General Relativity discusses the gravity problem, and to see these relativistic effects the following ratio must be comparable to $1$ $$ GM/(R c^{2}), $$ where $G$ Newton's constant, $M$ the mass of the stellar body (sun, earth, black hole, ....), $R$ the corresponding radius and $c$ the speed of light. Therefore, the relativistic effects manifest when the stellar body is a denser medium (as black holes) (small radius and/or big mass).

I have some questions:

  • Where does this ratio come from? How is it derived? Why does this ratio tell us when relativistic effects matter and not some other number?
  • Black holes are a very dense medium. What is the order of magnitude of their radii?
  • 2
    $\begingroup$ How do you prove a ratio? $\endgroup$
    – bemjanim
    Apr 19, 2020 at 21:02
  • $\begingroup$ This was my question? $\endgroup$ Apr 19, 2020 at 21:03
  • 4
    $\begingroup$ Mercury is much farther away from the sun than the half Schwarzschildradius and still has a precession of the periapsis, so your question does not have an answer because you can not prove something that's wrong. You can only prove that with very large r the Einstein reduces to Newton $\endgroup$
    – Yukterez
    Apr 19, 2020 at 21:30
  • $\begingroup$ I've edited the question to ask about the origin of that ratio rather than "proving" it. This is how I interpreted it in my answer. The OP can roll back the edits, if they think the question has changed from their intent $\endgroup$
    – Paul T.
    Sep 28, 2021 at 20:44

2 Answers 2


The Schwarzschild metric describes the spacetime of a black hole. A common way to express a metric is through the corresponding line element, which tells us how to measure distance in the spacetime:

$$ds^2 = -(1+2\varphi)\,c^2dt^2 + (1+2\varphi)^{-1}dr^2 + r^2 d\Omega^2,$$

where $\varphi = -\frac{GM}{c^2\,r}$ is the ratio in the question. And $-\frac{GM}{r}$ is the Newtonian gravitational potential.

In the limit where $\varphi\rightarrow 0$ the Schwarzschild metric reduces to the Minkowski metric for flat spacetime (in spherical coordinates).

It is common to compute a series expansion of the Schwarzschild metric for small $\varphi$. This is a weak gravity limit, as the gravitational potential $\varphi$ acts like a small perturbation on flat, Minkowski space. In this limit the effects of the black hole's gravity reproduce the dynamics of Newtonian gravity to lowest order. Relativistic corrections to Newtonian gravity appear as terms of $\varphi^2$ or higher order.

If $\varphi\ll 1$, then the lowest order approximation, i.e. Newtonian gravity, is a good approximation to the dynamics. If $\varphi\sim 1$, then Newtonian gravity will not be a good fit to the dynamics.

Depending on how accurately you need to model the dynamics, relativistic corrections can be important for much smaller $\varphi$s than you might think! For the sun, $\frac{GM}{c^2} \approx 1.5$ km and Mercury's closest approach to the sun during its orbit is $r\approx 46\times 10^6$ km. Despite the very small $\varphi$ for Mercury's orbit, we measure its dynamics well enough to notice the difference between Newtonian gravity and GR.

To address the second point, the ratio isn't quite density. The density of an object is $\rho = M/r^3$, and the ratio is proportional to $M/r$. The Schwarzschild radius of a black hole is proportional to its mass $r_s = \frac{2GM}{c^2}$. If we take the size of a black hole to be $r_s$, then we can calculate the average density of everything inside the event horizon. For a $10$ solar mass black hole, similar in size to several LIGO has detected, $$\varphi = \frac{GM}{c^2\,r_s} = \frac{1}{2}, \quad\quad \rho = \frac{M}{{r_s}^3} \approx \frac{(10)(2\times 10^{30}\,\mathrm{kg})}{((10)(3\, \mathrm{km}))^3} = 7\times 10^{14} \, \mathrm{g/cm}^3. $$ The super massive black hole at the center of M87, that's in the Event Horizon Telescope's famous image, has a mass of about $6.5$ billion solar masses. For this black hole $$\varphi = \frac{GM}{c^2\,r_s} = \frac{1}{2}, \quad\quad \rho = \frac{M}{{r_s}^3} \approx \frac{(6.5\times 10^9)(2\times 10^{30}\,\mathrm{kg})}{((6.5\times 10^9)(3 \,\mathrm{km}))^3} = 0.0017 \, \mathrm{g/cm}^3. $$

Relativistic effects are equally important near the event horizon of both, but M87's black hole has an average density on the order of atmospheric air.

  • $\begingroup$ I highly agree with this answer. The relevance of the ratio is exactly that it appears in the Schwarzschild metric and linear approximations depend on the ratio being small. $\endgroup$
    – Alwin
    Sep 28, 2021 at 1:13
  • $\begingroup$ This is a good answer, but it is missing an important part, which is why the Schwarzschild $r$ coordinate coincides with that of the Newtonian $r$ coordinate. $\endgroup$ Sep 28, 2021 at 2:57
  • $\begingroup$ No black hole has the atmospheric density of air. $\endgroup$
    – Tivity
    Oct 31, 2023 at 17:11

The quantity you wrote is equal to ${r_s \over 2R}$, where $r_s$ is the Schwarzschild radius, the radius of the horizon of no escape: the escape velocity from it would have to be the upper bound, namely the speed of light, $c= \sqrt{2GM/r_s}$.

If R were the radius of your celestial body, it is a black hole, as it is wholly within its very own Schwarzschild radius. Is this what you are worrying about?


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