I've always wondered, in which frame of reference does Newton's law $$ \boldsymbol{g} = -\frac{GM}{r^2} \widehat{\boldsymbol{r}} $$ actually apply? In general it can't be the one in which the the mass $M$ is not rotating, since otherwise geostationary satellites would fall to earth. So, how is the intended frame of reference rotating with respect to the mass $M$? I'm looking for an answer that could be used to make predictions, so simply working backwards from a measured orbital period and Newton's law won't do.

I'm sure that the Einstein Field Equation $$ R_{\mu\nu} - \frac{1}{2} R g_{\mu\nu} = \kappa T_{\mu\nu} $$ is key, since the stress energy tensor $T_{\mu\nu}$ changes if the body starts rotating. (Although this is only the case inside the body; outside it is still zero.) In the book A Most Incomprehesible Thing the author derives the Schwarzchild Metric and states that the assumptions include

  1. at $r = 0$ there is a non-rotating mass
  2. the region of space for which we are solving is empty, so $T_{\mu\nu} = 0$
  3. the metric is spherically symmetric
  4. the metric is static, i.e. $\partial_{t}g_{\mu\nu} = 0$

I sort of followed the derivation, but my problem is that it doesn't appear to use the non-rotation assumption in (1) . What he does is to use (2) to show that the Ricci Tensor is zero, turning the EFE into $R_{\mu\nu} = 0$ . Then he uses (3) and (4) to simplify that equation. This results in

$$ g_{tt} = 1 - \frac{R_s}{r}\\ g_{rr} = -\frac{1}{1-\frac{R_s}{r}}\\ g_{\theta\theta} = -r^2\\ g_{\phi\phi} = -r^2sin^2\theta\\ $$

Finally he uses Newton's law of gravitation show that $R_s = \frac{2GM}{c^2}$. However, the fact that the body at $r = 0$ was not rotating was never used. And it seems like cheating to use Newton's law to solve the EFE when the goal was to demonstrate compatibility between the EFE and Newton's law.

Hopefully my question isn't as confused as I am!

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    $\begingroup$ Hint: How did you deduce that $g_{rt}=0$? $\endgroup$ – Qmechanic Mar 4 '18 at 17:34
  • $\begingroup$ GR only has local frames of reference, not global ones, so you can't use GR to answer this question. Newtonian mechanics says you can use any intertial frame. $\endgroup$ – user4552 Mar 4 '18 at 18:16
  • $\begingroup$ @Qmechanic the book glosses over that by implicitly assuming that in the solution $g_{\mu\nu}$ has diagonal entries only. I presume that $g_{rt} = 0$ because $r$ is defined s.t. for light emanating from the centre $dr^2 = dt^2$ (this leaves open how $t$ is defined...). I guess $dr^2 = dt^2$ isn't true for light emanating from earth in the frame in which earth is not rotating. Um, am I on the right track...? $\endgroup$ – Alex Zeffertt Mar 4 '18 at 20:04
  • $\begingroup$ @BenCrowell would the question be more correct if it said "coordinate system" instead of "frame of reference"? $\endgroup$ – Alex Zeffertt Mar 4 '18 at 20:07
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    $\begingroup$ The newton’s law of gravitation is valid in all initial frames of reference (not taking into account relativistic effects) $\endgroup$ – Aniansh Mar 4 '18 at 20:11

I've been pondering Qmechanic's hint for some time and I now think I have the answer. (But please provide a better answer if you think I've got this wrong.)

In the derivation of the Schwarzchild metric in A Most Incomprehesible Thing it was assumed - not deduced - that the off diagonal components in the solution were zero. It was also assumed that $g_{tt}$ and $g_{rr}$ were functions of $r$ alone. These assumptions imply that the mass is not rotating with respect to falling bodies, as I will try to demonstrate below:

Newton's 1st law of motion "in the absence of external forces a body will continue to move in a straight line" can be expressed in an arbitrary coordinate system with the following tensor equation $$ \frac{d^2x^{\alpha}}{d\lambda^2} = -\Gamma^{\alpha}_{\beta\gamma}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda} $$

Now, suppose you have a coordinate system $t, r, \theta, \phi$ and a body is falling such that $\frac{d\theta}{d\lambda} = \frac{d\phi}{d\lambda} = 0$, i.e. $t$ and $r$ are changing with $\lambda$ but not $\theta$ or $\phi$. We can ask the question: will it start turning? This is equivalent to asking whether $\frac{d\theta^2}{d\lambda^2}$ and $\frac{d\phi^2}{d\lambda^2}$ are non-zero. Taking the latter as an example, this is equivalent to asking whether $$ \Gamma^{\phi}_{\beta\gamma}\frac{dx^{\beta}}{d\lambda}\frac{dx^{\gamma}}{d\lambda} $$

is non-zero. But since we have already asserted that $\theta$ and $\phi$ are not currently changing with $\lambda$ this implies that one of

$$ \Gamma^{\phi}_{rr}, \Gamma^{\phi}_{tt}, \Gamma^{\phi}_{rt} $$

is non-zero. However, these are all zero because the off-diagonal components of $g$ are zero and because $g_{tt}$ and $g_{rr}$ are functions of $r$ alone. (You need the definition of the Christoffel symbol to hand in order to see this.)

Thus, the initial assumptions about the solution $g$ impose a restriction on the coordinate system discussed, namely that in it, any body that is falling with $\theta$ and $\phi$ constant continues to fall with $\theta$ and $\phi$ constant.

Thanks Qmechanic, you are a great teacher!


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