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

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The discussion critiques the use of the light clock thought experiment in explaining time dilation in special relativity (SR) and its application to the Twin Paradox. It argues that acceleration plays a crucial role in resolving the paradox, as the twins do not remain in identical inertial frames due to one twin's acceleration. The conversation highlights that while SR can handle acceleration, the light clock's functionality is compromised under significant acceleration, leading to bending of the light beam. Additionally, it emphasizes that the proper time experienced by each twin can be calculated based on their unique paths through spacetime, illustrating that the traveling twin will always have aged less upon reunion. Ultimately, the resolution of the Twin Paradox requires a careful consideration of both SR and general relativity (GR) principles.
  • #151


JesseM said:
the instantaneous rate of a clock at a single instant, the rate is solely a function of its instantaneous velocity.
Passionflower said:
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.
 
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  • #152


matheinste said:
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.
 
  • #153


kev said:
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.
 
  • #154


kev said:
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?
 
  • #155


starthaus said:
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:
JesseM said:
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":
starthaus said:
JesseM said:
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:
starthaus said:
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:
kev said:
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:
kev said:
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:
starthaus said:
kev said:
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 Ta 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 Ta 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 Ta is the total turnaround time, the Ta/2 is the time needed to accelerate from a speed of 0 to the final cruising speed v. So, we have:

v = aTa / (2*sqrt[1 + (aTa/2c)2])

So, we have:

v2 = a2Ta2 / (4 + 4a2Ta2/4c2)

4v2 + (v2a2Ta2/c2) = a2Ta2

4v2 + (v2a2Ta2/c2) = c2a2Ta2/c2

4v2 = Ta2(c2a2 - v2a2)/c2

Ta2 = 4v2c2/[a2(c2 - v2)]

Ta2 = 4v2/[a2(1 - v2/c2)]

Ta = 2v/(a*sqrt[1 - v2/c2])

Now, kev actually did make one minor error in the above, if Ta 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.
 
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  • #156


starthaus said:
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.
 
  • #157


kev said:
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.
Passionflower said:
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:
kev said:
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.
 
  • #158


kev said:
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.
 
  • #159


JesseM said:
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
 
  • #160


JesseM said:
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:

Ta = 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.
 
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  • #161


starthaus said:
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?
starthaus said:
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 Ta should correlate with a. You need to actually think about the physical assumptions of the problem and not forget them when evaluating various equations.
 
  • #162


JesseM said:
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 Ta 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 .
 
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  • #163


starthaus said:
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?
 
  • #164


JesseM said:
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.
 
  • #165


starthaus said:
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 Ta, 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 Ta is necessarily brief because the terminal velocity v is reached very quickly, the second term goes to zero
 
  • #166


JesseM said:
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:
starthaus said:
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.
 
  • #167


JesseM said:
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 clocks:observed 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?
 
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  • #168


JesseM said:
No, we are talking about limits as acceleration approaches infinity and time approaches zero.
starthaus said:
Which is unphysical.
But in post #148 you said this was OK when Moller did it:
starthaus said:
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?
starthaus said:
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.
starthaus said:
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.
starthaus said:
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 Ta also approaches zero, but Ta 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 Ta 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?
 
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  • #169


JesseM said:
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 Ta also approaches zero, but Ta 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 Ta 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?
 
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  • #170


starthaus said:
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.
 
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  • #171


Passionflower said:
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.
 
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  • #172


starthaus said:
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).
starthaus said:
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?
starthaus said:
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 vc 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 Ta on a and vc (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 Ta as a function of a and vc 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 Ta is the total turnaround time, the Ta/2 is the time needed to accelerate from a speed of 0 to the final cruising speed vc. So, we have:

vc = aTa / (2*sqrt[1 + (aTa/2c)2])

So, we have:

vc2 = a2Ta2 / (4 + 4a2Ta2/4c2)

4vc2 + (vc2a2Ta2/c2) = a2Ta2

4vc2 + (vc2a2Ta2/c2) = c2a2Ta2/c2

4vc2 = Ta2(c2a2 - vc2a2)/c2

Ta2 = 4vc2c2/[a2(c2 - vc2)]

Ta2 = 4vc2/[a2(1 - vc2/c2)]

Ta = 2vc/(a*sqrt[1 - vc2/c2])

Now, kev actually did make one minor error in the above, if Ta 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.
JesseM said:
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?
starthaus said:
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 (vc 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.
starthaus said:
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 vc is fixed, v(T, a) during the acceleration phase. The acceleration phase is always defined to end at Ta/2, and we pick the value of Ta/2 in such a way as to ensure that v(Ta/2, a) = vc.
starthaus said:
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=Ta/2 to equal to the cruising speed vc, which implies a relation between Ta and a and vc, namely Ta = 2vc/(a*sqrt[1 - vc2/c2]). With this relation, it will be true that v(Ta/2, a) = vc, even in the limit as a approaches infinity and Ta/2 approaches 0.
JesseM said:
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 said:
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?
starthaus said:
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.
starthaus said:
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.
starthaus said:
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.
 
  • #173


JesseM said:
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 Ta 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.
 
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  • #174


Passionflower said:
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:
JesseM said:
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.
Passionflower said:
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.
 
  • #175


JesseM said:
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.

JesseM said:
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.

JesseM said:
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.

JesseM said:
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.
 
  • #176


starthaus said:
Good, then we are done. Because scenario 1 has nothing to do with reality whereas scenario
2 is the one encountered in real life.
Good, so I guess you retract your claim that kev and I have made any "error", since of course lack of realism is not an "error" in a theoretical example, the only possible errors in theoretical examples are in mathematics or in steps that violate the basic laws of physics.
starthaus said:
Pretty bad as in unrealistic.
Good, no actual error then. By the way, if you complain about "unrealistic" examples you'll need to complain about virtually every example of a spacetime that appears in a textbook which discusses general relativity, they pretty much always contain unrealistic assumptions like the notion that the spacetime is asymptotically flat, or eternally static (as in the external Schwarzschild solution), or that the distribution of matter is completely uniform (as in cosmological solutions), etc.
starthaus said:
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.
You can't take the limit of that expression if you don't know how Ta depends on a. Were you just assuming that Ta is a constant? But then in the limit as a approaches infinity, the final cruising speed will approach c, and you are left with the conclusion that the total elapsed time for the entire trip approaches 0 as well--a much weirder assumption physically than just the assumption that the acceleration time is very small compared to the time spent moving inertially!
JesseM said:
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.
starthaus said:
Good, then there is nothing further to argue about.
Glad you see that you now understand you have been attacking a strawman this whole time. In future, if you want to avoid a lot of wasted time, try to avoid leaping to uncharitable conclusions about what people meant, instead if you think someone is saying something clearly erroneous stop and consider if there may be an alternate interpretation of their words, and ask politely for clarification instead of rushing into attack immediately.
 
  • #177


JesseM said:
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?
Passionflower said:
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.
When I used the word "contribute" above, I was referring to a numerical contribution only, not to the conceptual or causal significance of the acceleration. If we break down the total elapsed proper time into (proper time elapsed in inertial phase) + (proper time elapsed in acceleration phase), then as the coordinate time of the acceleration phase gets very small, (proper time elapsed in acceleration phase) gets very small too, so this term makes very little numerical contribution to the total elapsed proper time. That's all I meant, apologies if the language was ambiguous.
 
  • #178


JesseM said:
If we break down the total elapsed proper time into (proper time elapsed in inertial phase) + (proper time elapsed in acceleration phase), then as the coordinate time of the acceleration phase gets very small, (proper time elapsed in acceleration phase) gets very small too...
Up to here I fully agree.

JesseM said:
, so this term makes very little numerical contribution to the total elapsed proper time.
The term as in "the elapsed proper time during acceleration" yes. But "this term" and "acceleration" are not synonymous. The rate and duration of this acceleration phase has a very big impact on the second term.

No need for apologies as language is almost always ambiguous.
 
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  • #179


JesseM said:
Good, so I guess you retract your claim that kev and I have made any "error", since of course lack of realism is not an "error" in a theoretical example, the only possible errors in theoretical examples are in mathematics or in steps that violate the basic laws of physics.

Sure you did, your language was ambigous. It took tens of posts for you to admit it by separating the two cases. You were talking about a didactical case, I was talking about the real life case. Your didactical approach does not apply to real life.


.
 
  • #180


starthaus said:
Sure you did, your language was ambigous. It took tens of posts for you to admit it by separating the two cases.
Presumably you are referring to the "two cases" I distinguished in this part of post #172:
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.
So you really think the problem was that I took "tens of posts" before finally "admitting" that I was not advocating 2? It couldn't be that you just rushed to an uncharitable interpretation and then failed to read my posts carefully enough? Read again what I wrote back in post #155:
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)).
And in that post I also responded to statement you made to kev that a certain calculation "clearly contradicts your [kev's] earlier claim that the contribution of the acceleration period is negligible" by quoting his post #13:
kev said:
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
The distinction I made there between a claim that time elapsed during the acceleration phase "can be reduced" a negligible amount by an appropriate choice of problem to analyze, vs. the idea that it "always reduces to a negligible amount in all problems" which was a claim kev didn't make, corresponds precisely to the distinction between claims 1) and 2) which I made in the more recent post #172.

But in case you missed that distinction I made in the long #155, I made the same distinction again in the very short post #166:
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.
And, let's see, I also repeated the point again in my next post #168:
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.
So, looks like I have been pretty clear about this distinction all along.

By the way, I also want to point out that from the very beginning when I asked for examples of errors in post #137, I specifically said I wanted clear mathematical or physical errors, not just complaints about unrealistic simplifications of the kind routinely used in textbooks:
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.

...

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.
So either you initially thought kev and I had made errors on a physical/mathematical level and have changed your mind, or you didn't read that initial request for examples carefully enough to realize that I wasn't just asking for examples where kev and I had made simplifications that would be unrealistic in actual experiments.
 
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