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Originally Posted by Flaney
Tensor wrote:
The slow down of time is not a concept of SR. There are two postulates of SR(which has been mentioned to you before):

1) The laws of physics are the same in all inertial frames.
2) The speed of light is the same as observed in all inertial frames

Which of the above don't you agree with and why?
Originally Posted by Flaney
2)
How does the reciept of a photon translate into speed? It is an event with no other information (except energy). Subsequent photon receptions build an event history where derived speed can be calculated (based on local clocks). Any event chain variations can be attributed to relative velocity differences.
What about single photon emission and detection? If we know the emitted time, detection time, and the distance we can get the velocity of a single photon.

Originally Posted by Flaney
Photons I emit are not detectble by me any more. As I travel along, I release an additional photon, which I have carried away some distance; it will be farther from the previous photon I emitted by vt. These photons get detected at some point as events that occur in a dilated history: the transmit events no longer coincide with detection events.
Sure they do. All it requires is a frame transformation.

Originally Posted by Flaney
Should the reciever think the emitters clock is running slow because the events come less frequently?
Nope. Because if doesn't matter if the emitter is moving away or approaching, the clock of the emitter is running slow either way. By your logic, the clock should appear to be running faster, if approaching, because events come more frequently when approaching.

Originally Posted by Flaney
To measure a photon's speed, I need to measure it's position at two places and the times it reaches each position. But once I have detected it, it is gone. I have to detect some other photon that was moving near it at the second location, where it subsequenly vanishes. Only if we assume all photons are equal can a specific speed be determined.
Do you have a reference showing all photons are not the same (except for energy)? How are photons of monochromatic light different?

Originally Posted by Flaney
This does not need to mean photons are racing around with different speeds. How quickly does the photon speed drop when traversing a non-vacua medium?
That would depend on the index of refraction for that medium, wouldn't it?

Originally Posted by Flaney
Space itself might 'force' photons to a local C in a similar manner.
Might? You have a reference for this happening?

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Originally Posted by kilopi
So, CM, in your opinion, does my Ann/Bob page describe anything wrongly?
I know you didn't ask me Kilopi, but your page described it very well, IMHO.

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Just in case you forgot Sam:

Originally Posted by Tensor
Celestial Mechanic's post (in the previous thread) using the math demonstrates this. You have never commented on those posts. If you believe there is a paradox, could you show us exactly where Celestial Mechanic is wrong?

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Originally Posted by Flaney
To measure a photon's speed, I need to measure it's position at two places and the times it reaches each position.
Right, exactly.

To actually measure the speed of starlight at the earth, I think we would have to find a star that blinks, such as a pulsar. Then we would have to set up two telescopes with light receptors, one on a mountain top, and another in-line with that one, down in a valley. Then we would have to measure the time it takes between when the first receptor sees the blink and when the second one sees it, and we need to know the exact distance between the first and second receptor. And we would have to calculate the different rates of the two atomic clocks due to altitude differences.

But I suspect it would be measured at “c” here on earth, because I think our gravitational field regulates light speed here and it also regulates the rates of the atomic clocks, different rates at different altitudes.

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Originally Posted by Tensor
Just in case you forgot Sam:

Originally Posted by Tensor
Celestial Mechanic's post (in the previous thread) using the math demonstrates this. You have never commented on those posts. If you believe there is a paradox, could you show us exactly where Celestial Mechanic is wrong?
Tensor,

Thanks. I haven’t had time to get to those posts yet. It takes me a lot of time to figure out the exact errors in some of the other paradox resolution thought experiments. Plus, I’ve got to get out a lot so my feet won’t go dead on me.

I think it would almost be easier if everyone just forgot about the 1905 paradox, that can not be resolved, and they should go on to study the 1911 paper. That paper adds a gravitational field and that takes away the paradox, since in the 1911 paper, only one clock actually slows down, it really slows down, and both observers agree. The slow observer “sees” the fast one running fast, and the fast one “sees” the slow one running slow. In this, there is no paradox.

He left out “fields” and “acceleration” in the 1905 theory, and this is one reason he got the paradox.

The little error in the 1905 theory wasn’t bad in and of itself. What makes it seem bad are all the guys who claim it is not an error. They are, in effect, trying to “prove” Einstein’s 1905 error to be correct and not an error. And that drives some of us up the wall, those of us who have studied many of his other time-related papers. He actually corrects the error in the 1911 paper, but his wording is not very easy to understand.

He was a theoretical physicist, so he couldn’t come right out and say in his 1911 paper, “Opps, I made a blooper back in my 1905 paper, and I’d like to correct it now.” If he had said that, no publisher would have trusted his 1911 paper. That’s why he had to be careful how he worded the 1911 paper, so it wouldn’t strongly admit the blooper in the 1905 paper. That’s why he said in 1911:

“The principle of the constancy of the velocity of light holds good according to this theory in a different form from that which usually underlies the ordinary theory of relativity.”

See? He couldn’t say outright, “I was wrong in part of my 1905 paper, and this paper includes my correction.” No theoretical physicist can say that in a subsequent paper and still expect his latest papers to be taken seriously and be published.

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Originally Posted by Flaney
Space itself might 'force' photons to a local C in a similar manner. Fast photons would not pass up its slower brothers if some depth or maximum transmission transition is exceeded. Beyond such a point, the events would be compressed, but all would be finally zipping along at C.

Hey! That’s something like what I said earlier! I think it is “the fields” of space, not just “empty space”, that control the speed of light through different areas of space. Sort of like an airplane flying in air. Some air is turbulent. Some air is moving one way. Some air is moving the other way. So, the GROUND SPEED changes, but the AIR SPEED does not. In the case of light I think the “space speed” changes in different places in space, but the “ground speed” is determined by the gravitational potential at the surface of the astronomical body where the local light photon speed is measured by means of an atomic clock that is also “geared” to the local gravitational potential.

7. Originally Posted by Sam5
Sean,

You:
”The first reference frame has B remaining stationary at x=3 and A moving from x=0 to x=3

The second reference frame has A remaining stationary at x=0 and B moving from x=3.75 to x=0.’

Me:
This is incorrect.

A the beginning of the thought experiment, points A and B are stationary in K and K1. Since no one is moving, K and K1 overlap along their x axes.

Distance AB = BA

So, you can’t change the distance from B=x3 to B=x3.75 in the middle of the thought experiment.
I'm going to bed now, and I will probably sleep in (I think I might be coming down with the flu). Before I do, I'm going to say this:

Some time ago in this thread, I offered you the opportunity to simply say, "I believe that SR's constant-speed-of-light postulate is flawed, and thus the theory will necessarily reach the wrong conclusions, but the logic from the postulate to the conclusions is sound." You refused to accept it, because you believe that SR is internally flawed.

Now, you originally stated that this flaw was apparent in the clock paradox, because it predicts different times for the two clocks from the two observers. I just showed you how the math in SR (Einstein's own equations!) show that there is no paradox.

So now you say "But you can't change distances." I'm sorry, but the non-absoluteness of time and space is another part of SR. I will show you exactly how SR says the distance does change. Then you'll go back another step up the logic chain and say, "But you can't do this." and then I'll show you how SR leads to that.

Eventually, we'll get all the way back up to the beginning, and you'll say "But the speed of light is not a constant for all observers, even though everything after that works!"

Why don't you just say it now?

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Originally Posted by Tensor
Originally Posted by Flaney
Photons I emit are not detectble by me any more. As I travel along, I release an additional photon, which I have carried away some distance; it will be farther from the previous photon I emitted by vt. These photons get detected at some point as events that occur in a dilated history: the transmit events no longer coincide with detection events.
Sure they do. All it requires is a frame transformation.
How would the photon know (or the receiver) what the state of the source is in order to establish a frame transformation?

Originally Posted by Tensor
Originally Posted by Flaney
Should the reciever think the emitters clock is running slow because the events come less frequently?
Nope. Because if doesn't matter if the emitter is moving away or approaching, the clock of the emitter is running slow either way. By your logic, the clock should appear to be running faster, if approaching, because events come more frequently when approaching.
So, the event chain of photon detections won't be compressed when the source is approaching?

Originally Posted by Tensor
Originally Posted by Flaney
To measure a photon's speed, I need to measure it's position at two places and the times it reaches each position. But once I have detected it, it is gone. I have to detect some other photon that was moving near it at the second location, where it subsequenly vanishes. Only if we assume all photons are equal can a specific speed be determined.
What about single photon emission and detection? If we know the emitted time, detection time, and the distance we can get the velocity of a single photon.
Doesn't this assume information transfer between reference frames beyond photon communication?

Originally Posted by Tensor
Originally Posted by Flaney
This does not need to mean photons are racing around with different speeds. How quickly does the photon speed drop when traversing a non-vacua medium?
That would depend on the index of refraction for that medium, wouldn't it?

Originally Posted by Flaney
Space itself might 'force' photons to a local C in a similar manner.
Might? You have a reference for this happening?
No reference. But it seems to follow from refraction indexes. What might the refraction index of the vacua between various locales?

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Sean,

The kinematical part of the paper has several flaws in it. First, he claims “no ether”, but he uses two of them in the theory. The c-regulator for K and the separate c-regulator for K1. These c-regulators serve as light-propagating “ethers” or “media”, and they are both moving relatively when the frames move relatively. It is with this media that he sets his light-clocks, by sending a beam between A and B and reflecting it back to A.

When each medium is “stationary” with an observer, either B or A, then the times for the light to travel from A to B and back from B to A are the same. But if the observers attempt to measure the travel times for the beams in a moving medium, then the times are not the same. The different path times in moving and non-moving media are reflected in these equations of his:

This is where the relative c – v and c + v photon speeds turn up, as they must turn up with observers and emitters moving within a stationary medium.

Next, the kinematical part doesn’t consider acceleration or gravitational fields. That was a mistake. It’s my understanding that there is no place in observable space where there are “no fields”. In 1911 he switched over to "atomic clocks" rather than light clocks, and atomic clock rates are also affected by acceleration and fields, which reveals his more realistic approach to this subject in 1911.

Third, he is the one who published the symmetry principle that says both observers will see the same thing, but in reverse, when they look at each other.

Four, he is the one who left out the opinion of the A observer in the “peculiar consequence” thought experiment.

You are the one who has his mind fixed on the Bob and Ann example. That’s why you wind up with these four conflicting end times: 12.5, 10, 8, 6.4, whereas Einstein only wound up with two conflicting end times.

You change B from x=3 to x=3.75 before any motion begins.

Tell me, when A and B are stationary, what is the distance from A to B and from B to A? When A moves, why does B’s position remain at x=3, but when B moves, you have him suddenly start from x=3.75? Start him from x=3 and see what happens. Remember, before the motion, AB = BA.

No motion:

A----------B

Ok, now, move B to A and tell me what A sees, and move A to B and tell me what B sees.

Einstein corrected his errors in 1911. He knew more in 1911 than he knew in 1905. You need to study the 1911 paper and forget about Bob and Ann.

10. Kilopi:

I sat down and carefully read your explanations of the Twin Paradox, both the original and the "Redux". They are correct and not "amateurish". If I had done more than skim them initially, I might not have wound up spelling "Ann" as "Anne" in my examples!

11. Some questions for Sam5:

1) What is the value of c in the "c-regulator" in the gravity field of the Earth? Of the Sun? Of a galaxy with an observed redshift of z=3? Of a quasar with an observed redshift of z=3?

2) What is the value of c away from gravitational fields, such as in one of the "voids"? Assuming, for the sake of argument, that the voids really are empty.

3) In our various Ann and Bob examples, where does Ann's "c-regulator" end and Bob's "c-regulator" begin?

4) Can you calculate what the value of c is from reasonable input, say, the mass contained within a volume of space, or something else?

12. Originally Posted by Flaney
How does the reciept of a photon translate into speed? It is an event with no other information (except energy).[Snip!]
You are forgetting about momentum and therefore direction. Very important pieces of information!
Originally Posted by Flaney
Photons I emit are not detectble by me any more.[Snip!]
Have you ever tried shining a flashlight in a mirror? Many of the classic methods of measuring c involved reflected light signals over sizable distances, such as mountain-top to mountain-top.
Originally Posted by Flaney
[Snip!]SR is time series information propagation.
And it works, too!

Edited to add: I forgot about polarization, that is another piece of information that a photon has.

13. Thanks, CM, for the comments. Tensor, yours too. I'll wait for your PM, SeanF! And thanks J*, I haven't had this much fun since the last time.

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Originally Posted by Celestial Mechanic
Some questions for Sam5:

1) What is the value of c in the "c-regulator" in the gravity field of the Earth? Of the Sun? Of a galaxy with an observed redshift of z=3? Of a quasar with an observed redshift of z=3?

2) What is the value of c away from gravitational fields, such as in one of the "voids"? Assuming, for the sake of argument, that the voids really are empty.

Based on what I’ve studied of Einstein’s work and what I’ve studied of experiments and real observation, as published in reputable mainstream papers, it seems that the speed of light in a gravitational field, when measured by an observer using an atomic clock resting on the surface of an astronomical body is “c”, ie 186000 mps, but only when the photons are traveling at the observer and the atomic clock. When the photons are traveling someplace else, the value of c changes, when measured by that same observer and clock.

It seems to me, based on what I’ve studied, that we must use an atomic clock, since they change their rates at different gravitational potential, and light changes its speed at different gravitational potential, so it seems that the local speed of light, at each atomic clock resting in a gravitational field is 186000 mps, when the photons hit, leave, or pass right by that clock.

But if the atomic clock is in deep space (where its tick rate speeds up, due to lack of a strong gravitational potential where it is), and if the clock is moving toward a light emitter (a star), then the measurement of the speed of light photons being absorbed at the clock would be c + v, and if the clock is moving away from the star, the speed of the light absorbed would be c – v, at that clock.

So it seems that the speed is regulated at c only on the surface of a body, when the clock is resting on the surface of the same body. So, at the sun, an atomic clock at the sun would measure c, locally, for the local photons at the sun, but at the earth an earth clock would measure slower than c for the photons when they are moving at the sun, but when sun photons finally reach the earth, they would speed up somewhere in space between the earth and the sun and be measured arriving on earth at c, when absorbed at an earth-based atomic clock, but that clock would have to be resting on the earth, not flying around in the air or in space.

Now, about the galaxy and quasar. Evidently this same rule applies. When the photons are moving around inside that galaxy or at that quasar the same rules would apply locally at them.

If they are moving away from the earth at high speeds, when they first emit the light, it is aimed in the direction of the earth but moving away from the earth. This is sometimes described as “expanding space carrying the photons away from us for a while”. As the photons work their way through the gravitational fields of space that are closer to our own galaxy, they speed up relative to the earth, but the fields in space control the speed of light in space to roughly “c” relative to the bodies that are generating those fields.

At the surfaces of those bodies, when measured with an atomic clock resting on those surfaces, the speed would be c when the photons arrive at or depart from the local clock, but out in space from those bodies and from a local clock, the speed would be somewhat different. An individual atomic clock apparently has to be resting on the surface of a body, and the photons apparently have to be absorbed at that clock, for the “c” rule to apply.

If we’ve got an observer at a star saying his local atomic clock measured a photon leaving his star at c, and if the star is moving away from the earth, then the photon starts out traveling at c – v relative to the earth. It travels through space roughly at c relative to the “background field” of many space bodies, and as it gets near the earth it speed up to c relative to the earth observer and his stationary earth-based clock. When it is absorbed by his clock, the photon is going at c relative to the earth clock, but c + v in the direction away from the star that emitted it and relative to that star.

This is difficult to figure out and describe, because the light propagating medium is invisible. But this seems to work the same way sound works, except that the “media” (mediums) that regulate the speed of light are of three kinds or three “orientations”: 1) the local medium at the star, 2) the space medium, 3) the local medium at the earth. And I think the “media” are the gravitational fields. Near a star it is generated by that star. Out in deep space the fields are generated by many bodies and are “blended” in deep space. Near the earth, the local dominant field is generated by the earth. If you think of the light propagating media being “visible” like “air” or a physical “atmosphere” that extends from a body and into space all the way to another body, then the light propagating media would look like an “atmosphere” that is “thick” at each body, and that thins out between bodies and is “thick” again at the next body. The light would change its speed in just the opposite way that sound changes speed. In “thick” media, light slows down, and in “thin” media, light speeds up. So, looking from a wide view, light slows down near astronomical bodies and speeds up in the space between them.

Deep space would represent the “voids” you mentioned, but I think of it this way: Light going from the sun to a star 10 light years away in our galaxy would travel through this part of our galaxy. If we back off and look at our galaxy as a whole overall unit, then our galaxy is carrying with it all the local fields in our area of the galaxy that are generated by the astronomical bodies in our local area, so the light speed in this area would be an average of c in deep space, with its speed regulated by these local deep-space fields. Relative to a distant high z galaxy, our galaxy is carrying its own local medium (or media) through space with it. Once some of our photons work their way to a distant galaxy, then the fields of that galaxy control the local speed of the photons while they are moving through that galaxy.

15. Originally Posted by Sam5
But if the atomic clock is in deep space (where its tick rate speeds up, due to lack of a strong gravitational potential where it is), and if the clock is moving toward a light emitter (a star), then the measurement of the speed of light photons being absorbed at the clock would be c + v, and if the clock is moving away from the star, the speed of the light absorbed would be c – v, at that clock.
And you still maintain that this is not Against the Mainstream physics??

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Originally Posted by Celestial Mechanic
Some questions for Sam5:

3) In our various Ann and Bob examples, where does Ann's "c-regulator" end and Bob's "c-regulator" begin?

4) Can you calculate what the value of c is from reasonable input, say, the mass contained within a volume of space, or something else?

I think the c-regulators are disregarded in that thought experiment. There is nothing in that thought experiment that controls the speed of light in space. The tick-rates of atomic clocks are not considered, and so the local “speed of light” can not be measured. That thought experiment is based on a misunderstanding of the kinematical part of the SR theory. The theory itself contains errors, and the Ann and Bob experiment contains other errors.

In Einstein’s thought experiments in 1905, there are 2 c-regulators, each of which are stationary with each of 2 frames and each of 2 clocks. When a clock “moves” it moves through the other clock’s c-regulator, but it doesn’t move through its own. It carries its own along with it when it moves. In the “peculiar consequence” thought experiment, the A clock is moving through the B clock’s c-regulator. That’s why in that thought experiment the A clock length contracts and time dilates.

But, based on what Einstein said earlier about the A clock’s own “stationary” point of view, with the B clock “moving”, then the A observer would “see” the B clock moving through the A clock c-regulator, and would see the B clock length contract and time dilate. This is what causes the clock paradox in that theory. Einstein talks about the A clock being "stationary", near the end of Section 3 when he introduces the K’ system, the “third” system or “frame”. He says:

“For this purpose we introduce a third system of co-ordinates K’ , which relatively to the system k is in a state of parallel translatory motion parallel to the axis of X, such that the origin of co-ordinates of system k moves with velocity -v on the axis of X.”

He changes the names of his coordinate systems and his clocks all throughout the paper, so in this example, the k system would be what Sean and I have been calling the K1 system, and this is the system in which the A clock is stationary. Note that he has the third K’ system moving relative to the k (K1) system. Then a few sentences later he says:

“Since the relations between x', y', z' and x, y, z do not contain the time t, the systems K and K’ are at rest with respect to one another,”

That means K’ and K are the same system. That means that the B clock that is in K is the same B clock that is in K’. This means that the A clock is stationary relative to the c-regulator in the K1 (same as the k) system, and the B clock is stationary relative to the c-regulator in the K (same as the K’) system. It also means that A “sees” the B clock and the K system “moving”. His earlier thought experiments say that the B clock “sees” the K1 system and the A clock “moving”. So, there is no “absolute motion” in the theory, only the “relative motion” of the A and B clocks, inside the K1 and K systems, and when they move relatively, their c-regulators move relatively. And that produces the paradox.

The “peculiar consequence” occurs when the B observer “sees” the A clock “really” time dilate, due only to “relative motion”. But, it would have been even more “peculiar” if Einstein had told what the A observer “sees”, since the A clock would “see” the B clock move, length contract, and time dilate in the same amount, and when the A and B observers meet, they would disagree about which clock is “lagging behind”. They would be right there together, touching each other, but “seeing” different times on each other’s clocks, and this is an impossible situation.

This problem was straightened out in 1911, when Einstein had one atomic clock on the sun, another on the earth, and he had the local gravitational field regulating the tick rate of each clock, which conforms to true scientific observation. So, the sun clock runs slow, and the earth clock runs fast. No paradox.

However, as far as I can tell, Einstein never completely straightened out the situation about “moving” clocks. Resting clocks are one thing, but “moving” clocks are something else. He seems to have always relied on his original SR theory to describe the tick rates of “moving clocks”, so the paradox keeps cropping up over the years. The best description of reality (regarding motion) he ever gave, is when he talked (in his 1916 book) about the train moving along the surface of the earth and encountering two light beams, one at c + v, and the other at c – v. In this example, the earth generates the local gravitational field, while the train moves through it. So the earth-based observer, using an earth-based atomic clock, would measure “c” for the speed of light at the earth, while the train observer moving through the gravitational field would measure c + v for light coming toward the front of the train, and c – v for light coming from the rear. In this case, the real c-regulator at the surface of the earth would be its local gravitational field.

17. Originally Posted by Sam5
Einstein said in the paper: “It is clear that the same results hold good of bodies at rest in the “stationary'' system, viewed from a system in uniform motion.”
Yes, but he said that early in Section 4, immediately after discussing the length contraction and before he discusses the "peculiar consequence." So there's no reason to assume that that "peculiar consequence," which hasn't even been mentioned yet, would be included in the "same results" he's referring to here. And, in fact, if you understood the math of SR, you would see that it's not included.

Originally Posted by Sam5
He also said:

“whence it follows that the time marked by the clock (viewed in the stationary system) is slow by seconds per second”

From B’s point of view, B is in “the stationary system”. From A’s point of view, A is in “the stationary system”.
Oh, no. See, at the very beginning, where he first uses the term "stationary system," he says:

Originally Posted by Einstein
Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the "stationary system.''
"Stationary system" does not mean "the system that an observer considers to be stationary," it just means "the specific system that I identified at the beginning." The reciprocity of SR, that either of two moving observers can be considered to be at rest, does not mean that either observer can be considered to be in that specific system (named the "stationary system"). It just means that either that specific system or the other system (or any other system) can be considered to be at rest.

So, when he says that clocks start out at A and B in the "stationary system" (K) and then "the clock at A is moved with the velocity v along the line A-B to B", you can't pretend that the clock at A stays in K (the "stationary system") and B gets put in k (the "moving system"). All you can do is pretend that either system K or system k is actually at rest.

Again, this is what happens when you try to look at the words instead of the math. You:

Originally Posted by Sam5
The guy was a GENIUS!

But I think he could have used a good manuscript editor, because the way he worded things is really confusing to the rest of us.
The way he worded things has certainly confused you. Learn the math, and then you will understand what the words mean.

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Originally Posted by SeanF
Oh, no. See, at the very beginning, where he first uses the term "stationary system," he says:

Originally Posted by Einstein
Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the "stationary system.''
"Stationary system" does not mean "the system that an observer considers to be stationary," it just means "the specific system that I identified at the beginning." The reciprocity of SR, that either of two moving observers can be considered to be at rest, does not mean that either observer can be considered to be in that specific system (named the "stationary system"). It just means that either that specific system or the other system (or any other system) can be considered to be at rest.
I disagree with that. With relative motion only, it is the A and B observers who both identify their own systems as being “stationary”. For a couple of reasons. First, they “see” their system as stationary, while they “see” the other system as “moving”. Their own c-regulator is stationary with their own system, while the other c-regulator is stationary with the other system and “moves” with it. If there are multiple clocks in their system, all the clocks tick at the same rate. And both of two observers have equal rights to claim this, since this motion is only “relative” in the SR thought experiments, especially since neither experience acceleration and neither has a gravitational field. This situation comes about because in the theory he leaves out too many other laws of physics, such as acceleration and gravitational fields.

If you come along and say one system is “stationary” while the other is not, then all you are doing is saying that you are “fixed” inside one of the frames, and that is the frame that you call “stationary”.

That’s why in the Ann and Bob thought experiment, Ann is “fixed” on the earth. This has psychological effect of causing readers to think, “Well, Ann really IS ‘fixed’ and ‘stationary’, since she is on the surface of the earth.” See? It’s a psychological trick of the Ann and Bob thought experiment. It is Bob that you first think of as “zooming off”, so you get the mental impression that “Bob really is moving” while “Ann really is stationary”. That’s why you always want to have Bob seeing his own clock at 4 and Ann seeing her own clock at 5 at the turn-around. Why? Because Ann “really is stationary” while Bob “really is moving.” And so you determine Bob's turn-around time based on what "stationary Ann" says it is. And you claim Bob "agrees", but in SR theory, he doesn't agree.

Einstein played this very same trick on himself in the A and B example, because he defined A as “moving” and he defined B as “stationary”, so he “saw” only A as “moving”, but not B. That’s because he fixed his own mind as being stationary with B.

If you work your equations the other way, from the real point of view of A, then you will have A seeing B move. But you must NOT fall for the error of A “seeing” A “move” from the point of view of “stationary B”. You must see A as being completely stationary from A’s point of view, and B being completely stationary ONLY from B’s point of view.

I think Einstein confused himself by thinking of A as sort of having B’s permission to call A “stationary”, while Einstein himself continued to think of A as “moving”.

This is what you do with Ann and Bob. You determine Bob’s turn-around time by what Ann “sees” on Bob’s clock. Ann “sees” 4, so you say it is 4. But it’s not. As far as Bob is concerned, it is Ann that moves.

Put Ann on the earth, and put Bob on Earth2, solidly fixed inside another star system. He sees Ann and her whole star system move 3 light years at .6c in 5 years. During that time he sees Ann’s clock tick off 4 years.

But you can’t say that he sees Ann move 3 light years at .6c in 4 years by his clock, because when you do that, you have him seeing his own clock time dilated, and he never sees that.

Originally Posted by SeanF
So, when he says that clocks start out at A and B in the "stationary system" (K) and then "the clock at A is moved with the velocity v along the line A-B to B", you can't pretend that the clock at A stays in K (the "stationary system") and B gets put in k (the "moving system"). All you can do is pretend that either system K or system k is actually at rest.

At the start, before the relative motion begins, B is ALREADY in k. And A is in K. Because, before the motion begins, K and k are stationary relative to one another, so the x and x’ axes overlap, and the beginning of the coordinate systems coincide at x=0 and x’=0. They are essentially the SAME system.

When the relative motion begins, A sees B and the K frame move and sees B move toward A at v. B sees A and the k frame move and sees A move toward B at v.

The distance is the same, the velocity is the same, each sees their frame as “stationary”, and each sees the other frame as “moving”. Since we have no big mass like the earth, since we have no acceleration, and since we have no gravitational field, the motion is relative only, and that’s why both can equally consider themselves to be in the “stationary” frame.

If you “see” only A move, then you are fixing your mind only in the K frame. You must learn to switch orientations and fix your mind only in the k frame, and then only in the K frame, and when you can do that, then you can see the paradox.

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Sean,

You have convinced yourself that only Bob “moves” in the Bob and Ann example, and that only “A” moves in the A and B example. As long as you believe that, then you will always try to manipulate the math to make it appear that that is true.

That’s how you wind up with the “infinitely disappearing clock dilation” phenomenon.

First you say Ann sees Bob move for 5 years and sees his clock dilate to 4. Then you say Bob moves for 4 years and sees Ann’s clock dilate to 3.2. Then you can carry this on out further and say, “When Ann’s clock is at 3.2, she sees Bob’s clock at 2.56. When Bob’s clock is at 2.56, he sees Ann’s clock at 2.048. When Ann’s clock is at 2.048 she sees Bob’s clock at 1.6384, etc, etc, etc”, until all the clocks stop.

5 x .8 = 4

4 x .8 = 3.2

3.2 x .8 = 2.56

2.56 x .8 = 2.048

2.048 x .8 = 1.6386

You should never have Bob turning around at 4 by his clock in the first place, since he sees Ann move 3 light years at .6c in 5 years by his own clock.

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Sean,

You are also being guided by this philosophy:

The kinematical part of SR theory is TRUE.

Therefore, all the math MUST WORK OUT.

So, anything I do to get the math to work out and prove the theory is TRUE, is ok, no matter how ridiculous the thought experiments become. If they become very ridiculous, then that is just because of so many “peculiar consequences” of the theory itself, and since the theory is TRUE, any “peculiar consequences” that turn up must be ok.

This is what everybody does in the “twins paradox resolution” thought experiments. If anything odd, unusual, impossible, or bizarre turns up in their own thought experiments, they just claim that is a “peculiar consequence” of the theory itself. They get away with attributing their own mistakes to Einstein.

That's why there are so many DIFFERENT kinds of "resolutions". All mistakes in them are attributed to Einstein and his "peculiar consequences".

Just as you said when you came up with impossible oddities in your own experiments, “That’s relativity!” That’s how you hide your own mistakes. You blame them on Einstein.

21. Originally Posted by Sam5
Originally Posted by SeanF
Oh, no. See, at the very beginning, where he first uses the term "stationary system," he says:

Originally Posted by Einstein
Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced hereafter, we call it the "stationary system.''
"Stationary system" does not mean "the system that an observer considers to be stationary," it just means "the specific system that I identified at the beginning." The reciprocity of SR, that either of two moving observers can be considered to be at rest, does not mean that either observer can be considered to be in that specific system (named the "stationary system"). It just means that either that specific system or the other system (or any other system) can be considered to be at rest.
I disagree with that.
WHAT?! Einstein specifically defines the term "stationary system" as a name for a specific system, a "verbal distinguishment." You can't then go to a later part in the paper and say, "here he says 'stationary system' which means whatever the observer wants to call stationary."

Originally Posted by Sam5
That’s why in the Ann and Bob thought experiment...
Have you noticed that ever since I said "I'll agree to talk about only the 1905 paper if you will" that you have consistently been bringing other things into the discussion, including that "amateur internet" stuff you said we should avoid? Why is that?

Originally Posted by Sam5
Originally Posted by SeanF
So, when he says that clocks start out at A and B in the "stationary system" (K) and then "the clock at A is moved with the velocity v along the line A-B to B", you can't pretend that the clock at A stays in K (the "stationary system") and B gets put in k (the "moving system"). All you can do is pretend that either system K or system k is actually at rest.
At the start, before the relative motion begins, B is ALREADY in k. And A is in K. Because, before the motion begins, K and k are stationary relative to one another, so the x and x’ axes overlap, and the beginning of the coordinate systems coincide at x=0 and x’=0. They are essentially the SAME system.
Originally Posted by Einstein
"If at the points A and B of K there are stationary clocks..."
A and B define points in space in the system K (A and B are not the clocks, they are points in space).

Originally Posted by Einstein
"...[T]he clock at A is moved with the velocity v along the line AB to B..."
The clock is moving relative to the points A and B, which are defined in the system K. Ergo, the clock is moving relative to the system K.

Can that clock be considered stationary, and the other clock to be in motion? Yes. I showed that math in this post. Can the first clock be considered to be motionless in system K? No, it is moving relative to A and B! Can the second clock be considered to be in motion in system K? No, it is at rest relative to A and B!

22. Originally Posted by Sam5
The “peculiar consequence” occurs when the B observer “sees” the A clock “really” time dilate, due only to “relative motion”. But, it would have been even more “peculiar” if Einstein had told what the A observer “sees”, since the A clock would “see” the B clock move, length contract, and time dilate in the same amount, and when the A and B observers meet, they would disagree about which clock is “lagging behind”. They would be right there together, touching each other, but “seeing” different times on each other’s clocks, and this is an impossible situation.
The above paragraph is completely wrong. It is a personal misinterpretation, and has nothing whatever to do with Einstein's 1905 paper. There is nothing within the paper, or Einstein's theory, that would lead anyone to conclude that two people standing beside each other looking at the same clock would see two different times upon that clock.

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Originally Posted by SeanF
WHAT?! Einstein specifically defines the term "stationary system" as a name for a specific system, a "verbal distinguishment."
He said, “Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others...”

Well, he could do that simply by saying “System K”, “System k”, or “Planet A”, “Planet B”, etc.

But he actually fools himself by saying that “System K is the Stationary System”.

Even though he has already said, “The introduction of a “luminiferous ether'' will prove to be superfluous inasmuch as the view here to be developed will not require an “absolutely stationary space'' provided with special properties, nor assign a velocity-vector to a point of the empty space in which electromagnetic processes take place.”

So he says there is no “absolutely stationary space” in space, then down at the beginning of Section 3 he says, “Let us in “stationary'' space take two systems of co-ordinates....", so he proclaims one frame to be “stationary in stationary space”. First he says there is no "stationary space", then he says there is.

You’ve got to watch his wording, it’s very tricky, and that’s how he wound up fooling himself in that theory.

He says he would not use a “luminiferous ether....... in which electromagnetic processes take place”, yet he wound up using two of them, the two c-regulators, one for each of the two systems, where “electromagnetic processes take place” just like they do here on the earth, with a fixed velocity for c, just like at the earth, provided an observer is “fixed” inside the system where the “electromagnetic processes” are taking place. If the observer is not fixed in a system in which the c-regulator is fixed, then “electromagnetic processes take place” differently when an observer is moving relative to that other system and that other ether (that other c-regulator), and that's how he winds up with “peculiar consequences”. The "peculiar consequences" are only in the theory. The velocity-vectors of the two ethers is “0” relative to the clocks that are fixed inside each system, but the velocity vectors of the ethers vary with “v”, relative to the clocks in the other system.

Originally Posted by SeanF
You can't then go to a later part in the paper and say, "here he says 'stationary system' which means whatever the observer wants to call stationary."
Sure I can, because that's what he does.

Near the end of Section 3, he says, “For this purpose we introduce a third system of co-ordinates K’, which relatively to the system k is in a state of parallel translatory motion parallel to the axis of X....” That’s when system k becomes the “stationary” system.

And a few sentences later he says, “Since the relations between x', y', z' and x, y, z do not contain the time t, the systems K and K’ are at rest with respect to one another, and it is clear that the transformation from K to K’ must be the identical transformation.”

That means K’ is K.

So, to k it is k that is “stationary”, and it is K that is “moving”. But to K it is K that is "stationary" while it is k that is "moving". Basically, in the “relative motion” it is K that moves relatively to the left while k moves relatively to the right.

It’s very simple. But it takes re-reading the first 4 sections several times and underlining certain of his statements before one can understand where his mistakes are. If you just skim over all that important stuff, you’ll miss it.

24. Originally Posted by Sam5
So, to k it is k that is “stationary”, and it is K that is “moving”. But to K it is K that is "stationary" while it is k that is "moving". Basically, in the “relative motion” it is K that moves relatively to the left while k moves relatively to the right.

It’s very simple. But it takes re-reading the first 4 sections several times and underlining certain of his statements before one can understand where his mistakes are. If you just skim over all that important stuff, you’ll miss it.
You're calling that a mistake? That's the whole point of relativitiy, isn't it? That either of those systems can be assumed stationary? You can't blame that "mistake" on Einstein, you're going to have to go back to Newton, or Galileo.

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Sean,

If you will take two pieces of graph paper and draw a short horizontal line on both of them and then put a dot in the middle of each line, then label one dot “B” on one piece of paper and label the other dot “A” on the other piece of paper, then you’ll have the two systems.

If you get a photographic light box to back-light the pages. Then hold one page firm in your right hand with your hand resting on the light box, and move the other page left and right with your left hand, then the one in your right hand is the “stationary” system.

Now, rest your left hand on the light box and move the page in your right hand left and right. Then the page in your left hand is the “stationary” system.

Then take both your hands off the light box and move the pages relatively at equal speeds (relative to the light box).

That Relativity!

The background grid on each sheet is the c-regulator for that sheet.

If you try to measure the travel time for a beam of light to move from one point on a sheet to a different point on the same sheet, and then be reflected back, all at the same speed of c, you’ll get equal times for the two different directions.

Now, if you have a Point A and Point B on one sheet, and a Point A’ and Point B’ on the other sheet, the same distance apart on each sheet, and then move the sheets relatively, while you imagine a photon of light traveling across one sheet at c, starting at let’s say “A”, using the A background grid as the c-regulator, then you reflect the photon back to A after it bounces off the moving Point B’, then you’ll find the travel times from stationary Point A to moving Point B’ and back to stationary Point A are NOT the same as when you moved the photon at c from A to B and back to A.

Your principle of the “constancy of the velocity of light” has not been violated, and you are using only one of the c-regulators to regulate the speed of light to c, yet you get different travel times when: 1) you send a beam from a stationary A to a stationary B, where it is reflected back to a stationary A, and 2) when you send a beam from a stationary A to a moving B’, where it is reflected back to a stationary A. That’s because when you reflect the beam moving from a stationary A to a moving B’ and back to a stationary A, the travel distances “shrivel” up, while they do not when you send the light from a stationary A to a stationary B, where it reflects back to a stationary A.

26. Originally Posted by Sam5
Originally Posted by SeanF
WHAT?! Einstein specifically defines the term "stationary system" as a name for a specific system, a "verbal distinguishment."
He said, “Let us take a system of co-ordinates in which the equations of Newtonian mechanics hold good. In order to render our presentation more precise and to distinguish this system of co-ordinates verbally from others...”

Well, he could do that simply by saying “System K”, “System k”, or “Planet A”, “Planet B”, etc.
Yes, he could have. But he didn't.

Originally Posted by Sam5
But he actually fools himself by saying that “System K is the Stationary System”.
He didn't fool himself - you're the only one who is "fooled."

Originally Posted by Sam5
Even though he has already said, “The introduction of a “luminiferous ether'' will prove to be superfluous inasmuch as the view here to be developed will not require an “absolutely stationary space'' provided with special properties, nor assign a velocity-vector to a point of the empty space in which electromagnetic processes take place.”

So he says there is no “absolutely stationary space” in space, then down at the beginning of Section 3 he says, “Let us in “stationary'' space take two systems of co-ordinates....", so he proclaims one frame to be “stationary in stationary space”. First he says there is no "stationary space", then he says there is.
He says there is no absolutely stationary space. Then, in Section 3, he puts the word "stationary" in quotation marks precisely because he knows it's not absolutely true. You do know what the use of quotation marks in a context like this is, don't you? It means "I'm going to refer to it as such even though it is not absolutely such."

Originally Posted by Sam5
Originally Posted by SeanF
You can't then go to a later part in the paper and say, "here he says 'stationary system' which means whatever the observer wants to call stationary."
Sure I can, because that's what he does.
He does? Are you going to present a cite where he uses the term "stationary system" to refer to another system than the one he identified initially?

Originally Posted by Sam5
Near the end of Section 3, he says, “For this purpose we introduce a third system of co-ordinates K’, which relatively to the system k is in a state of parallel translatory motion parallel to the axis of X....” That’s when system k becomes the “stationary” system.

And a few sentences later he says, “Since the relations between x', y', z' and x, y, z do not contain the time t, the systems K and K’ are at rest with respect to one another, and it is clear that the transformation from K to K’ must be the identical transformation.”
Nope, looks like you're not even going to provide a cite where he uses the term "stationary system" at all. He doesn't even use the word "stationary" in your cites.

Originally Posted by Sam5
That means K’ is K.

So, to k it is k that is “stationary”, and it is K that is “moving”. But to K it is K that is "stationary" while it is k that is "moving". Basically, in the “relative motion” it is K that moves relatively to the left while k moves relatively to the right.
But when Einstein says "stationary system," he's referring to K. Always. Even when he considers k to be at rest and K to be in relative motion to k, K is still "the stationary system," by name.

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Originally Posted by SeanF
He says there is no absolutely stationary space. Then, in Section 3, he puts the word "stationary" in quotation marks precisely because he knows it's not absolutely true. You do know what the use of quotation marks in a context like this is, don't you? It means "I'm going to refer to it as such even though it is not absolutely such."

Right. I pointed that out to you earlier, both in the English translation and in the original German.

That also means that the k system also has equal right to think of itself as being “stationary”, while it thinks of the K system as moving. That’s the point I made just two posts up.

Originally Posted by SeanF
Are you going to present a cite where he uses the term "stationary system" to refer to another system than the one he identified initially?

Sure, I already did, but here’s another place, at the beginning of Section 3:

“Now to the origin of one of the two systems (k) let a constant velocity v be imparted in the direction of the increasing x of the other stationary system (K), “

See? They are BOTH “stationary” systems to the observers that are fixed within them. And when they are not moving relatively, they are both stationary to all observers.

When they are moving relatively, they are “stationary” to the observers that are fixed within them. He doesn’t say k “blasts off”. He says, “a constant velocity v be imparted”. He is not using a real “force” here, just an imaginary force. That’s because we have to leave acceleration out of the thought experiments. k moves relative to K as K moves relative to k. The motion is relative not absolute.

And in this separate part, quoted below, the observers in k see k as “stationary” and K’ as “moving”, and since K’ is fixed with K, that means the k observers see K as “moving”.

“For this purpose we introduce a third system of co-ordinates K', which relatively to the system k is in a state of parallel translatory motion parallel to the axis of X, “

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Originally Posted by SeanF
But when Einstein says "stationary system," he's referring to K. Always. Even when he considers k to be at rest and K to be in relative motion to k, K is still "the stationary system," by name.

I’ve already said several times that that is where you are making your big mistake. Because of that attitude, you always fix your mind as thinking that K actually is the ONLY stationary system in all the thought experiments.

But, since this is “relative motion” with no “absolute space” and no “acceleration”, BOTH systems can equally be thought of as the “stationary” system by observers fixed inside each of the two systems.

So observer A in k ALWAYS thinks of k as the “stationary” system. That’s why he said, “It is clear that the same results hold good of bodies at rest in the “stationary'' system, viewed from a system in uniform motion.” This means that “viewed from” the k system, it is the K frame’s bodies that “length contract” while the k bodies don’t length contract at all.

And that’s why he didn’t seek A’s opinion in the “peculiar consequence” thought experiment. He only wanted B’s opinion. He knew that if the A observer expressed an opinion, then A would disagree with the B observer and the paradox would be obvious. The paradox is not so obvious as long as he seeks the opinion of only the B observer.

29. Originally Posted by Sam5
Originally Posted by SeanF
He says there is no absolutely stationary space. Then, in Section 3, he puts the word "stationary" in quotation marks precisely because he knows it's not absolutely true. You do know what the use of quotation marks in a context like this is, don't you? It means "I'm going to refer to it as such even though it is not absolutely such."
Right. I pointed that out to you earlier, both in the English translation and in the original German.

That also means that the k system also has equal right to think of itself as being “stationary”, while it thinks of the K system as moving. That’s the point I made just two posts up.

Originally Posted by SeanF
Are you going to present a cite where he uses the term "stationary system" to refer to another system than the one he identified initially?
??? As I pointed out, the term "stationary system" does not occur in the cites you gave. You're blatantly lying now, aren't you?

Originally Posted by Sam5
but here’s another place, at the beginning of Section 3:

“Now to the origin of one of the two systems (k) let a constant velocity v be imparted in the direction of the increasing x of the other stationary system (K), “

See? They are BOTH “stationary” systems to the observers that are fixed within them. And when they are not moving relatively, they are both stationary to all observers.
Well, they are both the same system until the velocity is imparted to the one. I can't help but wonder if that might not be a translation error anyway. Doesn't make much sense to refer to the "other stationary system" when you didn't refer to "one stationary system" or "one of the two stationary systems." It'd certainly make more sense with a comma, as "the other, stationary system." Doesn't matter, though, because it's still K that's called "the stationary system."

Originally Posted by Sam5
When they are moving relatively, they are “stationary” to the observers that are fixed within them. He doesn’t say k “blasts off”. He says, “a constant velocity v be imparted”. He is not using a real “force” here, just an imaginary force. That’s because we have to leave acceleration out of the thought experiments. k moves relative to K as K moves relative to k. The motion is relative not absolute.

And in this separate part, quoted below, the observers in k see k as “stationary” and K’ as “moving”, and since K’ is fixed with K, that means the k observers see K as “moving”.

“For this purpose we introduce a third system of co-ordinates K', which relatively to the system k is in a state of parallel translatory motion parallel to the axis of X, “
I don't understand where you think you're going with this. Do you think since any system can be considered "at rest" and thus "stationary," that you can then call it "the stationary system" (even though Einstein didn't), and then proceed as if whenever Einstein says something about "the stationary system" he's talking about yours? Language doesn't work that way.

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Sean,

It’s really difficult to figure all this stuff out without using the graph paper to represent the two frames.

Once you make up the graph paper, and when you start moving the pages around, you’ll see what I’m talking about. Then you’ll see their equality.

If you glue one graph paper to your wall and only move the other graph paper, then you won’t see their equality, because you’ll be fixing the glued paper to the frame of the earth. That’s what people do in the Ann and Bob thought experiments. They try to glue Ann to the frame of the solid earth, and then they try to tell you that what she “sees” is “real” because she is “stationary” and Bob is “moving”, and therefore Bob “must” agree with Ann. But if you use the two-graph paper sheets method, you’ll see that both systems are perfectly equal.

It doesn’t matter if we call K “stationary” or k “stationary”, since K is always “stationary” to the observers fixed inside the K frame and k is always stationary to the observers fixed inside the k frame. Both frames are completely equal and interchangeable in this regard, since we are dealing with relative motion.

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