In 1939, Albert Einstein published a paper entitled "On a Stationary System with Spherical Symmetry Consisting of Many Gravitating Masses." In it, he considers the problem of whether it is possible to reach a spacetime metric containing singularities in a real physical system – that is, starting from actual gravitating masses.

Consider a system of many small objects, with equal masses, which move under the influence of the gravitational field of the entire system. Furthermore suppose that the particles move in concentric, randomly oriented circular orbits, so that the overall gravitational field is approximately static and spherically symmetric. Then the spacetime metric is

$$ds^2 = -a(r)(dx^2 + dy^2 + dz^2) + b(r)\, dt^2 ~$$

where $a(r)$ and $b(r)$ are functions of the radius $r^2 = x^2 + y^2 + z^2$. By substituting this into the Einstein field equation

$$G_{\mu\nu} = R_{\mu\nu} - \frac12 g_{\mu\nu}R + \kappa T_{\mu\nu} = 0$$

one can obtain differential equations for $a(r)$ and $b(r)$:

$$\alpha'' + \frac{\alpha'}{r} + \frac14 \alpha'^2 - \frac1{r^2} + \kappa mn a^{-1/2} \left( \frac{\alpha'^2}{\frac32 \alpha'^2 - \frac2{r^2}} \right)^{1/2} = 0$$

$$\beta' = \frac1{\alpha'} \left( \frac2{r^2} - \frac12 \alpha'^2 \right)$$

where $\alpha = \ln (r^2 a)$; $\beta = \ln b$; $m$ is the mass of each particle; and $n$ is the particle density.

At first glance, these equations do not look very tractable. However, the idealized limiting case where the gravitating particles are concentrated within an infinitesimally thin spherical shell of radius $r_0 \pm \Delta$ is relatively simple. Einstein solves this case and shows that:

$$r_0 > \frac\mu2 (2 + \sqrt3)$$

where $\mu/2 = \frac12 \left( \frac{\kappa}{8\pi} mN \right)$ is the Schwarzschild radius, with $mN$ being the total mass of the system.

Since this lower bound is above the Schwarzschild radius, Einstein reasons, a system of masses in circular orbits cannot form a black hole. He also generalizes this result to the case of continuous particle density and obtains a similar lower bound. Physically, as $r_0$ decreases towards the bound (that is, as the system of masses becomes smaller and smaller), the kinetic energy of the system goes to infinity. Intuitively, one would expect that the same reasoning should apply to other systems as well, though the fully general case is not addressed rigorously in the paper.

Einstein concludes:

The essential result of this investigation is a clear understanding as to why the "Schwarzschild singularities" do not exist in physical reality. Although the theory given here treats only clusters whose particles move along circular paths, it does not seem to be subject to reasonable doubt that more general cases will have analogous results. The "Schwarzschild singularity" does not appear for the reason that matter cannot be concentrated arbitrarily. And this is due to the fact that otherwise the constituting particles would reach the velocity of light.

Today, we know that black holes exist; hence, the argument above must be wrong. But where was Einstein's mistake? Was it the assumption of stable circular orbits?

I saw this question: Einstein and the existence of Black Holes However, the question does not discuss the actual argument made by Einstein himself, and the currently accepted answer simply states that Einstein's arguments are no more than heuristics and intuition.

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    $\begingroup$ @safesphere Strong statements require strong proof. Please provide a reference. $\endgroup$
    – my2cts
    Feb 28, 2021 at 11:53
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    $\begingroup$ @safesphere I was looking for a reference more specific than "wiki". Wiki: "The first modern solution of general relativity that would characterize a black hole was found by Karl Schwarzschild in 1916". There is no statement that he was wrong. $\endgroup$
    – my2cts
    Feb 28, 2021 at 15:46
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    $\begingroup$ @safesphere And what would the thin spherical shell observe? We know that free falling particles reach the singularity in finite proper time. $\endgroup$
    – Javier
    Feb 28, 2021 at 19:14
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    $\begingroup$ @safesphere The question is valid whether or not one believes that strict black holes exist in nature. Your comment says "You would observe that a thin spherical shell collapses to within the Planck length of its Schwarzschild radius". Einstein's conclusion asserts that a thin shell of matter can't get that close to the Schwarzschild radius, so your own comment agrees that Einstein's argument must be wrong. The question is, where exactly is the mistake? $\endgroup$ Feb 28, 2021 at 19:22
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    $\begingroup$ @ExpertNonexpert At least now people are not burned alive for contradicting the establishment :) $\endgroup$
    – safesphere
    Mar 1, 2021 at 5:15

2 Answers 2


Einstein's calculation is correct, but his conclusion about its significance is wrong.

The calculation shows that circular orbits cannot exist below a minimum radius, and it shows that this minimum radius is greater than the Schwarzschild radius. This is well-known today,$^*$ and it has a name: innermost circular orbit (ICO).

By the way, Einstein did not assume that the orbits had to be stable. The innermost stable circular orbit (ISCO) is at an even larger radius.

In any case, I don't know why Einstein thought the existence of an innermost circular orbit (stable or not) should imply that matter can't collapse to smaller radii. Why didn't he consider radially infalling matter? Or matter spiraling inward? Radial and spiralling timelike geodesics are not limited by the ICO, and I didn't find any clear clues in Einstein's paper about why he chose not to consider them. He even referred to the circular-orbit assumption as a "special assumption" in the introduction on page 923.

$^*$ This well-known result is usually expressed in a different coordinate system, one that might be more familiar. In the more familiar coordinate system, the Schwarzschild radius is $r=2m$, and the ICO is $r=3m$ (see section 2.3.1 in https://arxiv.org/abs/1410.4481). In Einstein's coordinate system, the Schwarzschild radius is $r_E=m/2$, and the ICO is $r_E=(m/2)(2+\sqrt{3})$. The two coordinate systems are related to each other by $r_E(1+m/2r_E)^2=r$, as shown in another question.

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    $\begingroup$ Based on his general attitude towards physics, I would speculate that he felt that he already knew what nature did and was looking for calculations to justify his intuition. It is always tempting to stop when you are convinced that you have found the correct answer. $\endgroup$
    – Rococo
    Mar 1, 2021 at 0:31
  • $\begingroup$ In Schwarzschild coordinates, the radius $r = 3m$ is associated to the photon sphere. Do you say this is the same as ICO ? The innermost stable circular orbit (actually limit of stability) is located at $r = 6m$ (three times the Schwarzschild radius). $\endgroup$
    – Cham
    May 24, 2022 at 15:19

I have not read the paper but from the quotations you draw from it, I would say the lesson is that the physics associated with a black hole horizon is very counter-intuitive.

Einstein treats circular orbits and I think his mathematical reasoning is correct. He goes on to say that similar results will apply to other orbits. Let's look now at the quotation:

"The "Schwarzschild singularity" does not appear for the reason that matter cannot be concentrated arbitrarily. And this is due to the fact that otherwise the constituting particles would reach the velocity of light."

I am not sure whether he is referring here to a coordinate singularity which appears at the horizon if one adopts coordinates which are singular there, or the curvature singularity. If he is referring to the curvature singularity then the first sentence is, for all we know, correct. I mean that in the limit where a singularity would be present according to classical GR, very likely classical GR is no longer valid and so a true mathematical singularity does not appear. This is not known. The second sentence, however, is, I think, a reference to the fact that the relative velocity between an infalling particle and the worldline of a particle maintaining a fixed Schwarzschild coordinate just outside the horizon tends to $c$ as the location of the latter tends to the location of the horizon. However it does not follow from this that the infalling particle cannot fall in. What follows is that no force will be strong enough to keep a particle at a fixed Schwarzschild coordinate $r$ as $r \rightarrow r_S$.

Finally, on the question whether it takes an infinite time for a black hole to form. This is an ill-posed question because there is no such thing as a single temporal duration between two events in spacetime. There are a number of measures of temporal duration between two given events. Among them, for example, there is the proper time along some given worldline, and there is also the coordinate time according to some given system of coordinates. The proper time for particles to gather sufficiently for a horizon to form is finite. The Schwarzschild coordinate time for this to happen is infinite. But the spacetime itself is continuous across the horizon so it is reasonable to assert that the part of spacetime over the horizon is a bona-fide part. The counter-intuitive feature is that one may, if one chooses, assert that events in the part of spacetime at the horizon are in the infinite future of events at our instruments. One does not have to say that, but one may: there exists a perfectly good assignment of coordinates in space and time which leads to that statement (i.e. the Schwarzschild coordinates). But it is also valid to adopt a reference frame falling through a horizon and consider timing relative to the proper time along worldlines of particles fixed in such a frame. In that frame the horizon is reached in a finite time. You can pick whichever clock you prefer.

Predictions of GR for what is observed in instruments such as telescopes on planet Earth do not depend on what frame is adopted for measuring time near a horizon. What GR predicts is the worldlines of the signals travelling from near a horizon out to other events. In particular, it is predicted that the signals from objects falling to the horizon will vanish in a finite time, in the sense that the signal strength diminishes exponentially with time on the local clock on planet Earth. You can argue afterwards about whether or not you think the horizon has formed yet. No observation will depend on what opinion you take. You would just be arguing about which spacelike surface to call "simultaneous" and the universe does not care.


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