# What is the physical meaning of the Schwarzschild radius for objects that aren't black holes?

Earth has a Schwarzschild radius of a little less than a centimeter. What does this mean for the matter of Earth's core that is within this radius?

A related question comes up for what happens when an almost black hole accretes matter and slowly becomes a black hole. Prior to the moment of the Schwarzschild radius crossing the boundary of the object, what does the matter within the radius experience?

(1) The Schwarzschild radius for an object that is not a black hole is the size down to which you would need to compress the object before it becomes a black hole.

(2) By the equivalence principle, absolutely nothing special happens locally at the event horizonâ€”there's no (local) way to even tell that you've crossed it. So nothing unusual happens as the boundary of the object crosses the Schwarzschild radius.

• When people say you can't tell if you've passed the event horizon, do they mean nothing will immediately happen? Or do they mean that you can't tell exactly where the horizon is? – Cole Johnson Aug 27 '16 at 3:09
• @ColeJohnson Both. – tparker Aug 27 '16 at 4:20

What does this mean for the matter of Earth's core that is within this radius?

Nothing, since the mass outside the radius does not contribute to the force (see Newton's shell theorem).

A related question comes up for what happens when an almost black hole accretes matter and slowly becomes a black hole.

In the system of the coordinate bookkeeper the velocity of the infalling matter not only slows down but converges to zero when it approaches the Schwarzschild radius because of the gravitational time dilation. So there is never enough mass inside the Schwarzschild radius to form a true horizon in a finite coordinate time. The combined mass of the initial body and the infalling material will therefore be larger than before, but so is the volume over which the total mass is spread out. Therefore in the system of the coordinate bookkeeper the radius that contains the mass will always be larger than the Schwarzschild radius of the mass.

Prior to the moment of the Schwarzschild radius crossing the boundary of the object, what does the matter within the radius experience?

The observer that crossed the horizon in a finite proper time took an infinite amount of coordinate time to even reach the horizon, so there is no longer a connection to the outside world. He will also experience spaghettification before he inevitably ends in the singularity.

The answers by tparker and ĐˇĐ¸ĐĽĐľĐ˝ Đ˘Ń‹Ń€Đ°Đ˝ work well enough. There is though I think a bit more. Suppose you put a black cloak around a gravitating body so you could not probe beneath it. The body's gravitation would be the same if it were a star of some mass, or the same mass collapsed into a black hole. So from that perspective if you were standing on Earth or this cloak at one Earth radius around an Earth-mass black hole there would be no difference with regards to gravity.

Now if we strip away the cloak we now see a difference. Both have gravitational fields that are Schwarzschild up to a certain radius. For the Earth that ends at its surface, but for the black hole it continues all the way into the black hole and up to the singularity. For a material body, the vacuum solution ends, and there is a continuity condition that has to be established between the vacuum solution and the non-vacuum solution in the material bulk of the body. In the interior one has to work with Ricci tensors for source terms and stress-energy tensors. In general this is the Birkhoff problem, which of course is applied to stars, white dwarfs and neutron stars. In general this is a computationally difficult problem to work with. It is not so hard if you have an idealized body with a constant density.

• Strictly speaking, that's only true if the gravitating body is static and spherically symmetric, right? – Brian Bi Oct 14 '16 at 23:49