Calculating Christoffel symbols for a spherical shell in 2-space

[TheMan112]

Hello everybody, this is my first post here.

I need help with a task in General Relativity.

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Given we have a spherical shell in 2-space of radius \rho. With the line element:

ds^{2} = g_{ab} dx^{a}dx^{b} = \rho d \theta^{2} + \rho^2 sin^2(\theta) d \phi^2

(a,b \in 1,2)

Then calculate the following:

g^{ab}, \Gamma^c_{ab}, R^1_{212}, R^2_{121}, R_{11}, R_{22}, R
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To compute the metric, I'm immediately considering the Schwarzschild solution since this is a spherically symmetric problem.

The covariant metric will then be:

g_{ab} = diag(e^v, -e^\lambda, -\rho^2, -\rho^2 sin^2 \theta)

But since:

e^v dt^2 = 0 and e^\lambda d \rho^2 = 0

Then I'm not sure what to do. However, since:

(a,b \in 1,2)

Can I then set the following?

g_{ab} = diag(-\rho^2, -\rho^2 sin^2 \theta)

If not, then how should I attack the problem?

----

Moreover, I'm not sure how to calculate the Christoffel symbol from the metric, could anyone give me an example how to perform the calculation in my case based on this equation:

\Gamma^c_{ab} = \frac{1}{2} g^{cl} (g_{la,b} + g_{lb,a} - g_{ab,l})

Thanks in advance.

/TheMan112

 

[CompuChip]

To compute the metric, I'm immediately considering the Schwarzschild solution since this is a spherically symmetric problem.

I'm not sure what you mean here. It's given that
g_{ab} dx^{a}dx^{b} = \rho d \theta^{2} + \rho^2 sin^2 d \phi^2
with x^1 = \theta, x^2 = \phi, right?
So can't you just read of the g_{ab}? Then you will see that it is extremely easily invertible and you can write down g^{ab} right away. Also applying the formula for getting the Christoffel symbols you quoted is not hard then.

 

[HallsofIvy]

Hello everybody, this is my first post here.

I need help with a task in General Relativity.

-------
Given we have a spherical shell in 2-space of radius \rho. With the line element:

ds^{2} = g_{ab} dx^{a}dx^{b} = \rho d \theta^{2} + \rho^2 sin^2 d \phi^2

Did you mean
ds^{2} = g_{ab} dx^{a}dx^{b} = \rho d \theta^{2} + \rho^2 sin^2(\phi) d \phi^2
Then by definition the metric tensor is
\left(\begin{array}{cc} \rho & 0 \\ 0 & \rho^2 sin^2(\phi) \end{array}\right)

 

[TheMan112]

Did you mean
ds^{2} = g_{ab} dx^{a}dx^{b} = \rho d \theta^{2} + \rho^2 sin^2(\phi) d \phi^2
Then by definition the metric tensor is
\left(\begin{array}{cc} \rho & 0 \\ 0 & \rho^2 sin^2(\phi) \end{array}\right)

Ah, sorry, no, there should be a theta there:

ds^{2} = g_{ab} dx^{a}dx^{b} = \rho^2 d \theta^{2} + \rho^2 sin^2(\theta) d \phi^2

Ok, so g_{12}= 0 and g_{21}=0 because they are combinations of d \theta^{2} and d \phi^2 (and there are no such defining the line element). Have I understood right?

g_{ab} = \left( \begin{array}{cc} \rho^2 & 0 \\ 0 & \rho^2 sin^2(\theta) \end{array} \right)

 

[TheMan112]

I'm not sure what you mean here. It's given that
g_{ab} dx^{a}dx^{b} = \rho d \theta^{2} + \rho^2 sin^2 d \phi^2
with x^1 = \theta, x^2 = \phi, right?
So can't you just read of the g_{ab}? Then you will see that it is extremely easily invertible and you can write down g^{ab} right away. Also applying the formula for getting the Christoffel symbols you quoted is not hard then.

Yeah... about that. Now that I've got the correct (I hope) metric I need to calculate the Christoffel symbols.

\Gamma^c_{ab} = \frac{1}{2} g^{cl} (g_{la,b} + g_{lb,a} - g_{ab,l}) = \frac{1}{2} \left( \begin{array}{cc} \rho^{-2} & 0 \\ 0 & \rho^{-2} sin^{-2}(\theta) \end{array} \right) \left( \left( \begin{array}{cc} 2 \rho & 0 \\ 0 & 2 \rho sin^{2}(\theta) \end{array} \right) + \left( \begin{array}{cc} 0 & 0 \\ 0 & \rho^2 sin(2 \theta) \end{array} \right) - 0 \right) = ... = \frac{1}{2} \left( \begin{array}{cc} 2 \rho^{-1} & 0 \\ 0 & 2 \rho^{-1} + 2 cot(\theta) \end{array} \right) \right)

I'm not all used to working with tensors, the method I use above is to differentiate g_{ab} with respect to \rho, \theta and \phi respectively. Is this right?

 

[pervect]

For the line element

rho^2*(d[theta]^2 + sin(theta)^2*d[phi]^2);

an automated calculation gets two non-zero Christoffel symbols:

<br /> \Gamma^\theta{}_{\phi\phi} = -\sin \theta \cos \theta<br />
<br /> \Gamma^\phi{}_{\phi\theta} = \Gamma^\phi{}_{\theta\phi} = \cot \theta<br />

 

[Rainbow Child]

Another way to calculate the Christoffel symbols, is the trick which uses the geodesics equation, i.e.

\ddot{x}^\kappa+\Gamma_{\phantom{k}\mu\nu}^\kappa\,\dot{x}^\mu\,\dot{x}^\nu=0 \quad (\ast)

From the line element d\,s^2=g_{\alpha\beta}\,d\,x^\alpha\,d\,x^\beta write the Lagrangian \mathcal{L}=g_{\alpha\beta}\,\dot{x}^\alpha\,\dot{x}^\beta and calculate the Euler-Lagrange equations, i.e.

\frac{d}{d\,\tau}\frac{\partial\,\mathcal{L}}{\partial\,\dot{x}^\kappa}-\frac{\partial\,\mathcal{L}}{\partial\,x^\kappa}=0

and bring them to the form of (\ast).
For the problem at hand

\ddot{\theta}-\frac{1}{2}\,\sin(2\,\theta)\,\dot{\phi}^2=0\Rightarrow \Gamma_{\phantom{k}\phi\phi}^\theta=-\frac{1}{2}\,\sin(2\,\theta)

\ddot{\phi}+2\,\cot\theta\,\dot{\theta}\,\dot{\phi}=0\Rightarrow \Gamma_{\phantom{k}\theta\phi}^\phi=\Gamma_{\phantom{k}\phi\theta}^\phi=\cot\theta

 

[TheMan112]

For the line element

rho^2*(d[theta]^2 + sin(theta)^2*d[phi]^2);

an automated calculation gets two non-zero Christoffel symbols:

<br /> \Gamma^\theta{}_{\phi\phi} = -\sin \theta \cos \theta<br />
<br /> \Gamma^\phi{}_{\phi\theta} = \Gamma^\phi{}_{\theta\phi} = \cot \theta<br />

Ok, how do I do that "manually"?

Another way to calculate the Christoffel symbols, is the trick which uses the geodesics equation, i.e.

\ddot{x}^\kappa+\Gamma_{\phantom{k}\mu\nu}^\kappa\,\dot{x}^\mu\,\dot{x}^\nu=0 \quad (\ast)

From the line element d\,s^2=g_{\alpha\beta}\,d\,x^\alpha\,d\,x^\beta write the Lagrangian \mathcal{L}=g_{\alpha\beta}\,\dot{x}^\alpha\,\dot{x}^\beta and calculate the Euler-Lagrange equations, i.e.

\frac{d}{d\,\tau}\frac{\partial\,\mathcal{L}}{\partial\,\dot{x}^\kappa}-\frac{\partial\,\mathcal{L}}{\partial\,x^\kappa}=0

and bring them to the form of (\ast).
For the problem at hand

\ddot{\theta}-\frac{1}{2}\,\sin(2\,\theta)\,\dot{\phi}^2=0\Rightarrow \Gamma_{\phantom{k}\phi\phi}^\theta=-\frac{1}{2}\,\sin(2\,\theta)

\ddot{\phi}+2\,\cot\theta\,\dot{\theta}\,\dot{\phi}=0\Rightarrow \Gamma_{\phantom{k}\theta\phi}^\phi=\Gamma_{\phantom{k}\phi\theta}^\phi=\cot\theta

Thanks, I'm afraid though that we're not at that point in the course where we use Lagrangians to solve the problems. You wouldn't know how to do the calculation "my way"?

 

[CompuChip]

Ah, sorry, no, there should be a theta there:

ds^{2} = g_{ab} dx^{a}dx^{b} = \rho^2 d \theta^{2} + \rho^2 sin^2(\theta) d \phi^2

Ok, so g_{12}= 0 and g_{21}=0 because they are combinations of d \theta^{2} and d \phi^2 (and there are no such defining the line element). Have I understood right?

g_{ab} = \left( \begin{array}{cc} \rho^2 &amp; 0 \\ 0 &amp; \rho^2 sin^2(\theta) \end{array} \right)

Yes, that's right. Basically, if you write out the sum on the left hand side in full, you get
g_{ab} dx^a dx^b = g_{\theta\theta} d\theta^2 + g_{\theta\phi} d\theta d\phi + g_{\phi\theta} d\phi d\theta + g_{\phi\phi} d\phi^2
(where technically, d\theta^2 stands for \mathrm{d}\theta \otimes \mathrm{d}\theta, but never mind that -- (almost) everyone is sloppy with this).
Comparing to the right hand side you get the matrix you gave, and since it is diagonal its inverse g^{ab} consists of just the inverses of the diagonal elements.

To calculate the Christoffel symbols is in general not a pretty exercise, but for diagonal metrics it is quite easy if you use the symmetries right.
You have
\Gamma^c_{ab} = \frac{1}{2} g^{cl} (g_{la,b} + g_{lb,a} - g_{ab,l})
Note that in the right hand side there is a sum over l, but because g^{ab} is diagonal, the only contributions will come from the terms with c = l. So, first consider all symbols with c = \theta. Then in the first two bracketed terms, you will have theta in one of the metric indices. What must the other two indices be to get a non-vanishing symbol? On the other hand, the last term will contribute only if g_{ab} has non-zero derivative w.r.t. theta. What a and b can you get? Now you have all the symbols with theta upstairs. Take c = phi and do the same thing again.

Also, note that they are symmetric in a and b.

PS Rainbow Child, that is a cool trick. Didn't know that one

 

[TheMan112]

Yes, that's right. Basically, if you write out the sum on the left hand side in full, you get
g_{ab} dx^a dx^b = g_{\theta\theta} d\theta^2 + g_{\theta\phi} d\theta d\phi + g_{\phi\theta} d\phi d\theta + g_{\phi\phi} d\phi^2
(where technically, d\theta^2 stands for \mathrm{d}\theta \otimes \mathrm{d}\theta, but never mind that -- (almost) everyone is sloppy with this).
Comparing to the right hand side you get the matrix you gave, and since it is diagonal its inverse g^{ab} consists of just the inverses of the diagonal elements.

To calculate the Christoffel symbols is in general not a pretty exercise, but for diagonal metrics it is quite easy if you use the symmetries right.
You have
\Gamma^c_{ab} = \frac{1}{2} g^{cl} (g_{la,b} + g_{lb,a} - g_{ab,l})
Note that in the right hand side there is a sum over l, but because g^{ab} is diagonal, the only contributions will come from the terms with c = l. So, first consider all symbols with c = \theta. Then in the first two bracketed terms, you will have theta in one of the metric indices. What must the other two indices be to get a non-vanishing symbol? On the other hand, the last term will contribute only if g_{ab} has non-zero derivative w.r.t. theta. What a and b can you get? Now you have all the symbols with theta upstairs. Take c = phi and do the same thing again.

Also, note that they are symmetric in a and b.

PS Rainbow Child, that is a cool trick. Didn't know that one

Thank you so much. Very complete and concise, now I got the right answer (based on pervect's computation) and fully understand it. I can't tell you how helpful this was.

Should I add these together in some way to form \Gamma^c_{ab} or should I write the two solutions for c = \theta and c = \phi apart?

 

[CompuChip]

You can write them as two matrices, for example:

{ \left( \Gamma_{\theta} \right)_b }^c = \begin{pmatrix} 0 &amp; 0 \\ 0 &amp; - \sin\theta \cos\theta \end{pmatrix},<br /> { \left( \Gamma_{\phi} \right)_b }^c = \begin{pmatrix} 0 &amp; \cot\theta \\ \cot\theta &amp; 0 \end{pmatrix}
or you can just list them all, which is equally good to me.

(Small word of warning: The fact that you can write them as matrices does still not mean they are genuine tensors. It's just convenient notation).