Need help understanding the twins

[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?

 

[cfrogue]

error in my calculations.
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.

Yes, agreed, I see your point.

 

[JesseM]

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

OK, I think I understand what you mean--in Twin1's rest frame, Twin2 starts from a velocity of v, then accelerates with constant proper acceleration a for proper time BT and ends up with a velocity of 0 in Twin1's rest frame. So, this is just the mirror image of starting from a velocity of 0 and accelerating to a velocity of v, which suggests we can use the same formula for elapsed time of Twin2's acceleration in Twin1's frame that we used for the elapsed time of Twin1's acceleration in the launch frame. Is that what you're saying?

If so, let's define these two events on twin1's worldline:
Event 4a: The event on Twin1's worldline that is simultaneous in Twin1's current frame with the event of Twin2 beginning to accelerate
Event 5a: The event on Twin1's worldline that is simultaneous in Twin1's current frame with the event of Twin2 ending his acceleration

In this case, I think you make a good argument that the proper time on Twin1's worldline between Event 4a and 5a would be (c/a)*sinh(a*BT/c) (and note that this does not contradict my earlier statement involving a gamma factor, because I was calculating the proper time between a different pair of events on Twin1's worldline that were simultaneous with the beginning and end of Twin2's acceleration in the launch frame). Furthermore, Event 5a is indeed simultaneous in their final rest frame with the event of Twin1 stopping his acceleration, so this is the right endpoint to choose if you want to compare their final ages in their final rest frame. However, there's still a problem with your calculation of the elapsed proper time on Twin1's worldline. Let's summarize all the events used to calculate the proper time on different segments of Twin1's worldline:

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
Event 4a: The event on Twin1's worldline that is simultaneous in Twin1's current frame with the event of Twin2 beginning to accelerate
Event 5a: The event on Twin1's worldline that is simultaneous in Twin1's current frame with the event of Twin2 ending his acceleration

So, I'd agree with the following:

Proper time for Twin1 between Event 1a and Event 2a: BT
Proper time for Twin1 between Event 2a and Event 3a: t/gamma
Proper time for Twin1 between Event 4a and Event 5a: (c/a)*sinh(a*BT/c)

But when summarized like this, it's not hard to spot the problem--you haven't calculated the proper time between Event 3a and Event 4a! Since the launch frame defines simultaneity differently than the way Twin1's current rest frame (which is also both twin's final frame) defines simultaneity, naturally Event 3a (which is simultaneous in the launch frame with Twin2 starting to accelerate) is a different event from Event 4a (which is simultaneous in Twin1's current frame with Twin2 starting to accelerate). If you draw a spacetime diagram with lines of simultaneity from each frame that cross through the event of Twin2 starting to accelerate, you'll see that there is a gap between where the two lines of simultaneity cross Twin1's worldline--I know you don't like drawing spacetime diagrams but this is why they're useful, to get a better intuition of what's going on. Anyway, the point is that a good-size chunk of proper time will elapse for Twin1 between 3a and 4a, and that has to be included in any calculation of his final age in their final rest frame.

 

[cfrogue]

But when summarized like this, it's not hard to spot the problem--you haven't calculated the proper time between Event 3a and Event 4a! Since the launch frame defines simultaneity differently than the way Twin1's current rest frame (which is also both twin's final frame) defines simultaneity, naturally Event 3a (which is simultaneous in the launch frame with Twin2 starting to accelerate) is a different event from Event 4a (which is simultaneous in Twin1's current frame with Twin2 starting to accelerate). If you draw a spacetime diagram with lines of simultaneity from each frame that cross through the event of Twin2 starting to accelerate, you'll see that there is a gap between where the two lines of simultaneity cross Twin1's worldline--I know you don't like drawing spacetime diagrams but this is why they're useful, to get a better intuition of what's going on. Anyway, the point is that a good-size chunk of proper time will elapse for Twin1 between 3a and 4a, and that has to be included in any calculation of his final age in their final rest frame.

Yea, my first attack at the integral shows at least an error of T*atanh(-v).

I am working on a general integral from T0 to T1 using V0 to V1 to see what I get.

Once I do it a few thousand times and get the same answer I'll post it and if you do not mind taking a look at it, I would appreciate it.

Oh, and thanks for your help BTW.


Also, I am sure you noticed a problem with t' in the other frame. I have a way to handle that and calculate it correctly.

Obviously, I cannot just write it down.

 

[JesseM]

Yea, my first attack at the integral shows at least an error of T*atanh(-v).

I am working on a general integral from T0 to T1 using V0 to V1 to see what I get.

Once I do it a few thousand times and get the same answer I'll post it and if you do not mind taking a look at it, I would appreciate it.

Oh, and thanks for your help BTW.

No problem. And instead of calculating the result of an integral in symbolic form which is likely to get very messy, you might consider approaching it as a numerical problem, plugging in specific values for variables like v and BT and a, then using that to get an answer. Either way, the most general approach is to find velocity as a function of time v(t) in some frame, then if you have the times t₀ and t₁ of the two events on the object's worldline you want to calculate the proper time between, you evaluate the integral $$\int^{t_1}_{t_0} \sqrt{1 - v(t)^2/c^2} \, dt$$

Also, I am sure you noticed a problem with t' in the other frame. I have a way to handle that and calculate it correctly.

Obviously, I cannot just write it down.

Why can't you write it down? Can you at least describe the set of events on each worldline that mark the edges of each "piece" in the sum, as I did?

 

[cfrogue]

No problem. And instead of calculating the result of an integral in symbolic form which is likely to get very messy, you might consider approaching it as a numerical problem, plugging in specific values for variables like v and BT and a, then using that to get an answer. Either way, the most general approach is to find velocity as a function of time v(t) in some frame, then if you have the times t₀ and t₁ of the two events on the object's worldline you want to calculate the proper time between, you evaluate the integral $$\int^{t_1}_{t_0} \sqrt{1 - v(t)^2/c^2} \, dt$$

It is not that simple to just evaluate it because it involves v(T) and at relativistic speeds, the SR velocity sum function must be used or the results will be way off.
In other words, when looking at dv/dt, we actually have (v2 - v1) / (t2 - t1) and thus the velocity sum function must be used.


So, that integral is written simply enough, but its actual calculation is far more complicated.

This providers a nice derivation of the intgegral.
http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html

So, I have been through the integral and I am able to set a start time for this operation and call it 0.

It is c/a sinh(BT*a/c) + t.

This is known and decidable in the accelerating twin2 frame. Thus, there is the start point, 0, and the endpoint is BT.

In addition, the start velocity is v and the terminating velocity is 0. So, in effect, we have a -v, but since -sinh(x) = sinh(-x), then the results are the same.

Thus, I am able to conclude, the elapsed proper time for the "at rest" twin1 is c/a sinh(BT*a/c).

Why can't you write it down? Can you at least describe the set of events on each worldline that mark the edges of each "piece" in the sum, as I did?

Oh, I meant, I cannot just simply say it without proof.

I need an effective procedure for deciding t'.

Here it is.

Once O' stops its burn, it immediately sends out a light pulse. O records the time it receives the light pulse. Then, O performs the round trip speed of light calculation to decide the distance between the two ships. Once the distance D between the two ships is known, O subtracts D/c from the time it received the light pulse from O' and then has a correct endpoint to the experiment that matches the endpoint of O'. Thus, O and O' share a common start point and end point to the experiment.

 

[JesseM]

It is not that simple to just evaluate it because it involves v(T) and at relativistic speeds, the SR velocity sum function must be used or the results will be way off.
In other words, when looking at dv/dt, we actually have (v2 - v1) / (t2 - t1) and thus the velocity sum function must be used.

Yes, finding the actual nature of the v(t) function can be difficult if accelerating is involved.

So, that integral is written simply enough, but its actual calculation is far more complicated.

This providers a nice derivation of the intgegral.
http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html

So, I have been through the integral and I am able to set a start time for this operation and call it 0.

It is c/a sinh(BT*a/c) + t.

This is known and decidable in the accelerating twin2 frame. Thus, there is the start point, 0, and the endpoint is BT.

In addition, the start velocity is v and the terminating velocity is 0. So, in effect, we have a -v, but since -sinh(x) = sinh(-x), then the results are the same.

Thus, I am able to conclude, the elapsed proper time for the "at rest" twin1 is c/a sinh(BT*a/c).

So you're just trying to prove the claim that decelerating at a from v to 0 will take the same time at accelerating at a from 0 to v? Deriving this may be an interesting exercise, but I already agreed that this assumption was fine, the problem with your previous calculation lies elsewhere.

I need an effective procedure for deciding t'.

Here it is.

Once O' stops its burn, it immediately sends out a light pulse. O records the time it receives the light pulse. Then, O performs the round trip speed of light calculation to decide the distance between the two ships. Once the distance D between the two ships is known, O subtracts D/c from the time it received the light pulse from O' and then has a correct endpoint to the experiment that matches the endpoint of O'. Thus, O and O' share a common start point and end point to the experiment.

OK, but this just a variant of the standard physical procedure for defining simultaneity in a given frame, if we know the coordinate position of both twins as a function of time in the launch frame, we can just use the Lorentz transformation to figure out which point on the worldline of Twin1/O is simultaneous with the event of Twin2/O' finishing his acceleration, in the frame where both are at rest after ending their acceleration.

 

[cfrogue]

Yes, finding the actual nature of the v(t) function can be difficult if accelerating is involved.

So you're just trying to prove the claim that decelerating at a from v to 0 will take the same time at accelerating at a from 0 to v? Deriving this may be an interesting exercise, but I already agreed that this assumption was fine, the problem with your previous calculation lies elsewhere.

Please explain.
We have a decidable start and thus can call that 0 and a decidable end, BT.

OK, but this just a variant of the standard physical procedure for defining simultaneity in a given frame, if we know the coordinate position of both twins as a function of time in the launch frame, we can just use the Lorentz transformation to figure out which point on the worldline of Twin1/O is simultaneous with the event of Twin2/O' finishing his acceleration, in the frame where both are at rest after ending their acceleration.

OK, so it is not necessary to go through all the mechanics to deciding the endpoint as being simultaneous for both disjoint frames.

I did not know there was a shortcut.

 

[JesseM]

Please explain.
We have a decidable start and thus can call that 0 and a decidable end, BT.

I was just referring to the problem with your earlier calculations that I already pointed out--that you did not account for the proper time along the segment of Twin1's worldline between Event 3a and 4a.

 

[cfrogue]

I was just referring to the problem with your earlier calculations that I already pointed out--that you did not account for the proper time along the segment of Twin1's worldline between Event 3a and 4a.

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

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

Are these what you are talking about 3a and 4a?
To make sure I understand,
3a Twin1's proper time that twin2 launched.
4a Twin1's proper time that twin2 stipped the burn.

 

[JesseM]

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

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

Are these what you are talking about 3a and 4a?
To make sure I understand,
3a Twin1's proper time that twin2 launched.
4a Twin1's proper time that twin2 stipped the burn.

No, I was referring to how I defined them in post 122 (after I figured out that the last term (c/a)*sinh(a*BT/c) in your sum was supposed to be the time in Twin1's frame for Twin2 to accelerate from v to 0) where I wrote:

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
Event 4a: The event on Twin1's worldline that is simultaneous in Twin1's current frame with the event of Twin2 beginning to accelerate

So, both are events on Twin1's worldline that are simultaneous with Twin2 launching, but simultaneous in two different frames.

 

[cfrogue]

No, I was referring to how I defined them in post 122 (after I figured out that the last term (c/a)*sinh(a*BT/c) in your sum was supposed to be the time in Twin1's frame for Twin2 to accelerate from v to 0) where I wrote:

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
Event 4a: The event on Twin1's worldline that is simultaneous in Twin1's current frame with the event of Twin2 beginning to accelerate

So, both are events on Twin1's worldline that are simultaneous with Twin2 launching, but simultaneous in two different frames.

Yes, but (c/a)*sinh(a*BT/c) should be the correct elapsed time for twin1 as calculated by the acceleration equations.


The actual worldline events cannot be decided until they meet. That is what I meant by twin2 sending out a light pulse to twin1. This establishes the correct endpoint for twin2 when twin1 stopped the burn.

Then, it must be the case that BT is the elapsed time of twin1's burn. Then (c/a)*sinh(a*BT/c) is twin2's burn and finally what is left over is t'.

Then, one can apply the worldlines.

This is similar to Einstein's derivation of LT.

He did a translation of the simultaneity convention in the moving frame with a yet to be defined function Tau.

From the origin of system k let a ray be emitted at the time along the X-axis to x', and at the time be reflected thence to the origin of the co-ordinates, arriving there at the time ; we then must have , or, by inserting the arguments of the function Tau and applying the principle of the constancy of the velocity of light in the stationary system

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

 

[JesseM]

Yes, but (c/a)*sinh(a*BT/c) should be the correct elapsed time for twin1 as calculated by the acceleration equations.

That's the elapsed time for twin1 between event 4a and 5a, as I defined them in post 122. I was talking about the total elapsed time from event 1a to event 5a, which you had as BT + t/gamma + (c/a)*sinh(a*BT/c)--my point in post 122 is that BT was the time from 1a to 2a, t'/gamma was the time from 2a to 3a, and (c/a)*sinh(a*BT/c) was the time from 4a to 5a...but nowhere in the sum did you include the time from 3a to 4a, so this is not the correct value for the elapsed time that twin1 ages from 1a to 5a.

The actual worldline events cannot be decided until they meet. That is what I meant by twin2 sending out a light pulse to twin1. This establishes the correct endpoint for twin2 when twin1 stopped the burn.

Do you mean the correct endpoint for twin1 when twin2 stopped the burn? Twin2 is the one who accelerates second, and thus that moment in their final rest frame should be when they compare ages. Assuming this is what you meant, I already defined 5a as the event on twin1's worldline that is simultaneous in their final rest frame with the event of twin2 stopping the burn, so as I pointed out before, the definition of simultaneity in relativity ensures that if they use your light pulse method they'll identify 5a as the endpoint of twin1's worldline where they want to compare his age to that of twin2 as he stops the burn.

Then, it must be the case that BT is the elapsed time of twin1's burn. Then (c/a)*sinh(a*BT/c) is twin2's burn and finally what is left over is t'.

Sure, I agree with this calculation for twin2's elapsed time, assuming we define t' as the time in the launch frame between twin1 ending his acceleration and twin2 starting his own. But in post 122 I was talking about the calculation for twin1, not twin2.

Then, one can apply the worldlines.

I don't know what you mean by "apply the worldlines". Anyway, please look over the way I have defined events 1a through 5a in post 122, and tell me whether you agree or disagree that BT is the proper time for twin1 between 1a and 2a, that t'/gamma is the proper time for twin1 between 2a and 3a, and that (c/a)*sinh(a*BT/c) is the proper time fro twin1 between 4a and 5a, given the way I defined all these events.

This is similar to Einstein's derivation of LT.

He did a translation of the simultaneity convention in the moving frame with a yet to be defined function Tau.

From the origin of system k let a ray be emitted at the time along the X-axis to x', and at the time be reflected thence to the origin of the co-ordinates, arriving there at the time ; we then must have , or, by inserting the arguments of the function Tau and applying the principle of the constancy of the velocity of light in the stationary system

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

Sure, but this was a derivation of the Lorentz transformation itself, once you already have them in hand you can trust that any events with the same t-coordinate in a given frame will also be defined as simultaneous if observers at rest in that frame use light signals to check which events were simultaneous.

 

[cfrogue]

That's the elapsed time for twin1 between event 4a and 5a, as I defined them in post 122. I was talking about the total elapsed time from event 1a to event 5a, which you had as BT + t/gamma + (c/a)*sinh(a*BT/c)--my point in post 122 is that BT was the time from 1a to 2a, t'/gamma was the time from 2a to 3a, and (c/a)*sinh(a*BT/c) was the time from 4a to 5a...but nowhere in the sum did you include the time from 3a to 4a, so this is not the correct value for the elapsed time that twin1 ages from 1a to 5a.

OK, I think this is really the issue.

In my mind, I take this to be not logically decidable using this path you give. Yet, the problem must have a solution.

I do not know of any way to decide 3a to 4a from your methods without running into Dingle's problems.

So, I left this as a open question in the frame of twin1 and moved forward.

To decide the entire sequence, it cannot be denied that all parties agreed twin1 elapsed BT for its burn.

Also, while twin2 burns, it is clear, twins will elapse (c/a)*sinh(a*BT/c).

Still, at this point the problem is not logically decidable.

Thus, upon entering the frame, twin2 pulses light to twin 1.

Twin1 makes an artificial end to the experiment T. Why is it not really the end? It takes time for light to travel from twin2 to twin1.

Thus, twin1 needs to figure this out.

So, the round trip speed of light calculation is used by twin1 to determine the distance to twin2, call it D.

Now, twin1 can actually decide its time that twin2 entered the frame by subtracting D/c from its artificial end of the experiment time T.

I agree I am circumventing the normal methods of SR, but I have not violated any rules.

It is at the point that I can construct world lines for twin1 from twin2 in this somewhat contrived way.

Thus, 3a-4a goes away using this method.

 

[JesseM]

OK, I think this is really the issue.

In my mind, I take this to be not logically decidable using this path you give. Yet, the problem must have a solution.

Well, it is decidable, if you make use of the way different frames define simultaneity according to the Lorentz transformation. To state the rule in a very general way, suppose we have some event E1 which occurs at x1,t1 in frame A, and some inertial observer O who is moving at speed v in this frame, and who at time t1 is at position x1+d (so if we define event E2 to have coordinates x1+d,t1, then this is an event on O's wordline which is simultaneous in frame A with E1), so the function for his worldline in frame A must be x(t) = v*t - v*t1 + x1 + d. If an observer O has velocity v in frame A, then a trick to know is that according to the Lorentz transformation a line of simultaneity in O's frame will look like an FTL worldline with velocity (c^2/v) when represented in A's frame, so a line of simultaneity in O's frame that crosses through event E1 would be given in A's frame by x(t) = (c^2/v)*t - (c^2/v)*t1 + x1. That means that O's line of simultaneity crossing through E1 would cross O's own worldline when v*t - v*t1 + x1 + d = (c^2/v)*t - (c^2/v)*t1 + x1, so solving this for t gives t = [-(c^2/v)*t1 + x1 + v*t1 - x1 - d]/[v - (c^2/v)] = t1 - d/[v - (c^2/v)] = t1 + d*v/(c^2 - v^2). Then you can plug that value for time into the original equation for O's worldline to get x = v*t1 + d*v^2/(c^2 - v^2) - v*t1 + x1 + d = x1 + d*(1 + [v^2/(c^2 - v^2)]) = x1 + d*c^2/(c^2 - v^2). That means if you define a new event E3 to have position coordinate x = x1 + d*c^2/(c^2 - v^2) and time coordinate t = t1 + d*v/(c^2 - v^2), then E3 is simultaneous with E1 in O's rest frame, and lies along O's worldline. You can verify this using the Lorentz transformation, which shows that in O's rest frame, the original event E1 has the time coordinate:

t' = gamma*(t1 - v*x1/c^2)

And E3 has the time coordinate:

t' = gamma*(t1 + d*v/(c^2 - v^2) - v/c^2*[x1 + d*c^2/(c^2 - v^2)]) =
gamma*(t1 + d*v/(c^2 - v^2) - v*x1/c^2 - d*v/(c^2 - v^2)) =
gamma*(t1 - v*x1/c^2)

So E1 and E3 have the same time coordinate in O's rest frame, and we know E3 lies along O's worldline, so E3 must be the event on O's worldline that is simultaneous with E1 in O's rest frame. And E2 was the event on O's worldline that was simultaneous with E1 in frame A. In A's frame the difference in time coordinates between E3 and E2 is:

t1 + d*v/(c^2 - v^2) - t1 = d*v/(c^2 - v^2) = (d*v/c^2)*[1/(1 - v^2/c^2)] = (d*v/c^2)*gamma^2

And we know that the proper time elapsed on O's worldline between E2 and E3 will just be 1/gamma * the coordinate time in frame A between E2 and E3, so this shows the proper time for O between these events is gamma*d*v/c^2.

So, now we can just treat twin1 as the observer O in this proof, treat the launch frame as frame A in this proof, treat the event of twin2 beginning to accelerate as event E1 in this proof, treat 3a as event E2 in this proof, and treat 4a as event E3 in this proof. Then the proof tells us that if twin1 has velocity v in frame A, and twin1 has a distance d from twin2 when twin2 begins to accelerate in this frame, then the proper time for twin1 between 3a and 4a will be gamma*d*v/c^2.

I do not know of any way to decide 3a to 4a from your methods without running into Dingle's problems.

Who's Dingle?

So, I left this as a open question in the frame of twin1 and moved forward.

To decide the entire sequence, it cannot be denied that all parties agreed twin1 elapsed BT for its burn.

Also, while twin2 burns, it is clear, twins will elapse (c/a)*sinh(a*BT/c).

Still, at this point the problem is not logically decidable.

Thus, upon entering the frame, twin2 pulses light to twin 1.

Twin1 makes an artificial end to the experiment T. Why is it not really the end? It takes time for light to travel from twin2 to twin1.

Thus, twin1 needs to figure this out.

So, the round trip speed of light calculation is used by twin1 to determine the distance to twin2, call it D.

Now, twin1 can actually decide its time that twin2 entered the frame by subtracting D/c from its artificial end of the experiment time T.

I agree I am circumventing the normal methods of SR, but I have not violated any rules.

It is at the point that I can construct world lines for twin1 from twin2 in this somewhat contrived way.

Thus, 3a-4a goes away using this method.

You mean, figure out the time coordinate T in the final rest frame that twin1 will receive the pulse, then subtract D/c from that to get the time coordinate T - D/c in the final rest frame that twin2 stopped accelerating? Of course the time coordinate that twin2 stopped accelerating in this frame is not the same as twin1's age at the moment twin2 stopped accelerating in this frame, because twin1 was not at rest in this frame since t=0 in this frame. But I suppose you can say that in this frame, twin1 was initially accelerating for (c/a)*sinh(a*BT/c) of coordinate time, so that twin1 has been at rest in this frame for (T - D/c) - (c/a)*sinh(a*BT/c). And during the initial period twin1 was accelerating, twin1 aged BT, so twin1's total elapsed time at (T - D/c) in the final frame should be BT + (T - D/c) - (c/a)*sinh(a*BT/c).

This method should work, although if you know the coordinates in the launch frame that twin2 stopped accelerating (which I tried derive in post 109, not sure if I did all the algebra right), it would be a lot simpler to just find the time T' that twin2 stops accelerating in the final rest frame by applying the Lorentz transformation to the coordinates in the launch frame, and since relativistic simultaneity is already based on light pulses, you are guaranteed that T' will be equal to your (T - D/c). Then using the same logic as above, twin1's total elapsed time at T' in the final frame should be BT + T' - (c/a)*sinh(a*BT/c).

 

[Al68]

Who's Dingle?

Just in case you're serious, from Wiki (http://en.wikipedia.org/wiki/Herbert_Dingle" ):

Herbert Dingle (2 August 1890–4 September 1978), an English physicist and natural philosopher, who served as president of the Royal Astronomical Society from 1951 to 1953, is best known for his opposition to Albert Einstein's special theory of relativity and the protracted controversy that this provoked.

 

[cfrogue]

Well, it is decidable, if you make use of the way different frames define simultaneity according to the Lorentz transformation. To state the rule in a very general way, suppose we have some event E1 which occurs at x1,t1 in frame A, and some inertial observer O who is moving at speed v in this frame, and who at time t1 is at position x1+d (so if we define event E2 to have coordinates x1+d,t1, then this is an event on O's wordline which is simultaneous in frame A with E1), so the function for his worldline in frame A must be x(t) = v*t - v*t1 + x1 + d. If an observer O has velocity v in frame A, then a trick to know is that according to the Lorentz transformation a line of simultaneity in O's frame will look like an FTL worldline with velocity (c^2/v) when represented in A's frame, so a line of simultaneity in O's frame that crosses through event E1 would be given in A's frame by x(t) = (c^2/v)*t - (c^2/v)*t1 + x1. That means that O's line of simultaneity crossing through E1 would cross O's own worldline when v*t - v*t1 + x1 + d = (c^2/v)*t - (c^2/v)*t1 + x1, so solving this for t gives t = [-(c^2/v)*t1 + x1 + v*t1 - x1 - d]/[v - (c^2/v)] = t1 - d/[v - (c^2/v)] = t1 + d*v/(c^2 - v^2). Then you can plug that value for time into the original equation for O's worldline to get x = v*t1 + d*v^2/(c^2 - v^2) - v*t1 + x1 + d = x1 + d*(1 + [v^2/(c^2 - v^2)]) = x1 + d*c^2/(c^2 - v^2). That means if you define a new event E3 to have position coordinate x = x1 + d*c^2/(c^2 - v^2) and time coordinate t = t1 + d*v/(c^2 - v^2), then E3 is simultaneous with E1 in O's rest frame, and lies along O's worldline. You can verify this using the Lorentz transformation, which shows that in O's rest frame, the original event E1 has the time coordinate:

t' = gamma*(t1 - v*x1/c^2)

And E3 has the time coordinate:

t' = gamma*(t1 + d*v/(c^2 - v^2) - v/c^2*[x1 + d*c^2/(c^2 - v^2)]) =
gamma*(t1 + d*v/(c^2 - v^2) - v*x1/c^2 - d*v/(c^2 - v^2)) =
gamma*(t1 - v*x1/c^2)

So E1 and E3 have the same time coordinate in O's rest frame, and we know E3 lies along O's worldline, so E3 must be the event on O's worldline that is simultaneous with E1 in O's rest frame. And E2 was the event on O's worldline that was simultaneous with E1 in frame A. In A's frame the difference in time coordinates between E3 and E2 is:

t1 + d*v/(c^2 - v^2) - t1 = d*v/(c^2 - v^2) = (d*v/c^2)*[1/(1 - v^2/c^2)] = (d*v/c^2)*gamma^2

And we know that the proper time elapsed on O's worldline between E2 and E3 will just be 1/gamma * the coordinate time in frame A between E2 and E3, so this shows the proper time for O between these events is gamma*d*v/c^2.

So, now we can just treat twin1 as the observer O in this proof, treat the launch frame as frame A in this proof, treat the event of twin2 beginning to accelerate as event E1 in this proof, treat 3a as event E2 in this proof, and treat 4a as event E3 in this proof. Then the proof tells us that if twin1 has velocity v in frame A, and twin1 has a distance d from twin2 when twin2 begins to accelerate in this frame, then the proper time for twin1 between 3a and 4a will be gamma*d*v/c^2.

So why does your method diverge from mine.

Yours seems to involve the exchange of light signals ie events, which is not a part of this problem.

 

[cfrogue]

You mean, figure out the time coordinate T in the final rest frame that twin1 will receive the pulse, then subtract D/c from that to get the time coordinate T - D/c in the final rest frame that twin2 stopped accelerating? Of course the time coordinate that twin2 stopped accelerating in this frame is not the same as twin1's age at the moment twin2 stopped accelerating in this frame, because twin1 was not at rest in this frame since t=0 in this frame. But I suppose you can say that in this frame, twin1 was initially accelerating for (c/a)*sinh(a*BT/c) of coordinate time, so that twin1 has been at rest in this frame for (T - D/c) - (c/a)*sinh(a*BT/c). And during the initial period twin1 was accelerating, twin1 aged BT, so twin1's total elapsed time at (T - D/c) in the final frame should be BT + (T - D/c) - (c/a)*sinh(a*BT/c).

This method should work, although if you know the coordinates in the launch frame that twin2 stopped accelerating (which I tried derive in post 109, not sure if I did all the algebra right), it would be a lot simpler to just find the time T' that twin2 stops accelerating in the final rest frame by applying the Lorentz transformation to the coordinates in the launch frame, and since relativistic simultaneity is already based on light pulses, you are guaranteed that T' will be equal to your (T - D/c). Then using the same logic as above, twin1's total elapsed time at T' in the final frame should be BT + T' - (c/a)*sinh(a*BT/c).

This method runs into Dingle's false paradox and does not logically isolate the issue.

http://sheol.org/throopw/dingle-paradox01.html

 

[JesseM]

So why does your method diverge from mine.

Which method of yours? Are you talking about the old method of yours which calculated an elapsed time of BT + t'/gamma + (c/a)*sinh(a*BT/c) for twin1, or the new method I thought you might be talking about which I described in the second-to-last paragraph of my previous post, or some other method? I am still not clear on how your "light pulse" approach is supposed to work, so it would help if you'd tell me if I got it right in that second-to-last paragraph or if I misunderstood.

Yours seems to involve the exchange of light signals ie events, which is not a part of this problem.

Where do you get the idea that my method involves the exchange of light signals? I'm just picking five events on twin1's worldline--1a, 2a, 3a, 4a, and 5a--and calculating the proper time between each pair, using standard SR methods.

This method runs into Dingle's false paradox and does not logically isolate the issue.

http://sheol.org/throopw/dingle-paradox01.html

How does it "run into" a paradox which isn't real? As you say, Dingle's "paradox" was false, there is no genuine paradox there. And what issue do you think needs to be logically isolated?

According to SR, it is certainly true that if you have two observers at rest in a given frame at different positions, then if observer #1 sends a light signal to observer #2 at the moment some event E occurs on the worldline of observer #1, and observer #2 receives the signal at time coordinate T in this frame, then if he calculated T - D/c (where D is the distance between the two observers in this frame), it will be equal to the time coordinate T' of the original event E in this frame, and also the time coordinate of the event on observer #2's worldline which is simultaneous with E in this frame. Do you disagree with that?

 

[cfrogue]

Which method of yours? Are you talking about the old method of yours which calculated an elapsed time of BT + t'/gamma + (c/a)*sinh(a*BT/c) for twin1, or the new method I thought you might be talking about which I described in the second-to-last paragraph of my previous post, or some other method? I am still not clear on how your "light pulse" approach is supposed to work, so it would help if you'd tell me if I got it right in that second-to-last paragraph or if I misunderstood.

Sorry JesseM, which post #?

 

[JesseM]

I specified which post in that same paragraph:

Which method of yours? Are you talking about the old method of yours which calculated an elapsed time of BT + t'/gamma + (c/a)*sinh(a*BT/c) for twin1, or the new method I thought you might be talking about which I described in the second-to-last paragraph of my previous post, or some other method? I am still not clear on how your "light pulse" approach is supposed to work, so it would help if you'd tell me if I got it right in that second-to-last paragraph or if I misunderstood.

It was the previous post before the one where I wrote that, i.e. post #134.