0
$\begingroup$

Consider a merry go round with a radius of 1 light year and a tangential velocity of 0.9c. The rider is in a capsule on the edge of the merry go round. The observer is just a guy who is the same age as the rider but just floating in space. The merry go round start's spinning. The rider is in a couch in the capsule and at 0.9 c experiences a very tolerable 7.6 m/s^2 centrifugal acceleration.

From the rider's point of view, the observer is like an orbiting planet with an eccentric orbit, albeit at relativistic speed. So the rider will calculate that the observer will age more slowly.

But the observer will see the rider rotating very fast and will calculate the she will age more slowly.

When the merry go round runs for a few days and then stops, who is the one who experienced time dilation?

I read these recent threads:

How to tell who is experiencing time slower and who faster when travelling at different speed?

On The Twin Paradox The Symmetry Remains

Clearly, we humans continue to be puzzled at this hundred year old story. From reading the threads above, PM2ring's comment is enlightening:

"In Special Relativity, speed causes time dilation, but with constant speed the situation is symmetrical. If observers A & B have a constant relative velocity then A measures B's clock to be running slow by a factor of γ, and B measures A's clock to be running slow by a factor of γ.

To break the symmetry, (at least) one of the observers needs to make one or more changes of reference frame. It's not so much that the acceleration causes time dilation, it's merely the mechanism whereby the reference frame is changed."

This tells me that in my model above, the rider will age more slowly since she is not in an inertial frame of reference. But consider if the observer is on an identical merry go round, just shifted a little in the z direction. If both merry go rounds are spinning in the same direction, they are covariant. Each rider will see the other as stationary. Each will calculate that Neither rider will age slower.

But if the merry go rounds spin in opposite directions, each rider will see the other moving very fast and will calculate that the other should age more slowly. When the merry go rounds stop, they will be surprised to see they are the same age, even though each saw the other whipping by.

These two situations put the riders in symmetrical reference frames. Yet in one direction they see each other age similarly, and when spinning opposite, they see each other age differently.

And it doesn't require a merry go round. If both riders are on rocket sleds light years apart and travel towards each other at 0.9c, each will see the other age more slowly. Yet when they meet, they should be the same age since their reference frames are symmetrical.

I guess it would be nice to have a mechanistic explanation of what is going on. How is it that traveling fast changes the fundamental nature of aging.

$\endgroup$
4
  • $\begingroup$ Reminds me of the Hafele–Keating experiment. $\endgroup$
    – M. Enns
    May 22 '21 at 23:27
  • $\begingroup$ If that experiment were done on the moon which is tidally locked and not really rotating much, then the clocks would not differ from each other and only be slower than a moon based clock (ignore the gravitational relativity please and never mind that planes would not work-use rocket powered drones ). Also, whether the drones fly together or opposite should give the same time dilation relative to the moon based clock. Yet according to SR, the clocks should be the same when the drones fly together but differ when the drones fly opposite. $\endgroup$ May 23 '21 at 1:26
  • $\begingroup$ I wish there were someone who could do for relativity what Maxwell and Boltzmann did for thermodynamiics. Before kinetic theory, it was all about heat moving from point A to point B and the work that could be extracted. They invented this property 'entropy' but really did not know why it was happening. Just as today we do not know why time dilation occurs. The kinetic theory allowed the application of discrete math to calculate particle velocities. But as Feynemann said, we never really know why. Answering a why question just reframes the question with smaller questions. $\endgroup$ May 23 '21 at 1:38
  • 2
    $\begingroup$ @aquagremlin : we do know why time dilation occurs, there's a simple geometric explanation. Clocks measure the lengths of the spacetime paths they follow. Inertial observers follow the longest possible paths. Observers that are non-inertial (that accelerate) follow shorter paths. $\endgroup$
    – Eric Smith
    May 23 '21 at 11:34
3
$\begingroup$

From the rider's point of view, the observer is like an orbiting planet with an eccentric orbit, albeit at relativistic speed. So the rider will calculate that the observer will age more slowly.

But if the merry go rounds spin in opposite directions, each rider will see the other moving very fast and will calculate that the other should age more slowly. When the merry go rounds stop, they will be surprised to see they are the same age, even though each saw the other whipping by.

Neither of these are correct. You are trying to apply a calculation that only applies for an inertial frame to a frame that is not inertial.

For an arbitrary frame that has a time coordinate the time dilation is given by $$\frac{1}{\gamma}=\frac{d\tau}{dt}$$

For a rotating frame $$d\tau^2 c^2=-ds^2= \left(1-\frac{ \omega^2 r^2}{c^2} \right) \left( c \ dt - \frac{ r^2\omega/c}{ 1-\omega^2 r^2/c^2} d\phi \right)^2 - dr^2 - \frac{r^2}{ 1-\omega^2 r^2/c^2} d\phi^2 -dz^2$$ see https://arxiv.org/abs/0904.4184 for details. Evaluating this using units where $c=1$ gives $$\gamma=\left( 1- r^2 \omega^2 - \dot r^2 - r^2 \dot \phi^2 - 2 r^2 \omega \dot \phi - \dot z^2 \right)^{-1/2}$$

In the first scenario, where the inertial observer is seen from the rider’s perspective, the inertial observer is characterized by $\dot z =0$, $\dot r=0$, and $\dot \phi = -\omega$. Plugging that into the above expression gives $\gamma =1$, so the observer is not dilated in the rider’s frame. In contrast, the rider has $\dot \phi=0$, so $\gamma= 1/\sqrt{1-r^2 \omega^2}$. Thus even from the rider’s perspective the observer’s clock goes faster.

In the second scenario the opposite rider has $\dot \phi = -2\omega$, so $\gamma=1/\sqrt{1-r^2\omega^2}$. Thus each counter-rotating rider agrees that they are equally time dilated with the other.

So when you use the correct time dilation formula for the non-inertial frame then all of the correct symmetries are preserved and none of the incorrect ones are retained. The actual math works out as it should.

I guess it would be nice to have a mechanistic explanation of what is going on. How is it that traveling fast changes the fundamental nature of aging.

The mechanism is geometry. Space and time are not separate things but are different directions in the same space that we call spacetime. Everything about the geometry of spacetime is captured in $ds^2$, which is called the metric, or the spacetime interval. Time dilation is the spacetime equivalent of the familiar triangle inequality.

$\endgroup$
2
$\begingroup$

The person who travels always ages less the the one who does not. That is the only answer in Special relativity. Learn about the spacetime interval.

$\endgroup$
0
$\begingroup$

In all of your examples, one or both parties is accelerating. Even with the rocket sleds example, if the riders are to meet then they must become at rest with respect to one another, which means one or both must accelerate.

Each party can each measure their own rate of acceleration by doing experiments. If their accelerations are different then the situation is not symmetric, and one will experience more time dilation than the other. If their accelerations are the same then they will both experience the same amount of time dilation (relative to the external, non-accelerated, universe).

$\endgroup$
0
$\begingroup$

I guess it would be nice to have a mechanistic explanation of what is going on. How is it that traveling fast changes the fundamental nature of aging.

You're looking at this the wrong way: there's no such thing as "traveling fast" in an absolute sense, velocity is always relative.

The explanation for differential aging is geometric, rather than mechanical. Clocks are measuring instruments, and the reading on a clock shows the length of the path the clock takes through spacetime. If different clocks follow different paths, then naturally they can have different readings on them.

Inertialy moving bodies take the longest possible path (not the shortest, because we're measuring with time and time has the opposite sign to space in the spacetime metric). A clock that accelerates therefore takes a shorter path than one that doesn't, and will show a smaller reading when the two are brought together.

Relating this back to the carousel: the observer's clock is "at rest" (never accelerates) so it will have a longer spacetime path than a clock that's on the edge of the merry-go-round and constantly accelerating.

$\endgroup$
0
$\begingroup$

The choice of merry-go-rounds complicates the issues somewhat and masks the underlying principle, so consider the simpler case you mention of two people approaching each other.

Let's make the two people twins of the same age who start the experiment a distance apart and stationary relative to each other. As they travel toward each other, they will each consider the other to be ageing more slowly, yet when they meet they will find they are still the same age as each other. The paradox is resolved when you remember to take in to account the relativity of simultaneity- each twin will believe they started their journey earlier than the other did. When you factor-in the perceived difference in the start times, they restore the overall symmetry.

To take an extreme example to illustrate the point. Suppose you and I are walking toward each other. I set off on Monday and arrive at our meeting on Thursday, having aged three days. You walk towards me, leaving at what you consider to be Monday and meeting me on Thursday after ageing three days.

Owing to time dilation, from my perspective, you only aged two days while you were walking, not the three you experienced yourself, but owing to the relativity of simultaneity, I consider that you didn't start walking until Tuesday instead of the Monday departure you experienced in your reference frame. So the reduction in ageing which I observed during your walk is balanced by your later start date- you have still aged by three days when you meet me. You see the reciprocal effects.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.