# How to calculate the Riemann curvature at r=2GM?

1. Aug 12, 2018

### Bishal Banjara

As the coordinate singularity at r=2GM doesn't mean a physical singularity as Riemann curvature tensor is smooth although [g][/rr] metric behaves oddly in the case of Schwarzschild solution. Do somebody tell me an authentic reference how is this value is calculated? I have references writing different values directly without proper calculation. I found a document dealing with the problem (@gary Oas ) but it will result finally [R][/11]=0. See page 5 and insert all values thus derived in https://web.stanford.edu/~oas/SI/SRGR/notes/SchwarzschildSolution.pdf

2. Aug 12, 2018

### martinbn

$R_{11}$ is a component of the Ricci tensor, not the Riemann tensor. Which one are you trying to compute? If the coordinates are not good there, then you need to choose different coordinates.

3. Aug 12, 2018

### Bishal Banjara

Hello Martinbn, thank you for your consideration. Yes, in a definite sense it is Ricci Tensor. I particularly want to know how the values inserted for Riemann curvatures components at pp. 821 of Gravitation book (1st Edn) taking the reference to equations (14.50 and 14.51) are calculated. And I see the problem with Ricci Tensors Calculation in a way that the full insertion of all values in all such tensors individually finally result in zero, like R_00=R_11....=0. I think such individually zero is not true, am I wrong?

4. Aug 12, 2018

### e-pie

This one?

5. Aug 12, 2018

### Bishal Banjara

Hello e-pie, that's one.

6. Aug 12, 2018

### e-pie

Google Scholar has some papers on this. You will have to browse through them because I am not an expert in GR. I certainly can understand some of the math but not enough.

I will link them one by one.

But what book? Author?

7. Aug 12, 2018

### Ibix

Note that "Gravitation" is usually known as Misner, Thorne and Wheeler, or just MTW.

Page 821 notes that you substitute $e^{2\Phi}=e^{-2\Lambda}=1-2M/r$ into 14.51 (p 360/361). You'll need to calculate derivatives of $\Phi$ and $\Lambda$, although fortunately they have no time dependency. Then you can read off the Riemann components from 14.50. Note that this gives the Riemann in a mixed form, $R^{\mu\nu}{}_{\alpha\beta}$, whereas page 821 uses the completely lowered form, $R_{\mu\nu\alpha\beta}$, so you'll have to lower two indices to get the values given there. Note also that the book says that there are more non-zero components of the Riemann than those listed - you can use its symmetries to determine which ones.
Each component of the Ricci tensor should be zero in this case. How did you calculate it?

I'm guessing you missed one or other of the index raising/lowerings, or the components of the Riemann that MTW doesn't state explicitly, if you are having problems.

Last edited: Aug 12, 2018
8. Aug 12, 2018

### e-pie

Attributed to Louis Leestemaker of Amsterdam City College

See pages 18 to 20 of pdf file.

From http://www.uva.nl

I think the pdf is free-for-all. So I am uploading it.

Edit. I see @lbix has joined who I think is an expert on GR. You can refer to him.

<< Mentor Note -- PDF deleted; please follow the link to the text. Thanks >>

Last edited by a moderator: Aug 12, 2018
9. Aug 12, 2018

### Bishal Banjara

Hello e-pie, I prefer all.

10. Aug 12, 2018

### e-pie

Thanks. But no need for modesty, you should really refer to @lbix.

A study on Riemann-Christoffel tensor isn't complete without a physical meaning of the equations which I cleary lack and thus have created a huge hole in my physics knowledge. This is mostly because I did not have the time to go through physics at all for the last 6 to 7 years as I was working as a part time teacher for guiding maths olympiad enthusiasts(mostly teenagers between 14 to 16).

I will be free after 2 more years.

11. Aug 12, 2018

### Bishal Banjara

You mean like here in "http://web.mit.edu/klmitch/classes/8.033/Schwarzschild.pdf"? Here, also, I got a different value than what it is expected. May be I am wrong. I had inserted all values for Riemann curvature components at pp.9 with the derivative of Φ and Λ as $$\frac{M}{r(r-2M)}$$ and $$-\frac{M}{r(r-2M)}$$ for outside or vacuum solution respectively. I got $$\frac{2M}{r^3-2Mr^2}$$ for $${R^t}_{rtr}$$ and $$-\frac{M}{r}$$ for $${R^t}_{\theta t \theta}$$.
The Ricci Tensor components also result in zeros (in the document attached above).

Last edited: Aug 12, 2018
12. Aug 12, 2018

### e-pie

My link isn't sufficient. Can I give a direct link google?

13. Aug 12, 2018

### Bishal Banjara

Of course.

14. Aug 12, 2018

### Ibix

Presumably you are. It's a little difficult to know since you haven't said what values you got. You can use LaTeX to insert formulas in posts - see https://www.physicsforums.com/help/latexhelp/ for details of how to make this work, and quote my post to see how I did the index placement if you aren't familiar with LaTeX.
Yes, they should do so.

15. Aug 12, 2018

### e-pie

I asked the concerned mentor.

Last edited: Aug 12, 2018
16. Aug 12, 2018

### Bishal Banjara

I have made the corrections with my last post. Please review it once.

17. Aug 12, 2018

### e-pie

The rendered equations are still not visible. Use the preview button

Thanks.

18. Aug 12, 2018

### e-pie

$$\frac{M}{r(r-2M)}$$

19. Aug 12, 2018

### Bishal Banjara

??

20. Aug 12, 2018

### Bishal Banjara

I hope you could give me the http link by directly copying.