# Error in schwarzchild metric

1. Oct 18, 2004

### jimbo007

hi
recently i attended a lecture where a current researcher from my university was talking about black holes and the schwarzchild metric. basically he was saying no current theory predicts black holes and the schwarzchild solution is not actually correct, his solution was accepted because apparently he was a highly influential mathematician.
he said that the r term in the schwarzchild metric should actually be
r->(r^3+a^3)^(1/3) where a = 2GM/c^2
as you see the a term is quite small (G=gravitation constant and c is the speed of light).
my understanding of the majority of his talk was a bit shakey but he didnt seem like a nut as there were a few other professors attending the lecture also and they couldnt point out any flaws in his argument.
he reckons that no one else has pointed this mistake out yet and has submitted his report to be published in a journal.

2. Oct 18, 2004

### franznietzsche

1) Which journal?

2) Assuming you're presenting that correctly then the r term reduces to r for flat space time, which is wrong. It should reduce to 1, the value for the minkowski metric. Just to check you're saying that according to him it should be:
$$\left( \begin{array}{cccc}-(1-\frac{2M}{r}) & 0 & 0 & 0 \\ 0 & r^2 & 0 & 0 \\ 0 & 0 & r^2sin^2(\theta) & 0 \\ 0 & 0 & 0 & (r^3 + (\frac{2GM}{c^2})^3)^(1/3) \end{array} \right)$$

Which means that for space far removed from any source of mass, i.e., falt space time we get:
$$\left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & r^2 & 0 & 0 \\ 0 & 0 & r^2sin^2(\theta) & 0 \\ 0 & 0 & 0 & r \end{array}\right)$$

Which is not the Minkowski metric, however we know, from all the experimental verification of SR that the minkowksi metric is valid, so he cannot be right. There are problems with relativity, but not those kinds. The very fact that this guy claimed the Scwarzschild's solution was only accepted because of his personal influence is very big crackpot give away.

Last edited: Oct 18, 2004
3. Oct 18, 2004

### da_willem

I think he meant:

$$\left( \begin{array}{cccc}1-( \frac{2MG}{c^2 (r^3 + (\frac{2GM}{c^2})^3)^{1/3}}) & 0 & 0 & 0 \\ 0 & -(1-\frac{2MG}{c^2 (r^3 + (\frac{2GM}{c^2})^3)^{1/3}})^{-1} & 0 & 0 \\ 0 & 0 & -(r^3 + (\frac{2GM}{c^2})^3)^{2/3} & 0 \\ 0 & 0 & 0 & -(r^3 + (\frac{2GM}{c^2})^3)^{2/3} sin( \theta )^2 \end{array} \right)$$

So everywhere in the Schwarzschild metric replace r with the new r.

This does yield the flat metric for large r. It also means a correction to Newtons law yielding a potential V=MG/(r^3+2GM/c^2)^(1/3). The metric is still spherically symmetric. I can't find any reason why this cannot be correct.

It is correct that this solution has no r for which a term diverges (only r=0) so it would does not automatically imples a black hole.

I would like to see why the original solution cannot be correct, or why this one is better, so I'm also interested in the article...But I'm not sure this resultobey's Einsteins field equations. Anybody an idea?

Last edited: Oct 18, 2004
4. Oct 18, 2004

### kenhcm

I have just checked it, it does not satisfy the Einstein's field equations: $$R_{\mu\nu}=0$$.

Best regards,
Kenneth

5. Oct 18, 2004

### da_willem

I guess the problem is solved then....