Twin Paradox: Aging Slower with Continuous Acceleration?

[vibhuav]

TL;DR Summary

Twin paradox with travelling twin going around in circles

I understand that the travelling twin (T, say) is subjected to acceleration and deceleration while the stay-at-home twin (S) is in inertial frame all the time. It is this asymmetry which results in the travelling twin aging less than the other, when they two meet up.

Since acceleration is the key for time dilation and hence reduced aging, I thought, what will happen if the travelling twin goes around the earth at close-to-lightspeed, instead of to a distant star, and back? Then there is continuous acceleration because the direction of speed is continuously changing. I gave up on that thought because going around the earth, the rocket is in freefall and therefore an inertial reference frame. The acceleration and deceleration to get up there and be back is all that contributes to the reduced aging almost similar to the original twin paradox problem. (Right?)

If so, what will happen if the travelling twin goes around in a circle at high speed on a carousel? (The experiment can be done on a moon to eliminate the retarding effects of the atmosphere.) Now the travelling twin is not in an inertial ref frame at all throughout the "flight" but is continuously accelerated. Will he age much more slower? Being physically close to the "stay-at-home" twin, can both peer inside the other's ref frame and observe how the other's time is passing?

 

[PeterDonis]

I understand that the travelling twin (T, say) is subjected to acceleration and deceleration while the stay-at-home twin (S) is in inertial frame all the time. It is this asymmetry which results in the travelling twin aging less than the other, when they two meet up.

For that particular scenario, this rule of thumb works. But it doesn't generalize. Please read this article:

https://www.physicsforums.com/threads/when-discussing-the-twin-paradox-read-this-first.1048697/

After you've read that, you will need to rethink your scenarios.

what will happen if the travelling twin goes around the earth at close-to-lightspeed, instead of to a distant star, and back? Then there is continuous acceleration because the direction of speed is continuously changing. I gave up on that thought because going around the earth, the rocket is in freefall and therefore an inertial reference frame.

The rocket certainly won't be in free fall if its orbital speed around Earth is close to the speed of light.

Also, for this kind of scenario, you have to take into account the effects of altitude as well as speed; briefly, clocks run faster at higher altitudes in a gravitational field. So there are two effects, altitude and speed, and you have to take both into account.

You should be able to see from the above that the simple "acceleration causes reduced aging" rule of thumb does not work for this kind of scenario.

The acceleration and deceleration to get up there and be back is all that contributes to the reduced aging almost similar to the original twin paradox problem. (Right?)

No. See above.

what will happen if the travelling twin goes around in a circle at high speed on a carousel?

Here it's simpler because there is no altitude effect, it's just speed. Experiments like this have been run on muons circling around a storage ring, to confirm that the time dilation they experience is due entirely to speed; acceleration has no effect on it (acceleration is of course required to keep the muons going around in the circle, but by itself it has no effect on time dilation, only speed does).

Being physically close to the "stay-at-home" twin, can both peer inside the other's ref frame and observe how the other's time is passing?

There is no such thing as "peer inside the other's ref frame". The effects involved are comparisons between clocks (or phenomena such as the half-life of muon decay that are equivalent to clocks), not reference frames.

 

[Nugatory]

Being physically close to the "stay-at-home" twin, can both peer inside the other's ref frame and observe how the other's time is passing?

Yes. In principle we could use a powerful telescope to look directly at the face of the other person's wristwatch and directly observe the rate at which the hands are moving. The flashes of light used in the Doppler Shift Analysis from the FAQ @PeterDonis linked are effectively the same thing.

 

[PeterDonis]

In principle we could use a powerful telescope to look directly at the face of the other person's wristwatch

This isn't "peering inside the other's reference frame". It's just looking at the other clock.

 

[Nugatory]

This isn't "peering inside the other's reference frame". It's just looking at the other clock.

Of course, but it seems to be what OP is reaching for here....

 

[Dale]

TL;DR Summary: Twin paradox with travelling twin going around in circles

If so, what will happen if the travelling twin goes around in a circle at high speed on a carousel?

This was already mentioned, but this experiment has been done with muons in a circular containment ring. This is a highly relativistic carousel. It was performed in the late 70’s and was published here:

Bailey et al., "Measurements of relativistic time dilation for positive and negative muons in a circular orbit," Nature 268 (July 28, 1977) pg 301.

Their results were consistent with SR and confirmed both the twin paradox and the clock hypothesis with accelerations on the order of $10^{18} \ g$.

 

[Vanadium 50]

, but this experiment has been done with muons in a circular containment ring.

And a version is running right now at Fermilab, the g-2 experiment.

Unfortunately for the anti-relativity crowd, it is an engineering fact of life.

 

[Vanadium 50]

As a PS, this g-2 experiment and its predecssors (there have been at least 2) rely on the EM fields transforming according to the Lorentz Transformation. If either of these were not correcr, the experiment would not work at all. I don't mean "wrong answer". I mean "no answer".

 

[Ibix]

If so, what will happen if the travelling twin goes around in a circle at high speed on a carousel?

The elapsed time for a person undergoing any motion in flat spacetime can be written as the elapsed time in some inertial frame divided by the Lorentz gamma factor of the person as measured in that frame. This will typically involve integration if the speed isn't constant, but in your carousel example speed is constant. So the elapsed time is $\sqrt{1+(v/c)^2}\Delta t$, where $\Delta t$ is the time elapsed for someone standing beside the carousel and $v$ is the linear velocity of the person on the carousel, equal to the radius of their circle times the angular velocity.

what will happen if the travelling twin goes around the earth at close-to-lightspeed, instead of to a distant star, and back? Then there is continuous acceleration because the direction of speed is continuously changing. I gave up on that thought because going around the earth, the rocket is in freefall and therefore an inertial reference frame. The acceleration and deceleration to get up there and be back is all that contributes to the reduced aging almost similar to the original twin paradox problem. (Right?)

Wrong. You can ignore the launch and land phase, at least for a first pass, simply by having the orbiting phase be much longer. In this case the answer is actually the same as the carousel multiplied by the gravitational time dilation factor between an observer on Earth and an observer hovering at the altitude of the orbiting one. As noted by others this orbit will not be a freefall orbit if you want to orbit the Earth at large fractions of $c$, but that actually doesn't matter. The same formula applies. (Note that the maths rapidly gets more complex if you consider non-circular orbits, the rotation or non-sphericity of the Earth, or the effect of other astronomical objects on orbits - but this is all negligible here.)

Note that acceleration hasn't entered into the maths at all. The acceleration in the regular twin paradox does two things - first it allows the twins to meet again so they can compare ages unambiguously, and second it means that the twin who turns round cannot naively use the time dilation formula because that assumes that the person treating themselves as "at rest" is inertial at all times. Who is older has very little to do with acceleration - it only appears to be the deciding factor in specific situations like the standard twin paradox setting.

 

[vanhees71]

As a PS, this g-2 experiment and its predecssors (there have been at least 2) rely on the EM fields transforming according to the Lorentz Transformation. If either of these were not correcr, the experiment would not work at all. I don't mean "wrong answer". I mean "no answer".

I don't know, what the $(g-2)$ experiments have to do with time dilation or the twin paradox. All direct highly precise measurements of time dilation with instable particles/nuclei in storage rings indicate that the "clock hypothesis" is correct even for particles/nuclei at very high accelerations.

Of course the very accurate prediction of the value of $(g-2)$ by the Standard Model is based on relativistic QFT and thus a possible violation of these predictions might indicate that something is wrong with relativistic spacetime models, but it's more likely that the Standard Model is incomplete.

However, despite the apparent discrepancy between theory and experiment for quite some time close to the $5 \sigma$ significance level, more recent research on the theory side indicates that there might be no discrepancy after all. The most severe theoretical problem are the QCD corrections to the magnetic moment of the muon. One way is to extrapolate accurate measurements of $e^+ + e^- \rightarrow \text{hadrons}$ to the needed off-shell contributions via dispersion relations, which is not so simple. Another way is to use accurate lattice-QCD calculations to evaluate these corrections, and more recent research by various lQCD groups hint at the possibility that there are no discrepancies between the Standard Model and the measurement at the Fermilab (former BNL) experiment.

 

[Vanadium 50]

g-2 runs muons in a ring. at γ = 29.3. It is the OP's "lets measures this in a circle" experiment. And the muon lifetime is increased to the expected 65 microseconds.

 

[vanhees71]

Sure, but $(g-2)$ has nothing directly to do with the twin paradox or time dilation. An example of an experiment testing the time dilation, using atomic transition rates, is here:

https://arxiv.org/abs/1409.7951

 

[Vanadium 50]

The fact that the experiment is named "g-2" is irrelevant, as is the experiment's ultimate goal. The important thing is that they have fast muons going in a circle in a storage ring.

 

[Freixas]

TL;DR Summary: Twin paradox with travelling twin going around in circles

I understand that the travelling twin (T, say) is subjected to acceleration and deceleration while the stay-at-home twin (S) is in inertial frame all the time. It is this asymmetry which results in the travelling twin aging less than the other, when they two meet up.

Since acceleration is the key for time dilation and hence reduced aging,

to confirm that the time dilation they experience is due entirely to speed; acceleration has no effect on it

Peter, of course, has the right answer. But if it helps, the standard twin paradox can be tweaked to eliminate acceleration, so the assumption that the acceleration of one twin and not the other is what resolves the twin paradox is incorrect.

Consider three observers, A, B, and C. Relative to A, B is moving toward A and C and C is moving toward B and A. As B passes A, they synchronize their clocks to 0. As B passes C, C sets his clock to match B's. As C then passes A, they compare clocks.

Everyone in this scenario moves inertially. The time difference between A and C will be very close to what it would be in the standard twin paradox scenario if the time the traveling twin accelerates is kept very short. The shorter the acceleration time, the closer the values will be. Acceleration doesn't resolve the twin paradox; the different reference frames used for the outbound and inbound legs does.