Here's the deal, Sam5. The time dilation is not some all-on-its-own equation that Einstein threw in there. It comes out of those same transformation equations at the end of Section 3 - but it's not all that comes out of those transformation equations.

See, if, when viewed from the K frame, one clock starts out a coordinates (0,0,0,0) and the other starts out at coordinates (3,0,0,0) and they both end at (3,0,0,5), we conclude that the first clock moved (x-value changed from 0 to 3) at velocity 0.6c (it took 5 time units to move the 3 distance units).

However, that clock saw itself as stationary. So, if we apply the transformation equations to its starting and ending coordinates, we find that the starting point remains (0,0,0,0), but the ending point becomes (0,0,0,4). It stayed in the same place, but the time only took 4 units.

But we can't just apply those transformation equations to two of our relevant sets of coordinates. We have to apply them to the third set as well - the other clock's starting point. When we apply the equations to (3,0,0,0), we get (3.75,0,0,-2.25).

That's how we end up without a paradox. You want to simply apply the 80% time dilation without actually applying the transformation equations, and thus get a paradox. Because the 80% time dilation is a result of the transformation equations (but only one of the many results of the transformation equations), you can't do it that way.

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I don’t get a result different from what the paper predicted. I get a result that is exactly what the paper predicts, and that is a clock paradox. If he had reported what the Clock 1 observer “saw” during the relative motion in the “peculiar consequence” thought experiment, then he would have reported his predicted paradox. But in that thought experiment, he only reported what the Clock 2 observer “saw”, and you are unable to figure out on your own what the Clock 1 observer “saw”, and what he “saw” is based on the “predictions” in the paper itself. The paper predicts the paradox.

You specifically said:

”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'.”

When you put distance A and B = A’ and B’ on your two pieces of paper, your dimensions on the paper don’t “shrivel” up when you move the paper.

The dimensions on the paper won’t ever “shrivel up” even if you move the two pieces relativity at .6c. He admitted that in his 1907 relativity paper.

Nothing happens to the paper or the distances between A and B on one sheet and A’ and B’ on the other if you move them relatively in space.

In his theory, it is the light signals that go between his two moving frames that distort, and observers in both frames see exactly the same kind of distortion. But that distortion doesn’t conform to reality. To find out what light-signal distortions do conform to reality, consult the Doppler theory.

You mean they aren’t the same length on the pieces of paper? Sure they are. Their geometrical paper lengths don’t change. And if you tried to change the lengths, which of the two lengths would you change??

In his theory, it is the light signals between the frames that distort, causing both sets of frame observers to “see” (by means of the light signals) distortions of the lengths in each other’s frames, but not in their own frames. This is a visual effect, not a real geometrical-change effect.

Since this “seen” distortion is mutual and in the same amount, both sets observers “see” the same amount of distortion when they look into each other’s frames. It is the “seen” distortion that appears to make the “seen” clocks slow down, when they look into each other’s frames, but the slowdown is only an illusion too, since the “seen” distortions are of the same amount, and, therefore the “seen” slowdowns would be of the same amount as “seen” in each frame when the other frame is looked into, thus leading to a clock paradox when the clocks get together at the end.

And his paper “predicts” this, although his “peculiar consequence” thought experiment doesn’t come right out and say this, since in that experiment he is only telling you what the B observer (the clock 2 observer) “sees”, while he is not telling you what the A’ observer (the clock 1 observer) “sees”.

I’m beginning to think that the “peculiar consequence” he was actually referring to had nothing to do with the one clock appearing to slow down, but to the paradox itself, with both clocks appearing to slow down the same amount during the relative motion but with both observers disagreeing, when they unite, about what their clock read-outs really are. Clocks slowing down is not so unusual, but two clocks slowing down in exactly the same amount, but only one being really “lagged behind” when they unite is really peculiar. So he was probably talking about the “peculiar consequence” of the paradox itself.

If I have any number or x/x’ typos in this post, please let me know.

I think you are applying the equations incorrectly, and I think you are failing to understand the meaning of the symmetry principle in the theory. Sometimes you almost have it, but then you lose your hold on it.

When you say they start out with one clock at (0,0,0,0) and the other at (3,0,0,0), and they both end up at (3,0,0,5). You are using the K frame’s coordinate numbers and clock, as seen by all the K frame observers. And your end-time for the K1 clock is what the K frame observers "see" on it, but not what the K1 observers "see" on it.

The “3” is the K frame’s x number position, 3x. That never changes for the K frame. And the 5 is the time on the K frame’s Clock 2, as seen by K frame observers.

K frame:

A0x-----1x-----2x-----B3x

To make matters more clear for the other readers, let’s express that Clock 1 (in the K1 frame) starts out at (0x, 0y, 0z, 0t) as seen by the K frame observers. While Clock 2 (in the K frame) starts out at (3x, 0y, 0z, 0t).

At the end, the K frame observers have “seen” the relative motion appearing to be Clock 1 moving to Clock 2, and the K frame observers see Clock 1 as being at (3x, 0y, 0z, 5t) at the end.

But the K1 frame’s fixed Clock 1 x’ number is 3x’, and the K1 frame sees the B point and Clock 2 (in K) move from (0x’, 0y’, 0z’, 0t’) to (3x’, 0y’, 0z’, 5t’). The K1 frame observers “see” the relative motion as Clock 2 moving toward Clock 1.

A’3x’-----2x’-----1x’-----B’0x’

The K1 frame has its own coordinate system, and the K1 observers see Clock 2 move relative to it.

If you want to reverse the K1’s coordinate system numbers, you can:

A’0x’-----1x’-----2x’-----B’3x’

That doesn’t matter, because the K1 observers would still see Clock 2 move toward Clock 1. If you use these reversed numbers, the K1 observers would see Clock 2 move from 3x’ to 0x’ at .6 c in 5t’ time units on the K1 frame’s clocks. Remember, to all K1 frame observers, their own frame does not length contract or time dilate.

If you only do the calculations from the point of view of the K system, all you are going to turn up is what the K observers “see”, but not what the K1 observers “see”.

You’ve got to remember that what the K observers “see” is not what the K1 observers “see”. They both “see” the same phenomena when they look into each other’s systems, but the numbers are reversed and the clock that appears to be “moving” is reversed. K1 observers “see” Clock 2 as “moving”, while K observers “see” Clock 1 as “moving”. While K observers think Clock 1 is moving toward B, K1 observers think Clock 2 is moving toward A’.

While K observers don’t see the AB distance shrinking during the relative motion, they (according to the theory) “see” the A’B’ distance as being “shorter”.

But while the K1 observers don’t see the A’B’ distance shrinking during the relative motion, they (according to the theory) “see” the AB distance as being “shorter”.

Both system’s observers “see” the other system’s clock slow down in exactly the same amount. So when they finally unite, they disagree about what each other’s clock read-outs say.

And that is a very “peculiar consequence” of the theory.

And even more in 1915, when he replaced those earlier general relativity papers.

I always own up to my mistakes, Sam5. I don't just ignore it when they're pointed out and hope nobody else notices, like you do.

I am applying the equations correctly, and it is you who does not understand the symmetry of SR.

You're right up until the last sentence. All observers will agree on what the K1 clock says at the end - and none of them think it says "5".

I'm not sure what you mean when you say "'3' is the K frame’s x number position". A reference frame does not have a single x coordinate (or a single y, or z, or t). 3 is Clock2's x-coordinate in the K frame. Everything you say after that seems to be okay, though. The K frame observers see both clocks as being at (3,0,0,5) at the end.

Woah, hold on a second now. Where does this come from? How did Clock 1 all of a sudden get on the other side of Clock 2? Clock 1's starting x coordinate needs to be less than Clock 2's starting x coordinate.

Sure, there you go. But here's the deal. You're not doing SR. Like I said before, you're doing your own little theory. In your theory KA=(0,0,0,0) and KB=(3,0,0,0) transforms to K'A=(0,0,0,0) and K'B=(3,0,0,0). But that's not how SR works.

If you do it your way, you're going to end up with no length contraction, no time dilation, and no constant speed of light. But that doesn't prove SR is wrong.

If you do it SR's way, than KA=(0,0,0,0) and KB=(3,0,0,0) transforms to K'A=(0,0,0,0) and K'B=(3.75,0,0,-2.25). Under SR, then, you end up with length contraction, with time dilation, and with a constant speed of light, and no paradoxes, because Clock 1 ends up behind Clock 2 either way.

Sam5, your problem is that you're stuck in the "absolute space and time" mindset. You think that if two events are 5 years apart, then they are actually and absolutely 5 years apart. Therefore, you think if one observer sees that duration as 5 years, then the other observer will also see that duration as 5 years. That's why when you hear that they both see the other clock running slowly, you think that means they would both see the other clock saying 4 and their own clock saying 5. That's not the way SR works, and you won't understand all the consequences of SR (peculiar or otherwise) until you understand that.

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O dear, I seem to have lost control of my thread. How embarrassing!

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Oh phooey. You wouldn't even notice your own mistakes if I didn't point them out to you.


Who, me??

Ok, ok. I just give up.

Huh, what did I do now?

Only from your own prejudiced K frame point of view. You can call your frame numbers anything you want.

Yes I am.

You are just making math mistakes. You apparently don't know how to calculate how long it takes someone to go 3 ly at .6c.

It doesn't matter what I call them. This is MY frame. If you use your own numbers in your own frame to try to calculate what you think I see on MY clock, then you will continue to make your usual mistake.

I see you moving 3 ly at .6c and by all my clocks I see that taking you 5 years, although you seem to shrink up a little when you are moving. By the SR theory, I see all your clocks time dilate, I see your x axis shrink, and I see you moving in my direction for 5 years. If you claim that I see you move for 4 years, that's only what you "see" on my clock, because you "see" my clock as being "time dilated". But I don't. My clock says 5 years. Before you start out, the distance between us is 3 ly. Then you start moving. You move at .6c, and that takes you 5 years.

No, not so. I saw you move for 5 years. You went 3 ly at .6c. I saw your clock read 4 at the end, while my clock read 5. Something must have happened to your clock while you moved. As far as the speed of light is concerned, I live on the surface of the earth, so I measure the local speed to be "c" in a vacuum.

No, sorry.

My clock ended up reading 5. It was your clock that read 4. I saw you travel 3 light years at .6c, and that took you 5 years.

Nope.

Look, I saw you go 3 ly at .6c and that took you 5 years. All my hundreds of clocks agree. Before the motion started, we measured out the distance with our measuring rods. It was 3 ly. Now you claim that before the motion started, my distance to you was different than your distance to me.

I'm telling you what I saw. 3 ly / .6c = 5 years. All of us K1 frame people saw you move for 5 years.

I do my own reading of Einstein theories, and I do my own thinking. I don't go to Ann and Bob type K-8 school websites, where you K frame guys obviously get all your science information. Sorry

Sam5, in Section 3 of Einstein's 1905 paper, he says:

Then, at the end of Section 3, he says:

Now, I would like to know what you think is the significance of these equations. What does SR use them for? Specifically make note of the fact that if the values x=3, y=0, z=0, t=0, and v=0.6c are entered into them, the results are =3.75, =0, =0, and =-2.25.

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SeanF

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CM and Tensor both looked at the Ann and Bob website and said it fit with their understanding of SR. They're not getting their info from it.

But, apparently, you can't even understand the K-8 websites.

I haven't had much time of late to post (and I think most questions I might have answered have been addressed sufficiently well by others), but I've certainly been reading as well, and I think you've done one of the better jobs I've seen of explaining this. In particular, I think that it's useful to include the redshift calculations, allowing one to determine what each observer actually sees, makes everything clearer.

Celestial Mechanic,

You seem to be a reasonable and sensible person, so maybe you could help me with something. I don’t seem to be able to explain to Sean what I am interested in learning. So I will try asking you.

What I have seen on this thread so far, in my opinion, is everyone telling me what the K frame observers see, and then telling me what the K frame observers see the k frame observers see, and even what the K frame observers see the k frame observers see the K frame observers see. I think this is how Sean wound up with his four sets of end-time numbers for the two Ann and Bob clocks: 12.5, 10, 8, and 6.4.

It seems to me that everyone here (but me) is considering the K frame to be the “stationary one” and the k frame to be the “moving” one. That’s ok for a while, but we already have hundreds of examples of that on the internet already. What I’ve been trying to find out is what people here would think the k frame observers would see, if we considered the k frame to be “stationary system”, from the very beginning of Section 1 of the Kinematical Part of Einstein’s paper, all the way through to the end of Section 4. And I want to know what the k frame observers see the K frame observers see, if we all consider the K frame to be the “moving” one, from the beginning of Section 1 to the end of Section 4.

I understood the basic original point of view years ago. But I’ve been trying to get people to think of the k frame as “stationary” from the very beginning of the paper.

So, maybe I haven’t been making myself clear, because whenever I ask people to consider the k frame to be “stationary”, they just tell me the K frame observer’s opinions of what the k frame observers would see, but they never make the k frame the original true “stationary one”. They always keep the K frame as the stationary one and the k frame as the moving one.

Do you understand what I’m asking?

Let’s take the k frame as being the “stationary frame”, and follow it all the way through the first 4 sections, following these basic rules of the paper:

“the same laws of electrodynamics and optics will be valid for all frames of reference for which the equations of mechanics hold good.”

“the view here to be developed will not require an “absolutely stationary space” provided with special properties,”

“Let us in “stationary” space take two systems of co-ordinates, i.e. two systems,”

“Thus, whereas the Y and Z dimensions of the sphere (and therefore of every rigid body of no matter what form) do not appear modified by the motion, the X dimension appears shortened in the ratio



i.e. the greater the value of v, the greater the shortening. For v=c all moving objects--viewed from the “stationary'' system--shrivel up into plane figures”

“It is clear that the same results hold good of bodies at rest in the “stationary'' system, viewed from a system in uniform motion.”

The purpose being, so we can see if this theory in any way could match reality, which it certainly purports to do.

If you want simplify the process, you could simply change all the original designations of K to k, and k to K, and t to t’ and t’ to t, and make other appropriate changes so that you will be taking “a system of co-ordinates in which the equations of Newtonian mechanics hold good,” and in order to render our “presentation more precise and to distinguish this system of co-ordinates verbally from others which will be introduced”, we can call k the “stationary system.”

Then, if you could do this for me, maybe you can tell me what the k frame observers would report “seeing” when Clock 1 (the A’ clock in the “peculiar consequence” thought experiment in Einstein’s paper, which is fixed in k) meets up with Clock 2 (the B clock in the K frame).

Do you understand what I am trying to find out?

Isn't that the situation that I describe on this webpage, or am I misunderstanding your question?

Yep! He's killed one thread already, taken over this one and started in on a third. :roll: ](*,)

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Yahbut that happened clear back when.

You're misunderstanding his question (but then, so is he). Einstein says, "If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous;"

He's wondering about considering k the "stationary system" in this sense. I think he's still a little unclear about Einstein using "stationary system" as a name for K, not a general concept.

If the clocks are synchronized in K, then they're not in k. This will be true regardless of whether K or k is considered "at rest," because it's how the experiment is set up. If the clocks were synchronized in k, then they wouldn't be in K - in that case, the 2nd Clock would lag behind the 1st when they met.

Sam5's misunderstanding of the "symmetry principle" causes him to think that's a paradox because he thinks K being "the stationary system" (i.e. the clocks are synchronized) and k being "the sta