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?

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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:

$$\Gamma^\theta{}_{\phi\phi} = -\sin \theta \cos \theta$$
$$\Gamma^\phi{}_{\phi\theta} = \Gamma^\phi{}_{\theta\phi} = \cot \theta$$

 

[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:

$$\Gamma^\theta{}_{\phi\phi} = -\sin \theta \cos \theta$$
$$\Gamma^\phi{}_{\phi\theta} = \Gamma^\phi{}_{\theta\phi} = \cot \theta$$

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?

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

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:

$${ \left( \Gamma_{\theta} \right)_b }^c = \begin{pmatrix} 0 & 0 \\ 0 & - \sin\theta \cos\theta \end{pmatrix},<br /> { \left( \Gamma_{\phi} \right)_b }^c = \begin{pmatrix} 0 & \cot\theta \\ \cot\theta & 0 \end{pmatrix}$$
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).