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

In summary, the "light clock" thought experiment is often used to illustrate time dilation in special relativity. However, the Twin Paradox and other arguments show that using special relativity alone is not sufficient to resolve this thought experiment. Instead, one must turn to general relativity and consider the effects of acceleration on the clock. However, there are still debates about the role of acceleration in resolving the paradox, with some arguing that it is not necessary and others claiming that it is crucial.
  • #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 [tex]a->oo[/tex] of the expression that doesn't mix in the cruising speed (i.e. [tex]\frac{2c}{a} \, \, arcsinh(a T_a / 2c)[/tex],) 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.
 
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  • #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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  • #181


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

.

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

From my own direct experience in a number of previous threads I have seen both JesseM and kev recognize and
admit errors of various kinds.
This is absolutely not true wrt certain other well known members whom need not be named.
 
  • #182


JesseM said:
Presumably you are referring to the "two cases" I distinguished here:

So you really think the problem was that I took "tens of posts" before finally "admitting" that I was not advocating 2?

Precisely. Otherwise, you would have not taken up to post 172 to admit that acceleration plays a key role in the difference in elapsed time. Both Passionflower and I jhave asked you this direct question multiple times but post 172 the first time you made a clear statement of your position on the issue.


So, looks like I have been pretty clear about this distinction all along.

You didn't come clean until post 172. Had you admitted earlier, we would not have spent another 90 posts on the issue. Not that they were wasted posts, I worked out the complete case for asymettric accelerations at Passionflower's suggestion.
 
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  • #183


kev said:
I know it is true intuitively, but the maths might get rather involved. Here is my shot at it.

Let us say that rocket B accelerates to 0.8c in one second and then cruises, and exactly 10 years after B departed, A accelerates to 0.8c in one second (as measured by an observer C that remains at rest in the frame that A and B were originally at rest) and now both A and B are rest in new frame that has velocity 0.8c relative to the original frame.

From the accelerating rocket equations of Baez, that I gave earlier, the proper acceleration of a rocket with terminal velocity 0.8c (as measured in the unaccelerated frame) after a time of 1 second (as measured in the unaccelerated frame) is given as:

a = (v/t)/sqrt[1-(v/c)^2] = 0.8*0.6 = 0.48

The proper elapsed time (T) of the rocket during the acceleration phase is given by Baez as:

T = (c/a)*asinh(at/c) = (1/0.48)*asinh(0.48) = 0.9651 seconds.

using units of c=1. Note that the proper elapsed time is not much less than the 1 second measured by the unaccelerated observer C.

During the cruise phase of B's journey the elapsed proper time of B's clock according to C is 10years*0.6 = 6 years so the total elapsed proper time of B's clock is 6years + 0.9651 seconds in C's frame.

The total elapsed proper time of rocket A from the time B took of to the time A joined B in the new rest frame is 10 years + 0.9651s seconds according to C.

In frame C the elapsed time of A is obviously much greater than the elapsed time of B.

Now we look at the times measured by an observer that was at rest in a frame (D) that was always moving with velocity 0.8c relative to frame C. i.e frame D is the final rest frame of observers A and B.

In frame C the elapsed time from B taking off, to A joining B in frame D, was 10 years + 1 second, so in frame D the elapsed time between the two events is (10y + 1s)/0.6 = 16.6667 years + 1.66667 seconds. Other than the initial 1.6666 seconds that rocket B initially took to accelerate to rest in frame D, rocket B has been at rest in frame D for 16.6667 years so the total elapsed proper time of B's clock, according to C is 16.6667 years (plus the 0.9651 seconds of proper time B spent accelerating). The elapsed proper time of rocket A, according to observer D is 10 years (plus the 0.9651 seconds of proper time A spent accelerating).

So in frame D, the time that elapses between B taking off and A joining B in frame D is 10 years + 0.9651 seconds proper time as measured by clock A and 16.6667 years + 0.9651 seconds proper time as measured by clock B. In frame D much more proper time has elapsed on clock B while in the original frame C, much more proper time elapses on clock A.

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

1.Time dilation due to acceleration is not insignificant, it can play a major role in real life, as in https://www.physicsforums.com/blog.php?b=1954 the results of the Haefele-Keating or Vessot experiments.

2. You got an "insignificant" value because of the way you contrived your thought experiment.

3. Try calculating what value acceleration you need in order to ramp up to 0.8c in 1s and you'll see how far off from reality your thought experiment is. As a reference, accelerations of 10-20g are the top of the achievable scale.
 
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  • #184


starthaus said:
Precisely. Otherwise, you would have not taken up to post 172 to admit that acceleration plays a key role in the difference in elapsed time.
Uhhh, I just quoted some earlier posts of mine where I clearly said that the acceleration phase would make a significant contribution in experiments where it lasts a significant time, like post #168:
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.
starthaus said:
Both Passionflower and I jhave asked you this direct question multiple times but post 172 the first time you made a clear statement of your position on the issue.
Point to a single post where you or Passionflower asked a "direct question" about the time elapsed during accelerations of significant length where I did not say that in this case the proper time during the accelerating phase would be significant. If you're going to bring Passionflower's posts into it, I also made clear in posts to him that even when acceleration is quasi-instantaneous, the acceleration also plays a crucial conceptual role in understanding why one twin ages less, even if the proper time elapsed during the acceleration phase is negligible (see post #120 for example).
 
  • #185


JesseM said:
Uhhh, I just quoted some earlier posts of mine where I clearly said that the acceleration phase would make a significant contribution, like post #168:

Ok, so it didn't take up to post 172 to admit that acceleration plays a key role in the time dilation, it took you up to post 168.
 
  • #186


starthaus said:
Ok, so it didn't take up to post 172 to admit that acceleration plays a key role in the time dilation, it took you up to post 168.
Yes, because post #167 was the first post where you brought up the Hafele-Keating experiment as a "rebuttal", before that you gave no clear indication that you were misreading me in such a bizarre way as to think I would disagree that in some experiments the proper time during the accelerating phase would make a large contribution. I note that you didn't answer my request to back up your statement that you and Passionflower had ever asked me "direct questions" about my opinion on whether the accelerating phase would make a significant contribution in cases other than the one that were originally the topic of discussion, namely kev's example where the acceleration becomes very large so the time needed to reach the cruising velocity becomes very small. Why should I randomly volunteer the fact that the acceleration phase can have significant proper time in cases other than the one we were discussing, when it had no relevance to the issue of "errors" in kev's argument, and when you didn't ask me about this? Besides, I think basic reading comprehension would tell you that when in post #155 I quoted kev saying that the time dilation can be reduced to a negligible error, and then commented:
"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).
...the clear implication was that I understand the time dilation during the accelerating phase would not "always reduce to a negligible amount in all problems". But if there was any doubt in your mind that this was my implication, you could have actually done what you misremembered doing and asked a "direct question" about the issue.
 
  • #187


JesseM said:
Yes, because post #167 was the first post where you brought up the Hafele-Keating experiment as a "rebuttal", before that you gave no clear indication that you were misreading me in such a bizarre way as to think I would disagree that in some experiments the proper time during the accelerating phase would make a large contribution. I note that you didn't answer my request to back up your statement that you and Passionflower had ever asked me "direct questions" about my opinion on whether the accelerating phase would make a significant contribution in cases other than the one that were originally the topic of discussion, namely kev's example where the acceleration becomes very large so the time needed to reach the cruising velocity becomes very small. Why should I randomly volunteer the fact that the acceleration phase can have significant proper time in cases other than the one we were discussing, when it had no relevance to the issue of "errors" in kev's argument, and when you didn't ask me about this? Besides, I think basic reading comprehension would tell you that when in post #155 I quoted kev saying that the time dilation can be reduced to a negligible error, and then commented:

...the clear implication was that I understand the time dilation during the accelerating phase would not "always reduce to a negligible amount in all problems". But if there was any doubt in your mind that this was my implication, you could have actually done what you misremembered doing and asked a "direct question" about the issue.

Both I and passionflower asked you "direct questions" and you kept answering in such a fashion that led us to believe that you negated the role of acceleration. The first time you gave a straight answer to the question was post 168. The first time you gave a complete answer was post 172.
 
  • #188


starthaus said:
Both I and passionflower asked you "direct questions"
OK so, give an example of a post where you or Passionflower asked such a "direct question" about this issue which I didn't answer (I suspect if you had any you probably would have brought them up by now). As far as I can recall, you didn't ask questions about whether I disputed that the acceleration phase would make a significant contribution to the proper time differential in other examples besides kev's (like the Hafele-Keating experiment), you just jumped to the conclusion I was saying it never would based on your determination to believe I was making some stupid blunder.
starthaus said:
and you kept answering in such a fashion that led us to believe that you negated the role of acceleration.
"Negated the role of acceleration" is a totally vague phrase which allows you to equivocate between different meanings--I did say that the acceleration phase didn't contribute significantly to the total proper time differential in kev's example (claim 1 from post 172), but saying we can negate the contribution of the accelerating phase in that example doesn't somehow imply we can negate it in all examples (claim 2 from post 172). Since you brought up my posts to Passionflower, perhaps you didn't see my recent post where I reminded Passionflower that as far back as post #120 I did emphasize that acceleration plays an important role in a conceptual sense, regardless of the actual length of the acceleration phase, so that's another sense in which acceleration cannot be "negated":
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).
I would suggest you try to think more carefully and precisely about what your opponents are actually saying and not use vague and over-broad formulations which could have many possible meanings, like accusing people of denying that "the time differential is a function of acceleration" or of believing that "acceleration can be negated". You have a constant tendency to assume everyone you disagree with is stupid/ignorant and to therefore read everything they say in the most uncharitable way (i.e., if an interpretation of someone's words occurs to you that would make them out to be taking a stupid/ignorant position, you immediately jump to the conclusion that this is the correct interpretation and don't consider that there might be others or even ask questions to clarify what they're saying).
starthaus said:
The first time you gave a straight answer to the question was post 168.
Straight answer to what question? Still waiting for an example of an earlier post where you asked some question related to this issue and I didn't give a straight answer, seems to me your memory is playing tricks on you.
 
  • #189


JesseM said:
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).

OK, let's look at post 120 (still 40 posts away from post #80) so we can consider this as your first admission that acceleration plays a role in the differential elapsed time. Yet, you don't come out straight and saying this, instead , you make a claim that acceleration

"explains why v1 on the outbound leg may be different than v2 on the inbound leg "

The difference in elapsed time has nothing to do with the difference between the outbound speed v1 and the inbound speed v2, it has everything to do with the presence of acceleration during the journey. The fact that the two speeds are different plays no role.
 
  • #190


starthaus said:
Actually this statement is pretty amusing, speaking of blunders. [tex]T_a[/tex] is the duration the rocket accelerates (see the wiki page), there is no reason whatsoever why [tex]T_a[/tex] should correlate to [tex]a[/tex].

JesseM has not made a mistake here. Here is a quote from the wikipedia page you mentioned stating the parameters of the scenario: http://en.wikipedia.org/wiki/Twin_p...lt_of_differences_in_twins.27_spacetime_paths

Phase 1: Rocket (with clock K') embarks with constant proper acceleration a during a time Ta as measured by clock K until it reaches some velocity V.

Ta is determined by the terminal velocity V and acceleration a.

You are being your usual unhelpful self by stating that JesseM's equation:

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

is incorrect without saying why or providing a corrected equation. As far as I can tell there is no error in his equation when he has defined Ta as the total round trip time measured in the inertial frame. For a one way trip from zero to terminal velocity v (or vice versa) the time measured in the inertial frame is simply Ta/2 = v/(a*sqrt[1 - v2/c2]).


starthaus said:
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 [tex]a->oo[/tex]? You are forgetting that , according to your starting point (see the formula for v), [tex]v->c[/tex] so, the limit is undetermined (not that this even the correct way of calculating [tex]T_a[/tex]) . Your math has deserted you today.
Nope, JesseM is right. See below.


JesseM said:
, you get [tex]\frac{2c}{a} \, \, arcsinh(\frac{v}{c*\sqrt{ 1 - v^2/c^2}})[/tex]. Clearly, this does approach 0 in the limit as a approaches infinity, since a appears only in the denominator of the expression outside.
starthaus said:
Not so fast.

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

so, [tex]a->oo[/tex] implies [tex]v->c[/tex], meaning [tex]\gamma->oo[/tex]

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


Actually, [tex]a->oo[/tex] does not imply [tex]v->c[/tex] although I too initially made that mistaken assumption. Consider the case whan a particle accelerates from zero to 0.8c in zero seconds (or an infinitesimal time interval if you prefer), then that implies infinite acceleration, but the terminal velocity is less than c.

Using the equation I gave earlier for the proper time t of an accelerating particle in terms of terminal velocity v and coordinate time T spent accelerating in the inertial frame:

[tex]t = T \sqrt{(1-v^2)}/v \, \, arcsinh(v/\sqrt{(1 - v^2)})[/tex]

then in the limit as T goes to zero, t also goes to zero, for finite terminal velocity -1<v<1.
 
  • #191


yuiop said:
As far as I can tell there is no error in his equation when he has defined Ta as the total round trip time measured in the inertial frame.

So, according to you, a rocket would decelerate from +0.8c to 0 , turn around and accelerate to -0.8c in zero time, right?

Actually, [tex]a->oo[/tex] does not imply [tex]v->c[/tex] although I too initially made that mistaken assumption. Consider the case whan a particle accelerates from zero to 0.8c in zero seconds

I considered it for exactly zero seconds and...I discarded it as unphysical.

You never answered the question I asked you: what is the necessary acceleration to reach 0.8c in 1s in your original scenario? Could you answer , please?
 
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  • #192


starthaus said:
So, according to you, a rocket would decelerate from +0.8c to 0 , turn around and accelerate to -0.8c in zero time, right?
You asked jesseM what would happen if the particle/rocket had infinite acceleration. I was answering that question. I recon infinite acceleration would be sufficient to accelerate a particle from +0.8c to -0.8c. Wouldn't you recon that was sufficient?

starthaus said:
You never answered the question I asked you: what is the necessary acceleration to reach 0.8c in 1s in your original scenario? Could you answer , please?
That would be trivial to calculate, but I can not be bothered. The acceleration would be huge, but that does not make it in principle impossible. Your objections are petty as JesseM has pointed out. This forum of full of hypothetical questions about rockets traveling at 0.9c plus, when in practice our best rockets have only achieved less than 0.0001c. That does not mean a rocket could not in principle travel at 0.9c plus, although I expect there would be some practical difficulties. JesseM also gave the example of the barn and pole paradox. A man cannot run at 0.9c, but that does not mean we can not discuss the problem. You could criticize just about every thread in this forum, if we are not allowed to discuss hypothetical situations that have not been achieved in reality. Cut it out and grow up.

I could also point out that particles in particle accelerators are subjected to huge accelerations. If you want to change the the words "twins" and "rockets" to particles then that is fine.

The PF FAQ gives the example of muons in a cyclotron accelerated to 1,000,000,000,000,000,000 times the acceleration of gravity. That discredits your argument that we cannot discuss accelerations of greater than the practical limit of 10 or 20g.

Your objection also means that we are not allowed to discuss distances of the order of light years, because in practice actual manned missions have never gone beyond the Moon and we are barely out the back gate in terms of space exploration. Do you agree that despite the fact that no manned mission has ever gone much further than the Moon, that we could in principle go further?

Are you just deliberately trying to be disruptive with your inane, petty and juvenile objections to everything?
 
  • #193


yuiop said:
You asked jesseM what would happen if the particle/rocket had infinite acceleration.

No, I asked you how realistic is such a thing. How realistic is it?

That would be trivial to calculate, but I can not be bothered.

I suggest you do, so you can learn how unphysical is the scenario you proposed at post #24.

Your objection also means that we are not allowed to discuss distances of the order of light years, because in practice actual manned missions have never gone beyond the Moon and we are barely out the back gate in terms of space exploration. Do you agree that despite the fact that no manned mission has ever gone much further than the Moon, that we could in principle go further?

Of course we can. As a matter of fact, the experiments that verified the twins paradox were quite "local". Yet, the experimenters, in constructing the theory of their experiments, applied math correctly, without hacks in the style of instantaneous turnaround (see Vessot) and they considered the effects of acceleration as non-negligible (see Vessot, Haefele-Keating). In a word, they did the physics the right way, they did not try to reduce it to a cartoon exercise.
Are you just deliberately trying to be disruptive with your inane, petty and juvenile objections to everything?

You are back to your ad-hominems, please keep it civil.
 
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  • #194


starthaus said:
OK, let's look at post 120 (still 40 posts away from post #80) so we can consider this as your first admission that acceleration plays a role in the differential elapsed time.
It is of course silly to portray this as an "admission" when I volunteered it unprompted, and there was nothing in my previous posts to suggest I would think otherwise. You may have fantasized I had been saying otherwise, but that's just because you aren't very good at reading what people actually say rather than over-generalizing narrow statements about specific situations into ridiculously broad claims like "acceleration doesn't matter" which you can then ridicule. If you think anything I said prior to #120 suggests I was making some broad claim like that, rather than simply a narrow claim about the meaning of the clock hypothesis or the fact that the acceleration phase can be left out of a calculation of total elapsed proper time if it is very brief, then please provide a quote (my posts on this thread begin with #81 so there aren't too many to read through).

Also, if you do a search for previous posts by me that use the words "twin" and "acceleration", you can see I usually emphasize that in a conceptual sense, acceleration does play a crucial role since the twin that accelerates to turn around will always have aged less than the twin that moves inertially--for example:

https://www.physicsforums.com/showthread.php?p=2692235#post2692235
https://www.physicsforums.com/showthread.php?p=2338031#post2338031
https://www.physicsforums.com/showthread.php?p=2253255#post2253255
starthaus said:
Yet, you don't come out straight and saying this, instead , you make a claim that acceleration

"explains why v1 on the outbound leg may be different than v2 on the inbound leg "

The difference in elapsed time has nothing to do with the difference between the outbound speed v1 and the inbound speed v2, it has everything to do with the presence of acceleration during the journey. The fact that the two speeds are different plays no role.
It does help explain the difference in a situation where the acceleration lasts a very brief time compared to the time spent moving inertially (for example, if the acceleration lasts a few days but the inertial legs last years), which is the type of situation I had been discussing with Passionflower, because in that case the total elapsed time for the traveling twin will be approximately equal to [tex]T_1 * \sqrt{1 - v_1^2/c^2} + T_2 * \sqrt{1 - v_2^2/c^2}[/tex] while the total elapsed time for the inertial twin is approximately equal to [tex]T_1 * \sqrt{1 - v_0^2/c^2} + T_2 * \sqrt{1 - v_0^2/c^2}[/tex], where v0 is the speed of the inertial twin in the frame you're using and T1 and T2 are the times of the two inertial legs of the traveling twin's journey in this frame with speeds v1 and v2 respectively (so the sum of T1 and T2 is about equal to the total time between the twins departing and reuniting in this frame). As long as you don't use a frame where the two twins are moving at different speeds in the same direction during the outbound leg, then in order for the traveling twin to catch up with the inertial twin again after they've been moving apart, the speed v2 of the traveling twin must be greater than the inertial twin's speed v0 throughout the return leg of the trip, so the elapsed time of the traveling twin during this phase of the trip, [tex]T_2 * \sqrt{1 - v_1^2/c^2}[/tex], will be less than the elapsed time of the inertial twin [tex]T_2 * \sqrt{1 - v_0^2/c^2}[/tex] (if you did use a frame where they were moving at different speeds in the same direction during the outbound leg, then in that case the traveling twin's speed during the outbound leg would have to be higher, so the traveling twin's elapsed proper time during the outbound leg would be smaller). Of course this is not sufficient to explain why the total elapsed time for the traveling twin is less (since v0 may be greater than v1 during the outbound leg), nor is it a general argument that covers cases where the acceleration is not brief compared to the inertial legs, but it does help to understand conceptually why you can analyze the twin paradox from different frames and still get the conclusion that the twin that changes velocities has aged less in total, especially when paired with some specific numerical example as I did in post 63 from this thread for example (stevmg found this type of explanation helpful on this thread as you can see from post #51 here, and I think this approach to explaining things has been helpful to others in the past).
 
  • #195


JesseM said:
It is of course silly to portray this as an "admission" when I volunteered it unprompted, and there was nothing in my previous posts to suggest I would think otherwise.

So, you admit that your claim in post 120 about the time differential being due to the difference between [tex]v_1[/tex] and [tex]v_2[/tex] is incorrect?
You may have fantasized I had been saying otherwise, but that's just because you aren't very good at reading what people actually say rather than over-generalizing narrow statements about specific situations into ridiculously broad claims like "acceleration doesn't matter" which you can then ridicule.

You have not admitted the role of acceleration until post 172.
If you think anything I said prior to #120 suggests I was making some broad claim like that, rather than simply a narrow claim about the meaning of the clock hypothesis or the fact that the acceleration phase can be left out of a calculation of total elapsed proper time if it is very brief, then please provide a quote (my posts on this thread begin with #81 so there aren't too many to read through).

I simply think that your post 120 clearly reflects your misconception that the difference between the speeds [tex]v_1[/tex] and [tex]v_2[/tex] is the root of the time difference. I cited your claim exactly.

It does help explain the difference in a situation where the acceleration lasts a very brief time compared to the time spent moving inertially (for example, if the acceleration lasts a few days but the inertial legs last years), which is the type of situation I had been discussing with Passionflower, because in that case the total elapsed time for the traveling twin will be approximately equal to [tex]T_1 * \sqrt{1 - v_1^2/c^2} + T_2 * \sqrt{1 - v_2^2/c^2}[/tex] while the total elapsed time for the inertial twin is approximately equal to [tex]T_1 * \sqrt{1 - v_0^2/c^2} + T_2 * \sqrt{1 - v_0^2/c^2}[/tex], where v0 is the speed of the inertial twin in the frame you're using and T1 and T2 are the times of the two inertial legs of the traveling twin's journey in this frame with speeds v1 and v2 respectively (so the sum of T1 and T2 is about equal to the total time between the twins departing and reuniting in this frame). As long as you don't use a frame where the two twins are moving at different speeds in the same direction during the outbound leg, then in order for the traveling twin to catch up with the inertial twin again after they've been moving apart, the speed v2 of the traveling twin must be greater than the inertial twin's speed v0 throughout the return leg of the trip, so the elapsed time of the traveling twin during this phase of the trip, [tex]T_2 * \sqrt{1 - v_1^2/c^2}[/tex], will be less than the elapsed time of the inertial twin [tex]T_2 * \sqrt{1 - v_0^2/c^2}[/tex] (if you did use a frame where they were moving at different speeds in the same direction during the outbound leg, then in that case the traveling twin's speed during the outbound leg would have to be higher, so the traveling twin's elapsed proper time during the outbound leg would be smaller). Of course this is not sufficient to explain why the total elapsed time for the traveling twin is less (since v0 may be greater than v1 during the outbound leg), nor is it a general argument that covers cases where the acceleration is not brief compared to the inertial legs, but it does help to understand conceptually why you can analyze the twin paradox from different frames and still get the conclusion that the twin that changes velocities has aged less in total, especially when paired with some specific numerical example as I did in post 63 from this thread for example (stevmg found this type of explanation helpful on this thread as you can see from post #51 here, and I think this approach to explaining things has been helpful to others in the past).

Note that you are still insisting on dealing with the limiting cases of zero acceleration period and that none of your formulas includes the terms expressed in acceleration. All your formulas deal with the extreme cases where acceleration is not relevant.
 
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  • #196


starthaus said:
So, you admit that your claim in post 120 about the time differential being due to the difference between [tex]v_1[/tex] and [tex]v_2[/tex] is incorrect?
"Due to" is a pretty imprecise phrase, but I explained above why I think that, in the specific situation I was talking about with Passionflower (where the acceleration period was negligible), it is helpful conceptually to think about the fact that the velocity of the traveling twin must be greater than that of the inertial twin on at least one of the two legs of the trip, regardless of what frame you use to define velocities. What's more if you look at my wording in post 120, what I said was that "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". In other words, here I was primarily referring to the relevance of acceleration to the calculation for elapsed time which used the formula [tex]T_1 * \sqrt{1 - v_1^2/c^2} + T_2 * \sqrt{1 - v_2^2/c^2}[/tex], not making a more general claim that the only conceptual significance of acceleration is that it allows the velocities to be different. In fact right after that sentence I immediately went on to discuss the broader conceptual significance of acceleration, saying "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". In this comment, the "conceptual role" played by acceleration had nothing to do with different velocities, but rather to do with the fact that inertial paths are geodesics and geodesics always maximize the proper time in SR.
starthaus said:
You have not admitted the role of acceleration until post 172.
Once again you use overly-broad phrases which carelessly lump together quite different notions. By "the role of acceleration", you could either be talking about A) the fact that the elapsed proper time during the acceleration phase must be taken into account to get an accurate the total elapsed proper time (which is true in experiments with long non-inertial phases like the Hafele-Keating experiment, but not true in cases where the time of the acceleration phase is negligible), or B) the fact that acceleration plays an important conceptual role in understanding why the non-inertial clock elapses less than the inertial one (true in all possible experiments involving a pair of clocks that are separated and later reunited, where one clock moves inertially between the separation and the reunion). If you are talking about B), I didn't actually talk about the conceptual importance of acceleration in post #172 at all, in fact I didn't bring up the conceptual importance of acceleration in a post to you until post #184 (because you never gave any indication that you wanted to discuss this topic), where I directed you to read my post #120 to Passionflower which did discuss the conceptual importance.

On the other hand if you are talking about A), then you have a short memory, since in post #185 you already admitted I had made clear this is true in some experiments earlier:
Ok, so it didn't take up to post 172 to admit that acceleration plays a key role in the time dilation, it took you up to post 168.
And of course in my reply in post #186 I pointed out that I had already made this pretty damn clear in post #155:
Besides, I think basic reading comprehension would tell you that when in post #155 I quoted kev saying that the time dilation can be reduced to a negligible error, and then commented:
"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).
...the clear implication was that I understand the time dilation during the accelerating phase would not "always reduce to a negligible amount in all problems".
...which you didn't give any substantive response to. I also pointed out in post #186 that the reason I didn't say anything more direct about experiments with significant non-inertial periods until post 168 was that I didn't even realize you were concerned with such cases, given that our discussion until then had been about what happens in kev's example where the acceleration is large and brief:
Yes, because post #167 was the first post where you brought up the Hafele-Keating experiment as a "rebuttal", before that you gave no clear indication that you were misreading me in such a bizarre way as to think I would disagree that in some experiments the proper time during the accelerating phase would make a large contribution.
You didn't give any substantive response to this either, you just made the false claim that 'Both I and passionflower asked you "direct questions" and you kept answering in such a fashion that led us to believe that you negated the role of acceleration.' I guess you now tacitly admit this was a false memory on your part, since you didn't even attempt to find an example of an earlier post where you or Passionflower had asked me any "direct question" about experiments where the non-inertial phase lasted a non-negligible time (or about the conceptual importance of acceleration, i.e. interpretation B above).
starthaus said:
I simply think that your post 120 clearly reflects your misconception that the difference between the speeds [tex]v_1[/tex] and [tex]v_2[/tex] is the root of the time difference.
No, post 120 said that the conceptual importance of acceleration is that geodesics maximize proper time, and the inertial path is a geodesic in SR while the path with an acceleration is not. But of course it is also true that acceleration's relevance to the calculation [tex]T_1 * \sqrt{1 - v_1^2 / c^2} + T_2 * \sqrt{1 - v_2^2 /c^2}[/tex] is that it explains why v1 and v2 can in general be different, whereas the corresponding calculation of the inertial twin's elapsed time would involve only a single velocity (but as I said above in #194, the fact that one formula involves two velocities v1 and v2 and the other only involves a single velocity v0 is not sufficient to prove that one formula will yield a smaller total elapsed time, even if you add some reasoning to show that either v1 or v2 must be greater than v0. Like I said though, I think if you add the latter observation it is helpful in understanding how it can be true that different frames all agree the non-inertial twin elapsed a smaller time, especially when combined with some specific scenario analyzed from the perspective of two different frames as an illustration).
starthaus said:
Note that you are still insisting on dealing with the limiting cases of zero acceleration period
No I'm not:
It does help explain the difference in a situation where the acceleration lasts a very brief time compared to the time spent moving inertially (for example, if the acceleration lasts a few days but the inertial legs last years), which is the type of situation I had been discussing with Passionflower, because in that case the total elapsed time for the traveling twin will be approximately equal to [tex]T_1 * \sqrt{1 - v_1^2/c^2} + T_2 * \sqrt{1 - v_2^2/c^2}[/tex]
And of course, the reason I am "insisting on" dealing with a case where the acceleration is small enough to be ignorable in an approximate calculation is because that is "the type of situation I had been discussing with Passionflower", so if you insist on this increasingly-desperate attempt to show I said something wrong in my previous posts, you'll have to deal with the case I was actually discussing at the time.
 
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  • #197


Almost forgot this question!
Austin0 said:
yuiop said:
O.K, you are right that this is how things would look from D's point of view, but that is just D's point of view and D is not a final arbitrator.
Yu are right D is not the final arbiter just an exception to agreement , on the other hand what frame is the final arbiter if there is not frame independent agreement?

There is no final arbiter for the elapsed times on A and D's clocks relative to each other when they never started and finished in the same place.

Austin0 said:
yuiop said:
However, for the first leg of the journey, when B is going away from A, D sees A coming towards him at 0.5c (same as in the final leg) but B is going away from him at (0.8-0.5)/(1-0.8*0.5)=0.92857c so the time dilation of B's clock on the outward journey is much greater than the time dilation of A's clock, according to D. The first leg of the experiment took 14 years of A's proper time in D's frame. This equates to 14/sqrt(1-0.5^2)=16.1658 years by D's clock. D calculates that the proper time of B's clock on the outward journey "during the same time" is 16.1658*sqrt(1-0.92857^2)=6 years. Overall, D says 20 years of proper time elapsed on A's clock and 12 years of proper time elapsed on the combined times of B and C and all is good.
Well maybe not quite 100% just yet. . :-p

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

Here is one way to picture it. Imagine a 8 ly long rod that extends from where B starts to where B finishes in A's frame. In D's frame this rod is moving at 0.5c to the right and B is chasing after it. In 16.166 yrs the end of the rod B is chasing moves 16.166*0.5=8.66 lys. Add this to the head start the end of the rod had and the total distance traveled by B in D's frame is 8.66+6.93=15.56 lys. This equates to a velocity of 15.56/16.166=0.928c.

You can do the calculations more formally, by considering the two events (A,B) and (B,C) in A's frame in x,t terms as (x1,t1) = (0,0) and (x2,t2) = (8,10) and converting those coordinates of the two events to the point of view of D using the Lorentz transformations. Just let me know if you need that clarifying.
 
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  • #198


yuiop said:
Here is one way to picture it. Imagine a 8 ly long rod that extends from where B starts to where B finishes in A's frame. In D's frame this rod is moving at 0.5c to the right and B is chasing after it. In 16.166 yrs the end of the rod B is chasing moves 16.166*0.5=8.66 lys. Add this to the head start the end of the rod had and the total distance traveled by B in D's frame is 8.66+6.93=15.56 lys. This equates to a velocity of 15.56/16.166=0.928c.

You can do the calculations more formally, by considering the two events (A,B) and (B,C) in A's frame in x,t terms as (x1,t1) = (0,0) and (x2,t2) = (8,10) and converting those coordinates of the two events to the point of view of D using the Lorentz transformations. Just let me know if you need that clarifying.

Mentz has just reminded me another thread about a program he created that allows you construct scenarios in a space time diagram and transform from one frame to another. Might be fun to try out Mentz's program and set up the scenario we were discussing and see if it adds up. See https://www.physicsforums.com/showpost.php?p=2864155&postcount=15
 
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  • #199


yuiop said:
Almost forgot this question!


There is no final arbiter for the elapsed times on A and D's clocks relative to each other when they never started and finished in the same place.



Here is one way to picture it. Imagine a 8 ly long rod at rest in frame A extending from B starts to where B finishes in A's frame. in D's frame this rod is moving at 0.5c to the right and B is chasing after it. In 16.166 yrs the end of the rod B is chasing moves 16.166*0.5=8.66 lys. Add this to the head start the end of the rod had and the total distance traveled by B in D's frame is 8.66+6.93=15.56 lys. This equates to a velocity of 15.56/16.166=0.928c.

You can do the calculations more formally, by considering the two events (A,B) and (B,C) in A's frame in x,t terms as (x1,t1) = (0,0) and (x2,t2) = (8,10) and converting those coordinates of the two events to the point of view of D using the Lorentz transformations. Just let me know if you need that clarifying.
I can see my bad habit of hasty mental appraisals may have struck again. :-)
I should tattoo on my forhead DO THE MATH before speaking.
In any case I need to refresh myself with this scenario as I have lost track with all the irrelevant dust in the air of this thread
 
  • #200


starthaus said:
You never answered the question I asked you: what is the necessary acceleration to reach 0.8c in 1s in your original scenario? Could you answer , please?
Hmm, out of curiosity I checked it, when we have:

[tex]\alpha = 329,356,000 m/s^2[/tex]

[tex]\tau = 1 s[/tex]

I get a coordinate velocity of 0.8 and a coordinate acceleration of 7488461 m/s2 about 2.3% or the proper acceleration.

It is rather high but on the other hand if we consider the maximum acceleration from QT perspective then we should get to somewhere in the 5.56 x 1051 m/s2 region.

Plenty of playroom it seems :-p
 
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  • #201


Passionflower said:
Hmm, out of curiosity I checked it, when we have:

[tex]\alpha = 329,356,000 m/s^2[/tex]

[tex]\tau = 1 s[/tex]It is rather high but on the other hand if we consider the maximum acceleration from QT perspective then we should get to somewhere in the 5.56 x 1051 m/s2 region.

Plenty of playroom it seems :-p

You mean, for a rocket carrying atomic clocks ?
That has to turn around from +0.8c to -0.8c?
Do you think anything would be left out of the rocket and the clocks? :-)
I get a coordinate velocity of 0.8 and a coordinate acceleration of 7488461 m/s[sup2[/sup] about 2.3% or the proper acceleration.

This is incorrect since [tex]a=\frac{\alpha}{\gamma^3}[/tex] where [tex]\gamma=1/0.6[/tex]. Your result is off by one order of magnitude.
 
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  • #202


starthaus said:
This is incorrect since [tex]a=\frac{\alpha}{\gamma^3}[/tex] where [tex]\gamma=1.33[/tex]. Your result is off by one order of magnitude.
Oh? I get a gamma of 1 2/3 not 1 1/3

How did you get 1 1/3?
The proper velocity is 1 1/3
 
  • #203


Passionflower said:
Oh? I get a gamma of 1 2/3 not 1 1/3

How did you get 1 1/3?
The proper velocity is 1 1/3
[tex]\beta=0.8[/tex]
[tex]\gamma=1/0.6[/tex]
 
  • #204


starthaus said:
[tex]\gamma=1/0.6[/tex]
Wait, hold on...

Yes that is 1 2/3 not 1 1/3 right?
 
  • #205


Passionflower said:
0.6? You mean 0.8 right?

No, I mean exactly what I wrote above.
 
  • #206


starthaus said:
No, I mean exactly what I wrote above.
No ignore that.

1/0.6 = 1 2/3 right?
 
  • #207


Passionflower said:
No ignore that.

1/0.6 = 1 2/3 right?

Yes, whatever 1/0.6 comes out to be. Your coordinate acceleration is off by a factor of 10.
 
  • #208


starthaus said:
Yes, whatever 1/0.6 comes out to be. Your coordinate acceleration is off by a factor of 10.
Hold on, so now we agree that gamma is 1 2/3

Ok, you are right, I see what happened the coordinate acceleration is in ly/y^2 and alpha is in m/s^2.

In ly/y^2:

Alpha is: 34,668,889 ly/y^2
A is: 7488460.769 ly/y^2

With a ratio of 21,6%

Good spotting!
 
  • #209


Passionflower said:
Hold on, so now we agree that gamma is 1 2/3

Ok, you are right, I see what happened the coordinate acceleration is in ly/y^2 and alpha is in m/s^2.

In ly/y^2:

Alpha is: 34,668,889 ly/y^2
A is: 7488460.769 ly/y^2

With a ratio of 21,6%

Good spotting!

I knew we'll agree.
 
  • #210


starthaus said:
I knew we'll agree.
Yes in math there is no middle between true and false!
 

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