Twin paradox explained for laymen

[FactChecker]

That's correct. The post I was responding to there was, IIRC, not limiting itself to SR scenarios.
First, as @jbriggs444 points out, the twin is lower in a gravitational potential well, which makes him age slower, not faster.

Second, the turnaround in this case, just as in the usual SR twin paradox scenario, can be made to occupy a negligibly short portion of the traveling twin's entire worldline, so the aging during it can be ignored. Its primary effect is simply to allow the traveling twin to turn around, i.e., to make his worldline such that he can come back to meet up with the stay-at-home twin again.

Yes. I misspoke in the earlier post when I said that the traveling twin would age faster during the slingshot. He would age slower. The twin on Earth would age more. But that is just substituting actual gravity for acceleration (pseudo-gravity). Either one makes the situation of the twins non-symmetric and alleviates the "paradox" complaint.

 

[vanhees71]

Not always. The case I described is a counterexample, if we have the stay-at-home twin floating in free space instead of sitting on Earth's surface. Then the stay-at-home twin's worldline is a geodesic, and so is the traveling twin's (since his turnaround is accomplished by a free-fall slingshot maneuver around a distant planet or star), but only the former's worldline is a worldline of maximal proper time between the starting and ending events.

Of course there are simpler examples possible: for example, an astronaut in a spaceship orbiting Earth launches a probe radially outward, with just the right velocity such that the probe returns to the ship after the ship has made exactly one orbit. The ship's and the probe's worldlines are both geodesics, but only the latter's worldline is one of maximal proper time between the starting and ending events.

You are right. It's not so clear, for general geodesics connecting the same spacetime points. So one has to do the calculation. Perhaps your latter example is a nice exercise for testbodies in a Schwarzschild spacetime (e.g., one in a circular orbit, the other being shot radially out in the and falling back as described. I'll try that!

 

[jbriggs444]

Not always. The case I described is a counterexample, if we have the stay-at-home twin floating in free space instead of sitting on Earth's surface.

If intuition serves, a geodesic will be a path that yields a local maximum for elapsed time among a set of "similar" paths.

 

[PeterDonis]

that is just substituting actual gravity for acceleration

Not really. The traveling twin in the slingshot scenario has zero proper acceleration (which is of course the point of the scenario, to have a turnaround with zero proper acceleration). The phrase you use here is usually used to describe, for example, standing on Earth's surface as compared to standing in a rocket accelerating at 1 g in free space, i.e., the proper acceleration is the same in both cases, only the spacetime geometry, and hence the reason for the proper acceleration being present, is different.

 

[PeterDonis]

a geodesic will be a path that yields a local maximum for elapsed time among a set of "similar" paths

First, we should be restricting attention to timelike geodesics, since those are the only ones for which the "maximum" heuristic is valid in spacetime anyway.

Second, I don't think even the "local maximum" rule always works; a timelike geodesic will always be a local extremum, but I think there are cases where the extremum is a local saddle point, not a local maximum.

 

[Sagittarius A-Star]

One thing is true: there is only a paradox for those who do not understand SR. Mostly the general public. You would be struggling to find a physicist who thinks there is a paradox that needs an elaborate, pseudo-gravitational explanation.

The connection between time-dilation and the artificial gravity, that an apple-tree in an uniformly accelerated rocket experiences, can be visualized by this video (A reference to GR or the principle of equivalence is not required):

 

[PAllen]

First, we should be restricting attention to timelike geodesics, since those are the only ones for which the "maximum" heuristic is valid in spacetime anyway.

Second, I don't think even the "local maximum" rule always works; a timelike geodesic will always be a local extremum, but I think there are cases where the extremum is a local saddle point, not a local maximum.

I believe there is a theorem I’ve seen quoted in a number of GR texts and papers, which says, to the best of my memory, that for a sufficiently small causal diamond, for any two points in it that can be connected by a timelike path, there is a unique timelike geodesic contained in the causal diamond, and that this geodesic maximizes proper time among all paths contained in the diamond.This, among other things, rules out saddle points over small scales. This theorem is also what makes rigorous Synge’s notion of a World function.

This theorem also implies that for any timelike geodesic, for any sufficiently small neighborhood of any event on it, the geodesic is the maximizing path between two points on it contained in that neighborhood, among paths contained in that neighborhood.

 

[PAllen]

That's a remarkably uncovariant explanation.

Not really. General covariance means that despite different coordinate descriptions, calculation of invariants come out the same. And for physics, that means all direct observables must be invariants. However, it also entails the notion that any coordinate description is as good as any other.

Thus, for example, for muons created in the upper atmosphere reaching the ground, the Earth frame explanation of time dilation is neither more nor less valid than the muon frame explanation of the short distance between the upper atmosphere and ground.

In the twin scenario in SR, we have that in any inertial coordinates, all differences in clock rates are purely dependent on velocity, and this will be sufficient to explain any differences in proper time along paths (which are, of course invariant). However, in non-inertial coordinates, e.g. in which a non-geodesic path is a stationary position coordinate, coordinate velocity alone does not explain the relation of proper time to coordinate time. The difference can be described as pseudogravity. In any such noninertial coordinates, there is coherent explanation of proper time differences along paths that is due to a mix of velocity and the effects of pseudogravity (manifested mathematically as a non-trivial metric and nonvanishing connection). General covariance states that an explanation in any such general coordinates is equally as valid as that in inertial coordinates. None of this has anything to do with GR as it currently understood. It simply uses methods that became common with GR.

Of course I agree with you that all of this is a wildly overcomplicated way of understanding the twin scenarios. However, I also don't like the notion of "changing frames", because in my view a frame is something used by people to describe physics, and particles or bodies do not have frames - they simply may be described in any frame or coordinates.

 

[PeroK]

In the twin scenario in SR, we have that in any inertial coordinates, all differences in clock rates are purely dependent on velocity, and this will be sufficient to explain any differences in proper time along paths (which are, of course invariant). However, in non-inertial coordinates, e.g. in which a non-geodesic path is a stationary position coordinate, coordinate velocity alone does not explain the relation of proper time to coordinate time. The difference can be described as pseudogravity. In any such noninertial coordinates, there is coherent explanation of proper time differences along paths that is due to a mix of velocity and the effects of pseudogravity (manifested mathematically as a non-trivial metric and nonvanishing connection). General covariance states that an explanation in any such general coordinates is equally as valid as that in inertial coordinates. None of this has anything to do with GR as it currently understood. It simply uses methods that became common with GR.

That is most certainly NOT the simplest way to resolve the twin paradox! We were looking for the simplest explanation.

 

[PAllen]

That is most certainly NOT the simplest way to resolve the twin paradox! We were looking for the simplest explanation.

I strongly agree.

 

[PeterDonis]

for a sufficiently small causal diamond

The key is that "sufficiently small". There is no general formula for how to pick out the sufficiently small causal diamonds in any spacetime; it depends on the particular spacetime.

 

[Sagittarius A-Star]

That is most certainly NOT the simplest way to resolve the twin paradox! We were looking for the simplest explanation.

The calculation in the rest frame of the traveling twin can be made faily easy, if the scenaio is set-up accordingly and only an approximational calculation is done.

Let's take the specific example from Wikipedia: distance d = 4 light years, at a speed v = 0.8c. The "stationary" twin gets older by 10 years, the traveling twin by 6 years, γ = 5/3.

The traveling twin approches the star after 3 years proper time and calculates, the the "stationary" twin must have aged then by 3 years /γ = 1.8 years. So while both inertial travel phases, the "stationary" twin must get older by 2 * 1.8 years = 3.6 years. The "gap" to 10 years is (10 - 3.6) years = 6.4 years. This "gap" has to be filled by pseodo-gravitational time-dilation while "turnaround":

Δt₁ = time "gap" of the "stationary" twin's aging while "turnaround"
Δt₂ = duration of "turnaround" (defined as arbitrarily short, neglectable)
Φ = pseudo-gravitational potential
a = proper acceleration of frame while "turnaround"
h = distance of twins * γ
Δv = velocity change by turnaround = 2 * v = 1.6c

Approximate calculation:
Δt₁ = Δt₂ (1 + Φ /c²) = Δt₂ (1 + a*h /c²) = Δt₂ (1 + (Δv/Δt₂)*h /c²) = Δt₂ + (1.6c * 4LY /c²)
= Δt₂ + (1.6 * 4LY /c) = Δt₂ + (1.6 * 4Y) = 6.4 Years.

PeroK told me, that Δt₂ := a few minutes. That can be neglected.

Derivation of the used time-dilation formula:
https://www.physicsforums.com/threa...idered-to-be-accelerating.991333/post-6367026

Remark:
I find the factor γ in h in several papers (linked in posting #129), but I don't yet know the reason for it.

 

[PeterDonis]

The calculation in the rest frame of the traveling twin

There are two "rest frames" involved here (if we leave out the turnaround), the outgoing one and the returning one. Both have the same magnitude of $v$ relative to the stay-at-home frame, so the time dilation factor is the same for both, but the $v$ is in opposite directions.

 

[Sagittarius A-Star]

There are two "rest frames" involved here

I would say, that the rest frame changes from one inertial frame via the turnaround to another inertial frame.

 

[PeterDonis]

I would say, that the rest frame changes from one inertial frame via the turnaround to another inertial frame.

There is no "the" rest frame for the traveling twin unless you mean a non-inertial frame. There is no single inertial frame in which the traveling twin is at rest for the entire scenario.

In fact, even with regard to non-inertial frames, the phrase "the rest frame changes", while it is indeed a common one, is IMO misleading. "The" rest frame cannot change; the whole point of defining "the" rest frame is to have a single frame in which the chosen observer is always at rest. If you're not going to do that, you might as well drop the "rest frame" idea altogether and just do the calculation in the most convenient inertial frame for the problem, which in this case is the stay-at-home twin's frame.

 

[PeroK]

Approximate calculation:
Δt₁ = Δt₂ (1 + Φ /c²) = Δt₂ (1 + a*h /c²) = Δt₂ (1 + (Δv/Δt₂)*h /c²) = Δt₂ + (1.6c * 4LY /c²)
= Δt₂ + (1.6 * 4LY /c) = Δt₂ + (1.6 * 4Y) = 6.4 Years.

There is a lot of theoretical baggage in that calculation. Not least that you have used a "height" of $4LY$ for the pseudo-gravitational potential. Where does $4LY$ come from? That's the distance between the Earth and the accelerating traveller in the rest frame of the Earth! The very frame you are at pains to avoid using.

If you are going to use the Earth frame to do your gravitational calculations, why not use it to measure the length of a couple of flat spacetime intervals?

 

[Sagittarius A-Star]

There is no "the" rest frame for the traveling twin unless you mean a non-inertial frame.

Yes, I mean a non-inertial rest frame, which is most of the time inertial, only while turn-around not.

 

[PeterDonis]

I mean a non-inertial rest frame, which is most of the time inertial, only while turn-around not.

Then the rest frame is not changing, but you said it was.

I think you have not fully thought through the approach you are trying to use.

 

[PeterDonis]

Approximate calculation:
Δt₁ = Δt₂ (1 + Φ /c²) = Δt₂ (1 + a*h /c²) = Δt₂ (1 + (Δv/Δt₂)*h /c²) = Δt₂ + (1.6c * 4LY /c²)
= Δt₂ + (1.6 * 4LY /c) = Δt₂ + (1.6 * 4Y) = 6.4 Years.

In addition to the issue @PeroK has raised with this, there is another issue: during the turnaround, the stay-at-home twin is not at a constant distance from the traveling twin; he is free-falling in the pseudo-gravitational field. So his gravitational potential/time dilation is not constant either.

 

[Sagittarius A-Star]

So his gravitational potential/time dilation is not constant either.

That can be avoided by defining the scenarion, that the turnaround time is short enough and therefore the acceleration of the frame great enough, that the distance can be regarded as almost constant.

 

[Sagittarius A-Star]

Where does $4LY$ come from?

That's a good point. In the German Wikipedia they write explicitely, that x' = x * γ shall be used in the formula, but they do not write, why. So I have to figure this out.
German Source:
https://de.wikipedia.org/wiki/Zwillingsparadoxon#Variante_mit_Beschleunigungsphasen

 

[PeroK]

That's a good point. In the German Wikipedia they write explicitely, that x' = x * γ shall be used in the formula, but they do not write, why. I have to figure out this.
German Source:
https://de.wikipedia.org/wiki/Zwillingsparadoxon#Variante_mit_Beschleunigungsphasen

It's fairly obvious. During the acceleration phase, the distance from the traveller back to Earth is not well-defined in the traveller's frame. Naively it varies from $2.4$ to $4$ light years, then back down to $2.4$ again. It's only by using the Earth frame that you can describe an approximately constant pseudo-gravitational potential difference.

 

[PeterDonis]

It's only by using the Earth frame that you can describe an approximately constant pseudo-gravitational potential difference.

In the Earth frame there is no pseudo-gravitational field to begin with, so this doesn't make sense.

What would make sense would be to describe the turnaround using Rindler coordinates in which the time $t = 0$ corresponded to the instant at which the traveling twin is momentarily at rest with respect to the stay-at-home twin. The turnaround would then start at some Rindler coordinate time $t = - T$ and end at $t = + T$. Increasing the proper acceleration $a$ would then just decrease $T$; in fact it should be easy to see that the product $a T$ must be constant (assuming that the traveling twin's speed relative to the stay-at-home twin on the outbound and inbound inertial legs of the trip remains the same).

I have not had a chance to do the math yet for this, to check whether in fact the free-fall distance covered by the stay-at-home twin during the turnaround can be made arbitrarily small by letting $a$ increase without bound. I'm actually not sure that will be the case.

 

[Sagittarius A-Star]

It's fairly obvious. During the acceleration phase, the distance from the traveller back to Earth is not well-defined in the traveller's frame. Naively it varies from $2.4$ to $4$ light years, then back down to $2.4$ again. It's only by using the Earth frame that you can describe an approximately constant pseudo-gravitational potential difference.

I found another source with the same problem. They say, that the distance in Andrew's frame is used ("stationary" twin), but not, why:

Finally, for the turn around period 3), Bob is at a lower gravitational potential than Andrew, so that Bob’s clock runs slower according to (22), with dAB=d, the distance between Andrew and the distant star in Andrew’s frame.

Source (page 7 of 10):
https://www.hilarispublisher.com/op...the-twinsin-the-paradox-2090-0902-1000218.pdf

The same is valid for equation (8) with the un-contracted length L₀ in:
https://arxiv.org/ftp/arxiv/papers/1002/1002.4154.pdf

 

[PAllen]

In the Earth frame there is no pseudo-gravitational field to begin with, so this doesn't make sense.

What would make sense would be to describe the turnaround using Rindler coordinates in which the time $t = 0$ corresponded to the instant at which the traveling twin is momentarily at rest with respect to the stay-at-home twin. The turnaround would then start at some Rindler coordinate time $t = - T$ and end at $t = + T$. Increasing the proper acceleration $a$ would then just decrease $T$; in fact it should be easy to see that the product $a T$ must be constant (assuming that the traveling twin's speed relative to the stay-at-home twin on the outbound and inbound inertial legs of the trip remains the same).

I have not had a chance to do the math yet for this, to check whether in fact the free-fall distance covered by the stay-at-home twin during the turnaround can be made arbitrarily small by letting $a$ increase without bound. I'm actually not sure that will be the case.

I think in this case the Rindler coordinate distance increases then decreases. I agree with your intuition that can’t make this coordinate distance change small. The correct approach is: so what? It’s a simple integral and must work out right. I seem to recall an old thread here where someone posted calculations from a text by Moeller filling in the details of Einstein’s approach rigorously, demonstrating that it worked out exactly, as it must.

 

[FactChecker]

I'm afraid that in the acceptance of the known mathematical proof, the real issue is being missed. The Twins Paradox has been designed to make the situation of the two twins appear identical. The rational for saying that the "traveling" twin is younger needs to be addressed.

Suppose you are given these facts:
There are two twins, t1 and t2, each with a reference frame that remains centered at him.
1) They are initially together and at rest wrt each other.
2) After 1 year, they both agree that they are 0.5 light-years apart.
3) After two years, they are back together and at rest wrt each other.

Is one younger? If so, which one and why?
In order to apply SR, we must determine which one accelerated and turned around or if both did, to some extent. You can not use purely mathematical logic here. You must bring something more into the problem and justify that. When you identify one twin as the traveling and accelerating one, you must justify that. Feeling acceleration is one way. Selecting a particular reference frame and calling in inertial is another.

 

[PeterDonis]

Suppose you are given these facts:
There are two twins, t1 and t2, each with a reference frame that remains centered at him.
1) They are initially together and at rest wrt each other.
2) After 1 year, they both agree that they are 0.5 light-years apart.
3) After two years, they are back together and at rest wrt each other.

Is one younger? If so, which one and why?

Your statement of the scenario is ambiguous, because "after 1 year" and "after two years" do not say whether those times are coordinate times or proper times, or which twin they refer to. Also, "0.5 light-years apart" does not say how that distance is being determined.

If you resolve those ambiguities, you will find that you have also given enough additional information to give definite answers to your questions.

For example, if you say that 2) means "each twin agrees that, after 1 year by his clock, the other twin is 0.5 light-years from him in his own rest frame", and 3) means "each twin agrees that, after two years by his clock, they are back together again", then the answers to your questions are obvious.

You can not use purely mathematical logic here.

You can if you specify an unambiguous scenario. But you haven't.

 

[PeterDonis]

When you identify one twin as the traveling and accelerating one, you must justify that. Feeling acceleration is one way. Selecting a particular reference frame and calling in inertial is another.

No, these are not two different ways, they're the same way. You can't just pick any reference frame and call it inertial. You have to demonstrate that it's inertial, by showing that an object at rest in the frame feels zero acceleration.

 

[FactChecker]

No, these are not two different ways, they're the same way. You can't just pick any reference frame and call it inertial. You have to demonstrate that it's inertial, by showing that an object at rest in the frame feels zero acceleration.

Then you do need acceleration to even begin and apply the SR equations. You use it to determine which twin turns around. It is not a large step to want to see what role acceleration directly plays in making the accelerating twin younger.

 

[PeterDonis]

you do need acceleration to even begin and apply the SR equations. You use it to determine which twin turns around

That is one way to specify it, but not the only way.

For example, here's another way:

Twin A sees light signals coming from twin B to be redshifted for about half his (Twin A's) trip, then blueshifted for the other half.

Twin B sees light signals coming from twin A to be redshifted for almost all his (Twin B's) trip, then blueshifted for a very short time at the end of his trip.

This information, all by itself, is sufficient to tell which twin has aged more when they meet again.

 

[FactChecker]

That is one way to specify it, but not the only way.

For example, here's another way:

Twin A sees light signals coming from twin B to be redshifted for about half his (Twin A's) trip, then blueshifted for the other half.

Twin B sees light signals coming from twin A to be redshifted for almost all his (Twin B's) trip, then blueshifted for a very short time at the end of his trip.

This information, all by itself, is sufficient to tell which twin has aged more when they meet again.

Yes, there are many ways to confirm something that is true. But the definition of "inertial" and acceleration are closely linked. It is very natural to ask if there is a direct connection between the clock effects when a twin turns around and the acceleration that occurs at exactly the same time that a reference frame centered at him ceases to be inertial (defined in terms of acceleration). It is less natural to ask the same question regarding red shifts.

 

[PAllen]

Yes, there are many ways to confirm something that is true. But the definition of "inertial" and acceleration are closely linked. It is very natural to ask if there is a direct connection between the clock effects when a twin turns around and the acceleration that occurs at exactly the same time that a reference frame centered at him ceases to be inertial (defined in terms of acceleration). It is less natural to ask the same question regarding red shifts.

Clock rates and red shifts are inextricably linked. The frequency of a wave is a clock rate. In particular, the relation between the direct observation of some clock compared to ones own clock is always ( all cases in SR and GR) exactly the same as the frequency shift factor.

 

[PeterDonis]

the definition of "inertial" and acceleration are closely linked.

Yes.

It is very natural to ask if there is a direct connection between the clock effects when a twin turns around and the acceleration that occurs at exactly the same time that a reference frame centered at him ceases to be inertial (defined in terms of acceleration).

The "clock effects" you are talking about are coordinate-dependent. Trying to treat coordinate-dependent quantities as if they were "real things" is always problematic.

Also, while there is a general rule in flat spacetime that, if two twins separate and meet again and have aged differently, at least one of them must have had nonzero proper acceleration in between, that rule does not generalize to curved spacetimes.

It is less natural to ask the same question regarding red shifts.

It might be less natural for someone unfamiliar with SR, but it's well worth the effort to retrain your intuitions to make it more natural. For one thing, using directly observed redshifts to predict differential aging between twins that separate and then meet again is a general rule that works in any spacetime.

 

[PeroK]

I'm afraid that in the acceptance of the known mathematical proof, the real issue is being missed. The Twins Paradox has been designed to make the situation of the two twins appear identical. The rational for saying that the "traveling" twin is younger needs to be addressed.

Suppose you are given these facts:
There are two twins, t1 and t2, each with a reference frame that remains centered at him.
1) They are initially together and at rest wrt each other.
2) After 1 year, they both agree that they are 0.5 light-years apart.
3) After two years, they are back together and at rest wrt each other.

Is one younger? If so, which one and why?

The answer to that question is that you study the motion of both twins in any available IRF and use the fact that spacetime distance (proper time) along a worldline is invariant. Neither twin needs to measure anything. The differential ageing when they meet is the difference between the lengths of the spacetime paths they have taken.

 

[vanhees71]

I believe there is a theorem I’ve seen quoted in a number of GR texts and papers, which says, to the best of my memory, that for a sufficiently small causal diamond, for any two points in it that can be connected by a timelike path, there is a unique timelike geodesic contained in the causal diamond, and that this geodesic maximizes proper time among all paths contained in the diamond.This, among other things, rules out saddle points over small scales. This theorem is also what makes rigorous Synge’s notion of a World function.

This theorem also implies that for any timelike geodesic, for any sufficiently small neighborhood of any event on it, the geodesic is the maximizing path between two points on it contained in that neighborhood, among paths contained in that neighborhood.

Yes, right, and that's what got me wrong, because in @PeterDonis example we deal with large-scale geodesics, and there the connection of two points (or one where you end up at the same point in a round-trip scenario we need for the discussion of the twin paradox) by a time-like geodesic curve doesn't need to be unique.

That's easy to visualize on the sphere: The geodesics are the great circles. Now take the example of the geodesic from the equator to the north pole along the zero meridian. You can go "directly" along a quarter circle or the detour over the south pole along a 3/4-circle.