Strange Twins Paradox: A Crackpot Physics Mystery | Discover the Solution Here!

[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.

 

[Nugatory]

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.

 

[JohnWisp]

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.

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.

 

[tom.stoer]

Introduce a third inertial frame w.r.t. which both twins are moving; then use the ideas from

https://www.physicsforums.com/showpost.php?p=4337470&postcount=4

 

[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.

 

[Dale]

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

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

 

[JohnWisp]

Introduce a third inertial frame w.r.t. which both twins are moving; then use the ideas from

https://www.physicsforums.com/showpost.php?p=4337470&postcount=4

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

So, can you please explain why your post solves this problem?

 

[JohnWisp]

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.

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?

 

[JohnWisp]

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

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.

 

[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.

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?

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?

 

[Dale]

Can you please tell me how to use equations here?

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&#039;,x&#039;,y&#039;,z&#039;)=(\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.

 

[JohnWisp]

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?

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

Do you?

Further, if you read
http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

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.

 

[Dale]

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.

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.

 

[JohnWisp]

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 (\lambda_A,0,0,0) and the worldline of the "moving" twin is (\lambda_B,\lambda_B v,0,0), where v is the velocity and the \lambda are parameters for their respective worldlines.

Frame 1.
t&#039;=(t-vx/c^2)\gamma

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

x=vt.

Substitute

t&#039;=(t-v(vt)/c^2)\gamma
t&#039;=(1-v^2/c^2)t\gamma
t&#039;=t/\gamma

So, this is the time dilation for frame 1.

Now, for frame 2

t=(t&#039;+vx&#039;/c^2)\gamma

Since this frame is moving the negative direction,

x'=-vt'.

Substitute

t=(t&#039;+v(-vt&#039;)/c^2)\gamma
t=(1-v^2/c^2)t&#039;\gamma
t=t&#039;/\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?

 

[JohnWisp]

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.

OK, let us consider this link.

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

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 1This 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

 

[ghwellsjr]

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

Do you?

Further, if you read
http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

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.

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

 

[JohnWisp]

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

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?

 

[ghwellsjr]

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?

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?

 

[JohnWisp]

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?

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.

 

[ghwellsjr]

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.

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

 

[JohnWisp]

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

Why don't you show where I said the clocks were synched at the end of the experiment?

 

[Dale]

x=vt.
...
x'=-vt'

You have introduced a bit of notational confusion, so I am not sure what you think you are calculating here. After I first posted and before you responded I cleared up some of the notation. You may want to look at it again.

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

Are you saying this is false under the relativity?

No, you are correct, each frame does see the other clocks as time dilated. So far this doesn't lead to the problem you mention in your OP.

 

[Dale]

For any constant acceleration,

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

That is only for constant acceleration. Your OP had non constant acceleration. The symmetry does not cancel out for the scenario in the OP.

However, this symmetry discussion is really irrelevant since what you are actually interested in doesn't require acceleration to illustrate.

 

[JohnWisp]

You have introduced a bit of notational confusion, so I am not sure what you think you are calculating here. After I first posted and before you responded I cleared up some of the notation. You may want to look at it again.

No, you are correct, each frame does see the other clocks as time dilated. So far this doesn't lead to the problem you mention in your OP.

1) what is the problem with the other frame as being x'=-vt'? That is special relativity.

2) The problem in the OP is that each frame viewing the other as time dilated causes a clock comparison issue where the clocks will have actual real number times.

I have already explained this. Since the clock comparison produces actual times on the clocks to compare, we have t1<t2, t1>t2 or t1=t2. Are you saying this is false?

 

[JohnWisp]

That is only for constant acceleration. Your OP had non constant acceleration.

Where did I have this?

Please show me.

 

[Dale]

Where did I have this?

Please show me.

steps 2 and 4

 

[JohnWisp]

steps 2 and 4

This is a method approved by Einstein. It means ignore the acceleration.

2) T2 instantly acquires some v.

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

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 $\frac{1}{2}tv^2/c^2$(up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.

http://www.fourmilab.ch/etexts/einstein/specrel/www/

 

[Dale]

1) what is the problem with the other frame as being x'=-vt'? That is special relativity.

Not exactly a problem. As I said, it is a notational confusion. When you say x=vt you are writing the equation of a worldline. So x and t are no longer general coordinates, but only points on a worldline. So I prefer to use the parametric representation of the line with a clearly distinct parameter for each worldline (see post 11).

When you make the substitution you are no longer talking about general t or t', but only specific ones on that line. So the first equation gives you the relationship between t and t' on one worldline and the second equation gives you the relationship between t and t' on a completely different worldline. So there is no contradiction in the fact that different relationships hold on different lines. The only place where both relationships must hold is where the lines intersect. And that is correct since both hold at t=t'=0.

2) The problem in the OP is that each frame viewing the other as time dilated causes a clock comparison issue where the clocks will have actual real number times.

I have already explained this. Since the clock comparison produces actual times on the clocks to compare, we have t1<t2, t1>t2 or t1=t2. Are you saying this is false?

Yes, this is false. There is no clock comparison issue. You haven't used the proper time equation yet, so you haven't shown what the clocks actually read. Whenever you specify when a reading is taken on a clock then you will get one answer independent of the frame.

 

[Dale]

This is a method approved by Einstein. It means ignore the acceleration.

2) T2 instantly acquires some v.

I have no problem with that, but the fact remains that steps 2 and 4 are non constant accelerations so the symmetry for a constant acceleration is not applicable. Your assertion of symmetry is unwarranted.

 

[ghwellsjr]

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

Why don't you show where I said the clocks were synched at the end of the experiment?

Oops, I goofed. Somehow my mind interpreted the phrase "they have exact time values" as meaning they were exactly the same. I apologize and to make peace, I offer you some diagrams to illustrate your scenario for the case where v=0.6c and t=12 months. First, a diagram showing the only frame you mentioned in your first post, that of the original rest state of the two twins:

The dots represent one-month intervals of Proper Time for each twin. Please note how both twins are time dilated after they both acquire a speed of 0.6c.

I have shown the synchronization process that twin 2 carries out at his Proper Time of 8 months. His signal travels over to twin 1 at his Proper Time of 14 months and he returns the signal. This return signal arrives at twin 2 at his Proper Time of 26 months. One-half of the round trip signal time is 9 months so he adds 9 months to 8 months and determines that twin 1's clock read 14 months when his own clock read 17 months. This is a difference of 3 months.

We can easily determine this difference if we realize in this scenario that gamma is the ratio of the Coordinate Time to the Proper Time for twin 1 at the moment he accelerates. Since the Coordinate Time is the same as the Proper Time for twin 2, we can say;

γ = CT/PT1
CT = PT2
γ = PT2/PT1
γ-1 = PT2/PT1 - PT1/PT1
γ-1 = (PT2-PT1)/PT1

Now we note that t = PT1 so:

t(γ-1) = PT2-PT1

Now I'm going to transform the above frame into one in which both twins are at rest after they both accelerate:

Now you can clearly see that the Proper Time of 17 months which is also the Coordinate Time is equal to the Proper Time of 14 months for twin 1.