1. Aug 21, 2014

What if Schwatzchild has made the same mistake as Newton?

( 1 - 2MG / c2 r ) dt

can be view as a binomial expansion to first order of

1 / ( 1 + 4MG / C2 r )1/2 dt

then would the event horizon be at the singularity?

I've been reviewing Susskind's online lecture series, and would like to explore the derivation of the Schwatzchild in more detail. Links to any relevant papers appreciated.

2. Aug 21, 2014

### atyy

Historically, Schwarzschild did indeed make a mistake about the event horizon, and was not able to distinguish the coordinate singularity seen in some coordinate systems as the event horizon is approached, from the true singularity. A good way to see that the event horizon is real is to use a coordinate system which smoothly covers the event horizon. One example are Kruskalâ€“Szekeres coordinates

http://preposterousuniverse.com/grnotes/grnotes-seven.pdf [Broken]
http://www.blau.itp.unibe.ch/newlecturesGR.pdf (Section 26.6)

Last edited by a moderator: May 6, 2017
3. Aug 21, 2014

### Staff: Mentor

The Schwarzchild spacetime is an exact solution to the Einstein field equations of general relativity. It's also (provably) the only solution to these equations for a spherically symmetric stationary mass distribution. Thus, the only way that it could need additional correction terms would be if the field equations themselves were wrong - which is certainly possible - but so far no one has found any theory that works better than general relativity.

4. Aug 21, 2014

Thanks

Both posts are a great help.

The spherically symmetric stationary mass distribution assumption seems quite reasonable, I'm happy with the explanations of that. In atty referrence 1, Sean Carrol sites the proof that the Schwartzchild solution is the only solution, as Birkhoff's Theorem. I'll study that in more detail as well as the constraints imposed by the field equations themselves.

5. Aug 21, 2014

### atyy

It's in Carroll's notes that "the Schwarzschild solution is the unique spherically symmetric solution to Einsteinâ€™s equations in vacuum", but I just wanted to highlight that the uniqueness for the Schwarzschild solution is under the assumption of a vacuum solution, so there is no matter in the Schwarzschild solution. Also, the uniqueness does not depend on the solution being stationary.

Because the Schwarzschild solution represents vacuum, it can be used outside a spherically symmetric star, being joined to something else that describes matter in the star. The solution with the event horizon is a fully vacuum solution, and if people want to be clear they say something like the "maximally extended vacuum Schwarzschild solution".

6. Aug 21, 2014

### Staff: Mentor

More precisely, the fact that the solution is stationary (in fact, static) is a *result* of the solution, not an assumption that needs to be made at the start in order to find the solution. (Even more precisely, the solution has a 4th Killing vector field in addition to the three associated with spherical symmetry; but the solution does not require that that KVF is timelike everywhere, so the term "stationary", strictly speaking, does not necessarily apply everywhere.)