- #1

George Keeling

Gold Member

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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.

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.