I Inverting the metric coefficients in the Schwarzschild line element

[PeterDonis]

"I want to know whether I am doing wrong though metric is independent to the final result or it doesn't reconcile, naturally", what should we be concluded?

I'm not sure what you're asking.

If you're asking whether what you did in the OP of this thread is correct, I have already said that it's wrong, and explained why. See my posts #10 and #11.

If you are asking whether you can obtain the inverse metric $g^{\alpha \beta}$ by raising both indexes on the metric $g_{\alpha \beta}$, it should be obvious that you can't, since in order to raise indexes you need to already know the inverse metric $g^{\alpha \beta}$. You obtain the inverse metric by considering the metric as a matrix and obtaining its matrix inverse.

 

[Bishal Banjara]

I'm not sure what you're asking.

If you're asking whether what you did in the OP of this thread is correct, I have already said that it's wrong, and explained why. See my posts #10 and #11.

If you are asking whether you can obtain the inverse metric $g^{\alpha \beta}$ by raising both indexes on the metric $g_{\alpha \beta}$, it should be obvious that you can't, since in order to raise indexes you need to already know the inverse metric $g^{\alpha \beta}$. You obtain the inverse metric by considering the metric as a matrix and obtaining its matrix inverse.

The only way I could make my question very simple, be like, what if inverting the metric coefficients $g_{oo}$ and $g_{rr}$ of the usual Schwarzschild solution for the final result calculation of Einstein's tensor components $G_{oo}$ and $G_{rr}$? Does this final result after inverting the metrics coincide to the initial result of the original Schwarzschild solution?

 

[PeterDonis]

inverting the metric coefficients $g_{oo}$ and $g_{rr}$

Is a meaningless, wrong thing to do. It makes no sense.

Does this final result after inverting the metrics coincide to the initial result of the original Schwarzschild solution?

No. It is just nonsense. See above.