Need help understanding the twins

[cfrogue]

That last equation doesn't make any sense, you said in the original problem that twin2 accelerates in an identical manner to twin1, so if twin1 accelerates for time BT then so should twin2. Also, are these times stated in the frame where the twins were originally at rest at the start? If so, in the rest of your calculations you don't even consider the new definition of simultaneity in the rest frame of O and O' after the acceleration, showing that you've missed the point entirely.

I said they accelerate the same way.

Where do you bring in R of S?

 

[matheinste]

the math is above.

let me know

How would you like to talk us through the mathematics, step by step, explaining each step and then there can be no misunderstandings.

Matheinste.

 

[cfrogue]

How would you like to talk us through the mathematics, step by step, explaining each step and then there can be no misunderstandings.

Matheinste.

Sure, let me deal with Jesse's R of S first.

 

[JesseM]

I said they accelerate the same way.

So doesn't that mean they accelerate for the same amount of time?

Where do you bring in R of S?

We want to know who will be older according to the definition of simultaneity in the final rest frame of O and O', do we not?

 

[Jorrie]

I have a simpler twins experiment

O, O', twin1 and twin2 are in the same frame and sync their clocks.

O and twin1 accelerate for some agreed upon time BT at a.

Then for a long period of time there exists relative velocity.

Finally, O' and twin2 perform the same acceleration for BT as twin1 and in the same exact direction.

I'm contemplating converting your scenario to a non-accelerated, three inertial frame situation for discussion, but first I must understand your scenario.

Your definition of frames O and O' is a bit confusing. At the beginning and at the end, O and O' are the same frames, whereas for the 'middle part', before Twin2 has completed acceleration, they are not the same.

I think you must define O' as the original inertial frame, with both twins at rest and O as the final inertial frame, where both twins are at rest again. You can obviously have a Twin1 frame and a Twin2 frame as well, but they are not inertial and hence not the same as O and O'. I suspect that the math that you posted later may be influenced by such frame definitions.

 

[Fredrik]

I can only interpret #49 as saying that the two twins start out at rest at the same location in some inertial frame. Then A accelerates with constant proper acceleration a for a proper time t, and then he stops. Some time later B does the exact same thing.

If that's what cfrogue meant, what happens is that when B has stopped, he's at the exact same location as A and has aged exactly the same. This follows immediately from the definition of proper time and the fact that you can obtain B's world line by adding the same constant vector to every point on A's world line.

 

[Jorrie]

If that's what cfrogue meant, what happens is that when B has stopped, he's at the exact same location as A and has aged exactly the same. This follows immediately from the definition of proper time and the fact that you can obtain B's world line by adding the same constant vector to every point on A's world line.

No, I think he meant that A stopped accelerating, not stopped in motion relative to B. A and B will eventually be a long distance away from each other, but in the same (new) inertial frame again.

 

[Fredrik]

OK, in that case, at any time in the original rest frame, the one who started accelerating first is younger. They'll be a distance v/t apart, where t is the time between the moments when they started accelerating, and v is their final velocities (which are both the same). The age difference will be t-t/\gamma.

 

[JesseM]

OK, in that case, at any time in the original rest frame, the one who started accelerating first is younger.

I didn't think he wanted the answer in the original rest frame, but rather in the new rest frame of O and O' who have traveled along with the twins. In this case the twin who accelerated second will be younger.

 

[ExecNight]

O and twin1 accelerate for some agreed upon time BT at a.

Then for a long period of time there exists relative velocity.

Finally, O' and twin2 perform the same acceleration for BT as twin1 and in the same exact direction.

The accelerations are symmetric and so they end up in the same frame again but at a distance.

I don't know how i can be more clear when saying that, they CAN NOT make a symmetric acceleration starting from different t coordinates in a circular motion based, three dimensional universe.

So even if they sign every treaty to do the same things. They will end up in different frames.

If i accept the assumption that they accelerate symetrically and reach the same frame, still which twin will be younger can not be calculated because the energy used in the acceleration process is an unknown for the other twin. The reason being explained in the first paragraph and the twin that used more energy in the process ages less.

;)

 

[Jorrie]

If i accept the assumption that they accelerate symetrically and reach the same frame, still which twin will be younger can not be calculated because the energy used in the acceleration process is an unknown for the other twin. The reason being explained in the first paragraph and the twin that used more energy in the process ages less.

;)

Cfrogue's specifications are not very clear, but surely, if they had identical coordinate acceleration profiles in the original rest frame (or identical proper-acceleration profiles in their own frames respectively), then they will end up in the same inertial frame after the accelerations stopped. This is despite the fact that the accelerations happen at different coordinate times and they will be spatially separated. Then all inertial frames will agree on who is younger... [Edit: Oops, this last statement is wrong. The relative aging will be frame dependent, because the twins are not colocated.]

 

[Jorrie]

Total elapsed proper time calculation of O for twin2 c/a sinh(aBT/c) + t'/λ + BT
Conclusion of O, twin1 is older.
...
Total elapsed proper time calculation of O' for twin2 c/a sinh(aBT/c) + t + BT
Conclusion of O', twin1 is younger.
...
However, both O' and O cannot be correct with their SR conclusions when presented with the factual data of the clock sync.*

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

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

Interestingly, if both twins are accelerated in such a way that they are brought together (colocated) in the original frame (O') again, they would be the same age again. If only one is accelerated again to bring them together (colocated) in the new frame (O), then the one being accelerated again would be younger, in an absolute sense (a-la the classical twins).

You agree with all this?

 

[ExecNight]

Interestingly, if both twins are accelerated in such a way that they are brought together (colocated) in the original frame (O) again, they would be the same age again.

Wrong! Circular motion. Accelerating at different t coordinate. Repeat with me please.
Circular motion. Different end velocities with same amount of energy cosumption.

If only one is accelerated again to bring them together (colocated) in the new frame (O'), then the one being accelerated again would be younger, in an absolute sense (a-la the classical twins).

Not necessarily right. The one accelerated might very well relatively slowed down so that the other twin can catch up ;) Well that makes things complicated. When you try to use acceleration or velocity as the factor but not the energy consumption..

Listen to the Jester for he might have something of value to share. But don't take him seriously as in the end he is just the jester. Think and make up your mind :)

 

[Jorrie]

Wrong! Circular motion. Accelerating at different t coordinate. Repeat with me please.
Circular motion. Different end velocities with same amount of energy cosumption.

I have no clue what you are talking about! There is no circular motion here (using the common meaning of the words).

 

[ExecNight]

Hmm take a 50 cm long stick, tie it up with another 30 cm long stick perpendicularly. And attach a weight to the short stick using a 20 cm long string.

Now circular motion time.. Spin the stick on its empty edge.

So tell me, what will happen if i push the weight towards a certain point at different times using same energy.


:)

 

[Jorrie]

So tell me, what will happen if i push the weight towards a certain point at different times using same energy.
:)

No, sorry, you tell me what the heck has this got to do with the twins scenario that we are discussing...

 

[ExecNight]

If you take circular motion reality out of the scenario, then you have an invalid experiment at hand. Which will give you unquestionably invalid results. :)

Well at least they are as valid as Netwon's equations in quantum mechanics. Short range experiments will give you good results. On the other hand if you are talking about c speeds and traveling the universe, you are in trouble.

 

[cfrogue]

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

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

Interestingly, if both twins are accelerated in such a way that they are brought together (colocated) in the original frame (O') again, they would be the same age again. If only one is accelerated again to bring them together (colocated) in the new frame (O), then the one being accelerated again would be younger, in an absolute sense (a-la the classical twins).

You agree with all this?

I have other models that can bring them together.

I am still trying to decide if the equations are accurate though.

The reason for this is basically the Bell's paradox. These two start at the same frame when O accelerates, but later O' accelerates and they are not in the same frame. I am still not perfectly sure they end up in the same frame when O' is finished accelerating though I cannot see a reason why they would not be yet.

However, the outcome if they are in the same frame is interesting.

Anyway, they could be brought together by a common accel/decel phase for both but because of SR's simultaneity convention, I do not think this is necessary.

What do you think?

 

[cfrogue]

I'm contemplating converting your scenario to a non-accelerated, three inertial frame situation for discussion, but first I must understand your scenario.

Your definition of frames O and O' is a bit confusing. At the beginning and at the end, O and O' are the same frames, whereas for the 'middle part', before Twin2 has completed acceleration, they are not the same.

I think you must define O' as the original inertial frame, with both twins at rest and O as the final inertial frame, where both twins are at rest again. You can obviously have a Twin1 frame and a Twin2 frame as well, but they are not inertial and hence not the same as O and O'. I suspect that the math that you posted later may be influenced by such frame definitions.

The reason for the symmetric acceleration phases is to isolate the relative motion phase.

I do not think a third frame is necessary since we have independent observers O and O'.

Also, the math I posted is based on SR's acceleration equations.

The only problem is that the acceleration, of value a, must be in the context of the launch frame or negative launch frame.

That is kind of artificial.

 

[cfrogue]

That last equation doesn't make any sense, you said in the original problem that twin2 accelerates in an identical manner to twin1, so if twin1 accelerates for time BT then so should twin2. Also, are these times stated in the frame where the twins were originally at rest at the start? If so, in the rest of your calculations you don't even consider the new definition of simultaneity in the rest frame of O and O' after the acceleration, showing that you've missed the point entirely.

Anyway, the problem is easiest to consider if they both accelerate instantaneously. Suppose they are initially at rest in frame #1, both at position x=0 in this frame. Then at t=0, when both twins are aged 20, twin1 instantaneously accelerates to 0.6c, and continues at that velocity. After 10 years, twin2 accelerates to 0.6c in an identical manner.

So, twin1 accelerated at x=0,t=0 (age 20), and twin2 accelerated at x=0,t=10 (age 30). So, let's apply the Lorentz transformation to find when these acceleration events occurred in frame #2 which is moving at 0.6c relative to frame #1, where both twins come to rest after acceleration. The Lorentz transformation is:

x' = gamma*(x - vt)
t' = gamma*(t - vx/c^2) (and we can ignore the c if we use units where c=1, like light years and years).
gamma = 1/(1 - v^2/c^2)

So with v=0.6c, we have gamma=1.25. If we plug in x=0 and t=0 into the above transformation, we get back x'=0 and t'=0. If we plug in x=0 and t=10 into the above, we get:

x' = 1.25*(0 - 0.6*10) = -7.5
t' = 1.25*(10 - 0.6*0) = 12.5

So, in this frame--the frame where O and O' will be at rest after both twins have accelerated--twin 1 accelerated at x'=0, t'=0, after which twin 1 was at rest and twin 2 was moving at 0.6c, then twin 2 accelerated at x'=-7.5, t'=12.5, after which twin 2 was at rest. During the 12.5 years between twin 1 accelerating and twin 2 accelerating, twin 1 was at rest so he aged 12.5 years, while twin 2 was moving at 0.6c so he only aged 12.5*sqrt(1 - 0.6^2) = 12.5*0.8 = 10 years, so he was 2.5 years younger than twin 1 at the moment he accelerated. Then after twin 2 accelerated too, twin 2 was at rest as well, so they both aged at the same rate after that. Thus at all times after twin 2 accelerates, twin 2 is 2.5 years younger than twin 1 in this frame.

That last equation doesn't make any sense, you said in the original problem that twin2 accelerates in an identical manner to twin1, so if twin1 accelerates for time BT then so should twin2. Also, are these times stated in the frame where the twins were originally at rest at the start? If so, in the rest of your calculations you don't even consider the new definition of simultaneity in the rest frame of O and O' after the acceleration, showing that you've missed the point entirely.

BT is in the context of the accelerating frame.

new definition of simultaneity

The acceleration equations integrate with gamma. This already is designed under the context of the R of S.

But, please continue with this so I make sure I am correctly seeing what you are saying.

 

[cfrogue]

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

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

Interestingly, if both twins are accelerated in such a way that they are brought together (colocated) in the original frame (O') again, they would be the same age again. If only one is accelerated again to bring them together (colocated) in the new frame (O), then the one being accelerated again would be younger, in an absolute sense (a-la the classical twins).

You agree with all this?

Assume they are in the same frame after acceleration.

They can use SR's round trip speed of light calculation to decide their distance.

They can then both apply a common accel/decel to end of together based on this distance.

Since the acceleration phases are again symmetric, then the original decision is not altered.

What do you think?

 

[JesseM]

That last equation doesn't make any sense, you said in the original problem that twin2 accelerates in an identical manner to twin1, so if twin1 accelerates for time BT then so should twin2. Also, are these times stated in the frame where the twins were originally at rest at the start? If so, in the rest of your calculations you don't even consider the new definition of simultaneity in the rest frame of O and O' after the acceleration, showing that you've missed the point entirely.

BT is in the context of the accelerating frame.

What accelerating frame? You didn't mention anything about one, and "accelerating frame" is an ill-defined phrase anyway, there are an infinite number of distinct ways to construct a non-inertial frame where an accelerating observer is at rest. Anyway, when you wrote the equation BT = c/a asinh(ata'/c), is this supposed to be your own version of the equation on the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html t = (c/a) sh(aT/c) ? Do you understand that in this relativistic rocket equation, the symbol t stands for time in the inertial frame where the rocket was initially at rest before it began accelerating, while T stands for the proper time measured by a clock aboard the rocket? If so, shouldn't BT in your equation have the same meaning as t in this equation, and ta' in your equation have the same meaning as T in this equation?

new definition of simultaneity

The acceleration equations integrate with gamma. This already is designed under the context of the R of S.

What does "integrate with gamma" mean? And "this already is designed under the context of the R of S" doesn't tell me which frame you want to compare the ages of the two twins at the end of the experiment--is it supposed to be the rest frame of O and O'? Do you even have any well-defined idea of which frame you're comparing their final ages in?

 

[cfrogue]

What accelerating frame? You didn't mention anything about one, and "accelerating frame" is an ill-defined phrase anyway, there are an infinite number of distinct ways to construct a non-inertial frame where an accelerating observer is at rest. Anyway, when you wrote the equation BT = c/a asinh(ata'/c), is this supposed to be your own version of the equation on the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html t = (c/a) sh(aT/c) ? Do you understand that in this relativistic rocket equation, the symbol t stands for time in the inertial frame where the rocket was initially at rest before it began accelerating, while T stands for the proper time measured by a clock aboard the rocket? If so, shouldn't BT in your equation have the same meaning as t in this equation, and ta' in your equation have the same meaning as T in this equation?

What does "integrate with gamma" mean? And "this already is designed under the context of the R of S" doesn't tell me which frame you want to compare the ages of the two twins at the end of the experiment--is it supposed to be the rest frame of O and O'? Do you even have any well-defined idea of which frame you're comparing their final ages in?

Do you understand that in this relativistic rocket equation, the symbol t stands for time in the inertial frame where the rocket was initially at rest before it began accelerating,



What did I write that was not consistent?


OK, they start in the same frame. I assume this is well defined.

Then, O and twin1 accelerate BT in their frame at acceleration a as viewed from the launch frame.

Let's first agree or disagree here.

 

[cfrogue]

What accelerating frame? You didn't mention anything about one, and "accelerating frame" is an ill-defined phrase anyway, there are an infinite number of distinct ways to construct a non-inertial frame where an accelerating observer is at rest. Anyway, when you wrote the equation BT = c/a asinh(ata'/c), is this supposed to be your own version of the equation on the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html t = (c/a) sh(aT/c) ? Do you understand that in this relativistic rocket equation, the symbol t stands for time in the inertial frame where the rocket was initially at rest before it began accelerating, while T stands for the proper time measured by a clock aboard the rocket? If so, shouldn't BT in your equation have the same meaning as t in this equation, and ta' in your equation have the same meaning as T in this equation?

What does "integrate with gamma" mean? And "this already is designed under the context of the R of S" doesn't tell me which frame you want to compare the ages of the two twins at the end of the experiment--is it supposed to be the rest frame of O and O'? Do you even have any well-defined idea of which frame you're comparing their final ages in?

They end up in the same frame and compare their ages there.

 

[JesseM]

Do you understand that in this relativistic rocket equation, the symbol t stands for time in the inertial frame where the rocket was initially at rest before it began accelerating,



What did I write that was not consistent?

Again, can you answer whether your equation is just supposed to be a rewritten version of that equation? If so, shouldn't BT in your equation play the same role as t in that equation, and refer to the time in an inertial frame, not an accelerating one?

OK, they start in the same frame. I assume this is well defined.

Then, O and twin1 accelerate BT in their frame at acceleration a as viewed from the launch frame.

Let's first agree or disagree here.

If BT is intended to play the same role as t in that equation, it cannot refer to the time in "their frame", it must refer to the time in the launch frame just like t does.

 

[cfrogue]

What accelerating frame? You didn't mention anything about one, and "accelerating frame" is an ill-defined phrase anyway, there are an infinite number of distinct ways to construct a non-inertial frame where an accelerating observer is at rest. Anyway, when you wrote the equation BT = c/a asinh(ata'/c), is this supposed to be your own version of the equation on the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html t = (c/a) sh(aT/c) ? Do you understand that in this relativistic rocket equation, the symbol t stands for time in the inertial frame where the rocket was initially at rest before it began accelerating, while T stands for the proper time measured by a clock aboard the rocket? If so, shouldn't BT in your equation have the same meaning as t in this equation, and ta' in your equation have the same meaning as T in this equation?

What does "integrate with gamma" mean? And "this already is designed under the context of the R of S" doesn't tell me which frame you want to compare the ages of the two twins at the end of the experiment--is it supposed to be the rest frame of O and O'? Do you even have any well-defined idea of which frame you're comparing their final ages in?

http://www.ejournal.unam.mx/rmf/no521/RMF52110.pdf

Please look at equation 3.

 

[cfrogue]

Again, can you answer whether your equation is just supposed to be a rewritten version of that equation? If so, shouldn't BT in your equation play the same role as t in that equation, and refer to the time in an inertial frame, not an accelerating one?

If you look at the link you posted, I wrote it the same way.

If BT is intended to play the same role as t in that equation, it cannot refer to the time in "their frame", it must refer to the time in the launch frame just like t does.

Normally, it is
tau = c/a asinh( (at)/c )
Agreed?

 

[JesseM]

http://www.ejournal.unam.mx/rmf/no521/RMF52110.pdf

Please look at equation 3.

How does that answer my question of what you meant by "integrate with gamma"? That equation is calculating the proper time of an accelerating particle, and the g in that equation is not gamma, it's the proper acceleration. You can't use that equation to deal with questions of simultaneity, it has nothing to do with that. None of the relativistic rocket equations have anything to do with the question of simultaneity in the final rest frame of the twins--the only times that appear in the relativistic rocket equations are the time coordinate in the launch frame, and the proper time as measured by a clock aboard the rocket.

 

[cfrogue]

How does that answer my question of what you meant by "integrate with gamma"? That equation is calculating the proper time of an accelerating particle, and the g in that equation is not gamma, it's the proper acceleration. You can't use that equation to deal with questions of simultaneity, it has nothing to do with that. None of the relativistic rocket equations have anything to do with the question of simultaneity in the final rest frame of the twins--the only times that appear in the relativistic rocket equations are the time coordinate in the launch frame, and the proper time as measured by a clock aboard the rocket.

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

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

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

 

[JesseM]

Normally, it is
tau = c/a asinh( (at)/c )
Agreed?

Ah, I see, you're using the inverse equation T = (c/a) sh^-1(at/c). So you mean your BT is supposed to correspond to T, the proper time measured aboard the rocket, yes? It's better to refer to "proper time" than to "time in their frame", since as I said before, accelerating observers don't have any unique rest frame.

OK, but given that I agree that the proper time BT during the acceleration phase is related to the coordinate time ta' in the launch frame by BT = c/a asinh(a*ta'/c), and likewise I agree that c/a sinh(a*BT/c) = ta'. However, the next step in your proof doesn't make much sense:

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

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