# Schwarzschild metric assistance

1. Jun 28, 2005

### lurathis

Hello, I've enjoyed reading these forums for a while now as they have lots of great insight! Today I decided to register so I can ask a question that's been bothering me for a few days now. (By the way, I'm moving onto my senior year as a physics major)

So yea, to the question. I have been trying to derive the Schwarzschild metric (a static, spherically symmetric, vacuum solution) in a more rigorous fashion than I have seen documented in a few books and I'm having a little trouble. I have been able to make all $g_{\mu\nu} = 0$ for all $\mu \neq \nu$ to give me a metric of the form $$g_{\mu \nu} = A(r) dr^2 + C(r, \theta, \phi) d\theta^2 + D(r,\theta,\phi) d\phi^2 + B(r) dt^2$$.
I then defined my r coordinate such that when keeping $\theta$ and $\phi$ constant, the metric is just $$g_{\mu \nu} = r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2$$ (i.e. Setting r such that it is not physical radius but so that the area of the 2-sphere at r is $4 /pi r^2$)

Then I have:
$$g_{\mu \nu} = A(r) dr^2 + r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2 + B(r) dt^2$$

I know that I must next use the vacuum field constraint so that the Ricci tensor, $R_{\mu \nu} = 0$, so I computed all of the Christoffel symbols and have them to my disposal (if you wish to see them just let me know and I'll post them up... luckily there are only 13 nonzero ones). Now here is the problem: I'm not very comfortable yet with tensor algebra, so I don't know exactly how to use my Christoffel symbols to put them into my vacuum field constraint. I know the equation of the Riemann curvature tensor (posted at the end). I tried taking the Riemann curvature tensor $R^\omega{}_{\beta \gamma \lambda}$ such that $\lambda = \gamma$ and we have $R^\gamma{}_{\beta \gamma \lambda}=R_{\beta \lambda}$. So I know I only need two equations to solve for A(r) and B(r) so I tried solving for $R_{1 1}$ and $R_{4 4}$ (where the $r, \theta, \phi, t \rightarrow 1, 2, 3, 4$) the long way and actually plugging in all the Christoffel symbols for all $R^\gamma{}_{1 \gamma 1}$(quite long... 2 free indeces to sum over which gives 16 sets of sums with 4 terms each). I did the same for $g_{4 4}$, and got the following out of the two:

$$\frac{6}{r^2} + \frac{7}{4} \frac{(\frac{dB(r)}{dr})^2}{B(r)} - 2\frac{\frac{d^2B(r)}{dr^2}}{B(r)} + \frac{1}{r} \frac{\frac{dA(r)}{dr}}{A(r)} + \frac{1}{4} \frac{\frac{dA(r)}{dr}}{A(r)} \frac{\frac{dB(r)}{dr}}{B(r)} = 0$$

and

$$2\frac{\frac{dA(r)}{dr} \frac{dB(r)}{dr}}{A^2(r)} - 2\frac{\frac{d^2A(r)}{dr^2}}{A(r)} - \frac{\frac{dA(r)}{dr} \frac{dB(r)}{dr}}{4 A^2(r)} - \frac{\frac{dB(r)}{dr}}{r A(r)} - \frac{(\frac{dB(r)}{dr})^2}{4 A(r) B(r)} = 0$$

Apart from taking out a factor of $\frac{1}{A(r)}$ from the second equation, I really don't know how to solve for A(r) and B(r). Am I doing something wrong? I'm concerned that my lack of familiarity with tensors is leading me to incorrect equations. So basically, am I using the $R_{\mu \nu} = 0$ requirement correctly?

Sorry for the long post, but any help would be greatly appreciated. Thanks!

Riemann curvature tensor:

$$R^\omega{}_{\beta \gamma \lambda} = \Gamma^\omega{}_{\beta \lambda, \gamma} - \Gamma^\omega{}_{\beta \gamma, \lambda} + \Gamma^\omega{}_{\gamma \sigma} \Gamma^\sigma{}_{\beta \lambda} - \Gamma^\omega{}_{\lambda \sigma} \Gamma^\sigma{}_{\beta \gamma}$$

(where the ',' in the subscript denotes the usual notation for derivative)

2. Jun 28, 2005

### dextercioby

I'll be optimistic in this case and say that,apparently,you're not doing anything wrong.You may wonder whether your system of equations in A and B is first of all correct and,if so,then possibly solvable exactly.I haven't seen your calculations,so there's still a nonzero chance for it to be wrong.In case it's right,the system doesn't look solvable exactly,due to the nonlinearity.So the "less rigurous" approaches used in books (the Dirac's one is among the simplest possible) at least have the advantages of leading to exact solutions.

You may look for the original article by Schwarzschild.

Daniel.

3. Jun 28, 2005

### pervect

Staff Emeritus
Given the Ricci R_tt and R_rr (that's R11 and R44 in your notation)

we know that
R_tt = R_rr = 0

which means that the difference

A(r)*R_tt - B(r)*R_rr = 0

This should simplify enormously to

A(r) d/dr B(r) + B(r) d/dr A(r) = 0

This implies that A(r)*B(r) = constant, as it is d/dr (A(r)*B(r)) = 0

GRTensorII gives me with your metric

$$R_{tt} = -1/4\,{\frac {- \left( {\frac {d}{dr}}B \left( r \right) \right) \left( {\frac {d}{dr}}A \left( r \right) \right) B \left( r \right) r+2\, \left( {\frac {d^{2}}{d{r}^{2}}}B \left( r \right) \right) A \left( r \right) B \left( r \right) r- \left( {\frac {d}{dr}}B \left( r \right) \right) ^{2}A \left( r \right) r+4\, \left( {\frac {d}{dr}}B \left( r \right) \right) A \left( r \right) B \left( r \right) }{ \left( A \left( r \right) \right) ^{2}B \left( r \right) r}}$$

$$R_{rr} = -1/4\,{\frac {- \left( {\frac {d}{dr}}B \left( r \right) \right) ^{2} A \left( r \right) r+2\, \left( {\frac {d^{2}}{d{r}^{2}}}B \left( r \right) \right) A \left( r \right) B \left( r \right) r- \left( { \frac {d}{dr}}B \left( r \right) \right) \left( {\frac {d}{dr}}A \left( r \right) \right) B \left( r \right) r-4\, \left( {\frac {d}{ dr}}A \left( r \right) \right) \left( B \left( r \right) \right) ^{ 2}}{ \left( B \left( r \right) \right) ^{2}A \left( r \right) r}}$$

A(r)*R_tt - B(r)*R_rr = 0 =
$$-{\frac { \left( {\frac {d}{dr}}B \left( r \right) \right) A \left( r \right) + \left( {\frac {d}{dr}}A \left( r \right) \right) B \left( r \right) }{rA \left( r \right) }}$$

This means you can then substitute B(r) = K/A(r). The texts suggest that you can subsume the factor K into a scale factor for the time, which means you can make K=1 and B(r) = 1/A(r).

4. Jun 28, 2005

### robphy

Maybe you should also chat with rickphysics
who recently asked a related question

These might be useful as well
http://www.mathpages.com/rr/s6-01/6-01.htm
http://scholar.uwinnipeg.ca/courses/38/4500.6-001/Cosmology/Schwarzschild_Metric.htm [Broken]
http://ocw.mit.edu/NR/rdonlyres/Physics/8-033Fall2003/134C70E4-DDF9-41E9-B359-990AC0205747/0/schwarzschild03.pdf [Broken]

Here's a nice set of notes
http://www.roma1.infn.it/teongrav/Relativity_notes.html [Broken]
http://www.roma1.infn.it/teongrav/VALERIA/CORSORELATIVITA/CAP10_Schwarzschild_Solution.pdf [Broken]

Last edited by a moderator: May 2, 2017
5. Jun 28, 2005

### lurathis

Thanks very much for the prompt responses, I did look at the article by Schwarzschild and found his take very interesting. After letting that clear my mind, I referred back to my Christoffel terms and realised that two had A instead of $A^{-1}$. This is of course throws everything off, so let me go back and see what my new results are and I'll get back to you all.

By the way, just for interest (to avoid another embarassing situation) I just wanted to verify that my equations for $R_{1 1}$ and $R_{4 4}$ are correct:

$$R_{1 1} = -4(\Gamma^2{}_{1 2, 1} + \Gamma^3{}_{1 3, 1} + \Gamma^4{}_{1 4, 1}) + \Gamma^1{}_{1 1}(\Gamma^2{}_{2 1} + \Gamma^3{}_{3 1} + \Gamma^4{}_{4 1}) - (\Gamma^2{}_{1 2})^2 - (\Gamma^3{}_{1 3})^2 - (\Gamma^3{}_{1 3})^2$$.

$$R_{4 4} = 4(\Gamma^1{}_{4 4, 1}) + \Gamma^1{}_{4 4}(\Gamma^1{}_{1 1} + \Gamma^2{}_{2 1} + \Gamma^3{}_{3 1} + \Gamma^4{}_{4 1})$$

(Also, I am curious as to whether this calculation has to be actually churned out term by term the way I am doing it, or whether via some more advanced tensor algebra that I am unfamiliar with, this is simpler. I'm just wondering since writing out the entire 64 term equation for both $R_{1 1}$ and $R_{4 4}$ and then looking to which terms are zero is really tedious. Not that I'd know how to do it, I'm just curious as to whether later on in my studies I'll be able to do this in a more concise fashion)

6. Jun 28, 2005

### lurathis

So the two Christoffel symbols I had incorrect didn't end up being present in the equations for $R_{1 1}$ and $R_{4 4}$ anyway so no progress there. Nevertheless, I am going back and doing the whole thing over to see if I can figure something out.

7. Jun 28, 2005

### lurathis

Whilst redoing it on my own, and trying to simplify, I actually did A(r)*R_tt - B(r)*R_rr = 0. (Only to then come back to the forum and realise that is what you suggest). However, I end up with:

$$\frac{6}{r^2} + \frac{(B')^2}{4B}(7 + \frac{1}{B}) + \frac{1}{r}(\frac{B'}{B} + \frac{A'}{A}) - \frac{3}{2} \frac{A'B'}{AB} = 0$$

I'm not sure how to get this of the form A'B + B'A = 0...

Last edited: Jun 28, 2005
8. Jun 28, 2005

### pervect

Staff Emeritus
Well, I cheat and use a program called GRTensor II to calculate all this nasty stuff, rather than do it by hand.

http://grtensor.phy.queensu.ca/

Unfortunately, to run it, you'll need to buy either a version of Maple (preferred) or Mathematica (to run the GrTensor M version). Even a very old version of Maple will work, though, it doesn't have to be the latest version to run GRTensorII.

There are some public domain symbolic algebra programs available

http://web.usna.navy.mil/~wdj/opensource_math.html [Broken]

unfortunately these aren't compatible with the GRTensor package. I don't have any experience with any of them, Maxima looks good on paper though. It's got to be easier than trying to not make any mistakes doing it all by hand - but you'll have to do more work when going this route, as these programs should have general symbolic algebra capability, but won't have the calculation of the Ricci and Riemann built-in the way GRTensorII does.

It looks like the latest version of Maxima might be a reasonable choice, it apparently has packages that deal with GR, and they are apparently being maintained

http://www.vttoth.com/maxima.html

One thing to beware of if you do use GRTII is that it has a funny idea of what Christoffel symbols are. This is easily fixed when you are aware of the problem, just define a new symbol, which I call CC, to represent the Christoffel symbols as follows:

grdef(CC{ ^a b c} := Chr{b c ^a});

Last edited by a moderator: May 2, 2017
9. Jun 28, 2005

### lurathis

Pervect, this is what you got using my form of the metric, right? Do you also use the same Christoffel symbols I've been using?

$$\Gamma^1{}_{1 1} = \frac{A'}{2A}$$
$$\Gamma^2{}_{2 1} = \Gamma^3{}_{3 1} = \Gamma^2{}_{1 2} = \Gamma^3{}_{1 3} = \frac{1}{r}$$
$$\Gamma^3{}_{3 2} = \Gamma^3{}_{2 3} = \cot\theta$$
$$\Gamma^4{}_{4 1} = \Gamma^4{}_{1 4} = \frac{B'}{2B}$$
$$\Gamma^1{}_{2 2} = \frac{-r}{A}$$
$$\Gamma^1{}_{3 3} = \frac{-r}{A} \sin^2\theta$$
$$\Gamma^1{}_{4 4} = \frac{-B'}{2A}$$
$$\Gamma^2{}_{3 3} = -\sin\theta \cos\theta$$

Because I keep getting what I wrote in post #7, but if you put it into the computer and it gives what you wrote in #3 using the same Christoffel symbols and metric and everything, I'm going to have to take another good look at it (I'm kinda burnt out at the moment, but maybe later tonight).

I looked at the websites you suggested and I'll keep them in mind for future calculations. However, this one has now become what I call personal and I would like to try and get it by hand. So if you could reconfirm the results you posted before, that would be great. Thanks

10. Jun 28, 2005

### pervect

Staff Emeritus
OK, the metric is
ds^2 = B(r)*d*t^2 +A(r) * dr^2 + r^2*d*theta^2 + r^2*sin(theta)^2* dphi^2

The computer-generated Christoffel symbols are

Code (Text):

CC(up,dn,dn)

d
-- B(r)
r               dr
CC   [t t] = -1/2 -------
A(r)

d
-- B(r)
t              dr
CC   [r t] = 1/2 -------
B(r)

d
-- B(r)
t              dr
CC   [t r] = 1/2 -------
B(r)

d
-- A(r)
r              dr
CC   [r r] = 1/2 -------
A(r)

theta
CC       [theta r] = 1/r

phi
CC     [phi r] = 1/r

theta
CC       [r theta] = 1/r

r                     r
CC   [theta theta] = - ----
A(r)

phi                cos(theta)
CC     [phi theta] = ----------
sin(theta)

phi
CC     [r phi] = 1/r

phi                cos(theta)
CC     [theta phi] = ----------
sin(theta)

2
r                r sin(theta)
CC   [phi phi] = - -------------
A(r)

theta
CC       [phi phi] = -sin(theta) cos(theta)

The computer-generated Riemann is
Code (Text):

/  / 2      \
t                 |  |d       |             /d      \2
R    [r t r] = -1/4 |2 |--- B(r)| B(r) A(r) - |-- B(r)|  A(r)
|  |  2     |             \dr     /
\  \dr      /

\
/d      \ /d      \     |   /      2
- |-- B(r)| |-- A(r)| B(r)|  /  (B(r)  A(r))
\dr     / \dr     /     | /
/

/d      \
|-- B(r)| r
t                         \dr     /
R    [theta t theta] = -1/2 -----------
B(r) A(r)

/d      \             2
|-- B(r)| r sin(theta)
t                     \dr     /
R    [phi t phi] = -1/2 -----------------------
B(r) A(r)

/  / 2      \
r                |  |d       |             /d      \2
R    [t t r] = 1/4 |2 |--- B(r)| B(r) A(r) - |-- B(r)|  A(r)
|  |  2     |             \dr     /
\  \dr      /

\
/d      \ /d      \     |   /      2
- |-- B(r)| |-- A(r)| B(r)|  /  (A(r)  B(r))
\dr     / \dr     /     | /
/

/d      \
|-- A(r)| r
r                        \dr     /
R    [theta r theta] = 1/2 -----------
2
A(r)

/d      \             2
|-- A(r)| r sin(theta)
r                    \dr     /
R    [phi r phi] = 1/2 -----------------------
2
A(r)

d
-- B(r)
theta                    dr
R        [t t theta] = 1/2 -------
r A(r)

d
-- A(r)
theta                     dr
R        [r r theta] = -1/2 -------
r A(r)

2
theta                    sin(theta)  (A(r) - 1)
R        [phi theta phi] = ----------------------
A(r)

d
-- B(r)
phi                  dr
R      [t t phi] = 1/2 -------
r A(r)

d
-- A(r)
phi                   dr
R      [r r phi] = -1/2 -------
r A(r)

phi                        A(r) - 1
R      [theta theta phi] = - --------
A(r)

Not all the components are listed: R^a_bcd = -R^a_bdc, i.e. permutting the last two elements of R just changes the sign.

The replies I've made to you use your metric, some other replies in other threads have used the first poster's metric.

Hopefully this helps.

11. Jun 29, 2005

### pervect

Staff Emeritus
I redid it with Maxima, the output, attached as a text file to this post, might be more convenient.

Some notes:

The christoffel symbols are given by
$$\Gamma^k{}_{ij} = MCS_{i,j,k}$$

note the strange ordering where the first subscript of gamma is the last subsript of MCS.

AR is $$\frac{d A}{d R}$$

The Riemann tensor is $$R_{i,j,k,l}$$ - this is standard ordering, but no indices have been raised yet. The usual symmetries of the Riemann apply when exchanging indices.

R_abcd = R_bacd = -R_bacd = -R_abdc

This is important because some of the nonzero components need to be found from the symmetry relations. If you download the program from sourceforge and run it with the inputs shown, you can get it to spit out any component you want to make sure you are not missing one by accident.

The Ricci tensor is $$LR_{i,j}$$

You get a simple expression from B * LR[1,1] - A * LR[4,4], which has to be multipled further by A*R to get it in the form $$A B_R + B A_R$$

#### Attached Files:

• ###### Schws.txt
File size:
4.4 KB
Views:
79
12. Jun 29, 2005

### lurathis

That looks fantastic, thanks. I'm not sure why my way kept getting a strange result, but now that I have this to go by I can do some tests.

Thanks a lot Pervect. This program looks neat

13. Jun 29, 2005

### lurathis

Okay, Maxima seems correct as it would end up giving you the correct Schwarzschild metric. So I've come up with a system of checkpoints to see where I'm making my mistake. If you could possibly flag the checkpoint that is wrong, that would really help me try to fix my error.

Maxima agrees with all of my Christoffel symbols, so no problems there. The problem must be then in how I calculate $R_{\mu\mu}$ (Due to the fact that I'm relatively new to this tensor business). The way I did it was to set $\omega = \gamma$ in the Riemann curvature tensor $R^\omega{}_{\beta \gamma \lambda}$ such that $R^\omega{}_{\beta \gamma \lambda}=R^\gamma{}_{\beta \gamma \lambda}$ because $R^\gamma{}_{\beta \gamma \lambda}=R_{\beta \lambda}$ ([Checkpoint 1]this is true, right? [ /Checkpoint 1]).

So basically I'm just taking $R^\gamma{}_{\i \gamma \i}$ for i = 1 for $R_{1 1}$ and i = 4 for $R_{4 4}$. So according to the Riemann curvature equation (I posted before), I should be summing over $\gamma$ and $\sigma$, right?
So then (sorry for writing the whole thing out, but I think I'm doing something fundamentally incorrect and would really like to know what) the general form of $R_{1 1}$ is:

[Checkpoint 2]
$$R_{1 1} =$$
$$\Gamma^1{}_{1 1, 1} - \Gamma^1{}_{1 1, 1} + \Gamma^1{}_{1 1} \Gamma^1{}_{1 1} - \Gamma^1{}_{1 1} \Gamma^1{}_{1 1} + ...$$
$$\Gamma^2{}_{1 1, 2} - \Gamma^2{}_{1 2, 1} + \Gamma^2{}_{2 1} \Gamma^1{}_{1 1} - \Gamma^2{}_{1 1} \Gamma^1{}_{1 2} + ...$$
$$\Gamma^3{}_{1 1, 3} - \Gamma^3{}_{1 3, 1} + \Gamma^3{}_{3 1} \Gamma^1{}_{1 1} - \Gamma^3{}_{1 1} \Gamma^1{}_{1 3} + ...$$
$$\Gamma^4{}_{1 1, 4} - \Gamma^4{}_{1 4, 1} + \Gamma^4{}_{4 1} \Gamma^1{}_{1 1} - \Gamma^4{}_{1 1} \Gamma^1{}_{1 4} + ...$$

$$\Gamma^1{}_{1 1, 1} - \Gamma^1{}_{1 1, 1} + \Gamma^1{}_{1 2} \Gamma^2{}_{1 1} - \Gamma^1{}_{1 2} \Gamma^2{}_{1 1} + ...$$
$$\Gamma^2{}_{1 1, 2} - \Gamma^2{}_{1 2, 1} + \Gamma^2{}_{2 2} \Gamma^2{}_{1 1} - \Gamma^2{}_{1 2} \Gamma^2{}_{1 2} + ...$$
$$\Gamma^3{}_{1 1, 3} - \Gamma^3{}_{1 3, 1} + \Gamma^3{}_{3 2} \Gamma^2{}_{1 1} - \Gamma^3{}_{1 2} \Gamma^2{}_{1 3} + ...$$
$$\Gamma^4{}_{1 1, 4} - \Gamma^4{}_{1 4, 1} + \Gamma^4{}_{4 2} \Gamma^2{}_{1 1} - \Gamma^4{}_{1 2} \Gamma^2{}_{1 4} + ...$$

$$\Gamma^1{}_{1 1, 1} - \Gamma^1{}_{1 1, 1} + \Gamma^1{}_{1 3} \Gamma^3{}_{1 1} - \Gamma^1{}_{1 3} \Gamma^3{}_{1 1} + ...$$
$$\Gamma^2{}_{1 1, 2} - \Gamma^2{}_{1 2, 1} + \Gamma^2{}_{2 3} \Gamma^3{}_{1 1} - \Gamma^2{}_{1 3} \Gamma^3{}_{1 2} + ...$$
$$\Gamma^3{}_{1 1, 3} - \Gamma^3{}_{1 3, 1} + \Gamma^3{}_{3 3} \Gamma^3{}_{1 1} - \Gamma^3{}_{1 3} \Gamma^3{}_{1 3} + ...$$
$$\Gamma^4{}_{1 1, 4} - \Gamma^4{}_{1 4, 1} + \Gamma^4{}_{4 3} \Gamma^3{}_{1 1} - \Gamma^4{}_{1 3} \Gamma^3{}_{1 4} + ...$$

$$\Gamma^1{}_{1 1, 1} - \Gamma^1{}_{1 1, 1} + \Gamma^1{}_{1 4} \Gamma^4{}_{1 1} - \Gamma^1{}_{1 4} \Gamma^4{}_{1 1} + ...$$
$$\Gamma^2{}_{1 1, 2} - \Gamma^2{}_{1 2, 1} + \Gamma^2{}_{2 4} \Gamma^4{}_{1 1} - \Gamma^2{}_{1 4} \Gamma^4{}_{1 2} + ...$$
$$\Gamma^3{}_{1 1, 3} - \Gamma^3{}_{1 3, 1} + \Gamma^3{}_{3 4} \Gamma^4{}_{1 1} - \Gamma^3{}_{1 4} \Gamma^4{}_{1 3} + ...$$
$$\Gamma^4{}_{1 1, 4} - \Gamma^4{}_{1 4, 1} + \Gamma^4{}_{4 4} \Gamma^4{}_{1 1} - \Gamma^4{}_{1 4} \Gamma^4{}_{1 4}$$
[ /Checkpoint 2]

which, after crossing out all zero terms, reduces down to:

[ Checkpoint 3]
$$R_{1 1} = -4(\Gamma^2{}_{1 2, 1} + \Gamma^3{}_{1 3, 1} + \Gamma^4{}_{1 4, 1}) + \Gamma^1{}_{1 1}(\Gamma^2{}_{2 1} + \Gamma^3{}_{3 1} + \Gamma^4{}_{4 1}) - (\Gamma^2{}_{1 2})^2 - (\Gamma^3{}_{1 3})^2 - (\Gamma^3{}_{1 3})^2$$.
[ /Checkpoint 3]

If you plug in the Christoffel symbols at this point then it gives what I put in post #1 with the $\frac{1}{r^2}$ terms and other such things that Maxima does not give.

Any ideas?

Thanks for all the help.

14. Jun 29, 2005

### pervect

Staff Emeritus
This is correct, of course you do have to sum over the repeated index.

Thus $R_{11} = R^1{}_{111}+R^2{}_{121}+R^3{}_{131}+R^4{}_{141}$

I'm suspicious of checkpoint 2. Let's look at $R^1{}_{111}$ for example.

It will be equal to zero, but it will be

$\Gamma^1{}_{11,1}-\Gamma^1{}_{11,1}+ 4*[\Gamma^1{}_{11} \Gamma^1{}_{11} - \Gamma^1{}_{11} \Gamma^1{}_{11}]$

Because you have to sum over repeated indices and in

$$R^u{}_{vab} = \partial_a \Gamma^u{}_{vb} - \partial_b\Gamma^u{}_{va}+ \Gamma^u{}_{pa}\Gamma^p{}_{vb} - \Gamma^u{}_{pb}\Gamma^p{}_{va}$$

p is a repeated index, so the terms with a p appear 4 times, the terms without a p appear only once.

You should also know or be told how to get R^1{}_{111} out of R_{1111} by raising an index. You do that in general by the tensor equation

$$R^a{}_{cde} = g^{ab} R_{bcde}$$

In the diagional metric case, this is very easy to do. While you have a repeated index b, g^ab is nonzero only when a=b, and it reduces to a straightfowards multiplication by g^aa.

Note that g^ab is just the inverse of g_ab, which is also easy for a diagonal matrix, you take the reciprocals of all the diagonal elements.

The end result of these statemetns is that for a diagonal metric (as we have)

R^1_{abc} is just 1/g_11 R_{1abc}

15. Jun 29, 2005

### lurathis

I see. That would explain a lot then because I was summing the entire thing because of that extra repeated index 'p' (not just summing the terms with p).

Wait, since nothing is fixing p, don't you mean

$R^1{}_{111} = \Gamma^1{}_{11,1}-\Gamma^1{}_{11,1} + (\Gamma^1{}_{11} \Gamma^1{}_{11} - \Gamma^1{}_{11} \Gamma^1{}_{11}) + (\Gamma^1{}_{12} \Gamma^2{}_{11} - \Gamma^1{}_{12} \Gamma^2{}_{11}) + ...$
$(\Gamma^1{}_{13} \Gamma^3{}_{11} - \Gamma^1{}_{13} \Gamma^3{}_{11}) + (\Gamma^1{}_{14} \Gamma^4{}_{11} - \Gamma^1{}_{14} \Gamma^4{}_{11})$

By the way, thanks very much for those much needed tips on raising or lowering the indices.

16. Jun 29, 2005

### pervect

Staff Emeritus
Yep, I blew it - you've got the right idea though. Stuffing this into GRT2, there should be 24 nonzero terms which are the product of two Christoffel symbols, along with the 8 partial derivatives in the expression for R_11.

Here I represent $\Gamma^1{}_{23}$ as g123 to get the 24 terms below.

g212*g111-g211*g112+g222*g211-g221*g212+g232*g311-g231*g312+g242*g411-g241*g412+g313*g111-g311*g113+g323*g211-g321*g213+g333*g311-g331*g313+g343*g411-g341*g413+g414*g111-g411*g114+g424*g211-g421*g214+g434*g311-g431*g314+g444*g411-g441*g414

17. Jun 29, 2005

### lurathis

Yup, fabulous. After doing all that and solving for my A(r) and B(r), as well as setting a constant equal to c^2 to accomodate the minkowski metric as r approaches infinity, I end up with:

$$d\tau^2 = (1+\frac{1}{Kr})^{-1}dr^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2) + c^2 (1+\frac{1}{Kr})dt^2$$

Now for the last and final push, I don't really understand how to equate this to Newton's gravitational laws in order to solve for my constant K. The only source I've found (Wolfram at http://scienceworld.wolfram.com/physics/SchwarzschildBlackHole.html) says that $g_{44} = -1-2\phi$. I have no idea where they got this from (of course it does give the correct answer).
This is the final question, I promise

18. Jun 29, 2005

### pervect

Staff Emeritus
To accomodate the Minkowski metric, you really need to make the coefficient of the dt^2 term negative. We know that A(r)B(r) = constant, I think I may have said (incorrectly) to make the constant +1 somewhere, but actually -1 is the choice you need to get the Minkowski metric.

The degree of freedom you have left is to show that -1/2K is the mass of the object. There are a couple of ways you can approach this. The first is to look at how Newtonian gravity can be expressed in curved space-time. If you know the geodesic equations, the argument for this can be very brief.

I think you said you were reading Hartle or some other textbook? If so, look for "Cartan" in the index. If you have MTW, look at chapter 12, "Newtonian gravity in the language of curved space-time".

I didn't see much on the internet with a quick google on the topic.

The geodesic equations are

$$\frac{d^2 x^a}{d \lambda^2} + \Gamma^a{}_{bc} \frac{dx^b}{d\lambda}\frac{dx^c}{d\lambda} = 0$$

These equations describe the path an object with no external force will take in curved space-time.

We can make lambda equal to proper time if we desire (lambda=tau). Let us do so.

To get the Newtonian equations we have
$$\frac{d^2 x^i}{d \tau^2} = 0$$

for i=0,2,3 (x^0=t, x^2=theta, x^3=phi)

$$\frac{d^2 x^i}{d \tau^2} -GM/r^2 = 0$$
only when i=1.

The case when i=1, such that x^1 = r (the radial coorinate), is the only case where we have "acceleration". We get the same behavior if this acceleration is due to a force, or if it is due to space-time curvature. The result of this is that we can get Newtonian gravity with curved space-time just by making a single Christoffel symbol non-zero. Because the "force" of gravity is not velocity dependent in Newtonian thoery,
$$\Gamma^r{}_{tt}$$ is the only possiblity for the origin of the force term.

You may wonder about the relationship
d^2 t / dtau^2 = 0

We are assuming low velocities and no time dilation (Newtonian theory has no time dilation, after all).

Thus to get Newtonian gravity in the limit of large r, we need only
$$\Gamma^r{}_{tt}$$ to be -GM/r^2 for large r. When we do, an object will follow the same geodesics that it does with a "force" of -GmM/r^2 acting on it (an acceleration of -GM/r^2.

Plugging this into the equation gives the relationship between K, an arbitrary constant, and M, the "mass" of the body.

I've made sly use of the fact that g_rr (and for that matter, g_tt) are nearly unity at r=infinity, without really pointing out the need for this assumption. Congratulations to you if you can spot why this is necessary, I'm not quite sure how to explain it concisely.

I'm afraid this isn't as clear as it could be, but I hope it helps.

19. Jun 30, 2005

### lurathis

That was extremely helpful, thanks. Oh and that was meant to be a negative sign in front of the c^2... just mistakenly left it out of the post.

I'm not sure if I followed you to the letter, but I read through your post and then ran with the idea (perhaps deviating a little).

Since my $g_{44}$ is already of the form $-c^2 (1+\frac{1}{Kr})$ then it's as if (well in my mind at least) we have the Minkowski flat space minus whatever deviation caused by the mass. So in other words, my metric can be written of the form $g_{\mu\nu}=m_{\mu\nu} + \epsilon \gamma_{\mu\nu}$ where $\epsilon$ is a nonzero constant, $m_{\mu\nu}$ is the Minkowski term, and $\epsilon \gamma_{\mu\nu}$ is my deviation from the flat Minkowski metric.

The only thing I did not really understand (but took for granted to see where it took me, was the reason for choosing $\Gamma^1{}_{4 4}$. But in any case, I used that to say that, ignoring the terms of order greater than $\epsilon$
$$\Gamma^1{}_{4 4} = \frac{-\epsilon}{2} g^{11} \gamma_{44,1}$$
This is, of course, where you use your $g_{11}=1$ approximation to say that
$$\Gamma^1{}_{4 4} = \frac{-\epsilon}{2} \gamma_{44,1}$$

Then the geodesic equation becomes
$$\frac{d^2 x^1}{d \tau^2} = -\Gamma^1{}_{4 4} = \frac{\epsilon}{2} \gamma_{44,1} = -\nabla \Phi$$

So then $\Phi=\frac{-\epsilon}{2} \gamma_{44}=\frac{GM}{r}$. So we solve for $\epsilon \gamma_{44}$ and we plug that into $g_{44}$ to get
$$g_{44}=-c^2 + \frac{2GM}{r}$$.

We compare that to my term for $g_{44}$ where
$$g_{44}=-c^2(1+\frac{1}{Kr}) = -c^2 + \frac{2GM}{r}$$
and find that $K=\frac{-c^2}{2GM}$

Then plugging in my K I get:

$$d\tau^2 = (1-\frac{2GM}{c^2r})^{-1}dr^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2) + c^2 (1-\frac{2GM}{c^2r})dt^2$$

Cheer!

P.S. If you could clarify for me why I used $\Gamma^1{}_{44}$ as opposed to say $\Gamma^1{}_{ii}$ where i is any other number that would be great. I know the superscript has to be one to match with the $\nabla \Phi$. Thanks.

Last edited: Jun 30, 2005
20. Jun 30, 2005

### pervect

Staff Emeritus
Let's try a change of notation. Let's re-write the geodesic deviation equation for d^2x^r/dt^2 in more familiar vairables, r, theta, phi, t, and of course tau. We'll replace dr/dtau with v_r, dr/dtheta with v_theta, dr/dphi with v_phi. And dt/dtau gets replaced with one.

Then we have

$$\frac{d^2 r}{d \tau^2} + \Gamma^1{}_{11} v_r^2 + 2\Gamma^1{}_{12} v_r v_{theta} + 2\Gamma^1{}_{13} v_r v_{phi} + 2\Gamma^1{}_{14} v_r + \Gamma^2{}_{22} v_{phi}^2 + \Gamma^2{}_{23} v_{phi} v_{theta}+2 \Gamma^2_{24} v_{phi} + ... ...... + \Gamma^1{}_{44}=0$$

I got tired of writing down the terms, because the only point I wanted to make is that only one term does not get multiplied by a velocity or the square of a velocity, and that's the last term, $\Gamma^1{}_{44}$.

That's just what we want to mimic Newtonian gravity, a constant acceleration. We don't want a force that's proportional to velocity, nor do we want a force that's proportional to velocity^2.

If we re-did the problem in a cartesian set of coordinates, some of the points about the potential would be clearer. My statement was that acceleration occured only in the 'r' direction, and was independent of velocity, and I didn't mention the potential. I'm assuming you read about the potential in a textbook (that's the approach my textbok uses too).

The statement of Newtonian gravity in cartesian coordinates would be that the acceleration has the magnitudes

d Phi / dx, d Phi / dy, d Phi/dz, where Phi is the potential function

Thinking aboutf Newtonian gravity in generalized coordinates would be an even better match, if you're familiar with the Lagrangain

L = T-V

Let's let the coordinates be x1,x2,x3 and the velocities be x1', x2', x3' (yet another change of notation).

For cartesian coordinates, T = .5*m*(x1'^2+x2'^2+x3'^2), and V=Phi

$$\frac{d}{dt}(\frac{\partial L}{\partial x_i'})= \frac{\partial L}{\partial x_}$$

wraps up the equations of motion we want to mimic, by using geometry instead of forces.

Last edited: Jun 30, 2005