Need help understanding the twins

[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.

 

[cfrogue]

I specified which post in that same paragraph:

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

OK, thanks, please give me some time.

You are very thorough.

 

[cfrogue]

I specified which post in that same paragraph:

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

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 does not make sense to me.

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

I do not care of they register the same proper time. I only care that they establish a common stop point in their own proper times.

I am not sure how you could refute this.

 

[JesseM]

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 does not make sense to me.

Do you disagree, or you don't understand?

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

I do not care of they register the same proper time. I only care that they establish a common stop point in their own proper times.

I didn't say anything about registering the same proper time, I was talking about coordinate times. If x,t are the coordinates in the launch frame of twin1 receiving the light signal sent by twin2 at the moment he stopped accelerating, then T = gamma*(t - vx/c^2) is the time coordinate in the final rest frame of twin1 receiving this signal. So, if the final distance between twin1 and twin2 in the final rest frame is D, then that shows that whatever age twin1 was at (T - D/c), this must be the point on his worldline that was simultaneous with twin2 stopping his acceleration, according to the final rest frame's definition of simultaneity. You do want to use the final rest frame's definition of simultaneity to compare their ages, do you not?

If you don't want to use the method above where we figure out the x,t coordinates of twin1 receiving the signal in the launch frame and use it to get the time coordinate T of this event in the final rest frame, what do you want to do? Is T supposed to represent twin1's own proper time when he gets the signal, rather than the time coordinate of his getting the signal in the final rest frame? If so, how do you propose to actually find the value of T if not by dividing his worldline into pieces and calculating the proper time on each one? (which requires that you include the 3a-4a piece)

 

[cfrogue]

Originally Posted by 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 does not make sense to me.

Do you disagree, or you don't understand?

Well, I guess I do not understand.

Perhaps, you could explain it differently.

I didn't say anything about registering the same proper time, I was talking about coordinate times. If x,t are the coordinates in the launch frame of twin1 receiving the light signal sent by twin2 at the moment he stopped accelerating, then T = gamma*(t - vx/c^2) is the time coordinate in the final rest frame of twin1 receiving this signal. So, if the final distance between twin1 and twin2 in the final rest frame is D, then that shows that whatever age twin1 was at (T - D/c), this must be the point on his worldline that was simultaneous with twin2 stopping his acceleration, according to the final rest frame's definition of simultaneity. You do want to use the final rest frame's definition of simultaneity to compare their ages, do you not?

If you don't want to use the method above where we figure out the x,t coordinates of twin1 receiving the signal in the launch frame and use it to get the time coordinate T of this event in the final rest frame, what do you want to do? Is T supposed to represent twin1's own proper time when he gets the signal, rather than the time coordinate of his getting the signal in the final rest frame? If so, how do you propose to actually find the value of T if not by dividing his worldline into pieces and calculating the proper time on each one? (which requires that you include the 3a-4a piece)

You like worldlines too much.

They cannot be used to solve unknowns.

This is why I do not like them. I feel caged.

Try to ignore your world line thing for the moment and think about the recursive decision process I described.
We have some unknowns are are trying to decide them.

The process I came up with decides them.

Do you disagree?

 

[JesseM]

You like worldlines too much.

Worldlines are simply functions defining the coordinates an object passes through in a given frame. Are you saying you think the problem can be solved even if we don't know what coordinates the twins pass through?

They cannot be used to solve unknowns.

Sure they can. For example, if you know coordinate position as a function of coordinate time, you can use to solve for the proper time.

Try to ignore your world line thing for the moment and think about the recursive decision process I described.
We have some unknowns are are trying to decide them.

The process I came up with decides them.

Do you disagree?

I have no idea what the process is because you don't answer my questions about it. Again, is T supposed to be the coordinate time in the final rest frame that twin1 receives the signal from twin2, or is it supposed to be twin1's proper time at the moment he receives the signal, or something else? If it's the proper time, then how do you propose to actually decide the value of this variable T?

 

[cfrogue]

Worldlines are simply functions defining the coordinates an object passes through in a given frame. Are you saying you think the problem can be solved even if we don't know what coordinates the twins pass through?

Sure they can. For example, if you know coordinate position as a function of coordinate time, you can use to solve for the proper time.

I have no idea what the process is because you don't answer my questions about it. Again, is T supposed to be the coordinate time in the final rest frame that twin1 receives the signal from twin2, or is it supposed to be twin1's proper time at the moment he receives the signal, or something else? If it's the proper time, then how do you propose to actually decide the value of this variable T?

The process I came up with decides them.

Do you disagree?

O' stops accelerating and immediately sends a light signal to O.

O receives the signal.

O calls this a pseudo end of the experiment.

The problem is that it took time for the light to travel to O from O'.
Thus, O must decide the distance light traveled.
O does the round trip speed of light calculation and is able to decide the distance to O'.
Thus, O then knows the in its own proper time when O' stopped accelerating.
Now, O has BT and t', the relative motion phase.
Since acceleration is absolute motion, O knows (c/a)*sinh(a*BT/c) transpired for the burn of twin2.
Thus, t' is logically decidable from algebra and logic but not from the restrictive logic of world lines.

 

[JesseM]

The process I came up with decides them.

Do you disagree?

I can't agree or disagree if you won't answer my questions about what your "process" actually is. Again, is T the coordinate time when twin1/0 receives the signal, or the proper time of twin1/O, or something else?

 

[cfrogue]

I can't agree or disagree if you won't answer my questions about what your "process" actually is. Again, is T the coordinate time when twin1/0 receives the signal, or the proper time of twin1/O, or something else?

Twin1 receives it at the proper time of twin1.

You know what, I see there is a flaw.

Twin1 needs to then communicate this time logic of its proper time to twin2 for its correct calculation for validation.

No matter, the result is the same.

But, I want to think about this more.

 

[JesseM]

Twin1 receives it at the proper time of twin1.

Duh, I know that. But that still doesn't tell me whether the variable T that you wrote before is supposed to refer to twin1's proper time, or to the coordinate time in their final rest frame.

In any case, my more basic question is this: how do you propose to actually solve for twin1's proper time at the moment he receives the signal? If you don't know the actual value of his proper time when he receives the signal, then this is no use in determining if he is younger or older than twin2 was at the moment twin2 stopped accelerating.

 

[cfrogue]

Duh, I know that. But that still doesn't tell me whether the variable T that you wrote before is supposed to refer to twin1's proper time, or to the coordinate time in their final rest frame.

In any case, my more basic question is this: how do you propose to actually solve for twin1's proper time at the moment he receives the signal? If you don't know the actual value of his proper time when he receives the signal, then this is no use in determining if he is younger or older than twin2 was at the moment twin2 stopped accelerating.

OK you are funny.