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I'm sorry if this sounds like a silly question. I think I am jsut missing some key insight. But take the Schwarzschild solution, which describes the spacetime around a static spherically symmetric mass (but I also want ask about any other kind of spacetime). The metric is $$ ds^2 = -(1-\frac{r_s}{r})dt^2 + \frac{r^2}{1-r_s/r} + r^2d\Omega^2.$$ My question is, whose coordinates are these $(t,r,\theta,\phi)$? Are these the coordinates associated with an 'observer' at the center of the gravitating mass? But is that problematic at all because of the singularity at $r=0$?

Maybe this can further explain my question. Suppose someone is on a circular orbit. Their proper time elapsed between each orbit is shorter than the coordinate time $t$. In special relativity, it's been said that time dilation and length contraction for a moving observer in our frame aren't literally happening (because in their frame, it is us whose is contracted) but rather we just see that a clock that they're carrying is slow compared to our clock, or a ruler they're carrying is shorter compared to our ruler. So, in this case, who is the observer using the clock with coordinate time $t$ who sees an orbiting person's clock run slower compared to theirs? Because surely if these coordinates are the coordinates of someone orbiting, they should see that the elapsed coordinate time $t$ is their proper time, right?

Then, what if we had a complicated mass distribution and we solved for the metric in some coordinate system? How would we know which or what kind of observer those coordinates belong to?

EDIT: I was asked to clarify my question. I'm not sure if this helps, but another way I thought of this is as follows. Let's say everyone around a star is using the Schwarzschild coordinates, and a person in a circular orbit at fixed radius $R$ starts their orbit at $t_1$ and closes the orbit at $t_2$. Since everyone is using these coordinates, would they all say the time elapsed for an orbit at that radius is $t_2-t_1$? And what if there were another person orbiting at a greater radius $R'$ but moving faster, such that they both start at $t_1$ and end at $t_2$. They should measure different proper times, but why wouldn't everyone agree that the time elapsed is $t_2-t_1$ for both of the orbiters since everyone is using Schwarzschild coordinates? Where does proper time (which changes depending on the radius) come into this?

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  • $\begingroup$ en.wikipedia.org/wiki/Schwarzschild_coordinates $\endgroup$ – G. Smith Nov 3 '20 at 4:17
  • $\begingroup$ Please see my answer to a related question: physics.stackexchange.com/a/552874/123208 $\endgroup$ – PM 2Ring Nov 3 '20 at 7:42
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    $\begingroup$ You may be confusing a coordinate system with a frame of reference. A frame of reference is a coordinate system attached to a physical observer. So every frame has a coordinate system, but not every coordinate system is a frame. For example, the cosmological coordinate system of the expanding universe cannot be attached to any observer and thus is not a frame of reference. Perhaps you should clarify what exactly you are asking. $\endgroup$ – safesphere Nov 3 '20 at 18:20
  • $\begingroup$ Thanks for your comment! That helps a little. I added an edit that I hope might explain my question better. $\endgroup$ – UrsaCalli79 Nov 3 '20 at 22:20
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    $\begingroup$ "everyone around a star is using the Schwarzschild coordinates, and a person in a circular orbit at fixed radius R starts their orbit at $t_1$ and closes the orbit at $t_2$" - The time he measures on his clock is his proper time, not the Schwarzschild time. $\endgroup$ – safesphere Nov 4 '20 at 20:18
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I think the insight that you're missing is that coordinates are meaningless. You can calculate any physically meaningful quantity using any coordinate system and you'll get the same result, because they all describe the same world. As such, there's no such thing as the coordinate system "of" any object or person.

In special relativity there's a convention of identifying inertially moving objects with inertial reference frames that cover all of Minkowski space, as though every person walking along the street had an infinite network of metersticks and synchronized clocks attached to them. That coordinate system is called "their" reference frame, and many people even seem to have the idea that you must use that frame when solving problems involving that person.

You can't do that in general relativity. There is no well defined global rest frame associated with any object. You're therefore forced in GR to do what you should also do in SR: just pick whatever coordinates are most convenient for the problem you're solving. In the case of the Schwarzschild geometry, the Schwarzschild coordinates are usually the most convenient, unless you care about the event horizon and black hole interior.

who is the observer using the clock with coordinate time $t$ who sees an orbiting person's clock run slower compared to theirs?

It can be whoever you want. If you establish a listening post at a fixed $(r,θ,φ)$, with $r\gg 2m$, then its $dt/ds$ will be constant and close to $1$. But there doesn't need to be someone out there to give meaning to $t$. The meaning of $t$ is completely fixed by the metric.

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  • $\begingroup$ Thanks for your answer! This makes more sense now. I guess a follow up would be, let's say everyone around this star is using the Schwarzschild coordinates, and a person in a circular orbit starts at $t_1$ and ends the orbit at $t_2$. Since everyone is using these coordinates, would they all say the time elapsed for an orbit at that radius is $t_2-t_1$? Then where does proper time (which changes depending on the radius) come into this? $\endgroup$ – UrsaCalli79 Nov 3 '20 at 22:15
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For example: if the observer is monitoring a falling object on the planet's surface, $t$ is what his clock measures, $r$ is $R + h$ according to some measuring device. He has previously chosen a north pole, and a "Greenwich meridian", to record the the evolution of $\theta$ and $\phi$, and must have calculated the radius $R$ of the planet.

Using that values, the metric delivers the corresponding $d\tau$, the time elementary increment of a clock at the object.

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