There is no twin paradox - mathematical proof

[JesseM]

It is using the other twin as the reference (and as a consequence as a reference frame) that matters, not the acceleration per se. One twin maintains a velocity relative to the other twin, then obtains and maintains an equal but opposite velocity relative to the other twin.

There is no obligation that you perform the calculation in the rest frame of the inertial twin, you can do it in any inertial frame you want. For example, you could choose a frame where the "traveling" twin is at rest for the first phase of the journey while the inertial twin is moving at velocity v; then after the turnaround, the traveling twin would be traveling at 2v/(1 + v^2/c^2), while the inertial twin would continue to move at v. You'd get exactly the same answer for the elapsed time on each twin's clock when they reunite in this case.

By the way, do you understand that the question of which twin accelerates has an objective answer--that in any inertial frame, whichever twin accelerates will change velocity (speed, direction, or both) before and after the acceleration, while the inertial twin will have constant speed? And that we can define the notion of an "inertial frame" physically by saying that any object whose position-coordinate is constant in this frame will be experiencing no G-forces (whereas an accelerating object always knows it's objectively accelerating because of the G-forces it feels)?

 

[neopolitan]

There is no obligation that you perform the calculation in the rest frame of the inertial twin, you can do it in any inertial frame you want. For example, you could choose a frame where the "traveling" twin is at rest for the first phase of the journey while the inertial twin is moving at velocity v ...

v relative to what? at "rest" relative to what? turned around and maintained a new velocity relative to what?

You use a "cheat" by giving the "travelling" twin a higher velocity for a similar period, which then comes to the same answer. Of course that is going to work out.

I am fully aware that you can use whatever frame you like. I am fully aware that a velocity in an observed inertial frame is transformed when considering it from another reference frame. But you are basically talking about three frames now. The frame relative to the moving object being considered (in which the moving object can be considered to be at rest), the observed inertial frame in which that moving object can be considered to have velocity v and the third reference inertial frame, which considers the observed inertial frame to have a velocity of its own.

You consistently misused these frames in another thread, by adding a third frame to the question without acknowleging it.

The relative velocities of two objects which are considered to be at rest in their own frames will always be equal and opposite. The equation you threw in for Equation 4 in the light clock thread pertained to observations from a third frame.

But I will desist, before you lock the thread and prevent others from commenting. If anyone wants to discuss this further, please check my public profile from which you can send me email. Alternatively there are contact details available via the link in the very first post in this thread.

cheers,

neopolitan

 

[JesseM]

v relative to what? at "rest" relative to what? turned around and maintained a new velocity relative to what?

Relative to any inertial observer you like (as measured by that observer's own rulers and clocks).

You use a "cheat" by giving the "travelling" twin a higher velocity for a similar period, which then comes to the same answer. Of course that is going to work out.

That's not a cheat, it's simply a requirement in order for them to move apart and later reunite. There's no possible set of velocities you can come up with such that the speed and direction of the inertial twin is constant, but the speed of the the "traveling" twin is always less, yet the distance between them is increasing during the first phase of the trip and decreasing during the second phase. Go ahead, try to find a counterexample! You are free to pick any speed and direction for the inertial twin, and any speed and direction for each phase of the traveling twin's trip, provided that they are moving farther apart from one another for the first phase and moving closer together for the second phase. No matter what combination of speeds you assume, you will always find that if each twin's clock is assumed to be slowed down by a factor of $$\sqrt{1 - v^2/c^2}$$ relative to an observer at rest in this frame, then the twin that changed velocities will show less elapsed time when they reunite than the twin that moved inertially. Do you disagree?

I am fully aware that you can use whatever frame you like. I am fully aware that a velocity in an observed inertial frame is transformed when considering it from another reference frame. But you are basically talking about three frames now. The frame relative to the moving object being considered (in which the moving object can be considered to be at rest), the observed inertial frame in which that moving object can be considered to have velocity v and the third reference inertial frame, which considers the observed inertial frame to have a velocity of its own.

What do you mean by the "observed" frame? There is no inertial frame where the twin that accelerates is at rest during both the outbound phase and the inbound phase. What I was doing was considering a single inertial observer who is at rest relative to the traveling twin during the outbound phase, but then continues to move on inertially when the traveling twin accelerates, so that afterwards this same observer will measure the traveling twin to have some substantial velocity during the inbound phase (which must be greater in magnitude than that of the other twin in order to ensure that the traveling twin is able to catch up and they can reunite). There is no third frame here, just a single frame in which the traveling twin changes speed before and after the acceleration.

If this doesn't clear things up, please specify what you think the third frame is. If frame #1 (which I think is the only frame I used) is the one where the traveling twin was at rest during the outbound phase, what exactly is the velocity of frame #1 relative to this other frame you think I have used?

You consistently misused these frames in another thread, by adding a third frame to the question without acknowleging it.

I think whenever I introduced a new frame I specified pretty clearly what I was doing. Can you point me specifically to where you think I introduced a frame without acknowledging it?

The relative velocities of two objects which are considered to be at rest in their own frames will always be equal and opposite. The equation you threw in for Equation 4 in the light clock thread pertained to observations from a third frame.

No, again, that was the single inertial frame where K' was at rest during the outbound phase, but not at rest during the inbound phase. Since K' accelerates, it is impossible to find a frame where K' is at rest both before and after the acceleration. And if K' had a speed of v in both the inbound and outbound phase as measured in the frame of K, that means that in this second frame where K' was at rest in the outbound phase (where the speed of K is v, consistent with your 'equal and opposite' comment), the speed of K' in the inbound phase must be 2v/(1 + v^2/c^2) according to the velocity addition formula (which can itself be derived from the length contraction and time dilation formulas if you like).

But I will desist, before you lock the thread and prevent others from commenting.

I have no power to lock threads, only the "mentors" can do that. But if you are open-minded about actually considering my arguments rationally instead of looking for excuses to ignore the points I am making, then please either continue the discussion here or let me know if you would prefer to continue the discussion by email or PM.

 

[jostpuur]

The use of the prime is consistent between equations 8 and 9, because as you say they are equivalent but only in so much as they are the same equation from the perspectives of two different frames of reference. Please check the use to which the equations are then used, taking these two perspectives from two different frames of reference and then using them in one single frame of reference. That is the inconsistency. I don't claim there is an inconsistency before that usage.

Ok. I apologize for assuming you didn't know the equations 8 and 9 being equivalent. Your choice of words "The use of prime is clearly not consistent" is wrong anyway, because the inconsistency is far from clear. If an equation is true, in the sense that the left and right sides have the same numbers, I don't think it means anything to talk about equation being "used in some frame". To me it seems that you are seeing a non existing problem. You should try to explain more clearly where you see the problem.

That is to say, I am not so much assuming that "Pythagorean distance in four space" is invariant in boosts, I merely not assuming that it isn't invariant.
...
A single photon can only make one journey, irrespective of how it is perceived by different observers. If you want to say that makes Pythagorean distance in four space invariant, then I cannot argue against it.
...
Without taking refuge in other derivations, can you explain why Pythagorean distance in four space must not (or should not, or perhaps cannot) be invariant? The mathematics works for my derivation without that requirement.

The Pythagorean distance in spacetime is not invariant. Instead the spacetime interval

$$\Delta s^2 = (c\;\Delta t)^2 - \Delta x^2$$

is, and you can verify with almost any example, that your magnitude of journey

$$\mu^2 = (c\;\Delta t)^2 + \Delta x^2$$

is not.

On the page 7 of your lightclock.doc, you compute something like this.

In one frame the spatial distance is L, and the consumed time is $\frac{1}{2}t$, and then the magnitude of journey is

$$L^2 + (\frac{1}{2}ct)^2$$

In the other frame the spatial distance is $\sqrt{L^2 + (\frac{1}{2}vt)^2}$, and the consumed time $\frac{1}{2}t'$, and then the magnitude of journey is

$$L^2 + (\frac{1}{2}vt)^2 + (\frac{1}{2}ct')^2$$

Then you assume these to be equal, and get

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

(edit: Stupid mistake. Square root was forgotten)

which seems to be some kind of time dilation without the square root. Right here you were assuming, that the Pythagorean distance is invariant. Your argument "single photon in the light clock can only travel along one single path" was right, but the mathematical interpretation was wrong.

In fact I did not fully understand the derivation. IMO there should have been $(\frac{1}{2}vt')^2$, but it probably doesn't matter very much, because the result is wrong anyway.

(edit: I think this requires more attention now)

I haven't been following your explanation about twin paradox very closely, because I'm not sure if the explanation is using this assumption about Pythagorean spacetime. Has it been using?

 

[neopolitan]

...

In the other frame the spatial distance is $\sqrt{L^2 + (\frac{1}{2}vt)^2}$, and the consumed time $\frac{1}{2}t'$, and then the magnitude of journey is

$$L^2 + (\frac{1}{2}vt)^2 + (\frac{1}{2}ct')^2$$

Then you assume these to be equal, and get

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

which seems to be some kind of time dilation without the square root.

I noticed that typo only last night. I need to be on another machine to fix it. You can surely do the maths from the previous step to see that the square root was omitted in error?

cheers,

neopolitan

 

[neopolitan]

The Pythagorean distance in spacetime is not invariant. Instead the spacetime interval

$$\Delta s^2 = (c\;\Delta t)^2 - \Delta x^2$$

is, and you can verify with almost any example, that your magnitude of journey

$$\mu^2 = (c\;\Delta t)^2 + \Delta x^2$$

is not.

On the page 7 of your lightclock.doc, you compute something like this.

In one frame the spatial distance is L, and the consumed time is $\frac{1}{2}t$, and then the magnitude of journey is

$$L^2 + (\frac{1}{2}ct)^2$$

In the other frame the spatial distance is $\sqrt{L^2 + (\frac{1}{2}vt)^2}$, and the consumed time $\frac{1}{2}t'$, and then the magnitude of journey is

$$L^2 + (\frac{1}{2}vt)^2 + (\frac{1}{2}ct')^2$$

Then you assume these to be equal, and get

$$t' = t(1-v^2/c^2)^\frac{1}{2}$$

I fixed a problem from the original, in recognition of the fact that it was an obvious typo, which will remain in the original document until I can fix it on another machine.

Right here you were assuming, that the Pythagorean distance is invariant. Your argument "single photon in the light clock can only travel along one single path" was right, but the mathematical interpretation was wrong.

Let's do it with space-time invariance:

In one frame the spatial distance is L, and the consumed time is $\frac{1}{2}t$, and then the magnitude of journey is

$$-L^2 + (\frac{1}{2}ct)^2$$

In the other frame the spatial distance is $\sqrt{L^2 + (\frac{1}{2}vt)^2}$, and the consumed time $\frac{1}{2}t'$, and then the magnitude of journey is

$$-L^2 + (\frac{1}{2}vt)^2 + (\frac{1}{2}ct')^2$$

Since these now refer to space-time invariance, which is ok, we get

$$t' = t.(1-v^2/c^2)^\frac{1}{2}$$

So, is the mathematical interpretation better now? I will look into it more closely and modify my paper as necessary. In the example I discuss, I do think that pythagorean distance is equally as invariant as the space-time interval.

Thanks for the input.

cheers,

neopolitan

 

[jostpuur]

I noticed that typo only last night. I need to be on another machine to fix it. You can surely do the maths from the previous step to see that the square root was omitted in error?

cheers,

neopolitan

Oh yes it was my mistake

Funny...

You were earlier in the document telling that there was something wrong in the usual time dilation. This must have confused me.

 

[jostpuur]

In one frame the spatial distance is L, and the consumed time is $\frac{1}{2}t$, and then the magnitude of journey is

$$-L^2 + (\frac{1}{2}ct)^2\quad\quad\quad\quad(1)$$

In the other frame the spatial distance is $\sqrt{L^2 + (\frac{1}{2}vt)^2}$, and the consumed time $\frac{1}{2}t'$, and then the magnitude of journey is

$$-L^2 + (\frac{1}{2}vt)^2 + (\frac{1}{2}ct')^2\quad\quad\quad\quad(2)$$

Since these now refer to space-time invariance, which is ok, we get

$$t' = t.(1-v^2/c^2)^\frac{1}{2}\quad\quad\quad\quad(3)$$

Let us keep our heads cool. This is now going forward.

I have difficulty seeing which one of the t and t' is time in the spaceship's frame, and which one in the other frame.

By equation (1) it looks like that t is in the spaceship's frame. But isn't the final result (3) then wrong way?

To me it seems there is two mistakes in the equation (2). You should have

$$-L^2 - (\frac{1}{2}vt')^2 + (\frac{1}{2}ct')^2,$$

and then the result would be

$$t = t'\sqrt{1-v^2/c^2}.$$

btw. was it your intention to now use the space time interval (ct)^2-x^2 ? I assumed so since you had put minus signs in front of both L^2.

 

[neopolitan]

Let us keep our heads cool.

Indeed. We should also not conflate two separated issues within the single thread. The equations in your post are from my paper, which you looked at and commented on, not from the my original material about the twin paradox. So you are rightly confused about what relates to any spaceship's frame.

btw. was it your intention to now use the space time interval (ct)^2-x^2 ? I assumed so since you had put minus signs in front of both L^2.

Yes, it was. I will need to review this.

To me it seems there is two mistakes in the equation (2). You should have

$$-L^2 - (\frac{1}{2}vt')^2 + (\frac{1}{2}ct')^2,$$

and then the result would be

$$t = t'\sqrt{1-v^2/c^2}.$$

Damn the Christmas preparations.

You will have to go back to original paper and recheck the scenario, as I did. In that scenario there is absolutely no reason to assume your invariance equation. Simple geometry gives the pythagorean equation. Is it not unreasonable to expect that a first principles derivation must rely on post derivation consequences to work?

This may mean that the light clock cannot be used to provide a satisfactory derivation of the relativistic equations. For me this is of no consequence since my primary, preferred derivation is actually the one in http://geocities.com/neopolitonian/sr.doc. The light clock derivation is just me being "clever".

For my part, I will need to review the time-space interval invariance equation and also my primary derivation.

For new readers, jostpuur and I are discussing equations on page 5 and 6 of http://geocities.com/neopolitonian/lightclock.doc - this is really a side issue and not particularly closely related to the twin paradox thread.

cheers,

neopolitan

 

[jostpuur]

I first thought that the problem with the geometry of space time would also be the source of confusion with the twin paradox issue, but now I'm not so sure if that's really the case.

Although I'm quite sure that it is the source of confusion in many places elsewhere, because the Pythagorean distance in space time simply doesn't work like that.

 

[Jorrie]

For new readers, jostpuur and I are discussing equations on page 5 and 6 of http://geocities.com/neopolitonian/lightclock.doc - this is really a side issue and not particularly closely related to the twin paradox thread.

The moderators must be in a festive mood, because I find it very strange that they allow these references to and discussions about an unpublished (slightly off-mainstream, it seems) 'paper' to carry on!