Rotation and circular motion under different simultaneity conventions

1. Jun 9, 2014

johnny_bohnny

A interesting question came up to my mind, and it's related to the concept of different synchonization methods. If we change the synchronization parameter from 1/2 to some other value, light will have different velocities in opposite directions, which implies that two velocities with opposite directions but same speed value under 1/2 parameter will differ in value while using another synchronization parameter.

My question is the following: if we take a rotating object or some other object that is undergoing uniform circular motion, it will change direction at each instant but the speed will be the same. For instance, all points on the equator have the same speed because the radius from the center is same or all of them, but they have different velocities since their direction is different. Under some other synchronization parameter, what will happen with the rotation, because it seems that now a point will also change its speed while changing its directions, since different direction implies diffrent speed under non-standard synchronization. So how will the rotation under 1/2 parameter look like under some other parameter?

And this is maybe off-topic, but I think nonetheless it might be related. If we specify the speed of light in opposite direction, let's say on the x-axis, how can we specify its speed in all other directions, since there are infinite number of possible directions, and if light isn't isotropic how can we 'handle' each of them while defining our one-way velocity for each direction?

2. Jun 9, 2014

WannabeNewton

Your question is a bit ill-formed because the very notion of a circular trajectory through space requires a synchronization convention to have been adopted to start with. It does not make sense to talk about a circular trajectory in a given inertial frame with a certain synchronization convention and then switch synchronization conventions and ask about that same circular trajectory because it won't be invariant under a change of clock synchronization. Rather one must ask the following: if I choose Einstein synchronization in a given inertial frame and under this synchronization have a particle moving in a circular trajectory about an observer at rest in the frame's origin, what happens to the particle's trajectory under a change of clock synchronization? Keep in mind when I say "circular trajectory" I mean a trajectory in space for which the distance from said observer remains constant in time; it's clear that this is a synchronization dependent notion.

If the change in synchronization convention is simply a change from $\epsilon = \frac{1}{2}$ to $\epsilon = \text{const.} \neq \frac{1}{2}$ then this is very easy to answer. In this case the simultaneity planes in the inertial frame will be tilted at an angle to the standard simultaneity planes, that is the ones under Einstein synchronization, and instead of a circular trajectory we get an elliptic trajectory. The particle will then no longer have the same velocity at each instant relative to the observer but rather will have one that varies in accordance with the elliptic trajectory.

3. Jun 9, 2014

johnny_bohnny

So in the case of an elliptic trajectory, a particle won't have the same instantaneous velocity during the whole trip, like in the circular motion case. For instance, Earth travels in an elliptical orbit around the sun, does that mean that its orbital velocity varies depeding on the position on its elliptical trajectory, and if it does, how does the acceleration occur and why during the elliptical movement?

4. Jun 9, 2014

Staff: Mentor

You don't need any relativity to answer that question, the classical solution for planetary motion (and central force problems in general) works just fine.

Yes, the speed of the earth relative to the sun changes as it moved around in its elliptical orbit; it picks up speed as it falls closer to the sun as it moves from aphelion to perihelion, and then slows down again as it climbs back to aphelion.

You can get this result from conservation of energy. The total energy, which is the sum of the gravitational potential energy and the kinetic energy, is constant. The potential energy is less when the earth is closer to the sun so the kinetic energy and speed must be greater.

5. Jun 10, 2014

johnny_bohnny

@Wannabe Newton, thank you very much for your detailed answer. I would like to ask you one more thing, which really gives me issues, and it's connected to non-standard synchronization method. Let's suppose that we take some other synchronization parameter other than 1/2 and under those conditions define our speed of light on the x-axis as three times greater in one direction than in another. How will we then define the one-way speeds of light under the same synchonization parameter in every other direction, since there are infinitely many directions. Could you give me an example of using this method?

@Nugatory Thank you also for the well-formed answer. If Earth changes its speed during the eliptical movement, it means that is undergoing acceleration during the path. Now my other question is from a relativistic perspective. If the Earth was on a clear circular path it would stay rigid, or to say it better its circular orbit would be considered as an example of a rigid body motion. Does this change when considering the ellipical orbit, since it changes its velocity during the elliptical trajectory, or to say it better, do distances remain constant?

6. Jun 10, 2014

Staff: Mentor

The speed of the earth relative to the sun is only a few kilometers per second, nowhere near fast enough for any relativistic effects to show up.

Nonetheless, the earth changes shape and experiences internal stresses. This has nothing nothing to do with relativity, Born rigidity, and all that. It's a purely classical effect; because the earth is not a point particle, at any moment some parts of the earth are closer to the sun and the moon than others, so the different parts of the earth experience different gravitational effects from them. That's what causes tides, and it happens on both circular and elliptical orbits.

Two other important things:
1) An object moving in a circle isn't changing its speed, but it is accelerating. The magnitude of the acceleration is $v^2/r$, and it points directly inward. In the case of the earth's orbit around the sun, the gravitational attraction between them provides the force that causes this acceleration; without it the earth would travel in a straight line and eventually leave the solar system.
2) So far I've been talking classical. In relativity the changes of speed and direction caused by gravity are fundamentally different than those caused by other forces, and shouldn't even be considered as accelerations at all. That's a General Relativity discussion, and there are plenty of threads here; a search for the keywords "curvature gravity" will find them.

Last edited: Jun 10, 2014
7. Jun 10, 2014

johnny_bohnny

8. Jun 10, 2014

Staff: Mentor

Velocity is a vector, so changes of direction are also changes of velocity. An object in a circular orbit around the sun has a constant speed relative to the sun; an object in an elliptical orbit around the sun does not. Neither has a constant velocity relative to the sun.

Either way, the earth is not undergoing rigid body motion. It changes shape and develops internal stresses. However, any relativistic contribution to this effect is completely swamped by the classical tidal effects.

9. Jun 10, 2014

johnny_bohnny

But relativistic effects still count? If we eliminate gravity and imagine an object undergoing motion on an elliptic path would that be a rigid motion?

10. Jun 10, 2014

Staff: Mentor

Elliptic or circular, or in a straight line with changing speed

11. Jun 10, 2014

johnny_bohnny

I'm pretty sure that forum member DaleSpam mentioned that during circular motion a body stays rigid.
So I'm wondering is the same happening during elliptical motion, that's the whole point.

12. Jun 10, 2014

WannabeNewton

I'm not understanding what your confusion is exactly, with regards to the above. Sure there are infinitely many directions but there are only three independent directions in space. You need only worry about the one-way speed of light along these three directions. All other directions are determined from these three.

A body whose points are in elliptical motion does not undergo Born rigid motion. For any time-like congruence in flat space-time that has non-vanishing vorticity, the congruence will be Born-rigid if and only if the congruence is parallel to a time-like Killing field of flat space-time. More precisely, if $\xi$ is the tangent field to the congruence and $\xi^{\flat}$ its dual, then $\xi^{\flat}\wedge d\xi^{\flat} \neq 0$ implies that $\xi$ describes Born rigid motion if and only if $\xi = \sum \alpha_i \partial_i$ where $\alpha_i$ are constants and $\partial_i$ are flat space-time Killing fields in an adapted coordinate system. For example for circular motion we have $\xi = \partial_t + \omega \partial_{\phi}$ with $\omega$ the angular velocity and $\partial_{\phi}$ the axial Killing field (the vector field that generates azimuthal rotations).

It is easy to see that a time-like congruence describing points in elliptical motion relative to some inertial frame will not have the above form. In fact we know that the 4-velocity is in general $u = (\frac{dt}{d\tau}, \frac{dr}{d\tau}, 0,\frac{d\phi}{d\tau}) = \frac{dt}{d\tau}(1, \frac{dr}{dt}, 0, \frac{d\phi}{dt})$ so that $\xi = \partial_t + \frac{dr}{dt}\partial_r + \frac{d\phi}{dt}\partial_{\phi}$ and $\gamma \equiv \frac{dt}{d\tau} = \frac{1}{\sqrt{1 - (\frac{dr}{dt})^2 - r^2(\frac{d\phi}{dt})^2}}$; for an elliptic trajectory neither $\frac{dr}{dt}$ nor $\frac{d\phi}{dt}$ are constants.

13. Jun 11, 2014

johnny_bohnny

Could you give me an example of this for better understanding? Let's say I specify that the speed of light is three times faster in one direction then in another on the x-axis. So what will the speeds be in other directions. I basically understand your point but I would like to get a concrete example to see how this non-standard synchronization works.

So basically this means that all objects on Earth and Earth itself is changing its shape at any instant, by a slight value, but still non-rigidity is there.

14. Jun 11, 2014

WannabeNewton

You haven't given enough information to construct an example but that's alright, let me construct a very simple one. The $\epsilon$ simultaneity formulation says as long as $0< \epsilon < 1$, the simultaneity convention $t = t_1 + \epsilon (t_2 - t_1)$ is valid; here $t_1, t_2$ are the local emission and reception times of a light signal and $t$ is the simultaneous distant time. Consider then an inertial frame with cylindrical coordinates $(r,\theta,z)$. We can define $\epsilon(\theta) = \frac{1}{4}(1 + \cos^2 \theta)$; note that $0 \leq \cos^2 \theta \leq 1$ so $\frac{1}{4} \leq \frac{1}{4}(1 + \cos^2 \theta) \leq \frac{1}{2}$ for any value of $\theta$ which means $\epsilon(\theta)$ defines a valid synchronization convention. Now $\frac{c_{\text{out}}}{c_{\text{in}}} = \frac{1 - \epsilon}{\epsilon} = \frac{3 - \cos^2 \theta}{1 + \cos^2 \theta}$ so this will tell you the relationship between the outgoing and incoming one-way speeds of light for this particular synchronization convention. In this convention the two speeds have a different ratio along different directions lying in planes parallel to the $x$-$y$ plane but have no $z$ or $r$ dependence. For $\theta = 0,2\pi$ we have $c_{\text{out}} = c_{\text{in}}$ and for $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$ we have $c_{\text{out}} = 3c_{\text{in}}$.

What I stated above was for flat space-time. The definition of Born-rigidity is the same in curved space-times as it is in flat space-time but the theorem that rotational time-like congruences are Killing congruences iff they are Born-rigid does not hold for curved space-times in general, even if they do possess Killing fields.

15. Jun 11, 2014

johnny_bohnny

@WBN, thanks for the answer, I'll take my time in analyzing it, but I get the basics at first.

In your first post you mentioned that a circular path will become an elliptical one using a different convention. So what can be said about a rotating object, which is of course rotating under 1/2 synchronization, if we use another convention? Will we get some sort of a 'elliptical' rotation in this case, what will be the shape of that object?

If I am at rest wrt to Earth, will I continue to be at rest if we change the convention?

16. Jun 11, 2014

WannabeNewton

The angular velocity of rotation is entirely independent of clock synchronization but the shape of a rotating object can change under a change of synchronization convention.

Yes but that is a special case. In general velocities are not invariant under a change of synchronization convention.