Twin paradox explained for laymen

[Dale]

Acceleration does not directly effect the different aging. But it has an indirect effect on it,

I agree. To me, the acceleration does not cause the time dilation, but it does resolve the paradox. The paradox is not about calculating the amount of time dilation, it is about the symmetry.

The confused student has learned that “motion is relative” and therefore thinks that the effects should be the same for each twin because of the symmetry. The acceleration breaks that symmetry and thereby resolves the paradox. Any proposal to avoid the acceleration always introduces some other asymmetry.

 

[FactChecker]

I have to admit that this is a good point. But I think that to say that turning around causes the effect but that acceleration does not is quibbling about semantics. Turning around is acceleration.

There is more to it than this. Using only the relative positions of the twins, there is no way to mathematically determine which twin is moving and which is stationary. Some reference to an external object or force is necessary to break the mathematical symmetry of the two twins. The most obvious way is to say that the traveling twin feels acceleration when he turns around.

 

[Sagittarius A-Star]

That may be a true statement, but it would make for a complicated law as to if and when you got a speeding ticket.

I would ague with relativity, that I should get no ticket. In my rest frame, I had speed Zero. Only the speed-measurement equipment moved too fast backwards towards me.

 

[PeroK]

Using only the relative positions of the twins, there is no way to mathematically determine which twin is moving and which is stationary. Some reference to an external object or force is necessary to break the mathematical symmetry of the two twins. The most obvious way is to say that the traveling twin feels acceleration when he turns around.

As has been emphasised many times, you get the same results without acceleration, by simply measuring the spacetime distance along two joined inertial paths. And that is simply spacetime geometry.

The analogy with the triangle inequality applies. Suppose you wanted to confirm that the spatial distance along two edges of a triangle is larger than the distance along the remaining edge. In order to follow the non-straight path, you must accelerate to change direction at the intermediate vertex.

You would then indirectly attribute the triangle inequality to acceleration. The amount of acceleration would define the change in angle and, indirectly, tell you the length of the overall path.

This overlooks the simpler explanation that it was nothing to do with the acceleration at the vertex: it was a simple case of $|AB| + |BC| > |AC|$.

 

[FactChecker]

As has been emphasised many times, you get the same results without acceleration, by simply measuring the spacetime distance along two joined inertial paths. And that is simply spacetime geometry.

Using only the relative positions, there is no way to mathematically determine which twin stays on one inertial path and which changes to a different inertial path. Something else is needed to distinguish between the twins. Saying that one twin turns around and switches to a different inertial path is just another way to say that he accelerated.

 

[FactChecker]

As has been emphasised many times, you get the same results without acceleration, by simply measuring the spacetime distance along two joined inertial paths. And that is simply spacetime geometry.

Can you give me an example where an observer did not follow an inertial path and yet did not accelerate?

 

[PeroK]

Using only the relative positions, there is no way to mathematically determine which twin stays on one inertial path and which changes to a different inertial path. Something else is needed to distinguish between the twins. Saying that one twin turns around and switches to a different inertial path is just another way to say that he accelerated.

If you define your geometry to be hyperbolic, flat spacetime then you can show mathematically that for timelike paths in hyperbolic geometry you have $|AC| < |AB| + |BC|$.

That's the mathematics. There are no complications in using hyperbolic geometry.

To do a physical experiment you need some way to establish an inertial reference frame. Any one will do. You could pick any frame in which the Earth and distant planet are moving inertially. The Earth frame is the simplest, but it works out equally in all inertial reference frames. That's where physics enters. Mapping the mathematical hyperbolic geometry to a physical scenario.

It's true that a physical object in flat spacetime cannot change frames without accelerating, but that constraint can be removed if you simply measure time along spacetime paths (swapping physical clocks at the turnaround).

 

[Motore]

Can you give me an example where an observer did not follow an inertial path and yet did not accelerate?

(And in fact, there are versions of the scenario where even the turnaround does not require acceleration: for example, the traveling twin could pass close enough to some large, distant planet or star to "slingshot" around it and be heading back towards Earth, and stay in free fall the whole time.)

 

[PeroK]

Can you give me an example where an observer did not follow an inertial path and yet did not accelerate?

If you want to remove acceleration from the experiment, you need to swap physical clocks at the turnaround. Or, you simply study the problem mathematically using hyperbolic geometry.

 

[vanhees71]

As has been emphasised many times, you get the same results without acceleration, by simply measuring the spacetime distance along two joined inertial paths. And that is simply spacetime geometry.

The analogy with the triangle inequality applies. Suppose you wanted to confirm that the spatial distance along two edges of a triangle is larger than the distance along the remaining edge. In order to follow the non-straight path, you must accelerate to change direction at the intermediate vertex.

You would then indirectly attribute the triangle inequality to acceleration. The amount of acceleration would define the change in angle and, indirectly, tell you the length of the overall path.

This overlooks the simpler explanation that it was nothing to do with the acceleration at the vertex: it was a simple case of $|AB| + |BC| > |AC|$.

But it is indeed true that in order to have different curves connecting the same two spacetime points at least one of the worldlines cannot be a straight line in Minkowski space, i.e., at least one must be accelerated against the inertial reference frames.

But it's also generally true: If both observers are accelerated wrt. the inertial reference frames, along worldlines with the same initial and final points, in general measure different proper times it takes for their travel.

The best analogon to Euclidean geometry is indeed that there of course also the lengths of curves connecting the same two points may be of different length.

The largest proper time you get always for the straight line, as one sees easily by using the variational principle. In a sense you have also a triangle rule in Minkowski space, but the straight line here is the longest not the shortest connection (for time-like curves only of course).

 

[FactChecker]

And in fact, there are versions of the scenario where even the turnaround does not require acceleration: for example, the traveling twin could pass close enough to some large, distant planet or star to "slingshot" around it and be heading back towards Earth, and stay in free fall the whole time.) So it is not true to say that the acceleration of the traveling twin is "far greater" than that of the twin who remains on earth.

It seems to me that slingshotting around a distant star brings an entirely different geometry into the situation. I don't think that SR applies in that situation.

 

[vanhees71]

I also think that's GR. But also in GR the maximum proper time among all time-like curves connecting to points is reached for a geodesic connecting the two points.

 

[PeroK]

But it is indeed true that in order to have different curves connecting the same two spacetime points at least one of the worldlines cannot be a straight line in Minkowski space, i.e., at least one must be accelerated against the inertial reference frames.

But it's also generally true: If both observers are accelerated wrt. the inertial reference frames, along worldlines with the same initial and final points, in general measure different proper times it takes for their travel.

The best analogon to Euclidean geometry is indeed that there of course also the lengths of curves connecting the same two points may be of different length.

The largest proper time you get always for the straight line, as one sees easily by using the variational principle. In a sense you have also a triangle rule in Minkowski space, but the straight line here is the longest not the shortest connection (for time-like curves only of course).

What I'm saying is this:

First, in Euclidean geometry (with the usual definition of distance) we have the triangle inequality. Take three points in the plane: $A = (0,0), \ B = (1, 1), \ C = (0, 2)$. We have:
$$|AB| = \sqrt 2, \ \ |BC| = \sqrt 2, \ \ |AC| = 2$$
And we see that, indeed:
$$|AC| < |AB| + |BC|$$
This is generally called the triangle inequality.
Now, in hyperbolic geometry with $ds^2 = dt^2 - dx^2$ we have three points:
$$A = (0, 0), \ \ B = (t, vt), \ \ C = (2t, 0), \ \ (0 < v < 1)$$
Then we have:
$$|AB| = t\sqrt{1 - v^2} < t, \ \ |BC| = t\sqrt{1 - v^2} < t, \ \ |AC| = 2t$$
And, we see that:
$$|AC| > |AB| + |BC|$$
This is generally called the twin paradox.

Now, you could demonstrate the triangle inequality (to show that we have approximately Euclidean geometry on the surface of the Earth) in a real experiment by pacing out the sides of a triangle. And you could, if you wanted, attribute the difference in lengths measured to the acceleration when you changed direction at the intermediate vertex. Rather than accepting the underlying Euclidean geometry of space.

Likewise, you could demonstrate the hyperbolic geometry of spacetime by doing a real twin paradox experiment. And, if you wanted, you could again attribute the difference in times measured to the acceleration when you changed direction. Again, rather than accepting the underlying hyperbolic geometry of spacetime.

 

[FactChecker]

I also think that's GR. But also in GR the maximum proper time among all time-like curves connecting to points is reached for a geodesic connecting the two points.

And we know that the twin who slingshots around a large mass would age faster, so it is not immediately clear to me that the twins would not have identical ages when they get back together.

None of that is correct. You can set this up entirely with people looking at clocks through windows without anyone accelerating.

I think that those scenarios only add a complication to the paradox, they do not resolve it. In an identically symmetric way, the "traveling" twin can imagine the Earth twin switches his clock to another inertial frame that comes back to the "traveling" twin and makes the Earth twin younger. So ther is still a paradox. Once again, the simplest way to break the logical symmetry is to say that the traveling twin feels acceleration.

 

[martinbn]

I agree. To me, the acceleration does not cause the time dilation, but it does resolve the paradox. The paradox is not about calculating the amount of time dilation, it is about the symmetry.

The confused student has learned that “motion is relative” and therefore thinks that the effects should be the same for each twin because of the symmetry. The acceleration breaks that symmetry and thereby resolves the paradox. Any proposal to avoid the acceleration always introduces some other asymmetry.

In my post I gave an example where the accelerations are the same for the two. The point was that it cannot be the acceleration that makes the difference.

 

[FactChecker]

In my post I gave an example where the accelerations are the same for the two. The point was that it cannot be the acceleration that makes the difference.

Your example is flawed. You ignore that the distance between the observers has a great effect on the acceleration. The greatest effect is when the observers are widely separated.

 

[Dale]

In my post I gave an example where the accelerations are the same for the two. The point was that it cannot be the acceleration that makes the difference.

Even then the acceleration profiles were different. Both twins agree which twin is the early accelerating and which is the late accelerating twin. The acceleration still eliminates the symmetry.

Again, (proper) acceleration does not cause the differential aging, but it does break the symmetry.

 

[PeroK]

So ther is still a paradox.

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.

 

[Nugatory]

It seems to me that slingshotting around a distant star brings an entirely different geometry into the situation. I don't think that SR applies in that situation.

The calculation is the same regardless of the geometry: We integrate the quantity $g_{uv}dx^udx^v$ along the path. The difference is that in the flat spacetime of SR (and if we choose our coordinates in the most natural way) the $g_{uv}$ are all either 1 or -1 so we don't bother to write them down and the integral becomes so trivial that we don't notice it.

 

[jbriggs444]

And we know that the twin who slingshots around a large mass would age faster

Huh?

The twin who slingshots around a large mass is lower in a potential well. So he ages more slowly. And he is moving faster. So he ages more slowly.

Both of these are coordinate-dependent heuristics for time dilation, but the result should hold for differential aging: Shorter elapsed time for the slingshotting twin.

 

[martinbn]

Even then the acceleration profiles were different. Both twins agree which twin is the early accelerating and which is the late accelerating twin. The acceleration still eliminates the symmetry.

Again, (proper) acceleration does not cause the differential aging, but it does break the symmetry.

Of course there is no symmetry, but the experienced acceleration is the same for the two. It is not the acceleration that makes it asymmetric.

 

[FactChecker]

The twin who slingshots around a large mass is lower in a potential well. So he ages more slowly. And he is moving faster. So he ages more slowly.

Both of these are coordinate-dependent heuristics for time dilation, but the result should hold for differential aging: Shorter elapsed time for the slingshotting twin.

Sorry. I stand corrected.

 

[FactChecker]

Of course there is no symmetry, but the experienced acceleration is the same for the two. It is not the acceleration that makes it asymmetric.

The relative distances when the acceleration occurs is different. So it is the acceleration and the timing/relative distance of it that breaks the symmetry.

 

[PeroK]

The relative distances when the acceleration occurs is different. So it is the acceleration and the timing/relative distance of it that breaks the symmetry.

That's a remarkably uncovariant explanation.

 

[martinbn]

The relative distances when the acceleration occurs is different. So it is the acceleration and the timing/relative distance of it that breaks the symmetry.

What do you mean the relative distences are different? Distances to what?

 

[Dale]

Of course there is no symmetry, but the experienced acceleration is the same for the two. It is not the acceleration that makes it asymmetric.

I disagree that the acceleration is the same. Acceleration is a vector valued function of time and those functions are not the same. The acceleration is indeed asymmetric.

Again, my claim isn’t that the acceleration causes the aging, just that it breaks the symmetry. With no other information besides their accelerometer readings the twins agree who accelerated early and who accelerated late. So the acceleration itself does break the symmetry.

 

[FactChecker]

What do you mean the relative distences are different? Distances to what?

The relative distance between two points is the distance from one point to the other.

 

[PAllen]

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.

Einstein was a strong proponent of the pseudogravity explanation for most of his life. He never considered there was paradox, but his preferred explanation was pseudogravity. It could be made to work for every smooth (continuous second derivative, specifically) variant in SR, and generalizes to GR.

I do not like this approach, but I find it nonsensical to claim it is invalid.

 

[jbriggs444]

The relative distance between two points is the distance from one point to the other.

A "point" in four dimensional space time is stretched into a chain of "events" known as a "world line".

Measuring the distance between two events is unambiguous. Measuring the distance between two world lines is ambiguous -- it depends on which pair of events you pick out.

The process of picking a particular world line through an event to be the "point" associated with that event is also a potential source of ambiguity.

 

[vanhees71]

Huh?

The twin who slingshots around a large mass is lower in a potential well. So he ages more slowly. And he is moving faster. So he ages more slowly.

Both of these are coordinate-dependent heuristics for time dilation, but the result should hold for differential aging: Shorter elapsed time for the slingshotting twin.

Now, I'm confused. The geodesic equation is derived from the free-particle action principle with the Lagrangian
$$L=-m c \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}},$$
i.e., it's the curve between fixed initial and final points which minimizes
$$A=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda L=-mc \tau.$$
So $\tau$ is maximal for the geodesic (free-fall worldline) connecting two points, right?

 

[PeroK]

Einstein was a strong proponent of the pseudogravity explanation for most of his life. He never considered there was paradox, but his preferred explanation was pseudogravity. It could be made to work for every smooth (continuous second derivative, specifically) variant in SR, and generalizes to GR.

I do not like this approach, but I find it nonsensical to claim it is invalid.

Touche. Obviously the great man was a law unto himself!

 

[PeterDonis]

It seems to me that slingshotting around a distant star brings an entirely different geometry into the situation. I don't think that SR applies in that situation.

That's correct. The post I was responding to there was, IIRC, not limiting itself to SR scenarios.

we know that the twin who slingshots around a large mass would age faster

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.

 

[PeterDonis]

in GR the maximum proper time among all time-like curves connecting to points is reached for a geodesic connecting the two points.

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.

 

[PeterDonis]

The relative distances when the acceleration occurs is different.

A better way to state this would be that the time elapsed on the traveling twin's clock between the starting event (where he leaves the stay-at-home twin) and when the acceleration occurs is different. That makes it clear that the difference is invariant.

 

[PeroK]

I do not like this approach, but I find it nonsensical to claim it is invalid.

I'm not sure whether I'd say it's invalid, but here are the two things I dislike most:

1) It avoids using the simple geometry of flat spacetime - that explanation should be a lightbulb moment for any student of SR.

2) It fails to dispel the myth that GR is required to explain the twin paradox. Reading the 100+ posts in this thread you may come to the conclusion that whether GR and gravity are required to explain the twin paradox is still an open question.