Revisiting the Flaws of the Light Clock in Special and General Relativity

[Austin0]

My apologies if my reply was irrelevant. I was commenting on the above part of your post.
I'm afraid when I come accoss the words "Twin paradox" in the same paragraph as "no real consensus as to the proper resolution" I cannot help myself, its an automatic reaction. Most regular posters here have developed a mechanism for overcoming such reactions. I must try harder to do the same

Matheinste.

No aps neccessary. If you misinterpreted my words as meaning there was any question at all about there being a satisfactory resolution I can understand your reaction.
As I understood the OP he was also not questioning that there was a resolution but only if there was a resolution that did not involve acceleration as a deciding factor.

 

[matheinste]

If the scenario is the usual one, twin leaves earth, SAME twin returns to Earth then acceleration cannot be avoided.

In other proposed scenarios such as the one where we have an observer who remains on earth, a traveller passing Earth outbound, setting his clock to Earth time in passing, meeting a third observer traveling inbound, setting clocks to be the same when they meet at the agreed turnaround point, and this third observer not stopping at Earth on the way back but comparing clocks in passing, then acceleration by none of the observers is involved. But the out and inbound travellers do not share the same inertial frame and do not agree about the age of the earthly observer when they meet at what would have been the turnaround point in the original scenario. The result will of couse be the similar, the inbound observer's clock will have a reading less than the Earth clock.

Whatever scenario is proposed in SR, the solution is availible and follows logically from the axioms.

Matheinste.

 

[Austin0]

If the scenario is the usual one, twin leaves earth, SAME twin returns to earth then acceleration cannot be avoided.

In other proposed scenarios such as the one where we have an observer who remains on earth, a traveller passing Earth outbound, setting his clock to Earth time in passing, meeting a third observer traveling inbound, setting clocks to be the same when they meet at the agreed turnaround point, and this third observer not stopping at Earth on the way back but comparing clocks in passing, then acceleration by none of the observers is involved. But the out and inbound travellers do not share the same inertial frame and do not agree about the age of the earthly observer when they meet at what would have been the turnaround point in the original scenario. The result will of couse be the similar, the inbound observer's clock will have a reading less than the Earth clock.
Whatever scenario is proposed in SR, the solution is availible and follows logically from the axioms.

Matheinste.

Acceleration cannot be avoided to create the actual occurences but it is not neccessarily crucial to the analysis or resolution. It has been done with reciprocal dilation throughout and the final outcome attributed to relative simultaneity.
Here again you have asserted that the conclusion would be the same regarding purely inertial frames but have not examined or countered my explicit exception to this claim.
Why is this? It is vey simple and right there. DO you not understand it , was I not clear enough perhaps?

 

[matheinste]

Acceleration cannot be avoided to create the actual occurences but it is not neccessarily crucial to the analysis or resolution. It has been done with reciprocal dilation throughout and the final outcome attributed to relative simultaneity.
Here again you have asserted that the conclusion would be the same regarding purely inertial frames but have not examined or countered my explicit exception to this claim.
Why is this? It is vey simple and right there. DO you not understand it , was I not clear enough perhaps?

I have lost track a bit here, my fault. If you propose a single question or counter example, or guide me to the particular post, I would be happy to try and answer or at least explain why I cannot answer it.

Matheinste.

 

[CPL.Luke]

I agree with Matheinste,

The point I was trying to make is that if you pretend that the spaceship instantly accelerates in the opposite direction, then the problem is symmetrical and there is a paradox.

however once you acknowledge that the spaceship must undergo acceleration, then you can just use the rindler metric, which is

ds²=-a²x²c²dt²+dx²

the problem is no longer symmetrical, and once you calculate it out both parties should agree on the amount of proper time which has elapsed.

alternatively if your careful with how the observers will observe each others clocks you'll get the right answer. Just as Austin0 described.

 

[Austin0]

I have lost track a bit here, my fault. If you propose a single question or counter example, or guide me to the particular post, I would be happy to try and answer or at least explain why I cannot answer it.

Matheinste.

It is in post #64 but here is the question:

Originally Posted by kev
However, I can make a definitive statement about the elapsed proper time between any two timelike events.

Let us say A remains at rest in frame A. All references to coordinate measurements will mean measurements made by observers at rest in frame A. B passes A at coordinate time zero at a coordinate velocity of +0.8c. After a coordinate time of 10 years, B passes C who is going in the opposite direction with a coordinate velocity of -0.8c. The coordinate distance between event (B passing A) = event(B,A) and event(B,C) is 8 lightyears. Other observers in different reference frames will disagree with the coordinate times, distance and velocities measured by A and will also disagree on what clock A reads at event(B,C), but all observers will agree that 6 years of proper time elapses on clock B between events (A,B) and (B,C). (Definitive statement 1.) Eventually clock C passes A at event(C,A).

All observers agree that 6 years of proper time passes on clock C between events (B,C) and (C,A). (Definitive statement 2).

All observers will agree that 20 years of proper time elapses on clock A between events (B,A) and (C,A). (Definitive statement 3). All observers agree that the combined elapsed proper time between the 3 events (B,A), (B,C) and (C,A) is 12 years (Definitive statement 4) and that this proper time interval is less that the proper time interval between events (B,A) and (C,A). (Definitive statement 5).

Originally Posted by Austin0
Suppose there is a frame D such that A and C are traveling relative to it, at 0.5c and -0.5c
respectively. At event (B,C) ..in D.. t'''=0 and the spatial interval between A and C is dx''' with observed clock times in A of t₀ and in C of t''₀

As observed in D ...A and C meet at dx'''/2 with observed clock times in A of (T) and in C of (T'') event (C,A)
For both frames dx'''*0.5/ 0.5c = dt = dt'' from this it would seem to follow that dt = T-t₀ =T''-t''₀= dt'' Wouldn't this be equal elapsed proper time observed in both frames between events (B,C) and (C,A) ?

Do you see some reason why this would not apply??

 

[Austin0]

Hi kev I kept waiting for a responce and then sort of forgot but not completely, ;-)
Nor have I forgotten your telling me that if I could demonstrate non-reciprocal time dilation between inertial frames it would mean demonstrating real motion. So I am hoping for some clarification here.

However, I can make a definitive statement about the elapsed proper time between any two timelike events.

Let us say A remains at rest in frame A. All references to coordinate measurements will mean measurements made by observers at rest in frame A. B passes A at coordinate time zero at a coordinate velocity of +0.8c. After a coordinate time of 10 years, B passes C who is going in the opposite direction with a coordinate velocity of -0.8c. The coordinate distance between event (B passing A) = event(B,A) and event(B,C) is 8 lightyears. Other observers in different reference frames will disagree with the coordinate times, distance and velocities measured by A and will also disagree on what clock A reads at event(B,C), but all observers will agree that 6 years of proper time elapses on clock B between events (A,B) and (B,C). (Definitive statement 1.) Eventually clock C passes A at event(C,A).
All observers agree that 6 years of proper time passes on clock C between events (B,C) and (C,A). (Definitive statement 2).

All observers will agree that 20 years of proper time elapses on clock A between events (B,A) and (C,A). (Definitive statement 3). All observers agree that the combined elapsed proper time between the 3 events (B,A), (B,C) and (C,A) is 12 years (Definitive statement 4) and that this proper time interval is less that the proper time interval between events (B,A) and (C,A). (Definitive statement 5).

It would seem that you have demonstrated that frame C was actually moving relative to frame A as the inertial clock located in C and moving toward A had less elapsed proper time than inertial clock A moving toward clock C at the same relative v.
Explanation?

A unprimed... C '' double primed
Suppose there is a frame D'''---- such that A and C are traveling relative to it, at A(v)= 0.5c and C(v)= (- 0.5 ) c
respectively. At event (B,C) .. as observed iin D.. t'''=0 and the spatial interval between A and C is dx''' with observed clock times (from D) in A of t₀ and in C of t''₀

As observed in D ...A and C meet at dx'''/2 with observed clock times in A of (T) and in C of (T'') this is event (C,A) as per kev.
For both frames,( A and C) ---- dt and dt'' = dx'''*0.5/ 0.5c from this it would seem to follow that
dt = T-t₀ = T''- t''₀= dt'' Wouldn't this be equal elapsed proper time observed in both frames between events (B,C) and (C,A) ?

Do you see some reason why this would not apply??

 

[yuiop]

Hi kev I kept waiting for a responce and then sort of forgot but not completely, ;-)
Nor have I forgotten your telling me that if I could demonstrate non-reciprocal time dilation between inertial frames it would mean demonstrating real motion. So I am hoping for some clarification here.

Sorry about the delay. I have a number of threads I would like to analyse and a number of projects I am working on such as working out the equation for the time dilation of an accelerating light clock and doing a java program to demonstrate 2D relativistic transformations for light clocks, barn-pole paradox, Thomas rotation etc. but the spare time I have for such things seems to be taken up defending unjustified attacks on my work from you-know-who. :tongue:

It would seem that you have demonstrated that frame C was actually moving relative to frame A as the inertial clock located in C and moving toward A had less elapsed proper time than inertial clock A moving toward clock C at the same relative v.

Explanation?

The trouble here is that different observers will disagree with where A was located spatially relative to event(B,C) and what time was showing on A's clock at event(B,C), because A was not located at that event.

A unprimed... C '' double primed
Suppose there is a frame D'''---- such that A and C are traveling relative to it, at A(v)= 0.5c and C(v)= (- 0.5 ) c
respectively. At event (B,C) .. as observed iin D.. t'''=0 and the spatial interval between A and C is dx''' with observed clock times (from D) in A of t0 and in C of t''0

As observed in D ...A and C meet at dx'''/2 with observed clock times in A of (T) and in C of (T'') this is event (C,A) as per kev.
For both frames,( A and C) ---- dt and dt'' = dx'''*0.5/ 0.5c from this it would seem to follow that
dt = T-t0 = T''- t''0= dt'' Wouldn't this be equal elapsed proper time observed in both frames between events (B,C) and (C,A) ?

Do you see some reason why this would not apply??

O.K, you are right that this is how things would look from D's point of view, but that is just D's point of view and D is not a final arbitrator. At event(B,C), D will consider A to be 3.46 ly away to the left and C to be 3.46 ly to the right. From D's POV , C is coming towards him at 0.5c so C arrives at D's location in 3.46/0.5 = 6.9282 years by D's clock and A arrives at the same time. Because B is moving at 0.5c relative to D, D will work out the the proper time elapsed on B's clock is 6.9282*sqrt(1-0.5^2) = 6 years. So yes, for this leg of the journey D considers the elapsed times on A and C's clocks to be the same. However, for the first leg of the journey, when B is going away from A, D sees A coming towards him at 0.5c (same as in the final leg) but B is going away from him at (0.8-0.5)/(1-0.8*0.5)=0.92857c so the time dilation of B's clock on the outward journey is much greater than the time dilation of A's clock, according to D. The first leg of the experiment took 14 years of A's proper time in D's frame. This equates to 14/sqrt(1-0.5^2)=16.1658 years by D's clock. D calculates that the proper time of B's clock on the outward journey "during the same time" is 16.1658*sqrt(1-0.92857^2)=6 years. Overall, D says 20 years of proper time elapsed on A's clock and 12 years of proper time elapsed on the combined times of B and C and all is good.

As you can see from the above calculations, D says A was 3.46*2=6.92 lys away from event(B,C) "at the time" B and C crossed paths, while A says he was 8 lys away from B and C "at the time" B and C crossed paths, so A and D differ on where A was "at the time" B and C crossed and what time was showing on A's clock "at the time" B and C crossed.

Hope that helps.

 

[Austin0]

Hi kev
Well you know I agree completely with you about the problems with the acceleration explanation and have said some of the same things you just posted , so i would be happy if you came up with a totally inertial scenario.

You didn't mean to make a suggestion of absolute motion, but if there is any way of measuring non-reciprocal time dilation, then that certainly implies a way of measuring absolute motion.

So you went directly from saying the above to yhr brlow

All observers will agree that 20 years of proper time elapses on clock A between events (B,A) and (C,A). (Definitive statement 3).While only 12 years elapse on B and C

It would seem that you have demonstrated that frame C was actually moving relative to frame A as the inertial clock located in C and moving toward A had less elapsed proper time than inertial clock A moving toward clock C at the same relative v.

You ignored this one.If you were successful with this exercise, it would be demonstrating real motion would it not?

O.K, you are right that this is how things would look from D's point of view, but that is just D's point of view and D is not a final arbitrator. So yes, for this leg of the journey D considers the elapsed times on A and C's clocks to be the same.

Yu are right D is not the final arbiter just an exception to agreement , on the other hand what frame is the final arbiter if there is not frame independant agreement?

However, for the first leg of the journey, when B is going away from A, D sees A coming towards him at 0.5c (same as in the final leg) but B is going away from him at (0.8-0.5)/(1-0.8*0.5)= 0.92857c
Well you didn't like my advice ,what can I say, "subtraction of velocities equation"?
so the time dilation of B's clock on the outward journey is much greater than the time dilation of A's clock, according to D. The first leg of the experiment took 14 years of A's proper time in D's frame. This equates to 14/sqrt(1-0.5^2)=16.1658 years by D's clock. D calculates that the proper time of B's clock on the outward journey "during the same time" is 16.1658*sqrt(1-0.92857^2)=6 years. Overall, D says 20 years of proper time elapsed on A's clock and 12 years of proper time elapsed on the combined times of B and C [B and all is good.

Well maybe not quite 100% just yet. . :tongue:

How exactly does B take 16.1658 years on D's clock to travel approx. 6.9 ly at 0.92867c ?

After a coordinate time of 10 years, B passes C who is going in the opposite direction with a coordinate velocity of -0.8c. .

So this time we have 14 years of proper time for the first leg instead of 10 , but why quibble because more importantly frame E disagrees with this completely.
Frame E Moving 0.5c --> + x relative to A and ( - 0.5c) relative to B and origen colocated with both at ( B,A)
At event (B,C) in E at t'''' and x₂''''' and position of A at t'''', at x₁''''' with dx'''''=x₂''''' - x₁''''' etc etc

I am sure you see the rest of this written on the wall. yes??

As you can see from the above calculations, D says A was 3.46*2=6.92 lys away from event(B,C) "at the time" B and C crossed paths, while A says he was 8 lys away from B and C "at the time" B and C crossed paths, so A and D differ on where A was "at the time" B and C crossed and what time was showing on A's clock "at the time" B and C crossed.

He said , she said :-)

 

[starthaus]

It can also be seen that I have taken the time dilation due to acceleration into account and it is insignificant compared to the velocity time dilation and an unnecessary complication.

This is false, the only reason that there is such a disparity between the values is that you have chosen a very long "cruising" time and a very short period of acceleration. In reality, if you had chosen comparable values to the "cruise" time $$T_c$$ vs. the acceleration time $$T_a$$ you would have found that the difference in proper time between the two twins is equally spread between the two effects and, that both depend on acceleration $$a$$:

$$d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} arcsinh(aT_a/c)$$

for the accelerating twin

vs.$$dt=2T_c+4T_a$$

for the inertial twin. If you make $$T_a$$ closer to $$T_c$$ than you made them in your skewed example, you will find, to your surprise, that the effects are comparable.
One more thing, the above has nothing to do with the clock "hypothesis"

 

[JesseM]

This is false, the only reason that there is such a disparity between the values is that you have chosen a very long "cruising" time and a very short period of acceleration. In reality, if you had chosen comparable values to the "cruise" time $$T_c$$ vs. the acceleration time $$T_a$$ you would have found that the difference in proper time between the two twins is equally spread between the two effects and, that both depend on acceleration $$a$$:

$$d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} arcsinh(aT_a/c)$$

for the accelerating twin

It depends what you mean by "depend on acceleration a". It's true that you can express the time elapsed as a function of the proper acceleration as well as the time spent accelerating, but it is also always possible to express it as a function only of the velocity as a function of time v(t) for the traveling twin in some inertial frame. For any brief interval of time dt where the traveling twin has velocity v in that interval, the traveling twin will age by $$\sqrt{1 - v^2/c^2} \, dt$$, so if t₀ and t₁ are the times that the traveling twin departs from and reunites with the inertial twin, the total elapsed time for the traveling twin can be calculated with the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$. You can see that this integral depends only on the velocity as a function of time, not the acceleration, and this is exactly what is meant by the clock hypothesis.

 

[starthaus]

It depends what you mean by "depend on acceleration a". It's true that you can express the time elapsed as a function of the proper acceleration as well as the time spent accelerating, but it is also always possible to express it as a function only of the velocity as a function of time v(t) for the traveling twin in some inertial frame. For any brief interval of time dt where the traveling twin has velocity v in that interval, the traveling twin will age by $$\sqrt{1 - v^2/c^2} \, dt$$, so if t₀ and t₁ are the times that the traveling twin departs from and reunites with the inertial twin, the total elapsed time for the traveling twin can be calculated with the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$. You can see that this integral depends only on the velocity as a function of time, not the acceleration, and this is exactly what is meant by the clock hypothesis.

I know perfectly well how the proper time us calculated, so please, explain the term in $$\frac{c}{a} arcsinh(aT_a/c)$$. Is this a function of $$v$$ only?

 

[JesseM]

I know perfectly well how the proper time us calculated, so please, explain the term in $$\frac{c}{a} arcsinh(aT_a/c)$$. Is this a function of $$v$$ only?

No, but you don't need to include that term in order to calculate the elapsed time on the accelerating clock, instead you can just use the clock's v(t). Again, the clock hypothesis just says it is possible to calculate the elapsed proper time in the way I described, not that it is the only way to calculate it.

 

[starthaus]

No, but you don't need to include that term in order to calculate the elapsed time on the accelerating clock,

This is false. The term you are trying to explain way is the time elapsed during acceleration. So, one more time, please explain the term $$\frac{c}{a} arcsinh(aT_a/c)$$ using $$v$$ only.

 

[JesseM]

This is false. The term you are trying to explain way is the time elapsed during acceleration.

My claim was that you don't need to include that term if you want to calculate "the time elapsed during acceleration" (although that's one way of calculating it), instead you can find the function giving velocity as a function of time v(t) for the accelerating clock in some inertial frame, and then calculate the elapsed proper time between the moment of the start of acceleration at t₀ and the moment of the end of acceleration at t₁ using the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$. Do you disagree that this is another valid way to calculate the elapsed proper time?

 

[starthaus]

My claim was that you don't need to include that term if you want to calculate "the time elapsed during acceleration" (although that's one way of calculating it), instead you can find the function giving velocity as a function of time v(t) for the accelerating clock in some inertial frame, and then calculate the elapsed proper time between the moment of the start of acceleration at t₀ and the moment of the end of acceleration at t₁ using the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$. Do you disagree that this is another valid way to calculate the elapsed proper time?

.

Yes, the derivation is all based on the $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$, this is well known but that is not the point, the final answer is a function of acceleration as shown in post #80 . Do you disagree with that?

The other point that I was making to kev (that you also missed) is that he has inadvertenly ignored the difference due to the acceleration by making the acceleration period very small wrt the cruising period. If the two periods are comparable, the acceleration effect cannot be neglected. Do you disagree with that?

 

[JesseM]

.

Yes, the derivation is all based on the $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$, this is well known but that is not the point

Well it's my point, you can define it in terms of a function that includes a dx/dt term but no $$d^2 x / dt^2$$ term. Perhaps the confusion is that according to the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html a, v(t) does itself include the term a (if the velocity is zero at t=0 then we have v(t) = at / sqrt[1 + (at/c)^2]) so the result when you evaluate the integral will include an a term too, but this a is just a constant which tells you the proper acceleration, it is not the same as the time-derivative of the coordinate velocity v(t) (or the second time derivative of the coordinate position x(t)) i.e. $$d^2 x / dt^2$$. The "clock hypothesis" just says that the total elapsed time can be found by an integral where the only derivative of of x(t) in the integrand is the first derivative i.e. dx/dt.

The other point that I was making to kev (that you also missed) is that he has inadvertenly ignored the difference due to the acceleration by making the acceleration period very small wrt the cruising period. If the two periods are comparable, the acceleration effect cannot be neglected. Do you disagree with that?

I would say that talking about the "acceleration effect" is a way of speaking which is apt to confuse people if they have also heard of the clock hypothesis which says that time dilation is only a function of velocity (which is understood to mean that the elapsed time can be calculated using an integral of the form I gave above). It would be less confusing to say that if the two periods are comparable, you have to take into account the way the velocity was changing continuously during the period of acceleration if you want to calculate the elapsed time.

 

[starthaus]

but this a is just a constant which tells you the proper acceleration,

Proper acceleration, far from being "just a constant" is a measurable physical entity. The final result, the difference in the elapsed proper time of the twins is a proper function of proper acceleration.

it is not the same as the time-derivative of the coordinate velocity v(t) (or the second time derivative of the coordinate position x(t)) i.e. $$d^2 x / dt^2$$. The "clock hypothesis" just says that the total elapsed time can be found by an integral where the only derivative of of x(t) in the integrand is the first derivative i.e. dx/dt.

True. As true as the fact that , when expressing v as a function of a, the final result is not a useful function of v but a function of a.

I would say that talking about the "acceleration effect" is a way of speaking which is apt to confuse people if they have also heard of the clock hypothesis which says that time dilation is only a function of velocity (which is understood to mean that the elapsed time can be calculated using an integral of the form I gave above).

Only if such people are unaware of basics physics.

It would be less confusing to say that if the two periods are comparable, you have to take into account the way the velocity was changing continuously during the period of acceleration if you want to calculate the elapsed time.

In order to be precise, one always needs to take into account the accelerated period.

 

[Passionflower]

All these "acceleration is not relevant" arguments are rather silly because if we do not take proper (or inertial) acceleration into account we cannot possibly measure which twin ages more!

Watch how ridiculous the "paradox" becomes when we consider acceleration to be irrelevant:

Have two twins co-located with a relative velocity of zero. They separate with an instant velocity of v, when they are a distance x removed from each other, their relative velocity instantly becomes -v. As soon as they meet their instant velocity becomes 0 again. Which twin ages more?

The "paradox" is only a paradox for those who consider acceleration to be irrelevant!

 

[starthaus]

All these "acceleration is not relevant" arguments are rather silly because if we do not take proper (or inertial) acceleration into account we cannot possibly measure which twin ages more!

Watch how ridiculous the "paradox" becomes when we consider acceleration to be irrelevant:

Have two twins co-located with a relative velocity of zero. They separate with an instant velocity of v, when they are a distance x removed from each other, their relative velocity instantly becomes -v. As soon as they meet their instant velocity becomes 0 again. Which twin ages more?

The "paradox" is only a paradox for those who consider acceleration to be irrelevant!

Yes, it is the typical approach for the people that prefer dumbed down versions of reality beacuse the math necessary to describe the realistic situations, is "too hard".

 

[yuiop]

This is false, the only reason that there is such a disparity between the values is that you have chosen a very long "cruising" time and a very short period of acceleration. In reality, if you had chosen comparable values to the "cruise" time $$T_c$$ vs. the acceleration time $$T_a$$ you would have found that the difference in proper time between the two twins is equally spread between the two effects and, that both depend on acceleration $$a$$:

$$d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} asinh(aT_a/c)$$

for the accelerating twin

I know perfectly well how the proper time us calculated, so please, explain the term in $$\frac{c}{a} asinh(aT_a/c)$$. Is this a function of $$v$$ only?

The correct equation for the proper time $$d \tau$$ of the traveling twin with cruising time $$T_c$$ and acceleration time $$T_a$$ (both as measured in the inertial stay at home twins rest frame) can be stated as a function of v only like this:

$$d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, asinh(v \gamma/c)$$

Where v is the cruising velocity and $$\gamma$$ is $$1/\sqrt{1-v^2/c^2}$$.

For a numerical example let us say that the total cruising time $$T_c$$ is 10 years and the total acceleration time $$T_a$$ is also 10 years, (both measured in the rest frame of the stay at home twin) so that the inertial twin ages by 20 years during the round trip of the travelling/accelerating twin. The cruising velocity of the traveling twin is 0.8c relative to the stay at home twin and the gamma factor is 1/0.6. Just to be absolutely clear, the traveling twin goes away from home at 0.8c for 5 yers, then spends 5 years slowing down to a stop, another 5 years accelerating to 0.8c and finally the last 5 years cruising at 0.8c for a round trip time of 20 years, all as measured by the inertial twin.

The proper time that elapses on the traveling twins clock in years is:

$$d \tau=10*0.6+10*0.6/0.8* \,asinh(0.8/0.6) = 6+8.23959 = 14.2396$$

Note that most of the time dilation happens during the cruise phase even though equal times are spent cruising and accelerating according to the inertial twin. This is because on average the instantaneous relative velocity during the acceleration phase is less than the cruising relative velocity and the acceleration has not increased the time dilation.

Your equation is wrong and predicts the proper time of the traveling twin to be longer than the stay at home twins' proper time. Basically you screwed up the factors when hacking your derivation. You didn't get it from any textbook did you?

 

[JesseM]

Proper acceleration, far from being "just a constant" is a measurable physical entity.

Of course it's measurable, but it's a constant in the equation for elapsed proper time of an observer with constant proper acceleration, which is where the equation you wrote down came from (that equation wouldn't be used for calculating the elapsed proper time for an observer whose proper acceleration was varying). And if the proper acceleration is varying, you can still calculate v(t) and then derive the elapsed proper time from that.

True. As true as the fact that , when expressing v as a function of a, the final result is not a useful function of v but a function of a.

OK, but a is not a variable in the equation v(t) = at / sqrt[1 + (at/c)^2]) (or in the equation for proper time as a function of coordinate time which you wrote down), the only variable that v is a function of is the coordinate time t. Regardless of the rocket's motion (even if it has a varying proper acceleration), you can always express v as function where the only variable is t, though the function may of course include some numerical constants. And I understand the "clock hypothesis" to mean, first of all, that in the limit as we look at arbitrarily short time intervals dt, the amount of time elapsed on a clock during that interval will be equal to $$\sqrt{1 - v^2/c^2} \, dt$$, so even if two different clocks are at different positions or have different instantaneous accelerations, as long as they have the same instantaneous velocity they will have the same instantaneous rate of ticking in a given inertial frame. And second, as a corollary of the above, I understand the clock hypothesis to mean that if you know v(t) for a given clock between t₀ and t₁, the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$ will give the elapsed proper time for this clock between t₀ and t₁. Both of these are nontrivial statements about how the laws of physics work in our universe, for example we could write down an alternate set of hypothetical laws where the instantaneous time dilation did depend on the instantaneous acceleration, so two clocks with the same instantaneous velocity could have different instantaneous rates of ticking. Hopefully you don't disagree with any of the above? Assuming you don't, my point was just that talking about the time dilation as a function of acceleration could potentially confuse someone who doesn't fully understand the clock hypothesis, so from a pedagogical point of view it's better not to talk in that way, even though nothing you said was incorrect on a technical level.

Only if such people are unaware of basics physics.

Well, one of the points of the forum is to teach about basic physics to people who aren't necessarily aware of it. You brought up the clock hypothesis (or were you responding to someone else having brought it up earlier), so I just wanted to clarify for people reading what it means to say in the context of the clock hypothesis that time dilation is not a function of acceleration, which might confuse someone who just saw you saying that the elapsed time was a function of a.

In order to be precise, one always needs to take into account the accelerated period.

True, but before you said:

The other point that I was making to kev (that you also missed) is that he has inadvertenly ignored the difference due to the acceleration by making the acceleration period very small wrt the cruising period. If the two periods are comparable, the acceleration effect cannot be neglected. Do you disagree with that?

I took this to mean you were speaking about what was needed to get an approximately correct figure for the elapsed time, since even if the two periods are not comparable and the acceleration period is very small, any finite period of acceleration needs to be taken into account if you want a perfectly precise figure for the elapsed time (say, in a theoretical word-problem where the length of the acceleration period is known with total precision).

 

[Passionflower]

Well, one of the points of the forum is to teach about basic physics to people who aren't necessarily aware of it.

Then the first thing to teach should be that the $\gamma$ factor between two observers is determined by their acceleration both proper and inertial. If we know that we can calculate the time dilation as $\gamma$ is a factor in this. And so of course the total dilation depends on acceleration, as $\gamma$ is determined by acceleration. Or that if we know the time dilation we can calculate the acceleration. These are useful things.

However if one starts with "oh, they are moving but we don't know how this started" then all bets are off since one cannot calculate time dilation from that. All one can do is dwell on completely inconsequential observed time dilation and length contraction using completely unphysical planes of simultaneity. "They keep going forever with a relative speed x, both observers measure the other observers clock to go slower and their length to be shorter So what? You think that should be the basics?

 

[yuiop]

All these "acceleration is not relevant" arguments are rather silly because if we do not take proper (or inertial) acceleration into account we cannot possibly measure which twin ages more!

Watch how ridiculous the "paradox" becomes when we consider acceleration to be irrelevant:

Have two twins co-located with a relative velocity of zero. They separate with an instant velocity of v, when they are a distance x removed from each other, their relative velocity instantly becomes -v. As soon as they meet their instant velocity becomes 0 again. Which twin ages more?

The "paradox" is only a paradox for those who consider acceleration to be irrelevant!

Hi Passionflower. You have a sort of point and it is just as confusing to say that relative proper times is only a function of relative velocities. It is better to say the twins paradox is resolved by the asymmetry of path lengths in spacetime as analysed from any inertial reference frame. Mostly the arguments here are semantic and philosophical as others have mentioned. The paradox can be analysed either in terms of velocity and spacetime path lengths or in terms of acceleration. Both methods are equally valid mathematically, but the acceleration explanation can be confusing to newcomers. You can calculate the time dilation of an accelerating particle in terms of the sum of its instantaneous velocities, but then you might ask how much additional time dilation should we factor in for the acceleration and the answer is none! On the other hand you can calculate the the time dilation of the same particle in terms of its acceleration and distance away at any instant and arrive at the same answer and in this case no additional time dilation is attributed to instantaneous velocity. They are two sides of the same coin. Basically $$v\gamma=aT_a$$ and you can either use $$v\gamma$$ or $$aT_a$$ in the calculations (but not both).

I do agree that any differential ageing involves acceleration, but in some situations where both twins experience equal proper acceleration, you also have to factor in where and when in spacetime the acceleration occurred and this basically boils down to comparing path lengths in space time.

Starthaus gives an equation where the proper time of the constant velocity leg of the journey is a function of the acceleration at the turn around (which is mathematically correct when he gets his equation right), but philosophically it is not ideal, because it implies that the proper time that elapses during the constant velocity leg is applied retrospectively once we find out what the acceleration is at the turn around. Let us say that the twin launches away from Earth with constant velocity and passes Alpha Centauri at constant velocity. What is the proper time that elapses on the traveling twins clock between leaving Earth and passing Alpha Centauri? The acceleration explanation can not give an answer, because it does not know where,when or how quickly the twin is going to turn around, unless the clock has some sort of predictive ability. Note that there is a definitive proper time elapsed on the traveling clock between the events "passing Earth" and "passing Alpha Centauri" even if the traveling clock was always moving at constant velocity and no acceleration was involved.

 

[Passionflower]

The paradox can be analysed either in terms of velocity and spacetime path lengths or in terms of acceleration. Both methods are equally valid mathematically, but the acceleration explanation can be confusing to newcomers.

Really? So then please explain to me mathematically how without any acceleration (proper or inertial) you can say which twin ages more. Use any example you like.

Mostly the arguments here are semantic and philosophical as others have mentioned.

They become philosophical, and frankly irrelevant, as soon as one completely ignores acceleration as there is absolutely nothing to calculate in that case.

 

[JesseM]

The correct equation for the proper time $$d \tau$$ of the traveling twin with cruising time $$T_c$$ and acceleration time $$T_a$$ (both as measured in the inertial stay at home twins rest frame) can be stated as a function of v only like this:

$$d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, asinh(v \gamma/c)$$

Where v is the cruising velocity and $$\gamma$$ is $$1/\sqrt{1-v^2/c^2}$$.

It depends on how you choose to express things, your method avoids having to know the value of the proper acceleration while starthaus' equation assumed we did know it. If we do know the value of the proper acceleration a, we also know the rocket will be decelerating from v to 0 for T_a/2 and accelerating from 0 to v for T_a/2, and according to the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html , the proper time for a rocket that accelerates with constant proper acceleration a starting from a speed of 0 would be given by tau = (c/a)*arcsinh(aT/c), so for T=T_a/2 we get tau = (c/a)*arcsinh(aT_a/2c), and by the symmetry we can double that to get the total proper time elapsed from the beginning of the acceleration to the end, giving tau = (2c/a)*arcsinh(aT_a/2c). So, I get a total elapsed time of:

$$d \tau=\frac{T_c}{\gamma}+\frac{2c}{a} \, \, arcsinh(a T_a / 2c)$$

If we want to express it without reference to a, another equation on the relativistic rocket page says that velocity v as a function of coordinate time T would be v(T) = aT / sqrt[1 + (aT/c)^2], so if we know it takes a time of T_a/2 to go from 0 to the cruising speed v, that means that v = aT_a / 2 * sqrt[1 + (aT_a/2c)^2], which means:

v^2 = a^2 T_a^2 / 4*[1 + (a^2*T_a^2 / 4c^2)]
4v^2 + 4*v^2*a^2*T_a^2 / 4c^2 = a^2 * T_a^2
4v^2 + (v^2/c^2)*a^2*T_a^2 = a^2*T_a^2
4v^2 = a^2*T_a^2*(1 - v^2/c^2)
a^2 = 4v^2 / (T_a^2 * (1 - v^2/c^2))
a = 2v*gamma/T_a

Substituting that into my equation above:

$$d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, arcsinh(v\gamma /c)$$

...which is the same as the equation you wrote down.

edit: whoops, I read your post hastily and thought you were objecting to the part of starthaus' post where he wrote the equation $$\frac{c}{a} arcsinh(aT_a/c)$$ for the proper time during the acceleration phase, but now I think you probably objecting to this other equation starthaus posted:

$$d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} asinh(aT_a/c)$$

Maybe in this equation starthaus is actually letting T_a represent half the acceleration period, which would explain why he uses asinh(aT_a/c) rather than asinh(aT_a/2c) as I did, but why is there a 4 out in front? I don't know where the first term comes from either...

 

[starthaus]

The correct equation for the proper time $$d \tau$$ of the traveling twin with cruising time $$T_c$$ and acceleration time $$T_a$$ (both as measured in the inertial stay at home twins rest frame) can be stated as a function of v only like this:

$$d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, asinh(v \gamma/c)$$

Where v is the cruising velocity and $$\gamma$$ is $$1/\sqrt{1-v^2/c^2}$$.

If you do the proper calculations, you'll find out that $$v\gamma=aT_a$$

Your equation is wrong

Err, no.

and predicts the proper time of the traveling twin to be longer than the stay at home twins' proper time.

Err, also no.

Basically you screwed up the factors when hacking your derivation. You didn't get it from any textbook did you?

No, I didn't hack it and I didn't get it from a textbook.

The proper time that elapses on the traveling twins clock in years is:

$$d \tau=10*0.6+10*0.6/0.8* \,asinh(0.8/0.6) = 6+8.23959 = 14.2396$$

...which clearly contradicts your earlier claim that the contribution of the acceleration period is negligible.

 

[starthaus]

$$d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} asinh(aT_a/c)$$

I guess in this equation starthaus is actually letting T_a represent half the acceleration period, so the second term is correct. Not sure where he got the first term though...

...from knowing math and physics

 

[JesseM]

Then the first thing to teach should be that the $\gamma$ factor between two observers is determined by their acceleration both proper and inertial.

Not necessarily, what about two observers who pass one another moving inertially in space and note their ages as they pass? And even if you assume they started at rest relative to one another, if you assume acceleration was quasi-instantaneous (as is commonly assumed in these kinds of problems to make the math simpler), then the question of which one accelerated initially is irrelevant to figuring out their elapsed time, all that matters is which one accelerated to turn around once they were a significant distance apart. Then all you have to do is pick a frame, find the velocity of the one that turned around both before and after the turnaround (as well as the time between departure and turnaround in the frame you chose), as well as the velocity of the one that moved inertially between departure and reunion, and from that you can figure out how much elapsed time both have experienced between departure and reunion.

 

[yuiop]

Substituting that into my equation above:

$$d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, arcsinh(v\gamma /c)$$

...which is the same as the equation you wrote down.

edit: whoops, I read your post hastily and thought you were objecting to the part of starthaus' post where he wrote the equation $$\frac{c}{a} arcsinh(aT_a/c)$$ for the proper time during the acceleration phase, but now I think you probably objecting to this other equation starthaus posted:

$$d \tau=2T_c/ \sqrt{1+(aT_a/c)^2}+4\frac{c}{a} asinh(aT_a/c)$$

I guess in this equation starthaus is actually letting T_a represent half the acceleration period, so the second term is correct. Not sure where he got the first term though...

If you start with my equation:

$$d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, arcsinh(v\gamma /c)$$

and substitute $$aT_a$$ for $$v\gamma$$ and $$\sqrt{1+(aT_a/c)^2}$$ for $$\gamma$$ you end up with the equation Starhaus should have ended up with if hadn't blundered:

$$d \tau=T_c/ \sqrt{1+(aT_a/c)^2}+\frac{c}{a} arcsinh(aT_a/c)$$

Even if you allow for the fact that Starhaus might of meant T_c and T_a were the the cruising time and accelerating time of the the outwrad leg only, rather than the round trip time, the equation he should have ended up with is:

$$d \tau= 2T_c/ \sqrt{1+(aT_a/c)^2}+2\frac{c}{a} arcsinh(aT_a/c)$$

Either way, Starthaus messed up.

However, he is right that it does produce the mathematical oddity that the proper time of the outward leg is apparently determined by the acceleration at the end of the leg, which just shows you have to be careful in how you physically interpret the equations.

 

[JesseM]

...from knowing math and physics

I edited my post, the second term appears to be incorrect even if you let T_a be half the acceleration period. Were you indeed making that assumption? In that case the total proper time elapsed during the acceleration period should be $$2\frac{c}{a} \, arcsinh (aT_a/c)$$, not $$4\frac{c}{a} \, arcsinh (aT_a/c)$$.

 

[starthaus]

the equation he should have ended up with is:

$$d \tau= 2T_c/ \sqrt{1+(aT_a/c)^2}+2\frac{c}{a} arcsinh(aT_a/c)$$

Either way, Starthaus messed up.

Err, you are wrong again. The equation is correct, read here.

However, he is right that it does produce the mathematical oddity that the proper time of the outward leg is apparently determined by the acceleration at the end of the leg,

I never claimed such nonsense. You need to understand my post.

which just shows you have to be careful in how you physically interpret the equations.

Yes, you also need to be careful in getting your math right.

 

[starthaus]

I edited my post, the second term appears to be incorrect even if you let T_a be half the acceleration period. Were you indeed making that assumption? In that case the total proper time elapsed during the acceleration period should be $$2\frac{c}{a} \, arcsinh (aT_a/c)$$, not $$4\frac{c}{a} \, arcsinh (aT_a/c)$$.

You are making the same error(s) as kev. My formula is correct.

 

[JesseM]

If you start with my equation:

$$d \tau=\frac{T_c}{\gamma}+\frac{c T_a}{v \gamma} \, \, arcsinh(v\gamma /c)$$

and substitute $$aT_a$$ for $$v\gamma$$ and $$\sqrt{1+(aT_a/c)^2}$$ for $$\gamma$$

Why would you do that, though? According to my calculations (which ended up with the same equation you got), a = 2v*gamma/Ta, so that implies v*gamma = a*T_a/2, not aT_a. Did I make an error?

 

[JesseM]

You are making the same error(s) as kev. My formula is correct.

Care to point out the error? Do you agree that if we look at the second half of the acceleration period, the rocket starts from a speed of 0 and accelerates for a coordinate time of T_a in your notation? If so, the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html page says that if a rocket starts from 0 and accelerates for a coordinate time of t, its elapsed proper time (which they denote with T) is:

T = (c/a) sh^-1(at/c)

Do you disagree with that equation? If not, you should presumably agree that during the second half of the acceleration the rocket experiences a proper time of (c/a)*arcsinh(aT_a/c), so by symmetry the total elapsed proper time during the acceleration phase is just double that.