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

[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.

 

[starthaus]

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?

Yes, you did, you need to think about the definition of "cruising" speed.

 

[starthaus]

Care to point out the error?

Please read my answer to kev's alleged finding an error in my formula. You are both missing two periods of acceleration.
The formulas in post #80 are correct. Yours and kev's are not.

 

[Passionflower]

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.

Obviously because you measure the elapsed time between the event when they pass with a relative speed and when they return. What happens before is irrelevant.

all that matters is which one accelerated to turn around once they were a significant distance apart.

Exactly, acceleration again!

 

[JesseM]

Please read my answer to kev's alleged finding an error in my formula. You are both missing two periods of acceleration.

OK, so you're saying that the rocket also initially accelerated for a time T_a to achieve the cruising speed of v, and then at the end decelerated for a time of T_a as well. Still, I trust you have no objection to my and kev's equations in a different scenario where the rocket just passed by the "stationary" observer (i.e the one who remains inertial and has a velocity of zero in the frame we're considering) already moving at speed v, continues to cruise for time T_c/2, then accelerates to turn around for time T_a, then cruises back towards the "stationary" observer, passing him after another period of T_c/2 (and they compare clocks as they pass).

Anyway, given the scenario you are considering with initial acceleration and final deceleration, I have no objection to the second term, let me think about the first term...but before I do that, can you tell me if T_c represents the total cruising time in both directions, or just the cruising time in one direction?

 

[starthaus]

Anyway, given the scenario you are considering with initial acceleration and final deceleration, I have no objection to the second term, let me think about the first term...but before I do that, can you tell me if T_c represents the total cruising time in both directions, or just the cruising time in one direction?

one direction, this is why it gets doubled in the final formula

 

[yuiop]

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?

I think it only because you are considering a,v and gamma for half the acceleration phase while I am considering the total acceleration phase. We end up with the same results.

 

[yuiop]

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

I was talking about a different scenario where the acceleration phase was much more extreme and took place in seconds. With very extreme acceleration with the acceleration phase period tending to zero, the time dilation due to acceleration becomes negligable. The greater the acceleration is, the more you can ignore it.

 

[JesseM]

one direction, this is why it gets doubled in the final formula

OK, thanks. Then the starting equation in your scenario and with your definitions would be:

d \tau=\frac{2T_c}{\gamma}+\frac{4c}{a} \, \, arcsinh(a T_a / c)

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 to go from 0 to the cruising speed v, that means that v = aT_a / sqrt[1 + (aT_a/c)²], so:

v² = a²*T_a² / (1 + a²*T_a²/c²)

v² = c² * a² * T_a² / (c² + a²*T_a²)

v²/c² = a² * T_a² / (c² + a²*T_a²)

(1 - v²/c²) = [(c² + a²*T_a²) - (a² * T_a²)] / (c² + a²*T_a²) = c² / (c² + a²*T_a²) = 1 / (1 + a²*T_a²/c² )

gamma = 1/sqrt(1 - v²/c²) = sqrt(1 + (aT_a/c)²)

So plugging that into the top equation,

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

...which agrees with what you had.

 

[Passionflower]

The greater the acceleration is, the more you can ignore it.

Sorry but this is just absurd, you cannot ignore acceleration, great or small, in the twin experiment. Without acceleration there is no absolute time dilation.

I am still waiting for an example of the twin experiment without acceleration but "in terms of velocity and spacetime path lengths", which, if I understand correctly, in your view does not require acceleration.

 

[yuiop]

It is obvious that the equations given by Starthaus are symbol for symbol copied from wikipedia, so it would have have been helpful if he credited wikipedia and gave a link to where the wikipedia scenario is described less vaguely than Starthaus described it.

In the wikipedia scenario there are 4 acceleration periods and 2 cruising periods so that explains why our equations differ, because the scenarios and definitions of the periods differ.

I think you will find if you use my equations with T_a and T_c defined as total times for the whole journey, you can use the same equation for any of number of acceleration and cruising phases, without having to have different equations for different twins scenarios, as long as the acceleration is the same for all the acceleration phases.

 

[yuiop]

Sorry but this is just absurd, you cannot ignore acceleration, great or small, in the twin experiment. Without acceleration there is no absolute time dilation.

I am still waiting for an example of the twin experiment without acceleration but "in terms of velocity and spacetime path lengths", which, if I understand correctly, in your view does not require acceleration.

In an earlier reply to you I acknowledged that in any situation where differential ageing occurs then acceleration is involved, but you can calculate the time dilation without using the acceleration. That is not the same as saying acceleration is not required.

 

[Passionflower]

In the wikipedia scenario there are 4 acceleration periods and 2 cruising periods so that explains why our equations differ, because the scenarios and definitions of the periods differ.

I have no desire to be squeezed between Starthaus and the others in this argument but that is the situation in the standard twin "paradox", you need 4 accelerations and optionally two cruising periods..

 

[JesseM]

Sorry but this is just absurd, you cannot ignore acceleration, great or small, in the twin experiment. Without acceleration there is no absolute time dilation.

But if the acceleration is instantaneous, you don't actually need to consider the acceleration phase when calculating the elapsed proper time. If you know the ship cruised at speed v₁ for a coordinate time of t₁, then instantaneously accelerated to turn around so it had a speed of v₂ in the other direction and took an additional coordinate time of t₂ to reunite with the inertial observer, then the elapsed time for the observer that turned around is just t₁*sqrt(1 - v₁²/c²) + t₂*sqrt(1 - v₂²/c²). So in this sense, the acceleration can be ignored in your calculations.

I have no desire to be squeezed between Starthaus and the others in this argument but that is the situation in the standard twin "paradox", you need 4 accelerations and optionally two cruising periods..

Most textbook discussions of the twin paradox assume instantaneous accelerations.

 

[Passionflower]

So in this sense, the acceleration can be ignored in your calculations.

Acceleration is essential in the twin experiment. Without acceleration there is no absolute time dilation.

Show me a calculation where you completely ignore acceleration (proper or inertial) that shows an absolute time dilation and I show you where you made a mistake!

 

[JesseM]

Acceleration is an essential in the twin experiment. Without acceleration there is no absolute time dilation.

Show me a calculation where you completely ignore acceleration (proper or inertial) that shows an absolute time dilation and I show you where you made a mistake!

I don't know what you mean by "completely ignore", but by "ignore" I only mean that you don't have to consider the contribution that the accelerating phase makes to the total elapsed proper time, since you are treating the acceleration as instantaneous (I think my meaning was fairly clear from the context, especially given my comment 'you don't actually need to consider the acceleration phase when calculating the elapsed proper time'). In that sense the calculation I already gave you in my last post ignores acceleration, though of course acceleration plays a role in that it explains why v₁ on the outbound leg may be different than v₂ on the inbound leg (and since inertial paths are geodesics and geodesics always maximize proper time, it plays a conceptual role in understanding why the inertial twin is always the one who ages more than the one who turns around, similar to the idea that a straight line between two points in Euclidean geometry always has a shorter length than any bent path between the same points).

 

[Passionflower]

it plays a conceptual role in understanding why the inertial twin is always the one who ages more than the one who turns around).

Indeed, that is why it very important in STR education to stress that fact!

 

[yuiop]

Sorry but this is just absurd, you cannot ignore acceleration, great or small, in the twin experiment. Without acceleration there is no absolute time dilation.

The proper time of the traveling twin can be calculated using

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

Where T_c and Ta are total cruise and acceleration times, then when a is very large and Ta is necessarily brief because the terminal velocity v is reached very quickly, the second term goes to zero and this is why some textbooks assume instantaneous acceleration and the error due to ignoring the the time dilation during the acceleration phase is negligible. The equation can not be reversed to consider the point of view of the accelerating twin, because it is only valid for an inertial observer.

 

[starthaus]

I was talking about a different scenario where the acceleration phase was much more extreme and took place in seconds.

I know, I pointed out to you that doing so is incorrect.
Not only that your math was incorrect but your physics was unrealistic. Calculate the acceleration that woulg get your rocket to your 0.8c in 1s . What do you get?

With very extreme acceleration with the acceleration phase period tending to zero, the time dilation due to acceleration becomes negligable.

You can't get high speeds in short time (see the exercise above)

The greater the acceleration is, the more you can ignore it.

Err, this one is in the category "not even wrong".

 

[starthaus]

The proper time of the traveling twin can be calculated using

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

Where T_c and Ta are total cruise and acceleration times, then when a is very large and Ta is necessarily brief because the terminal velocity v is reached very quickly, the second term goes to zero [/color]

No, it doesn't. Do the exercise I gave you and you'll find how false it is.
As a simple alternative, use a=10g (a huge number) and evaluate the second term. What did you get?

The equation can not be reversed to consider the point of view of the accelerating twin, because it is only valid for an inertial observer.

This is false as well. There are papers on this subject. Try to get your facts straight.

 

[JesseM]

Indeed, that is why it very important in STR education to stress that fact!

If you're talking about the conceptual explanation for why one twin ages less than the other, of course. But from the context of kev's comments it was clear he was just saying the acceleration phase didn't always need to be considered in actual calculations of elapsed proper time. Also, when you originally objected to "acceleration is not relevant" type arguments in post #89, it followed on the heels of a discussion I was having with starthaus where we were talking about the clock hypothesis, and there it is important to emphasize the reverse, that acceleration is indeed completely irrelevant if you want to know the instantaneous rate of a clock at a single instant, the rate is solely a function of its instantaneous velocity.

 

[JesseM]

I know, I pointed out to you that doing so is incorrect.
Not only that your math was incorrect but your physics was unrealistic. Calculate the acceleration that woulg get your rocket to your 0.8c in 1s . What do you get?

Virtually all textbook calculations involve simplifications, and instantaneous accelerations (which imply infinite proper acceleration) are routinely assumed in textbooks on SR. Anyway this sort of simplification would be a reasonable approximation if the acceleration phase only lasted a few days or weeks while the inertial legs lasted for years, which might be realistic for an interstellar journey.

The equation can not be reversed to consider the point of view of the accelerating twin, because it is only valid for an inertial observer.

This is false as well. There are papers on this subject. Try to get your facts straight.

I assume kev meant that the equation you posted wouldn't accurately calculate elapsed proper time if we were using the time coordinate of a non-inertial frame, which is correct.

 

[Passionflower]

the instantaneous rate of a clock at a single instant, the rate is solely a function of its instantaneous velocity.

Velocity with respect to what?

If your answer is with the prior instant, than it is the acceleration between the two instants that changed the rate! So what causes clock rates to change? Acceleration!

If the answer is different I am happy to await your further explanation.

 

[starthaus]

I was having with starthaus where we were talking about the clock hypothesis, and there it is important to emphasize the reverse, that acceleration is indeed completely irrelevant if you want to know the instantaneous rate of a clock at a single instant, the rate is solely a function of its instantaneous velocity.

No one is disputing the "clock hypothesis". What I and passionflower are disputing is the incorrect statements made by kev.

 

[starthaus]

Virtually all textbook calculations involve simplifications, and instantaneous accelerations (which imply infinite proper acceleration) are routinely assumed in textbooks on SR. Anyway this sort of simplification would be a reasonable approximation if the acceleration phase only lasted a few days or weeks while the inertial legs lasted for years, which might be realistic for an interstellar journey.

It would be good if you stopped defending kev's hacks for a while and you let him answer,. By stepping in for him every time you preclude him from learning.

I assume kev meant that the equation you posted wouldn't accurately calculate elapsed proper time if we were using the time coordinate of a non-inertial frame, which is correct.

This is not what he claimed.

 

[JesseM]

Velocity with respect to what?

An inertial coordinate system.

If your answer is with the prior instant

How could you talk about a "velocity" with respect to an "instant"? I'm guessing you meant to ask "rate of a clock with respect to what"? If so, the answer would again be relative to the time coordinate of an inertial frame (i.e. d\tau /dt). The clock hypothesis says that this only depends on the instantaneous velocity in that frame (it's always true that d\tau /dt = \sqrt{1 - v^2/c^2}), it doesn't depend on instantaneous acceleration or instantaneous position or anything else.

 

[JesseM]

Virtually all textbook calculations involve simplifications, and instantaneous accelerations (which imply infinite proper acceleration) are routinely assumed in textbooks on SR. Anyway this sort of simplification would be a reasonable approximation if the acceleration phase only lasted a few days or weeks while the inertial legs lasted for years, which might be realistic for an interstellar journey.

It would be good if you stopped defending kev's hacks for a while and you let him answer,. By stepping in for him every time you preclude him from learning.

Well, perhaps I object to your silly arguments because you use a lot of the same type of silly arguments on me, like how the irrelevant practical concerns you raise about realistic accelerations here resemble the irrelevant practical concerns about the size of real gaps in train tracks in the train-on-a-bridge thread. If you cracked open a real physics textbook once in a while instead of thinking you can learn everything about physics from from wikipedia, you might see that it's universal to include all sorts of simplifications in order to illustrate the basic physical concepts involved.

Irrelevant practical concerns aside, can you point to a single error in math or physics kev has made in this discussion?

I assume kev meant that the equation you posted wouldn't accurately calculate elapsed proper time if we were using the time coordinate of a non-inertial frame, which is correct.

This is not what he claimed.

It isn't? What do you think he meant by "the point of view of the accelerating twin", if not a non-inertial rest frame for the accelerating twin?

 

[Passionflower]

An inertial coordinate system.

A change in velocity in an inertial coordinate system is equivalent with acceleration. Do you actually agree with me that a change in velocity in an inertial coordinate system causes a change in clock rate?

 

[JesseM]

A change in velocity in an inertial coordinate system is equivalent with acceleration. Do you actually agree with me that a change in velocity in an inertial coordinate system causes a change in clock rate?

Sure, why wouldn't I? The clock hypothesis says the instantaneous clock rate depends on the instantaneous velocity (so the instantaneous acceleration is irrelevant), which means if you're considering some extended interval of time where the clock is accelerating, the instantaneous velocity will be different at the beginning of the interval than the end, so the clock rate will be different too.

 

[starthaus]

Well, perhaps I object to your silly arguments because you use a lot of the same type of silly arguments on me,

Perhaps this is because the two of you share the same misconceptions and , occasionally, the same errors in computations.

like how the irrelevant practical concerns you raise about realistic accelerations here resemble the irrelevant practical concerns about the size of real gaps in train tracks in the train-on-a-bridge thread.

...and the same penchant for dragging out past discussions where we disagreed.. This shows a severe smallness of character.

If you cracked open a real physics textbook once in a while instead of thinking you can learn everything about physics from from wikipedia, you might see that it's universal to include all sorts of simplifications in order to illustrate the basic physical concepts involved.

I happen to have a lot of books and I have cited them repeatedly, so your remark is both insulting and unwarranted. I have cited Rindler repeatedly on GR. My favorite book that I cite from is Moller precisely because he doesn't resort to any of the hacks that you and kev seem to favor.

Irrelevant practical concerns aside, can you point to a single error in math or physics kev has made in this discussion?

Read the thread, I pointed them out as he made them.

It isn't? What do you think he meant by "the point of view of the accelerating twin", if not a non-inertial rest frame for the accelerating twin?

It means that the calculation , in kev's incorrect opinion, cannot be made from the perspective of the accelerated twin. This is false.

 

[starthaus]

Sure, why wouldn't I? The clock hypothesis says the instantaneous clock rate depends on the instantaneous velocity (so the instantaneous acceleration is irrelevant), which means if you're considering some extended interval of time where the clock is accelerating, the instantaneous velocity will be different at the beginning of the interval than the end, so the clock rate will be different too.

...meaning that the elapsed proper time is a function of proper acceleration, as demonstrated by the formula I've posted. To make matters even more interesting, when one does the calculations from the perspective of the accelerating twin, the elapsed time definitely depends on acceleration. There is an excellent paper by H.Nikolic (who happens to be the excellent physics forums science advisor called Demystifier) on this subject on this subject. There is another one by Minguzzi and one by Iorio, I suggest that you read them, they would go a long way in correcting your misconceptions about the acceleration role on the elapsed proper time. Here, start with the Nikolic one. I am attaching it such that you don't continue to claim that I do my learning from wiki.

 

[Passionflower]

Sure, why wouldn't I?

First of all I do not know why you constantly raise the clock hypothesis, as far as I see nobody is contesting this here.

Ok, good, so let me ask you this: Would you then agree that \gamma in this case directly depends on the rate and duration of the acceleration? And if so, do you readily see that since \gamma is a factor in the formula to calculate the absolute time dilation that effectively the rate and duration of acceleration is a factor, not the only factor, but nevertheless a factor, of the absolute time dilation?

 

[JesseM]

Perhaps this is because the two of you share the same misconceptions and , occasionally, the same errors in computations.

Point to any "misconceptions" or "errors" I have made. The only reason for the initial difference between your equation and mind (and kev's) is that we made different physical assumptions, I assumed the only acceleration was at the turnaround while you assumed (or copied your equation from a wikipedia page which assumed) an initial and final acceleration as well. If you think I have made any physical or mathematical errors aside given my physical assumptions, please point them out instead of just making vague accusations.

...and the same penchant for dragging out past discussions where we disagreed.. This shows a severe smallness of character.

I just like to point out how you use the same irritating tactics in thread after thread.

I happen to have a lot of books and I have cited them repeatedly, so your remark is both insulting and unwarranted. I have cited Rindler repeatedly on GR.

And do you think that's a book that avoids simplifying assumptions, of the kind you call "hacks"?

My favorite book that I cite from is Moller precisely because he doesn't resort to any of the hacks that you and kev favor.

I don't know which book you're referring to, what's the title?

Read the thread, I pointed them out as he made them.

I only saw you object on the basis that he made different physical assumptions than you, or that he used "hacks" (simplifications of the type that are routinely used in textbook discussions), or that you interpreted an English statement by him in a silly uncharitable way (like the one below). Again, show me a single clear error in his actual calculations.

It means that the calculation , in kev's incorrect opinion, cannot be made from the perspective of the accelerated twin. This is false.

No, he didn't say a calculation of elapsed time couldn't be made, he just said you couldn't use the same equation ('The equation can not be reversed') to calculate the elapsed time from the point of view of the accelerated twin, where I presume by "point of view" he meant a non-inertial frame (especially since he went on to say that the equation can't be reversed 'because it is only valid for an inertial observer').

 

[starthaus]

And do you think that's a book that avoids simplifying assumptions, of the kind you call "hacks"?

Most of the time.

I don't know which book you're referring to, what's the title?

C.Moller "The Theory of Relativity"

I only saw you object on the basis that he made different physical assumptions than you, or that he used "hacks" (simplifications of the type that are routinely used in textbook discussions), or that you interpreted an English statement by him in a silly uncharitable way (like the one below). Again, show me a single clear error in his actual calculations.

Look thru the thread, there are plenty. You can look for the "Err,no".

No, he didn't say a calculation of elapsed time couldn't be made, he just said you couldn't use the same equation ('The equation can not be reversed') to calculate the elapsed time from the point of view of the accelerated twin, where I presume by "point of view" he meant a non-inertial frame (especially since he went on to say that the equation can't be reversed 'because it is only valid for an inertial observer').

You tend to "presume" a lot.

 

[JesseM]

First of all I do not know why you constantly raise the clock hypothesis, as far as I see nobody is contesting this here.

I didn't say anyone was contesting it. But starthaus brought it up in post #80, and I was just making the point in the next few posts after that that it could be misleading to both bring up the clock hypothesis and at the same time make statements like the one that the elapsed time will "depend on the acceleration a". See my summary of the point I was making in post #92

Ok, good, so let me ask you this: Would you then agree that \gamma in this case directly depends on the rate and duration of the acceleration?

Gamma in what case? Gamma at any given instant of course is just a function of the velocity at that instant, so presumably you mean "depends on" in some other way, like that the velocity at a given instant itself depends on the clock's history of past accelerations (in that sense one could even say that gamma for a man in a rocket 'depends on' his decision 20 years earlier to enter the space academy and become an astronaut)

And if so, do you readily see that since \gamma is a factor in the formula to calculate the absolute time dilation that effectively the rate and duration of acceleration is a factor, not the only factor, but nevertheless a factor, of the absolute time dilation?

What does "absolute time dilation" mean? Are you talking about elapsed proper time rather than instantaneous clock rates? And by "factor", do you just mean some sort of causal factor, or do you mean something more specific like that we must include a variable representing acceleration in our calculations?

 

[Passionflower]

If a formula is wrong it should be called out without fear of personal retribution, if it is called incorrectly wrong it should be called out as well again without fear of personal retribution.

Could we please get back on track and leave personal remarks out?

 

[Passionflower]

Gamma at any given instant of course is just a function of the velocity at that instant, so presumably you mean "depends on" in some other way, like that the velocity at a given instant itself depends on the clock's history of past accelerations (in that sense one could even say that gamma for a man in a rocket 'depends on' his decision 20 years earlier to enter the space academy and become an astronaut)

If a person undergoes constant proper acceleration in an inertial frame of reference his gamma increases. You in all honesty do not see any connection of the increase in gamma with acceleration?

Does the formula:

\Delta \gamma = \Delta xa_p

Mean anything to you?

Another interesting relation between gamma and acceleration is:

\gamma = (a_pa_c)^{1/3}

Where a_p is proper acceleration and a_c is coordinate acceleration.

 

[JesseM]

Most of the time.

So for example, would you complain about the lack of realism in the scenario of a man running towards a barn at 0.866c in the barn-pole paradox on p. 63? Is a rocket that takes a very brief time to accelerate from relativistic velocity in one direction to a relativistic velocity in the opposite direction significantly more unrealistic than a man running towards a barn at 0.866c?

C.Moller "The Theory of Relativity"

Conveniently this book is available online in pdf form. Looking at the section where he discusses what he calls the "clock paradox", note that on p. 260 of the book (p. 278 of the pdf file), he considers the limit as the acceleration goes to infinity and "the velocity v is attained nearly instantaneously", and derives some equations for this case. He again talks more about infinite acceleration on pp. 261-262 (279-280 of the pdf). I suppose you would lecture Moller that this is a pointless exercise, since "realistically" no ship could withstand such great accelerations?

Look thru the thread, there are plenty. You can look for the "Err,no".

I looked back over all the posts from p. 6 on (starting with post #81), the only "err, no"'s you wrote in response to actual equations or quantitative statements kev made were when kev posted your equation and said it was wrong, but he thought it was wrong because he misunderstood the scenario you were considering (which you never spelled out), not realizing you were supposing an initial and final acceleration as well as a turnaround. So, looks like no actual physical/mathematical errors, just a misunderstanding of what scenario was being analyzed. And I note you also didn't respond to my request to back up your claim that I had made errors.

No, he didn't say a calculation of elapsed time couldn't be made, he just said you couldn't use the same equation ('The equation can not be reversed') to calculate the elapsed time from the point of view of the accelerated twin, where I presume by "point of view" he meant a non-inertial frame (especially since he went on to say that the equation can't be reversed 'because it is only valid for an inertial observer').

You tend to "presume" a lot.

So you weren't "presuming" when you interpreted him to be saying "the calculation ... cannot be made from the perspective of the accelerated twin"? Do you have some infallible mind-reading powers I don't know about? If not, hopefully now you can see that you were jumping to conclusions, and that my alternate interpretation above is consistent with the words he used. But perhaps kev himself can settle the matter and tell us which interpretation is closer to what he meant.

 

[JesseM]

If a person undergoes constant proper acceleration in an inertial frame of reference his gamma increases. You in all honesty do not see any connection of the increase in gamma with acceleration?

You didn't say anything about "increase in gamma" or any other change in gamma with time, you just said "gamma in this case [which I took to mean the value of gamma at some instant] directly depends on the rate and duration of the acceleration". Your reading of my statements is really pretty uncharitable if you think I don't know perfectly well that a continuous acceleration will produce an increase in speed over time which will produce an increase in gamma over time. I think I was pretty clear on the fact that I was talking about the value of gamma at any particular moment, which is just a function of the velocity at that moment and doesn't depend on acceleration or past history (i.e. two clocks that have the same instantaneous velocity at some time also, I'm sure you'd agree, have the same value for gamma and the same instantaneous rate of ticking at that time, even if their histories are different and they are accelerating differently)

Does the formula:

\Delta \gamma = \Delta xa_p

Mean anything to you?

I haven't seen that one before, no. If a rocket is undergoing constant proper acceleration in the +x direction, and it started at x=0 at time t=0 when its velocity was also zero, then one of the http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html says that its position at time t will be:

x = (c²/a) (sqrt[1 + (at/c)²] - 1) = (c²/a)*sqrt[1 + (at/c)²] - (c²/a)

And also

gamma = sqrt[1 + (at/c)²]

So, it doesn't seem to me that your equation is correct, at least not if we are considering the change in x and gamma from t=0 to some later time. It would rather be true in this case that:

\Delta \gamma = (a/c^2) * (\Delta x + (c^2 / a))

But even this equation might not continue to be true if we considered the change in gamma and x between two times where the first time wasn't t=0 when the rocket had an initial velocity of 0.

Another interesting relation between gamma and acceleration is:

\gamma = (a_pa_c)^{1/3}

Where a_p is proper acceleration and a_c is coordinate acceleration.

Hmm, the coordinate acceleration at time t would be the derivative with respect to t of the coordinate velocity, which is given by v = at * (1 + (at/c)²)^-1/2. By the product rule, the derivative of this is equal to:

a * (1 + (at/c)²)^-1/2 + at * d/dt (1 + (at/c)²)^-1/2

And by the chain rule, d/dt (1 + (at/c)²)^-1/2 = (-1/2)*(1 + (at/c)²)^-3/2 * 2(a/c)²t = -t*(a/c)²*(1 + (at/c)²)^-3/2

So plugging that back in, the coordinate acceleration looks to be:

a*(1 + (at/c)²)^-1/2 - a*t²*(a/c)²*(1 + (at/c)²)^-3/2

= a*(1 + (at/c)²)^-1/2*[1 - (at/c)²*(1 + (at/c)²)^-1]

= a*(1 + (at/c)²)^-1/2*[(1 + (at/c)²) - (at/c)²]/(1 + (at/c)²)

= a*(1 + (at/c)²)^-1/2/(1 + (at/c)²)

= a*(1 + (at/c)²)^-3/2

So to get back gamma I think you would have to take the coordinate acceleration, divide by the proper acceleration, and then put the result to the -1/3 power, i.e.

\gamma (t) = (a_c (t) / a_p)^{-1/3}

Note that this equation would only hold under the specialized conditions of constant proper acceleration and a velocity of 0 at t=0, though. Two clocks with the same coordinate acceleration and proper acceleration at a given instant might have different values for gamma if they hadn't both had the same constant proper acceleration since t=0 and/or hadn't started from a velocity of 0 at t=0.

 

[matheinste]

C.Moller "The Theory of Relativity"

The book of that title and author that I have, dated 1953, although very good, treats the basics of SR pretty much the same as most of the other texts, and I have at least 25 of them to compare it with. One little complaint I have about it is that it is one of the few texts that claims that the twin paradox cannot be resolved within SR. Perhaps that is due to the era in which the book was written, but if this is the only text a beginner read, although unlikely in the present day, he or she would be misled in this respect.

Matheinste.

 

[starthaus]

And I note you also didn't respond to my request to back up your claim that I had made errors.

Sure I did, read post 135. If, after reading the paper, you will continue to maintain that the elapsed time differential is not dependent on acceleration, I will provide you with more papers published in peer reviewed journals that contradict your misconception. Read the paper by Nikolic first, he's one of the best science advisors in this forum (next to DrGreg).

 

[Passionflower]

So to get back gamma I think you would have to take the coordinate acceleration, divide by the proper acceleration, and then put the result to the -1/3 power, i.e.

\gamma (t) = (a_c (t) / a_p)^{-1/3}

Yes you are correct, I missed the division operator in the latex.

 

[starthaus]

I looked back over all the posts from p. 6 on (starting with post #81), the only "err, no"'s you wrote in response to actual equations or quantitative statements kev made were when kev posted your equation and said it was wrong, but he thought it was wrong because he misunderstood the scenario you were considering (which you never spelled out), not realizing you were supposing an initial and final acceleration as well as a turnaround. So, looks like no actual physical/mathematical errors, just a misunderstanding of what scenario was being analyzed.

This is not true, look again at posts 97 (last entry) and 124.

 

[starthaus]

Conveniently this book is available online in pdf form. Looking at the section where he discusses what he calls the "clock paradox", note that on p. 260 of the book (p. 278 of the pdf file), he considers the limit as the acceleration goes to infinity and "the velocity v is attained nearly instantaneously", and derives some equations for this case. He again talks more about infinite acceleration on pp. 261-262 (279-280 of the pdf). I suppose you would lecture Moller that this is a pointless exercise, since "realistically" no ship could withstand such great accelerations?

Conveniently, you left out the fact that Moller, before taking the limit, calculates the exact formula of elapsed proper time:

\tau'_2=\frac{c}{g} sinh^{-1}(\frac{g \Delta T}{c}) (eq 158)

where g represents ... acceleration. So, Moller doesn't cut any corners, he derived the exact formula. Selective quoting makes for very bad science, or no science at all.
I have no objection when a true scientist derives an exact formula only to take it to the limit afterwards. I have very strong objections when someone uses the truncated formula as a starting point. This is why I disagree with you and kev.

 

[starthaus]

One little complaint I have about it is that it is one of the few texts that claims that the twin paradox cannot be resolved within SR.

I looked on page 260 and he uses the hyperbolic motion treatment, consistent with SR.He doesn't claim that the paradox can be treated using GR only I like his book very much, he's a model of rigor.

 

[matheinste]

I looked on page 260 and he uses the hyperbolic motion treatment, consistent with SR.He doesn't claim that the paradox can be treated using GR only I like his book very much, he's a model of rigor.

Look at page 49. It is not a simple "we cannot" statement so perhaps I have misinterpreted it. What do you think of it?

Matheinste.