1. 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.

2. Order of Kilopi
Join Date
Nov 2003
Posts
6,197
Originally Posted by SeanF
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.
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.

Originally Posted by SeanF
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'.
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.

3. Order of Kilopi
Join Date
Nov 2003
Posts
6,197
Originally Posted by SeanF
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.
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.

4. Originally Posted by Sam5
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.
And even more in 1915, when he replaced those earlier general relativity papers.

5. Originally Posted by Sam5
Originally Posted by SeanF
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.
If I have any number or x/x’ typos in this post, please let me know.
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.

Originally Posted by Sam5
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.
I am applying the equations correctly, and it is you who does not understand the symmetry of SR.

Originally Posted by Sam5
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.
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".

Originally Posted by Sam5
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.
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.

Originally Posted by Sam5
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.
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.

Originally Posted by Sam5
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’
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.

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

7. Order of Kilopi
Join Date
Nov 2003
Posts
6,197
Originally Posted by SeanF
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.
Oh phooey. You wouldn't even notice your own mistakes if I didn't point them out to you.

Originally Posted by SeanF
I am applying the equations correctly, and it is you who does not understand the symmetry of SR.
Who, me??

Originally Posted by SeanF
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".
Ok, ok. I just give up.

Originally Posted by SeanF
Woah, hold on a second now.
Huh, what did I do now?

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

Originally Posted by SeanF
Sure, there you go. But here's the deal. You're not doing SR.
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.

Originally Posted by SeanF
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).
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.

Originally Posted by SeanF
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.
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.

Originally Posted by SeanF
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.
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.

Originally Posted by SeanF
Sam5, your problem is that you're stuck in the "absolute space and time" mindset.
Nope.

Originally Posted by SeanF
You think that if two events are 5 years apart, then they are actually and absolutely 5 years apart.
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.

Originally Posted by SeanF
Therefore, you think if one observer sees that duration as 5 years, then the other observer will also see that duration as 5 years.
I'm telling you what I saw. 3 ly / .6c = 5 years. All of us K1 frame people saw you move for 5 years.

Originally Posted by SeanF
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.
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

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

Originally Posted by Einstein
To any system of values x, y, z, t, which completely defines the place and time of an event in the stationary system, there belongs a system of values , , , , determining that event relatively to the system k, and our task is now to find the system of equations connecting these quantities.
Then, at the end of Section 3, he says:

Originally Posted by Einstein
...[T]he transformation equations which have been found become

where
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.

9. Originally Posted by Sam5
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
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.

10. Originally Posted by Celestial Mechanic
BTW, thank you for reading my posts, Tensor. It's good to know that someone is paying attention.
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.

11. Order of Kilopi
Join Date
Nov 2003
Posts
6,197
Originally Posted by Celestial Mechanic
I am going to try to dispel a few more misconceptions by giving coordinates for 6 events
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?

12. Originally Posted by Sam5
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.
Isn't that the situation that I describe on this webpage, or am I misunderstanding your question?

13. Originally Posted by Eroica
O dear, I seem to have lost control of my thread. How embarrassing!
Yep! He's killed one thread already, taken over this one and started in on a third. :roll: ](*,)

14. Originally Posted by Eroica
O dear, I seem to have lost control of my thread. How embarrassing!
Yahbut that happened clear back when.

15. Originally Posted by kilopi
Originally Posted by Sam5
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.
Isn't that the situation that I describe on this webpage, or am I misunderstanding your question?
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 stationary system" (i.e. the clocks are synchronized) should be the same thing. They're not. K being "at rest" and k being "at rest" are the same thing. The clocks are only synchronized in one frame, though, and that's not relative.

16. Order of Kilopi
Join Date
Nov 2003
Posts
6,197
Originally Posted by Kaptain K
Yep! He's killed one thread already, taken over this one and started in on a third. :roll: ](*,)

I haven’t “killed” any threads. Nobody has to respond to any of my posts. You post your opinion and I post my opinion. That’s how this thread started, with several people posting their opinion, and then I saw the thread and I posted my opinion.

I take it that no one on this thread has any PhD in Relativity Physics, so we all are posting our own opinions about this subject.

But when I post my opinion back to you, I sometimes get receive several troll posts responding to my post to you, calling me stupid, uneducated, and other nasty things. Those are the trolls, not me.

I don’t understand how someone, like on the “light” thread, can make a long post of relativity gibberish, that is not based in any way on scientific fact, observation, or experimentation, then when I point out a flaw in the logic of that post, the poster gets mad at me for pointing out that what he posted – while pretending to be an authority on the subject – was gibberish.

17. Originally Posted by Sam5
But when I post my opinion back to you, I sometimes get receive several troll posts responding to my post to you, calling me stupid, uneducated, and other nasty things. Those are the trolls, not me.
Are those PMs?

18. Order of Kilopi
Join Date
Nov 2003
Posts
6,197
Originally Posted by SeanF
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.
EXACTLY! YOU’VE FINALLY GOT IT!! YOU FINALLY SAID IT!!!! THAT’S THE PARADOX I’VE BEEN TALKING ABOUT ALL THESE YEARS!!!!!!!!

19. Originally Posted by Sam5
Originally Posted by SeanF
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.
EXACTLY! YOU’VE FINALLY GOT IT!! YOU FINALLY SAID IT!!!! THAT’S THE PARADOX I’VE BEEN TALKING ABOUT ALL THESE YEARS!!!!!!!!
THERE IS NO PARADOX!!!! You get different results if the clocks are synchronized in K than you do if the clocks are not synchronized in K. How is that a paradox?

20. Order of Kilopi
Join Date
Nov 2003
Posts
6,197
Originally Posted by SeanF
You get different results if the clocks are synchronized in K than you do if the clocks are not synchronized in K.
Right. This is basically what I've been tring to say for the past 4 years.

There is a paradox IF you synchronize the clocks in K and k at the same time, using the same synchronization process he outlined, and by using it in the different frames as “seen” from each frame’s point of view, that is, as seen from the k frame’s “stationary” point of view and as seen from the K frame’s “stationary” point of view, and then if you have the relative motion take place and then ask the K observers what they “see”, and ask the k observers what they “see”, from their own unique points of view.

In other words, we as “scientists” must be “objective”, and we must ascertain BOTH points of view.

But we will get false results in the SR theory if we only report what K observers “see” and what they think the k observers “see”. But, in SR theory, if we allow both K observers and k observers to synchronize their own clocks in their own systems, while considering their own systems to be the “stationary” system, then we will get conflicting results and a paradox, when we ask the K observers and separately ask the k observers what they “see”. So, in either case, SR theory fails as an objective “scientific” theory.

Now, let’s consider the true physics transformation:

In reality, k frame observers would actually synchronize their atomic clocks in k, just as earth observers would synchronize their atomic clocks on earth.

K frame observers would synchronize their atomic clocks in K, just as sun observers would synchronize their atomic clocks on the sun.

In reality, earth observers see sun atomic clocks running slow, while sun observers see earth atomic clocks running fast, and both sets of observers agree about this. They agree that the sun clocks are running slower than the earth clocks.

But in SR theory (while not considering acceleration or field effects on clocks), and by synchronizing the clocks for only one frame, the K frame, he wound up with K observers seeing themselves as being inside the “stationary system”, and they see k clocks run slow. In the “peculiar consequence” thought experiment, he has the K clocks running faster than the k clocks, and the k clocks running slower than the K clocks, and if that were true in reality, he would have the k observers “seeing” the K clocks run “fast”, not “slow”. But in the theory, if the k observers really did consider their frame to be the “stationary” one, and if they synchronized their clocks accordingly, which he says they have a perfect right to do, then the k observers would see the K clocks run “slow”, not “fast”. And that is the paradox of the SR theory.

What he does in the “peculiar consequence” thought experiment is report ONLY what the K observers “see”, and so, from their point of view, they “see” the k clocks running “slow”, and his math equations claim that if the K observers “see” the k frame’s clocks running “slow”, then the k frame observers must “see” the K frame’s clocks running “fast”. But, if we really asked the k observers what they “see”, then they would say, “First we synchronized our clocks with our frame being the ‘stationary’ frame, just as Einstein told us to do, because all of us in this frame are ‘stationary’ relative to one another and relative to our own clocks, then the relative motion began, and we saw the K frame clocks slow down.”

So, if Einstein had reported what the k frame observers actually “saw” during the “peculiar consequence” thought experiment, after they synchronized their own “stationary clocks” using his “stationary frame” method of synchronization, then they would have reported the K frame clocks “slowing down”, not “speeding up”. So, if we look upon his thought experiments in SR theory from a completely objective point of view, just as we should look upon nature while trying to figure out the true “laws of physics”, we would see the paradox in the theory. But if we look upon the thought experiments ONLY from the point of view of the K system, we will see only the k frame clocks as “slowing down”, and our point of view will not be objective, because in the theory, if the k observers’ clocks really do “slow down”, then they would “see” the K clocks “speeding up”, rather than “slowing down.” And when we look at the thought experiments from the k observers point of view, with them synchronizing their clocks while considering themselves to be in the “stationary” frame, then they would see the K clocks “slowing down”, not “speeding up”, and this reveals the basic paradox of the SR theory.

As I mentioned, in 1911 he did report what the sun observers actually “see” and what the earth observers actually “see”, from an objective point of view. Thus, everyone agrees that the sun clocks are running “slow”, while the earth clocks are running “fast” (running faster than he sun clocks), and all observers agree, both at the sun, at the earth, and everywhere else, including real observers and imaginary ones.

21. Originally Posted by Sam5
Originally Posted by SeanF
You get different results if the clocks are synchronized in K than you do if the clocks are not synchronized in K.
Right. This is basically what I've been tring to say for the past 4 years.

There is a paradox IF you synchronize the clocks in K and k at the same time...
When K and k are in relative motion to each other, the clocks cannot be synchronized for both frames. Absolutely, positively, can not happen. They can only be synchronized for one frame, which is why Einstein specifically says that they are synchronized for the K frame.

Your interpretation of SR and the thought experiment is flawed from this point on.

22. Originally Posted by Sam5
There is a paradox IF you synchronize the clocks in K and k at the same time, using the same synchronization process he outlined, and by using it in the different frames as “seen” from each frame’s point of view, that is, as seen from the k frame’s “stationary” point of view and as seen from the K frame’s “stationary” point of view, and then if you have the relative motion take place and then ask the K observers what they “see”, and ask the k observers what they “see”, from their own unique points of view.
As SeanF points out, Einstein would say that it would not be possible to synchronize the clocks in both K and k. As SeanF has pointed out to me, he says, in that 1905 paper, section 1.2, "Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous." So, since you can't synchronize the clocks in both systems, it would seem that special relativity doesn't result in a paradox.

23. Order of Kilopi
Join Date
Nov 2003
Posts
6,197
Originally Posted by SeanF
When K and k are in relative motion to each other, the clocks cannot be synchronized for both frames.
That is why the clocks must be synchronized before the motion begins, just as Einstein has them being synchronized in the K frame in the “peculiar consequence” thought experiment, before the motion begins in that experiment.

That’s why I told you to make the line AB the same length as A’B’ on your two pieces of graph paper, before you started moving the two pieces of paper relatively.

24. Originally Posted by Sam5
Originally Posted by SeanF
When K and k are in relative motion to each other, the clocks cannot be synchronized for both frames.
That is why the clocks must be synchronized before the motion begins, just as Einstein has them being synchronized in the K frame in the “peculiar consequence” thought experiment, before the motion begins in that experiment.

That’s why I told you to make the line AB the same length as A’B’ on your two pieces of graph paper, before you started moving the two pieces of paper relatively.
Doesn't matter. You can't get Clock 1 from A to B while it remains in K. The synchronization is lost for Clock 1 as soon as it's out of K and into k.

25. Order of Kilopi
Join Date
Nov 2003
Posts
6,197
Originally Posted by SeanF
You can't get Clock 1 from A to B while it remains in K. The synchronization is lost for Clock 1 as soon as it's out of K and into k.
Right. That’s what I’ve been trying to tell you. When clock 1 moves, it is no longer in K, it is in K1, which is moving relative to K. Clock 1 is moving along line AB, but it is in the K1 frame. Just as clock 2 is moving along line B'A', but it is in the K frame.

That’s why the SR clock “slow-downs” don’t show up until the relative motion begins. That’s why relative “v” is so important in the theory. If you have no “v”, you’ve got no clock slow-downs. The higher the “v” in the theory, the more the clock’s rates appear to “slow down”. The “rate” of the “slow-downs” is due only to the value of the relative “v”. The overall duration of the observed slow-downs is determined by the local clocks as seen by the observers in their own frame, looking at their own clocks, and they always see their own clocks and their own frame as being “stationary”.

The paradox comes about when we ask both sets of observers what they “see”. They will both “see” the same thing in the “slow-down” of the “other frame’s” clocks, but they will never see a slow-down in their own frame’s clocks. And that is why it is important to synchronize all the clocks first, before the motion begins, while every clock is “stationary” relative to every other clock, and while all observers in both frames agree on this.

26. Originally Posted by Sam5
Originally Posted by SeanF
You can't get Clock 1 from A to B while it remains in K. The synchronization is lost for Clock 1 as soon as it's out of K and into k.
Right. That’s what I’ve been trying to tell you. When clock 1 moves, it is no longer in K, it is in K1, which is moving relative to K. Clock 1 is moving along line AB, but it is in the K1 frame. Just as clock 2 is moving along line B'A', but it is in the K frame.

That’s why the SR clock “slow-downs” don’t show up until the relative motion begins. That’s why relative “v” is so important in the theory. If you have no “v”, you’ve got no clock slow-downs. The higher the “v” in the theory, the more the clock’s rates appear to “slow down”. The “rate” of the “slow-downs” is due only to the value of the relative “v”. The overall duration of the observed slow-downs is determined by the local clocks as seen by the observers in their own frame, looking at their own clocks, and they always see their own clocks and their own frame as being “stationary”.

The paradox comes about when we ask both sets of observers what they “see”. They will both “see” the same thing in the “slow-down” of the “other frame’s” clocks, but they will never see a slow-down in their own frame’s clocks. And that is why it is important to synchronize all the clocks first, before the motion begins, while every clock is “stationary” relative to every other clock, and while all observers in both frames agree on this.
I told you what both observers see, in this post. What you don't understand is that SR is coordinate transformations, not "clock slow-downs."

Clock 1 ticking "12:00:00" is a four dimensional point in space-time. Clock 2 ticking "12:00:00" is a four dimensional point in space-time. In the K frame, those two points in space-time are simultaneous - the difference in their t-coordinate values is zero. That is what Einstein means when he says, "there are stationary clocks which, viewed in the stationary system, are synchronous."

In the k frame, however, those two points in space-time have a non-zero difference in their t-coordinates - they are not simultaneous. So the very instant Clock 1 is in relative motion to K, still at the Clock 1="12:00:00" event, it is no longer simultaneous with Clock 2's "12:00:00" event. Clock 2, still in K, still sees the two "12:00:00" events as simultaneous.

That is why SR does not predict that the Clock 1 observer will see Clock 2 the same as the Clock 2 observer sees Clock 1. They will both perceive the other running "slow", but they will not both perceive the other starting that slow running at "12:00:00"!

27. Order of Kilopi
Join Date
Nov 2003
Posts
6,197
Originally Posted by SeanF
Clock 1 ticking "12:00:00" is a four dimensional point in space-time. Clock 2 ticking "12:00:00" is a four dimensional point in space-time. In the K frame, those two points in space-time are simultaneous - the difference in their t-coordinate values is zero. That is what Einstein means when he says, "there are stationary clocks which, viewed in the stationary system, are synchronous."

In the k frame, however, those two points in space-time have a non-zero difference in their t-coordinates - they are not simultaneous. So the very instant Clock 1 is in relative motion to K, still at the Clock 1="12:00:00" event, it is no longer simultaneous with Clock 2's "12:00:00" event. Clock 2, still in K, still sees the two "12:00:00" events as simultaneous.
“Synchronous”, as he used the word, means “running at the same rate”. All clocks in the SR theory “run at the same rate” when there is no motion taking place. If all clocks are not running at the same rate in SR theory, then the two frames are already in motion. Since he says, “If at the points A and B of K there are stationary clocks,” that means one clock is stationary at A and the other is stationary at B. This means the “A” clock (that you call Clock 1) has not yet begun to move relative to the K frame, and it also means that clock “B” (which you call Clock 2) has not yet begun to move relative to the K1 frame. Clock 1 is absolutely stationary at Point A in the K frame (and at Point A’ in the K1 frame). Since observers at Point A in K and Point B in K agree on this, that means the two frames and all the clocks are “stationary” together long enough for signals to go from Points A and B and Points A’ and B’, which are the same distance apart, since while Clock 1 is stationary in K, Point A overlaps Point A’ and Point B overlaps Point B’, while all frames and clocks are stationary relative to one another.

When the clocks 1 and 2 start to move relatively, then that’s when K observers “see” the dilation on the K1 clocks and that’s when the K1 observers “see” the dilation on the K clocks.

So, that’s why BOTH clocks are “synchronous”, before the motion begins, meaning they are running at the same rate, and it also means the relative motion has not yet begun.

Your term “simultaneous”, as you use it here, is similar to the term “synchronized” as used in some old WW II movies, when the Sergeant says, “Let’s all synchronize our watches.” That doesn’t mean all the soldiers take their watches apart and make them all “synchronous”, ie, make them run at the same rate. No, when they “synchronize” their watches, all they do is set all their watches at the same time, which is usually the Sergeant’s time on his watch. So, even though all their watches are still ticking at different rates, and are not “synchronous”, if they “synchronize” them to a certain time, they will remain “reasonably synchronized” during the next hour or so, so all the soldiers will know exactly when to move in unison. But of course, just because they are “synchronized” one time, that doesn’t mean they are “synchronous” thereafter.

It’s like these new “atomic clocks” that you can buy at Wal-Mart. They are just regular clocks, but they contain a radio receiver, and they receive a radio signal that is based on a time pulse that is generated by a master atomic clock in Boulder, CO. The cheap clocks are “synchronized” to that atomic clock at least once a day, ie, their time reading is reset at least once a day, by means of that radio signal. Even though they are never running “synchronously” with the Boulder master atomic clock, they are at least reset once a day to the time signal coming from that Boulder atomic clock.

28. Order of Kilopi
Join Date
Nov 2003
Posts
6,197
Originally Posted by SeanF
They will both perceive the other running "slow", but they will not both perceive the other starting that slow running at "12:00:00"!
Well that doesn’t matter. It doesn’t matter if they start seeing Planet X time running slow starting at New York time 10 PM, or Los Angeles time 7 PM or UTC time or GMT time.

Since Clock 1 and Clock 2 in the thought experiment are some distance apart, they won’t see each other’s local time as being the same even before the motion begins. But they will see each other’s clocks as being “synchronous” as long as they both are stationary. Its just like all points on earth see their clocks as being “synchronous”, since all observers use the single earth’s rotation rate to determine this. Of course now all the scientists use atomic clocks, but they are supposed to follow the earth’s single rotation rate, and that’s why they have to take away “leap seconds” from the atomic clocks every now and then.

29. Originally Posted by Einstein
Thus with the help of certain imaginary physical experiments we have settled what is to be understood by synchronous stationary clocks located at different places, and have evidently obtained a definition of "simultaneous," or "synchronous," and of "time." The "time" of an event is that which is given simultaneously with the event by a stationary clock located at the place of the event, this clock being synchronous, and indeed synchronous for all time determinations, with a specified stationary clock.
Also note that the first definition of "synchronous" at dictionary.com is "Occurring or existing at the same time."

You must keep in mind that what you are reading is an English translation of a German original document (and that it's nearly 100 years old). While "synchronized" would certainly be a clearer choice than "synchronous," it is obvious both from context and from the math what Einstein meant.

But you don't understand the math, do you? You never did answer my question about what you think those equations mean.

30. Order of Kilopi
Join Date
Nov 2003
Posts
6,197
Originally Posted by Einstein
“The ‘time’ of an event is that which is given simultaneously with the event by a stationary clock located at the place of the event, this clock being synchronous, and indeed synchronous for all time determinations, with a specified stationary clock.”
The two clocks are “synchronous” in the theory, ie “running at the same rate”, because in the theory they are both “stationary” relative to one another.

While moving relatively, Clock 1 and Clock 2 are not “synchronous”.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•
here