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.

 

[georgir]

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?

Both twins think the other's clock is dilated, only during the period between the first's acceleration and the second's acceleration. But they can not use clock synchronization during that time.

After the second's acceleration, both think their clocks are running at the same rate. And they can measure the difference in their readings.

That difference will match the difference that the first twin thinks accumulated between their accelerations, not what the second twin thinks. This is because before and after the acceleration, the second twin will have a different definition of "now" for locations other than his own.

The very act of acceleration will appear to cause a shift in time, a redefinition of "now" at all distant locations. Locations in the direction of the acceleration will seem to shift into the future, the further away they are the bigger the shift. And likewise locations in the other direction will seem to shift into the past.

So in this specific setup the twin accelerating second will think this for the other's clock: it starts to run slow, but when he suddenly launches it jumps forward in time, and not only makes up the difference but reverses it. Then they both agree, the second twin's clock is behind.

EDIT: This describes the setup with instant acceleration. I didn't read other replies in detail and am not sure why references to constant acceleration were made in some of them, or if the OP is interested in discussing such an alternative scenario.

 

[ghwellsjr]

That difference will match the difference that the first twin thinks accumulated between their accelerations, not what the second twin thinks. This is because before and after the acceleration, the second twin will have a different definition of "now" for locations other than his own.

The very act of acceleration will appear to cause a shift in time, a redefinition of "now" at all distant locations. Locations in the direction of the acceleration will seem to shift into the future, the further away they are the bigger the shift. And likewise locations in the other direction will seem to shift into the past.

So in this specific setup the twin accelerating second will think this for the other's clock: it starts to run slow, but when he suddenly launches it jumps forward in time, and not only makes up the difference but reverses it. Then they both agree, the second twin's clock is behind.

This interpretation is only one of many and there are other interpretations that do not involve any jumps in time on any clocks. For example, if both twins use the radar method to establish their own non-inertial reference frames for the entire scenario, they will not have any reason to believe that the other ones clock jumped in time. In fact both clocks make smooth, monotonic, ever-increasing times on their clocks, although they tick at different rates during the scenario.

You should always indicate that your interpretation is just one of many instead of making it sound like jumping in times on remote clocks is the only way to explain the scenario.

 

[georgir]

ghwellsjr, this is not a matter of interpretation.
Just add the two lines of simultanety right before the acceleration and right after the acceleration on your graphs and it will become much clearer.
Probably I should have done that, but there are plenty of charts that show it already, i.e.:
http://upload.wikimedia.org/wikipedia/commons/b/b1/Relativity_of_Simultaneity.svg
(from http://en.wikipedia.org/wiki/Relativity_of_simultaneity)
The line marked x represents what a stationary observer at point A calls "now". The lines marked x' or x'' represent what differently moving observers at point A call "now". If you suddenly accelerate, your definition of "now" will change, it is a fact, not interpretation.

 

[Nugatory]

ghwellsjr, this is not a matter of interpretation.
Just add the two lines of simultanety right before the acceleration and right after the acceleration on your graphs and it will become much clearer.
... If you suddenly accelerate, your definition of "now" will change, it is a fact, not interpretation.

Terrific, now we're arguing about the correct interpretation of "interpretation"

I think Ghwellsjr's point is that the entire notion of "now" for spacelike-separated events is itself an interpretation of a physical observation about the path of a light signal from an observer whose world line passes through one event, to the other event, and then back to the observer.

This is a somewhat sterile argument in the SR context, because every inertial observer has an obvious definition of "now" right under his nose (basically, "has the same t coordinate"); the temptation to assign physical significance to this simultaneity convention is nearly irresistible. However the argument becomes very important in GR; and saves some grief even in SR when working with non-inertial frames.

 

[ghwellsjr]

ghwellsjr, this is not a matter of interpretation.
Just add the two lines of simultanety right before the acceleration and right after the acceleration on your graphs and it will become much clearer.

I don't know what you mean by adding two lines of simultaneity. On either of my two graphs, every horizontal line is a line of simultaneity and there are no jumps in the clocks of either twin. Maybe you could copy one of my graphs and mark it up to show what you mean because it is not clear what you are talking about.

Probably I should have done that, but there are plenty of charts that show it already, i.e.:
...
The line marked x represents what a stationary observer at point A calls "now". The lines marked x' or x'' represent what differently moving observers at point A call "now". If you suddenly accelerate, your definition of "now" will change, it is a fact, not interpretation.

There is no standard definition of "now" for accelerating observers. There is only a standard definition of "now" in Special Relativity for Inertial Reference Frames (IRF) and by extension for an inertial observer at rest presumably at the origin of an IRF but if an observer accelerates, then there are many different definitions of "now" for a non-inertial rest frame of an observer. Just because you like one doesn't mean it is any more a fact than the ones you don't like. They are all "facts" according to their different definitions. Those definitions are no more a fact than Einstein's second postulate that light propagates at c in all directions in an IRF.

 

[JohnWisp]

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.

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.

1) As far as x-vt or x=-vt', you may consult section 4 of Einstein's paper. I am only adhering to his standard.

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

2) As far as the clock comparison and your claim it is false, you need to show why it is false to back your claim. All of my claims are backed with the specific math or I do not make claims. I expect you will do the same.

 

[JohnWisp]

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.

This is false.

Einstein wrote the following,

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

The two clocks stated in the same frame exactly as this experiment.

So, the OP did exactly the same as Einstein suggested. This is therefore, mainstream and not subject to contradiction.

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

Then, someone wanted to see another interpretation. So, I brought up constant acceleration.

This is absolutely not necessary since Einstein did not include any acceleration factors in his paper.

 

[JohnWisp]

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.

Your drawings are wrong.

Acceleration is absolute under and not subject to worldlines.

I already gave the acceleration equations and I suggest you refute the mainstream to prove your case since my equations are nothing more than mainstream.

 

[Nugatory]

This is false.

Einstein wrote the following: "If the clock at A is moved with the velocity v..."

Einstein did not write that; it was written by whoever translated the German that Einstein did write into English. This process introduces enough ambiguity that you cannot take the English translations completely literally, and you are reading much more into that word "moved" than can be justified.

Instead, you really have to understand the math, which is way more precise and language-independent.

 

[JohnWisp]

Both twins think the other's clock is dilated, only during the period between the first's acceleration and the second's acceleration. But they can not use clock synchronization during that time.

After the second's acceleration, both think their clocks are running at the same rate. And they can measure the difference in their readings.

That difference will match the difference that the first twin thinks accumulated between their accelerations, not what the second twin thinks. This is because before and after the acceleration, the second twin will have a different definition of "now" for locations other than his own.

The very act of acceleration will appear to cause a shift in time, a redefinition of "now" at all distant locations. Locations in the direction of the acceleration will seem to shift into the future, the further away they are the bigger the shift. And likewise locations in the other direction will seem to shift into the past.

So in this specific setup the twin accelerating second will think this for the other's clock: it starts to run slow, but when he suddenly launches it jumps forward in time, and not only makes up the difference but reverses it. Then they both agree, the second twin's clock is behind.

EDIT: This describes the setup with instant acceleration. I didn't read other replies in detail and am not sure why references to constant acceleration were made in some of them, or if the OP is interested in discussing such an alternative scenario.

1) The acceleration shift in time is absolute under special relativity. In fact, it is slowed. So, any accelerated frame will appear to shift into the past. And, the distance it no important unless you have specific acceleration equations that prove your statement.

2) The rest of your post does not matter until you understand accelerating frame tick slower.

The clock postulate generalizes this to say that even when the moving clock accelerates, the ratio of the rate of our clocks compared to its rate is still the above quantity. That is, it only depends on v, and does not depend on any derivatives of v, such as acceleration. So this says that an accelerating clock will count out its time in such a way that at anyone moment, its timing has slowed by a factor (γ) that only depends on its current speed; its acceleration has no effect at all.

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

 

[JohnWisp]

Einstein did not write that; it was written by whoever translated the German that Einstein did write into English. This process introduces enough ambiguity that you cannot take the English translations completely literally, and you are reading much more into that word "moved" than can be justified.

Instead, you really have to understand the math, which is way more precise and language-independent.

I understand the math.

I also understand German. It is hard to translate motion incorrectly between the two languages.

Now, if you have the original German and can show the problem you suggest, then please post your proof.

 

[Passionflower]

Einstein did not write that; it was written by whoever translated the German that Einstein did write into English. This process introduces enough ambiguity that you cannot take the English translations completely literally, and you are reading much more into that word "moved" than can be justified.

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.

 

[Dale]

1) As far as x-vt or x=-vt', you may consult section 4 of Einstein's paper. I am only adhering to his standard.

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.

2) As far as the clock comparison and your claim it is false, you need to show why it is false to back your claim. All of my claims are backed with the specific math or I do not make claims. I expect you will do the same.

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.

 

[Dale]

This is false.

Einstein wrote the following,

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

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.

 

[Passionflower]

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

Identical acceleration profiles of two observers at different x values is not a symmetric situation. Think in this respect about Bell's spaceship paradox.

 

[georgir]

I don't know what you mean by adding two lines of simultaneity. On either of my two graphs, every horizontal line is a line of simultaneity and there are no jumps in the clocks of either twin.

Yes, every horizontal line is a line of simultanety for the observer that is considered at rest in the particular graph. That is, before the acceleration on graph 1, and after the acceleration on graph 2.

Now transform the horizontal lines from graph 2 to graph 1. For example, through the red point marked "14" on your graph, it will be a line connecting red "14" and blue "17". That is what the red twin thinks is his "now" at that moment, not the horizontal line.

The interesting lines of simultanety are through red point "12". Before acceleration, it is the horizontal line. After acceleration, it is the line connecting red "12" and blue "15".

There is no standard definition of "now" for accelerating observers. There is only a standard definition of "now" in Special Relativity for Inertial Reference Frames (IRF) and by extension for an inertial observer at rest presumably at the origin of an IRF but if an observer accelerates, then there are many different definitions of "now" for a non-inertial rest frame of an observer. Just because you like one doesn't mean it is any more a fact than the ones you don't like. They are all "facts" according to their different definitions. Those definitions are no more a fact than Einstein's second postulate that light propagates at c in all directions in an IRF.

We are not interested in "now" for accelerating observers in this setup. There is no accelerating reference frames. There is only instant acceleration, and inertial reference frames for before and after the acceleration, which are completely well defined.

 

[georgir]

1) The acceleration shift in time is absolute under special relativity. In fact, it is slowed. So, any accelerated frame will appear to shift into the past. And, the distance it no important unless you have specific acceleration equations that prove your statement.

2) The rest of your post does not matter until you understand accelerating frame tick slower.
The clock postulate generalizes this to say that even when the moving clock accelerates, the ratio of the rate of our clocks compared to its rate is still the above quantity. That is, it only depends on v, and does not depend on any derivatives of v, such as acceleration. So this says that an accelerating clock will count out its time in such a way that at anyone moment, its timing has slowed by a factor (γ) that only depends on its current speed; its acceleration has no effect at all.

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

You are missing the point... What you are talking about is how an accelerated clock looks to an inertial observer. What I am talking about is how things look to an accelerating observer... And not even during the acceleration itself, just right before it and right after it, both completely inertial cases. Re-read my posts.

 

[pervect]

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.

 

[ghwellsjr]

I don't know what you mean by adding two lines of simultaneity. On either of my two graphs, every horizontal line is a line of simultaneity and there are no jumps in the clocks of either twin. Maybe you could copy one of my graphs and mark it up to show what you mean because it is not clear what you are talking about.

Yes, every horizontal line is a line of simultanety for the observer that is considered at rest in the particular graph. That is, before the acceleration on graph 1, and after the acceleration on graph 2.

Now transform the horizontal lines from graph 2 to graph 1. For example, through the red point marked "14" on your graph, it will be a line connecting red "14" and blue "17". That is what the red twin thinks is his "now" at that moment, not the horizontal line.

The interesting lines of simultanety are through red point "12". Before acceleration, it is the horizontal line. After acceleration, it is the line connecting red "12" and blue "15".

OK, is this what you mean?

And yes, that line of simultaneity through red point "12" is very interesting. Do you want to explain what happened to blue points "10" through "14"? Do you want to show how signals between red and blue propagate at the speed of light in the diagram as they cross the seam between the two frames?

There is no standard definition of "now" for accelerating observers. There is only a standard definition of "now" in Special Relativity for Inertial Reference Frames (IRF) and by extension for an inertial observer at rest presumably at the origin of an IRF but if an observer accelerates, then there are many different definitions of "now" for a non-inertial rest frame of an observer. Just because you like one doesn't mean it is any more a fact than the ones you don't like. They are all "facts" according to their different definitions. Those definitions are no more a fact than Einstein's second postulate that light propagates at c in all directions in an IRF.

We are not interested in "now" for accelerating observers in this setup. There is no accelerating reference frames. There is only instant acceleration, and inertial reference frames for before and after the acceleration, which are completely well defined.

I don't want to get hung up on terminology, I just want to show you something that you seem to be unaware of and that is the radar method of establishing an observer's rest frame. I'm going to repost the second diagram in which blue remains at rest (after his acceleration) because I have already put in the radar method for him to measure how far away the red twin is at blue's time 17.

He can't establish that distance in real time, he has to be making measurements all during his trip. Every month he sends a radar signal to the red twin which propagates on a 45-degree angle and the red twin reflects it back toward the blue twin along with his current time. So at blue's time 8 he sends a radar signal to red which arrives at red's time 14 and is received back at blue time 26. Blue then calculates one half of the delta between 8 and 26 which is 9 and applies that as a distance at the time which is the average of 8 and 26 which is 17. So he says that when his time was 17, red was 9 light-months away and the time on red's clock was 14 (blue reads this at his time 26 or else red sends him the information along with the radar echo).

Once you understand this process, you can verify it for every month along blue's path. And since blue was already at rest in this IRF, the radar method will exactly agree with the IRF.

I have redrawn the first diagram from earlier making it taller and removed the blue's radar measurement:

Please copy this diagram to your computer and print it or open it in Paint and use it to make a table of how red will calculate how far away blue is at every one of red's months. Also keep track of the corresponding times on blue's clock, just like I showed you earlier. When you get all done, you can sketch out the results on the same diagram and post it for us to see. Can you do that? I promise you, you will find it interesting, maybe even more interesting than what you said about red's point 12.

 

[georgir]

OK, is this what you mean?

No, and since it is such a simple thing I am now sure you are just trolling me.
Regardless, here is the diagram, and I am out of this thread.