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

• Bishal Banjara
In summary: 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
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

##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.

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?

This one?

#### Attachments

• IMG_20180812_213632.jpg
5.8 KB · Views: 380
Hello e-pie, that's one.

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?

Bishal Banjara said:
pp. 821 of Gravitation book (1st Edn) taking the reference to equations (14.50 and 14.51) are calculated

Bishal Banjara said:
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.
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.
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?
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:
Bishal Banjara
Attributed to Louis Leestemaker of Amsterdam City College

See pages 18 to 20 of pdf file.

From http://www.uva.nl

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

Last edited by a moderator:
Hello e-pie, I prefer all.

Bishal Banjara said:
Hello e-pie, I prefer all.

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.

Ibix said:
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.
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:
e-pie said:

e-pie said:
Of course.

Bishal Banjara said:
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.
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.
Bishal Banjara said:
The Ricci Tensor components also result in zeros (in the document attached above).
Yes, they should do so.

Bishal Banjara
Bishal Banjara said:
Of course.

Last edited:
Ibix said:
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.
I have made the corrections with my last post. Please review it once.

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

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

e-pie said:
$$\frac{M}{r(r-2M)}$$
??

Bishal Banjara said:
??
I hope you could give me the http link by directly copying.

You need two $signs or two # signs to delimit LaTeX, not one. Bishal Banjara Bishal Banjara said: 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). This is your #11, with LaTeX delimiters corrected. Bishal Banjara$\frac{M}{r(r-2M)}$Add one$ each start and end. Use the preview button.

Bishal Banjara said:
I hope you could give me the http link by directly copying

You mean the file I uploaded. I cannot unless a mod gives me permission. Maybe it was a copyright violation. So you have to wait until I clarify it via PM.

Ibix said:
This is your #11, with LaTeX delimiters corrected.
Yes I have corrected and have some questions for you. Please revisit the post again.

Bishal Banjara said:
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).
I agree about ##\Phi'## and ##\Lambda'## and the two components you've calculated, but there are twelve non-zero components of the Riemann in total (though only four six independent ones, I think). So you have a bit more work to do...

Last edited:
Ibix said:
I agree about ##\Phi'## and ##\Lambda'## and the two components you've calculated, but there are twelve non-zero components of the Riemann in total (though only four independent ones, I think). So you have a bit more work to do...
But there are only six components listed both in the above document and in the MTW of which I have calculated only two those are different than what they should be. Am I wrong? And I am in search of getting the same values listed in MTW.

Ibix said:
twelve non-zero components of the Riemann

@Ibix Aren't there 36 in 4d Space. This condition shows that the all the components whose 1st and 2nd or 3rd 4th indices are equal must be zero. Therefore square of 4C2=36.

#### Attachments

• IMG_20180812_231914.jpg
11.5 KB · Views: 597
Bishal Banjara said:
But there are only six components listed both in the above document and in the MTW of which I have calculated only two those are different than what they should be. Am I wrong?
MTW says on page 821 that there are other non-zero components obtainable by symmetries of the Riemann. I'm not sure about the all-lower-index Riemann that they are using, but there are twelve non-zero components in the (1,3) form you seem to be using (although on a re-read I'm not certain you have the lower indices in the order I expected - I think what you've called ##R^t{}_{rtr}## is usually ##R^t{}_{rrt}##, although it's possible I've got lost in the different notations used by different sources).

Bishal Banjara
If arranged in a 6 x 6 matrix, there are still 21 independent components.

e-pie said:
@Ibix Aren't there 36 in 4d Space. This condition shows that the all the components whose 1st and 2nd or 3rd 4th indices are equal must be zero. Therefore square of 4C2=36.View attachment 229214
Most of them are zero in the Schwarzschild metric. It's only the remaining ones we care about. And there are twenty independent components in general: http://mathworld.wolfram.com/RiemannTensor.html

Ibix said:
And there are twenty independent components in general:

Bianchi Identity?

4C4=1 21-1=20 ?

Ibix said:
Most of them are zero in the Schwarzschild metric. It's only the remaining ones we care about. And there are twenty independent components in general: http://mathworld.wolfram.com/RiemannTensor.html
To put this in a slightly different way: The spherical symmetry of the Schwarzschild spacetime means additional constraints on the components of the Riemann tensor. For the general N-dimensional case (without symmetries in the spacetime) is ##N^2(N^2-1)/12##, which follows from first checking the number of components allowed by the symmetries of the tensor and then removing some due to the Bianchi identity.

Bishal Banjara and Ibix
Ibix said:
MTW says on page 821 that there are other non-zero components obtainable by symmetries of the Riemann. I'm not sure about the all-lower-index Riemann that they are using, but there are twelve non-zero components in the (1,3) form you seem to be using (although on a re-read I'm not certain you have the lower indices in the order I expected - I think what you've called ##R^t{}_{rtr}## is usually ##R^t{}_{rrt}##, although it's possible I've got lost in the different notations used by different sources).
Yeah I got the point and answer here at https://physics.stackexchange.com/q...he-riemann-tensor-of-the-schwarzschild-metric

Orodruin said:
To put this in a slightly different way: The spherical symmetry of the Schwarzschild spacetime means additional constraints on the components of the Riemann tensor. For the general N-dimensional case (without symmetries in the spacetime) is N2(N2−1)/12N2(N2−1)/12N^2(N^2-1)/12, which follows from first checking the number of components allowed by the symmetries of the tensor and then removing some due to the Bianchi identity.
That means, ... Detail not shown after using Bianchi and swapping

Rnmlj + Rnljm + Rnjml= - Rnmjl - Rnlmj - Rnjlm = 0

Therfore 20 components and further restriction give 12 independent.

• Special and General Relativity
Replies
32
Views
3K
• Special and General Relativity
Replies
1
Views
1K
• Special and General Relativity
Replies
42
Views
2K
• Special and General Relativity
Replies
14
Views
5K
• Special and General Relativity
Replies
2
Views
2K
• Special and General Relativity
Replies
11
Views
2K
• Special and General Relativity
Replies
9
Views
1K
• Special and General Relativity
Replies
7
Views
1K
• Special and General Relativity
Replies
7
Views
2K
• Differential Geometry
Replies
2
Views
2K