The contracting relations on the Christoffel symbols

[Arman777]

Homework Statement

The contracting relations on the Christoffel symbols

Relevant Equations

Tensor Equations

I am trying to find $$\Gamma^{\nu}_{\mu \nu} = \partial_{\mu} log(\sqrt{g})$$ but I cannot.

by calculations, I manage to find

$$\Gamma^{\nu}_{\mu \nu} = \frac{1}{2}g^{\nu \delta}\partial_{\mu}g_{\nu \delta}$$

and from research I have find that $$det(A) = e^{Tr(log(A))}$$ but still I cannot make the connection. Any ideas?

where $g = det(g_{\alpha \beta})$

 

[vanhees71]

First of all you must always have $\sqrt{-g}$, because $g<0$ due to the signature of the Lorentzian manifold.

Further think about, how (the matrix) $g^{\mu \nu}$ is related to (the matrix) $g_{\mu \nu}$ and what this has to do with determinants!

 

[Arman777]

First of all you must always have $\sqrt{-g}$, because $g<0$ due to the signature of the Lorentzian manifold.

Further think about, how (the matrix) $g^{\mu \nu}$ is related to (the matrix) $g_{\mu \nu}$ and what this has to do with determinants!

I ll try and get back to you

 

[Arman777]

First of all you must always have $\sqrt{-g}$, because $g<0$ due to the signature of the Lorentzian manifold.

Further think about, how (the matrix) $g^{\mu \nu}$ is related to (the matrix) $g_{\mu \nu}$ and what this has to do with determinants!

I can see that $det(g^{\mu \nu })= 1/g$

 

[vanhees71]

Then further note that
$$\mathrm{det} (g_{\mu \nu})=g=\frac{1}{4!} \Delta^{\mu_1\ldots \mu_4} \Delta ^{\nu_1 \ldots \nu_4} g_{\mu_1 \nu_1} \cdots g_{\mu_4 \nu_4}.$$
Here and in the following I use the notation
$$\Delta^{\mu_1 \ldots \mu_4}=\Delta_{\mu_1 \ldots \mu_r}$$
for the Levi-Civita symbols (which are NOT tensor components), i.e., the totally antisymmetric symbols with $\Delta^{0123}=\Delta_{0123}=+1$.

Now take the derivative $\partial_{\mu}$ which gives
$$\partial_{\mu} g = \frac{1}{3!} (\partial_{\mu} g_{\mu_1 \nu_1}) \Delta^{\mu_1\ldots \mu_4} \Delta ^{\nu_1 \ldots \nu_4} g_{\mu_2 \nu_2} \cdots g_{\mu_4 \nu_4}=:\tilde{g}^{\mu_1 \nu_1} \partial_{\mu} g_{\mu_1 \nu_1}.$$
But now let's think about
$$\tilde{g}^{\mu_1 \nu_1} = \frac{1}{3!} \Delta^{\mu_1\ldots \mu_4} \Delta ^{\nu_2 \ldots \nu_4} g_{\mu_2 \nu_1} \cdots g_{\mu_4 \nu_4}$$
is. To that end we evaluate
$$g_{\alpha \mu_1} \tilde{g}^{\mu_1 \nu_1}= \frac{1}{3!} \Delta^{\mu_1\ldots \mu_4} \Delta ^{\nu_1 \ldots \nu_4} g_{\mu_1 \nu_1} \cdots g_{\mu_4 \nu_4} g_{\mu_1 \alpha}.$$
In the last step I used the symmetry of the metric, $g_{\mu_1 \alpha}=g_{\alpha \mu_1}$. Now do the contraction wrt. the $\mu_k$, which leads to
$$g_{\alpha \mu_1} \tilde{g}^{\mu_1 \nu_1}=g \frac{1}{3!} \Delta_{\alpha \nu_2\cdots \nu_1} \Delta ^{\nu_1 \ldots \nu_4} = g \delta^{\nu_1}_{\alpha}.$$
From this you have
$$\tilde{g}^{\mu_1 \nu_1}=g g^{\mu_1 \nu_1}.$$
So you get
$$\partial_{\mu} g=g g^{\mu_1 \nu_1} \partial_{\mu} g_{\mu_1 \nu_1}.$$
So finally
$$\partial_{\mu} \ln (\sqrt{-g})=\frac{1}{2} \partial_\mu \ln(-g)=\frac{1}{2g} \partial_{\mu} g=\frac{1}{2} g^{\alpha \beta} \partial_{\mu} g_{\alpha \beta},$$
which is what you wanted to prove.

 

[Arman777]

Then further note that
$$\mathrm{det} (g_{\mu \nu})=g=\frac{1}{4!} \Delta^{\mu_1\ldots \mu_4} \Delta ^{\nu_1 \ldots \nu_4} g_{\mu_1 \nu_1} \cdots g_{\mu_4 \nu_4}.$$
Here and in the following I use the notation
$$\Delta^{\mu_1 \ldots \mu_4}=\Delta_{\mu_1 \ldots \mu_r}$$
for the Levi-Civita symbols (which are NOT tensor components), i.e., the totally antisymmetric symbols with $\Delta^{0123}=\Delta_{0123}=+1$.

Now take the derivative $\partial_{\mu}$ which gives
$$\partial_{\mu} g = \frac{1}{3!} (\partial_{\mu} g_{\mu_1 \nu_1}) \Delta^{\mu_1\ldots \mu_4} \Delta ^{\nu_1 \ldots \nu_4} g_{\mu_2 \nu_2} \cdots g_{\mu_4 \nu_4}=:\tilde{g}^{\mu_1 \nu_1} \partial_{\mu} g_{\mu_1 \nu_1}.$$
But now let's think about
$$\tilde{g}^{\mu_1 \nu_1} = \frac{1}{3!} \Delta^{\mu_1\ldots \mu_4} \Delta ^{\nu_2 \ldots \nu_4} g_{\mu_2 \nu_1} \cdots g_{\mu_4 \nu_4}$$
is. To that end we evaluate
$$g_{\alpha \mu_1} \tilde{g}^{\mu_1 \nu_1}= \frac{1}{3!} \Delta^{\mu_1\ldots \mu_4} \Delta ^{\nu_1 \ldots \nu_4} g_{\mu_1 \nu_1} \cdots g_{\mu_4 \nu_4} g_{\mu_1 \alpha}.$$
In the last step I used the symmetry of the metric, $g_{\mu_1 \alpha}=g_{\alpha \mu_1}$. Now do the contraction wrt. the $\mu_k$, which leads to
$$g_{\alpha \mu_1} \tilde{g}^{\mu_1 \nu_1}=g \frac{1}{3!} \Delta_{\alpha \nu_2\cdots \nu_1} \Delta ^{\nu_1 \ldots \nu_4} = g \delta^{\nu_1}_{\alpha}.$$
From this you have
$$\tilde{g}^{\mu_1 \nu_1}=g g^{\mu_1 \nu_1}.$$
So you get
$$\partial_{\mu} g=g g^{\mu_1 \nu_1} \partial_{\mu} g_{\mu_1 \nu_1}.$$
So finally
$$\partial_{\mu} \ln (\sqrt{-g})=\frac{1}{2} \partial_\mu \ln(-g)=\frac{1}{2g} \partial_{\mu} g=\frac{1}{2} g^{\alpha \beta} \partial_{\mu} g_{\alpha \beta},$$
which is what you wanted to prove.

Just one question why $\partial_{\mu} g \neq 0$ ? or can we say that $\partial_{\mu}\partial^{\mu} g = 0$

 

[vanhees71]

Why should $\partial_{\mu} g=0$? Of course if you are in SR you can always choose a global Minkowski-orthonormal basis. Then $g_{\mu \nu}=\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)=\text{const}$ and $\partial_{\rho} g_{\mu \nu}=0$ and thus also $\partial_{\mu} g=0$. In this case of course the covariant derivative coincides with the partial derivative, because all Christoffel symbols vanish too.

 

[Arman777]

In this case of course the covariant derivative coincides with the partial derivative, because all Christoffel symbols vanish too.

So its true that $\partial_{\mu}\partial^{\mu}g = 0$

 

[vanhees71]

How do you come to this conclusion?

 

[Arman777]

How do you come to this conclusion?

I did not actually :p Well I was doing another mathematical calculation about the $\nabla_{\mu}\nabla^{\mu} \Phi$ ,where $\Phi$ is a scalar function. In order to obtain the desired result it seems that $\partial_{\mu}\partial^{\mu}g$ must be 0 ? If you want I can share a separate question about it as well.

 

[vanhees71]

This we can answer pretty easily with what we've derived above so far. First take an arbitrary vector field $V^{\mu}$ and calculate the "divergence". We have
$$\nabla_{\nu} V^{\mu} = \partial_{\nu} V^{\mu} +{\Gamma^{\mu}}_{\nu \rho} V^{\rho}.$$
From that we find taking the trace
$$\nabla_{\mu} V^{\mu} = \partial_{\mu} V^{\mu} + {\Gamma^{\mu}}_{\mu \rho} V^{\rho}.$$
Now we use our formula for the contraction of the Christoffel symbol
$$\nabla_{\mu} V^{\mu} =\partial_{\mu} V^{\mu} + (\partial_{\rho} \ln (\sqrt{-g})] V^{\rho} = \partial_{\rho} V^{\rho} + \frac{1}{\sqrt{-g}} V^{\rho} \partial_{\rho} \sqrt{-g} = \frac{1}{\sqrt{-g}} \partial_{\rho} (\sqrt{-g} V^{\rho}).$$
Further we have
$$V_{\mu}=\partial_{\mu} \Phi \; \Rightarrow \; V^{\rho}=g^{\rho \mu} \partial_{\mu} \Phi.$$
From this we get,
$$\nabla_{\rho} \nabla^{\rho} \Phi=\frac{1}{\sqrt{-g}} \partial_{\rho} (\sqrt{-g} g^{\rho \mu} \partial_{\mu} \Phi).$$
So you get the so-called Laplace-Beltrami operator.

 

[Arman777]

In my calculations I did something like this

but it seems wrong ... But I see your point.

 

[vanhees71]

I don't understand, how you get from the 2nd to the 3rd expression. Of course $\nabla_{\nu} \Phi=\partial_{\nu}\ Phi$, and then you get the correct expression, the Laplace-Beltrami operator.

 

[Arman777]

2nd to the 3rd expression.

how so ?
By using this I am getting

$$\nabla_{\mu}(g^{\mu \nu}\nabla_{\nu}\Phi) = \frac{1}{\sqrt{g}}\partial_{\mu}[g^{\mu \nu}\partial_{\nu}(\sqrt{g}\Phi)]$$

but you are writing

$$\nabla_{\mu}(g^{\mu \nu}\nabla_{\nu}\Phi) = \frac{1}{\sqrt{g}}\partial_{\mu}[g^{\mu \nu}\sqrt{g}\partial_{\nu}(\Phi)]$$

there is a $\sqrt(g)$ difference. In my its inside the partial and in yours its outside..

 

[vanhees71]

As I said, I don't understand your step. Can you prove it?

 

[Arman777]

As I said, I don't understand your step. Can you prove it?

I have just put the definiton of the $\nabla_{\nu}\Phi = \frac{1}{\sqrt{g}} \partial_{\nu}(\sqrt{g}\Phi)$ inside the paranthesis ?

 

[Arman777]

https://en.wikipedia.org/wiki/List_...try#Christoffel_symbols,_covariant_derivative

The last equation is also interesting

 

[vanhees71]

I have just put the definiton of the $\nabla_{\nu}\Phi = \frac{1}{\sqrt{g}} \partial_{\nu}(\sqrt{g}\Phi)$ inside the paranthesis ?

No! The definition of $\nabla_{\mu} \phi=\partial_{\mu} \Phi$. $\Phi$ is a scalar field!

 

[Arman777]

oh shi*t. I understand it now okay. I was also saying there's something wrong but could not figure out why. Okay some things make more sense now after your approach in #11. Thanks a lot