B Twin paradox explained for laymen

[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.

 

[FactChecker]

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.

The entire point is to theoretically predict and explain why there should be any difference at all. But without reference to some other fact (acceleration, light color shift, external objects, etc.), there is no way to define an IRF.

 

[PeroK]

The entire point is to theoretically predict and explain why there should be any difference at all. But without reference to some other fact (acceleration, light color shift, external objects, etc.), there is no way to define an IRF.

That is an entirely different problem! If we can't define an IRF, then we can't do much physics at all. That has nothing to do with the twin paradox, per se.

 

[PAllen]

The entire point is to theoretically predict and explain why there should be any difference at all. But without reference to some other fact (acceleration, light color shift, external objects, etc.), there is no way to define an IRF.

The conventional way to determine if a closed box is moving inertially ( thus defining a local inertial frame) is to determine whether Newton’s laws of motion apply in the nonrelativistic limit. Most simply, release a ball in the middle of the box. If it stays put, the box is inertial.

 

[FactChecker]

That is an entirely different problem!

I don't see that. The point that I have been trying to make this entire time is that there is a logical symmetry between the two twins unless some other fact (acceleration, light color shift, external objects, some other physical fact) is brought into the problem.

 

[FactChecker]

The conventional way to determine if a closed box is moving inertially ( thus defining a local inertial frame) is to determine whether Newton’s laws of motion apply in the nonrelativistic limit. Most simply, release a ball in the middle of the box. If it stays put, the box is inertial.

Yes. Measure the acceleration.

 

[PeroK]

I don't see that. The point that I have been trying to make this entire time is that there is a logical symmetry between the two twins unless some other fact (acceleration, light color shift, external objects, some other physical fact) is brought into the problem.

The asymmetry is defined using an assumed IRF. One twin changes direction. Like all SR problems, we have an assumed IRF in which the coordinates associated with the experiment are given. When we say that "planet X is $4$ light years from Earth", that statement only makes sense in an implied IRF.

It's also implied or stated explicitly that one twin moves inertially the whole time (the stay at home twin). That is the basis of the scenario.

 

[PeroK]

@FactChecker part of the problem(which I think is evident from your posts) is that you see statements like:

"Planet X is 4 light years from Earth", or "at time $t$ the distance between twins A and B is $d$" as having some absolute meaning.

These statements have no meaning other than as specified in an (implied) IRF. That you do not grasp that is the source of the confusion.

 

[FactChecker]

I guess my position can be summarized thus:
Essential parts (like inertial reference frame) of the Twins Paradox can not even be defined without talking about acceleration or some representation of it. So the most a person can say about the solution is that you do not have to use acceleration AGAIN to explain it. But very good scientists (including Einstein) used acceleration to explain the twin paradox. And it is exactly at the point of the turn-around that one twin departs from an inertial reference frame due to acceleration that distinguishes him from the other twin. So it is natural to examine the consequences of acceleration at that point.

 

[PeterDonis]

If we can't define an IRF, then we can't do much physics at all.

Yes, you can. Look at my post #135. I said nothing whatever about IRFs, yet there is enough information to know which twin ages more.

The asymmetry is defined using an assumed IRF.

It doesn't have to be. See above.

 

[PeroK]

Yes, you can. Look at my post #135. I said nothing whatever about IRFs, yet there is enough information to know which twin ages more.
It doesn't have to be. See above.

Okay. So, what is the statement of the twin paradox? Assuming no data from an implied IRF?