- #1
Radiohannah
- 47
- 0
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
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
