B Twin paradox explained for laymen

[Ibix]

Given that the second watch was switched off during the critical acceleration phases, how do you explain this? Using your pseudo-gravitational explanation?

The point is to use pseudo-gravity to regard the stay-at-home twin as above the traveller during the turnaround. Thus the stay-at-home's clock runs fast and elapses more time than the traveller.

I don't think the explanation is particularly intuitive because you need to be quite careful by what you mean by "during" the turnaround. That goes double for instantaneous turnaround - what's the pseudo-gravitational potential associated with an infinite acceleration? But if you are willing to do the mucking around it ought to work.

 

[vanhees71]

Ether theory is physically distinct from special relativity. The theories agree only up to first order in $\beta=v/c$ (roughly speaking). As you point out yourself ether theory has been ruled out at for at least one century ago with many experiments (Michelson-Morley, Trouton-Noble, Doppler effect on light,...). A nice review can be found here:

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

 

[Vanadium 50]

This whole "pseudo-gravity" explanation seems supremely unhelpful. Telling someone who is struggling with special relativity that understanding it requires understanding general relativity, kinda-sorta, does not sound like the easiest path. The fact that thousands of people get it without GR or GR-like or pseudo-GR establishes this.

(It might possibly be usable in reverse - starting from SR to get to GR as motivation - but it's certainly not often adopted this way)

"Acceleration" cannot be the answer to "why is the aging different" because it is possible to set this up with multiple travelers looking at clocks through windows with nobody accelerating at all. The answer is that these are different paths through spacetime, and they are different just like different paths between A and B in space (say LA to San Francisco via Baltimore) have different lengths.

"Acceleration" is only an answer to the question "I don't want to do this with windows and clocks - I want to compare two clocks, one staying home, and one that goes on the trip. How do I tell them apart?"

 

[robphy]

That's all fine, if you teach relativity to physics students. But the OP asks for a laymen explanation of the typical "twin paradox" scenario and states wrongly, that a scenario without real gravity would be symmetrically.

I think, for laymen an explanation with 4-vectors and 4D-spacetime may be too difficult to understand. An explanation comparing the empirical viewpoint of both twins maybe didactically better for an explanation to laymen.

The diagrams actually are intended for the lay person (after some introduction),
but the words I used for my reply weren't necessarily addressed to the layperson.
I see now I used "pseudo-riemannian" inappropriately to make my point
since Galilean is also pseudo-riemannian.

My main point is that
the way we use the position vs time graph in PHY 101 is only an approximation to what is really going on.
In fact, it is not well-appreciated that the position-vs-time graph is already a non-euclidean geometry.

Here's are the real measurements on a position-vs-time graph.
In this graph, the red lines are perpendicular to each other (in all three geometries mentioned earlier) and the blue segment is the hypotenuse. You can introduce units to redefine variables to make the sides have the same units... but that doesn't change the underlying geometrical relationships.

In PHY 101, we treat the blue segment as having length 1 (already in violation of Euclidean geometry)
but it's really length (approximately) 1-(5\times 10^{-17}). (Someday, we'll have a wristwatch that will measure this.)
The event that is "Length 1 along the blue segment" occurs to the right of the red vertical line.
This is time-dilation.

 

[martinbn]

I might be repeating something that was aleardy said, but here is a scenario that might help with the issue of the acceleration. The two twins depart together in the same direction. One of them turns back first, the second later. They both return to the original place, the one that turns first will wait there for the second one. This way they accelerate in the begining, during the turn, and at the end in exactly the same way. So the accelerations were the same for both. Just one turns later. When they meet they compare age and it turns out that they are not the same age. So acceleration cannot be the reason for the different aging.

 

[FactChecker]

I might be repeating something that was aleardy said, but here is a scenario that might help with the issue of the acceleration. The two twins depart together in the same direction. One of them turns back first, the second later. They both return to the original place, the one that turns first will wait there for the second one. This way they accelerate in the begining, during the turn, and at the end in exactly the same way. So the accelerations were the same for both. Just one turns later. When they meet they compare age and it turns out that they are not the same age. So acceleration cannot be the reason for the different aging.

The observed effect on another observer depends on the distance between the observers. The greater the relative distance, the greater the effect on the clocks. This is why the accelerations at the start and end of the twin's travel has very small effect, but the accelerations when he turns around has a large effect. Otherwise, the acceleration effects would largely cancel out. So your example does not remove the proposed effect of acceleration.

 

[vanhees71]

I might be repeating something that was aleardy said, but here is a scenario that might help with the issue of the acceleration. The two twins depart together in the same direction. One of them turns back first, the second later. They both return to the original place, the one that turns first will wait there for the second one. This way they accelerate in the begining, during the turn, and at the end in exactly the same way. So the accelerations were the same for both. Just one turns later. When they meet they compare age and it turns out that they are not the same age. So acceleration cannot be the reason for the different aging.

It's also interesting that this question has been investigated experimentally:

http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html#Clock_Hypothesis

The upshot is that indeed even at very high proper accelerations against an inertial frame "proper clocks" like lifetimes of unstable particles (muons) show the "proper time", defined by $\mathrm{d} \tau=\sqrt{1-\beta^2} \mathrm{d} t$, i.e., independent of acceleration.

 

[Sagittarius A-Star]

So acceleration cannot be the reason for the different aging.

Acceleration does not directly effect the different aging. But it has an indirect effect on it, because for example in the inertial rest frame of the stationary twin, the integral over time of the acceleration of the traveling twin equals the change of his/her velocity.

As @FactChecker stated correctly in posting #66, in the standard "twin paradox" scenario, this indirect effect is greater, if the distance of the turnaround from Earth is greater:

1) Reason, described in 4D-spacetime: If this spatial distance is greater, then the proper time of the traveling twin is smaller, because of the minus-sign in the formula for invariant spacetime-distance.

2) Reason, described in the rest frame of the stationary twin: If this spatial distance is greater, then the traveling twin travels longer with an almost constant slow tick-rate of his/her watch (1/"Gamma").

3) Reason, described in the rest frame of the traveling twin: If this spatial distance is greater, then the pseudo-gravitational potential of the remote "stationary" twin is greater, and therefore also the tick-rate of his/her watch is greater - while the short "turnaround"-time.

 

[PeroK]

Acceleration does not directly effect the different aging. But it has an indirect effect on it, because for example in the inertial rest frame of the stationary twin, the integral over time of the acceleration of the traveling twin equals the change of his/her velocity.

As @FactChecker stated correctly in posting #66, in the standard "twin paradox" scenario, this indirect effect is greater, if the distance of the turnaround from Earth is greater:

1) Reason, described in 4D-spacetime: If this spatial distance if greater, then the proper time of the traveling twin is smaller, because of the minus-sign in the formula for invariant spacetime-distance.

2) Reason, described in the rest frame of the "stationary" twin: If this spatial distance if greater, then the traveling twin travels longer with an almost constant slow tick-rate of his/her watch (1/"Gamma").

3) Reason, described in the rest frame of the traveling twin: If this spatial distance if greater, then the pseudo-gravitational potential of the remote "stationary" twin is greater, and therefore also the tick-rate of his/her watch is greater - while the short "turnaround"-time.

That's a bit like being given a speeding ticket that, instead of saying you were driving too fast, provides an acceleration profile that implies you were drving too fast.

By that argument, acceleration is an indirect cause of speeding. That may be a true statement, but it would make for a complicated law as to if and when you got a speeding ticket.

 

[FactChecker]

By that argument, acceleration is an indirect cause of speeding. 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 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.

 

[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.