There is no twin paradox - mathematical proof

[neopolitan]

For the twin paradox to be considered a true paradox the framing of the scenario must be stringent, that is to say we cannot permit assumptions to be ignored. Therefore I must start with a short description of the twin paradox followed by identification of the inherent assumptions.

From Einstein Light (http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm):

Jane and Joe are twins. Jane travels in a straight line at a relativistic speed v to some distant location. She then decelerates and returns. Her twin brother Joe stays at home on Earth. The situation is shown in the diagram, which is not to scale (see link above).

Joe observes that Jane's on-board clocks (including her biological one), which run at Jane's proper time, run slowly on both outbound and return leg. He therefore concludes that she will be younger than he will be when she returns. On the outward leg, Jane observes Joe's clock to run slowly, and she observes that it ticks slowly on the return run. So will Jane conclude that Joe will have aged less? And if she does, who is correct? According to the proponents of the paradox, there is a symmetry between the two observers, so, just plugging in the equations of relativity, each will predict that the other is younger. This cannot be simultaneously true for both so, if the argument is correct, relativity is wrong.

(The author goes on to explain that asymmetry resolves the paradox, this is an not entirely satisfactory explanation.)

There are a few assumptions, which are perhaps only obvious when one takes time to search for them.

"(S)ome distant location" appears sufficiently vague as to avoid creating problems but an inherent assumption is that this location shares the same frame as Joe.

By placing Joe on Earth we hide the other assumption, which is that we also share the same frame as Joe.

The at-a-distance observations of each other's clock potentially hides an assumption of instantaneous transmission of information. I doubt the author intended that, but it must be remembered that information is not transmitted instantaneously.

"(Jane) decelerates and returns" is distracting. As the author correctly points out this is a point of asymmetry. But a similar scenario (to be shown shortly) while show that it doesn't matter which decelerates and changes direction - Jane or the entire universe. It is generally assumed that the period during Jane changes direction is insignificant enough to ignore.

Finally, "Jane travels in a straight line at a relativistic speed v" begs the question "relativistic speed v relative to what?" The assumed answer is "relative to both Joe and the distant location" (and we the readers). This is a direct consequence of the assumption that Joe and the distant location share the same frame (and that we also share that frame).

Let me provide an analogous scenario.

Joe floats in a space suit with two clocks.

Jane sits at one end of an extremely long structure with another two clocks. At the other end of the structure is a beacon. According to Jane, the structure has a length of L. Joe knows this.

Jane and Joe pass each other twice, at relativistic velocities of v and -v. Joe and the beacon pass each other twice, also at relativistic velocities of v and -v (Jane and the beacon are fixed to the same structure and hence share the same frame).

Four events are noteworthy:

1. Joe and Jane are collocated as they pass for the first time. Their clocks begin measuring time elapsed.

2. Joe and the beacon are collocated as they pass for the first time. Joe's clocks are paused and the beacon sends a message to Jane's clocks to pause.

3. Joe and the beacon are collocated as they pass for the second time. Joe's clocks restart measuring time elapsed and the beacon sends a message to Jane's clocks to resume measuring time elapsed.

4. Joe and Jane are collocated as they pass for the second time. Their clocks stop measuring time elapsed and Joe and Jane exchange one of their clocks. Neither consults the other as they each attempt to work out what the other's clock will read.

Observe that I do not say who reverses direction. For the purposes of the mind experiment, we can say that both Joe and Jane were anaethetised while one of them reversed direction, but neither knows which of them has now changed direction relative to the third observer (the reader).

There is an asymmetry in this scenario, but Jane and Joe cannot determine on whose part that asymmetry lies.

Jane's calculations:

Joe is in motion relative to Jane. Jane calculates that Joe must take a period of 2L/v to travel between events 1 and 2 and events 3 and 4. She further calculates that because Joe is in motion, his clocks will run slow and will show a time elapsed of γ.2L/v where γ = sqrt (1-v^2/c^2). She can check her clock and see that the first period elapsed (between event 1 and event 2) was L/v + L/c and the second period elapsed (between event 3 and event 4) was L/v-L/c for a total of 2L/v.

Joe's calculations:

Jane is in motion relative to Joe. Joe therefore calculates that Jane's structure is foreshorted by a factor of γ. Therefore the time elapsed while the entirety of the structure passes twice will be γ.2L/v. Sure enough, he checks his clock and sees that this is the case.

Working out what Jane's clock will read is a little more complex. Joe knows that not only is Jane's structure foreshortened, but her clocks will also run slow by a factor of γ.

The first period elapsed can therefore be calculated as follows (noting that Jane's relative motion is in the same direction as the message from the beacon to Jane's clocks):

t1=γ.(γ.L/v + γ.L/(c-v))=γ^2.(L/v + L/(c-v))
=γ^2.(L/v.(c^2-v^2)/(c^2-v^2) + L(c+v)/(c^2-v^2))
=γ^2.(c^2.L/v - Lv + Lc + Lv)/(c^2-v^2)
=γ^2.(c^2.L/v + Lc)/(c^2-v^2)

but since γ^2 = 1 - v^2/c^2 = (c^2 - V^2)/c^2,

t1=(c^2 - V^2)/c^2 . (c^2.L/v + Lc)/(c^2-v^2)
=(c^2.L/v + Lc)/c^2
=L/v + L/c

The same process can be used to calculate that the second period elapsed is L/v-L/c. The total time elapsed on Jane's clock, as calculated by Joe, will be 2L/v - precisely the same as calculated by Jane.



There is no disagreement and there is no paradox, merely a poorly frame scenario.

cheers,

neopolitan

(If you want to read more, try http://www.geocities.com/neopolitonian/lightclock.doc and http://www.geocities.com/neopolitonian/sr.doc)

 

[Fredrik]

I haven't read the whole post, so you will have to discuss the details of your proof with someone else. I would just like to add that no proof is needed. The special theory of relativity is just the claim that space-time can be mathematically represented by Minkowski space. This claim, or any other physical theory for that matter, can't be proved or disproved mathematically. We need experiments to tell us if a theory is an accurate model of reality or not.

The only time when this is not true, is when the "theory" contradicts itself, but in that case, it would be completely wrong to call it a theory. "Nonsense" would be a more appropriate word.

Is there any way that SR could be nonsense? Many people don't realize that since SR is the simple claim I mentioned above, the only way a logical inconsistency could be present in the theory is if it's already present in the mathematical definition of Minkowski space. But Minkowski space is just the set $\mathbb{R}^4$ along with a few functions. If you know what those functions are, you also know that they can't introduce contradictions into the theory. So if there are contradictions in SR, they would have to be present in the definition of real numbers. But real numbers can be constructed from rational numbers, rational numbers can be constructed from integers, and integers can be constructed from the axioms of set theory. So if there really was a twin paradox, it would be the downfall of pretty much all of mathematics.

I might as well take this opportunity to post a space-time diagram that I made for a discussion in another forum.

http://web.comhem.se/~u87325397/Twins.PNG (Why doesn't the "img" tag work?)

This is the Matlab code that generated the graphs, in case someone wants to play around with it:

clf; axis equal; axis([0 40 0 40]); ylabel('t (years)'); xlabel('x (light-years)'); hold on;

x=0:0.01:16;
plot(x,x,'k:'); plot(x,-x+36,'k:');
plot(x,1.25*x,'k'); plot(x,-1.25*x+40,'k');

A=6; plot(x,sqrt(x.^2+A^2),'g');
A=12; plot(x,sqrt(x.^2+A^2),'g');

z=x-16;
A=6; plot(x,20+sqrt(z.^2+A^2),'g');
A=12; plot(x,20+sqrt(z.^2+A^2),'g');

plot(x,0.8*x+7.2,'b');
plot(x,0.8*x+3.6,'b');
plot(x,-0.8*x+32.8,'r');
plot(x,-0.8*x+36.4,'r');

plot(x,6,'c'); plot(x,12,'c'); plot(x,20,'c'); plot(x,26,'c'); plot(x,32,'c');

s=20:0.01:40; plot(16,s,'k');

 

[neopolitan]

Goddag yxskaft, Fredrik!

You might want to read posts before replying, since I make no claim to prove or disprove your statement: "The special theory of relativity is just the claim that space-time can be mathematically represented by Minkowski space."

SR is entirely safe, I am just trying make people aware that there is no twin paradox and that you can demonstrate that fact mathematically.

cheers,

neopolitan

 

[Fredrik]

I read enough to understand that. My point is that you don't have to use a thought experiment to refute the twin "paradox". A thought experiment is always carried out within a theoretical framework, and in this case it's sufficient to only think about what the framework is to complete the proof.

 

[neopolitan]

I am a little unclear on whether you agree that the twin paradox is poorly stated, and hence irrelevant, or do you think that the traveling twin in the "paradox" can explain why she ages less than the nominally stationary twin. Remember the paradox revolves around the idea that both of the twins can consider themselves to be stationary and the other to be in motion. If you state that the traveling twin knows herself to be travelling, then you destroy the grounds of the paradox entirely.

Your image indicates that B is aware that she is in motion ("The rocket turned around a minute ago"). The image also seems to be fundamentally wrong. A ages 20 years on each leg and B ages 12 years on each leg. There is no 2 minute discontinuity in which A ages 25.6 years. I would appreciate other people's input on that, but there is no requirement that I am aware of for B to return to A in order for relativistic effects to manifest.

If merely changing directions made you age so significantly then pilots, ferry captains, postmen and bus and train drivers would all be dead and waving your hand would cause it to age significantly faster than the rest of your body. (Procreation would also represent a danger to parts of the male party.)

It seems that you might benefit from reading my post in entirety since I believe it addresses the twin paradox issue far more safely than the introduction such deadly discontinuities.

cheers,

neopolitan

 

[jcsd]

I am a little unclear on whether you agree that the twin paradox is poorly stated, and hence irrelevant, or do you think that the traveling twin in the "paradox" can explain why she ages less than the nominally stationary twin. Remember the paradox revolves around the idea that both of the twins can consider themselves to be stationary and the other to be in motion. If you state that the traveling twin knows herself to be travelling, then you destroy the grounds of the paradox entirely.

'stationery' and 'moving' in special relativity are frame-dependent qualities of the motion of an object/observer, but 'inertial' and 'non-inertial' are absolute qualities of the motion of an object/observer.

As you've noted which twin is stationery and which is not is pretty much irrelevant, however the twin paradox arises because one twin is inertial and the other is non-inertial.

Your image indicates that B is aware that she is in motion ("The rocket turned around a minute ago"). The image also seems to be fundamentally wrong. A ages 20 years on each leg and B ages 12 years on each leg. There is no 2 minute discontinuity in which A ages 25.6 years. I would appreciate other people's input on that, but there is no requirement that I am aware of for B to return to A in order for relativistic effects to manifest.

When, as viewed by an inertial frame, 'the rocket turns around' it means that the rocket is non-inertial. Whether the rocket is 'in motion' is a moot point, that the rocket is non-inertial is not. (btw the twin that 'turns round' will age less than the other twin).

When the ageing occurs again is a moot point it's enitrly dependent on how we compare the clocks when they are at a distance to each other and as we know in relativity simulatenity fails with disnatce.

Simulatneity is still absolute though when thereis no distance involved so as when the twins start and stop the clocks they are conincidnetal we can say absolutely that the clocks have measured different lengths of time between the same events

If merely changing directions made you age so significantly then pilots, ferry captains, postmen and bus and train drivers would all be dead and waving your hand would cause it to age significantly faster than the rest of your body. (Procreation would also represent a danger to parts of the male party.)

The relatvistic corrections for the motion of pilots, ferry captains, etc are small as to be insignicficant (plus as I said the twin who changes directions actually ages less)

It seems that you might benefit from reading my post in entirety since I believe it addresses the twin paradox issue far more safely than the introduction such deadly discontinuities.

Your post doesn't really address the twin paradox.

The introduction of discontinous velocites is in the twin paardox is unphysical yes, but it's simply done for ease of calculation as it doesn't qualitively effect the result. The twin paardox can be demsontsrated with finite accelartions, it's just slightly more difficult to model and the basic result is the same.

 

[neopolitan]

Try reading my first post, ie post #1, rather than my replies to Fredrik.

By the way, I think your definition of inertial is open to question. I deliberately tried to avoid using the term "inertial" because the "stationary" observer is equally as inertial as the "travelling" observer. Without an absolute frame of refence, you can't talk about absolute qualities.

But I repeat, try reading my first post, please respond to that rather than responding to my responses to the responses of others.

cheers,

neopolitan

 

[jcsd]

Try reading my first post, ie post #1, rather than my replies to Fredrik.

I have already read your first post.

By the way, I think your definition of inertial is open to question. I deliberately tried to avoid using the term "inertial" because the "stationary" observer is equally as inertial as the "travelling" observer. Without an absolute frame of refence, you can't talk about absolute qualities.

Why delibrately avoid using the term inertial? We're talking about special relativity and both the postulates of special relativity specifically reference 'inertial observers'. If you need me to expand on this I will do.

Some quantities/qualitities are frame dependent some are not, so it is not wrong per se to talk about absolutes.

Your first post is erroneous and does not address the twin paradox, precisley because you have not recgnised that both observers are not 'equally as inertial'.

I am sorry I do not want to go through your first post in fine comb detail, it is rather long. But the 'error' is clear - you haven't recognised you cannot escape the differemce between inertial and non-inertial observers. The equations you use apply to inertial frames only, whereas (at least) one of the twins must be non-inertial.

 

[jostpuur]

So we have chosen the frame so that the Earth is in origo and in rest, and we ignore the orbiting around sun, so that this is an inertial frame.

"(S)ome distant location" appears sufficiently vague as to avoid creating problems but an inherent assumption is that this location shares the same frame as Joe.

I guess you mean to point out, that the distant location $x\in\mathbb{R}^3$ in the spatial space is chosen so that it doesn't depend on time, in this frame. Or if we consider its world line $x(t)$ in the four dimensional space time, it is a straight line upwards.

In other frames the point $x(t)\in\mathbb{R}^3$ would change as function of time, and in your language "it would not share the same frame".

Okey, I understood you correctly?

By placing Joe on Earth we hide the other assumption, which is that we also share the same frame as Joe.

And then we note, that the Joe is in rest in this chosen frame too.

Finally, "Jane travels in a straight line at a relativistic speed v" begs the question "relativistic speed v relative to what?" The assumed answer is "relative to both Joe and the distant location" (and we the readers). This is a direct consequence of the assumption that Joe and the distant location share the same frame (and that we also share that frame).

We talk about the speed in the chosen frame. Since the Joe and the distant location are rest in this frame, talking about the speed in this frame, or about the speed relative to the Joe and the distance location, are obviously equivalent. This is so trivial claim, that I have difficulty seeing what should be considered as assumptions and what as consequences.

So far everything you have said seems to be fine, although it was a bit laborous to translate your explanations into more understandable form. I must tell you, that you are underestimating physicists. Your remarks contain nothing new to a person who knows relativity.

"(Jane) decelerates and returns" is distracting. As the author correctly points out this is a point of asymmetry. But a similar scenario (to be shown shortly) while show that it doesn't matter which decelerates and changes direction - Jane or the entire universe. It is generally assumed that the period during Jane changes direction is insignificant enough to ignore.

I seriously doubt this! It matters very much who is accelerating. Are you sure you are aware of the fact, that the Jane will be younger than Joe after the trip?

I think I'll post this now, and see rest of your post next. To avoid too long post.

 

[neopolitan]

By the long structure and the beacon you mean, that the other end of the long structure is attached to some massive object (that can be assumed to be stationary), and the long structure is rotating around it?

No, I never said the structure was attached to anything other than Jane and the beacon. I did later say they would move relative to each other in such a way that it is not possible for either to know which was moving. That pretty much eliminates the possibility of rotational movement on Jane's part.

Given this fundamental misunderstanding. I will not respond to anything else in your post.

The setup is such that it is analogous to moving between two predetermined, fixed points in space (as Earth and "some distant location" are assumed to be in standard formation of the twin paradox). You can assume that the universe is empty with the exception of Joe, Jane and the structure plus a mischievous god which gives them the relative motion described, if that makes it easier.

Cheers,

neopolitan

 

[neopolitan]

I seriously doubt this! It matters very much who is accelerating. Are you sure you are aware of the fact, that the Jane will be younger than Joe after the trip?

Check the scenario and the maths which follows, note that I completely eliminated the question of acceleration. Note further that despite eliminating acceleration such that we can never know who accelerated, Jane ends up younger than Joe.

Cheers,

neopolitan

 

[jostpuur]

Given this fundamental misunderstanding. I will not respond to anything else in your post.

I deleted the post now, so that it is not causing confusion anymore.

 

[neopolitan]

Why delibrately avoid using the term inertial? We're talking about special relativity and both the postulates of special relativity specifically reference 'inertial observers'. If you need me to expand on this I will do.

I avoid the term inertial because some people misunderstand it to mean being in motion. It doesn't. A stationary frame also has inertia - inertia is what makes it necessary to expend energy to set the frame in motion. So, the term "inertial observers" basically means that the observers won't change velocity. That velocity can be zero.

You don't need to expand on this, you need to check your terms.

cheers,

neopolitan

 

[jostpuur]

"Joe floats around in a space suit, Jane is located on a long structure, there is a beacon in the other end, and then Joe and Jane pass each others twice with relativistic velocities v and -v."

I think nobody can understand what is happening there. Explain it with coordinates

 

[jostpuur]

btw. I took today two difficult exams, and I feel like I've deserved a free evening that I can waste doing nothing useful.

(If you want to read more, try http://www.geocities.com/neopolitonian/lightclock.doc

Wwwhoooo... okey. One should always check links in the original post, it seems. Let's see if I manage post this before a lock hits the thread.

Page 6:

Observe that two sets of Lorentz transformations are employed, one from the perspective of the nominally stationary observer, S:

$$\Delta t' = \gamma(\Delta t - v\Delta x / c^2)\quad\quad\quad (8)$$
$$\Delta x' = \gamma(\Delta x - v\Delta t)$$

and one set from the perspective of the observer who is nominally in motion, M:

$$\Delta t = \gamma(\Delta t' + v\Delta x' / c^2)\quad\quad\quad (9)$$
$$\Delta x = \gamma(\Delta x' + v\Delta t')$$

In the derivation of time dilation, (8) – the temporal Lorentz transformation from the perspective of S – is used. For the derivation of length contraction, however, (9) – the temporal Lorentz transformation from the perspective of M – is brought to bear. The use of the prime is therefore clearly not consistent.

The use of prime is fully consistent, since the equations 8 and 9 are in fact equivalent, as you can verify by exercising your algebra skills by solving $\Delta x$ and $\Delta t$ out of the equations 8, or by solving $\Delta x'$ and $\Delta t'$ out of the equations 9.

Page 7:

The photon in a light clock in a “stationary” frame not only travels through a distance of 2L but also through a period of time t. The total magnitude of the journey, $\mu$, can be expressed:

$$\mu^2 = \cdots = 2L^2 + \frac{1}{2}(ct)^2$$

You have replaced the usual spacetime interval with a new concept of magnitude of journey, which seems to be the Pythagorean distance in four dimensional space, and then assume that this distance is invariant in boosts.

 

[neopolitan]

"Joe floats around in a space suit, Jane is located on a long structure, there is a beacon in the other end, and then Joe and Jane pass each others twice with relativistic velocities v and -v."

I think nobody can understand what is happening there. Explain it with coordinates

Whose coordinates?

According to Jane, she sits at (0,0,0,t) and at a distance L away from her at the other end of the structure is a beacon. To make it easier on herself, she nominates the length of the structure as her x-axis so that the beacon sits at (L,0,0,t).

According to Joe, he sits at (0,0,0,t).

Event 1 is at (0,0,0,t1).

Event 4 is at (0,0,0,t4).

Events 2 and 3 are at either

• (L,0,0,t2 according to Jane) and (L,0,0,t3 according to Jane) ), or
• (0,0,0,t2 according to Joe) and (0,0,0,t3 according to Joe),

depending on whether you listen to Jane or Joe. Note that L is the length of the structure at rest (which it can only be in Jane's frame).

The issue is that either

• Joe moves past Jane, along the structure and past the beacon with a speed of v and then later moves back past the beacon, along the structure and past Jane again, also with a speed of v, or
• Jane and her structure move past Joe in one direction with a speed of v, and then back again in the other direction also with a speed of v.

Neither of them can say with certainty which of them does the moving. Despite that, the time spent moving past is each other is such that less time elapses for Joe, irrespective of whether he moves or not.

People will argue about simultaneity, but the mathematics doesn't support that being a problem.

Try to work through the mathematics, don't trust that I got it right but don't just assume that I got it wrong either. Work it through and come to your own conclusion.

I hope this clarifies things a little.

neopolitan

 

[neopolitan]

The use of prime is fully consistent, since the equations 8 and 9 are in fact equivalent, as you can verify by exercising your algebra skills by solving $\Delta x$ and $\Delta t$ out of the equations 8, or by solving $\Delta x'$ and $\Delta t'$ out of the equations 9.

The use of the prime is consistent between equations 8 and 9, because as you say they are equivalent but only in so much as they are the same equation from the perspectives of two different frames of reference. Please check the use to which the equations are then used, taking these two perspectives from two different frames of reference and then using them in one single frame of reference. That is the inconsistency. I don't claim there is an inconsistency before that usage.

You have replaced the usual spacetime interval with a new concept of magnitude of journey, which seems to be the Pythagorean distance in four dimensional space, and then assume that this distance is invariant in boosts.

Not really, I start at the beginning and work forwards, identifying assumptions as I go and eliminating the ones that we all know not to be valid. What I don't do, and I can understand that some people get upset about it, is use the ramifications of other derivations as the basis for my derivation. That is to say, I am not so much assuming that "Pythagorean distance in four space" is invariant in boosts, I merely not assuming that it isn't invariant.

It was a while since I last worked on that paper, but as I recall I was discussing the motion of one photon and the different interpretations of the photons movement in two different frames of reference. A single photon can only make one journey, irrespective of how it is perceived by different observers. If you want to say that makes Pythagorean distance in four space invariant, then I cannot argue against it.

Without taking refuge in other derivations, can you explain why Pythagorean distance in four space must not (or should not, or perhaps cannot) be invariant? The mathematics works for my derivation without that requirement.

cheers,

neopolitan

 

[CPL.Luke]

neopolitan if i understand you correctly joe moves past jane with a velocity u (in janes frame) and somehow returns later though the mechanism for such reversal is left unkown.

the problem with this is that the only way for joe to return would be if there was a gravitational field to bring him back (the only way to keep him inertial), however in this instance joe would be effected by gravitational time dilation effects which would be unkown) and if joe ever underwent acceleration than the equations of special rleativity take on a different form, and his time would run slower.

 

[neopolitan]

Hi Luke,

Did you check the first post, in that I make clear that I am only measuring the time elapsed between two pairs of events. Anything that happens outside of the time which elapses between each of the pairs of events is immaterial. Yes, Joe or Jane may be affected by any gravitation used to turn them around, but their clocks will be stopped during that time.

The twin paradox is usually framed in such a way that you are only talking about two inertial journeys, one there and one back. I just make that blatantly obvious and show that you expect one of the twins to experience more time elapsing than the other one.

The mathematics, if you work it through, shows that the twins will not expect anything else.

cheers,

neopolitan

 

[jcsd]

I avoid the term inertial because some people misunderstand it to mean being in motion. It doesn't. A stationary frame also has inertia - inertia is what makes it necessary to expend energy to set the frame in motion. So, the term "inertial observers" basically means that the observers won't change velocity. That velocity can be zero.

You don't need to expand on this, you need to check your terms.

cheers,

neopolitan

'inertial observers' are referenced in the postulates of relativity so the term needn't be avoided. Inertial observers do not experince acceleration so do not change direction (as viewed by another inertial observer).

 

[jcsd]

Hi Luke,

Did you check the first post, in that I make clear that I am only measuring the time elapsed between two pairs of events. Anything that happens outside of the time which elapses between each of the pairs of events is immaterial. Yes, Joe or Jane may be affected by any gravitation used to turn them around, but their clocks will be stopped during that time.

The twin paradox is usually framed in such a way that you are only talking about two inertial journeys, one there and one back. I just make that blatantly obvious and show that you expect one of the twins to experience more time elapsing than the other one.

The mathematics, if you work it through, shows that the twins will not expect anything else.

cheers,

neopolitan

This is were you're falling down. Just because an observers total journey can be broken down into two journeys in which their motion is inertial it does not mean their total journey can be viewed as inertial (if we allow infinite accleration as we have done in this case).

 

[pervect]

"(Jane) decelerates and returns" is distracting. As the author correctly points out this is a point of asymmetry. But a similar scenario (to be shown shortly) while show that it doesn't matter which decelerates and changes direction - Jane or the entire universe. It is generally assumed that the period during Jane changes direction is insignificant enough to ignore.

Finally, "Jane travels in a straight line at a relativistic speed v" begs the question "relativistic speed v relative to what?" The assumed answer is "relative to both Joe and the distant location" (and we the readers). This is a direct consequence of the assumption that Joe and the distant location share the same frame (and that we also share that frame).

There *is* a well-defined meaning to traveling in a "straight line" in special relativity. And there is a definite difference between worldlines which are "straight" and "not-straight" in special relativity.

The twin paradox in special relativity can be understood in many ways, but one important way of understanding it is via the means of the principle of maximal aging. In a very slightly modified form (replacing maximal with extremal) this principle has the advantage of allowing a smooth transition to general relativity from special relativity.

This is described in Taylor's textbook, "Exploring black holes", sample chapters of which are available online at http://www.eftaylor.com/pub/chapter1.pdf, which I will quote in part because I think it is illuminating. I encourage people to read the entire chapter, of which I will only quote a (longish) part.

To get ready for curved spacetime (whatever that may mean), look further
at the motion of a free particle in flat spacetime, the arena of the free-float
frame (Section 8) in which special relativity correctly describes motion.
How does a free particle move in flat spacetime? We say: “What a ridiculous
question! Everyone knows that a free particle moves with constant
speed in a straight line—at least as observed in a free-float frame.” Ah yes,
but why does a free particle move straight with constant speed? What lies
behind this motion? Our answer for flat spacetime will be a trial run for
the description of motion in curved spacetime, the arena of general
relativity.

A deep description of motion arises from the famous Twin Paradox.
Recall that one identical twin relaxes on Earth while her twin sister frantically
travels to a distant star and returns. When the two meet again, the
stay-at-home twin has aged more than her traveling sister. (This outcome
can be predicted by extending Sample Problems 1 and 2 to include return
of the traveler to the point of origin.) Upon being reunited, the “identical
twins” are no longer identical. Very strange! But (almost) no one who has
studied relativity doubts the difference in age, and experiments with fastmoving
particles verify it.

Which twin has the motion we can call natural? Isaac Newton has a definition
of natural motion. He would say, “A twin at rest tends to remain at
rest.” So it is the stay-at-home twin who moves in the natural way. In contrast,
the out-and-back twin suffers the forces required to change her state
of motion—from outgoing motion to incoming motion—so that the two
sisters can meet again in person. The motion of the traveling twin is
forced, not natural.

Viewed from a second relatively moving free-float frame, the stay-at-home
twin moves with constant speed in a straight line. Hers is also natural
motion. Newton would say, “A twin in motion tends to continue this
motion at constant speed in a straight line.” So the motion of the stay-on-Earth twin is also natural from the viewpoint of a second frame in uniform
relative motion—or from any frame moving uniformly with respect to the
original frame. In any such frame, the time lapse on the wristwatch of the
stay-at-home twin can be calculated from the metric (equation [1]).

The lesson of the Twin Paradox is that the natural motion of a free object
between two events in flat spacetime is the one for which the wristwatch
worn by the object has a maximum time reading between those two
events. Purists insist that we say not maximum reading but rather extremal
reading: either maximum or minimum. This book contains only examples
of maximum wristwatch time for natural motion. Still, let’s try to keep the
purists happy! Replace the two words maximum and minimum with the
single word extremal. The result is the Principle of Extremal Aging.
Principle of Extremal Aging: The path a free object takes between two events
in spacetime is the path for which the time lapse between these events, recorded
on the object’s wristwatch, is an extremum.

It turns out that the Principle of Extremal Aging describes motion even
when spacetime is not flat. The Principle of Extremal Aging accompanies
us into curved spacetime, into the realm of general relativity. But for now
we stay in flat spacetime and use the Principle of Extremal Aging to derive
relativistic expressions for energy and momentum.

So in conclusion, given two events in the flat spacetime of special relativity, A and B, given that B is in the light-cone of A, there is one and only one trajectory ("a straight line") starting at A and ending at B which maximizes the age of a clock traversing it.

The difference between "extremal" and "maximal" becomes important in general relativity, but this thread is long enough already without getting into that distinction. For a complete technical disucssion see for instance

http://www.eftaylor.com/pub/Gray&TaylorAJP.pdf

 

[neopolitan]

This is were you're falling down. Just because an observers total journey can be broken down into two journeys in which their motion is inertial it does not mean their total journey can be viewed as inertial (if we allow infinite accleration as we have done in this case).

Hi jcsd,

I never claimed the journeys were inertial overall. Two inertial legs accords with the standard framing of the twin paradox. I am fully aware that a only each leg of the journey (event 1-event 2 and event 3-event 4) is inertial but what happens outside of each of those legs is carefully removed from consideration.

This is a standard experimentation procedure, you do your damnedest to isolate causes. In the scenario at post #1, I deliberately eliminate acceleration as a possible cause - we do not know which of the twins experienced acceleration! Even when we have eliminated acceleration, we still find that Joe ends up experiencing less elapsed time than Jane.

This means that acceleration has nothing to do with the phenomenon. This has already been shown elsewhere so I don't consider it news.

I repeat, for newcomers, please read the first post, work through the mathematics and then comment.

cheers,

neopolitan

 

[neopolitan]

There *is* a well-defined meaning to traveling in a "straight line" in special relativity. And there is a definite difference between worldlines which are "straight" and "not-straight" in special relativity.

Thanks for that Pervect,

It's encouraging to see that you read and responded to the original post :) Please note though that my original words were, inter alia:

Finally, "Jane travels in a straight line at a relativistic speed v" begs the question "relativistic speed v relative to what?" The assumed answer is "relative to both Joe and the distant location" (and we the readers). This is a direct consequence of the assumption that Joe and the distant location share the same frame (and that we also share that frame).

The concern I raise is not about the issue of straight lines (or geodesics) but rather about the under what terms we can say that Jane travels at a specific speed.

I have no argument about the concept of straight lines, geodesics nor extremal aging.

Note that in Taylor's description he assigns forced motion to one twin. My scenario deliberately makes that assignment impossible so Taylor's work, while interesting, doesn't apply.

Cheers,

neopolitan

 

[neopolitan]

I must have been tired yesterday.

Reviewing my original post I see that I may well have caused confusion. The scenarios presented are not analogous because in my scenario I assign Jane and Joe different roles from those assigned on the Einstein Light website. My Joe ages less than Joe Wolfe's Joe while my Jane ages more than Joe Wolfe's Jane.

To alleviate the probable confusion, I will restate my scenario with two new characters, Stewart (equivalent to the twin paradox' stay-at-home twin) and Trevor (equivalent to the twin paradox' traveling twin). I will also incorporate some clarifications that I didn't initially consider necessary.

Let me provide a scenario which is analogous to the scenario described in the twin paradox.

Trevor floats in a space suit with two clocks.

Stewart sits at one end of an extremely long structure which also floats in space, unattached to anything bar Stewart and the beacon with another two clocks. At the other end of the structure is a beacon. According to Stewart, the structure has a length of L. Trevor knows this.

Observe that Stewart represents "the Earth" and the beacon represents "some distant location" in the twin paradox. The assumption that "the Earth" and "some distant location" have a fixed separation is inherent in the twin paradox.

Stewart and Trevor are sufficiently distant from any masses as to be considered to be alone in the universe, with no gravitational field in effect. The gravitation exerted by Stewart and his structure on Trevor is negligible.

Stewart and Trevor pass each other twice, at relativistic velocities of v and -v. Trevor and the beacon pass each other twice, also at relativistic velocities of v and -v (Stewart and the beacon are fixed to the same structure and hence share the same frame).

Four events are noteworthy:

1. Trevor and Stewart are collocated as they pass for the first time. Their clocks begin measuring time elapsed.

2. Trevor and the beacon are collocated as they pass for the first time. Trevor's clocks are paused and the beacon sends a message to Stewart's clocks to pause.

3. Trevor and the beacon are collocated as they pass for the second time. Trevor's clocks restart measuring time elapsed and the beacon sends a message to Stewart's clocks to resume measuring time elapsed.

4. Trevor and Stewart are collocated as they pass for the second time. Their clocks stop measuring time elapsed and Trevor and Stewart exchange one of their clocks. Neither consults the other as they each attempt to work out what the other's clock will read.

Observe that I do not say who reverses direction. For the purposes of the mind experiment, we can say that both Trevor and Stewart were anaethetised while one of them reversed direction, but neither knows which of them has now changed direction relative to any third observer (such as the reader). The clocks are paused while any acceleration takes place and therefore no acceleration affects the measured time elapsed.

There is an asymmetry in this scenario, but Stewart and Trevor cannot determine on whose part that asymmetry lies.

Stewart's calculations:

Trevor is in motion relative to Stewart. Stewart calculates that Trevor must take a period of 2L/v to travel between events 1 and 2 and events 3 and 4. Stewart further calculates that because Trevor is in motion, his clocks will run slow and will show a time elapsed of γ.2L/v where γ = sqrt (1-v^2/c^2). Stewart can check his clock and see that the first period elapsed (between event 1 and event 2) was L/v + L/c and the second period elapsed (between event 3 and event 4) was L/v-L/c for a total of 2L/v.

Trevor's calculations:

Stewart is in motion relative to Trevor. Trevor therefore calculates that Stewart's structure is foreshorted by a factor of γ. Therefore the time elapsed while the entirety of the structure passes twice will be γ.2L/v. Sure enough, Trevor checks his clock and sees that this is the case.

Working out what Stewart's clock will read is a little more complex. Trevor knows that not only is Stewart's structure foreshortened, but that Stewart's clocks will also run slow by a factor of γ.

The first period elapsed can therefore be calculated as follows (noting that Stewart's relative motion is in the same direction as the message from the beacon to Stewart's clocks):

t1=γ.(γ.L/v + γ.L/(c-v))=γ^2.(L/v + L/(c-v))
=γ^2.(L/v.(c^2-v^2)/(c^2-v^2) + L(c+v)/(c^2-v^2))
=γ^2.(c^2.L/v - Lv + Lc + Lv)/(c^2-v^2)
=γ^2.(c^2.L/v + Lc)/(c^2-v^2)

but since γ^2 = 1 - v^2/c^2 = (c^2 - V^2)/c^2,

t1=(c^2 - V^2)/c^2 . (c^2.L/v + Lc)/(c^2-v^2)
=(c^2.L/v + Lc)/c^2
=L/v + L/c

The same process can be used to calculate that the second period elapsed is L/v-L/c. The total time elapsed on Stewart's clock, as calculated by Trevor, will be 2L/v - precisely the same as calculated by Stewart.

​

There is no disagreement and there is no paradox, merely a poorly framed scenario.

cheers,

neopolitan

(If you want to read more, try http://www.geocities.com/neopolitonian/lightclock.doc and http://www.geocities.com/neopolitonian/sr.doc)

 

[paw]

You say,

The clocks are paused while any acceleration takes place and therefore no acceleration affects the measured time elapsed.

in an attempt to eliminate acceleration from the argument. However, turning the clocks off simply stops the twins from recording elapsed time in their own frames. It doesn't stop the passage of time. The accelerating twins clock is paused for less proper time than the inertial twin. Or to put it another way, the events of pausing, and restarting, the clocks are not simultaneous in all frames. We could tell which twin accelerated by the different times their clocks were off (if you include a mechanism to do this).

The fact always remains; the twin who accelerated, or the twin with the longer world line if you prefer, is the twin with the shortest accumulated proper time.

 

[neopolitan]

Paw,

You merely make a statement, you don't provide any proof.

I provide evidence that the acceleration is not the cause, since the twin who appears to move along the structure experiences less elapsed time than the one who sits on the structure, irrespective of which twin changes direction.

This is a quite simple proof that there is no twin paradox. The proof can be shown mathematically. Almost everyone admits that there really isn't a twin paradox. The only point of contention seems to be what is the best explanation for why it isn't a paradox.

Your solution seems to be vague and complex. You might not think it is complex, but let's see your mathematical proof. Can you produce a mathematical proof which is more simple than mine?

Sorry if that comes over as aggressive, but physics isn't really about the exchange of opinions.

cheers,

neopolitan

 

[pervect]

Paw,

You merely make a statement, you don't provide any proof.

I'm not sure he needs to provide a proof to question your "proof". Your "proof" seems to be problematical to me as well.

I provide evidence that the acceleration is not the cause, since the twin who appears to move along the structure experiences less elapsed time than the one who sits on the structure, irrespective of which twin changes direction.

I originally thought both twins had a structure in your example, but on re-reading it I see that they don't. However, there is nothing significant to the physics of being "part of a structure". We can certainly add another structure to the other twin in the problem without changing anything.

Also, if the whole structure is to be accelerated, the issue of the rigidity (or rather, the lack o f rigidity) of the structure has to be addressed, something that you have not done.

This is a quite simple proof that there is no twin paradox. The proof can be shown mathematically. Almost everyone admits that there really isn't a twin paradox. The only point of contention seems to be what is the best explanation for why it isn't a paradox.

I agree with many points. The Twin paradox is not a true paradox. It is a "paradox" simply in the term of being a surprising or counter-intuitive result. Most people agree there is no true paradox. And there are certainly a lot of different explanations of the twin paradox out there. The sci.physics.faq presents at least three, for example.

Your solution seems to be vague and complex.

I feel much the same about your "solution" and "proof", unfortunately. The diagram is pretty clear, and seems to be correct. (Oops, I now see that was someone else's diagram! Sorry for the confusion, but that makes me even more leery about your "proof").

However, your post is murky, and I don't quite understand your point.

It's also a bit disturbing the way you appear to dismiss other respected and standard explanations of the twin paradox beside your own.

 

[JesseM]

Let me provide an analogous scenario.

Joe floats in a space suit with two clocks.

Jane sits at one end of an extremely long structure with another two clocks. At the other end of the structure is a beacon. According to Jane, the structure has a length of L. Joe knows this.

Jane and Joe pass each other twice, at relativistic velocities of v and -v. Joe and the beacon pass each other twice, also at relativistic velocities of v and -v (Jane and the beacon are fixed to the same structure and hence share the same frame).

Four events are noteworthy:

1. Joe and Jane are collocated as they pass for the first time. Their clocks begin measuring time elapsed.

2. Joe and the beacon are collocated as they pass for the first time. Joe's clocks are paused and the beacon sends a message to Jane's clocks to pause.

3. Joe and the beacon are collocated as they pass for the second time. Joe's clocks restart measuring time elapsed and the beacon sends a message to Jane's clocks to resume measuring time elapsed.

4. Joe and Jane are collocated as they pass for the second time. Their clocks stop measuring time elapsed and Joe and Jane exchange one of their clocks. Neither consults the other as they each attempt to work out what the other's clock will read.

Observe that I do not say who reverses direction. For the purposes of the mind experiment, we can say that both Joe and Jane were anaethetised while one of them reversed direction, but neither knows which of them has now changed direction relative to the third observer (the reader).

If you don't say who reverses direction, it's impossible to predict who will have aged more when they pass for the second time--although in your scheme, where they artificially stop their clocks at certain moments, you can say how much time has elapsed on these started-and-stopped clocks between meetings.

Jane's calculations:

Joe is in motion relative to Jane. Jane calculates that Joe must take a period of 2L/v to travel between events 1 and 2 and events 3 and 4. She further calculates that because Joe is in motion, his clocks will run slow and will show a time elapsed of γ.2L/v where γ = sqrt (1-v^2/c^2). She can check her clock and see that the first period elapsed (between event 1 and event 2) was L/v + L/c and the second period elapsed (between event 3 and event 4) was L/v-L/c for a total of 2L/v.

But L/v + L/c is not the time between event 1 and event 2 according to Jane, rather it's the time between event 1 and event 2b, namely:

event 2b: The signal that was sent out by the beacon at the moment Joe passed it the first time catches up with Jane and she stops her clock

event 2 and event 2b are two separate events, and no matter whose frame you choose, they occur at different times. The time assigned to a given event in some inertial frame is not the time that the light from the event is observed; for example, if in 2007 according to my clock, I see the light from a nova that is 100 light-years away according to a ruler at rest in my frame, then I retroactively say this event occurred at time-coordinate 1907 in my frame, since I assume the light took 100 years to travel the 100 light-years between the nova and myself.

Likewise, we could also define the event 3b--

event 3b: The signal that was sent out by the beacon at the moment Joe passed it the second time catches up with Jane and she stops her clock

...and then the time on Jane's clock between 3b and 4 will be L/v - L/c.

Joe's calculations:

Jane is in motion relative to Joe. Joe therefore calculates that Jane's structure is foreshorted by a factor of γ. Therefore the time elapsed while the entirety of the structure passes twice will be γ.2L/v. Sure enough, he checks his clock and sees that this is the case.

Working out what Jane's clock will read is a little more complex. Joe knows that not only is Jane's structure foreshortened, but her clocks will also run slow by a factor of γ.

The first period elapsed can therefore be calculated as follows (noting that Jane's relative motion is in the same direction as the message from the beacon to Jane's clocks):

t1=γ.(γ.L/v + γ.L/(c-v))=γ^2.(L/v + L/(c-v))
=γ^2.(L/v.(c^2-v^2)/(c^2-v^2) + L(c+v)/(c^2-v^2))
=γ^2.(c^2.L/v - Lv + Lc + Lv)/(c^2-v^2)
=γ^2.(c^2.L/v + Lc)/(c^2-v^2)

but since γ^2 = 1 - v^2/c^2 = (c^2 - V^2)/c^2,

t1=(c^2 - V^2)/c^2 . (c^2.L/v + Lc)/(c^2-v^2)
=(c^2.L/v + Lc)/c^2
=L/v + L/c

The same process can be used to calculate that the second period elapsed is L/v-L/c.

Yes, regardless of who accelerates, the time elapsed on Joe's clock between passing Jane and passing the beacon in both directions (i.e. the time on Joe's clock between event 1 and event 2 + the time on Joe's clock between event 3 and event 4) will be gamma*2L/v, and the total time elapsed on Jane's clock between the event of Joe passing her for the first time and her getting the signal from the beacon when Joe passed it the first time, plus the time elapsed on Jane's clock between the event of her getting the signal from the beacon when Joe passed it the second time and the event of Joe passing her the second time (i.e. the time on Jane's clock between event 1 and event 2b + the time on Jane's clock between event 3b and event 4) will be 2L/v. But of course there will be some significant gap between her receiving the first signal from the beacon and her receiving the second signal (unless Joe turns around right as he passes the beacon), and if you include this extra time instead of artificially stopping the clocks, then Jane's age will be greater if Joe was the one who accelerated to turn around, while Jane's age will be less if Jane was the one who accelerated.

There is no disagreement and there is no paradox, merely a poorly frame scenario.

But by artificially stopping the clocks in the way you do, you sidestep the issue of who has really aged more in total between their two meetings, which is the heart of the twin paradox! You might as well say that both stop their clocks at the moment they first pass, then start them again when they pass a second time, in which case both will show zero elapsed time between the two meetings--would you say that this shows that neither one aged between the meetings, and therefore the twin paradox is resolved??

 

[neopolitan]

From my corrected version:

Let me provide a scenario which is analogous to the scenario described in the twin paradox.

Trevor floats in a space suit with two clocks.

Stewart sits at one end of an extremely long structure which also floats in space, unattached to anything bar Stewart and the beacon with another two clocks. At the other end of the structure is a beacon. According to Stewart, the structure has a length of L. Trevor knows this.

Observe that Stewart represents "the Earth" and the beacon represents "some distant location" in the twin paradox. The assumption that "the Earth" and "some distant location" have a fixed separation is inherent in the twin paradox.

Stewart and Trevor are sufficiently distant from any masses as to be considered to be alone in the universe, with no gravitational field in effect. The gravitation exerted by Stewart and his structure on Trevor is negligible.

Stewart and Trevor pass each other twice, at relativistic velocities of v and -v. Trevor and the beacon pass each other twice, also at relativistic velocities of v and -v (Stewart and the beacon are fixed to the same structure and hence share the same frame).

Four events are noteworthy:

1. Trevor and Stewart are collocated as they pass for the first time. Their clocks begin measuring time elapsed.

2. Trevor and the beacon are collocated as they pass for the first time. Trevor's clocks are paused and the beacon sends a message to Stewart's clocks to pause.

3. Trevor and the beacon are collocated as they pass for the second time. Trevor's clocks restart measuring time elapsed and the beacon sends a message to Stewart's clocks to resume measuring time elapsed.

4. Trevor and Stewart are collocated as they pass for the second time. Their clocks stop measuring time elapsed and Trevor and Stewart exchange one of their clocks. Neither consults the other as they each attempt to work out what the other's clock will read.

Observe that I do not say who reverses direction.​

(Please see my revised post a little further back for the entire scenario.)

If you don't say who reverses direction, it's impossible to predict who will have aged more when they pass for the second time--although in your scheme, where they artificially stop their clocks at certain moments, you can say how much time has elapsed on these started-and-stopped clocks between meetings.

Let's not stop their clocks artificially then. Let's say that Trevor travels for an extremely long time at relative speed of v, where v is rather low. Then they are gassed so that the sleep, and only woken after one of them undergoes the necessary acceleration, because v is such a low speed, the twin undergoing acceleration will not be able to perceive having undergone it. This acceleration happens in a very short period of time compared to the traveling times of both leg.

Let's say Trevor and Stewart are of very long lived, extremely patient species, so that a journey of t=100,000 years is not a problem. Let's also say that Trevor (and his structure) and Stewart are hardy enough to survive accelerating from v to -v in one day without anyone noticing they have undergone it. The differencr between time elapsed on the two clocks as a result of one day's acceleration in 100,000 years will disappear into error margins of most clocks (the difference will not be anything close to a full day, it will be the difference between time elapsed on that day according to Trevor and time elapsed of that day according to Stewart which will be less than the difference on a normal day which is not much since v is so low). Even if the twins have superlatively accurate clocks, the time taken to reverse directions will be infinitesimal compared to the time spent traveling inertially.

Now argue it.

cheers,

neopolitan​

 

[neopolitan]

It's also a bit disturbing the way you appear to dismiss other respected and standard explanations of the twin paradox beside your own.

I really am only dismissing the ones which rely on acceleration being the explanation. I have done my best to eliminate acceleration from the scenario and found that the twin paradox is explained. Despite that, many say that you have to consider acceleration.

An analogy (I like analogies):

Why do clothes dry when you put them on the clothes line outside? One respected and standard explanation: "because of warm, dry winds from the desert".

Ok, what about where I live where it is thousands of kilometers from any desert and the air is just above freezing and there is no wind. I leave my clothes on the line and they dry anyway - so long as it doesn't rain or snow, naturally. It just takes longer.

If someone stuck to the claim that it is "because of warm, dry winds from the desert" would I not be justified in dismissing that explanation? (Irrespective of how good the explanation might be in Cairo.) It would occur to me that "warm, dry winds" is just part of the explanation.

In the same fashion, I believe that the acceleration is just part of the explanation. We need to look a little deeper.

cheers,

neopolitan

(PS: if you don't believe that wet clothes dry when left outside in air that is just above freezing, so long as it doesn't rain or snow, and you refuse to even try it out or think it through, well ... then I would not like to discuss physics with you, since you would be demonstrating a particularly unscientific mind-set.)

 

[JesseM]

(Please see my revised post a little further back for the entire scenario.)

Yes, I was already making these sorts of assumptions in my response.

Let's not stop their clocks artificially then. Let's say that Trevor travels for an extremely long time at relative speed of v, where v is rather low. Then they are gassed so that the sleep, and only woken after one of them undergoes the necessary acceleration, because v is such a low speed, the twin undergoing acceleration will not be able to perceive having undergone it. This acceleration happens in a very short period of time compared to the traveling times of both leg.

Let's say Trevor and Stewart are of very long lived, extremely patient species, so that a journey of t=100,000 years is not a problem. Let's also say that Trevor (and his structure) and Stewart are hardy enough to survive accelerating from v to -v in one day without anyone noticing they have undergone it. The differencr between time elapsed on the two clocks as a result of one day's acceleration in 100,000 years will disappear into error margins of most clocks (the difference will not be anything close to a full day, it will be the difference between time elapsed on that day according to Trevor and time elapsed of that day according to Stewart which will be less than the difference on a normal day which is not much since v is so low). Even if the twins have superlatively accurate clocks, the time taken to reverse directions will be infinitesimal compared to the time spent traveling inertially.

Now argue it.

Well, what's your point? If their clocks are not stopped, then your derivation of Stewart showing an elapsed time of 2L/v and Trevor showing an elapsed time of gamma*2L/v no longer works, because it was specifically based on Stewart starting and stopping his clocks on receiving a signal from the beacon. With no clock-stopping, whichever one does the accelerating will show significantly less time on their clock when the two reunite.

Perhaps you are assuming that the one-day acceleration happens immediately after Travor passes the beacon for the first time, so that your derivation will still be approximately correct? The problem is that in your derivation you assumed the time for the signal to get from the beacon to Stewart will be L/c, but if you want it to be Stewart rather than Trevor who accelerates, then this only works if Stewart does not begin to accelerate until after he has received the signal from the beacon, which in his frame will be 100,000 years after Trevor passed the beacon. I suppose you could have the beacon accelerate before Stewart accelerates, but if it accelerates as soon as it passes Trevor and then starts moving at speed v in the opposite direction in Trevor's frame, then after the light beam reaches Stewart and Stewart accelerates to the same speed, the distance between them will no longer be L in Stewart's new rest frame after he finishes accelerating.

On the other hand, if Stewart and the beacon accelerate simultaneously in Trevor's frame, then their separation will continue to be L in Stewart's new inertial rest frame after he accelerates. But in this case there will be a gap of many years between the moment Trevor passes the beacon the first time and the moment Trevor passes the beacon the second time, and if you take this into account you will find that Trevor has aged significantly more than Stewart when they reunite.

 

[neopolitan]

Perhaps you are assuming that the one-day acceleration happens immediately after Travor passes the beacon for the first time, so that your derivation will still be approximately correct? The problem is that in your derivation you assumed the time for the signal to get from the beacon to Stewart will be L/c, but this only works if Stewart does not begin to accelerate until after he has received the signal from the beacon, which in his frame will be 100,000 years after Trevor passed the beacon. I suppose you could have the beacon accelerate before Stewart accelerates, but if it accelerates as soon as it passes Trevor and then starts moving at speed v in the opposite direction in Trevor's frame, then after the light beam reaches Stewart and Stewart accelerates to the same speed, the distance between them will no longer be L in Stewart's new rest frame after he finishes accelerating.

On the other hand, if Stewart and the beacon accelerate simultaneously in Trevor's frame, then their separation will continue to be L in Stewart's frame after he accelerates. But in this case there will be a gap of many years between the moment Trevor passes the beacon the first time and the moment Trevor passes the beacon the second time, and if you take this into account you will find that Trevor has aged significantly more than Stewart when they reunite.

Exactly. It is not the acceleration that resolves the problem. It is the poorly framed scenario of the twin paradox.

My scenario just makes it obvious.

The information that Trevor has passed the beacon cannot be known instaneously by Stewart. If the mischievous god is at the beacon end and knocks out Trevor, it cannot simultaneously knock out Stewart, and both ends of the structure cannot be accelerated simultaneously.

The very framing of the twins paradox grants preferred status to the stay-at-home twin. It is _not_ the acceleration that resolves the paradox, but understanding that the framing of the paradox was faulty.

That's why I summarily reject other explanations, however respected.

cheers,

neopolitan

 

[JesseM]

Exactly. It is not the acceleration that resolves the problem. It is the poorly framed scenario of the twin paradox.

Er, yes it is. Whichever twin accelerates to turn around, that will be the twin whose total elapsed time (not time on a clock that starts and stops during the trip as in your scenario) is less. Do you disagree?

The information that Trevor has passed the beacon cannot be known instaneously by Stewart.

Who cares if Stewart knows it instantaneously or not? Why do you think this is relevant? Stewart can certainly calculate in retrospect when various events happened in any given frame, once there has been time from signals from each event to reach him, and the results should agree with the theory of relativity.

If the mischievous god is at the beacon end and knocks out Trevor, it cannot simultaneously knock out Stewart, and both ends of the structure cannot be accelerated simultaneously.

But if they plan their velocity changes in advance, you can certainly arrange things so that both Stewart and the beacon accelerate simultaneously in Trevor's frame (or whichever frame you choose).

The very framing of the twins paradox grants preferred status to the stay-at-home twin.

In what way?

It is _not_ the acceleration that resolves the paradox, but understanding that the framing of the paradox was faulty.

Again, do you agree that if two twins depart from a common location and later reunite, and one of them moves inertially while the other accelerates to turn around, it will always be the one who turned around that has aged less in total? If you do agree, then how can you say it is not the acceleration that resolves the paradox?

 

[neopolitan]

Er, yes it is. Whichever twin accelerates to turn around, that will be the twin whose total elapsed time (not time on a clock that starts and stops during the trip as in your scenario) is less. Do you disagree?

...

Again, do you agree that if two twins depart from a common location and later reunite, and one of them moves inertially while the other accelerates to turn around, it will always be the one who turned around that has aged less in total? If you do agree, then how can you say it is not the acceleration that resolves the paradox?

It is using the other twin as the reference (and as a consequence as a reference frame) that matters, not the acceleration per se. One twin maintains a velocity relative to the other twin, then obtains and maintains an equal but opposite velocity relative to the other twin.

I repeat: it is not the acceleration which resolves the paradox. The acceleration is merely another symptom, not the cause.

cheers,

neopolitan