Need help understanding the twins

[cfrogue]

Why would the total elapsed proper time involve a sum of the proper time BT during the acceleration phase and the coordinate time in the launch frame of the acceleration phase ta'? Also, to talk about an "elapsed time" you need to specify both an endpoint and a beginning point...the beginning point would be the point where the first twin departed from the second twin as he began the acceleration phase, but what would be the endpoint? If you want to compare their ages in the final rest frame of the two twins, you need to make sure the endpoints for each twin's worldline that you use in your proper time calculation are simultaneous in the final rest frame, but you don't appear to have done anything like that.

Yes, the beginning and endpoints are at issue.

But, since this is theoretical and the twins agreed on BT and both know the value, I am assuming I can do this.

More specifically, if BT is known by all parties, the calculations seem to point in that direction.

What is your view?

 

[JesseM]

Yes, but the orginal integral is gamma, if you look to the left part.

Also, it is not the case that g is the proper acceleration of the accelerating frame. It is the acceleration from the non-accelerating frame.

No, g is the proper acceleration. Note that the "proper acceleration" of an object at some point on its worldline is based on considering the inertial rest frame where it is instantaneously at rest at that point, and finding the coordinate acceleration in that frame at that point. And if you read the paper, they make clear that g refers to the acceleration in an instantaneous co-moving frame:

Therefore, for a particle that has a constant acceleration a'_x = g with respect to an inertial frame S' which instantaneously accompanies the particle

Also, if you look at the meaning of a on the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html (which plays the same role as g on the page you linked to), they also make clear that it refers to the acceleration in an instantaneously co-moving inertial frame:

First of all we need to be clear what we mean by continuous acceleration at 1g. The acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference traveling at the same instantaneous speed as the rocket (see relativity FAQ on accelerating clocks). This acceleration will be denoted by a.

You cannot integrate with a proper acceleration of the accelerating frame when operating in the launch frame which is what the integral does.

The integral is meant to calculate the elapsed proper time on a clock with constant proper acceleration, not constant coordinate acceleration.

Anyway, I guess I now better understand what you meant by "the acceleration equations integrate with gamma", but the following sentence "This already is designed under the context of the R of S" makes no sense. You can't assume that just because the derivation involves a relativistic gamma, that means it's magically going to take care of dealing with simultaneity issues between the launch frame and the rest frame of the rocket when it stops accelerating. The relativistic acceleration equation only involves two times, the coordinate time t in the launch frame and the proper time T measured aboard the rocket, it doesn't deal with coordinate times (or simultaneity, which just means equal coordinate times) in any other frame.

 

[JesseM]

Yes, the beginning and endpoints are at issue.

But, since this is theoretical and the twins agreed on BT and both know the value, I am assuming I can do this.

Do what? BT is just the proper time during the acceleration phase, if they both accelerate the same way they will both age by the same amount BT during their respective acceleration phases. Any difference in aging will have to come during the proper time they experience during the inertial phases of their journey (this is why it's simpler to just assume the acceleration is instantaneous as I did in my calculation in post 60). And there's no way to figure out how much each one ages during the inertial phases after the acceleration without picking endpoints, and making sure they are simultaneous in their final rest frame, which you didn't do. Instead you just said that the time for twin 1 in the "relative motion phase" was t' while the time for twin 2 in the "relative motion phase" was t'/gamma--where are you getting these values? Are they supposed to represent proper times, or times in some reference frame?

 

[cfrogue]

No, g is the proper acceleration. Note that the "proper acceleration" of an object at some point on its worldline is based on considering the inertial rest frame where it is instantaneously at rest at that point, and finding the coordinate acceleration in that frame at that point. And if you read the paper, they make clear that g refers to the acceleration in an instantaneous co-moving frame:

Take a look at this.

Bell considered two spaceships starting from rest in a Lorentz system S, and undergoing identical accelerations a(t) in that system.

Then, look at equation 7 and read past it a little and tell me what you think.
http://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1919v2.pdf

All integration is done from acceleration of the view of the launch frame in this paper.

Is this correct?

 

[cfrogue]

Do what? BT is just the proper time during the acceleration phase, if they both accelerate the same way they will both age by the same amount BT during their respective acceleration phases. Any difference in aging will have to come during the proper time they experience during the inertial phases of their journey (this is why it's simpler to just assume the acceleration is instantaneous as I did in my calculation in post 60). And there's no way to figure out how much each one ages during the inertial phases after the acceleration without picking endpoints, and making sure they are simultaneous in their final rest frame, which you didn't do. Instead you just said that the time for twin 1 in the "relative motion phase" was t' while the time for twin 2 in the "relative motion phase" was t'/gamma--where are you getting these values? Are they supposed to represent proper times, or times in some reference frame?

OK, I have decided I am in agreement with you.


The actual integral is all done in the acclerating frame and derived in the launch frame later.

http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf

Thus, the correct answer is

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

where t is the elapsed proper time of the non-accelerating frame.

 

[cfrogue]

Do what? BT is just the proper time during the acceleration phase, if they both accelerate the same way they will both age by the same amount BT during their respective acceleration phases. Any difference in aging will have to come during the proper time they experience during the inertial phases of their journey (this is why it's simpler to just assume the acceleration is instantaneous as I did in my calculation in post 60). And there's no way to figure out how much each one ages during the inertial phases after the acceleration without picking endpoints, and making sure they are simultaneous in their final rest frame, which you didn't do. Instead you just said that the time for twin 1 in the "relative motion phase" was t' while the time for twin 2 in the "relative motion phase" was t'/gamma--where are you getting these values? Are they supposed to represent proper times, or times in some reference frame?

As such, the argument presented for the twins is perfect.

 

[JesseM]

As such, the argument presented for the twins is perfect.

Are you responding to a specific sentence in my post, or the whole thing? And I just got through giving various criticisms of your approach, and asking questions like "where are you getting these values? Are they supposed to represent proper times, or times in some reference frame?" This one-line answer is not a substantive response.

 

[cfrogue]

Are you responding to a specific sentence in my post, or the whole thing? And I just got through giving various criticisms of your approach, and asking questions like "where are you getting these values? Are they supposed to represent proper times, or times in some reference frame?" This one-line answer is not a substantive response.

You clarified my thinking.

I agree, I mistakenly thought the acceleration was from the non-accelerating frame. Thank you. The mainstream has many comments about this one way of the other. But, you conversations led me to the actual correct answer.

But, now this twins paradox is decidable without debate.

O and O' disagree.

 

[Jorrie]

If you define the frames O and O' as final and original inertial frames respectively (as I https://www.physicsforums.com/showpost.php?p=2449151&postcount=65"), then it looks like your two conclusions are right (I have not checked your math). Your calculations seem to be for the two inertial frames, not from the (partly) accelerated points of view (you would then have needed different equations, I think).

Assume they are in the same frame after acceleration.

I think it came out in your exchanges with JesseM that your equations seem correct and hence your conclusions, but, you have some confusion in the choice of coordinate systems. Your equations seem to work in both the original launch frame (I call it O' to be compatible with your calculations) and in the final, common frame (O). There is absolutely no need to have the coordinate systems accelerate with the observers here - it just creates part of the confusion. Leave them as the launch frame and the final frame.

The constant acceleration in your equations is the proper acceleration as measured by each observer by accelerometers, or by momentarily comoving inertial frames, as JesseM has pointed out. It is not launch frame coordinate acceleration (unlike in Bell's paradox, where it is defined as a constant launch coordinate acceleration).

But, please take careful note of what I wrote here:

However, both inertial frames O and O' are correct in their conclusions, because they each made a frame dependent calculation/measurement of two events that are not spatially colocated. I think nobody can say, in absolute terms, which twin is younger here (unless they are brought together again). Your scenario does have a reciprocal aging situation, which BTW, does not exist in the classical twin paradox, or in the GPS system, for that matter.

 

[cfrogue]

Are you responding to a specific sentence in my post, or the whole thing? And I just got through giving various criticisms of your approach, and asking questions like "where are you getting these values? Are they supposed to represent proper times, or times in some reference frame?" This one-line answer is not a substantive response.

Here is the new math.

Since O and O' end up in the same frame, why do you ask what they'll conclude separately? Presumably you mean "what will they conclude about the ages of the twins in their own rest frame", and since they share the same rest frame, naturally they'll conclude exactly the same thing about the twin's ages. Anyway, the answer in this frame will be that the twin who accelerated second will be younger than the twin who accelerated first.

Here is the math.

Calculations of O for the twins

Elapsed proper time calculation for twin 1
twin1's acceleration phase
BT
twin1's relative motion phase
t'
twin2's acceleration phase
c/a sinh(aBT/c)

Total elapsed proper time calculation of O for twin1 BT + t' + c/a sinh(aBT/c)

Elapsed proper time calculation for twin 2

twin1's acceleration phase
c/a sinh(aBT/c)
twin2's relative motion phase
t'/ λ
twin2's acceleration phase
BT

Total elapsed proper time calculation of O for twin2 c/a sinh(aBT/c) + t'/λ + BT
Conclusion of O, twin1 is older.

Calculations of O' for the twins

Elapsed proper time calculation for twin 1
twin1's acceleration phase
BT
twin1's relative motion phase
t/λ
twin2's acceleration phase
c/a sinh(aBT/c)

Total elapsed proper time calculation of O' for twin1 BT + t/λ + c/a sinh(aBT/c)

Elapsed proper time calculation for twin 2
twin1's acceleration phase
c/a sinh(aBT/c)
twin2's relative motion phase
t
twin2's acceleration phase
BT

Total elapsed proper time calculation of O' for twin2 c/a sinh(aBT/c) + t + BT
Conclusion of O', twin1 is younger.

 

[cfrogue]

I think it came out in your exchanges with JesseM that your equations seem correct and hence your conclusions, but, you have some confusion in the choice of coordinate systems. Your equations seem to work in both the original launch frame (I call it O' to be compatible with your calculations) and in the final, common frame (O). There is absolutely no need to have the coordinate systems accelerate with the observers here - it just creates part of the confusion. Leave them as the launch frame and the final frame.

The constant acceleration in your equations is the proper acceleration as measured by each observer by accelerometers, or by momentarily comoving inertial frames, as JesseM has pointed out. It is not launch frame coordinate acceleration (unlike in Bell's paradox, where it is defined as a constant launch coordinate acceleration).

But, please take careful note of what I wrote here:

Agreed, I made the changes.

Please evaluate these new considerations.

 

[JesseM]

Calculations of O for the twins

Elapsed proper time calculation for twin 1
twin1's acceleration phase
BT
twin1's relative motion phase
t'

Are these two the proper times for twin 1 during his acceleration phase and his inertial phase, or is t' supposed to be coordinate time in some frame?

twin2's acceleration phase
c/a sinh(aBT/c)

This is the coordinate time in the launch frame, yes?

Total elapsed proper time calculation of O for twin1 BT + t' + c/a sinh(aBT/c)

Why would a calculation of the elapsed proper time for twin 1 include the factor c/a sinh(aBT/c)? As before, isn't that just the coordinate time for the acceleration phase in the launch frame?

Elapsed proper time calculation for twin 2

twin1's acceleration phase
c/a sinh(aBT/c)
twin2's relative motion phase
t'/ λ

Again, why do you say that if the inertial phase of twin 1's trip lasted t', the inertial phase of twin 2's phase should last t'/gamma? That doesn't make any sense--again, the only way to calculate the time the inertial phases last is to pick endpoints for each twin which are simultaneous in their final rest frame, and then figure out the proper time between the end of the acceleration and the endpoint for that twin, and add it to the proper time for the acceleration phase (and for twin 2 you also need to add the proper time for the inertial phase that happens after twin 1 departs but but before twin 2 accelerates).

 

[cfrogue]

Are these two the proper times for twin 1 during his acceleration phase and his inertial phase, or is t' supposed to be coordinate time in some frame?

t' is the proper time in the context of twin 1 and O.

twin2's acceleration phase
c/a sinh(aBT/c)

This is the coordinate time in the launch frame, yes?

Correct as interpreted by twin1.

Total elapsed proper time calculation of O for twin1 BT + t' + c/a sinh(aBT/c)

Why would a calculation of the elapsed proper time for twin 1 include the factor c/a sinh(aBT/c)? As before, isn't that just the coordinate time for the acceleration phase in the launch frame?

The equation c/a sinh(aBT/c) would be the elapsed proper time experienced by twin1 for the acceleration of twin2.

Again, why do you say that if the inertial phase of twin 1's trip lasted t', the inertial phase of twin 2's phase should last t'/gamma? That doesn't make any sense--again, the only way to calculate the time the inertial phases last is to pick endpoints for each twin which are simultaneous in their final rest frame, and then figure out the proper time between the end of the acceleration and the endpoint for that twin, and add it to the proper time for the acceleration phase (and for twin 2 you also need to add the proper time for the inertial phase that happens after twin 1 departs but but before twin 2 accelerates).

t' is in the context of O's frame.

But, your point is valid.

Since acceleration is absolute motion under SR, then the start and end points of the acceleration period are different in each frame's proper time but can be decided.

For example, when twin1 accelerates for BT in its proper time, we know absolutely that twin2 elapsed c/a sinh(aBT/c).

Normally, under SR, one must have time intervals absolute only with an inertial frame, but acceleration is also absolute.

Thus, we are actually comparing simultaneity but with different clock beats given acceleration.

This is the trick I am using.

 

[Jorrie]

Agreed, I made the changes.

Please evaluate these new considerations.

Your coordinate system definitions are still utterly confusing, as is evident from JesseM's remarks above. Please get a consistent definition and rewrite the equations. Again, you only have two inertial frames: the launch frame and the final frame, keeping in mind that the constant proper acceleration is not referring to either of them. I would use coordinate time t for frame O and t' for frame O', while you seem to have them mixed up somewhat.

 

[cfrogue]

Your coordinate system definitions are still utterly confusing, as is evident from JesseM's remarks above. Please get a consistent definition and rewrite the equations. Again, you only have two inertial frames: the launch frame and the final frame, keeping in mind that the constant proper acceleration is not referring to either of them. I would use coordinate time t for frame O and t' for frame O', while you seem to have them mixed up somewhat.

Here, I am modeling the solution based on a piecewise integral offered at wiki.

http://en.wikipedia.org/wiki/Twin_paradox

Einstein also used a piecewise integral for his solution.

keeping in mind that the constant proper acceleration is not referring to either of them

Please explain.

 

[JesseM]

The equation c/a sinh(aBT/c) would be the elapsed proper time experienced by twin1 for the acceleration of twin2.

But during the period twin2 is accelerating, twin1 is moving inertially, so isn't that section of twin1's worldline already included in t', which is supposed to be the proper time of twin1's inertial phase? Or are you defining t' to be only the proper time along the section of twin1's worldline where twin1 is moving inertially and twin2 is not simultaneously (according to the definition of simultaneity in twin1's inertial rest frame) accelerating?

But, your point is valid.

Since acceleration is absolute motion under SR, then the start and end points of the acceleration period are different in each frame's proper time but can be decided.

I'm not talking about the start and end points of the acceleration period. I'm talking about the start and endpoints of the worldline of each twin that you want to use to compare their elapsed proper time. Presumably the starting point is the point on each one's worldline when they first depart from one another at the same age (because twin1 begins to accelerate while twin2 continues to move inertially for a while)--note that this point on twin2's worldline lies well before the point where twin2 begins to accelerate himself. Then the endpoints have to be simultaneous in their final rest frame if you want to compare their ages in that frame, so even if you pick the point on twin2's worldline immediately after he stops accelerating, twin1 stopped accelerating much earlier so you'll have to pick a point on his worldline that lies well after he stopped accelerating (the part of his worldline between the end of his acceleration and the 'endpoint' is his inertial phase).

For example, when twin1 accelerates for BT in its proper time, we know absolutely that twin2 elapsed c/a sinh(aBT/c).

According to the definition of simultaneity in twin2's rest frame during that phase, yes. But If twin2's entire aging during the inertial phase after twin1 departed is already included in t', then you don't need to add c/a sinh(aBT/c) separately.

Normally, under SR, one must have time intervals absolute only with an inertial frame, but acceleration is also absolute.

Thus, we are actually comparing simultaneity but with different clock beats given acceleration.

This is the trick I am using.

Don't understand anything you said above. It would help if you'd address my question of why you think if twin1's inertial phase (or just the part of twin1's inertial phase when twin2 is not accelerating?) lasts for t', then twin2's inertial phase (or just the part of twin2's inertial phase when twin1 is not accelerating) lasts for t'/gamma. What is your reasoning?

 

[cfrogue]

But during the period twin2 is accelerating, twin1 is moving inertially, so isn't that section of twin1's worldline already included in t', which is supposed to be the proper time in twin1's inertial phase? Or are you defining t' to be only the proper time along the section of twin1's worldline where twin1 is moving inertially and twin2 is not simultaneously (according to the definition of simultaneity in twin1's inertial rest frame) accelerating?

I do not do the worldline thing because it is a geometric representation of LT.

The math of LT is all that is needed.

Under this context, there is relative motion and t' is the proper time of O for thisd and t'/gamma iis the interpretation of O for O'.

I'm not talking about the start and end points of the acceleration period. I'm talking about the start and endpoints of the worldline of each twin that you want to use to compare their elapsed proper time. Presumably the starting point is the point on each one's worldline when they first depart from one another at the same age (because twin1 begins to accelerate while twin2 continues to move inertially for a while)--note that this point on twin2's worldline lies well before the point where twin2 begins to accelerate himself. Then the endpoints have to be simultaneous in their final rest frame if you want to compare their ages in that frame, so even if you pick the point on twin2's worldline immediately after he stops accelerating, twin1 stopped accelerating much earlier so you'll have to pick a point on his worldline that lies well after he stopped accelerating (the part of his worldline between the end of his acceleration and the 'endpoint' is his inertial phase).

Can't do wordlines. They are a cartoon book view of SR.

Let's do LT.

According to the definition of simultaneity in twin2's rest frame during that phase, yes. But If twin2's entire aging during the inertial phase after twin1 departed is already included in t', then you don't need to add c/a sinh(aBT/c) separately.

According to the definition of simultaneity
What does this mean in the context of this experiment?

Don't understand anything you said above. It would help if you'd address my question of why you think if twin1's inertial phase (or just the part of twin1's inertial phase when twin2 is not accelerating?) lasts for t', then twin2's inertial phase (or just the part of twin2's inertial phase when twin1 is not accelerating) lasts for t'/gamma. What is your reasoning?

t' is just the time that elapses in the frame of O for the relative motion phase.

Now, given that value, t'/gamma is the value of O'.

 

[Jorrie]

Here, I am modeling the solution based on a piecewise integral offered at wiki.

According to your 'funny' definition of frames, your modeling is wrong. The equations you are using work fine from the launch frame POV, but not as you chose frames. When I said your equations are OK, I referred to the logical coordinate choice. I think JesseM is trying to make you realize your mistake yourself, so he is referring to your 'funny' coordinate choices when he questions your equations. Keep this difference in mind, to avoid further confusion.

Hint (JesseM has said so some time ago, but here it is again): as you defined O and O', they are (at the end) the exact same frame, so finding that they get different (reciprocal) elapsed times cannot be right.

"... keeping in mind that the constant proper acceleration is not referring to either of them"

Please explain.

My 'them' are the launch frame and the final frame, both inertial. The proper acceleration used in the Wiki-article is per the observers (accelerometers), not inertial coordinate accelerations, as explained before. [Edit: But, it is fine to use it in the equations for the original (or final) inertial frame calculations - the equations have been derived with that in mind.]

 

[JesseM]

I do not do the worldline thing because it is a geometric representation of LT.

But you haven't used the Lorentz transformation at all in your calculations! Anyway, if you want the nongeometric version of a spacetime diagram, you can instead give the space coordinate of each twin as a function of time, x(t), for each twin, and give the coordinates of the beginning and endpoints that you're calculating the proper time between. According to the relativistic rocket page, if you want to know the distance traveled in the launch frame as a function of time since the beginning of acceleration, it's d = (c^2/a)*[cosh(a*T/c) - 1] = (c^2/a)*(sqrt[1 - (a*t/c)^2] - 1). So, if the beginning point for each twin is the origin, x=0 and t=0, and then twin 1 accelerates for a proper time BT (a coordinate time in this frame of (c/a)*sinh(a*BT/c) ), that means twin 1's x(t) function between t=0 and t1 =(c/a)*sinh(a*BT/c) will be x(t) = (c^2/a)*(sqrt[1 - (a*t/c)^2] - 1). At time t1 when the acceleration ends in this frame, twin 1's position will be x = (c^2/a)*(sqrt[1 - (a*t1/c)^2] - 1). Also, using the equation for velocity as a function of time in the launch frame, twin 1's velocity at t1 will be v1 = a*t1 / sqrt[1 + (a*t1/c)^2]. Then after that twin 1 will coast inertially at this same velocity, starting from x = (c^2/a)*(sqrt[1 - (a*t1/c)^2] - 1) at time t1. So, twin 1's x(t) after t1 will be:

x(t) = v1*t - v1*t1 + (c^2/a)*(sqrt[1 - (a*t1/c)^2] - 1)

You can see that at t=t1, this gives x(t) = (c^2/a)*(sqrt[1 - (a*t1/c)^2] - 1).

So to sum up, the x(t) for twin 1 looks like this:

before t1: x(t) = (c^2/a)*(sqrt[1 - (a*t/c)^2] - 1)
after t1: x(t) = v1*t - v1*t1 + (c^2/a)*(sqrt[1 - (a*t1/c)^2] - 1)

where t1 = (c/a)*sinh(a*BT/c), and v1 = a*t1 / sqrt[1 + (a*t1/c)^2].

Now we can similarly calculate x(t) for twin 2. Twin 2 remains at the origin (x=0) for some time t2 in this frame before accelerating, so until time t2 we have x(t) = 0. Afterwards, twin2 accelerates at the same rate for proper time BT, so twin2's position as a function of time for this period will be x(t) = (c^2/a)*(sqrt[1 - (a*[t - t2]/c)^2] - 1). He will continue to accelerate for a proper time BT and a coordinate time (c/a)*sinh(a*BT/c), so the acceleration will end at time t3 = t2 + (c/a)*sinh(a*BT/c), at which point twin 2 will be at position x(t3) = (c^2/a)*(sqrt[1 - (a*[t3 - t2]/c)^2] - 1). Now if you want to compare twin 2's age with twin 1's at the moment twin 2 stops accelerating, in their new mutual rest frame, you have to find an event E which is on the path of twin1--so it lies along the line x(t) = v1*t - v1*t1 + (c^2/a)*(sqrt[1 - (a*t1/c)^2] - 1)--and which is simultaneous with the event of twin2 stopping accelerating at position (c^2/a)*(sqrt[1 - (a*[t3 - t2]/c)^2] - 1) and time t3 = t2 + (c/a)*sinh(a*BT/c). And once we have found this event E which is on the path of twin1, we must calculate the proper time experienced by twin1 between the start at x=0 and t=0 and the endpoint at event E, which will be the sum of the proper time experienced by twin1 between the beginning of his acceleration phase and the end of his acceleration phase, and the proper time experienced by twin1 between the beginning of his inertial phase and the event E. Meanwhile we calculate the proper time experienced by twin2 between the start and the end of his acceleration (which, as you remember, is simultaneous with E in their final rest frame), which is the sum of the proper time he experienced on his inertial phase between (x=0, t=0) and (x=0, t=t2) and the proper time experienced between the beginning of his accelerating phase and the end of it.

Do you disagree with this approach?

Under this context, there is relative motion and t' is the proper time of O for thisd and t'/gamma iis the interpretation of O for O'.

Proper time for what? If you want to calculate the proper time in a piecewise manner, you need to be more clear about the beginning and end of each piece. What event for twin1/O is the beginning of the piece, and what event for twin1/O is the end? For example, is t':

A) the proper time experienced by twin1/O between the moment twin1 stops accelerating and the moment (in twin1/O's final rest frame) that twin2 stops accelerating?

B) the proper time experienced by twin1/O between the moment twin1 stops accelerating and the moment (in twin1/O's final rest frame) that twin2 starts accelerating?

C) something else?

Can't do wordlines. They are a cartoon book view of SR.

First of all, spacetime diagrams are perfectly rigorous and can be found in any textbook. Secondly, the word "worldline" has nothing specifically to do with any sort of visual diagram, it just means the set of points in spacetime that a given object passes through. For example, if a given object is moving inertially at 0.6c in some inertial frame, and at t=0 years in this frame its position is x=10 light years, then the object's position as a function of time will be given by x(t) = t*(0.6c) + 10, and the set of all points x,t which satisfy this function (like x=16, t=10) will be the object's "worldline". You can't avoid talking about the set of coordinates an object passes through if you want to do a quantitative calculation of proper time in some coordinate system, and "worldline" is just a shorthand way of saying "set of points in spacetime that the object passes through".

According to the definition of simultaneity
What does this mean in the context of this experiment?

You want to compare their ages at the "same time" in their final frame, no? Then you have to find two events on their worldlines which are simultaneous in the final frame (they happen at the 'same time'), and integrate the proper time for each twin up to the event on their worldline.

t' is just the time that elapses in the frame of O for the relative motion phase.

"Relative motion phase" is too vague, they are in relative motion at all times from the start of twin1's acceleration until the end of twin2's acceleration. Perhaps you mean relative inertial motion? See my questions about t' above.

 

[cfrogue]

But you haven't used the Lorentz transformation at all in your calculations! Anyway, if you want the nongeometric version of a spacetime diagram, you can instead give the space coordinate of each twin as a function of time, x(t), for each twin, and give the coordinates of the beginning and endpoints that you're calculating the proper time between. According to the relativistic rocket page, if you want to know the distance traveled in the launch frame as a function of time since the beginning of acceleration, it's d = (c^2/a)*[cosh(a*T/c) - 1] = (c^2/a)*(sqrt[1 - (a*t/c)^2] - 1). So, if the beginning point for each twin is the origin, x=0 and t=0, and then twin 1 accelerates for a proper time BT (a coordinate time in this frame of (c/a)*sinh(a*BT/c) ), that means twin 1's x(t) function between t=0 and t1 =(c/a)*sinh(a*BT/c) will be x(t) = (c^2/a)*(sqrt[1 - (a*t/c)^2] - 1). At time t1 when the acceleration ends in this frame, twin 1's position will be x = (c^2/a)*(sqrt[1 - (a*t1/c)^2] - 1). Also, using the equation for velocity as a function of time in the launch frame, twin 1's velocity at t1 will be v1 = a*t1 / sqrt[1 + (a*t1/c)^2]. Then after that twin 1 will coast inertially at this same velocity, starting from x = (c^2/a)*(sqrt[1 - (a*t1/c)^2] - 1) at time t1. So, twin 1's x(t) after t1 will be:

x(t) = v1*t - v1*t1 + (c^2/a)*(sqrt[1 - (a*t1/c)^2] - 1)

You can see that at t=t1, this gives x(t) = (c^2/a)*(sqrt[1 - (a*t1/c)^2] - 1).

So to sum up, the x(t) for twin 1 looks like this:

before t1: x(t) = (c^2/a)*(sqrt[1 - (a*t/c)^2] - 1)
after t1: x(t) = v1*t - v1*t1 + (c^2/a)*(sqrt[1 - (a*t1/c)^2] - 1)

where t1 = (c/a)*sinh(a*BT/c), and v1 = a*t1 / sqrt[1 + (a*t1/c)^2].

OK, nice post BTW.

Let's just focus on O and first the calcs for the acceleration phase for twin1.

O will see twin1 elapsed BT since they are in the same ship and the burn times are in the context of the accelerating ship.

Now, for twin2, the elapsed time for the acceleration phase is (c/a)*sinh(a*BT/c).

Do you mind if you stop here and agree or disagree on the above?

 

[JesseM]

OK, nice post BTW.

Let's just focus on O and first the calcs for the acceleration phase for twin1.

O will see twin1 elapsed BT since they are in the same ship and the burn times are in the context of the accelerating ship.

Now, for twin2, the elapsed time for the acceleration phase is (c/a)*sinh(a*BT/c).

You mean, the elapsed proper time for twin2 between the event on twin2's worldline where twin1 departs (x=0 and t=0 in the above example) and the event on twin2's worldline that is simultaneous with the event of twin1 stopping his acceleration, according to the launch frame's definition of simultaneity? If so, yes I agree.

 

[cfrogue]

You mean, the elapsed proper time for twin2 between the event on twin2's worldline where twin1 departs (x=0 and t=0 in the above example) and the event on twin2's worldline that is simultaneous with the event of twin1 stopping his acceleration, according to the launch frame's definition of simultaneity? If so, yes I agree.

Yes, we are in agreement.

Likewise, O' sees the same elapsed proper time for each since acceleration is absolute for twin1's acceleration phase.

Twin1 = BT and twin2 = (c/a)*sinh(a*BT/c).

Agreed?

 

[JesseM]

Yes, we are in agreement.

Likewise, O' sees the same elapsed proper time for each since acceleration is absolute for twin1's acceleration phase.

Not sure what you mean. All frames always agree on the proper time along a worldline between a specific pair of events on that worldline, regardless of whether we're talking about acceleration or inertial motion--proper time is a frame-invariant quantity.

Twin1 = BT and twin2 = (c/a)*sinh(a*BT/c).

Yes, given the pair of events on each worldline that I mentioned, these are the proper times on each worldline between those events.

 

[cfrogue]

Not sure what you mean. All frames always agree on the proper time along a worldline between a specific pair of events on that worldline, regardless of whether we're talking about acceleration or inertial motion--proper time is a frame-invariant quantity.

Yes, given the pair of events on each worldline that I mentioned, these are the proper times on each worldline between those events.

OK, so both O and O' calculate the below as the correct answer for the acceleration of twin1 and O.

Twin1 = BT and twin2 = (c/a)*sinh(a*BT/c).

Agreed?

 

[cfrogue]

Not sure what you mean. All frames always agree on the proper time along a worldline between a specific pair of events on that worldline, regardless of whether we're talking about acceleration or inertial motion--proper time is a frame-invariant quantity.

Yes, given the pair of events on each worldline that I mentioned, these are the proper times on each worldline between those events.

Then O' waits some long period of time t, the relative motion phase, and then does exactly the same burn BT at a.

So, O' calculates the total as,

Twin1 = BT + t/λ + (c/a)*sinh(a*BT/c).
Twin2 = (c/a)*sinh(a*BT/c) + t + BT.

Agreed?

 

[JesseM]

Then O' waits some long period of time t, the relative motion phase, and then does exactly the same burn BT at a.

So, O' calculates the total as,

Twin1 = BT + t/λ + (c/a)*sinh(a*BT/c).
Twin2 = (c/a)*sinh(a*BT/c) + t + BT.

Agreed?

Total of what? If you're going to calculate proper time by adding segments, you need to specify what events constitute the endpoint of each segment. For example, pick the following events on Twin2's worldline:

Event 1: Twin1 departing, at x=0 and t=0
Event 2: The event on Twin2's worldline that is simultaneous with the event of Twin1 stopping his acceleration, according to the launch frame's definition of simultaneity
Event 3: The event of Twin2 beginning his acceleration
Event 4: The event of Twin2 stopping his acceleration

Then to find the total proper time between event 1 and 4, you add (proper time between Event 1 and 2) + (proper time between Event 2 and 3) + (proper time between Event 3 and 4)

If Twin1 accelerated for a proper time BT, then on Twin2's worldline (proper time between event 1 and 2) is (c/a)*sinh(a*BT/c)

If Twin2 waited a proper time t after Event 2 before beginning to accelerate, then (proper time between event 1 and 2) is t

If Twin2 accelerated for a proper time BT, then (proper time between Event 3 and 4) is BT.

So, does this fit the logic of why you said the time for Twin2 would have a total time of (c/a)*sinh(a*BT/c) + t + BT ? If so, I don't have a problem with this, but I think I do have a problem with the calculation for Twin1. The fact that the middle segment's time for Twin1 is supposed to be t/gamma suggests that you are using the launch frame's definition of simultaneity (also Twin2's before accelerating) to calculate the time elapsed on Twin1's clock between the moments of Event 2 and Event 3 on Twin2's worldline. So, let's define the following events on Twin1's worldine:

Event 1a: Twin1 begins to accelerate away from Twin2 at x=0 and t=0 in the launch frame
Event 2a: Twin1 stops accelerating
Event 3a: The event on Twin1's worldline that is simultaneous in the launch frame (not Twin1's own current frame) with the event of Twin2 beginning to accelerate

In this case, (proper time between Event 1a and Event 2a) = BT, and (proper time between Event 2a and Event 3a) = t/gamma. Again, does this fit with what you were thinking? But in this case it's not too clear where the third term of the sum you gave for Twin1's time, (c/a)*sinh(a*BT/c), is supposed to come from. This is obviously supposed to be some sort of time for Twin2's acceleration phase, but it looks like the coordinate time in the launch frame for Twin2's acceleration phase. Suppose we pick the following for the 4th event on Twin1's worldline:

Event 4a: The event on Twin1's worldline that is simultaneous in the launch frame (not Twin1's own current frame) with the event of Twin2 stopping his acceleration

In this case the coordinate time in the launch frame between Event 3a and 4a would be (c/a)*sinh(a*BT/c), but since Twin1's clock is slowed down by a factor of 1/gamma in this frame, the proper time between Event 3a and 4a would instead by [(c/a)*sinh(a*BT/c)]/gamma. Also, in this case Event 4a on Twin1's worldline would be simultaneous with Event 4 on Twin2's worldline in the launch frame, so you wouldn't be comparing their ages according to the definition of simultaneity used in their final rest frame as you were supposed to.

Alternately, it's possible you're imagining a different fourth event on Twin1's worldline like this--

Event 4b: The event on Twin1's worldline that is simultaneous in Twin1's current rest frame with Twin2 finishing his acceleration

But again, the proper time between 3a and 4b would not be (c/a)*sinh(a*BT/c) in this case. First of all, in Twin1's current rest frame, event 3a is not simultaneous with the beginning of Twin2's acceleration, it was specifically defined to be simultaneous with the beginning of Twin2's acceleration in the launch frame. Second of all, even if you did calculate the time elapsed in Twin1's current rest frame between the beginning and end of Twin2's acceleration, it would not be equal to (c/a)*sinh(a*BT/c)--that formula only works if you're using the frame where the accelerating object begins accelerating from a velocity of 0, whereas in Twin1's current rest frame, Twin2 already had some significant nonzero velocity before beginning to accelerate.

So no matter which way I look at it, your calculation for Twin1's proper time doesn't appear to make much sense. You really need to go back and give careful thought to which events are supposed to mark the beginning and end of each segment you want to add, and to make sure that the final events on the final segment of each twin's sum are actually simultaneous in their final rest frame, if that's the frame where you want to compare their final ages.

 

[cfrogue]

Total of what? If you're going to calculate proper time by adding segments, you need to specify what events constitute the endpoint of each segment. For example, pick the following events on Twin2's worldline:

Event 1: Twin1 departing, at x=0 and t=0
Event 2: The event on Twin2's worldline that is simultaneous with the event of Twin1 stopping his acceleration, according to the launch frame's definition of simultaneity
Event 3: The event of Twin2 beginning his acceleration
Event 4: The event of Twin2 stopping his acceleration

Then to find the total proper time between event 1 and 4, you add (proper time between Event 1 and 2) + (proper time between Event 2 and 3) + (proper time between Event 3 and 4)

If Twin1 accelerated for a proper time BT, then on Twin2's worldline (proper time between event 1 and 2) is (c/a)*sinh(a*BT/c)

If Twin2 waited a proper time t after Event 2 before beginning to accelerate, then (proper time between event 1 and 2) is t

If Twin2 accelerated for a proper time BT, then (proper time between Event 3 and 4) is BT.

So, does this fit the logic of why you said the time for Twin2 would have a total time of (c/a)*sinh(a*BT/c) + t + BT ? If so, I don't have a problem with this,

At this point, we are in perfect agreement.

but I think I do have a problem with the calculation for Twin1. The fact that the middle segment's time for Twin1 is supposed to be t/gamma suggests that you are using the launch frame's definition of simultaneity (also Twin2's before accelerating) to calculate the time elapsed on Twin1's clock between the moments of Event 2 and Event 3 on Twin2's worldline. So, let's define the following events on Twin1's worldine:

Event 1a: Twin1 begins to accelerate away from Twin2 at x=0 and t=0 in the launch frame
Event 2a: Twin1 stops accelerating
Event 3a: The event on Twin1's worldline that is simultaneous in the launch frame (not Twin1's own current frame) with the event of Twin2 beginning to accelerate

In this case, (proper time between Event 1a and Event 2a) = BT, and (proper time between Event 2a and Event 3a) = t/gamma. Again, does this fit with what you were thinking?

All agreed at this point

But in this case it's not too clear where the third term of the sum you gave for Twin1's time, (c/a)*sinh(a*BT/c), is supposed to come from. This is obviously supposed to be some sort of time for Twin2's acceleration phase, but it looks like the coordinate time in the launch frame for Twin2's acceleration phase. Suppose we pick the following for the 4th event on Twin1's worldline:

Event 4a: The event on Twin1's worldline that is simultaneous in the launch frame (not Twin1's own current frame) with the event of Twin2 stopping his acceleration

In this case the coordinate time in the launch frame between Event 3a and 4a would be (c/a)*sinh(a*BT/c), but since Twin1's clock is slowed down by a factor of 1/gamma in this frame, the proper time between Event 3a and 4a would instead by [(c/a)*sinh(a*BT/c)]/gamma. Also, in this case Event 4a on Twin1's worldline would be simultaneous with Event 4 on Twin2's worldline in the launch frame, so you wouldn't be comparing their ages according to the definition of simultaneity used in their final rest frame as you were supposed to.

Alternately, it's possible you're imagining a different fourth event on Twin1's worldline like this--

Event 4b: The event on Twin1's worldline that is simultaneous in Twin1's current rest frame with Twin2 finishing his acceleration

But again, the proper time between 3a and 4b would not be (c/a)*sinh(a*BT/c) in this case. First of all, in Twin1's current rest frame, event 3a is not simultaneous with the beginning of Twin2's acceleration, it was specifically defined to be simultaneous with the beginning of Twin2's acceleration in the launch frame. Second of all, even if you did calculate the time elapsed in Twin1's current rest frame between the beginning and end of Twin2's acceleration, it would not be equal to (c/a)*sinh(a*BT/c)--that formula only works if you're using the frame where the accelerating object begins accelerating from a velocity of 0, whereas in Twin1's current rest frame, Twin2 already had some significant nonzero velocity before beginning to accelerate.

So no matter which way I look at it, your calculation for Twin1's proper time doesn't appear to make much sense. You really need to go back and give careful thought to which events are supposed to mark the beginning and end of each segment you want to add, and to make sure that the final events on the final segment of each twin's sum are actually simultaneous in their final rest frame, if that's the frame where you want to compare their final ages.

These are some good questions.
This paper considers a twin that accel/decel and then comes back with accel/decel.

Even though the clock is "slowed", there is no gamma factor because it is already included with the integral.

http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf

Thus, O becomes sort of a negative launch frame from twin2 coming toward it with -v.

Wiki also solves it piecewise this way except all the calcs are done from the stay at home twin. It does not use a gamma factor in addition to the acceleration calculations. Also, please note the -v argument in the article.

http://en.wikipedia.org/wiki/Twin_paradox

Finally, since acceleration is absolute motion, the adjustments are decidable from each frame without adjustments. The only concern for the elapsed proper time will be BT and a and the accelerating clock beating slower and the non-accelerating clock beating faster with all parties agreeing how much for each.

What do you think?

 

[cfrogue]

So no matter which way I look at it, your calculation for Twin1's proper time doesn't appear to make much sense. You really need to go back and give careful thought to which events are supposed to mark the beginning and end of each segment you want to add, and to make sure that the final events on the final segment of each twin's sum are actually simultaneous in their final rest frame, if that's the frame where you want to compare their final ages.

Yea, I think I'll read that acceleration paper some more.

I am not 100% confident on that portion either.

Then, all I can do to support that calculation is point to some article and hope it is correct.

Looking at this,
http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf

Can you tell me where the Integral [e^Integral] comes from in equation 1?

The author just sticks it out there without prior justification.

 

[JesseM]

These are some good questions.
This paper considers a twin that accel/decel and then comes back with accel/decel.

Even though the clock is "slowed", there is no gamma factor because it is already included with the integral.

http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf

Please don't just rely on vague handwavy analogies with other papers, your reasoning needs to be more precise. The reason a gamma factor doesn't appear in this paper's equations, but did in my calculation of the proper time between events 3a and 4a, is because I was calculating the time elapsed on the clock of an observer moving at constant velocity in the launch frame between events on his worldline simultaneous with the beginning and end of another object's velocity in the launch frame. So, naturally, the time elapsed on the constant-velocity observer's clock is going to be 1/gamma times (coordinate time in the launch frame between the beginning and end of the acceleration)--do you disagree? The paper above isn't calculating anything like this, so the fact that 1/gamma doesn't appear in their equations proves nothing about an error in my calculations.

Wiki also solves it piecewise this way except all the calcs are done from the stay at home twin. It does not use a gamma factor in addition to the acceleration calculations.

Again, because they're not calculating the time elapsed on the clock of an observer moving at constant velocity between events on his worldline simultaneous in the launch frame with the beginning and end of another observer's acceleration. It's a pretty basic conclusion in SR that, for arbitrary events 1 and 2 (whether they represent the beginning and end of someone's acceleration or something else entirely), if you pick two events E1 and E2 on the worldline of of a clock C that are simultaneous with 1 and 2 in some inertial frame A, and in frame A clock C is moving at constant speed v between E1 and E2, then if the coordinate time between 1 and 2 in frame A is T, then the proper time on C's worldline between E1 and E2 must be T/gamma. Do you dispute this? If there is any doubt in your mind about this, then you really need to go back to basics in your study of SR, not go looking at papers with complicated derivations of acceleration equations which will only confuse you if you haven't mastered the basics.

Finally, since acceleration is absolute motion, the adjustments are decidable from each frame without adjustments.

Nope, acceleration is "absolute" in one sense alone: that all inertial frames agree on the simple yes/no question of whether a given object is accelerating at some point on its worldline. They certainly don't agree on the exact value of its coordinate acceleration at that point, or how fast its clock is ticking at that point, so you can't just make the handwavy verbal argument that because it's "absolute" in this simple yes/no sense, it's "absolute" in every other sense.

It is true, incidentally, that if one frame calculates the proper time for an object between the beginning and end of that object's acceleration, then all other frames will agree--but this is because the proper time on a worldline between two events on that worldline is always absolute, this would be just as true if you picked two events on an inertial section of an object's worldline and calculated the proper time for the object between those events.

 

[cfrogue]

Please don't just rely on vague handwavy analogies with other papers, your reasoning needs to be more precise. The reason a gamma factor doesn't appear in this paper's equations, but did in my calculation of the proper time between events 3a and 4a, is because I was calculating the time elapsed on the clock of an observer moving at constant velocity in the launch frame between events on his worldline simultaneous with the beginning and end of another object's velocity in the launch frame. So, naturally, the time elapsed on the constant-velocity observer's clock is going to be 1/gamma times (coordinate time in the launch frame between the beginning and end of the acceleration)--do you disagree? The paper above isn't calculating anything like this, so the fact that 1/gamma doesn't appear in their equations proves nothing about an error in my calculations.

Again, because they're not calculating the time elapsed on the clock of an observer moving at constant velocity between events on his worldline simultaneous in the launch frame with the beginning and end of another observer's acceleration. It's a pretty basic conclusion in SR that, for arbitrary events 1 and 2 (whether they represent the beginning and end of someone's acceleration or something else entirely), if you pick two events E1 and E2 on the worldline of of a clock C that are simultaneous with 1 and 2 in some inertial frame A, and in frame A clock C is moving at constant speed v between E1 and E2, then if the coordinate time between 1 and 2 in frame A is T, then the proper time on C's worldline between E1 and E2 must be T/gamma. Do you dispute this? If there is any doubt in your mind about this, then you really need to go back to basics in your study of SR, not go looking at papers with complicated derivations of acceleration equations which will only confuse you if you haven't mastered the basics.

Nope, acceleration is "absolute" in one sense alone: that all inertial frames agree on the simple yes/no question of whether a given object is accelerating at some point on its worldline. They certainly don't agree on the exact value of its coordinate acceleration at that point, or how fast its clock is ticking at that point, so you can't just make the handwavy verbal argument that because it's "absolute" in this simple yes/no sense, it's "absolute" in every other sense.

It is true, incidentally, that if one frame calculates the proper time for an object between the beginning and end of that object's acceleration, then all other frames will agree--but this is because the proper time on a worldline between two events on that worldline is always absolute, this would be just as true if you picked two events on an inertial section of an object's worldline and calculated the proper time for the object between those events.

A clock C is moving at constant speed v between E1 and E2, then if the coordinate time between 1 and 2 in frame A is T, then the proper time on C's worldline between E1 and E2 must be T/gamma. Do you dispute this?

No, but I need to show you the step by step process which I cannot do right now.

Naturally, from the burn of twin1 and the relative motion time dilation for twin1, there is no question for the result at that point for O' calculating twin1.

However, I need to be able to integrate using the proper time of twin2 and correctly calculate the elapsed time of twin1 showing the integral and the reasoning.

Naturally, this implies a start velocity of v and an ending velocity of 0 from the POV of O' and twins2

Also, this implies two times which are Tau1 and Tau1+BT.

Do you see any problem with this method?