What does comoving mean?

  • #1
George Keeling
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Summary:

What does comoving mean? With some of examples of usage. It seems to mean not moving!

Main Question or Discussion Point

I come across the adjective 'comoving' quite often. I understand comoving coordinates for the Universe. They are coordinates which expand with the expansion of the Universe(?) but I'm confused about what it means in essence. Here are some examples:

In Sean Carrol's book there is a question (5.5) which starts "Consider a comoving observer sitting at constant spatial coordinates ##\left(r_\ast\ ,\theta_\ast\ ,\ \phi_\ast\right)## around a Schwarzschild black hole of mass ##M##. The observer drops a beacon onto the black hole (straight down along a radial trajectory)....". That seems to mean the observer is not moving with respect to the black hole. So comoving means not moving with respect to something?

It gets worse, the second part of the question asks: "Calculate the proper speed of the beacon. That is, imagine there is a [second] comoving observer at fixed ##r##, with a locally inertial coordinate system set up as the beacon passes by, and calculate the speed as measured by the comoving observer." This other observer is in a locally inertial coordinate system with fixed ##r## so they might be in an orbit round the black hole. But if they are also comoving, that might mean they are also on the radial line ##\left(\theta_\ast\ ,\ \phi_\ast\right)##. Perhaps they were hovering there in their rocket and they turn engines off as the beacon comes by so they can now have inertial coordinates. I can calculate ##dr/d\tau## for the beacon easily enough in Schwarzschild coordinates but I am not sure if it what Carroll is after and what the other comover has to do with it.

Does comoving have an opposite like co-stationary, for example? But that would seem to mean the same! I hope you understand my confusion.
 

Answers and Replies

  • #2
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Summary:: What does comoving mean? With some of examples of usage. It seems to mean not moving!

I come across the adjective 'comoving' quite often. I understand comoving coordinates for the Universe. They are coordinates which expand with the expansion of the Universe(?) but I'm confused about what it means in essence.
It generally means 'having the same velocity' as something (a relation). As example, "A pair of comoving observers can sync their clocks in the frame in which they are both stationary".

The comoving coordinate system maps the entire universe. An object is said to be comoving if it is at rest with respect to any local worldline that traces a straight path to the big bang. Only relative to such comoving objects does the CMB appear isotropic. So yes, it is a relation to something, in your case, the black hole, with the observer maintaining some fixed coordinate position above the thing, which is best visualized by him sitting on a sturdy ring that surrounds the black hole at some distance, thus not requiring him to have orbital speed.

It gets worse, the second part of the question asks: "Calculate the proper speed of the beacon. That is, imagine there is a [second] comoving observer at fixed ##r##, with a locally inertial coordinate system set up as the beacon passes by, and calculate the speed as measured by the comoving observer." This other observer is in a locally inertial coordinate system with fixed ##r## so they might be in an orbit round the black hole.
Perhaps they were hovering there in their rocket and they turn engines off as the beacon comes by so they can now have inertial coordinates.
OK, you seem to realize that neither observer is in a locally intertial coordinate system. Yes, they maintain constant coordinate position relative to a very distant reference, but the gravity of the mass nearby gives them weight, and that weight shouldn't exist for an observer that is locally inertial (freefalling). This just might be a terminology nit. He seems to mean 'locally inertial' in the same way that my mailbox isn't going anywhere despite not being in freefall.
Briefly jumping off their vantage point (or turning off the engine) as the beacon passes by isn't going to change what they measure at that moment.

But if they are also comoving, that might mean they are also on the radial line ##\left(\theta_\ast\ ,\ \phi_\ast\right)##.
Yes, the beacon follows this line.

I can calculate ##dr/d\tau## for the beacon easily enough in Schwarzschild coordinates but I am not sure if it what Carroll is after and what the other comover has to do with it.
Point seems to be that the higher observer that dropped the beacon is going to measure a different (slower) speed of the beacon as it passes the lower observer than the lower observer would measure, and the exercise asks that it be computed. Suppose the beacon trails a low mass string behind it with meter-markings, which works to a point. Each observer can locally measure the marking going by at some rate, and they'll not measure the same value.

Does comoving have an opposite like co-stationary, for example? But that would seem to mean the same! I hope you understand my confusion.
No such term, and no single word of which I am aware that means 'not comoving'.
 
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  • #3
PeterDonis
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Here are some examples
IMO Carroll's use of the term "comoving" in this context is a bad choice of terminology.

The most common use of "comoving" in GR is to describe inertial observers, not observers who have nonzero proper acceleration, as the observers he is describing in the black hole scenario do. For example, comoving observers in an FRW spacetime (FRW spacetimes describe expanding universes such as ours) are inertial. A more usual term for an observer who is maintaining a constant altitude above a black hole (or any gravitating body) would be "hovering" or "stationary". If you try substituting either of those terms for "comoving" in Carroll's black hole scenario, I think it will make what is going on in the scenario clearer.
 
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PeterDonis
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This other observer is in a locally inertial coordinate system with fixed ##r##
There is no such thing. Here Carroll's terminology is fine, but you have to be careful about what it means.

This second observer is "hovering" or "stationary" (using the terms I suggested in my previous post instead of "comoving") at fixed ##r##, but no locally inertial coordinate system can be, because an object moving inertially will not remain at fixed ##r##. What Carroll is describing is this: imagine an inertially moving (i.e., freelly falling) object that happens to come to rest momentarily next to the hovering observer at the same instant that the beacon passes him (for example, a ball thrown upward in such a way that it reaches its maximum altitude next to the hovering observer at that instant). The locally inertial coordinate system in which that freely falling object is at rest is the one Carroll is describing. The hovering observer is also momentarily at rest in this coordinate system, at the instant the beacon passes him, but only momentarily.

The reason Carroll uses a locally inertial coordinate system, instead of just using coordinates like Schwarzschild coordinates in which the hovering observer remains at rest, is that he wants you to be able to use the laws of special relativity, as applied in an inertial frame, to calculate the answer to his question.
 
  • #5
George Keeling
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What Carroll is describing is this: imagine an inertially moving (i.e., freelly falling) object that happens to come to rest momentarily next to the hovering observer at the same instant that the beacon passes him (for example, a ball thrown upward in such a way that it reaches its maximum altitude next to the hovering observer at that instant).
Surely "ball thrown upward" achieves the same effect as the hovering observer switching engines off or jumping off the ring that goes around the black hole, where they were perching. Assuming that's correct let's call it the primed frame.

Now I have found a cheat sheet which says. "The spacetime interval between two events is the same in Schwarzschild coordinates and in a comoving observer's locally inertial reference [primed] frame:$$
ds^2=-d{\tau^\prime}^2+d{r^\prime}^2=-\left(1-\frac{R_s}{r}\right){dt}^2+\left(1-\frac{R_s}{r}\right)^{-1}{dr}^2
$$I agree with the initial statement but the first part of the equation is surely wrong. It should be $$
ds^2=-d{t^\prime}^2+d{r^\prime}^2=-\left(1-\frac{R_s}{r}\right){dt}^2+\left(1-\frac{R_s}{r}\right)^{-1}{dr}^2
$$The first part, ##ds^2=-d{t^\prime}^2+d{r^\prime}^2##, is now the nice simple Minkowski metric equation as desired. It is then possible to go on and calculate ##dr^\prime/d\tau## which is not as pretty as ##dr/d\tau## which I mentioned before.
 
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PeterDonis
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Surely "ball thrown upward" achieves the same effect as the hovering observer switching engines off or jumping off the ring that goes around the black hole
No, because the latter means the observer is no longer hovering. So you can only run the experiment once.

I have found a cheat sheet
Where? Please give a reference.

the first part of the equation is surely wrong
I can't tell without seeing how this "cheat sheet" defines ##\tau'##. If it's the proper time of the freely falling observer who is at rest in the locally inertial frame, then it's the same as what you are calling ##t'##, so your version is saying the same thing.
 
  • #9
George Keeling
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Ouch! I was confused by the metric equation for the comoving, inertial observer* which is$$
ds^2=-d{\tau^\prime}^2=-d{t^\prime}^2+d{r^\prime}^2
$$So for the stationary observer ##dr^\prime=0## so ##d\tau^\prime=dt^\prime##. The beacon flying past has ##dr^\prime\neq0## but one can still use ##d\tau^\prime=dt^\prime## for the observer's time measurements. Thanks!

* the observer is stationary w.r.t their inertial primed coordinate system and momentarily stationary w.r.t. the Schwarzschild coordinate system.
 

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