The Affine Connection components

[Radiohannah]

Hello!

I have a few questions about how the Affine connection works.

I know the geodesic equation;


$$\Gamma^{\lambda}_{\mu \nu} = \frac{\partial x^{\lambda}}{\partial \xi^{\alpha}} \frac{\partial^{2} \xi^{\alpha}}{\partial x^{\nu} \partial x^{\mu}}$$

So, for example, if I had the expression

$$\ddot{t} + \frac{A'}{A} \frac{dr}{d \tau} \frac{dt}{d \tau} = 0$$

I can read off

$$\Gamma^{t}_{rt} = \frac{A'}{2A}$$

Where the expression is divided by 2 because they are off diagonal terms, since the subscript indices are not the same.

I am not sure how this then applies to the actual


$$\Gamma_{t}$$


matrix. Should the


$$\Gamma^{t}_{rt} = \Gamma^{t}_{tr}$$ terms be exactly the same? Because in an example the

$$\Gamma^{t}_{tr} = \frac{A'}{2A}$$

Whereas the

$$\Gamma^{t}_{rt} = \frac{A'}{2B}$$

And I don't understand why there should be a different letter on the denominator, I had always thought that if they are off-diagonal terms they should be identical?? Is that not the case?

And finally, as for all of the other components in the matrix, would these all just be 0? I think that because, I know that the Affine connection is not a tensor, and so will not have diagonal "1" terms like the metric tensor, so all other terms must be 0? Is that true?
ie
$$\Gamma^{t}_{tt} = 0$$ ?

Many thanks in advance!

Hannah

 

[jfy4]

The Christoffell symbols are symmetric in the lower two indicies:

$$\Gamma^{a}{}_{bc}=\Gamma^{a}{}_{cb}$$

So they should be the same. Check the reference.

 

[Mentz114]

The Christoffell symbols are symmetric in the lower two indicies:

It follows from the first equation by the commutativity of partial differentiation.