Twin Paradox: Definitive Proof That It's SR?

[Sam5]

??? As I pointed out, the term "stationary system" does not occur in the cites you gave.

Well, let me ask you this. Let’s say the A observer is fixed inside the k system. Then which system does the A observer see as the “stationary system”?

[kilopi]

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.

Seek their opinion? It's a thought experiment--they're figments of his imagination.

[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.

Try again:

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2 tv^2/c^2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.

A and B are points in space in K (Einstein calls it "the stationary system, but we'll assume K to be in motion). There are clocks at A and B in K, which are synchronous when viewed in K. To keep clear that the Clocks and the Points are separate, we'll call the clock at A "Clock 1" and the clock at B "Clock 2". Since we consider the system K to be in motion, we also must consider Clocks 1 and 2 to be in motion.

Now, "...[T]he clock at A is moved with the velocity v along the line AB to B..." Since we consider K to already be in motion, how do we do this? We have to change the velocity of Clock 1. We can't change the velocity of Clock 2, because if we did it would no longer stay at point B. We want to move Clock 1 to Point B and meet up with Clock 2 there, so Clock 2 has to stay at Point B. So, we need to change Clock 1's velocity so that it has a velocity of v relative to K. So, let's say that K's original velocity was -v. This means we would need to change Clock 1's velocity to 0 in order for it to have a velocity of v relative to K.

Now, do you think the math for this is going to end up being in any way different than when I did it above? Even though we now have Clock 1 at rest and Clock 2 moving towards it, we're still going to have Clock 1 lagging behind because it is still Clock 1 that's no longer in K!

[kilopi]

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.

Don't try to tell me what people do in the Ann and Bob thought experiment--they're figments of my imagination.

You've latched onto the idea that Ann's frame of reference is symmetric with Bob's, as that is the basic idea of relatiivity. However, that is not true under Galilean relativity, and it is not true under special relativity. But Galileo (and Newton) recognized that, and Einstein recognized that--even pointing it out in the very paper in which he first proposed the theory of special relativit, as a "peculiar consequence." He did not call it a a paradox, and we know today that it is not a paradox. There are no instances of two people looking at the same thing and seeing two different things, in reality or theory.

That Einstein would recognize that, and want to take his "peculiar consequence" to the limit and produce general relativity is a testament to his "particular genius." But you can't fault him for that.

[Sam5]

Well, they are both the same system until the velocity is imparted to the one.

With no acceleration, with only an imaginary force, and with only relative motion, when a velocity is imparted to one, then a velocity is ALSO imparted to the other. If they used traction, or one pushed off using the other as an inertial mass from which to push off from, then Newton’s Third Law would apply, with an equal and opposite reaction.

But in his thought experiment, psychologically speaking, with you the reader thinking Einstein is “imparting” the motion to only the k frame, this causes you to think he is imparting the motion from his firm position here on earth. That makes you think only the k frame is really “moving”. This is a clever psychological maneuver on his part.

It took dozens of times for me to re-read that sentence, trying to figure out what is causing the motion to be imparted, what “force” is used, before I finally realized that he is leaving out Newton’s laws of motion. No rocket engine, no propellers, no wheels, no pulling or pushing, and no traction. This is a psychological maneuver in a thought experiment, and he uses an imaginary force that does not obey any of Newton’s three laws of motion.

When he says that K’ is moving relative to k, and when he says that K’ is the same as K (fixed with K), then he is saying that K is moving relative to k, and thus we know that observer A, who is fixed within the k system, looks upon the k system as being “stationary”. This is implicit in what he says in the first three sections. This means that K and k and their points of view are completely equal and interchangeable. It only depends on which frame you decide to fix yourself inside. Observers in k have just as much right to claim they are “fixed” and the K observers are “moving”.

He wants you to fix yourself inside the K frame, so you do, so you don’t see the paradox. I can fix myself inside either frame, and I see the paradox.

If we used radio communication between relatively moving A and B, A would say “Hey, you’re moving and time dilating”, while B would say, “No I’m not, you are.” So they would disagree right up to the very end, and they would disagree when they look at each other’s clocks when they unite.

[kilopi]

Well, they are both the same system until the velocity is imparted to the one.

With no acceleration, with only an imaginary force, and with only relative motion, when a velocity is imparted to one, then a velocity is ALSO imparted to the other. If they used traction, or one pushed off using the other as an inertial mass from which to push off from, then Newton’s Third Law would apply, with an equal and opposite reaction.

They are reference frames--they have no mass, and no physical reality. They are figments of the imagination. There is no ruler following me, you, Ann, or Bob around. There is no force necessary to impart a velocity to a reference frame--it has no mass!

[Sam5]

A and B are points in space in K

They are also two points in K1.

A and B are points in space in K (Einstein calls it "the stationary system, but we'll assume K to be in motion).

Relative to what?

There are clocks at A and B in K, which are synchronous when viewed in K.

And synchronous when viewed in K1 before the motion begins.

To keep clear that the Clocks and the Points are separate, we'll call the clock at A "Clock 1" and the clock at B "Clock 2".

We don't need to because they are already called clocks A and B.

Since we consider the system K to be in motion, we also must consider Clocks 1 and 2 to be in motion.

Relative to what?

Now, "...[T]he clock at A is moved with the velocity v along the line AB to B..." Since we consider K to already be in motion, how do we do this? We have to change the velocity of Clock 1

You haven't said what K is "in motion" relative to.

From A's point of view, A sees clock B move toward A in the K1 frame, so you have to "change" the velocity of Clock 2 relative to A.

[SeanF]

Well, they are both the same system until the velocity is imparted to the one.

With no acceleration, with only an imaginary force, and with only relative motion, when a velocity is imparted to one, then a velocity is ALSO imparted to the other. If they used traction, or one pushed off using the other as an inertial mass from which to push off from, then Newton’s Third Law would apply, with an equal and opposite reaction.

But in his thought experiment, psychologically speaking, with you the reader thinking Einstein is “imparting” the motion to only the k frame, this causes you to think he is imparting the motion from his firm position here on earth. That makes you think only the k frame is really “moving”. This is a clever psychological maneuver on his part.

It took dozens of times for me to re-read that sentence, trying to figure out what is causing the motion to be imparted, what “force” is used, before I finally realized that he is leaving out Newton’s laws of motion. No rocket engine, no propellers, no wheels, no pulling or pushing, and no traction. This is a psychological maneuver in a thought experiment, and he uses an imaginary force that does not obey any of Newton’s three laws of motion.

It wouldn't matter what the force is, because only the relative motion is important. As long as we end up with k moving at v relative to K, it doesn't matter how much of the momentum was imparted to k and how much to K. All that matters is that k's velocity relative to K is now v (and K's velocity relative to k is now -v).

When he says that K’ is moving relative to k, and when he says that K’ is the same as K (fixed with K), then he is saying that K is moving relative to k, and thus we know that observer A, who is fixed within the k system, looks upon the k system as being “stationary”. This is implicit in what he says in the first three sections. This means that K and k and their points of view are completely equal and interchangeable. It only depends on which frame you decide to fix yourself inside. Observers in k have just as much right to claim they are “fixed” and the K observers are “moving”.

He wants you to fix yourself inside the K frame, so you do, so you don’t see the paradox. I can fix myself inside either frame, and I see the paradox.

Sam5, take those two sheets of graph paper you have from your previous post. Hold the one with A and B on it with your left hand and the one with A' and B' with your right. Hold them so that A overlaps A' and B overlaps B'.

Now, move them so that A ends up overlapping B'. You can do this by moving your left hand, you can do it by moving your right, you can do it by moving both, but you have to move your hands together.

Now, start with A and A' lined up again. Now, move them so that B ends up overlapping A'. Again, you can do this by moving either hand (or both hands), but you have to move your hands apart.

See? Moving the clock at A to the point B is not the same as moving the clock at B to the point A.

If we used radio communication between relatively moving A and B, A would say “Hey, you’re moving and time dilating”, while B would say, “No I’m not, you are.” So they would disagree right up to the very end, and they would disagree when they look at each other’s clocks when they unite.

No, by the time they meet up, they'd agree on what time was on both clocks. Have to. It'd be a paradox otherwise, wouldn't it?

[SeanF]

A and B are points in space in K

They are also two points in K1.

But their coordinates may be different in K1.

A and B are points in space in K (Einstein calls it "the stationary system, but we'll assume K to be in motion).

Relative to what?

Doesn't matter. Whatever you want. Motion is relative. You were complaining that I couldn't consider the clock going from A to B to be at rest, so I did.

There are clocks at A and B in K, which are synchronous when viewed in K.

And synchronous when viewed in K1 before the motion begins.

To keep clear that the Clocks and the Points are separate, we'll call the clock at A "Clock 1" and the clock at B "Clock 2".

We don't need to because they are already called clocks A and B.

Where do you get that from? Einstein doesn't call them by those designations in his paper.

Since we consider the system K to be in motion, we also must consider Clocks 1 and 2 to be in motion.

Relative to what?

Again, doesn't matter.

Now, "...[T]he clock at A is moved with the velocity v along the line AB to B..." Since we consider K to already be in motion, how do we do this? We have to change the velocity of Clock 1

You haven't said what K is "in motion" relative to.

And it still doesn't matter.

From A's point of view, A sees clock B move toward A in the K1 frame, so you have to "change" the velocity of Clock 2 relative to A.

See, this is why you're getting confused. A is not the clock, A is the point in space. Clock 2 never moves relative to point A. That would be a different thought experiment, with different results.

[Sam5]

It wouldn't matter what the force is, because only the relative motion is important. As long as we end up with k moving at v relative to K, it doesn't matter how much of the momentum was imparted to k and how much to K. All that matters is that k's velocity relative to K is now v (and K's velocity relative to k is now -v).

Actually, their velocity relative to each other is “v”. It doesn’t matter on which side of them you place the “0” coordinate origination point on a graph. You can put it on the left side or the right side. Move the two pieces of graph paper relatively and you’ll see what I mean. K sees kas moving while k sees K as moving. So there is NO "stationary frame". There are only "relatively moving" frames, with observers fixed inside each frame seeing their own frame as "stationary". If you fix yourself inside k, then it is the K frame that the motion is "imparted" to.

[SeanF]

It wouldn't matter what the force is, because only the relative motion is important. As long as we end up with k moving at v relative to K, it doesn't matter how much of the momentum was imparted to k and how much to K. All that matters is that k's velocity relative to K is now v (and K's velocity relative to k is now -v).

Actually, their velocity relative to each other is “v”. It doesn’t matter on which side of them you place the “0” coordinate origination point on a graph. You can put it on the left side or the right side. Move the two pieces of graph paper relatively and you’ll see what I mean. K sees kas moving while k sees K as moving. So there is NO "stationary frame". There are only "relatively moving" frames, with observers fixed inside each frame seeing their own frame as "stationary". If you fix yourself inside k, then it is the K frame that the motion is "imparted" to.

Velocity is a vector, Sam5.

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.

See that negative sign there in front of the v? That's why K' ends up being the same as K. Because k has a positive velocity v relative to K and a negative velocity v relative to K'.

Velocity being a vector is also why you have to move your hands apart to do one thing but together to do the other, as I pointed out above.

[Sam5]

Sam5, take those two sheets of graph paper you have from your previous post. Hold the one with A and B on it with your left hand and the one with A' and B' with your right. Hold them so that A overlaps A' and B overlaps B'.

Now, move them so that A ends up overlapping B'. You can do this by moving your left hand, you can do it by moving your right, you can do it by moving both, but you have to move your hands together.

Now, start with A and A' lined up again. Now, move them so that B ends up overlapping A'. Again, you can do this by moving either hand (or both hands), but you have to move your hands apart.

See? Moving the clock at A to the point B is not the same as moving the clock at B to the point A.

In your four moving points graph example, you have 4 points, A and B, and A’ and B’.

Like this:

A----------B
A’---------B’

In the Einstein thought experiment you have only 2 points, A and B.

Like this:

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

In your graphs example, yes, if I move A and B’ together, then I move my hands toward each other, and if I move B to A’, I move them apart.

Unless, I change the paper to different hands. If I do that, then I can move B to A’ by moving my hands toward each other, and I can move A to B’ by moving my hands apart.

Hey, that’s relativity!

But when I have A on one piece of paper, and B on another, and they are “apart”, I can only move them “together” by moving my hands toward each other. And they are moving toward each other at the relative velocity of “v”.

In the case of just two points, A and B, moving A to B is the same as moving B to A.

[kilopi]

In the case of just two points, A and B, moving A to B is the same as moving B to A.

Yes, and if A then moves away from B, one of them must have been in a non-inertial frame. That's the broken symmetry that Einstein pointed out in his 1905 paper, and you insist is not broken--that somehow the symmetry must still apply because you think Einstein said it had to in that case.

Nothing could be farther from the truth.

[SeanF]

But when I have A on one piece of paper, and B on another, and they are “apart”, I can only move them “together” by moving my hands toward each other. And they are moving toward each other at the relative velocity of “v”.

But in Einstein's thought experiment, we have "points A and B of K". Both A and B need to be on the same sheet of paper. "...[T]the clock at A is moved with the velocity v along the line AB to B...". So, how do you get the clock from point A to point B with the clock at B staying at B? Simple. The clock at B goes on the same sheet of paper as points A and B, the clock at A goes on the other. Regardless of which sheet of paper you move, the clock goes from A to B and lags behind in time.

[Sam5]

But in Einstein's thought experiment, we have "points A and B of K". Both A and B need to be on the same sheet of paper. "...[T]the clock at A is moved with the velocity v along the line AB to B...". So, how do you get the clock from point A to point B with the clock at B staying at B?

You do it like this:

A/A’1--------------------B2/B’

The clock that is at A in K, remains at A’ in K1, but it moves from A to B in K. The clock that remains at B in K, moves from B’ to A’ in K1.

That’s because the whole frames are moving relatively. Like this at the start:

Twenty dashes apart:

A/A’1--------------------B2/B’

And then this about halfway through the motion:

Ten dashes apart:

A----------A’1>----------<B2----------B’

And at the end:

Together:

A--------------------A’1/B2--------------------B’

So, A never winds up overlapping B’ as you said in one of your thought experiments. There was no need for you to even mention A overlapping B’, since it does not happen in Einstein’s thought experiment. When Clock 1 and Clock 2 relatively move toward each other, A and B’ relatively move apart. They never move closer together.

If you had A and B’ moving closer together, then you would have Clock 1 and Clock 2 moving apart, not toward one another. You are getting your –v and +v confused.

If the two clocks move closer together, then their relative velocity is “v”. If they move apart, then their relative velocity is –v.

[Sam5]

The clock at B goes on the same sheet of paper as points A and B, the clock at A goes on the other. Regardless of which sheet of paper you move, the clock goes from A to B and lags behind in time.

Nope.

On one sheet of paper, 20 dashes apart, the K frame:

A--------------------B(clock2)

On the other sheet of paper, 20 dashes apart, the K1 frame:

A’(clock1)--------------------B’

Now, overlap the two frames and move Clock 1 and Clock 2 together, and you'll see A and B' move further apart.

The motion is relative. Both the K1 and K observers see the other frame's clock move and time dilate the same amount. The paradox still exists.

[SeanF]

The clock at B goes on the same sheet of paper as points A and B, the clock at A goes on the other. Regardless of which sheet of paper you move, the clock goes from A to B and lags behind in time.

Nope.

On one sheet of paper, 20 dashes apart, the K frame:

A--------------------B(clock2)

On the other sheet of paper, 20 dashes apart, the K1 frame:

A’(clock1)--------------------B’

Now, overlap the two frames and move Clock 1 and Clock 2 together, and you'll see A and B' move further apart.

The motion is relative. Both the K1 and K observers see the other frame's clock move and time dilate the same amount. The paradox still exists.

So the paper with A and B is reference frame K and the paper with A' and B' is reference frame k, correct? But if k and K are in relative motion and A matches up with A', then B won't match up with B'. That's what the four transformation equations at the end of Section 3 are for. If you leave those out, you're going to get a paradox.

Once you put those in, you find that moving A to B' doesn't put B at A', and vice versa.

[Sam5]

So the paper with A and B is reference frame K and the paper with A' and B' is reference frame k, correct? But if k and K are in relative motion and A matches up with A', then B won't match up with B'.

No, before the relative motion begins, the distances between A and B, and A’ and B’, and A and B’, and A’ and B are the same. He sets up the thought experiment that way, since he starts it out with none of the clocks moving relatively.

That’s why your distance from A to B on your paper should be the same as your distance A’ to B’ on the other paper.

Remember, the “length contraction” is only “seen” by the observers after the motion begins, and it is a light-signal-caused visual illusion.

And all observers in the K system “see” the K1 system as “length contracted”, and all observers in the K1 system “see” the K system contracted in the same amount, by this formula:



And, as I mentioned, he said in the 1907 paper that the geometrical lengths don’t really contract, so you can’t shrink either of your pieces of paper.

[SeanF]

So the paper with A and B is reference frame K and the paper with A' and B' is reference frame k, correct? But if k and K are in relative motion and A matches up with A', then B won't match up with B'.

No, before the relative motion begins, the distances between A and B, and A’ and B’, and A and B’, and A’ and B are the same. He sets up the thought experiment that way, since he starts it out with none of the clocks moving relatively.

That’s why your distance from A to B on your paper should be the same as your distance A’ to B’ on the other paper.

Remember, the “length contraction” is only “seen” by the observers after the motion begins, and it is a light-signal-caused visual illusion.

And all observers in the K system “see” the K1 system as “length contracted”, and all observers in the K1 system “see” the K system contracted in the same amount, by this formula:



And, as I mentioned, he said in the 1907 paper that the geometrical lengths don’t really contract, so you can’t shrink either of your pieces of paper.

Quit talking about anything other than the 1905 paper. You can't leave out the parts of that 1905 paper you don't like, thus creating a paradox, and then claim the paradox exists in the 1905 paper. The transformation equations are in the 1905 paper, and you have to use them. If you don't, you're just doing a thought experiment based on your own theory, not Einstein's. And if you end up with a paradox, it's because your own theory is flawed, not Einstein's.

[Sam5]

Quit talking about anything other than the 1905 paper.

Do whatever you want to do. If you want to make the distance on your graph paper from A to B different than from A’ to B’, then do what you want.

But then realize that you are talking about a “science fiction short story”, not a “physics theory”, and not anything close to "reality". That’s all it is, a science fiction short story.

He makes you think in the 1905 short story that “lengths contract”, but since he had that paper published in a physics journal, some physicists complained about it, so he had to admit in 1907 that none of the geometrical lengths actually contract. He’s playing tricks in the 1905 paper, and if he has tricked you into believing that only one clock really slowed down due to “relative motion”, then he disserves a Nobel Prize in Literature, not Physics.

[Tensor]

Kilopi, I think Sam is ignoring you. I also think he's ignoring CM also.

[Sam5]

I also think he's ignoring CM also.

I posted a couple of long detailed replies to CM, one page back. What purpose does it serve you to say that I ignored CM when I did not ignore CM? Why would you want to say I didn’t do something when I actually did do it? I’ve already got a couple of guys here who claim that I “believe” or I “maintain” this or that, when I don’t actually “believe” or “maintain” this or that the way they say I do. What’s the purpose of this misinformation?

[Celestial Mechanic]

[Snip!]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.

If I understand your theory correctly, your theory would predict that a Michelson-Morley experiment performed in space would work, but not on Earth or on any body with a reasonable mass.

How small a mass would cease to "regulate" c?

[Celestial Mechanic]

[Snip!]I think the c-regulators are disregarded in that thought experiment.

Of course "c-regulators" are disregarded in that experiment, as they are evidently gravitational in nature (as you define them) and thus not part of special relativity.

There is nothing in that thought experiment that controls the speed of light in space.

And in the absence of gravitational fields, even of the averaged out field of all the stars in the galaxy, the speed of light would be ... ?

The tick-rates of atomic clocks are not considered, and so the local “speed of light” can not be measured.[Snip!]

The first postulate states that the laws of physics are valid in all inertial frames. That means that the atomic clocks work just as well as they do on Earth and that the speed of light can be measured by experiment if either Ann or Bob choose to do so. Since their circumstances can be taken to be the same, namely both of them are in different space ships far from any gravitational fields, they must both measure the same value of the speed of light, whatever that may be in "c-unregulated space", regardless of their relative motion with respect to one another. What is this value of unregulated c in your theory?

[Sam5]

If I understand your theory correctly, your theory would predict that a Michelson-Morley experiment performed in space would work, but not on Earth or on any body with a reasonable mass.

Frankly, I don’t know. I think there are different ways to determine how much light changes speed while traveling through different parts of space, but I’m not sure how we could (or even if we could) detect an “ether wind”.

In the Michelson Morley experiment, the “ether theory” was rather simple in those days. The 19th Century physicists believed that the universe was “static”, and they beleived it contained one large overall ether that was stationary with the universe. They did not know of the high speed motions of the galaxies and the stars. So all Michelson Morley expected to find was evidence of the earth’s travel through the “fixed” ether at 18.6 mps in its motion around the sun. When they didn’t see any evidence of the earth traveling through an ether, the physicists of that era were perplexed.

By 1920 Einstein had begun to think of the “ether” and the electric, magnetic, and gravitational fields as maybe being one and the same, but in his 1920 paper he didn’t seem to think that these fields would be able to have any motion through space. I’m not sure why he didn’t think of them as moving, since the full-strength gravitational fields of astronomical bodies do move through space with the bodies and they are strongest right at the surfaces of the bodies, and the magnetic field of the earth moves through space with the earth.

I’m not positive that a Michelson Morley type of device could even detect an “ether wind” if there was one present, since it has light beams traveling in different directions through themselves, something like aiming a laser directly at a mirror and bouncing the beam back into the laser. It seems to me the best way to detect an ether wind would be to measure light flashes from a pulsar by means of two separate light detectors and then measuring the travel time of the light between the two detectors. If a spacecraft were moving through an ether, then in one direction of travel, the light would take more time than in the opposite direction of travel, but I’m not sure if we could synchronize two atomic clocks on a spacecraft, without the signals traveling between them being affected by the motion of the signals through the ether.

How small a mass would cease to "regulate" c?

I don’t know. I’ve wondered about that for years. I would think that light is regulated at c at the earth and starts being regulated by the earth’s fields at several hundred or several thousand miles from the earth. But suppose we had a large stone disk in space, like Michelson Morley used, then maybe it has enough of a field strength around it, close to it, so that light speed is regulated at that mass, starting a few mm from the mass. Perhaps the mass would have enough electrical or magnetic fields immediately around it to regulate the speed of light to c right at the mass and out to a couple of mm from it.

It’s strange to realize that no human has ever actually directly “seen” light travel at c, because it slows down to about the speed of light in water whenever it enters our eyes, and when it hits our rods and cones, it’s traveling at only about 130,000 mps.

[Sam5]

Of course "c-regulators" are disregarded in that experiment, as they are evidently gravitational in nature (as you define them) and thus not part of special relativity.

In our discussion, I’ve been trying to compare the kinematical part of the 1905 theory to “reality”. Does it reflect “reality” properly? No, it doesn’t, because “relative motion” can’t cause any clock to slow down, or any real time to “dilate”, or any real geometrical length to contract. When he developed his gravitational redshift theory in 1911, that one did conform to reality. So the first half of the 1905 paper is “fiction”, while the 1911 paper is an accurate physics theory.

Doppler’s theory contains illusional time dilation, but it also contains illusional time contraction, so it conforms to reality. When objects separate in space, their light signals seem to imply that their clocks are ticking more slowly, but, when the objects move back toward each other, their light signals seem to imply that their clocks are speeding up. So there is no paradox. This is a common every-day phenomenon that radio and TV engineers deal with. They don’t have to deal with the “relative motion” alleged clock dilations of SR theory, since they don’t actually occur in nature. So the theory is wrong, no matter how “beautiful” it is and no matter how good its math is or how well the math equations fit together.

And in the absence of gravitational fields, even of the averaged out field of all the stars in the galaxy, the speed of light would be ... ?

In 1920 Einstein said, “According to the general theory of relativity space without ether is unthinkable; for in such space there not only wonld be no propagation of light, but also no possibility of existence for standards of space and time (measuring-rods and clocks),”

I’ll go along with that idea.

But in SR theory, the “averaged out” fields of all the stars is not used. What is used is a c-regulator that is fixed with one frame and a separate c-regulator that is fixed with the other frame. This implies no “averaging out” of c-regulator fields in the space between the frames. That’s why he came up with a “c” limit for masses traveling relative to his c-regulators, yet we see distant galaxies that are apparently traveling faster than c, relative to the earth.

The first postulate states that the laws of physics are valid in all inertial frames.

Newton supposed that the “laws of physics” were valid everywhere. What evidence was there, before 1905, that this was not true? That postulate was already assumed by all the physicists. There was nothing "new" or "amazing" about it. The big trick is to find new laws that we don’t know about yet.

What is this value of unregulated c in your theory?

I don’t have an unregulated c in my theory. And anyway, my theory is not just my theory. I base what I say on what is reported in mainstream papers as being the result of experiments and observation. Yes, I try to fill in some of the gaps with a few speculations of my own, but I’m constantly comparing my speculations with what the mainstream physicists are saying they’ve observed and discovered.

My “ether field” hypothesis allowed me to predict, ten years ago, that some of the galaxies could actually be moving at higher than c speeds relative to the earth, back at a time when official university text books were saying that such a thing was impossible. I worked out the idea that light photons from a distant high-c galaxy could speed up on their way to the earth by being regulated in space by the local gravitational fields of the various astronomical bodies in space. So, by my theory, which I posted on the internet 3 to 3 1/2 years ago, the fields of bodies closer to the earth, that are not traveling away from the earth as fast as the most distant galaxies, could regulate the light photons as they travel through the fields of those closer bodies, and that would result in an earth-relative speed up in the photons. That way, a photon from a high-c galaxy could reach the earth.

Now, as of just the past couple of years, that basic idea has become a standard mainstream way of thinking of the situation, but with my “ether fields” being replaced by the “comoving space” term. But what is in that “comoving space” that could regulate the speed of the photons? Just “empty space”? No, of course not. The “fields” of the astronomical bodies that occupy that space.

The old myth, that was based on the 1905 theory, was published so many times over the years, and taught to so many physics students, even today a lot of amateurs refuse to believe there are any high-c galaxies. So, while my ideas are being supported more and more by mainstream cosmologists, their ideas are being supported less and less by mainstream cosmologists.

[Sam5]

So the paper with A and B is reference frame K and the paper with A' and B' is reference frame k, correct? But if k and K are in relative motion and A matches up with A', then B won't match up with B'. That's what the four transformation equations at the end of Section 3 are for. If you leave those out, you're going to get a paradox.

Once you put those in, you find that moving A to B' doesn't put B at A', and vice versa.

The numbers you come up with using Einstein’s equations might be correct, but you are using them too soon. Those numbers reflect the visual effects of the relative motion, but you are trying to use those numbers before any relative motion begins, so you can set the distances between A and B and A’ and B’ on your graph paper as being two different distances from the start, before the motion begins. But you can’t do that, and Einstein didn’t do that.

Before the motion begins, the length AB = A’B’

Then after the motion begins, the “length contraction” visual effect of his theory would causes the AB line to “shrink” in the eyes of the K1 observer, while the A’B’ line would shrink in the eyes of the K observer, so you still have the paradox. You can not cause any of the lines to “shrink” before any of the motion begins.

After the motion begins, his equations apply to someone in one frame thinking of himself as being “stationary”, while he “sees” the other frame moving. But either observer has the right to think this same way. And that leaves us with a paradox.

[Tensor]

I also think he's ignoring CM also.

I posted a couple of long detailed replies to CM, one page back. What purpose does it serve you to say that I ignored CM when I did not ignore CM? Why would you want to say I didn’t do something when I actually did do it?

Yes you did answer. It's my fault. I had a little time and was skimming the thead and I missed your answers within the post. My apologies.

I’ve already got a couple of guys here who claim that I “believe” or I “maintain” this or that, when I don’t actually “believe” or “maintain” this or that the way they say I do. What’s the purpose of this misinformation?

Well you seem to be insisting that gravity somehow regulates the speed of light. Which has nothing to do with SR, since gravity isn't part of it.
If I've misread this part of what you claim, could you correct me?

Again, my apologies for misreading your posts and claiming you were ignoring CM.

[Sam5]

Well you seem to be insisting that gravity somehow regulates the speed of light. Which has nothing to do with SR, since gravity isn't part of it.

If I've misread this part of what you claim, could you correct me?

Well, I got that idea from GR theory, Einstein’s 1911 theory. A standard explanation for the bending of a light beam as it passes the sun is that if we think of a light beam as a traveling plane wave, the portions of the wave nearest the sun slows down more than the portions that are a little further from the sun, so the light bends just as if it were moving from air into glass. You can find that information in his 1911 paper.

I think it is unreasonable to think that no scientist or one particular scientist should not be allowed to change his mind or his theories as time goes by and as he learns more about nature. I think he understood a little more about light in 1911 than he did in 1905.

If a local gravitational field can change the speed of light and change it more the closer the light travels past an astronomical body, then I think it is reasonable to conclude that the gravitational field of the body has something to do with regulating the speed of the light, at least as the photons are passing near the body.

[SeanF]

Do whatever you want to do.

I'm doing what the 1905 paper says I should do. You're the one doing "whatever you want to do," and you think that when you get different results than what the paper predicted, you've found a paradox in the paper. Uh-uh. You want to prove the paradox is in the paper, you have to do what the paper says to do.

So the paper with A and B is reference frame K and the paper with A' and B' is reference frame k, correct? But if k and K are in relative motion and A matches up with A', then B won't match up with B'. That's what the four transformation equations at the end of Section 3 are for. If you leave those out, you're going to get a paradox.

Once you put those in, you find that moving A to B' doesn't put B at A', and vice versa.

The numbers you come up with using Einstein’s equations might be correct, but you are using them too soon. Those numbers reflect the visual effects of the relative motion, but you are trying to use those numbers before any relative motion begins, so you can set the distances between A and B and A’ and B’ on your graph paper as being two different distances from the start, before the motion begins. But you can’t do that, and Einstein didn’t do that.

Before the motion begins, the length AB = A’B’

Then after the motion begins, the “length contraction” visual effect of his theory would causes the AB line to “shrink” in the eyes of the K1 observer, while the A’B’ line would shrink in the eyes of the K observer, so you still have the paradox. You can not cause any of the lines to “shrink” before any of the motion begins.

After the motion begins, his equations apply to someone in one frame thinking of himself as being “stationary”, while he “sees” the other frame moving. But either observer has the right to think this same way. And that leaves us with a paradox.

Those equations are not "length contraction visual effects." They are "coordinate transformations." And I'm not using them "too soon."

A and B are coordinates in K. A' and B' are coordinates in k. Once K and k are in relative motion, the coordinates transform. The transformation is the same regardless of which is "at rest" and which is "in motion." Because of the transformation, though, A-B is no longer the same as A'-B'.