Need help understanding the twins

[cfrogue]

Ok, here is my experiment then :)

Twins, A and B they set their clocks and use Swatch synchronization.

Now Twin B will not move and Twin A will fly past with his spaceship 10 kms away from Twin B(from B's POV),

Twin A uses the calculations(which are actually wrong) and flies as agreed.

Twin B on the other hand innocently watch Twin A as Twin A flies by the agreed coordinates. But then something happens and from Twin B's POV Twin A's spaceshift's energy level is so high that the spaceshift just dissappears from sight leaving behind a deadly gravitational wave, which distorts space-time on that frame for any object relatively traveling at or below c.

Because for Twin B, Twin A's pass just killed him as Twin A reached the required relative speed sucking him into a singularity. After a while Twin A returns back as agreed. Also getting sucked into the singularity. Then i go and find their watches and realize that they are fake swatches.


So maybe this time when they are out of the singularity, they will try making a valid experiment accepting the fact that they can't find each other again after reaching a certain relativistic speed :)

Hello cfrogue,

Two real objects can not be in the same coordinates so they will experience different paths. Ending up in different places using same calculations.

One of the twins will age less(the differnece increasing in direct proportion to their rest distance). Which twin will age less can not be calculated with the variables at hand.

what if I said humor is always good and this was funny.

 

[pervect]

Oh - it was a joke? He actually got the part about a gravitational wave right - though I doubt it would kill anyone, unless one of the twins was REALLY heavy. The rest was a bit muddled, but not that much more muddled than a lot of posts I see on here that are (supposedly, at least) serious :-) :-(.

 

[ExecNight]

Its not the twin that is heavy, he don't like Burger King! Its his spaceship..

Well the spaceship is quite heavy, around 4.2 SunMass. Actually it is not the spaceship that is heavy. It just contains one reactor to keep the charged singularity let's say under control. How did you think he could accelerate up to c speeds, by swiming in empty space? :P

 

[cfrogue]

post #49 Jesse.

 

[JesseM]

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.

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

They all perform Einstein's clock synchronization method to test the time differences of the clocks.

What will O conclude about the age ordinality of the twins?

What will O' conclude about the age ordinality of the twins?

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

 

[cfrogue]

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

Here is the math.

Calculations of O for the twins

Elapsed proper time calculation for twin 1
twin1's acceleration phase
BT
twin1's relative motion phase
t'
twin2's acceleration phase
ta'
But, as shown above in the links, BT = c/a asinh(ata'/c).
aBT/c = asinh(ata'/c)
sinh(aBT/c) = ata'/c
c/a sinh(aBT/c) = ta'
Please note this derivation will be used throughout the below also.
Total elapsed proper time calculation of O for twin1 BT + t' + ta' = BT + t' + c/a sinh(aBT/c)

Elapsed proper time calculation for twin 2

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

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

Calculations of O' for the twins

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

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

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

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

Since all are now in the same frame, they again use Einstein's clock synchronization method for the actual hard data of the experiment to test the clocks of the twins for age ordinality. This answer of the clock sync is to be accepted by all those in the frame because as Einstein said, "We assume that this definition of synchronism is free from contradictions". This clock sync can also establish the time ordinality of the clocks.

SR is supposed to be a reliable tool for the calculation of the proper time of one frame to another frame.

However, both O' and O cannot be correct with their SR conclusions when presented with the factual data of the clock sync.*

 

[matheinste]

------Since all are now in the same frame, they again use Einstein's clock synchronization method for the actual hard data of the experiment to test the clocks of the twins for age ordinality. This answer of the clock sync is to be accepted by all those in the frame because as Einstein said, "We assume that this definition of synchronism is free from contradictions". This clock sync can also establish the time ordinality of the clocks. -------

SR is supposed to be a reliable tool for the calculation of the proper time of one frame to another frame.



However, both O' and O cannot be correct with their SR conclusions when presented with the factual data of the clock sync.* ------

You cannot be seriouis. Entertaining though.


As they both end up in the same frame their clocks will be in synch. They will of course not show the same elapsed time. And of course both will agree on these elapsed times. But I suspect you know this

Matheinste.

 

[cfrogue]

------Since all are now in the same frame, they again use Einstein's clock synchronization method for the actual hard data of the experiment to test the clocks of the twins for age ordinality. This answer of the clock sync is to be accepted by all those in the frame because as Einstein said, "We assume that this definition of synchronism is free from contradictions". This clock sync can also establish the time ordinality of the clocks. -------

SR is supposed to be a reliable tool for the calculation of the proper time of one frame to another frame.



However, both O' and O cannot be correct with their SR conclusions when presented with the factual data of the clock sync.* ------

You cannot be seriouis. Entertaining though.


As they both end up in the same frame their clocks will be in synch. They will of course not show the same elapsed time. And of course both will agree on these elapsed times. But I suspect you know this

Matheinste.

the math is above.

let me know

 

[cfrogue]

------Since all are now in the same frame, they again use Einstein's clock synchronization method for the actual hard data of the experiment to test the clocks of the twins for age ordinality. This answer of the clock sync is to be accepted by all those in the frame because as Einstein said, "We assume that this definition of synchronism is free from contradictions". This clock sync can also establish the time ordinality of the clocks. -------

SR is supposed to be a reliable tool for the calculation of the proper time of one frame to another frame.



However, both O' and O cannot be correct with their SR conclusions when presented with the factual data of the clock sync.* ------

You cannot be seriouis. Entertaining though.


As they both end up in the same frame their clocks will be in synch. They will of course not show the same elapsed time. And of course both will agree on these elapsed times. But I suspect you know this

Matheinste.

Yes, but I avoid this correct answer.

It enrages the natives.

 

[JesseM]

Here is the math.

Calculations of O for the twins

Elapsed proper time calculation for twin 1
twin1's acceleration phase
BT
twin1's relative motion phase
t'
twin2's acceleration phase
ta'
But, as shown above in the links, BT = c/a asinh(ata'/c).

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.

 

[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.

 

[cfrogue]

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

Yes, the beginning and endpoints are at issue.

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

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

What is your view?

 

[JesseM]

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

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

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

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

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

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

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

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

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

 

[JesseM]

Yes, the beginning and endpoints are at issue.

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

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

 

[cfrogue]

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

Take a look at this.

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

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

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

Is this correct?

 

[cfrogue]

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

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


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

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

Thus, the correct answer is

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

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

 

[cfrogue]

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

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

 

[JesseM]

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

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

 

[cfrogue]

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

You clarified my thinking.

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

But, now this twins paradox is decidable without debate.

O and O' disagree.

 

[Jorrie]

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

Assume they are in the same frame after acceleration.

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

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

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

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

 

[cfrogue]

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

Here is the new math.

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

Here is the math.

Calculations of O for the twins

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

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

Elapsed proper time calculation for twin 2

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

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

Calculations of O' for the twins

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

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

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

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