Strange Twins Paradox

  • Thread starter JohnWisp
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  • #51
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Sorry but I think you are talking nonsense, do you actually speak German?

From the Urtext:

... und bewegt man die Uhr in A mit der Geschwindigkeit v auf der Verbindungslinie nach B, so gehen nach Ankunft dieser Uhr in B die beiden Uhren nicht mehr synchron, ......


There is nothing ambiguous about it: the clock in A is moved with a velocity of v and arrives at B.
Why are you attacking me?

My point was that Einstein assumed instantaneous acceleration or his conclusions may be false with his statement.

So, the translation does assume this.

That mean's instantaneous acceleration (moved from a rest state to v) is ignored in the calculations for Einstein's teachings on time dilation.

that is exactly what I was saying.
 
  • #52
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Often the seminal works are full of confusing notation, particularly since the notation was not standardized. One of the things that makes the great minds so great is that they were able to understand despite the poor notation of the time. Einstein was smart enough to use the confusing notation without becoming confused, you apparently are no Einstein.

If you are done trying to deflect the issue, you may want to go back and address the substantive criticism of post 28.

Exactly what math do you think backs up your claim? You have only demonstrated that the relationship between the t and t' coordinates on one worldline is different from the relationship between the t and t' coordinates on a different line. You have not demonstrated anything at all about the actual numbers read on the clocks, let alone that there is some contradiction among them.
I am confused by your post.

Are you claiming my assertions

1) x =vt
2) x'=-vt'

is not the appropriate logic for translation between the two frames?

Next, I do not know the actual times on the clocks at the Einstein clock sync method and never claimed I did.

I am claiming based on time dilation, once the clock values are determined, relativity can't be consistent since it claims one clock is less than the other and vice versa.

As far as drawing the wordlines, I do not know how to do it for this problem.

If I did, then there would be a solution everyone could agree on.
 
  • #53
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So what? Nowhere in Einstein's quote does he say that the acceleration was constant. The quote is irrelevant.

The acceleration described in the OP is not constant and the symmetry you assert does not exist. My statements are 100% true.

You are right. Einstein never claimed anything about acceleration calculations in his quote.

In fact, he did not include the effects in his conclusions and that is what the OP did.

But, some asked about the acceleration phase and I tried to provide the mainstream calculations for those questions.

The OP however is perfectly consistent with Einstein ignoring the acceleration effects.

So that is allowed unless you can show otherwise.
 
  • #54
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Identical acceleration profiles of two observers at different x values is not a symmetric situation. Think in this respect about Bell's spaceship paradox.
This is not applicable.

The Bell problem showed identical accelerations at different x values x1, x2.

Under non-relativity, the string would not break.

Under relativity, it does break.

So, this problem is not applicable given the different x values.
 
  • #55
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Some quick comments: the only space-time diagrams posted to this thread have been dismissed as "wrong", but the OP has not responded with a "correct" space-time diagram.

Alas, the verbal descriptions of the "paradox" seem rather ambiguous (having seen a few "paradoxes", I would guess it likely that this ambiguity is at the heart of the paradox.

I would place the burden on the OP to supply such a space-time diagram, as the alternative is to watch the thread spin around i circles forever.

I don't see how anybody else other than the original poster will feel motivated to go to the effort of creating and posting a space-time diagram to simply see it dismissed as "wrong" by the OP in a few lines of text, without apparent thought.

So if a space-time diagram is needed (and I think it is), it is pretty much up to the OP to step forwards and do one.
I do not have a world line solution.

If I did, then I would provide a solution to this problem which I have already confessed I do not have.

So, you are asking me to solve a problem I already confessed I cannot solve.
 
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  • #56
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  • #57
pervect
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I do not have a world line solution.

If I did, then I would provide a solution to this problem which I have already confessed I do not have.

So, you are asking me to solve a problem I already confessed I cannot solve.
Let's review and recap and rephrase a bit:

1) Assume T1 and T2 are in the same frame.

2) T2 instantly acquires some v.

3) Both remain in relative motion for some time

4) After t elapses in the T1 frame, T1 also acquires the same v.
Can you draw a space-time diagram for this much?
Can you interpret the space-time diagram given in

https://www.physicsforums.com/attachment.php?attachmentid=57752&stc=1&d=1365671314

I don't think I'll be able to help much if you can't get at least this far on your own.

Then, to complete the paradox, I would add the following:

"At some event, T2 stops moving"

q: what is the time of this event in the T2 frame - i.e. the proper time
q: what is the time of the event in the T1 frame - (we need to know this to place it on the diagram).
q: what is the position of the event in the T1 frame (this follows from the velocity and the answer to the previous question).

At this point, we can proceed with the "paradox" - if it still exists.

This is considerably less complicated than a lot of the references you've been using. Some familiarity with the Lorentz transform would be helpful in converting from the T1 frame to the T2 frame.
 
  • #58
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Are you claiming my assertions

1) x =vt
2) x'=-vt'

is not the appropriate logic for translation between the two frames?
Correct, the transform of x=vt is x'=0. In other words, x=vt is the same worldline as x'=0, while x'=-vt' is a completely different worldline.

Next, I do not know the actual times on the clocks at the Einstein clock sync method and never claimed I did.

I am claiming based on time dilation, once the clock values are determined, relativity can't be consistent since it claims one clock is less than the other and vice versa.
Such claims are not permitted here. Thread closed.

PS. You may open a new thread to ask your questions, but leave off the anti-relativity commentary. It is against the forum rules, and factually incorrect. The framework of Minkowski geometry guarantees that SR is self consistent. Furthermore, it is absurd for you to think that you have found an inconsistency when you, by your own admission, don't even know how to calculate the very numbers which you claim are inconsistent.
 
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  • #59
Fredrik
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I'm unlocking this thread at the request of ghwellsjr, because he wants to post some stuff he had already prepared when the thread got locked. We will probably lock the thread again tomorrow, so I don't encourage anyone else to continue the discussion. Note that the OP isn't even here anymore.
 
  • #60
ghwellsjr
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Thanks Fredrik.

This is a continuation from post #49 on this thread.

In that post I described how to calculate the coordinates of a non-inertial rest frame for the red twin. The first step is to draw in radar signals sent from the red twin every month and reflecting off the blue twin:

attachment.php?attachmentid=58392&stc=1&d=1367417803.png


Now we have to compile a list of information that the red twin collects.

In the first column are the Proper Times when he sent each signal and when he received the echo separated by commas.

The second column contains his calculations of the distance the blue twin was away from him and his Proper Time when the blue twin was there. The first calculation is one half of the difference between the two numbers in the first column and the second calculation is one half of the sum of those two numbers.

The third column is the Proper Time the red twin sees on the Blue Twin's clock whenever he receives an echo.

Code:
0,0	0.0 @ 0.0	0
1,4	1.5 @ 2.5	2
2,8	3.0 @ 5.0	4
3,12	4.5 @ 7.5	6
4,14	5.0 @ 9.0	8
5,16	5.5 @ 10.5	10
6,18	6.0 @ 12.0	12
7,20	6.5 @ 13.5	14
8,22	7.0 @ 15.0	16
9,24	7.5 @ 16.5	18
10,26	8.0 @ 18.0	20
11,28	8.5 @ 19.5	22
12,30	9.0 @ 21.0	24
13,31	9.0 @ 22.0	25
14,32	9.0 @ 23.0	26
15,33	9.0 @ 24.0	27
16,34	9.0 @ 25.0	28
17,35	9.0 @ 26.0	29
Now we plot the information in the second column and label the Proper Times for the third column, interpolating any that are missing:

attachment.php?attachmentid=58393&stc=1&d=1367417803.png


The advantage of this type of non-inertial frame is that the light signals continue to propagate at c and there are no discontinuities in the times on either twin.

Furthermore, we can reverse the process by having the blue twin perform the same radar measurements of red twin's positions as a function of his own Proper Time and recreate the second diagram shown on post #49. In other words, it continues to show, like Inertial Reference Frames what each observer actually sees and measures.

After you have done a few of these non-inertial frames, you can take some short cuts and only do the calculations where the discontinuities occur. Of course, real observers can't do this because they can't see the discontinuities until after they happen.
 

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