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

[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.

 

[yuiop]

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.

I think what JesseM meant was that the instantaneous rate of a moving clock relative to the clocks at rest in reference frame S, is solely a function of its instantaneous velocity relative to the clocks at rest reference frame S.

For example, if an clock is moving at constant velocity 0.8c relative to frame S, then its instantaneous clock rate relative to clocks in S is 0.6. A similar clock accelerating at 1,000,000g will also have an instantaneous clock rate of 0.6 relative to clocks at rest in S at the instant its velocity relative to S is 0.8c. The acceleration has no effect at all on its instantaneous relative clock rate. The "instant" is here defined as the time measured simultaneously by synchronised clocks at rest in frame of S.

 

[starthaus]

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.

You are right, he does claim that, nevertheless, on page 260 he treats the problem using hyperbolic motion just fine. Could it be that he considers accelerated motion as part of GR? This is why I only trust math when it comes to physics, I don't trust the literary part.

 

[starthaus]

I think what JesseM meant was that the instantaneous rate of a moving clock relative to the clocks at rest in reference frame S, is solely a function of its instantaneous velocity relative to the clocks at rest reference frame S.

For example, if an clock is moving at constant velocity 0.8c relative to frame S, then its instantaneous clock rate relative to clocks in S is 0.6. A similar clock accelerating at 1,000,000g will also have an instantaneous clock rate of 0.6 relative to clocks at rest in S at the instant its velocity relative to S is 0.8c. The acceleration has no effect at all on its instantaneous relative clock rate. The "instant" is here defined as the time measured simultaneously by synchronised clocks at rest in frame of S.

...all of which is perfectly correct, yet has no bearing on the discussion of the accelerated twins "paradox" where acceleration plays a key role in the final outcome.

 

[Passionflower]

I think what JesseM meant was that the instantaneous rate of a moving clock relative to the clocks at rest in reference frame S, is solely a function of its instantaneous velocity relative to the clocks at rest reference frame S.

You fail to see that a change in velocity is caused by acceleration?

 

[JesseM]

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.

I never expressed the "misconception" that "the elapsed time differential is not dependent on acceleration" (in the sense that you can find specific scenarios where the elapsed time is a function of acceleration--I would say that it's impossible to write a general expression for elapsed time as a function of acceleration regardless of the motion of the object, whereas you can write a general expression for elapsed time as a function of v(t)). Here is what you quoted me saying in post 135:

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.

Hmm, did I say anything faintly resembling "the elapsed time differential is not dependent on acceleration" there? Nope, try again to find an error in something I have actually said to you in this thread.

As for your claim about kev's supposed "error":

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.

Last entry of #97 was:

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.

But here you seem to be fantasizing about something kev said rather than correcting a real statement of his. I did a search and the only post on this thread where he used the word "negligible" prior to #97 was in #13:

That may be true, but I have shown in the last post, that the time dilation due to acceleration in the twins paradox can be reduced to a negligible error, e.g less than 4 seconds "lost" due to acceleration time dilation compared to 8 years "lost" due to constant velocity.

"Can be reduced" (by making the acceleration brief), not "always reduces to a negligible amount in all problems" (regardless of the length of the acceleration). And a later response by kev to your post #97 in post #112 also shows that this he was talking about choosing a problem where the acceleration was brief:

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.

So, strike one on finding a real technical error in kev's posts. The next post of yours which you point to is #124 where you say:

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 $$T_a$$ are total cruise and acceleration times, then when a is very large and $$T_a$$ is necessarily brief because the terminal velocity v is reached very quickly, the second term goes to zero

No, it doesn't. Do the exercise I gave you and you'll find how false it is.

This is actually an example of you making a pretty clear physics error--do you not understand that as the acceleration approaches infinity the time needed to change from one cruising velocity to another approaches zero (instantaneous acceleration), which means the proper time elapsed in the acceleration phase approaches zero too? It may not look like the second term approaches 0 from the form above, but if you take into account the physical dependence of T_a on a and v (which is what kev meant above when he said '$$T_a$$ is necessarily brief because the terminal velocity v is reached very quickly') and substitute an expression for T_a as a function of a and v into the above equation, you find that the second term does indeed go to zero in the limit as a approaches infinity. Just consider the formula for velocity as a function of coordinate time:

v(T) = aT / sqrt[1 + (aT/c)²]

So if T_a is the total turnaround time, the T_a/2 is the time needed to accelerate from a speed of 0 to the final cruising speed v. So, we have:

v = aT_a / (2*sqrt[1 + (aT_a/2c)²])

So, we have:

v² = a²T_a² / (4 + 4a²T_a²/4c²)

4v² + (v²a²T_a²/c²) = a²T_a²

4v² + (v²a²T_a²/c²) = c²a²T_a²/c²

4v² = T_a²(c²a² - v²a²)/c²

T_a² = 4v²c²/[a²(c² - v²)]

T_a² = 4v²/[a²(1 - v²/c²)]

T_a = 2v/(a*sqrt[1 - v²/c²])

Now, kev actually did make one minor error in the above, if T_a is the total acceleration time then as I said in post #96 the acceleration term should be $$\frac{2c}{a} \, \, arcsinh(a T_a / 2c)$$, which only differs from his expression by those factors of 2. Anyway, if you substitute in $$T_a = \frac{2v}{a*\sqrt{ 1 - v^2/c^2}}$$, you get $$\frac{2c}{a} \, \, arcsinh(\frac{v}{c*\sqrt{ 1 - v^2/c^2}})$$. Clearly, this does approach 0 in the limit as a approaches infinity, since a appears only in the denominator of the expression outside.

 

[yuiop]

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

$$\tau'_2=\frac{c}{\gamma v} \sinh^{-1}(\frac{g \Delta T}{c})$$ (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.

The equation I gave is also exact:

$$\tau=\frac{c}{\gamma v} \, sinh^{-1}(\frac{\gamma v}{c})$$

However, Moller (like me) has cut one corner. He has assumed that the initial or final velocity is zero and so in the general case, his equation is not correct.

For example consider the following case in units of c = 1 lyr/yr.

T is the coordinate time in inertial reference frame S.
The initial velocity of a rocket is zero at time T=0 in S.
The rocket has constant proper acceleration g = 1 lys/yr^2.
The elapsed proper time $$\tau$$ of the rocket is zero at time T=0 in frame S.

After 1 year in S the elapsed proper time of the rocket is asinh(1) = 0.88137 years.
After 2 years in S the elapsed proper time of the rocket is asinh(2) = 1.44363 years.

The difference in proper times of the rocket, between year 1 and year 2 in S is asinh(2)-asinh(1)=0.56226 years, while the Moller equation predicts 0.88137 years of elapsed proper time for the rocket between year 1 and year 2 in S, because it fails to take account of the fact that the initial velocity of the rocket at the start of the second period is not zero.

The more general equation is:

$$\Delta \tau =\frac{c}{g}\, asinh\left(\frac{g T_2}{c}\right) - \frac{c}{g}\, asinh\left(\frac{g T_1}{c}\right)$$

and this only reduces to:

$$\Delta \tau = \frac{c}{g}\, sinh^{-1}(\frac{g \Delta T}{c})$$

in the limited case where v=0 at T1 or T2.

This dependency on the inital and final velocities in the general case is clearer in the velocity version of the equation when written as:

$$\Delta \tau=\frac{c\sqrt{1-v_2^2/c^2}}{v_2 } \, \, asinh\left(\frac{v_2 }{c\sqrt{1-v_2^2/c^2}}\right) - \frac{c \sqrt{1-v_1^2/c^2}}{v_1 } \, \, asinh \left(\frac{v_1 }{c\sqrt{1-v_1^2/c^2}}\right)$$

When v1 or v2 is zero, the above equation reduces to:

$$\Delta\tau=\frac{c}{\gamma \Delta v } \, \, asinh(\gamma \, \Delta v/c)$$

Since you are the one that is always harping on about equations being incorect, beacuase they are not generalised enough, you should state the limiting assumptions you are using, when you are quoting an equation that is only valid for a special case.

 

[yuiop]

I think what JesseM meant was that the instantaneous rate of a moving clock relative to the clocks at rest in reference frame S, is solely a function of its instantaneous velocity relative to the clocks at rest reference frame S.

You fail to see that a change in velocity is caused by acceleration?

I never said that change in velocity is not caused by acceleration. That is trivially true. What I actually said was:

I think what JesseM meant was that the instantaneous rate of a moving clock relative to the clocks at rest in reference frame S, is solely a function of its instantaneous velocity relative to the clocks at rest reference frame S.

 

[starthaus]

The equation I gave is also exact:

$$\tau=\frac{c}{\gamma v} \, sinh^{-1}(\frac{\gamma v}{c})$$

However, Moller (like me) has cut one corner. He has assumed that the initial or final velocity is zero and so in the general case, his equation is not correct.

Of course that the final speed is zero, how else do you expect the rocket to turn around? By jumping frames?
You didn't answer any of my questions relative to realistic cases of acceleration being finite.

 

[starthaus]

But here you seem to be fantasizing about something kev said rather than correcting a real statement of his. I did a search and the only post on this thread where he used the word "negligible" prior to #97 was in #13:

Not true. Look at post #80

 

[starthaus]

you get $$\frac{2c}{a} \, \, arcsinh(\frac{v}{c*\sqrt{ 1 - v^2/c^2}})$$. Clearly, this does approach 0 in the limit as a approaches infinity, since a appears only in the denominator of the expression outside.

Not so fast.

$$v=\frac{at}{\sqrt{1+(at/c)^2}}$$

so, $$a->oo$$ implies $$v->c$$, meaning $$\gamma->oo$$

so you are rushing to conclusions. You have a lot more work to do in order to find the answer.

This is actually an example of you making a pretty clear physics error--do you not understand that as the acceleration approaches infinity the time needed to change from one cruising velocity to another approaches zero (instantaneous acceleration), which means the proper time elapsed in the acceleration phase approaches zero too?

Actually this statement is pretty amusing, speaking of blunders. $$T_a$$ is the duration the rocket accelerates (see the wiki page), there is no reason whatsoever why $$T_a$$ should correlate to $$a$$.

I see your error, it is a repaet of the error that I flagged above:

T_a = 2v/(a*sqrt[1 - v²/c²])

What happens when $$a->oo$$? You are forgetting that , according to your starting point (see the formula for v), $$v->c$$ so, the limit is undetermined (not that this even the correct way of calculating $$T_a$$) . Your math has deserted you today.

 

[JesseM]

Not so fast.

$$v=\frac{at}{\sqrt{1+(at/c)^2}}$$

so, $$a->oo$$ implies $$v->c$$, meaning $$\gamma->oo$$

That equation only applies during the accelerating phase, but kev was specifically assuming the ship was accelerating up to a fixed cruising speed, so the greater the value of a, the shorter the time t needed to reach this speed. As long as the ship only accelerates up to some fixed v and then coasts, then whatever that v happens to be, the proper time of the accelerating phase will be $$\frac{2c}{a} \, \, arcsinh(\frac{v}{c*\sqrt{ 1 - v^2/c^2}})$$, so if we hold v fixed and let a approach infinity, the proper time of the accelerating phase approaches zero. Do you disagree?

Actually this statement is pretty amusing, speaking of blunders. $$T_a$$ is the duration the rocket accelerates (see the wiki page), there is no reason whatsoever why $$T_a$$ should correlate to $$a$$.

If you are accelerating until you reach some fixed v and then stop, then yes, there is a very good reason T_a should correlate with a. You need to actually think about the physical assumptions of the problem and not forget them when evaluating various equations.

 

[starthaus]

That equation only applies during the accelerating phase, but kev was specifically assuming the time t of the accelerating phase was approaching zero. As long as the ship only accelerates up to some fixed v and then coasts, then whatever that v happens to be, the proper time of the accelerating phase will be $$\frac{2c}{a} \, \, arcsinh(\frac{v}{c*\sqrt{ 1 - v^2/c^2}})$$, so if we hold v fixed and let a approach infinity, the proper time of the accelerating phase approaches zero. Do you disagree?

Sure I disagree, you are talking about an unphysical situation, infinite acceleration in zero time. So, what is the resulting cruising value for v?

If you are accelerating until you reach some fixed v and then stop, then yes, there is a very good reason T_a should correlate with a. You need to actually think about the physical assumptions of the problem and not forget them when evaluating various equations.

I was wondering how you will try to get out of your latest two blunders .

 

[JesseM]

I was wondering how you will try to get out of your latest two blunders .

Great, a completely non-substantive response. Do you disagree that the ship is accelerating up to some fixed cruising velocity and then stopping, and that in this case the proper time approaches zero if a approaches infinity? Or are you imagining the ship is accelerating forever, completely contrary to the stated assumptions of the problem?

 

[starthaus]

Great, a completely non-substantive response. Do you disagree that the ship is accelerating up to some fixed cruising velocity and then stopping, and that in this case the proper time approaches zero if a approaches infinity? Or are you imagining the ship is accelerating forever, completely contrary to the stated assumptions of the problem?

You have been given the description of the experiment and the associated math before.
This is a realistic situation, not the unphysical one that you are trying to make up in order to cover your mistakes. The description contains nothing close to the nonsensical infinite acceleration/zero acceleration period you claim.

 

[JesseM]

Sure I disagree, you are talking about an unphysical situation, infinite acceleration in zero time. So, what is the resulting cruising value for v?

No, we are talking about limits as acceleration approaches infinity and time approaches zero. Do you imagine it is somehow less "unphysical" to consider the limit as acceleration approaches infinity but time does not approach zero? In this case the "cruising speed" in both directions would approach the speed of light in this limit, which is not a usual assumption in any textbook version of the twin paradox, even ones that do consider acceleration as quasi-instantaneous. In fact I bet if you looked at a large collection of textbooks on relativity, you would find many that do consider the limit as acceleration goes to infinity, but not a single one that didn't also assume that the time of the acceleration phase was approaching zero so the cruising velocity would have some fixed value. For example, try looking carefully at Moller's textbook where he assumes acceleration approaches infinity, on pages 260, 261 and 262 (278 - 280 of the pdf)--do you think he is not assuming the time approaches zero so the cruising speed is fixed at some value below c?

Once again, the problem is that while you have a decent grasp of the math, your physical intuitions concerning what assumptions are standard in physics discussions and which would be considered bizarre and unusual are totally off. But aside from this, the fact is that kev spelled out the fact that he was assuming a fixed terminal velocity so that increasing a would mean decreasing T_a, so even if you go against every physics textbook that considers the limit as acceleration approaches infinity and treat this as a "strange" assumption to make rather than an utterly commonplace one, you still cannot call kev's comment incorrect on a technical level since in the case he said he was considering, the proper time in the acceleration phase would indeed approach zero. Read again his words that you were reacting to in post #124:

then when a is very large and T_a is necessarily brief because the terminal velocity v is reached very quickly, the second term goes to zero

 

[JesseM]

But here you seem to be fantasizing about something kev said rather than correcting a real statement of his. I did a search and the only post on this thread where he used the word "negligible" prior to #97 was in #13:

Not true. Look at post #80

In post #80 you quote a statement by him made in post #24. If you look at the context of the statement that "I have taken the time dilation due to acceleration into account and it is insignificant compared to the velocity time dilation", it is obvious he is saying this is true for the particular numerical example he presented in that post, he wasn't saying the time dilation in the accelerating phase would always be insignificant in any possible version of the twin paradox.

 

[starthaus]

No, we are talking about limits as acceleration approaches infinity and time approaches zero.

Which is unphysical.
The experiment has been run already, it does not involve any weird accelerations either at takeoff or at landing. The results are known, the time dilation during the acceleration periods is non-neglible:

Hafele, J.; Keating, R. (July 14, 1972). "Around the world atomic clocksbserved relativistic time gains". Science 177 (4044):

Do you imagine it is somehow less "unphysical" to consider the limit as acceleration approaches infinity but time does not approach zero?

You are trying to twist things, what I am telling you is that infinite acceleration is a desperate attempt to save the unphysical scenario that you've created.
So, you haven't answered the question, if $$a->oo$$ and $$t->0$$ what is the crusing speed? Would you please answer the question?

 

[JesseM]

No, we are talking about limits as acceleration approaches infinity and time approaches zero.

Which is unphysical.

But in post #148 you said this was OK when Moller did it:

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.

This is exactly what I did, I started with an "exact" formula (i.e. one that didn't involve any limits):

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

...and then I took the limit as a approaches infinity while we keep v fixed. This is the idea kev was talking about too in the post you're objecting to now (although he made the minor error of forgetting to include the factors of 2 in the second term). Is it OK for Moller (and many other physics textbooks) to do this but not for me and kev? Or do you wish to take back what you said above about not having any objection to Moller taking the limit as the acceleration approaches infinity as long as he derived it from a formula not based on limits?

The experiment has been run already, it does not involve any weird accelerations either at takeoff or at landing. The results are known, the time dilation during the acceleration periods is non-neglible

You are attacking fantasy strawmen again, neither I nor kev ever said the elapsed time during the acceleration phase would be negligible in all possible experiments, certainly not in the Hafele-Keating experiment where the planes were accelerating for pretty much the entire trip. The point we're making is just that you can make the elapsed time during acceleration negligible if you consider a scenario where the acceleration is large and brief.

You are trying to twist things, what I am telling you is that infinite acceleration is a desperate attempt to save the unphysical scenario that you've created.

How is it "desperate" when it's exactly the situation that kev was considering, which you said was incorrect on a technical level? He said "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", the phrase "goes to zero" clearly indicates he was talking about the limit as a gets larger and larger.

So, you haven't answered the question, if $$a->oo$$ and $$t->0$$ what is the crusing speed? Would you please answer the question?

Again, I (and kev) assume a fixed cruising speed v and just consider the limit of the exact formula as a approaches infinity while v is kept fixed. It happens to be true that in this limit the time T_a also approaches zero, but T_a is just treated as a function of a and the fixed cruising speed v (this should be clear from my mathematical calculations in post #155), we aren't treating v as an unknown and trying to deduce its value in a double limit as a approaches infinity but T_a approaches 0. Once again this is a completely standard assumption made in any textbook that considers the limit as a approaches infinity in a twin paradox type problem--have you looked at what Moller does on pages 260-262? If so, do you disagree that he, too, is treating v as fixed while he takes the limit as acceleration approaches infinity, not treating v as unknown and trying to find its value in a double limit as acceleration approaches infinity and time approaches 0?

 

[starthaus]

But in post #148 you said this was OK when Moller did it:

In post 148 I showed that you cited selectively. I also showed that Moller first derived the general expression and then calculated the limit. Very different from your/kev hack.

This is exactly what I did, I started with an "exact" formula (i.e. one that didn't involve any limits):

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

...and then I took the limit as a approaches infinity while we keep v fixed.

You can't do that since v is a function of a. When a goes to infinity, v goes to c. This is unphysical enough to make you stop right here. We've been over this before.

Is it OK for Moller (and many other physics textbooks) to do this but not for me and kev? Or do you wish to take back what you said above about not having any objection to Moller taking the limit as the acceleration approaches infinity as long as he derived it from a formula not based on limits?

This is a red herring to detract from the earlier debate that the time differential is a function of acceleration.

You are attacking fantasy strawmen again, neither I nor kev ever said the elapsed time during the acceleration phase would be negligible in all possible experiments, certainly not in the Hafele-Keating experiment where the planes were accelerating for pretty much the entire trip.

Excellent, then neither of you, faced with experimental counterproof, will continue to claim that the time differential is not a function of acceleration.

Again, I (and kev) assume a fixed cruising speed v and just consider the limit of the exact formula as a approaches infinity while v is kept fixed.

You are using circular logic since the first formula in your "derivation" is:

$$v=\frac{at}{\sqrt{1+(at/c)^2}}$$

You can't "assume" v fixed when you are increasing a towards infinity (an unphysical attempt in itself).

I asked you three times to determine v from these conditions. I can only conclude that you realize that you can't do the calculation.

It happens to be true that in this limit the time T_a also approaches zero, but T_a is just treated as a function of a and the fixed cruising speed v (this should be clear from my mathematical calculations in post #155), we aren't treating v as an unknown and trying to deduce its value in a double limit as a approaches infinity but T_a approaches 0.

So, you realize that you are attempting to have $$a->oo$$ while $$t->0$$. What does this tell you about the product $$at$$?

Once again this is a completely standard assumption made in any textbook that considers the limit as a approaches infinity in a twin paradox type problem--have you looked at what Moller does on pages 260-262? If so, do you disagree that he, too, is treating v as fixed while he takes the limit as acceleration approaches infinity, not treating v as unknown and trying to find its value in a double limit as acceleration approaches infinity and time approaches 0?

Yes, I can see what he did, It is pretty bad, it doesn't mean that you need to copy it mindlessly, just because he dit it. If you want a realistic scenario, feel free to use the one that describes the Haefele-Keating experiment. It is pretty good, you should give it a try. Once you do that, you should feel a lot more confortable with the role played by acceleration in the difference between the elapsed proper times.


In an earlier post I provided you with three published papers by recognized scientists that explain the role played by acceleration in calculating elapsed time from the PoV of the accelerated twin. Have you read them?

 

[Passionflower]

Excellent, then neither of you, faced with experimental counterproof, will continue to claim that the time differential is not a function of acceleration.

I would not hold my breath.

For the live of me I cannot understand why JesseM and kev are so adamant to keep acceleration out of the picture.

If we take the simple example of a traveling twin going to a place of a distance x away and returning then the total time dilation depends completely on the rate and duration of the accelerations (4 if we assume the start and end at rest relative to each other). In fact we can make a formula without any need for a velocity or 'cruising' time. Here the time dilation is a function of acceleration, duration of acceleration, and total distance.

As an exercise for those who are interested look at the spacetime diagram in the two extremes, the first extreme is a triangle where the accelerations are instant and the other extreme is the smoothest possible curve where the twin starts deceleration half way at each leg of the trip. Express the total time dilation in terms of a and distance x. Can you make a single formula for all stages in-between?

More challenging would be to extend this to an arbitrary number of accelerations in arbitrary directions, e.g. a kind of 'zitterbewegung' formula between two points a distance x away.

 

[starthaus]

I would not hold my breath.

For the live of me I cannot understand why JesseM and kev are so adamant to keep acceleration out of the picture.

It has something to do with never admitting that they are wrong.

As an exercise for those who are interested look at the spacetime diagram in the two extremes, the first extreme is a triangle where the accelerations are instant and the other extreme is the smoothest possible curve where the twin starts deceleration half way at each leg of the trip. Express the total time dilation in terms of a and distance x. Can you make a single formula for all stages in-between?

More challenging would be to extend this to an arbitrary number of accelerations in arbitrary directions, e.g. a kind of 'zitterbewegung' formula between two points a distance x away.

I went ahead and https://www.physicsforums.com/blog.php?b=1954 , such that kev and JesseM can wonder where I "copied" it from. Hint: nowhere, I wrote it up, like the other posts in my blog.

 

[JesseM]

In post 148 I showed that you cited selectively.

You just pointed out that Moller first derives a non-limit formula, but I never said otherwise. I also first gave the non-limit formula $$d \tau=\frac{T_c}{\gamma}+\frac{2c}{a} \, \, asinh(a T_a / 2c)$$ and then used that make conclusions about the limit as the acceleration approaches infinity (while the cruising speed is kept constant).

I also showed that Moller first derived the general expression and then calculated the limit. Very different from your/kev hack.

How is it "very different"? Did I not first post the general expression above and then calculate the limit?

You can't do that since v is a function of a. When a goes to infinity, v goes to c.

You have to distinguish here between two uses of v--one is the cruising speed which is kept fixed, and one is the variable speed during the acceleration phase which is a function of t and a. I think the distinction was made fairly clear by my equations and discussion, but it might be more precise to distinguish them in the notation as the constant cruising speed v_c and the variable function v(t, a). Here, I'll repost my derivation in post #155 but with the better notation:

if you take into account the physical dependence of T_a on a and v_c (which is what kev meant above when he said '$$T_a$$ is necessarily brief because the terminal velocity v is reached very quickly') and substitute an expression for T_a as a function of a and v_c into the above equation, you find that the second term does indeed go to zero in the limit as a approaches infinity. Just consider the formula for velocity as a function of coordinate time:

v(T, a) = aT / sqrt[1 + (aT/c)²]

So if T_a is the total turnaround time, the T_a/2 is the time needed to accelerate from a speed of 0 to the final cruising speed v_c. So, we have:

v_c = aT_a / (2*sqrt[1 + (aT_a/2c)²])

So, we have:

v_c² = a²T_a² / (4 + 4a²T_a²/4c²)

4v_c² + (v_c²a²T_a²/c²) = a²T_a²

4v_c² + (v_c²a²T_a²/c²) = c²a²T_a²/c²

4v_c² = T_a²(c²a² - v_c²a²)/c²

T_a² = 4v_c²c²/[a²(c² - v_c²)]

T_a² = 4v_c²/[a²(1 - v_c²/c²)]

T_a = 2v_c/(a*sqrt[1 - v_c²/c²])

Now, kev actually did make one minor error in the above, if T_a is the total acceleration time then as I said in post #96 the acceleration term should be $$\frac{2c}{a} \, \, arcsinh(a T_a / 2c)$$, which only differs from his expression by those factors of 2. Anyway, if you substitute in $$T_a = \frac{2v_c}{a*\sqrt{ 1 - v_c^2/c^2}}$$, you get $$\frac{2c}{a} \, \, arcsinh(\frac{v_c}{c*\sqrt{ 1 - v_c^2/c^2}})$$. Clearly, this does approach 0 in the limit as a approaches infinity, since a appears only in the denominator of the expression outside.

Is it OK for Moller (and many other physics textbooks) to do this but not for me and kev? Or do you wish to take back what you said above about not having any objection to Moller taking the limit as the acceleration approaches infinity as long as he derived it from a formula not based on limits?

This is a red herring to detract from the earlier debate that the time differential is a function of acceleration.

So, I take it you don't actually dispute that Moller does take the limit as acceleration approaches infinity while keeping the final velocity (v_c in my notation, though he doesn't use the same notation) fixed? Do you think he is making an error on the level of theory or mathematics here?

Anyway, you are using a highly equivocal phrase when you say that I am arguing against the position that "the time differential is a function of acceleration". I'm sure you can see that the following two claims are distinct:

1) It is theoretically possible to come up with scenarios where the acceleration is so brief that the total elapsed time (and thus the 'time differential' with another clock) could be calculated very accurately by just adding the elapsed times on the inertial parts of the trip and ignoring the elapsed time during the accelerating phase

2) In all conceivable scenarios, the elapsed time during the accelerating phase can be ignored when calculating the total elapsed time (and thus the time differential)

Of course I have only argued for 1 above, I would never make an argument as boneheaded as 2. Your vague descriptions of my argument allow you to equivocate between 1 and 2 (so, for example, you present the Hafele-Keating experiment as a 'rebuttal' even though it would only be a rebuttal to 2, not 1). I trust you will keep this distinction in mind from now on, and make no further ridiculous suggestions that I have been making argument 2 above.

You are using circular logic since the first formula in your "derivation" is:

$$v=\frac{at}{\sqrt{1+(at/c)^2}}$$

You can't "assume" v fixed when you are increasing a towards infinity (an unphysical attempt in itself).

The cruising speed v_c is fixed, v(T, a) during the acceleration phase. The acceleration phase is always defined to end at T_a/2, and we pick the value of T_a/2 in such a way as to ensure that v(T_a/2, a) = v_c.

I asked you three times to determine v from these conditions. I can only conclude that you realize that you can't do the calculation.

If v = at / sqrt[1 + (at/c)²], simply saying that you want a to approach infinity while t approaches 0 does not provide enough information to give a well-defined limit for v without additional information on the relation between a and t. For example, if we added the information at t = 0.6c/a, in this case it would be true that in the limit as a approaches infinity t will approach 0, and we would have v=0.6c/sqrt[1.36]. But if we instead added the information that t = 0.8c/a, then we would have v=0.8c/sqrt[1.64]. So, your question as stated does not give enough information for a value of v. Fortunately my version does provide more information, I want the speed at time t=T_a/2 to equal to the cruising speed v_c, which implies a relation between T_a and a and v_c, namely T_a = 2v_c/(a*sqrt[1 - v_c²/c²]). With this relation, it will be true that v(T_a/2, a) = v_c, even in the limit as a approaches infinity and T_a/2 approaches 0.

Once again this is a completely standard assumption made in any textbook that considers the limit as a approaches infinity in a twin paradox type problem--have you looked at what Moller does on pages 260-262? If so, do you disagree that he, too, is treating v as fixed while he takes the limit as acceleration approaches infinity, not treating v as unknown and trying to find its value in a double limit as acceleration approaches infinity and time approaches 0?

Yes, I can see what he did, It is pretty bad,

"Pretty bad" just in the sense that you don't think it's physically realistic, or do you think it's "pretty bad" in the sense that he made an actual error in mathematics or theoretical physics? If you admit there is no error on a mathematical/theoretical level, but claim that there is such an error in my own derivation, what mathematical/theoretical error do you think I made that Moller didn't also make?

it doesn't mean that you need to copy it mindlessly,

Of course I didn't "copy it mindlessly", my actual equations are different than his, and I posted most of those equations before you even mentioned the Moller textbook.

If you want a realistic scenario, feel free to use the one that describes the Haefele-Keating experiment.

I never claimed that the scenario of very fast acceleration was supposed to be "realistic". It is a theoretical example that sacrifices realism for theoretical simplicity, something every physics textbook does routinely, if you have a problem with this sort of thing you have a bizarre attitude that differs from that of the physics community as a whole.

In an earlier post I provided you with three published papers by recognized scientists that explain the role played by acceleration in calculating elapsed time from the PoV of the accelerated twin. Have you read them?

No, because it's irrelevant to this discussion, nothing I have said would imply I disagree with the idea that you can calculate elapsed time in a non-inertial frame of the accelerated twin, and that in this frame the behavior of the two clocks during the accelerating phase would play a crucial role. Perhaps you are trying to imply that I said acceleration is always irrelevant and that this is a counterexample, but this again represents an equivocation between the claims 1 and 2 I listed above, of which only 1 represents anything I have actually argued on this thread.

 

[starthaus]

Anyway, you are using a highly equivocal phrase when you say that I am arguing against the position that "the time differential is a function of acceleration". I'm sure you can see that the following two claims are distinct:

1) It is theoretically possible to come up with scenarios where the acceleration is so brief that the total elapsed time (and thus the 'time differential' with another clock) could be calculated very accurately by just adding the elapsed times on the inertial parts of the trip and ignoring the elapsed time during the accelerating phase

2) In all conceivable scenarios, the elapsed time during the accelerating phase can be ignored when calculating the total elapsed time (and thus the time differential)

Of course I have only argued for 1 above, I would never make an argument as boneheaded as 2.

Good, then we are done. Because scenario 1 has nothing to do with reality whereas scenario
2 is the one encountered in real life.

"Pretty bad" just in the sense that you don't think it's physically realistic, or do you think it's "pretty bad" in the sense that he made an actual error in mathematics or theoretical physics? If you admit there is no error on a mathematical/theoretical level, but claim that there is such an error in my own derivation, what mathematical/theoretical error do you think I made that Moller didn't also make?

Pretty bad as in unrealistic, accelerations in real life are less than 20g even for the fastest rockets/test planes.

Now, kev actually did make one minor error in the above, if T_a is the total acceleration time then as I said in post #96 the acceleration term should be $$\frac{2c}{a} \, \, arcsinh(a T_a / 2c)$$, which only differs from his expression by those factors of 2. Anyway, if you substitute in $$T_a = \frac{2v_c}{a*\sqrt{ 1 - v_c^2/c^2}}$$, you get $$\frac{2c}{a} \, \, arcsinh(\frac{v_c}{c*\sqrt{ 1 - v_c^2/c^2}})$$. Clearly, this does approach 0 in the limit as a approaches infinity, since a appears only in the denominator of the expression outside.

Actually, if you took the limit when $$a->oo$$ of the expression that doesn't mix in the cruising speed (i.e. $$\frac{2c}{a} \, \, arcsinh(a T_a / 2c)$$,) you would be getting 0 by applying l'Hospital rule. Much cleaner than going through the circular argument about v.

No, because it's irrelevant to this discussion, nothing I have said would imply I disagree with the idea that you can calculate elapsed time in a non-inertial frame of the accelerated twin, and that in this frame the behavior of the two clocks during the accelerating phase would play a crucial role.

Good, then there is nothing further to argue about. Thank you for clarifying your position.

 

[JesseM]

For the live of me I cannot understand why JesseM and kev are so adamant to keep acceleration out of the picture.

This is way too broad a summary of my position, especially given that I already gave you a more nuanced explanation of what aspects of acceleration can be ignored and what aspects are vitally important back in post #120:

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).

If you do a search for previous threads where I discussed the twin paradox, you will see that I always emphasize that acceleration is crucial for understanding on a conceptual level why there is no symmetry in each twin's view of the other, and why the twin that accelerates will genuinely have aged less than the one that doesn't. So please don't continue to say that I am trying to "keep acceleration out of the picture" because it just isn't true. Rather, my point is just that when calculating the elapsed time of the non-inertial twin, in a theoretical problem of the kind that routinely appear in textbooks one can just assume that the accelerating phase is extremely brief, so that we don't need to worry about the time elapsed during the accelerating phase when calculating the total elapsed time. And I'm not in any way "adamant" that this is an assumption one should make when thinking about the twin paradox, just saying it's one you can make (and again, is routinely made in textbooks to simplify things), that one isn't making an error on theoretical or mathematical level by doing so. And I wouldn't have needed to spend so much time defending this if starthaus hadn't made a big show of claiming that kev and I are making an "error" when we talk about this scenario.

If we take the simple example of a traveling twin going to a place of a distance x away and returning then the total time dilation depends completely on the rate and duration of the accelerations (4 or 3 if we combine the turn around).

Not if you make the routine assumption that the time of the accelerations is very brief compared to the time of the constant-velocity outbound and inbound phases of the trip. Do you disagree that in this case the accelerations contribute very little to the elapsed time, and that in the limit as the coordinate time of the accelerations approaches zero (instantaneous accelerations) while the velocities of the inertial phases are kept constant, the total elapsed proper time approaches a simple sum of the elapsed proper time in the two inertial phases? If so that is the only point I have been making on this thread, and starthaus has been making a big stink over it for some reason.

 

[Passionflower]

This is way too broad a summary of my position, especially given that I already gave you a more nuanced explanation of what aspects of acceleration can be ignored and what aspects are vitally important back in post #120:

Ok, I accept that.

If you do a search for previous threads where I discussed the twin paradox, you will see that I always emphasize that acceleration is crucial for understanding on a conceptual level why there is no symmetry in each twin's view of the other, and why the twin that accelerates will genuinely have aged less than the one that doesn't. So please don't continue to say that I am trying to "keep acceleration out of the picture" because it just isn't true.

Ok.

Rather, my point is just that when calculating the elapsed time of the non-inertial twin, in a theoretical problem of the kind that routinely appear in textbooks one can just assume that the accelerating phase is extremely brief, so that we don't need to worry about the time elapsed during the accelerating phase when calculating the total elapsed time. And I'm not in any way "adamant" that this is an assumption one should make when thinking about the twin paradox, just saying it's one you can make (and again, is routinely made in textbooks to simplify things), that one isn't making an error on theoretical or mathematical level by doing so.

Ok, extremely brief is possible, but making it zero creates pathological conditions.

Not if you make the routine assumption that the time of the accelerations is very brief compared to the time of the constant-velocity outbound and inbound phases of the trip. Do you disagree that in this case the accelerations contribute very little to the elapsed time, and that in the limit as the coordinate time of the accelerations approaches zero (instantaneous accelerations) while the velocities of the inertial phases are kept constant, the total elapsed proper time approaches a simple sum of the elapsed proper time in the two inertial phases?

Do you understand I have trouble with the statement I highlighted?
Just by making the acceleration period shorter does not make it less important, acceleration is of the essence in the twin experiment. Acceleration sets the clock rate. Accelerating a little but for a long period can have the same effect on gamma as accelerating very much for a very short period.