1. Apr 8, 2013

### JohnWisp

Hi,

I did a search for "crackpot physics" and found https://www.physicsforums.com/showthread.php?t=111647 and then this forum.

A friend of mine found some strange twins paradox on the net but he can't remember where.

Neither of us can figure out the solution, but it must be crackpot.

So, here is the strange twins paradox.

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 t in the T1 frame.

4) After t elapses in the T1 frame, T1 also acquires the same v.

5) Both twins think each are time dilated. So, they conduct Einstein's clock synchronization method to compare their clocks.

6) After they compare, they have exact time values from the comparison. So, either t1<t2, t2 >t1 or t1=t2. Since these are actual time values, they are real numbers and must obey the law of trichotomy.

7) Yet, relativity says t2>t1 and t2<t1.

8) This violates the law of trichotomy of real numbers.

This must be wrong because relativity is always right.

2. Apr 8, 2013

### Staff: Mentor

You've overlooked relativity of simultaneity in step 4. At the exact moment that T1 speeds up, using the T1 frame, what time appears on T2's clock? It is not the time that T2 says is when T1 sped up.

This paradox is most easily resolved by drawing a space-time diagram though; and it's been extensively discussed in another thread here... I'll see if I can find that thread later.

3. Apr 8, 2013

### JohnWisp

I thought of that too, but both experience the same instantaneous acceleration, in other words, the acceleration is symmetric between the frames.

So, they both encounter the same effects of the relativity of simultaneity hence they cancel.

4. Apr 9, 2013

5. Apr 9, 2013

### ghwellsjr

John, your logic is like the following:

Suppose I have two identical wristwatches that each run on two battery cells. I have one on each wrist and I have synchronized them so they keep the same time. Then one day, one of the cells in the watch on my left wrist died and the watch started running slow. In fact, it lost one minute a day compared to the watch on my right wrist. One month later, when there was a half-hour difference between the two watches, the same thing happened to the watch on my right wrist and so both watches now tick at the same slow rate.

But do they now have the same time on them? Afterall, they both experienced the same thing.

6. Apr 9, 2013

### Staff: Mentor

This is not correct. Have you even tried to do the math and find out what relativity actually says?

7. Apr 9, 2013

### JohnWisp

I do not understand why you think this is a solution.

Your post is simply a recap of the special relativity constant acceleration equations

You can find a better explanation at the below link.

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

Last edited by a moderator: May 6, 2017
8. Apr 9, 2013

### JohnWisp

Can you please explain this post in the context of relative motion which demands that each frame views the other frame's clocks as time dilated?

9. Apr 9, 2013

### JohnWisp

Can you please tell me how to use equations here?

Then, I will post the math I think and you can comment on it.

Basically, what I am going to do is follow what Einstein did in his paper when he proves time dilation is a fact of special relativity.

I will do it for both frames.

10. Apr 9, 2013

### ghwellsjr

Yes, each frame views the other frame's clocks as time dilated and in your example, to the same extent, which only means they tick at the same rate, not that they are necessarily synchronized, which they won't be in your example for the same reason as my story in the quoted post illustrates, which is, that they didn't accelerate at the same time so they didn't change their tick rates at the same time. First one and then the other so why should they be synchronized?

11. Apr 9, 2013

### Staff: Mentor

The relevant equations are the Lorentz transformation and the proper time.

For ease of computation you should use units where c=1 and you should simplify your scenario. The point you are trying to make does not need any acceleration, simply have the two twins each be perpetually inertial and set t=0, x=0 at the time when they meet. Then in parametric form the worldline of the "at rest" twin is $(t,x,y,z)=(\tau_A,0,0,0)$ in the unprimed frame and the worldline of the "moving" twin is $(t',x',y',z')=(\tau_B,0,0,0)$ in the primed frame which is moving at a velocity v in the x direction wrt the unprimed frame, where the $\tau$ are proper times along their respective worldlines.

Last edited: Apr 9, 2013
12. Apr 9, 2013

### JohnWisp

I did not say the different frame clocks would be synchronized through the trip. I don't know what they are.

Do you?

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

you will find that acceleration is absolute in special relativity.

So, it dos not matter when or where the acceleration occurs, each frame elapses the same times for the acceleration period.

So, it is symmetric and cancels.

Last edited by a moderator: May 6, 2017
13. Apr 9, 2013

### Staff: Mentor

No, this is incorrect. SR is symmetric under spatial translations, time translations, rotations, and boosts. Because of those symmetries you can fix ONE acceleration such that it occurs at the origin (spatial translation) at t=0 (time translation) along the x-axis (rotation) starting from rest (boost). However, once you have fixed that you have used up all of the degrees of freedom in your symmetry. That means that the OTHER acceleration must be specified completely as to exactly when and where and which direction it occurs. There is no remaining symmetry.

14. Apr 9, 2013

### JohnWisp

Frame 1.
$t'=(t-vx/c^2)\gamma$

According to Einstein section 4
http://www.fourmilab.ch/etexts/einstein/specrel/www/

x=vt.

Substitute

$t'=(t-v(vt)/c^2)\gamma$
$t'=(1-v^2/c^2)t\gamma$
$t'=t/\gamma$

So, this is the time dilation for frame 1.

Now, for frame 2

$t=(t'+vx'/c^2)\gamma$

Since this frame is moving the negative direction,

x'=-vt'.

Substitute

$t=(t'+v(-vt')/c^2)\gamma$
$t=(1-v^2/c^2)t'\gamma$
$t=t'/\gamma$

So, this is the time dilation for frame 2.

So, each frame views the other clocks as time dilated.

Are you saying this is false under the relativity?

Last edited: Apr 9, 2013
15. Apr 9, 2013

### JohnWisp

OK, let us consider this link.

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

For any constant acceleration,

t = (c/a) sinh(aT/c)

a- acceleration as measured by the non-accelerating frame
T- Time as measured in the accelerating frame.
t - Time as measured in the non-accelerating frame.

Therefore, for the acceleration of twin T1, we have for the stay at home twin time as,

t = (c/a) sinh(aT/c)

and T for the accelerating twin.

Then, when the twin2 accelerates we have the same thing,

t = (c/a) sinh(aT/c)

So, for acceleration one, we have
T for twin1
t for twin 2

and for acceleration two we have
T for twin2
t for twin 1

This is symmetric and it cancels. that is a simple fact of the special relativity constant acceleration equations as supported again by

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

Last edited by a moderator: May 6, 2017
16. Apr 9, 2013

### ghwellsjr

In steps 5 & 6 you said the twins' clocks ended up synchronized, which is wrong.

Last edited by a moderator: May 6, 2017
17. Apr 9, 2013

### JohnWisp

Here is what I said.
5) Both twins think each are time dilated. So, they conduct Einstein's clock synchronization method to compare their clocks.

6) After they compare, they have exact time values from the comparison. So, either t1<t2, t2 >t1 or t1=t2. Since these are actual time values, they are real numbers and must obey the law of trichotomy.

Where in these statements do you think I said the clocks end up synchronized? In fact, I don't know how these clocks turn out. Do you?

18. Apr 9, 2013

### ghwellsjr

Yes, knowing v and t we can calculate the difference in the time values of the two clocks.

Since you obviously don't know what Einstein's clock synchronization method is, what did you think the bolded statements meant?

19. Apr 9, 2013

### JohnWisp

I wlll show how it is done.

F1 sends its time t1 to F2 but also records its time t1.

F2 reflects with its time t2 back to F1.

Now, F1 knows the round trip time tr.

It takes t1 + 1/2tr and compares it to t2.

That is the comparison.

20. Apr 9, 2013

### ghwellsjr

And if they are equal, the clocks are synchronized. So why did you deny that you said they were synchronized?