- #1

Bishal Banjara

- 42

- 2

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Bishal Banjara
- Start date

- #1

Bishal Banjara

- 42

- 2

- #2

martinbn

Science Advisor

- 2,884

- 1,210

- #3

Bishal Banjara

- 42

- 2

- #4

- #5

Bishal Banjara

- 42

- 2

Hello e-pie, that's one.

- #6

e-pie

- 129

- 18

I will link them one by one.

But what book? Author?

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

- #7

Ibix

Science Advisor

- 9,526

- 9,604

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

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?And I see the problem with Ricci Tensors Calculation in a way that the full insertion of all values in all such tensorsindividuallyfinally result inzero, like R_00=R_11....=0. I think such individually zero is not true, am I wrong?

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:

- #8

e-pie

- 129

- 18

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 >>**

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.

Last edited by a moderator:

- #9

Bishal Banjara

- 42

- 2

Hello e-pie, I prefer all.

- #10

e-pie

- 129

- 18

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.

- #11

Bishal Banjara

- 42

- 2

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 ofNote 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.

The Ricci Tensor components also result in zeros (in the document attached above).

Last edited:

- #12

e-pie

- 129

- 18

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

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

- #13

Bishal Banjara

- 42

- 2

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

- #14

Ibix

Science Advisor

- 9,526

- 9,604

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.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.

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

- #15

e-pie

- 129

- 18

Of course.

I asked the concerned mentor.

Last edited:

- #16

Bishal Banjara

- 42

- 2

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

- #17

e-pie

- 129

- 18

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

Thanks.

Thanks.

- #18

e-pie

- 129

- 18

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

- #19

Bishal Banjara

- 42

- 2

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

- #20

Bishal Banjara

- 42

- 2

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

- #21

Ibix

Science Advisor

- 9,526

- 9,604

You need two $ signs or two # signs to delimit LaTeX, not one.

- #22

Ibix

Science Advisor

- 9,526

- 9,604

This is your #11, with LaTeX delimiters corrected.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 ofΦandΛas $$\frac{M}{r(r-2M)}$$ 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).

- #23

e-pie

- 129

- 18

Add one $ each start and end. Use the preview button.

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.

- #24

Bishal Banjara

- 42

- 2

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

- #25

Ibix

Science Advisor

- 9,526

- 9,604

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 onlyYou 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 ofΦandΛas $$\frac{M}{r(r-2M)}$$ 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:

- #26

Bishal Banjara

- 42

- 2

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.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...

- #27

e-pie

- 129

- 18

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.

- #28

Ibix

Science Advisor

- 9,526

- 9,604

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).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?

- #29

e-pie

- 129

- 18

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

- #30

Ibix

Science Advisor

- 9,526

- 9,604

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 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

- #31

e-pie

- 129

- 18

And there are twenty independent components in general:

Bianchi Identity?

4C4=1 21-1=20 ?

- #32

e-pie

- 129

- 18

http://steventan.info/index.php/articles/2-the-schwarzschild-metric

#3 The Riemann Curvature Tensor.

- #33

- 19,577

- 9,972

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.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

- #34

Bishal Banjara

- 42

- 2

Yeah I got the point and answer here at https://physics.stackexchange.com/q...he-riemann-tensor-of-the-schwarzschild-metricMTW 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).

- #35

e-pie

- 129

- 18

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

Therfore 20 components and further restriction give 12 independent.

Share:

- Last Post

- Replies
- 2

- Views
- 308

- Replies
- 2

- Views
- 266

- Replies
- 4

- Views
- 517

- Replies
- 24

- Views
- 477

- Replies
- 5

- Views
- 594

- Replies
- 25

- Views
- 1K

- Replies
- 14

- Views
- 520

- Last Post

- Replies
- 5

- Views
- 347

- Last Post

- Replies
- 2

- Views
- 210

- Replies
- 42

- Views
- 1K