to consider the embankment to be moving and the railcar to be stationary. From that consideration, in the classical sense, the Doppler shift occurs at the source, which is what I said.


No, SeanF. It has been determined that light travels at “c” at the surface of the earth. And the earth IS moving through space, and light STILL travels at “c” at the surface of the earth. That’s what Einstein’s diagram shows. This is why the stationary observer sees the two flashes at the same time. That’s why the moving observer at the surface of the earth sees the B flash first and that’s why he sees the shift, and he sees the shift at himself. You should know there is no “ether wind” at the surface of the earth, so the Doppler shifts in this thought experiment can’t take place at the sources.

[Sam5]

When, in discussing SR, a phrase like "for the railcar observer" or "from the railcar observer's frame of reference" is used, that is understood to mean "from the frame of reference in which the railcar observer is motionless."

The rail car observer is “motionless” relative to the train, but not relative to the source A and B and not relative to the oncoming light waves and not relative to the embankment. The railcar observer is "moving" relative to the embankment and the sources.

Einstein says:

“Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.”

See? He says the moving observer will “see the beam of light emitted from B earlier than he will see that emitted from A”. Why, because he is “hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A”. This is correct. c + v and c – v. And because of the moving observer’s velocity, that’s why he sees the shifts at himself. They occur at him, not at the “sources”. I don't understand why you can't understand this. It's very simple.

[Sam5]

But what if I do not believe there is any light-propigating medium? Do I use imaginary paper to draw on, or am I allowed to use regular graph paper? Sooner or later you know the paper represents the medium, but that is really just your opinion. I can use the graph paper all day long and still beileve (read know if you want) that it does not represent and light-propigating medium.

Have you used the paper yet? Try it and see what happens.

You’ll find that if you have a stationary light observer and a moving light emitter, on the graph paper, when you make a graph of one light photon leaving the emitter, you’ll have to decide if you want it to travel at “c” relative to the observer or “c” relative to the emitter.

[Sam5]

Do a google search of particle decay time dilation; you'll get a bunch of hits. I looked at the fourth one down, a report of a study of decay rates of mu mesons produced by cosmic rays striking the upper atmosphere.

No, you said the time dilation happened inside an accelerator, without the particles accelerating, and I want your source for that.

[SeanF]

Sam5, you've still got it stuck in your head that the railcar is moving

Uhh, yes. Einstein said so right here in the book:


Just to make it clear to everyone, my complete quote was:

Sam5, you've still got it stuck in your head that the railcar is moving and the embankment is stationary. That is not absolutely true. It can be considered to be true in the classical sense, in which case the Doppler shift would occur at the observer.

Now, let's look at what Einstein says, shall we? The last sentence of the second paragraph of Section 3 of "Relativity - The Special and the General Theory" by Albert Einstein:


With the aid of this example it is clearly seen that there is no such thing as an independently existing trajectory (lit. "path-curve" - That is, a curve along which the body moves), but only a trajectory relative to a particular body of reference."

The key words there, Sam5, are "there is no such thing as an independently existing trajectory" - there is no such thing as absolute motion.



“Just when the flashes (as judged from the embankment) of lightning occur, this point M1 naturally coincides with the point M but it moves towards the right in the diagram with the velocity v of the train.”

Glad you put the footnote "as judged from the embankment" in there, Sam5. That's the "particular body of reference" which point M1's trajectory is relative to. (See Einstein's quote above)



Surely you aren’t going to dispute Einstein, are you?? He says the train “moves towards the right in the diagram” and I take his word for it. He also says:

“We suppose a very long train travelling along the rails with the constant velocity v and in the direction indicated in Fig 1. People travelling in this train will with a vantage view the train as a rigid reference-body (co-ordinate system); “

So, relative to the embankment, the train is “moving”. Relative to the embankment the two lightening strikes hit the embankment an equal distance from an observer at position M on the embankment. Relative to the train, the train observer is moving toward the origin point of flash B and he is moving away from the origin point of flash A. (Emphasis mine - SeanF)

That statement I bolded is most certainly not true. "People travelling in this train will with a vantage view the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train." Remember what Einstein said above. "[There is] only a trajectory relative to a particular body of reference." The observer on the train is not moving relative to the train, which is given as his body of reference.



Are you saying this is not true??

I don’t understand your confusion. It’s a simple thought experiment.

I'm not confused, Sam5, you are.

[Normandy6644]

The notion of the light travelling at c+v or c-v is wrong. The only reason the observer on the train sees the light travelling from point B first is because he is judging it from the reference point of on the train which has a velocity moving toward B.


A______|--person--|______B
->v

Nowhere in here does Einstein mention c+v, or c-v. I still don't know where you are getting them from. I agree with you that, at least in this thought experiment, the train must be moving relative to a fixed point not in this frame.

[daver]

Do a google search of particle decay time dilation; you'll get a bunch of hits. I looked at the fourth one down, a report of a study of decay rates of mu mesons produced by cosmic rays striking the upper atmosphere.

No, you said the time dilation happened inside an accelerator, without the particles accelerating, and I want your source for that.
Does that matter? Do you think that time dilation for decay products produced by cosmic rays fits in with your theory but time dilation for decay products from collisions from particle accelerators would somehow invalidate it?

Here's a link from SLAC: http://www2.slac.stanford.edu/vvc/theory/relativity.html ; it talks about tau particles about 4/5 of the way down. If you wanted you could hunt down a few of the more mainstream links and see that they say the same thing. I expect that if you were clever at google you could dredge up some actual papers on the subject.

[daver]

Hey, SeanF--it's your turn for a quote in the Sci Fi quotes thread. Do you want to relinquish your spot?

[Sam5]

Now, let's look at what Einstein says, shall we? The last sentence of the second paragraph of Section 3 of "Relativity - The Special and the General Theory" by Albert Einstein:


With the aid of this example it is clearly seen that there is no such thing as an independently existing trajectory (lit. "path-curve" - That is, a curve along which the body moves), but only a trajectory relative to a particular body of reference."


You'll have to tell me what chapter you are quoting. My book and the website has chapers and parts, but no "sections".

[SeanF]

In the same paragraph we've been discussing, the third paragraph of section nine of Einstein's book, Einstein says:



Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A.

See? From the embankment reference frame, the train observer hits the light from B first because he's moving into it. But from the train reference frame, the train observer hits the light from B first because B happened first.

Why have I been having to explain this to you (over and over again) when you've had Einstein's book in front of you all this time?

[Sam5]

The key words there, Sam5, are "there is no such thing as an independently existing trajectory" - there is no such thing as absolute motion.

I think you are getting confused. The "trajectory" of the train is left to right, relative to the surface of the earth, relative to the embankment. As far as the universe is concerned and other astronomical bodies, the train is going every which way.

[SeanF]

Now, let's look at what Einstein says, shall we? The last sentence of the second paragraph of Section 3 of "Relativity - The Special and the General Theory" by Albert Einstein:


With the aid of this example it is clearly seen that there is no such thing as an independently existing trajectory (lit. "path-curve" - That is, a curve along which the body moves), but only a trajectory relative to a particular body of reference."


You'll have to tell me what chapter you are quoting. My book and the website has chapers and parts, but no "sections".

They're usually referred to as chapters, but I've noticed that Einstein often refers to "the previous section" so I used that terminology. In my book, it's Part I - The Special Theory of Relativity, Section 3 - "Space and Time in Classical Mechanics". It starts out, "The purpose of mechanics is to describe how bodies change their position in space with 'time.'"

[SeanF]

The key words there, Sam5, are "there is no such thing as an independently existing trajectory" - there is no such thing as absolute motion.

I think you are getting confused. The "trajectory" of the train is left to right, relative to the surface of the earth, relative to the embankment. As far as the universe is concerned and other astronomical bodies, the train is going every which way.

Where's the confusion? You're absolutely right. But that means we can't know if the embankment is at "absolute rest" or the train is at "absolute rest" or neither. That's the whole point of Special Relativity. At any rate, he makes that statement in respect to a rock dropped off a bridge from a passing train, so it's certainly going to apply to the other thought experiment as well.

[SeanF]

Hey, SeanF--it's your turn for a quote in the Sci Fi quotes thread. Do you want to relinquish your spot?

Thanks, daver. I completely forgot I'd put an answer in over there! :o In Like Flint, wasn't it?

I'll be over with a quote shortly.

[Sam5]

Relative to the train, the train observer is moving toward the origin point of flash B and he is moving away from the origin point of flash A. (Emphasis mine - SeanF)



That statement I bolded is most certainly not true.

Sorry, I meant to say "relative to the embankment". Please forgive me.

[daver]

The notion of the light travelling at c+v or c-v is wrong. The only reason the observer on the train sees the light travelling from point B first is because he is judging it from the reference point of on the train which has a velocity moving toward B.


A______|--person--|______B
->v

Nowhere in here does Einstein mention c+v, or c-v. I still don't know where you are getting them from. I agree with you that, at least in this thought experiment, the train must be moving relative to a fixed point not in this frame.

Well, you can get c+v and c-v out of a moving traincar example if you want--the light isn't moving at that speed (it always moves at c) but you can get something that looks kind of like a speed of c +- v.

Let's say you have a mirror at the front and back of the train car, and a blip of light bouncing back and forth between the mirrors. When the light is moving from the rear mirror to the front mirror, the front mirror is receeding (according to our stationary observer) from the mirror at speed v, so it takes l/(c-v) seconds for the light pulse to reach the front of the car (where l is the length of the car). Similarly, it takes l(c+v) seconds for the pulse to return, or 2*l*c/(c*c - v*v) seconds for the round trip.

Some interesting things fall out of this; one is that "simultaneous" isn't well defined. To an observer on the train car, the amount of time for the pulse to travel from the rear of the car to the front of the car is the same as the amount of time for it to travel from the front of the car to the rear of the car. To the stationary observer the times can be very different. If you put a couple more mirrors in the car you can get the time dilation and length contraction formulae.

[Sam5]

The notion of the light travelling at c+v or c-v is wrong. The only reason the observer on the train sees the light travelling from point B first is because he is judging it from the reference point of on the train which has a velocity moving toward B.


A______|--person--|______B
->v

Nowhere in here does Einstein mention c+v, or c-v. I still don't know where you are getting them from. I agree with you that, at least in this thought experiment, the train must be moving relative to a fixed point not in this frame.


Ok, I’ll point it out to you in his quote:

“Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.”

That “hastening towards” means Einstein is counting the guy’s train velocity, plus the velocity of light moving toward the train from the front. That “riding on ahead” means Einstein is counting the guys train velocity, plus the velocity of the light moving toward the train from the rear. So, the train observer meets up with the B flash at the additive velocity of c + v, and he meets up with the A flash at the subtractive velocity of c – v.

You have to study Einstein’s text very carefully.

A >>>>>>>>>(>observer>)<<< B

[Sam5]

Where's the confusion? You're absolutely right. But that means we can't know if the embankment is at "absolute rest" or the train is at "absolute rest" or neither. That's the whole point of Special Relativity.

That’s not true. Physicists have finally discovered that when a train and the earth are moving relatively, it is the train that actually “moves”. That’s why the train has the “engine” and the rails don’t have any. They wondered about it for years, but they finally figured it out.

[SeanF]

Where's the confusion? You're absolutely right. But that means we can't know if the embankment is at "absolute rest" or the train is at "absolute rest" or neither. That's the whole point of Special Relativity.

That’s not true. Physicists have finally discovered that when a train and the earth are moving relatively, it is the train that actually “moves”. That’s why the train has the “engine” and the rails don’t have any. They wondered about it for years, but they finally figured it out.

Too bad you haven't.

Put the car on a treadmill and tell me if the engine makes it move.

[Sam5]

Put the car on a treadmill and tell me if the engine makes it move.

The treadmill would make the car move backwards. The "engine" is in the treadmill. You would have to mechanically link the car to the treadmill to keep the car from moving backwards. But then, the car's wheels would turn, but the car wouldn't go anywhere. Why would you want to put the car on a treadmill?

[SeanF]

Put the car on a treadmill and tell me if the engine makes it move.

The treadmill would make the car move backwards. The "engine" is in the treadmill. You would have to mechanically link the car to the treadmill to keep the car from moving backwards. But then, the car's wheels would turn, but the car wouldn't go anywhere. Why would you want to put the car on a treadmill?

But if the car is just sitting on the moving treadmill, it's going to move without running its engine and would need to run its engine in order to remain motionless.

Therefore, if the car is just sitting on a moving planet, it's going to move without running its engine and would need to run its engine in order to remain motionless.

So why are you so certain that "when a train and the earth are moving relatively, it is the train that actually 'moves'"? Clearly, a train running its engines on a moving planet could be remaining motionless, could it not?

[Sam5]

So why are you so certain that "when a train and the earth are moving relatively, it is the train that actually 'moves'"? Clearly, a train running its engines on a moving planet could be remaining motionless, could it not?

Relative to what?

[SeanF]

So why are you so certain that "when a train and the earth are moving relatively, it is the train that actually 'moves'"? Clearly, a train running its engines on a moving planet could be remaining motionless, could it not?

Relative to what?

Hmmm.



Where's the confusion? You're absolutely right. But that means we can't know if the embankment is at "absolute rest" or the train is at "absolute rest" or neither. That's the whole point of Special Relativity.

That’s not true. Physicists have finally discovered that when a train and the earth are moving relatively, it is the train that actually “moves”. That’s why the train has the “engine” and the rails don’t have any. They wondered about it for years, but they finally figured it out.

You claimed here that there was such a thing as absolute rest, and that it was detectable.

Why are you now insisting that describing the train as motionless must be "relative to" something?

[Sam5]

Ok, Sean, do you see that the waves on the right side of the drawing are leaving the emitter at a slower velocity than the waves on the left side of the picture?

LINK TO UCLA (http://www.astro.ucla.edu/~wright/doppler.htm)

The reason for it, the cause, in 19th Century terminology, was known as an “ether wind”. This is what Michelson and Morley tried to find at the surface of the earth in 1886.

[Sam5]

You claimed here that there was such a thing as absolute rest, and that it was detectable.

I don't think so. I think I said that we could tell an "absolute motion" if an object changes velocity and feels acceleration. This would be an absolute motion relative to his earlier velocity.

If there were only one object in the universe, we couldn’t tell if it were moving or not, but if it changed velocity (maybe by means of a rocket), I think we could say that we could tell that it experienced an “absolute motion” relative to its previous velocity.

[Sam5]

Why are you now insisting that describing the train as motionless must be "relative to" something?

I'm not insisting on anything. You are just nitpicking. If I ever said the “train is moving” without saying “relative” to what, then I was talking about the Chapter 9 thought experiment. If you say the train is running its engines on a moving planet and “remaining motionless”, I don’t know what you are talking about. Are you talking about “without the gears engaged”, or “relative to the earth”, or relative to some “distant galaxy”, or maybe you’ve put the train on a treadmill. I have no idea what you are getting at.

[SeanF]

Well, I certainly need some clarification then.



Where's the confusion? You're absolutely right. But that means we can't know if the embankment is at "absolute rest" or the train is at "absolute rest" or neither. That's the whole point of Special Relativity.

That’s not true. Physicists have finally discovered that when a train and the earth are moving relatively, it is the train that actually “moves”. That’s why the train has the “engine” and the rails don’t have any. They wondered about it for years, but they finally figured it out.

What, specifically, is "not true" in what I said?

[Sam5]

What, specifically, is "not true" in what I said?

Moan, groan.... We are talking about the Chapter 9 thought experiment, aren’t we? You used the word “embankment”, didn’t you? That means we are talking about the earth, the railroad tracks, the moving train, a stationary observer on the embankment, a guy on the train that is moving with the train. You said, “we can't know if the embankment is at "absolute rest" or the train is at "absolute rest" or neither”.

That’s wrong. As far as this thought experiment is concerned, the embankment is at rest and the train is moving.

This book was published in 1916. The thought experiment is very clear as to the conditions of what is going on. I think you are getting it mixed up with the 1905 paper, in which two “frames” were in space somewhere, and in that case, we had to refer to just “relative motion”.

If you meet up with a train in space, and you are in space, and you see “relative motion” between you and the train, then NO you can’t say which one is “moving” and which one is at “absolute rest”. And when that happens to you, please let me know. But if you will go down to the nearest Amtrak station, and ask the Station Master, “which is moving, the track or the train”, I’m sure he would be very happy to tell you. But to keep yourself from going bonkers, NEVER ask a physics professor that question.

[kilopi]

So GR says that the 3.6 years is due to the time dilation caused by the gravity field-equivalent acceleration? SR says the 3.6 years is due to the simultaneity difference between the reference frames. So I don't know that you've told me you need GR to explain the 3.6 years, have you?
It depends upon which frame of reference you choose. You absolutely need it if you're going to choose to have the "out and in" twin be considered "motionless". See? That is what is allowed in GR--all reference frames, even noninertial ones. As long as you restrict your analysis to inertial reference frames, you're probably not going to get in trouble, no matter how many inertial frames you use. I know JW disagrees with that, but it seems to be true.


First, mathematically, the equation breaks down if T=0, doesn't it?

Yes and no. What is the value of x/x at zero? If you're allowed to get as close as you like without really getting to zero?

Plus, that is not really a GR equation, but one derived from GR.


Secondly, in regards to the Dynamic calculations . . . in order to get the 3.6 years would you have the varying distance start at 3 light-years, end at 3 light-years, or average 3 light-years?
I'm not sure what you mean? Could you rephrase the question? :)

[Sam5]

SeanF,

Can you answer a question for me now, please? Can you tell me if you notice on the graph of the moving light emitter that the waves on the right side of the chart are moving away from the emitter more slowly than the waves on the left side of the chart? Can you explain why? Can you explain why this drawing shows an “ether wind” if there is no “ether”?

LINK TO UCLA (http://www.astro.ucla.edu/~wright/doppler.htm)

[Musashi]

It isn't moving faster or slower, it has just shifted frequencies.

[Sam5]

It isn't moving faster or slower, it has just shifted frequencies.

Can you copy and post your graph and show me what it looks like? I'll try to draw one up and show you what I'm talking about.

[Celestial Mechanic]

SeanF,

Can you answer a question for me now, please? Can you tell me if you notice on the graph of the moving light emitter that the waves on the right side of the chart are moving away from the emitter more slowly than the waves on the left side of the chart? Can you explain why? Can you explain why this drawing shows an “ether wind” if there is no “ether”?

LINK TO UCLA (http://www.astro.ucla.edu/~wright/doppler.htm)
There is no "ether wind" in the drawing. Each wave front that you see is a circle centered on the point where the emitter was when that wave front was emitted. In a frame where the emitter is stationary, all of these circles would be concentric. The arrow is used as an indicator of the motion of the emitter, not as an indicator of an "ether wind".

[Sam5]

There is no "ether wind" in the drawing. Each wave front that you see is a circle centered on the point where the emitter was when that wave front was emitted. In a frame where the emitter is stationary, all of these circles would be concentric. The arrow is used as an indicator of the motion of the emitter, not as an indicator of an "ether wind".

After a wave is “deposited” into the medium, the medium takes over and controls the speed of the light, which determines the rate of the expansion of the circles and the velocity of the expanding wavefronts.

In the meantime, the emitter moves to the right and emits another wave. Again the medium takes over and controls the speed of the wave, relative to the medium but not relative to the emitter.

If you measure the distance of the right wave crests from the emitter, and think of that part of the wave crest as being a single “wave” or “wave front”, you will find each crest to the right of the emitter is moving away from the emitter at a slower rate than each crest to the left of the emitter.

This is not taught in physics regarding light, and most students never notice it.

This means the right wave crests are moving away from the emitter at a velocity of c – v, while the left wave crests are moving away from the emitter at a velocity of c + v. With v being the velocity of the emitter through the medium, moving to the right.

Don’t measure the far right wave crest as being the distance from where it was emitted, because at the time we are seeing the wave front now, the emitter is no longer at that place. It has moved to the right while that light circle has been expanding. Measure the far right wave crest from the distance where the emitter is closest to it, the last dot on the far right, because that’s where the emitter is now when the far right wave crest is where it is now.

In a frame in which the medium remains stationary with the moving emitter, all the circles would be concentric. That’s why we measure the speed of light on earth at “c”. Because we are carrying our local medium along with the earth.

Remember we are seeing two frames here and one medium that is stationary with the observer. If the emitter were stationary with the observer and this medium, then yes, all the circles would be concentric and there would be no Doppler shift.

In the drawing, the arrow shows that the emitter is moving to the right, so the emitter would detect the ether wind moving to the left.

[freddo]

This is not taught in physics regarding light
Justice exists in this world?


In the meantime, the emitter moves to the right and emits another wave. Again the medium takes over and controls the speed of the wave, relative to the medium but not relative to the emitter.
:o
Why is this medium required? How does it know what to do to the speed of light based on the motion of the emitter?

[Sam5]

Celestial Mechanic,


If we look at the situation where the emitter appears stationary and all the circles are concentric, then if we want to cause a Doppler shift, we would have to move the observers. The observer on the right would move to the left toward the emitter and would receive the uncompressed waves at a velocity of c + v, causing him to see a blueshift. The observer on the left would be moving to the left away from the emitter and would be receiving the unexpanded waves at a velocity of c – v, causing him to see a redshift.

With a moving emitter or observer, you can't have the light waves both leave the emitter at c and arrive at the observer at c.

[Sam5]

This is not taught in physics regarding light
Justice exists in this world?


In the meantime, the emitter moves to the right and emits another wave. Again the medium takes over and controls the speed of the wave, relative to the medium but not relative to the emitter.
:o
Why is this medium required? How does it know what to do to the speed of light based on the motion of the emitter?


Once a wave or a part of one gets into the medium, it obeys whatever laws of physics that require it to move light at the speed of c in the medium. Like putting match boxes on a moving conveyor belt. Once you let go of the match box, the conveyor belt takes over. I don’t know what the medium is.

Even if light is “self propagating”, something has to tell the photons how fast to leave the emitter. If the emitter is moving and the wave are compressed on the right side, something has to tell the photons to move at a steady rate relative to the observer. This causes them to move at a slower rate relative to the moving emitter. So I don’t know what tells them that, but that’s what they are doing in this illustration. They are orienting their speed “c” with the observer, not with the emitter. Why? I don’t know.

This is the same thing that happens with sound in air. The air is the propagating medium of the sound. Something is the propagating medium of the light.

[Musashi]

Just go to the UCLA link. The light has shifted frequencies, there is nothing to indicate it is actualy moving faster or slower.

[Glom]

Sam5 and SeanF, this thread has been padded out extremely because you're both posting multiple points that you could have put into a single post.

[SeanF]

What, specifically, is "not true" in what I said?

Moan, groan.... We are talking about the Chapter 9 thought experiment, aren’t we? You used the word “embankment”, didn’t you? That means we are talking about the earth, the railroad tracks, the moving train, a stationary observer on the embankment, a guy on the train that is moving with the train. You said, “we can't know if the embankment is at "absolute rest" or the train is at "absolute rest" or neither”.

That’s wrong. As far as this thought experiment is concerned, the embankment is at rest and the train is moving.

This book was published in 1916. The thought experiment is very clear as to the conditions of what is going on. I think you are getting it mixed up with the 1905 paper, in which two “frames” were in space somewhere, and in that case, we had to refer to just “relative motion”.


Is the surface of Mars at absolute rest? Is the surface of Earth at absolute rest?

Please note that "absolute" means "It doesn't matter how you look at it."

[SeanF]

SeanF,

Can you answer a question for me now, please? Can you tell me if you notice on the graph of the moving light emitter that the waves on the right side of the chart are moving away from the emitter more slowly than the waves on the left side of the chart? Can you explain why? Can you explain why this drawing shows an “ether wind” if there is no “ether”?

LINK TO UCLA (http://www.astro.ucla.edu/~wright/doppler.htm)

Since sarcastic comments seem to be lost on you, I'll be blunt. A diagram drawn on a website to demonstrate a concept does not necessarily represent a literal representation of the physical reality, and I will not accept it as such.

But, since you like that website so much, why don't you read through this page (http://www.astro.ucla.edu/~wright/relatvty.htm)? Same website, same author.

[SeanF]

So GR says that the 3.6 years is due to the time dilation caused by the gravity field-equivalent acceleration? SR says the 3.6 years is due to the simultaneity difference between the reference frames. So I don't know that you've told me you need GR to explain the 3.6 years, have you?
It depends upon which frame of reference you choose. You absolutely need it if you're going to choose to have the "out and in" twin be considered "motionless". See? That is what is allowed in GR--all reference frames, even noninertial ones. As long as you restrict your analysis to inertial reference frames, you're probably not going to get in trouble, no matter how many inertial frames you use. I know JW disagrees with that, but it seems to be true.

But do you need to have the "out and in" twin be considered "motionless" in order to understand why he ends up younger and not the "stay at home" twin?




First, mathematically, the equation breaks down if T=0, doesn't it?

Yes and no. What is the value of x/x at zero? If you're allowed to get as close as you like without really getting to zero?

Plus, that is not really a GR equation, but one derived from GR.


Secondly, in regards to the Dynamic calculations . . . in order to get the 3.6 years would you have the varying distance start at 3 light-years, end at 3 light-years, or average 3 light-years?
I'm not sure what you mean? Could you rephrase the question? :)

You say the answer is independent of T, but not of d. You also say that if you use it dynamically with a varying d, you get the same answer. Obviously, if d varies from 3 to 4 you'll get a different answer than if d varies from 30 to 40, right?

So will you get 3.6 if d varies from 2 to 3, or from 2.5 to 3.5, or from 3 to 4? Or really, which of those three possibilities would be closest to 3.6?

[SeanF]

Sam5 and SeanF, this thread has been padded out extremely because you're both posting multiple points that you could have put into a single post.

Yeah, I've thought about trying to quote from multiple posts in a single response, but that gets kind of complicated. And while it would reduce the "post count" number, it wouldn't reduce the amount of actual text presented, so the thread wouldn't be any shorter to read, anyway.

Hey, what happened with the un-synchronized clock thing? Do you see where I think it's coming from yet, or have I been unclear (again)? :(

[Glom]

I still chewing it over.

[Sam5]

Just go to the UCLA link. The light has shifted frequencies, there is nothing to indicate it is actualy moving faster or slower.


The diagram indicates it. You can measure the distances with a millimeter ruler on your computer screen. On my screen the right side of the first wavefront is about 24 mm away from the emitter at the same time the left side of the first wavefront is about 39 mm away from the emitter. That means the photons on the left side of the emitter are moving away from the emitter at a faster velocity than the photons on the right side of the emitter.

The diagram you made on paper should show a wave emitted by an emitter that is moving away from an observer to be emitted toward the observer at c but away from the emitter at c + v, OR toward the observer at c – v and away from the emitter at c. You can’t have c relative to both the emitter and the observer in the same diagram.

I’ve made up a diagram to show what I’m talking about. Later I’ll copy it as a photo and I’ll post it here.

[SeanF]

Just go to the UCLA link. The light has shifted frequencies, there is nothing to indicate it is actualy moving faster or slower.


The diagram indicates it. You can measure the distances with a millimeter ruler on your computer screen. On my screen the right side of the first wavefront is about 24 mm away from the emitter at the same time the left side of the first wavefront is about 39 mm away from the emitter. That means the photons on the left side of the emitter are moving away from the emitter at a faster velocity than the photons on the right side of the emitter.
Should I use a magnifying glass when I do that? :)

Sorry, that's an "inside joke," and it doesn't really apply, but I'm sure some of our members will find it humorous.


The diagram you made on paper should show a wave emitted by an emitter that is moving away from an observer to be emitted toward the observer at c but away from the emitter at c + v, OR toward the observer at c – v and away from the emitter at c. You can’t have c relative to both the emitter and the observer in the same diagram.

Sure, you can. You just have to do it relativistically and realize that the amount of time that passes between the emission of one wave-front and the next (and the distance between the emission of the two wave-fronts) is relative and not absolute.

[Sam5]

A diagram drawn on a website to demonstrate a concept does not necessarily represent a literal representation of the physical reality, and I will not accept it as such.

Oh, come on. I’ll bet you’ve believed a lot of Minkowski diagrams in your lifetime without hesitation or skepticism. Then when you’re shown a simple basic realistic Doppler diagram, on a UCLA physics-light website, you don’t believe it. Do you think UCLA would fib to you about what is happening in their Doppler diagram? If it is not accurate, then why didn't they show a Minkowski diagram of the motion of the light? Why would they show a classical Doppler diagram if it is not correct? Why don't you draw us a diagram that shows the light from a moving emitter traveling at c relative to both the observer and the emitter?

[Sam5]

Sure, you can. You just have to do it relativistically and realize that the amount of time that passes between the emission of one wave-front and the next (and the distance between the emission of the two wave-fronts) is relative and not absolute.


Why don't you draw us a diagram that shows the light from a moving emitter traveling at c relative to both the observer and the emitter? I would like to see such a diagram. Maybe UCLA can replace their Doppler diagram with your diagram.

[SeanF]

A diagram drawn on a website to demonstrate a concept does not necessarily represent a literal representation of the physical reality, and I will not accept it as such.

Oh, come on. I’ll bet you’ve believed a lot of Minkowski diagrams in your lifetime without hesitation or skepticism. Then when you’re shown a simple basic realistic Doppler diagram, on a UCLA physics-light website, you don’t believe it. Do you think UCLA would fib to you about what is happening in their Doppler diagram? If it is not accurate, then why didn't they show a Minkowski diagram of the motion of the light? Why would they show a classical Doppler diagram if it is not correct? Why don't you draw us a diagram that shows the light from a moving emitter traveling at c relative to both the observer and the emitter?

Hmm. You still haven't answered me regarding your apparent understanding of the term "absolute rest" to mean "motionless relative to something else," and now I see you don't understand the term "necessarily."

A diagram can show a very accurate representation of reality. A diagram can also be set up to show a principle in a way that does not physically resemble reality at all.

And the reason they don't show a relativistic diagram of the Doppler effect (indicating a single emitter being both "at rest" and "in motion") is because it's not necessary for the point they are trying to make and would unnecessarily complicate the issue. They're not "fibbing" about Doppler with that diagram any more than NASA was "fibbing" about the Sun moving around the Earth on that sundial page I linked to.

As far as the diagram with an emitter both stationary and moving, I'll see what I can come up - but I virtually guarantee you will neither understand nor accept it.

[Sam5]

And the reason they don't show a relativistic diagram of the Doppler effect (indicating a single emitter being both "at rest" and "in motion").........would unnecessarily complicate the issue.

At this stage of the game, I don’t think anyone should be overly concerned about that.

[Diamond]

Sure, you can. You just have to do it relativistically and realize that the amount of time that passes between the emission of one wave-front and the next (and the distance between the emission of the two wave-fronts) is relative and not absolute.


Why don't you draw us a diagram that shows the light from a moving emitter traveling at c relative to both the observer and the emitter? I would like to see such a diagram. Maybe UCLA can replace their Doppler diagram with your diagram.

Both emitter and receiver measure the speed of the light to be c. They disagree on the frequency of the light. This is consistent wiht literally thousands of experiments including the original experiment of Michelson and Morley.

[SeanF]

And the reason they don't show a relativistic diagram of the Doppler effect (indicating a single emitter being both "at rest" and "in motion").........would unnecessarily complicate the issue.

At this stage of the game, I don’t think anyone should be overly concerned about that.

Oh, at this stage of our game, certainly. We've gotten into pretty complicated territory. But still, UCLA's website wasn't intended for "those BABB'ers who are discussing Relativity," it was intended for everybody.

[Sam5]

Both emitter and receiver measure the speed of the light to be c. They disagree on the frequency of the light. This is consistent wiht literally thousands of experiments including the original experiment of Michelson and Morley.

We don’t “measure” the “speed” of the incoming light from stars. It is assumed to be “c” at the surface of the earth, relative to the earth. It is also assumed to be c relative to the star that emitted it, near the surface of the star. But if we and the star are moving relatively, then somewhere in space the photons must change speed, both relative to the star or relative to the earth.

If the earth is moving toward a star that is fixed relative to the sun, we assume the light speed at the earth is “c”. But, we receive the light as blueshifted. That means that if we assume the light to be traveling at c in space relative to the star, then at some point in space, or in some area of space, between us and the star, the light photons are moving at c + v relative to the earth, and they must change speed relative to us so that we can receive them on earth at a speed of c relative to us.

[SeanF]

Both emitter and receiver measure the speed of the light to be c. They disagree on the frequency of the light. This is consistent wiht literally thousands of experiments including the original experiment of Michelson and Morley.

We don’t “measure” the “speed” of the incoming light from stars. It is assumed to be “c” at the surface of the earth, relative to the earth. It is also assumed to be c relative to the star that emitted it, near the surface of the star. But if we and the star are moving relatively, then somewhere in space the photons must change speed, both relative to the star or relative to the earth.

Well, actually we don't "assume" the speed of light to be c at the surface of the Earth. c is defined as 299,792,458 m/s. We actually measure the distance light travels in 1/299792458th of a second in a vacuum (granted, at the surface of the Earth) and we define one meter to be equal to that value.

Thus, the speed of light in a vacuum is known to be exactly c, not assumed, because c is specifically defined as "the speed of light in a vacuum."

[Sam5]

Thus, the speed of light in a vacuum is known to be exactly c, not assumed, because c is specifically defined as "the speed of light in a vacuum."


At the surface of the earth and as measured at the surface of the earth.

I think your problem is that you have been taught to think of the speed of light in two different ways, 1) as seen separately by the emitter, and 2) as seen separately by the observer. I am suggesting that you need to learn to see the speed of light from the point of view of both the emitter and the observer at the same time, because this is what happens in nature.

That is what the UCLA Doppler diagram represents, and it clearly shows the photons leaving the emitter at c – v on the right side, and c + v on the left side, while traveling toward two observers at c relative to the observers.

Actually, this diagram represents more like what would happen at the surface of the earth, with a moving emitter. In reality, if the emitter is a star, I think the diagram should show the light leaving the star at c relative to the star, and then slowing down (on the right) and speeding up (on the left) somewhere in space, relative to the star. Sometimes I forget to mention this phenomenon.

I think you should consider this possibility: Something at the star and that surrounds the star, regulates the speed of light to c, relative to the star, when it first leaves the star. Something at the earth and that surrounds the earth, regulates the speed of light to c, relative to the earth, when it arrives at the earth. But in deep space, the relative speed of the light, relative to the star and the earth, can be different from c. In fact, it must be different from c.

What needs to be considered in the diagram, if the emitter is a star, is the local c-regulator of the star and the local c-regulator of the earth. If the emitter is moving at the surface of the earth, then the diagram doesn’t need to consider this, since it would be moving through the earth’s own local c-regulator.

[Sam5]

SeanF,

To understand what I’m saying, you might consider what fish would measure the speed of light to be, since they are surrounded by their own local light-speed regulator, which is different from the one that surrounds us humans.

[SeanF]

Thus, the speed of light in a vacuum is known to be exactly c, not assumed, because c is specifically defined as "the speed of light in a vacuum."


At the surface of the earth and as measured at the surface of the earth.

I think your problem is that you have been taught to think of the speed of light in two different ways, 1) as seen separately by the emitter, and 2) as seen separately by the observer. I am suggesting that you need to learn to see the speed of light from the point of view of both the emitter and the observer at the same time, because this is what happens in nature.
You've never been to a distant star, so be careful making claims about "what happens in nature." You're not God. You can theorize about it, but so can we.

SR theorizes that the speed of light (in vacuo) is always the same relative to you, no matter who or where you are or who or where the light is. The light leaving Sirius is moving at c relative to Sirius, and it's moving at c relative to us, and it's moving at c relative to Alpha Centauri - all as soon as it leaves Sirius.


That is what the UCLA Doppler diagram represents, and it clearly shows the photons leaving the emitter at c – v on the right side, and c + v on the left side, while traveling toward two observers at c relative to the observers.

Actually, this diagram represents more like what would happen at the surface of the earth, with a moving emitter. In reality, if the emitter is a star, I think the diagram should show the light leaving the star at c relative to the star, and then slowing down (on the right) and speeding up (on the left) somewhere in space, relative to the star. Sometimes I forget to mention this phenomenon.

I think you should consider this possibility: Something at the star and that surrounds the star, regulates the speed of light to c, relative to the star, when it first leaves the star. Something at the earth and that surrounds the earth, regulates the speed of light to c, relative to the earth, when it arrives at the earth. But in deep space, the relative speed of the light, relative to the star and the earth, can be different from c. In fact, it must be different from c.

What needs to be considered in the diagram, if the emitter is a star, is the local c-regulator of the star and the local c-regulator of the earth. If the emitter is moving at the surface of the earth, then the diagram doesn’t need to consider this, since it would be moving through the earth’s own local c-regulator.

Local c-regulators have been considered by lots of people smarter than both of us, and pretty much uniformly dismissed. Most people who study physics all their lives have concluded that Relativity is most likely correct. All I can do is think about it logically, and I can (and do) conclude that Relativity is more logical and, quite frankly, more "cool." It's a beautiful thing.

[Sam5]

SR theorizes that the speed of light (in vacuo) is always the same relative to you, no matter who or where you are or who or where the light is. The light leaving Sirius is moving at c relative to Sirius, and it's moving at c relative to us, and it's moving at c relative to Alpha Centauri - all as soon as it leaves Sirius.

What about light leaving a galaxy that astronomers believe is receding from us at c? Is that light traveling at c relative to the galaxy and c relative to earth at the same time, at the same point in space? What do the smart experts say about that?

[kilopi]

What about light leaving a galaxy that astronomers believe is receding from us at c? Is that light traveling at c relative to the galaxy and c relative to earth at the same time, at the same point in space? What do the smart experts say about that?
I'm not quite sure, but I think you just described the basic idea behind special relativity. Surprising? You bet, but it's not a paradox.

The only thing in your comment that I'm not sure of is why you included the phrase "at the same point in space"--same point as what?

PS: I think I see now. The phrase is like "at the same time"--you're making sure that the measurement relative to the Earth and the measurement relative to the galaxy is made on one and the same object.

[SeanF]

SR theorizes that the speed of light (in vacuo) is always the same relative to you, no matter who or where you are or who or where the light is. The light leaving Sirius is moving at c relative to Sirius, and it's moving at c relative to us, and it's moving at c relative to Alpha Centauri - all as soon as it leaves Sirius.

What about light leaving a galaxy that astronomers believe is receding from us at c? Is that light traveling at c relative to the galaxy and c relative to earth at the same time, at the same point in space? What do the smart experts say about that?

I believe you're talking about galaxies that are so distant that the expansion of space between us becomes a noticeable effect.

In that case, I'd have to say that the rate at which we're separating isn't the same thing as the difference in our inertial frames. So, yes, the light is moving at c relative to both inertial frames, but as long the observers are separated enough for the expansion to contribute, it won't really seem like it.

How you like that answer? :)

I do believe that GR (itself an extension of SR) has been extended to account for this effect as well, but I may be wrong.

(By the way, the "astronomers" who "believe" these galaxies are receding from us at c [or even greater] by and large "believe" in Relativity as well)

[SeanF]

SR theorizes that the speed of light (in vacuo) is always the same relative to you, no matter who or where you are or who or where the light is. The light leaving Sirius is moving at c relative to Sirius, and it's moving at c relative to us, and it's moving at c relative to Alpha Centauri - all as soon as it leaves Sirius.

What about light leaving a galaxy that astronomers believe is receding from us at c? Is that light traveling at c relative to the galaxy and c relative to earth at the same time, at the same point in space? What do the smart experts say about that?

Oh, by the way, when viewed from the two different reference frames, the light isn't at the same point in space at the same point in time. Same time but different place, same place but different time. But wherever (and whenever) it is, it's traveling at c.

[kilopi]

I'm not quite sure
That's a bit different take on it, SeanF, you might be right. Sam5?

PS: I was talking about your first post, SeanF. As to the second, I think Sam5 can get around those ambiguities by asking that the measurements take place for the same point and time, from the point of view of a local observer.

[Sam5]

I believe you're talking about galaxies that are so distant that the expansion of space between us becomes a noticeable effect.

In that case, I'd have to say that the rate at which we're separating isn't the same thing as the difference in our inertial frames. So, yes, the light is moving at c relative to both inertial frames, but as long the observers are separated enough for the expansion to contribute, it won't really seem like it.

How you like that answer? :)


I think that’s the most brilliant vague answer I’ve ever read in my lifetime, and I think you deserve a Nobel Prize for it. I will put your name in nomination right away. That answer could qualify you for a position as a math professor at Princeton.

[Wally]

Let me ask your opinion about something. How do you suppose we see a blueshift in the starlight coming from a distant star that is fixed relative to the sun, when the earth moves toward that star every six months? Don’t you suppose it’s because of the c + v phenomenon, relative to the earth?

Wow! Sam! Man, quit while you can man! You've got a lot of learning to do before you argue a case as strongly as you have this one. SeanF is completely correct in everything I've read so far, and your arguments are starting to get a little bit silly (and I'm only on Monday!!!).

[Kaptain K]

I am suggesting that you need to learn to see the speed of light from the point of view of both the emitter and the observer at the same time, because this is what happens in nature.
And that is exactly what SR does! Light emitted from a star is travelling at "c" at all times, from all points of view. It is moving at "c" when it leaves the star, when it reaches the Earth and at all points in between, from the point of view of the star. It is moving at "c" when it leaves the star, when it reaches the Earth and at all points in between, from the point of view of the Earth. This is true no matter what the relative speeds of the star and Earth. The Lorentz equations adjust length and time to make this so. I am beginning to suspect that you cannot actually do the math. This would explain why you continue to quibble with the semantics of "thought experiments", while ignoring the theory itself.

One more time.

The Math is the Theory

Everything else is just window dressing.

[Sam5]

Oh, by the way, when viewed from the two different reference frames, the light isn't at the same point in space at the same point in time. Same time but different place, same place but different time. But wherever (and whenever) it is, it's traveling at c.


Doh

I know that. How could we see photon # 43958273467758399475 on earth if it was at some other place than earth at the time we are searching for it?

But, what happens is, that photon leaves the distant galaxy traveling at c relative to that galaxy, and less than c relative to the earth, and it speeds up while in route.

I think the terminology in physics and cosmology today refers to the phenomena as: “recessional velocity”, “peculiar velocity”, and “total velocity”. So, I’m talking about the “total velocity” changing while in route. “Recessional velocity” is related to the earth-relative radial velocity of the distant galaxy and the earth-relative velocity of the local “ether” through which the photon is traveling when it is first emitted.

This “ether” is generally known in physics today as the local “comoving space” of the photon, and the “recessional velocity” of this “comoving space”, is different at different distances from the earth. The photon travels through its own local area of “comoving” space (where ever it is traveling at the moment) at c, relative to that space, and this velocity is known as the “peculiar velocity” of the photon.

Relative to our own “local group” of galaxies and their collective “comoving space”, our own galaxy is believed to have a “peculiar velocity” of about 600 kps. This speed estimate is based on the measured anisotropy in the microwave background radiation.

Now, since the recessional velocity of the moving photon’s local comoving space (i.e. “local ether”) varies, relative to the earth, as the photon moves toward the earth – while the peculiar velocity of the photon always remains the same inside the local comoving space through which it is traveling – the net result is that the “total velocity” of the photon changes, both relative to the earth and relative to the galaxy that emitted it, as the photon moves in the direction of the earth.

If I have made any mistakes here, I’m sure you would be gracious enough to point them out to me.

[SeanF]

Oh, by the way, when viewed from the two different reference frames, the light isn't at the same point in space at the same point in time. Same time but different place, same place but different time. But wherever (and whenever) it is, it's traveling at c.


Doh

I know that. How could we see photon # 43958273467758399475 on earth if it was at some other place than earth at the time we are searching for it?

Nah, you miss my point. What I meant was that the amount of time that passes between the photon being emitted at the distant galaxy and that same photon reaching Earth (not to mention the distance the photon travels) is relative.

Simplifying it down to a simple emitter and receiver, whereas from one reference frame the photon hit the receiver one minute after being emitted and almost 18,000,000 kilometers away its emission point, from another reference frame the photon hit the receiver 75 seconds after being emitted and over 22,000,000 kilometers away!

However, my statement was not worded well. After all, everybody agrees that the photon gets nearly 300000 kilometers farther away from its emission point every second.



But, what happens is, that photon leaves the distant galaxy traveling at c relative to that galaxy, and less than c relative to the earth, and it speeds up while in route.

I think the terminology in physics and cosmology today refers to the phenomena as: “recessional velocity”, “peculiar velocity”, and “total velocity”. So, I’m talking about the “total velocity” changing while in route. “Recessional velocity” is related to the earth-relative radial velocity of the distant galaxy and the earth-relative velocity of the local “ether” through which the photon is traveling when it is first emitted.

This “ether” is generally known in physics today as the local “comoving space” of the photon, and the “recessional velocity” of this “comoving space”, is different at different distances from the earth. The photon travels through its own local area of “comoving” space (where ever it is traveling at the moment) at c, relative to that space, and this velocity is known as the “peculiar velocity” of the photon.

Relative to our own “local group” of galaxies and their collective “comoving space”, our own galaxy is believed to have a “peculiar velocity” of about 600 kps. This speed estimate is based on the measured anisotropy in the microwave background radiation.

Now, since the recessional velocity of the moving photon’s local comoving space (i.e. “local ether”) varies, relative to the earth, as the photon moves toward the earth – while the peculiar velocity of the photon always remains the same inside the local comoving space through which it is traveling – the net result is that the “total velocity” of the photon changes, both relative to the earth and relative to the galaxy that emitted it, as the photon moves in the direction of the earth.

If I have made any mistakes here, I’m sure you would be gracious enough to point them out to me.

In a universe with local c-regulators surrounding every spatial body, then what you said here would very likely all be correct.

But like I've said before, that ain't this universe.

[SeanF]

We don’t “measure” the “speed” of the incoming light from stars. It is assumed to be “c” at the surface of the earth, relative to the earth. It is also assumed to be c relative to the star that emitted it, near the surface of the star. But if we and the star are moving relatively, then somewhere in space the photons must change speed, both relative to the star or relative to the earth.


In your theory of light, where we have a light-propogating medium, which is not only carried through space by the local celestial body but also rotates with the local celestial body . . . why does the medium of every body, regardless of size or mass, limit the speed of light to the exact same value?

[Sam5]

In a universe with local c-regulators surrounding every spatial body, then what you said here would very likely all be correct.

But like I've said before, that ain't this universe.

Before you say that ain’t this universe, I think you’d better read over some of the latest “comoving space” and Vtot and Vrec papers.

What I am doing is reading all kinds of latest mainstream papers looking for evidence that what I am saying is correct.

Now, one might wonder why we don’t notice a 600 kps “ether wind” or “comoving space wind” here on earth if our own galaxy is moving through the “comoving space” of the local group? I suggest it is because we are experiencing the “moving fishbowl effect”. The fish will always measure the slower speed of light through water, even if their fish bowl is moved around at the earth’s surface, and no matter how fast the light is traveling outside the bowl, and up until the time it enters their water.

What this hypothesis suggests is that our own galaxy is surrounded by its own local “comoving space” that moves through the larger area of comoving space that surrounds our local group of galaxies. And to carry this hypothesis further, I suggest the same phenomenon occurs within our solar system and at (and near) the surface of the earth. So, it’s like sealed fishbowls moving around inside the ocean. Each one carries its own local comoving space with it as it moves, and inside each of those local comoving spaces, the speed of light is always measured, locally, inside each bowl, as being the same, but the speed of the light can vary, relatively, when moving between one bowl and another, because that speed is regulated in and by the overall water of the ocean. So let our own galaxy represent the ocean, while the solar system represents a fishbowl traveling around inside it, and the earth represents a smaller fishbowl traveling around inside the fishbowl of the solar system.

We might notice a redshift and a blueshift in the background radiation, but at the surface of the earth we will measure its radiation speed to be “c”, relative to the earth.

Ok, so, what Einstein did in SR theory was have a local “comoving space” stationary with his “stationary” inertial frame, and he had another separate “comoving space” that was “stationary” with the “moving” frame. That is why both sets of observers measured the velocity of light at “c” within their own inertial frames, but at c + v and c – v when they looked into each other’s frames. It was the “crossover” viewing that caused the theoretical visual “length contraction” when the observers looked into each other’s frames. And it was that “length contraction” amount (which was determined by the relative velocity “v” and the Lorentz Transformation) that told Einstein, by means of one of his own equations, at what “rate” to which the “moving clock” should be “adjusted” to produce the “time dilation” effect as “seen” when an observer in one comoving space looked at the clock in the other relatively moving comoving space (see paragraph 6, Section 4 of “On the Electrodynamics of Moving Bodies" for Einstein’s “adjustment” of the “moving clock's” tick rate).

So, the “traveling twin’s” clock doesn’t automatically “slow down” all by itself. We’ve got to “adjust it” so that it runs slow, based on Einstein’s calculations regarding the “length contraction” of its frame, as “seen” by us, when we look into its own relatively moving comoving frame, during the relative motion. And if we don’t make the same calculations from “Jack’s” point of view, when he looks at “Jill’s” clock while they are relatively moving, and if we don’t readjust Jack’s clock to “normal tick rate” and readjust Jill’s clock to “time dilated tick rate”, then of course, her clock will not tick slowly in the twins paradox thought experiment.

And so, I was right all along: In SR theory, “relative motion” can not possibly alter a clock’s tick rate. You have to physically adjust the tick rate of the clock if you want it to slow down, based on Einstein’s calculations of light travel time when one observer looks from his own comoving space into the relatively moving comoving space of the relatively moving other observer.

What you are trying to do with the moving train example is claim that the train observers are fixed inside their own separate “comoving space”. But they are not. They are clearly traveling through the comoving space of the embankment. But this can’t be said of the two observers in the two frames of the original SR theory, since Einstein set them up so that each frame is fixed inside its own independent and relatively moving comoving space. So, in the SR theory itself, we have two separate and relatively moving “comoving spaces”, while in the train example in his book, we have only one “comoving space” through which the train is traveling.

And I want to thank you for helping me to figure all of this out. Your understanding of the SR theory itself has helped me considerably. Most people I discuss this with on the internet get so mixed up, they aren’t much help at all.

[Sam5]

In your theory of light, where we have a light-propogating medium, which is not only carried through space by the local celestial body but also rotates with the local celestial body . . . why does the medium of every body, regardless of size or mass, limit the speed of light to the exact same value?

I’m not sure that it does. I think light might travel more slowly near the surface of more massive bodies. But when describing the overall situation, it is a little easier to leave that detail out. The concept is already complicated enough. Sometimes I say “approximately c” to account for this possible difference due to the mass of the body being discussed.

It might be that “c” is our own observed speed that we measure here on our earth, with its specific mass, inside its own specific solar system, inside its own galaxy, and with our own local atomic clocks. Of course we have the situation of atomic clock “time dilation”, so that if the speed is measured on the surfaces of other planets if different sizes and masses, then we’ve got to allow for changes in the local atomic clock rates, due to gravity considerations.

I’m assuming that on the surface of an average earth-sized planet orbiting an average sun-sized star, at about 1 AU distance from the star, and at about the same distance from the center of a same-sized galaxy, the local atoms will have the same size, shape, and mass and will react in the same way to similar environmental conditions, emitting light of the same local speed and frequency as that which we observe here on earth.

And on all the billions of same-sized planets in the universe, there are probably two guys like you and me arguing about this.

[daver]

It might be that ?c? is our own observed speed that we measure here on our earth, with its specific mass, inside its own specific solar system, inside its own galaxy, and with our own local atomic clocks. Of course we have the situation of atomic clock ?time dilation?, so that if the speed is measured on the surfaces of other planets if different sizes and masses, then we?ve got to allow for changes in the local atomic clock rates, due to gravity considerations.

Probably not, or we'd see differences in spectra from different mass suns (or differences between what is produced in earth labs and what is produced in the sun).

[kilopi]

Before you say that ain’t this universe, I think you’d better read over some of the latest “comoving space” and Vtot and Vrec papers.

What I am doing is reading all kinds of latest mainstream papers looking for evidence that what I am saying is correct.
Maybe, maybe not.

Definitely against the mainstream though. If you're using that belief to study Einstein's work, why do you call it wrong? It makes sense to a lot of people, and a lot of work has been done on it.

If your goal is to further your own theories, calling other peoples theories wrong, for no reason, doesn't seem to help.

[kilopi]

It may not be explicitly stated in any of Einstein's thought experiments, but it is implicitly part of the theory.
It's explicit. He doesn't call it the "twins paradox" but it's there (http://www.fourmilab.ch/etexts/einstein/specrel/www/#SECTION14) nonetheless.

You also know that if Jack and Jill both “blast off” in the opposite directions, go the same distance, then “turn around”, their atomic clocks will both slow down exactly alike, as per GR, but during their unaccelerated “relative motion” SR requires that they “see” each other’s clocks “slow down” the same amount, as per SR, and when they get back to earth SR also requires that only one of their clocks “lag behind” due to SR.
So you're saying that SR requires both for them to be the same, and also for them to be different. You're wrong.

[SeanF]

Hey, Sam5, I've got a question for you. GPS satellites in orbit, using atomic clocks, have to be adjusted because their clocks do not match time with those on Earth. You believe this difference to be a result of gravitational effects on the mechanical workings of the clock, is this correct?

[Sam5]

Hey, Sam5, I've got a question for you. GPS satellites in orbit, using atomic clocks, have to be adjusted because their clocks do not match time with those on Earth. You believe this difference to be a result of gravitational effects on the mechanical workings of the clock, is this correct?

Is this a trick question? If I say "yes" are you going to zap me with a zinger response??

Ok, I'll say "yes" and see what happens. For the same reason pendulum clocks slow down at high altitudes. For the same reason a thermodynamic clock slows down when you remove heat energy from it. A physical effect on the fundamental mechanism that regulates the "tick rates" of the clocks.

[russ_watters]

Ok, I'll say "yes" and see what happens. For the same reason pendulum clocks slow down at high altitudes. For the same reason a thermodynamic clock slows down when you remove heat energy from it. A physical effect on the fundamental mechanism that regulates the "tick rates" of the clocks. And what precisely is that effect?

[SeanF]

Hey, Sam5, I've got a question for you. GPS satellites in orbit, using atomic clocks, have to be adjusted because their clocks do not match time with those on Earth. You believe this difference to be a result of gravitational effects on the mechanical workings of the clock, is this correct?

Is this a trick question? If I say "yes" are you going to zap me with a zinger response??

Ok, I'll say "yes" and see what happens. For the same reason pendulum clocks slow down at high altitudes. For the same reason a thermodynamic clock slows down when you remove heat energy from it. A physical effect on the fundamental mechanism that regulates the "tick rates" of the clocks.

Well, here's your zinger.

General Relativity predicts that time will run faster higher in a gravitational field. But note that it does not merely make this as a generalized statement. You can take a specific altitude and plug it into the formulae of GR, and it will tell you exactly how much difference there will be between orbit at that altitude and sea level. This is a prediction of a change in time, based solely on the postulates of Relativity (constant speed of light and all that).

Now, the first GPS satellite was not launched until 60 years after Einstein wrote GR. The first artificial satellite of any kind was not launched until 40 years after Einstein wrote GR. The atomic clock was not even invented until 25 years after Einstein wrote GR.

But you are asking us to believe that gravity has some kind of previously unpredicted mechanical effect that, on the type of atomic clocks used in GPS satellites, just happens (by pure coincidence?) to exactly match the time difference predicted by GR? And GPS satellites need to be accurate to the millisecond, so I do mean exactly.

That's pretty amazing . . . don't you think?

[Sam5]

Hey, Sam5, I've got a question for you. GPS satellites in orbit, using atomic clocks, have to be adjusted because their clocks do not match time with those on Earth. You believe this difference to be a result of gravitational effects on the mechanical workings of the clock, is this correct?

Is this a trick question? If I say "yes" are you going to zap me with a zinger response??

Ok, I'll say "yes" and see what happens. For the same reason pendulum clocks slow down at high altitudes. For the same reason a thermodynamic clock slows down when you remove heat energy from it. A physical effect on the fundamental mechanism that regulates the "tick rates" of the clocks.

Well, here's your zinger.

General Relativity predicts that time will run faster higher in a gravitational field. But note that it does not merely make this as a generalized statement. You can take a specific altitude and plug it into the formulae of GR, and it will tell you exactly how much difference there will be between orbit at that altitude and sea level. This is a prediction of a change in time, based solely on the postulates of Relativity (constant speed of light and all that).

Now, the first GPS satellite was not launched until 60 years after Einstein wrote GR. The first artificial satellite of any kind was not launched until 40 years after Einstein wrote GR. The atomic clock was not even invented until 25 years after Einstein wrote GR.

But you are asking us to believe that gravity has some kind of previously unpredicted mechanical effect that, on the type of atomic clocks used in GPS satellites, just happens (by pure coincidence?) to exactly match the time difference predicted by GR? And GPS satellites need to be accurate to the millisecond, so I do mean exactly.

That's pretty amazing . . . don't you think?


Excuse me, but I think you need to read more about the history of atomic clocks and especially about 19th Century “natural” atomic clocks.

The idea about “time dilation” related to the internal harmonic oscillation rates of atoms was discussed by Lorentz in his relativity paper of 1904 and discussed by him in other electrodynamic papers of that era. In 1904 he said:

“Hence, in phenomena in which there is an acceleration in the direction of motion, the electron behaves as if it had a mass m1; in those in which the acceleration is normal to the path, as if the mass were m2.”

Lorentz recognized the variable internal vibration rates of atoms experiencing acceleration, and he also recognized that atomic time could change rates. He wrote in 1904: “The variable t’ may be called the ‘local time’.”

His 1904 paper was mainly about the inner workings of atoms when they experience motion through an ether or experience acceleration, and he said that accelerations cause the atoms to react differently from atoms that don’t experience accelerations.

In 1907 Einstein acknowledged Lorentz’s contribution to this topic of time (atomic time), and he said:

“One had only to realize that an auxiliary quantity introduced by H. A. Lorentz and named by him ‘local time’ could be defined as ‘time’ in general.”

Of course we all know that in 1873 Maxwell wrote:

“In the present state of science the most universal standard of length which we could assume would be the wave length in vacuum of a particular kind of light, emitted by some widely diffused substance such as sodium, which has a well-defined lines in its spectrum......

In astronomy a year is sometimes used as a unit of time. A more universal unit of time might be found by taking the periodic time of vibration of the particular kind of light whose wave length is the unit of length.”

So, physicists knew about “natural” atomic clocks long before they manufactured an atomic clock that actually had a clock-face read-out on it. The earliest kinds of atomic clocks were natural ones and their different “rates” were observed by means of spectrometers that observed the different freqencies of light the atoms emitted. It became more convenient to specifically manufacture man-made atomic clocks that had digital read-outs incorporated into their systems. Atomic clocks in GPS satellites emit radio signals and this, in effect, represents their clock rate "read-outs". The “time” rate changes due to gravity that take place inside GPS clocks are internal atomic harmonic oscillation rate changes caused by the variable gravitational potential at the individual clock. Einstein tended to think of these variable atomic clock rates as representing the flow-rate of all of “time”, at the clocks. As I mentioned earlier, biologists don’t think that way. They tend to think in terms of “thermodynamic time”.

So, in conclusion, what you call the “invention of the atomic clock” refers only to a self-contained device that makes use of natural atomic clocks (ie, the internal harmonic oscillation rates of atoms) that causes light of a certain frequency to be emitted, but that has a clock-face read-out attached to it, whereas Maxwell referred to the light frequency generated by natural atomic clocks as far back as 1873. Between 1873 and 1904 it was theorized that this emitted frequency could change if the atoms experienced different enviornmental conditions, such as more or less acceleration.

In 1895 Lorentz introduced the Lorentz Transformation equation, and in 1904 he applied it to the internal workings of atoms experiencing motion through an ether and/or acceleration.

[kilopi]

Excuse me, but I think you need to read more about the history of atomic clocks and especially about 19th Century “natural” atomic clocks.

::snip discussion of Lorentz::
That doesn't address SeanF's point at all, which is that the workings of GPS clocks seem to fall in line with mainstream physics, including GR (and so also SR).

[SeanF]

Interesting. But it doesn't change the point (as kilopi pointed out) that a theory based on a constant speed of light in vacuo which predicts a change in time itself matching that exact rate would be quite coincidental.

[Sam5]

Interesting. But it doesn't change the point (as kilopi pointed out) that a theory based on a constant speed of light in vacuo which predicts a change in time itself matching that exact rate would be quite coincidental.



No, one of the many changes in GPS clocks is a change due to acceleration changes at the clocks. Lorentz was one of the first people to study this phenomenon. He explained how atoms worked when subjected to acceleration. When they experience less acceleration, their harmonic oscillation rates speed up and they emit light of a higher frequency. With satellite-based atomic clocks, this is called “clock drift”. But there are also other environmental factors that cause clock drift in satellite clocks. Call the atomic clock place at Boulder Colorado and they will tell you how it works. Also, ask them about “oscillator error”, thermodynamic factors, and other causes of satellite clock drift.

And here, read this:

GPS CLOCK DRIFT (http://www.gisdevelopment.net/technology/gps/techgp0014.htm)


GPS CLOCK DRIFT 2 (http://216.239.57.104/search?q=cache:wuIwhOUpxK0J:tycho.usno.navy.mil/ptti/ptti2002/paper7.pdf+dod+MIL+gps+clock+drift&hl=en&ie=UTF-8)


See link below: Note: “environmental influences”, and there are many, not just changes in acceleration:

”Due to environmental influences on the satellite signals that these receivers use, small position errors are always present in GPS computed coordinates.”

GPS CLOCK DRIFT 3 (http://www.uaex.edu/Other_Areas/publications/HTML/FSA-1033.asp)

[Sam5]

SeanF,

I’ve already explained to you that scientists long ago, centuries ago, noticed that different environmental factors cause clock rate changes, and the changes are different for different kinds of clocks.

This has been known for centuries.

There are many factors that affect GPS clock drift rates.

The military people and scientists who operate and maintain the GPS clocks don’t just plug the Lorentz Transformation equation 1 : √1 – (v^2/c^2) into the elevation of the clocks and come up with stable clocks or a single formula for compensating for the satellite clock drift-rate errors. It’s just not done, and I think you need to enter the 21st Century and find out what is going on in modern physics if you are going to talk about this subject with any degree of authority and credibility.

[SeanF]

SeanF,

I’ve already explained to you that scientists long ago, centuries ago, noticed that different environmental factors cause clock rate changes, and the changes are different for different kinds of clocks.

This has been known for centuries.

There are many factors that affect GPS clock drift rates.

The military people and scientists who operate and maintain the GPS clocks don’t just plug the Lorentz Transformation equation 1 : √1 – (v^2/c^2) into the elevation of the clocks and come up with stable clocks or a single formula for compensating for the satellite clock drift-rate errors. It’s just not done, and I think you need to enter the 21st Century and find out what is going on in modern physics if you are going to talk about this subject with any degree of authority and credibility.

Yes, there are other corrections that need to be made, so they don't "just" plug in GR. But along with everything else they do, they have to "also" plug in GR. If you're suggesting that GR is wrong, then you're suggesting that there is some other factor affecting the clocks that nobody really knows what it is, because everybody's just "assuming" it's GR on the (pretty solid) grounds that it matches GR's predictions.

And those links you posted are mostly talking about correcting for transmission effects such as atmospheric refraction and reflection. None of them mentioned anything about the time corrections built into the satellite's clocks, which do include GR. (The second link does actually talk about the satellite clocks themselves, but seems to simply catalog the errors that have been noticed within the context of how old the clocks are, saying nothing about what causes the errors).

[Sam5]

If you're suggesting that GR is wrong,

I haven’t said GR is wrong. I’ve said GR is right. If you want to call the atomic clock rate slowdowns due to acceleration changes “GR”, then go ahead, although Lorentz was the first to propose them. But you can call it anything you want.

If you imply to people on this board that this is the only reason why GPS clocks drift, then I want to correct that error with more accurate information.

[russ_watters]

Damn, I had a post then I lost it due to a crash. SamF pretty much covered what I was saying though.

Sam5, the GPS system was designed and runs according to SR and GR. By this I mean the PEOPLE who designed it and the PEOPLE who run it used/use those two theories to make it work. The fact that it DOES work is a validation of SR and GR.

Also, none of the links you provided mentions relativity. Two of the links aren't even about time, but position. The quote you provided confirms that. They are quite simply not relevant to the disucssion.

It seems to me that by providing irrelevant links you are attempting to show that scientists agree with you. They do NOT.

Do a google search for what people DO say about GPS and relativity. I typed in "gps relativity" and this is the first link that came up:

http://www.metaresearch.org/cosmology/gps-relativity.asp


3. Does the GPS confirm the clock rate changes predicted by GR and SR?

...Therefore, we can assert with confidence that the predictions of relativity are confirmed to high accuracy...

4. Is the speed of light constant?

...the system has shown that the speed of radio signals (identical to the "speed of light") is the same from all satellites to all ground stations at all times of day and in all directions to within ±12 meters per second (m/s). I highly encourage you to read the article. It addresses a large number of your misconceptions about relativity including the whole twins paradox thing in the other thread, Lorenz, etc.

[Sam5]

D***, I had a post then I lost it due to a crash. SamF pretty much covered what I was saying though.

I think maybe you'd better go back to bed and sleep a little more.

[Sam5]

Also, none of the links you provided mentions relativity.

That's because they are real physics links written by real working physicists who work with the clocks and telling about how the clocks work and what causes the drifts.

Would you prefer that I link you to a high school physics class about "time dilation"?

[Musashi]

Sam, is there some reason you need to be insulting in your posts? Maybe that is how you get people around you to agree with you, or at least stop talking to you. Maybe it is you who needs to get some more sleep.

[SeanF]

If you're suggesting that GR is wrong,

I haven’t said GR is wrong. I’ve said GR is right. If you want to call the atomic clock rate slowdowns due to acceleration changes “GR”, then go ahead, although Lorentz was the first to propose them. But you can call it anything you want.
GR doesn't say "gravity affects clocks," GR says "gravity affects time." And since GR is an extension of SR, GR also says "relative motion affects time." Are you sure you think GR is right?


If you imply to people on this board that this is the only reason why GPS clocks drift, then I want to correct that error with more accurate information.

I'm not implying that GR is the only reason GPS clocks drift. I've never implied that relativistic time dilations are the only reason any clocks could ever "drift." Our disagreement is on whether or not relativistic time dilations do happen, not on whether or not they're the only thing that happens.

[SeanF]

Damn, I had a post then I lost it due to a crash. SamF pretty much covered what I was saying though.


Been there, done that.

By the way, who's SamF? I'm pretty sure I know what you meant, but . . . ;)

[Sam5]

Maybe it is you who needs to get some more sleep.

It is true. I do. Anyway, that was a joke. SeanF also joked about it.

[Sam5]

SeanF,

Ok, here’s a new one for you:

In the moving train thought experiment, let’s say the two observers don’t see the lightening flashes directly. Let’s say the position B light from the sky hits two mirrors that are turned at 45 degree angles. One mirror affixed to the train up near the front of the train, and one affixed to the embankment at position B. Just as the mirror on the train passes position B, the lightening flashes. So, the light hits the two mirrors and starts heading toward the two observers. The train observer sees the train mirror light and the ground observer sees the ground mirror light.

At what velocity will the light leave the surface of the train mirror, traveling through the air molecules, and at what velocity will the light leave the surface of the embankment mirror, traveling through same group of side-by side air molecules?

[SeanF]

SR deals with the speed of light in a vacuum, and so all of Einstein's thought experiments (remember, they are demonstrations) that deal with light are set up in vacuums. Chapter 7, Paragraph 3, Sentence 3. There is no air. The light always travels at c relative to the pertinent observer, whether it's coming directly from the flashes or reflected off mirrors.

But let's not change the subject just yet:




If you're suggesting that GR is wrong,

I haven’t said GR is wrong. I’ve said GR is right. If you want to call the atomic clock rate slowdowns due to acceleration changes “GR”, then go ahead, although Lorentz was the first to propose them. But you can call it anything you want.
GR doesn't say "gravity affects clocks," GR says "gravity affects time." And since GR is an extension of SR, GR also says "relative motion affects time." Are you sure you think GR is right?


I'd like you to answer this.

[SeanF]

Maybe it is you who needs to get some more sleep.

It is true. I do. Anyway, that was a joke. SeanF also joked about it.

My post was in clear reference to the typo, and was both affectionate and accepting. Your post could easily have been taken to refer to the lost post, and was both rude and dismissive.

I do enjoy the technical part of our discussion here, Sam5, but I'll thank you to not compare your conversational style with mine.

[russ_watters]

By the way, who's SamF? I'm pretty sure I know what you meant, but . . . ;) Sorry.
That's because they are real physics links written by real working physicists who work with the clocks and telling about how the clocks work and what causes the drifts.
Ok, what that quote SAYS is that most scientists don't accept relativity. Is that your point? If that is your point, its just plain wrong.

Let me ask another way: do you believe most scientists accept Einstein's relativity and just think they are wrong or do you believe most scientists do NOT accept Einstein's relativity?

[Sam5]

SR deals with the speed of light in a vacuum, and so all of Einstein's thought experiments (remember, they are demonstrations) that deal with light are set up in vacuums.

How can the train and embankment observers breathe in a vacuum? In Chapter 9, Einstein said it was a train on rails. He didn’t say anything about vacuum. I’ve already mentioned to you that this experiment is conducted at the surface of the earth. It’s not like the 1905 experiments that are conducted at unknown places in space. Einstein didn’t put the train in a vacuum, so why are you trying to? What you try to do is change the parameters of Einstein's own thought experiments to get the results you want.

Here is an animation that explains a little about the Doppler effect and binaries. The animation doesn’t take into account the travel time of the light.

BINARY DOPPLER ANIMATION (http://instruct1.cit.cornell.edu/courses/astro101/java/binary/binary.htm)

[russ_watters]

At what velocity will the light leave the surface of the train mirror, traveling through the air molecules, and at what velocity will the light leave the surface of the embankment mirror, traveling through same group of side-by side air molecules? Leave air out of it. Light travels at C in all cases and is OBSERVED to travel at C in all cases by all observers.

[russ_watters]

How can the train and embankment observers breathe in a vacuum? The air is not relevant to the thought experiment. And quite frankly you aren't ready to understand the effects of air on light. You first need to understand the way light behaves in a vaccum and it appears that you do not (still not sure though - you might just be acting like you don't understand).

[Sam5]

Ok, what that quote SAYS is that most scientists don't accept relativity. Is that your point? If that is your point, its just plain wrong.


No I didn’t say that at all. You apparently don’t understand that Einstein didn’t invent or copyright the word “relativity”. Galileo discussed it. Newton discussed it in the Principia. Doppler wrote about it. Several 19th and early 20th Century physicists and astronomers talked about it. Poincaré devoted a chapter in his 1902 book to relative motion.

Scientists of all kinds deal with “relativity” concepts every day, in every field. But they aren’t all “Einstein relativity” concepts and they don’t all use the Lorentz transformation equation.

[russ_watters]

Ok, what that quote SAYS is that most scientists don't accept relativity. Is that your point? If that is your point, its just plain wrong.


No I didn’t say that at all. You apparently don’t understand that Einstein didn’t invent or copyright the word “relativity”. Galileo discussed it. Newton discussed it in the Principia. Doppler wrote about it. Several 19th and early 20th Century physicists and astronomers talked about it. Poincaré devoted a chapter in his 1902 book to relative motion.

Scientists of all kinds deal with “relativity” concepts every day, in every field. But they aren’t all “Einstein relativity” concepts and they don’t all use the Lorentz transformation equation. Ok, let me clarify then (though I suspect you know what I meant). Do you believe that most sceintists accept or do not accept the validity of Einstein's Relativity?

For example, do you accept that the scientists who designed and the scientists who operate the GPS system utilize Einstein's Relativity in the operation of the GPS system?

The more coy you are, the more deceitful you appear. Thats why I'm asking the question(s). I want a clear understanding of where you stand. So if you could be so kind - a yes or no answer is what I am looking for.

[kilopi]

I think you need to enter the 21st Century and find out what is going on in modern physics if you are going to talk about this subject with any degree of authority and credibility.

Good one. Bears repeating:

I think you need to enter the 21st Century and find out what is going on in modern physics if you are going to talk about this subject with any degree of authority and credibility.

[Sam5]

SeanF,

Oh, moan, I think I’m responding to a post within a post within a post, and I think I’ve already answered that question several times. You can go back through the pages and find my answers. I realize you keep putting the question to me in ways in which you hope will evoke a “no” response, but I don’t think I’ve ever given you a “no” response to the basic question.

What you are missing is something we have never discussed, and that is the “Electrodynamical” part of the SR theory. I think that part is mostly correct, but I don’t think it contains any “twins paradox”. It deals, more realistically, with moving atoms and electrons and it follows the more realistic lines of the 1904 Lorentz paper. The GR theory, outside of the “cosmological constant” business, tends to be quite a lot about the inner workings of atoms and how they are affected in different ways due to environmental changes.

[Sam5]

The more coy you are, the more deceitful you appear.

I think that statement is rather rude and unwarranted.

If we take a whole board vote, I hardly think I would be voted into the “coy” category. I try to be precise and explicit in my answers. If you can't understand them, that's not my fault.

While we’re asking questions, why don’t you give me your “solution” to the twins paradox, in your own words, not anything you get from a Google search.

[russ_watters]

While we’re asking questions, why don’t you give me your “solution” to the twins paradox, in your own words, not anything you get from a Google search. Is that an attempt to avoid my question? A simple yes or no is all I asked.

[Sam5]

SeanF,

From your de Sitter link:

"he could think of no form of differential equation which could have solutions representing waves whose velocity depended on the motion of the source. In this case the emission theory would lead to phase relations such that the propagated light would be all badly "mixed up" and might even "back up on itself". He asked me, "Do you understand that?" I said no, and he carefully repeated it all. When he came to the "mixed up" part, he waved his hands before his face and laughed, an open hearty laugh at the idea!"

No, there won’t be any “mixed up” light because the light photons from the binaries are regulated in space as they travel, so there is no such thing as “fast light” overtaking “slow light”.

Does “fast sound” ever overtake “slow sound” in the air? No. Fast sound can overtake slow sound if the sound is moving to the observer via two different media, such as “air” and “metal” or “water” and “air”.

[Kaptain K]

How can the train and embankment observers breathe in a vacuum?
IT IS A THOUGHT EXPERIMENT!!! Things that are irrelevent are left out to get to the gist of the point that is being explained. Maybe the "observers" are solar powered inhabitants of the third moon of Aldeberan IV, who evolved in airless space and don't need oxygen! In other words, the trivialities such as the metabolism of the observer(s) is irrelevent to the hypothetical situation being discussed.

[Sam5]

How can the train and embankment observers breathe in a vacuum?
IT IS A THOUGHT EXPERIMENT!!!

Just please calm down. I don’t think there is any reason you need to raise the size of your letters.

I don’t think you understand the situation. SeanF wants to leave the air molecules out so I will not have a stationary background grid with which to measure the speed of the light from the moving train mirror to the moving observer.

[Musashi]

How are air molecules stationary?

[Sam5]

How are air molecules stationary?

The air at the surface of the earth is stationary relative to the embankment if there is no wind blowing. And we average out the molecular vibrations of the molecules due to heat energy and imagine them as fixed, like when you’ve got no “wind image” on your Doppler weather radar.

That gives us a good background grid that is the same over the train as over the embankment, so we can judge how fast the light is moving relative to the background grid. You know, like on graph paper.

[Kaptain K]

Light does not need a medium. There is no aether. "c" is measured by the observer relative to his frame of reference, what ever and where ever it is.

[Glom]

Light does not need a medium. There is no aether. "c" is measured by the observer relative to his frame of reference, what ever and where ever it is.

We all agree. Unfortunately, the crux of this thread is that Sam5 doesn't.

[Sam5]

Musashi,

When we have a background grid, we’ve got to start making some hard decisions and asking some difficult questions, such as, “Who’s comoving space grid (or “inertial frame grid”) do we use, that of the observer or that of the emitter?”

If you say “there is no grid”, then you will be very perplexed and you will try to separate the emitter and observer into two different worlds, and then you will wind up with two separate grids that run all the way from the emitter to the observer, but one grid is fixed with the emitter, and another grid is fixed with the observer. So you wind up with two separate “ethers”, even though you are trying to avoid having even a single ether.

But with the unrealistic two-grid system, what happens in one world either doesn’t happen in the other world, or conflicting things happen simultaneously and you wind up with an enigma and a big paradox.

Take for example, the chart below. Here we have an observer on the left and an emitter on the right. In the first 5 seconds, we have a wave leaving the emitter at c, while the emitter and the observer are both stationary relative to each other. So the wave moves to the left at c, and proceeds on toward the awaiting observer at c.

But then at second 6, we see that the emitter is moving to the right while emitting a new wave toward the left.

So, with which reference body do we fix the speed of the light at “c” at second 6? Relative to the observer? Or relative to the emitter?

In the graph I’ve fixed it at c relative to the emitter, and you clearly see that since the emitter is moving at c in the direction away from the observer, the observer will never receive the second wave.

But if I had fixed the second wave’s velocity at c with reference to the observer, then the new wave would have to leave the emitter at the velocity of 2c, relative to the emitter. You will notice that at time 6, the first wave is now moving away from the emitter at the velocity of 2c relative to the emitter, and we can’t have that.

So what do we do? How do we resolve this dilemma? This is the same dilemma that turns up in the university “Doppler Effect” drawings I posted links to. Most SR supporters just ignore the dilemma and pretend it doesn’t exist.

http://im1.shutterfly.com/procserv/47b3cc24b3127cce8339ab5d51d80000001610

[Musashi]

So you are trying to use the air molecules around the train and embankment to stand in for a grid? What if the air molecules at those places are moving at different rates? Not to mention, light does not need air to propigate. So, why not use a vacuum? In a vacuum with an embankment with rails, and a train, you can pick one (or the other) as stationary relative to the other, which will simplify the problem.

[SeanF]

SR deals with the speed of light in a vacuum, and so all of Einstein's thought experiments (remember, they are demonstrations) that deal with light are set up in vacuums.

How can the train and embankment observers breathe in a vacuum?
You know, I've got to admit, you surprised me with this - I expected you to ask how there could be lightning without an atmosphere.

If you don't know the answer to your question, though, you could ask a scuba diver or an astronaut. Not that it's really relevant.


In Chapter 9, Einstein said it was a train on rails. He didn’t say anything about vacuum.

That's because he made it a vacuum in Chapter 7 and never changed it back. You do realize that each chapter builds on previous chapters, right? Besides, since the whole theory is based on the speed of light in a vacuum, don't you think it's pretty obvious that the thought experiments would be taking place in vacuums?

I’ve already mentioned to you that this experiment is conducted at the surface of the earth. It’s not like the 1905 experiments that are conducted at unknown places in space. Einstein didn’t put the train in a vacuum, so why are you trying to? What you try to do is change the parameters of Einstein's own thought experiments to get the results you want.

No, you're the one trying to put things into the thought experiments that aren't there - like the atmosphere.

Here is an animation that explains a little about the Doppler effect and binaries. The animation doesn’t take into account the travel time of the light.

BINARY DOPPLER ANIMATION (http://instruct1.cit.cornell.edu/courses/astro101/java/binary/binary.htm)
What's your point? Do you still think de Sitter is about Doppler? By the way, that doesn't look to me like it's showing the light moving at c relative to the emitting stars . . .


I think I’ve already answered that question several times.

I assume your talking about the GR time-dilation question, and you haven't answered anything. I still don't know where you stand. You think that GR doesn't predict time dilation (not clock changes, time)? You think GR does make those predictions, but that part (and just that part) is wrong?


Does “fast sound” ever overtake “slow sound” in the air? No. Fast sound can overtake slow sound if the sound is moving to the observer via two different media, such as “air” and “metal” or “water” and “air”.

But your local c-regulator theory has the media moving relative to each other. If you have two stars moving in opposite directions relative to us (like the two binaries at the appropriate point in their orbit) and they both release photons at the exact same time, won't the photon released by the incoming star reach us first? After all:

Both photons will reach the edge of their star's "medium" at the same time (assuming the stars are the same), but both stars will have moved in that time. So one photon will be closer to us than the other when they enter the "interstellar" medium. That means that photon will hit the Earth-local "medium" first, which means it'll hit us first. QED.

Therefore, the stars won't appear to be where they were (relative to each other), and that distortion would be detectable.


SeanF wants to leave the air molecules out so I will not have a stationary background grid with which to measure the speed of the light from the moving train mirror to the moving observer.

No, Einstein left the air molecules out (Chapter 7, Paragraph 3, Sentence 3) because he's presenting a theory regarding the speed of light in a vacuum! You want to put them back in so that you can say, "Look, this guy's theory regarding the speed of light in a vacuum wouldn't really work when there's air around!" What's the point?

[Sam5]

We all agree. Unfortunately,


Glom, light might not need an “ether” in the traditional sense, but it needs a speed regulator. In SR theory, the speed regulator for the light is fixed with each of the relatively moving frames. There are two frames, there are two speed regulators, and they are moving relatively in the SR theory.

Have you studied the theory itself? Can we discuss the theory, Einstein’s own words?

The light-speed-variation “crossover” effect between the two relatively moving “ethers” or “c-regulators” first turns up in this equation in SR theory:

http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img11.gif

That’s why we have the c – v and c + v terms in that equation.

If light “always travels at c”, then why does Einstein have the c – v and c + v terms in that equation?

[Sam5]

You know, I've got to admit, you surprised me with this

Are you trying to pretend I was not joking?

Why don't we just forget the whole thing. I've got a lot of stuff I need to be doing.

[SeanF]

The light-speed-variation “crossover” effect between the two relatively moving “ethers” or “c-regulators” first turns up in this equation in SR theory:

http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img11.gif

That’s why we have the c – v and c + v terms in that equation.

If light “always travels at c”, then why does Einstein have the c – v and c + v terms in that equation?

What's the sentence right before that formula? "Taking into consideration the principle of the constancy of the velocity of the light we find that..."

The reason c - v and c + v are there because v is the velocity of the emitter and receiver. When there is no relative motion between the emitter/receiver and the observer, v = 0, and light is moving at c relative to stationary, so the light and the emitter are one light-minute apart in space after one minute in time.

When there is relative motion between the emitter/receiver and the observer, v > 0, and light is moving at c relative to stationary, so the light and the emitter are less than one light-minute apart in space after one minute in time - the distance is dependent on c (velocity of light) - v (velocity of emitter).

[Musashi]

Sam, you have misunderstood that formula. Just because it uses c-v and c+v does not mean that it is saying that light moves faster than or slower than c.

[Sam5]

But your local c-regulator theory has the media moving relative to each other. If you have two stars moving in opposite directions relative to us (like the two binaries at the appropriate point in their orbit) and they both release photons at the exact same time, won't the photon released by the incoming star reach us first? After all:

Both photons will reach the edge of their star's "medium" at the same time (assuming the stars are the same), but both stars will have moved in that time. So one photon will be closer to us than the other when they enter the "interstellar" medium. That means that photon will hit the Earth-local "medium" first, which means it'll hit us first. QED.

Therefore, the stars won't appear to be where they were (relative to each other), and that distortion would be detectable.


You are right about part of that, but no wave ever “outruns” another or “overtakes” another. This is a peculiarity of nature that we can observe in sound waves. If you revolve two electric bells around a center (such as putting them on the ends of a meter stick and revolving the stick in the middle around a pole fixed to the ground), then we will hear a changing Doppler shift from the two bells, and we will be hearing some waves from the near bell first, and some waves from the far bell later, but since this is a continuing process, all we notice are the Doppler shifts, and none of the sound waves ever “outrun” any other wave, not as long as they are traveling through the same medium.

You might hear blueshifted Bell-A wave #1200 from the left bell and redshifted Bell-B wave #1198 from the right bell at the same time, but those waves will reach you at the same time and travel through air at the same rate. Bell-A wave #1200 is NOT traveling faster than the normal speed of sound just because it’s bell was moving toward you when it was emitted. No. It is blueshifted “at the bell”, meaning, in this case, Bell-A wave #1200 leaves the bell traveling at 1,100 fps relative to you, BUT it leaves Bell-A at 1,100 fps minus the speed of the bell toward you, since the bell is moving through the air in your direction. This is why the blueshift takes place at the bell.

And this is what happens to the light at the revolving binaries.

But a little more detail about light. Since the binaries are apparently carrying their own local ether with them, I think the light would leave the binaries at about c relative to the binaries but then it would very soon begin to slow down and be regulated by the deep space medium through which the binaries are moving. Where that slowdown takes place, that’s where the shift takes place.

What de Sitter imagined was the “fast” and "forward" binary adding its speed to the light and that light maintaining that “fast” speed all the way to the earth, but as in the bell example, that doesn’t happen.

Look, if it would make you feel any better, I think the second half of the SR paper, the “Electrodynamical” part, is probably mostly right and I think it is from that part that the GR theory was eventually developed.

[Sam5]

The reason c - v and c + v are there because v is the velocity of the emitter and receiver. When there is no relative motion between the emitter/receiver and the observer, v = 0, and light is moving at c relative to stationary, so the light and the emitter are one light-minute apart in space after one minute in time.


Both observers are stationary relative to their own c-regulator. There is no absolutely “stationary” or absolutely “moving” observer in the theory. There are only TWO relatively moving observers and THEY see the speed of light travel at c only in their own frames.

He regulates the rates of the clocks while they are both relatively “stationary” by means of sending out a light signal from A to B and back to A again. The time from A to B is the same as the time from B to A, while the two frames are both stationary, OR as seen only in their own frames by observers in each frame. But there is a “crossover” of velocities when the two frames move relatively.

That’s why he keeps saying one frame is looking into the other frame. That’s why he says the “moving rods” have shrunk, “as seen from” the “stationary” frame. But both observers see the same phenomenon when they look into each other’s frames.

When there is a “crossover” between frames, that’s exactly what it means, because the c-regulator for observer A is stationary with observer A and the c-regulator for observer B is stationary with observer B, and the two c-regulators move in the theory relative to one another. He has two “ethers” in the theory that are moving relatively.

You keep claiming that only one frame is “really” stationary, but it’s not. Each observer “sees” his own frame as being “stationary” but the other as “moving”.

Look up the “third system” in Section 3 of the paper. That system is moving to the left relative to K1, and he says the “third system” is “at rest” with system K. So, in that example, the “moving” K1 system is seeing the "stationary" K system move, and that’s where he gets his idea that direction of motion doesn’t matter, the rods will “shrink” in both directions, as “seen” by either observer looking into each other’s frame.

The guy wasn’t trying to trick anyone. He tricked himself. He fooled himself.

[Sam5]

Sam, you have misunderstood that formula. Just because it uses c-v and c+v does not mean that it is saying that light moves faster than or slower than c.

Sorry but you are missing the crossover. The paper clearly says:


But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v, so that

http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img31.gif

[Sam5]

Musashi,

When a ray of light traveling in one system is measured in the other relatively moving system, the speed is clearly c - v or c + v, just as in the moving train example in his book Chapter 9.

You need to understand this. The speed of light is not "c relative to all systems", not even in the SR theory, not when the two systems are moving relatively.

This is why none of you can agree on what happens in the twins paradox. That's why none of you can resolve it. That's why all of you have different methods of trying to resolve it, yet you disagree among yourselves constantly. But you all jump on me, because you can all agree that you don't like me.

[Musashi]

You are right that I missed that. I don't even have the paper. I have worked out a graph similar to yours, but I have no where to put it. You are right that the speeds are not all c relative, in my graph. From an obersvation point that is not moving, I get two pulses travelling at c in opposite directions. If I change refrence frames to one of the pulses, I get the previous observation point moving away at c and the other pulse moving away at 2c. I know what the twin experiment is, but I haven't dealt with that at all here. Maybe the reason we can all disagree with each other, but all disagree with you is because you are wrong? Is that even a possibility? Maybe it is because of the way you address people in your posts? To me, it seems like you deliberatly ignore or misinterpret things in order to put your 'theory' in the best light. Sound and light are not analogous. The doppler pictures from UCLA do not show light speeding up or slowing down, etc.. You claim you are not being coy, but then you ignore the questions put to you. Whatever the case is, I will freely admit that I am out of my depth here. I am not a physicist, I have not studied physics for 10 years, so I don't mind being wrong. I will continue to follow this conversation, but I am not sure if I will be able to contribute anything else. I will try to understand more fully the physics involved, and maybe I will be of more help later.

[russ_watters]

I'll just repeat:
Do you believe that most scientists accept or do not accept the validity of Einstein's Relativity?

For example, do you accept that the scientists who designed and the scientists who operate the GPS system utilize Einstein's Relativity in the operation of the GPS system? Yes or no please.

[Kaptain K]

One more time!

There is no "c-regulator'" There is no aether. There is no medium. As long as you continue to try to compare light to sound (or ripples on a pond, or any other wave through a medium) you will continue to get the wrong answers. Light does not follow the same rules as waves through a medium. Light follows its own rules, and those rules are relativity. For nearly a century, experiment after experiment have confirmed the validity of relativity, to higher and higher precision. Until and unless you can either:
1) show that the math of relativity is incorrect.
or
2) Devise a (real, not thought) experiment that invalidates relativity.
All of your semantic nitpicks with thought experiments is just blowing smoke.

[Glom]

The problem with thought experiments is that they only give answers based on your own postulates. In other words, if you believe in c-regulators, then the result of your thought experiment will reflect that. Hence, to give the result of a thought experiment, that was based on the use of c-regulators, as proof of the existance of c-regulators, in circular.

[SeanF]

Sam, you have misunderstood that formula. Just because it uses c-v and c+v does not mean that it is saying that light moves faster than or slower than c.

Sorry but you are missing the crossover. The paper clearly says:


But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v, so that

http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img31.gif

Yes. That's because, when measured from the stationary system, the initial point of k is moving with a velocity of v, and the light is moving with a velocity of c - therefore, in the stationary system, the light and k are separating at a rate of c-v. But the light (when measured in the stationary system) is moving through "space," moving relative to "stationary," at exactly c.

In the system that is moving with the initial point k, the light and k are separating at a rate of c.


You need to understand this. The speed of light is not "c relative to all systems", not even in the SR theory, not when the two systems are moving relatively.

The speed of light is always c relative to the "stationary system," regardless of which system is considered to be "stationary."

[SeanF]

The problem with thought experiments is that they only give answers based on your own postulates. In other words, if you believe in c-regulators, then the result of your thought experiment will reflect that. Hence, to give the result of a thought experiment, that was based on the use of c-regulators, as proof of the existance of c-regulators, in circular.

And that's exactly what Sam5 is trying to do.

No, wait, not exactly - Sam5 is trying to disprove Relativity (which postulates no c-regulators) by dealing with the thought experiments under the "yes c-regulators" postulate of his own theory. He's trying to prove to us that Relativity is logically flawed, but he's doing so with a chain of logic that starts out with the assumption that Relativity is wrong.

Which is also circular reasoning.

[Sam5]

The problem with thought experiments is that they only give answers based on your own postulates. In other words, if you believe in c-regulators, then the result of your thought experiment will reflect that. Hence, to give the result of a thought experiment, that was based on the use of c-regulators, as proof of the existance of c-regulators, in circular.



No, not true. The c-regulator concept is in the SR theory itself. Einstein uses a different c-regulator for each of his two frames. Light travels at c within each regulator, but at c + v and c – v when the observers look over into each other’s frames, or when the different observers move within each other's frames, and this is what you deny, although he clearly states it in his paper and in his 1916 book. He even gives you the math equations telling you what they “see” when they look into each other’s frames.

The c-regulator merely regulates the speed of light to c, within each frame, but it’s not c when the observers look over into each other’s frames because the two c regulators are moving relatively.

In Chapter 9 of his book he said: “Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.”

In this example, he has the train traveling through the embankment’s c regulator, and that is why he says the train is “hastening towards” the beam from B, since the train and the beam are converging at a speed of c + v. And he says the train is “riding on ahead” of the beam coming from A, which means they are converging at a speed of c – v. This is because of the “crossover” viewing.

The embankment viewer is seeing the train moving through his own c-regulator toward the B flash, so he sees his own c-regulated light moving relative to the train in his frame at c + v.

Do you not understand this?

[Glom]

What does it mean to "look over into each other's frame"?

[Sam5]

And that's exactly what Sam5 is trying to do.

No, wait, not exactly - Sam5 is trying to.................


No, that’s not true. You just don’t understand what I’m talking about.

Why don’t you guys get together and solve the twins paradox riddle? Every one of you disagrees about how it should be done.

You are concentrating all your efforts on attacking me, just so you can avoid facing up to the fact that none of you can solve the twins paradox mystery. That’s why you’ve dropped that subject and you are all concentrating on attacking me.

[SeanF]

In Chapter 9 of his book he said: “Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A.”

In this example, he has the train traveling through the embankment’s c regulator, and that is why he says the train is “hastening towards” the beam from B, since the train and the beam are converging at a speed of c + v. And he says the train is “riding on ahead” of the beam coming from A, which means they are converging at a speed of c – v. This is because of the “crossover” viewing.

The embankment viewer is seeing the train moving through his own c-regulator toward the B flash, so he sees his own c-regulated light moving relative to the train in his frame at c + v.

Do you not understand this?

But the train viewer sees the light from both B and A moving at c relative to himself, not at c+v and c-v. He sees the light from B reach him first because, from this reference frame, the flash at B happened first.

Do you not understand this?

[Glom]

Why don’t you guys get together and solve the twins paradox riddle? Every one of you disagrees about how it should be done.

Read this (http://www.clavius.org/bibaulis2003.html). You are employing a similar tactic to Percy. You're saying that we can't agree and hence you're right. Fallacy.


You are concentrating all your efforts on attacking me, just so you can avoid facing up to the fact that none of you can solve the twins paradox mystery. That’s why you’ve dropped that subject and you are all concentrating on attacking me.

They're attacking what you say, not you.

[Sam5]

What does it mean to "look over into each other's frame"?


Well, if all observers are stationary, then there is only one frame, and the speed of light is “c” in that frame, so we’ve got no problem. But when one observer “starts to move”, if we see him “move”, then that means we are not in his frame, we are in our own frame looking into or at his frame move through our frame. When he moves relative to our frame, he is moving through our frame’s c regulator.

Our frame, since we see ourselves as “stationary”, contains our own c-regulator, so we always see the speed of light as c within our frame. But the “moving” observer is moving within and through our frame, so we see him as moving at c + v or c – v relative to our own c-regulated light in our own frame.

But the other observer sees the situation completely reversed.

Let’s look at the situation from his point of view. He has his own c-regulator and he sees us as moving through it. So he sees us as moving at c + v or c – v relative to his own light traveling in his own c-regulator.

When he sees us moving relatively, he sees us move through his c regulator, and when we see him moving relatively, we see him moving through our c regulator.

All of this takes place concurrently. He sees us “move” through his c regulator and sees us moving relative to his own regulated light at c + v or c – v, while we see him moving through our own c regulator and moving relative to our own regulated light at c + v or c – v.

This way of looking at things in the SR theory is not easy to understand at first, because Einstein tends to use mathematical equations to explain the effects, rather than using simple words like I have just done.

[Sam5]

Why don’t you guys get together and solve the twins paradox riddle? Every one of you disagrees about how it should be done.

Read this (http://www.clavius.org/bibaulis2003.html). You are employing a similar tactic to Percy. You're saying that we can't agree and hence you're right. Fallacy.


No, I’m not “right” just because you can’t agree. I’m right for the reasons I’ve explained to you in great detail, which you don’t understand. So you not only don’t understand the twins paradox, you don’t understand what I’m trying to tell you. But instead of you guys trying to figure out the twins paradox and present me with the correct solution, thereby proving that I am “wrong”, you just attack me over and over again. But that doesn’t make you “right” and me “wrong”. That just reveals that 1) you can’t solve the twins paradox, and 2) you can’t understand what I’m talking about. So, instead of admitting that you don’t understand the twins paradox OR what I’m talking about, you attack me in unison, and that makes you all feel better. Your best “solution” to this whole thing is for me to just go away, but that is not a proper solution to the twins paradox.

[Sam5]

But the train viewer sees the light from both B and A moving at c relative to himself, not at c+v and c-v. He sees the light from B reach him first because, from this reference frame, the flash at B happened first.


Then the train observer must calculate that the light is departing B at c – v and departing A at c + v.

Here’s how the train example is different from the relatively moving frames in the original SR theory.

First, in the SR theory the frames are identical. But in the train example, the embankment frame has great mass while the train frame does not. That’s why the train has to run its engine to move along the embankment frame.

If we tried this in the original SR examples, using some kind of traction to pull one frame along by grabbing onto the other frame, then both frames would move relatively due to Newton’s Third Law of motion. But this doesn’t happen in the train example, because of the great mass of the embankment frame when compared to the train frame. So what happens in the train example is, the train frame really moves while the embankment frame does not.

So in this example, you can’t have the train as being “stationary” while the “embankment moves”, unless you disobey some laws of physics, and if you disobey some laws of physics, then you've got a science fiction story and not a physics theory.

That is why Einstein does not say in the train example that the train observer sees the light moving at “c” relative to him. He only says the train observer sees the B flash first. And he tells why, because “in reality”, the train observer is moving through and relative to the embankment’s c-regulator, which is exactly what the embankment observer sees.

[russ_watters]

I'll just repeat:
Do you believe that most scientists accept or do not accept the validity of Einstein's Relativity?

For example, do you accept that the scientists who designed and the scientists who operate the GPS system utilize Einstein's Relativity in the operation of the GPS system? Yes or no please.

[Sam5]

Yes or no please.

After you give me your complete solution to the twins paradox, in your own words.

[SeanF]

That is why Einstein does not say in the train example that the train observer sees the light moving at “c” relative to him. He only says the train observer sees the B flash first. And he tells why, because “in reality”, the train observer is moving through and relative to the embankment’s c-regulator, which is exactly what the embankment observer sees.

So Einstein doesn't say that the train observer sees the light moving at c relative to him, does he?

Chapter 11, fifth paragraph:
"A co-ordinate system K then corresponds to the embankment, and a co-ordinate system K' to the train."

And the sixth paragraph:
"The relations must be so chosen that the law of the transmission of light in vacuo is satisfied for one and the same ray of light (and of course for every ray) with respect to K and K'."

And the last paragraph:
"We thus see that the velocity of transmission relative to the reference-body K' is also equal to c."

K' is the train, right? So he's saying the velocity of transmission relative to the train is c, isn't he? The train observer is motionless relative to the train, isn't he? Thus the speed of light is c relative to the train observer, isn't it?

It's been said numerous times on this thread, and I'm going to say it again. You do not understand the Theory of Special Relativity.

So, we'll define x as the spatial distance between the emission of the light and the reception of the light (viewed from the K frame), and y as the temporal distance between the emission of the light and the reception of the light (viewed from the K frame).

We shall also define x' as the spatial distance between the emission of the light and the reception of the light (viewed from the K' frame) and y' as the temporal distance between the emission of the light and the reception of the light (viewed from the K' frame)

x<>x'
y<>y'
x/y=c
x'/y'=c

Speed of light in a vacuum is constant.

And for Pete's sake, how many times do I have to tell you what that "in reality" referred to?!

[kilopi]

While we’re asking questions, why don’t you give me your “solution” to the twins paradox, in your own words, not anything you get from a Google search.
My solution (http://mentock.home.mindspring.com/twin2.htm) is basically; the same as Einstein's solution in his original 1905 paper--there is no paradox.

PS: Thanks Starbuck

[Starbuck]

While we’re asking questions, why don’t you give me your “solution” to the twins paradox, in your own words, not anything you get from a Google search.
My solution (http://mentock.home.mindspring.com/twin2.thm) is basically; the same as Einstein's solution in his original 1905 paper--there is no paradox.

I'm sure 3141.59265 will fix this shortly, but for the sake of the excruciatingly lazy, the url should be this (http://mentock.home.mindspring.com/twin2.htm)

[Sam5]

I'm sure 3141.59265 will fix this shortly, but for the sake of the excruciatingly lazy, the url should be this (http://mentock.home.mindspring.com/twin2.htm)

Excuse me, but I think you meant to say 3.14159265.

[Sam5]

Speed of light in a vacuum is constant.


So in the train example, we are to completely disregard all three of Newton’s Laws of Motion, and we are to believe that the train is “stationary” while the earth “moves”, when the train turns on its engine and starts its wheels to moving? I’m sorry, but I am not prepared to disregard those Laws, and I doubt if any working physicist in the entire world would either. A physics professor might, but a real physicist would not.

I think it would help you if you read more of Einstein’s actual papers, rather than relying on high school-level website thought experiments.

You can not “shrink” the embankment frame as seen by the train observer because in 1907 Einstein wrote:

“The shape of a body in the sense indicated we will call its ‘geometrical shape’. The latter obviously does not depend on the state of motion of a reference frame.”

Then in his 1916 book he says:

“The rigid rod is thus shorter when in motion than when at rest, and the more quickly it is moving, the shorter is the rod.”

At this point, the guy is telling us conflicting things. In 1905 he said the geometrical dimensions do shrink. In 1907 he said they do not shrink. In 1916 he said again that they do shrink.

He constantly changed the parameters of his theory to fit whatever point he was trying to make at the moment.

In 1905 and 1916 he said there was “no ether”, then in 1920 he said there WAS an ether.

He specifically said in 1920:

“Recapitulating, we may say that according to the general theory of relativity space is endowed with physical qualities; in this sense, therefore, there exists an ether. According to the general theory of relativity space without ether is unthinkable; for in such space there not only would be no propagation of light, but also no possibility of existence for standards of space and time (measuring-rods and clocks), nor therefore any space-time intervals in the physical sense.”

So, whatever you say here, is based on whatever you want to pick and choose out of what all he said over the years. And whenever you want to totally disregard any of the real Laws of Physics and of Nature to make your point, you just do it.

If you want no c-regulator in a frame, then you quote him saying there is “no ether”. But if you want a c-regulator, then you can quote his 1920 paper saying there is one. If you want “no geometrical shrinking” you quote him saying “nothing shrinks”, but if you want things to shrink, then you quote him saying “The rigid rod is thus shorter when in motion than when at rest.” If you need to make up some lost time at the end of your thought experiments, you just add that time to the beginning of your thought experiments.

You are constantly changing the parameters of your own thought experiments, just as he did. And you are violating the laws of physics when you do it, such as by having the train “stationary”, while the entire mass of the earth moves when the train turns on its engine and starts its wheels to moving, and by removing all the air at the surface of the earth.

By having the train as “stationary” and the earth “moving” when the train starts its engine and starts its wheels to rolling, you are violating Newton’s Third Law of motion by not having the train experience an equal and opposite reaction when its wheels turn. You are violating Newton’s Second Law, because the small “motive force” of the train’s wheels turning is causing the entire mass of the earth to move. You are violating Newton’s First Law of motion by ignoring the forces impressed upon the train when its wheels start to turn.

And when you violate so many laws of physics, you can make up any story you want, but all you’ve got as a result is a science fiction story. And you see nothing wrong with this or with your technique of violating the real laws of physics in your thought experiments

[Glom]

Excuse me, but I think you meant to say 3.14159265.

Starbuck was referring to kilopi, who got typed out his url incorrectly. kilopi = 1000×pi = 3141.59265

[Sam5]

Excuse me, but I think you meant to say 3.14159265.

Starbuck was referring to kilopi, who got typed out his url incorrectly. kilopi = 1000×pi = 3141.59265


Thank you.

[Sam5]

And for Pete's sake.......


SeanF, I don’t see how you can be such a non-mainstream guy. I mean, I’m following the mainstream cosmology trends on this, while you are sticking to the old myths.

Just within the past three or four years, mainstream papers have started referring to a body’s local “comoving space”. Whatever this stuff is, this seems to be the local speed regulator for light. You can look it up on google. Ned Wright mentions it too. The comoving space theory has something to do with the big bang theory, and on the large universal scale, the distant galaxies are fixed inside their own local “comoving space” as they move away from our own local “comoving space”. So we’ve got “space moving through space”, but they usually call it “expanding space”, except that space doesn’t ever seem to “expand” on a small local scale. Anyway, this "comoving space" acts like the "local ether".

I prefer to say that an astronomical body’s “fields” move through space with the body, and these “fields” might possibly be the elusive “ether”, with each local field group controlling the local speed of light locally.

Here, read this:

“One such concept of general relativity is that we should not think of galaxies rushing away from us through a fixed space, we should think of galaxies more or less standing still in their local space but that space itself is expanding. This is a difficult concept but it can be described mathematically. In this concept, light leaving a distant galaxy is traveling at velocity c toward us in that local space but since space itself is expanding and that local space is moving away from us, the velocity of a light packet toward us (defined in a certain way) is less than c until the packet gets to our local space and the velocity can even be negative for some early portion of the time light travels to us.”

www.cosmologymodels.com/general.html+star+comoving+space+light+velocity&hl =en&ie=UTF-8]comoving (http://216.239.57.104/search?q=cache:heYGdJo-NzQJ:

And look at what Ned Wright says:

“in these cosmological variables the speed of light is c with respect to local comoving observers”

www.astro.ucla.edu/~wright/cosmo_02.htm+wright+comoving+&hl=en&ie=UTF-8]LINK (http://216.239.57.104/search?q=cache:2VLA-v5k9zUJ:LINK (http://216.239.57.104/search?q=cache:oRu3MfTeCacJ:[url) TO SOURCE

This last statement relates our solar system to the entire universe. I think that inside our solar system, here locally, the sun is the center of our solar system’s “comoving space” and I think the earth is the center of our most local “comoving space”. So, on earth we measure the local speed of light to be “c”. I believe that every other astronomical body is the center of its own local “comoving space”, and each body would measure the local speed of light, at the body, to be “c”.

So I am very mainstream about this new concept, while you are not. You are sticking to the old incorrect way of thinking about how light travels around the universe, and, gradually, your point of view is becoming non-mainstream.

I am also very mainstream in my position that the known Laws of Physics can not be violated, while you violate them all the time in your thought experiments.

[SeanF]

Speed of light in a vacuum is constant.


So in the train example, we are to completely disregard all three of Newton’s Laws of Motion, and we are to believe that the train is “stationary” while the earth “moves”, when the train turns on its engine and starts its wheels to moving? I’m sorry, but I am not prepared to disregard those Laws, and I doubt if any working physicist in the entire world would either. A physics professor might, but a real physicist would not.

Don't the laws of physics say that if you're already moving, you need a force to stop? And don't the laws of physics say that if you're in contact with a moving embankment and friction is causing you to move with it, you will need to apply continuous force in order to override that friction and stay stopped?

(The answer to both those questions, is "Yes, they do." It is no violation of the laws of physics to say the embankment is moving and the train is at rest.)


I think it would help you if you read more of Einstein’s actual papers, rather than relying on high school-level website thought experiments.

You can not “shrink” the embankment frame as seen by the train observer because in 1907 Einstein wrote:

“The shape of a body in the sense indicated we will call its ‘geometrical shape’. The latter obviously does not depend on the state of motion of a reference frame.”

Then in his 1916 book he says:

“The rigid rod is thus shorter when in motion than when at rest, and the more quickly it is moving, the shorter is the rod.”

At this point, the guy is telling us conflicting things. In 1905 he said the geometrical dimensions do shrink. In 1907 he said they do not shrink. In 1916 he said again that they do shrink.

He constantly changed the parameters of his theory to fit whatever point he was trying to make at the moment.

In 1905 and 1916 he said there was “no ether”, then in 1920 he said there WAS an ether.

He specifically said in 1920:

“Recapitulating, we may say that according to the general theory of relativity space is endowed with physical qualities; in this sense, therefore, there exists an ether. According to the general theory of relativity space without ether is unthinkable; for in such space there not only would be no propagation of light, but also no possibility of existence for standards of space and time (measuring-rods and clocks), nor therefore any space-time intervals in the physical sense.”

So, whatever you say here, is based on whatever you want to pick and choose out of what all he said over the years. And whenever you want to totally disregard any of the real Laws of Physics and of Nature to make your point, you just do it.

If you want no c-regulator in a frame, then you quote him saying there is “no ether”. But if you want a c-regulator, then you can quote his 1920 paper saying there is one. If you want “no geometrical shrinking” you quote him saying “nothing shrinks”, but if you want things to shrink, then you quote him saying “The rigid rod is thus shorter when in motion than when at rest.” If you need to make up some lost time at the end of your thought experiments, you just add that time to the beginning of your thought experiments.

You are constantly changing the parameters of your own thought experiments, just as he did. And you are violating the laws of physics when you do it, such as by having the train “stationary”, while the entire mass of the earth moves when the train turns on its engine and starts its wheels to moving, and by removing all the air at the surface of the earth.

By having the train as “stationary” and the earth “moving” when the train starts its engine and starts its wheels to rolling, you are violating Newton’s Third Law of motion by not having the train experience an equal and opposite reaction when its wheels turn. You are violating Newton’s Second Law, because the small “motive force” of the train’s wheels turning is causing the entire mass of the earth to move. You are violating Newton’s First Law of motion by ignoring the forces impressed upon the train when its wheels start to turn.

And when you violate so many laws of physics, you can make up any story you want, but all you’ve got as a result is a science fiction story. And you see nothing wrong with this or with your technique of violating the real laws of physics in your thought experiments

Nobody's violating the laws of physics. At any rate, as kilopi has already pointed out, right now Relativity is part of the laws of physics.

I don't have access to all of Einstein's papers, but I can pretty easily guess that you're misinterpreting what you're reading. Considering your difficulty with "Now in reality," I can only imagine how you're dealing with "in the sense indicated" or "in this sense." I'll have to see if I can track down a book of his papers.

[kilopi]

So I am very mainstream about this new concept, while you are not.
Do you think it is mainstream to think that special relativity is inconsistent, or basically wrong somehow?

[AstroMike]

I'll have to see if I can track down a book of his papers.

This may help.

http://www.amazon.com/exec/obidos/tg/detail/-/0691084076/qid=1070838186/sr=1-23/ref=sr_1_23/102-3096083-1304140?v=glance&s=books

[Sam5]

Don't the laws of physics say that if you're already moving, you need a force to stop? And don't the laws of physics say that if you're in contact with a moving embankment and friction is causing you to move with it, you will need to apply continuous force in order to override that friction and stay stopped?

(The answer to both those questions, is "Yes, they do." It is no violation of the laws of physics to say the embankment is moving and the train is at rest.)



I don’t think you really want a real train to “stay stopped” in space in the sense you are talking about. It’s ok for the train to “stay stopped” relative to the surface of the earth, and it requires no energy to do that. Gravity and friction does that for the train.

But if you really want to really want to “stop” the train, relative to our area of space, you are going to have provide it with a tremendous amount of power to overcome: 1) the rotation of the earth (1000 mph at the equator), 2) the motion of the earth around the sun (18.6 mps), 3) the motion of our solar system around the center of the galaxy (250 mps), 4) and the motion of our galaxy through local space (about 600 kps). Now I ask you, do you really want to try to do this with a real train? I know you would do it in a second in one of your thought experiments, but would you try to do it with a real train?

[Sam5]

Mike, Sean,

When I click on the book image, I get a photo of Volume 8. I wonder if they are just selling that volume?

I’ve got Volume 2, “The Swiss Years 1900-1909” and it’s filled with great stuff. The guy really knew a lot about electrodynamics. He was an expert on the early atomic theories. This is the type of material such as is contained in the second half of “On the Electrodynamics of Moving Bodies”, and I think that section of his SR theory is basically correct. It is from this part of the theory that he developed some of his best and most successful theories, such as the E-mc^2 equation and the “gravitational redshift” theory.

This is very interesting, since the first half, the kinematical part, is flawed, but the second part is not. Why? Because in the first part he limited the “universal speed” of physical objects to “c” and he messed up the “clock dilation” stuff. But in the second part, he limited the “local speed” of physical objects to “c”, and he deduced what was going on inside atoms here at the earth. What he came up with in the second half of the theory were ideas that matched observation and experiment. Why? Because the observations and experiments were conducted right here on the surface of the earth.

So, if the distant galaxies are moving away from the earth at speeds faster than light, this makes the first part of his paper wrong, but since he limited the local speed of masses to “c”, and that is true and accurate at the surface of the earth, then the second half is correct and it matches experimental observations. This is the stuff that has become “laws of physics”, but the “time dilation due to relative motion” is certainly not a “law of physics”.

[SeanF]

Don't the laws of physics say that if you're already moving, you need a force to stop? And don't the laws of physics say that if you're in contact with a moving embankment and friction is causing you to move with it, you will need to apply continuous force in order to override that friction and stay stopped?

(The answer to both those questions, is "Yes, they do." It is no violation of the laws of physics to say the embankment is moving and the train is at rest.)



I don’t think you really want a real train to “stay stopped” in space in the sense you are talking about. It’s ok for the train to “stay stopped” relative to the surface of the earth, and it requires no energy to do that. Gravity and friction does that for the train.

But if you really want to really want to “stop” the train, relative to our area of space, you are going to have provide it with a tremendous amount of power to overcome: 1) the rotation of the earth (1000 mph at the equator), 2) the motion of the earth around the sun (18.6 mps), 3) the motion of our solar system around the center of the galaxy (250 mps), 4) and the motion of our galaxy through local space (about 600 kps). Now I ask you, do you really want to try to do this with a real train? I know you would do it in a second in one of your thought experiments, but would you try to do it with a real train?

All of those velocities you listed are relative. Even the 600kps for the Milky Way is relative to the CMB. You have no idea what the absolute velocity of the Earth's surface is - and that's kind of the whole point.

The bottom line is, you were wrong when you said the train must be moving because it's engine is running, you were wrong when you said that Einstein never claimed the light was moving at c relative to the train observer, you are wrong when you interpret Einstein's "Now in reality" to be declaring the stationary-train-moving-embankment reference frame unreal, you are wrong when you say that the term "co-moving space" refers to some kind of localized ether that follows every individual celestial body, and you are wrong when you say that SR is flawed.

You do not understand SR. From your insistence that Newton's laws demand that the train is moving, I suspect you don't understand Newton!

But let me ask you this - since you insist on harping on the air molecules around the train and the entire mass of the Earth meaning only the train could possibly be moving, does that mean that you accept that SR would work in deep space with two simple observers and nothing else?

[Sam5]

You do not understand SR.

In the link above, what do you think this means:

“In this concept, light leaving a distant galaxy is traveling at velocity c toward us in that local space but since space itself is expanding and that local space is moving away from us, the velocity of a light packet toward us (defined in a certain way) is less than c until the packet gets to our local space and the velocity can even be negative for some early portion of the time light travels to us.”

What do you think “the velocity can even be negative” means?

And what do you think this means:

“in these cosmological variables the speed of light is c with respect to local comoving observers”

What does Wright mean by “local”? Do you remember when I first used that term “local”, in reference to the “local aether fields” and the “local speed of light”?

[SeanF]

Oh, no, you don't, Sam. Not this time. I asked a question first.


But let me ask you this - since you insist on harping on the air molecules around the train and the entire mass of the Earth meaning only the train could possibly be moving, does that mean that you accept that SR would work in deep space with two simple observers and nothing else?

[russ_watters]

Yes or no please.

After you give me your complete solution to the twins paradox, in your own words. So you want an 'I asked you first, no I asked you first, no I asked you first, no I asked you first....' Loop? Little childish, isn't it?

Why won't you answer the question? Do you not like the implications of your answer?

Restated:
Do you believe that most scientists accept or do not accept the validity of Einstein's Relativity?

For example, do you accept that the scientists who designed and the scientists who operate the GPS system utilize Einstein's Relativity in the operation of the GPS system?
Do you think it is mainstream to think that special relativity is inconsistent, or basically wrong somehow? Good luck, kilopi - thats essentially the question I'm trying to get an answer to. Its looking more to me like he's just doing all this for kicks.

[Sam5]

Oh, no, you don't, Sam. Not this time. I asked a question first.


But let me ask you this - since you insist on harping on the air molecules around the train and the entire mass of the Earth meaning only the train could possibly be moving, does that mean that you accept that SR would work in deep space with two simple observers and nothing else?

First, let me clarify something. I think the Electrodynamical Part of SR is mostly correct. Why? Because in it Einstein discussed electrons moving around within a local c-regulator.

Now, in “deep space” with two simple observers? The Kinematical Part of the theory? No. Relative motion can not have any affect on clocks because the clocks don’t know the relative motion is taking place. They don’t “know” it because they feel no force at themselves or they feel no changing force.

Ok, now, please answer my questions.

[Sam5]

russ,

I will answer your question if you will answer my question.

I can answer your question, but you can’t answer my question. So if I answer your question first, I’ve got no way to compel you to attempt to answer my question. And if I answer your question first, you’ll probably make some excuse about why you still won’t answer my question. You’ll probably say, “You didn’t answer my question the way I wanted you to, so I’m not going to answer your question,” or something like that.

Now I see that you actually have three questions to my one.

I’ll make a deal with you. You answer my question, and I’ll answer two of your questions.

[Musashi]

You know Sam, you could just answer the questions and then if the others refuse, you could claim a moral victory...

[Sam5]

Oh, no, you don't, Sam. Not this time. I asked a question first.


Ok, so what do you think “the velocity can even be negative” means?

And what do you think this means:

“in these cosmological variables the speed of light is c with respect to local comoving observers”

What does Wright mean by “local”? Do you remember when I first used that term “local”, in reference to the “local aether fields” and the “local speed of light”?

[Sam5]

You know Sam, you could just answer the questions and then if the others refuse, you could claim a moral victory...

Lol, I've had plenty of those already, but no one here seems to notice them.

I would love to answer Russ' question. I've got a great answer for him. But I want to hear his solution to the twins paradox.

[kilopi]

answer your question if you will answer my question.
I guess I missed one of your questions. Which one?

[Sam5]

Sean,


Why do you suppose Ned is now using the term “local” rather than “all”? Do you remember when I referred to local aether fields and local observers within them? Or have you forgotten already?

[kilopi]

[I would love to answer Russ' question. I've got a great answer for him. But I want to hear his solution to the twins paradox.
He could just use mine. Turns out, it's the same as Einstein's, in his 1905 paper.

[Celestial Mechanic]

[Snip!]I would love to answer Russ' question. I've got a great answer for him.[Snip!]
But will it fit in the margin? 8)

[Sam5]

[Snip!]I would love to answer Russ' question. I've got a great answer for him.[Snip!]
But will it fit in the margin? 8)

Ok, let’s hear your resolution of the twins paradox? :D

[Sam5]

You know Sam...

Let's hear yours too. Now is the chance for you to prove me wrong. Solve the twins paradox for us. :D

[freddo]

Without your answer, how can you be proven wrong?

[Sam5]

Without your answer, how can you be proven wrong?

I've already given my response to the twins paradox several times, and I'd like to learn the opinion of others, such as yourself. How would you solve the twins paradox? You can find my answer on some of the earlier pages of this thread.

[freddo]

Ok, I found your 'solution.'

2 Problems:
- You are denying a change of reference frame
- You claim that this statement: "The speed of light in a vacuum (c) is the same for all observers in inertial frames." is erroneous.

I don't agree with either of these assertions - and I'm not about to regurgitate the contents of the thread to explain why.


Ah, no wonder we can't agree. You've just said that Einstein was talking bartsibrel!

No. A photon always travels at the same speed relative to all observers, but it's path can be changed by gravity or equivalent effects.

[Sam5]

Ok, I found your 'solution.'

- You claim that this statement: "The speed of light in a vacuum (c) is the same for all observers in inertial frames." is erroneous.

I don't agree with either of these assertions -

Well this physicist and Ned Wright of UCLA disagree with you and agree with me:


“One such concept of general relativity is that we should not think of galaxies rushing away from us through a fixed space, we should think of galaxies more or less standing still in their local space but that space itself is expanding. This is a difficult concept but it can be described mathematically. In this concept, light leaving a distant galaxy is traveling at velocity c toward us in that local space but since space itself is expanding and that local space is moving away from us, the velocity of a light packet toward us (defined in a certain way) is less than c until the packet gets to our local space and the velocity can even be negative for some early portion of the time light travels to us.”

www.cosmologymodels.com/general.html+star+comoving+space+light+velocity&hl =en&ie=UTF-8]comoving (http://216.239.57.104/search?q=cache:heYGdJo-NzQJ:

And look at what Ned Wright says:

“in these cosmological variables the speed of light is c with respect to local comoving observers”

www.astro.ucla.edu/~wright/cosmo_02.htm+wright+comoving+&hl=en&ie=UTF-8]LINK (http://216.239.57.104/search?q=cache:2VLA-v5k9zUJ:[url) TO SOURCE

The old "inertial frame" idea is obsolete.

[Sam5]

freddo,

So what's your resolution of the twins paradox?

[Kaptain K]

So what's your resolution of the twins paradox?

Mine is the same as Einstein's (and the rest of the gang here).

There is no paradox!!!!

[Musashi]

For me, it isn't about proving you wrong, it's about trying to figure out what you are saying. Above that, I have never addressed the issue of the twins paradox, but I will now. I do not know the answer to the paradox (if there is one). I do not have the book, so I cannot study the issue. I will freely admit that I have no answer to that. Maybe that is why I haven't addressed it. Do not expect to hear an answer from me about the twins paradox. I don't understand why it is so important to you to hear my answer to the paradox. If I wanted to talk to you about the paradox, I would have done so by now, surely. This is a busy time fo year for me. I cannot say how long it would take me to understand the twins paradox issue under normal circumstances, but I can gaurante that I will not be able to address that with total comprehension untill at least after the holidays. Furthermore, having a discussion with you is tiring because you refuse to stick to an issue. Instead, when you are asked to clarify your position or demonstrate your understanding, you make a snide comment, ignore the post, or move to a different subject. If you wish, I can address some points about the train experiment, but I am reluctant to do so because I doubt you will be able to contribute any criticism to them. Not because of your level of understanding, but because of your refusal to respond in a constructive or rational manner.

[kilopi]

Well this physicist and Ned Wright of UCLA disagree with you and agree with me:
And look at what Ned Wright says:

“in these cosmological variables the speed of light is c with respect to local comoving observers”
How does that particular quote agree with you? I thought you were advocating different values?

[SeanF]

You do not understand SR.

In the link above, what do you think this means:

“In this concept, light leaving a distant galaxy is traveling at velocity c toward us in that local space but since space itself is expanding and that local space is moving away from us, the velocity of a light packet toward us (defined in a certain way) is less than c until the packet gets to our local space and the velocity can even be negative for some early portion of the time light travels to us.”

What do you think “the velocity can even be negative” means?

And what do you think this means:

“in these cosmological variables the speed of light is c with respect to local comoving observers”


I believe you're talking about galaxies that are so distant that the expansion of space between us becomes a noticeable effect.

In that case, I'd have to say that the rate at which we're separating isn't the same thing as the difference in our inertial frames. So, yes, the light is moving at c relative to both inertial frames, but as long the observers are separated enough for the expansion to contribute, it won't really seem like it.

You'll note that that's almost identical to what your link says.

Your link says "[the light] is traveling at velocity c toward us in that local space." Get that? "Towards us"? "that local space?" This theory says that the light is moving at c towards us while it's still in the emitters "local c-regulator"! Does yours?

Your link also says "the speed of light is c with respect to local comoving observers.' Get that? The light's velocity is c relative to observers at the local galaxy as well!

Your link also says "since space itself is expanding and that local space is moving away from us, the velocity of a light packet toward us (defined in a certain way) is less than c until the packet gets to our local space." My post above says, "as long the observers are separated enough for the expansion to contribute, it won't really seem like it." Same thing.

BTW, the "negative velocity" simply means that the galaxy could be far enough away that the expansion of space would be greater than c, which would mean the light is actually receding from us until it overcomes the expansion.



What does Wright mean by “local”? Do you remember when I first used that term “local”, in reference to the “local aether fields” and the “local speed of light”?
See, Sam, this is why you need to deal with the math of the theories and not the words. You read that and you thought, "Gee 'local co-moving space' kind of sounds like 'local aether fields', so they must be the same thing!" No.

You both used the word "local." Whoop-dee-dee. Does your theory also apply to "local" phone calls and "local" television programming? After all, they both use the word "local," too.

[Wally]

Glom, you left out the part where, when George "turns around" to head back to earth, he observes Fred's clock immediately leap ahead of his own. During the return trip, he still perceives it to run slow, but it'll always still be ahead of his. Hence when he gets back to earth, there's no disagreement that Fred has aged more than he.

Ah, the enigmatic element shows itself. I still have to grasp the reason for this clock jump.

I've seen where all 3 reference frames were grafted on the same time/space graft, and it becomes immediately clear why the clock appears to jump ahead as soon as the 3rd FoR is entered. Wish I could remember the site (although I'm sure there's quite a few out there).

[Wally]

Glom, you left out the part where, when George "turns around" to head back to earth, he observes Fred's clock immediately leap ahead of his own. During the return trip, he still perceives it to run slow, but it'll always still be ahead of his. Hence when he gets back to earth, there's no disagreement that Fred has aged more than he.
That's not right. When George turns around to head back to Earth, he observes Fred's clock immediately speed up. During the return trip Fred's clock will appear to George to be running faster than George's clock. Fred, however, will only see George's clock speed up when Fred "sees" George turn for home (which will be some time after George actually turns for home).

Sorry I'm lagging so far behind in responding. . .

Eroica, you're incorrect. At no point do either Fred or George "see" the other's clock as running faster than theirs. Time dilation at relativistic speeds occurs regardless of the "direction" the travelers are traveling. The other's clock does in fact "jump ahead" when one of the observers changes reference frames, but he then notes the other's clock as again going slower than his own.

[Sam5]

For me, it isn't about proving you wrong, it's about trying to figure out what you are saying.

Ok, I’ll try to explain it more simply.

Any kind of clock “slowdown” is never caused only by “relative motion” because a clock experiences no “forces” impressed upon it by the “relative motion”. An “inertial frame” clock doesn’t even know it’s moving, by Newton’s First, Second, and Third laws of motion.

However, a moving clock can change rates if it experiences acceleration (since a “force” is placed upon the inner workings of the clock), or if some other kind of force is caused by that motion, such as when a clock moves “through something”, such as through a stationary magnetic or electric field.

For example, if you move a coil of wire near a stationary magnet, or if you move a magnet near a stationary coil of wire, you create an electron flow in the coil of wire. But this is NOT due to “relative motion”. This is due to the electrons in the coil of wire experiencing and “feeling” a changing magnetic “flux” which is just simply a changing magnetic field strength at the electrons.

If you place the magnet and the coil ten feet away from each other, and move either one of them around in a small circle, there is no electron flow produced in the coil of wire. Why? Because they are too far apart and the magnetic field is too week at a distance of 10 feet for the electrons in the coil to “feel” the changing flux.

So, here is proof that it is not the “relative motion” that causes the electron flow, but the changing strong magnetic field strength at the coil and at the electrons inside the coil that causes it, when the magnet is moving very near the coil of wire, or when the coil is moving very near the magnet, when the electrons inside the coil are moving through the magnetic field and feeling the changing field strength.

If you have a small electric clock that uses a small magnet and coil armature motor, that clock will tick at a fairly steady rate, but if you bring in a large industrial magnet and move it near the clock, the clock will change rates because the coil feels a stronger changing magnetic flux from the industrial magnet than it feels from the small magnet inside the motor of the clock. The field strength of the industrial magnet will overpower the field strength of the small magnet inside the clock and the movement of the large field near the small coil of wire inside the clock will change the rate of the clock, not because of the “motion” but because the electrons inside the coil feel a strong and changing magnetic field strength.

In this experiment, we can move either the large industrial magnet OR the clock, but it’s not the “relative motion” that changes the rate of the clock, it is the changing magnetic field strength that does it. So the clock and the magnet do not even know they are “moving”, since the acceleration is so very slight. All the electrons in the coil know is that powerful magnetic field is changing strength near them, and the electrons feel that changing field strength and that causes them to flow through the wire.

Regarding light, all light must have a speed regulator through which it travels. If the speed regulator travels with the body of reference, such as with the earth, then the light at the earth will travel at c at the surface of the earth. What is this “speed regulator” for light in space? I’m not sure, but I think it is a combination of the electric, magnetic, and gravitational fields of space, which are at their full field strengths at the surfaces of astronomical bodies and that travel with astronomical bodies. At the surface of the earth the earth’s fields are stronger than the fields of any other space object. At the surface of the sun, the sun’s fields are stronger than the fields of any other space object.

So, an observer will NOT measure the speed of light to always be c, everywhere, if the light photons are traveling somewhere other than very near the observer. In other words, an observer will measure the speed of the photons to be c at the surface of the earth, but he will calculate the speed of photons to be less than c when they are traveling near the surface of a massive body such as the sun. So, while an observer will measure the photon speed at the earth to be c, he will measure the photon speed at the sun to be less than c.

He can not “see” the photons while they are at the sun, since they are at the sun, but he can measure the deflection of a light beam that passes the sun and eventually arrives on earth, and when those photons finally reach the earth, even though they are traveling at c on earth, the observer can calculate what their speed was when they passed near the sun.

We can only “see” photons that are directly entering our eyes at the moment, but with all the scientific instruments we have today, we can determine that those photons were deflected when they passed massive bodies, and we can calculate at what speed they were traveling when they passed those bodies. Also, we can calculate at what speed they left their emitters. If the emitters were moving relative to the earth when the photons were emitted, the initial speed of the photons relative to the earth would be either c – v or c + v. We can determine this by means of calculations, even though the photons, when they finally reach our instruments and eyes, are traveling, locally, at us, at the speed of c. And if the photons left the emitter while traveling at c + v relative to the earth, they slowed down to c when they arrived at the earth. If they left their emitter at c – v relative to the earth, then they speed up to c when they arrive at the earth.

[Sam5]

So what's your resolution of the twins paradox?

Mine is the same as Einstein's (and the rest of the gang here).

There is no paradox!!!!


Ok, so, if, based on SR theory, both observers “see” each other’s clocks tick more slowly when compared to their own clocks, then how can only one of the two clocks “really” be “lagged behind” in time when the two clocks are united?

[Sam5]

There is no paradox!!!!

In science, you can’t just have one astronomer yelling another one, “The universe is NOT expanding”, while the other astronomer yells at the first one, “The universe IS expanding.” You’ve got to prove (at least in theory) why your point of view is the correct one.

[Sam5]

Time dilation at relativistic speeds occurs regardless of the "direction" the travelers are traveling. The other's clock does in fact "jump ahead" when one of the observers changes reference frames, but he then notes the other's clock as again going slower than his own.

Sorry, no. There is no "jumping ahead" in SR theory. The dilation only occurs when the motion is relaive, in either direction of travel, but it never "jumps". Einstein never said in the paper that the clock rate "jumps ahead". He said only that it changes and the amount of change depends only on the relative velocity, no matter what the direction is.

What you are doing is NOT quoting Einstein. You are quoting a popular book or website that contains a false resolution of the twins paradox.

[Sam5]

I've seen where all 3 reference frames were grafted on the same time/space graft, and it becomes immediately clear why the clock appears to jump ahead as soon as the 3rd FoR is entered. Wish I could remember the site (although I'm sure there's quite a few out there).

Stay away from those sites. Read the original theory yourself and figure it out for yourself. There is no "third frame of reference" in the SR theory except for the K' frame that overlaps the K frame and travels with it, so that third frame IS the K frame.