# A Calculate the Weyl and Ricci scalars for a given metric

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1. Apr 9, 2016

### suresh chand

hello dear,
I need to calculate the Weyl and Ricci scalars for a given metric. lets assume for a kerr metric. by using the grtensor-II package in maple i am not able to get the results. would anyone help me out. for ordinary metric like schwarzschild metric, However its not working for Kerr metric or metric having form like kerr metric. does anyone have any idea about this??

2. Apr 9, 2016

### vanhees71

Unfortunately I don't have maple (and also haven't ever used it). If it helps you, contact me, and I can send you a Mathematica notebook. I've just written a very simple code to evaluate all the stuff needed for a given metric (Christoffel symbols, Ricci Tensor, Einstein tensor, etc.).

3. Apr 9, 2016

### suresh chand

actually there is a problem with grtensor package i think, it does work well for some metric, however, like kerr metric it doesn't. yes it would be really helpful if u can send me that code. have u done for Weyl Scalars and Ricci Scalar for kerr?? actually i have calculated (Christoffel symbols, Ricci Tensor, Einstein tensor, etc.) for the same metric, however, i am having trouble for Weyl & Ricci scalars.
Thanks.
suresh

4. Apr 9, 2016

### vanhees71

I've never looked at the Kerr metric specifically. You find the Schwarzschild metric, where I used my simple code (just using the definitions of the objects by brute force Ricci calculus ;-)) here:

http://th.physik.uni-frankfurt.de/~hees/cosmo-SS15/

There shouldn't be a problem to just define the Kerr metric in the first few lines and then let run the notebook through Mathemica.

5. Apr 9, 2016

### Staff: Mentor

I know what the Ricci scalar is, but what is the Weyl scalar? I'm not aware of any such thing.

6. Apr 9, 2016

7. Apr 9, 2016

### Staff: Mentor

Ah, ok. This is just a repackaging of the Weyl tensor as five complex scalars, i.e., ten real scalars. I don't know which, if any, symbolic math packages handle the Newman-Penrose formalism, which is what would be needed to handle this.

8. Apr 9, 2016

### pervect

Staff Emeritus
You just do something like this:

grcalc(Ricciscalar);
grdisplay(_);
graleter(_);

select simplification method 1;

grdisplay(_);

you should get zero, it's a vacuum solution.

I see an option to calculate WeylSq, which may be the square of what you want, but I'm not positive, I'm not familiar with that. Anyay the procedure is the same as the above

grcalc(WeylSq);
grdisplay(_);

The result is too long to display until you simplify it

gralter(_);
grdisplay(_);

Then you get something like:

$$-48\,{\frac {{m}^{2} \left( \left( \cos \left( \theta \right) \right) ^{6}{a}^{6}-15\,{r}^{2}{a}^{4} \left( \cos \left( \theta \right) \right) ^{4}+15\,{r}^{4}{a}^{2} \left( \cos \left( \theta \right) \right) ^{2}-{r}^{6} \right) }{ \left( \cos \left( \theta \right) \right) ^{12}{a}^{12}+6\, \left( \cos \left( \theta \right) \right) ^{10}{r}^{2}{a}^{10}+15\,{a}^{8} \left( \cos \left( \theta \right) \right) ^{8}{r}^{4}+20\,{a}^{6} \left( \cos \left( \theta \right) \right) ^{6}{r}^{6}+15\,{a}^{4} \left( \cos \left( \theta \right) \right) ^{4}{r}^{8}+6\,{a}^{2} \left( \cos \left( \theta \right) \right) ^{2}{r}^{10}+{r}^{12}}}$$

which I have no idea if it's actually right or not, or whether it's the thing that you actually want to calculate.