- #1
WendysRules
- 37
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Let the metric be defined as ##ds^2=dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2##
Through some calculations, we then see that our connection one forms are ##\omega_{12} = -d \theta## and ##\omega_{21}= d\theta##, ##\omega_{13} = -sin\theta d \phi## and ##\omega_{31} = sin\theta d\phi## ##\omega_{23}=cos\theta d\phi## and ##\omega_{32} = -cos\theta d\phi## (for work, look at: https://www.physicsforums.com/threa...ctor-in-spherical-coords.903231/#post-5687781)
We can then use the fact that ##\omega^i_j = \Gamma^i_{jk} \sigma^k## to compute the Christoffel symbols.
Notice that if ##i = j## all the Christoffel symbols will vanish due to ##\omega^i_i = 0## thus, right off the bat we know that ##\Gamma^1_{11} = \Gamma^1_{12} = \Gamma^1_{13} = \Gamma^2_{21} = \Gamma^2_{22} = \Gamma^2_{23} = \Gamma^3_{31} = \Gamma^3_{32} = \Gamma^3_{33} = 0##
With metric compabitibility, we know that ##\Gamma^i_{jk} = -\Gamma^i_{kj}## thus... ## \Gamma^1_{21} = \Gamma^1_{31} = \Gamma^2_{12} = \Gamma^2_{32} = \Gamma^3_{13} = \Gamma^3_{23} = 0##
So onto the ones we can calculate, ##\omega^1_2 = -d \theta = \Gamma^1_{21} \sigma^1 + \Gamma^1_{22} \sigma^2 + \Gamma^1_{23} \sigma^3 = 0 + \Gamma^1_{22} r d\theta + \Gamma^1_{23} r\sin\theta d\phi##
The only way for this equation to be true is for##\Gamma^1_{22} = \frac{-1}{r} ## and ##\Gamma^1_{23} = 0 ## We then can set ##\Gamma^1_{32} = 0 ##
Next, ##\omega^1_3 = -sin\theta d \phi = \Gamma^1_{31} \sigma^1 + \Gamma^1_{32} \sigma^2 + \Gamma^1_{33} \sigma^3 = 0 + 0 + \Gamma^1_{33} r\sin\theta d\phi## therefore, we see that ##\Gamma^1_{33} = \frac{-1}{r}##Next, ##\omega^2_3 = cos\theta d\phi = \Gamma^2_{31} \sigma^1 + \Gamma^2_{32} \sigma^2 + \Gamma^2_{33} \sigma^3 = \Gamma^2_{31} dr + \Gamma^2_{33} r\sin\theta d\phi## To make this true, we then see that ##\Gamma^2_{31} = 0 ## and ##\Gamma^2_{33} = \frac{\cos\theta}{r\sin\theta} = \frac{\cot\theta}{r} ##
Which raises a red flag! I've never see a Christoffel symbol where we have that r in the denominator for ##\Gamma^{\theta}_{\phi \phi}## so if anyone can point out my error, i'd appreciate it.
Through some calculations, we then see that our connection one forms are ##\omega_{12} = -d \theta## and ##\omega_{21}= d\theta##, ##\omega_{13} = -sin\theta d \phi## and ##\omega_{31} = sin\theta d\phi## ##\omega_{23}=cos\theta d\phi## and ##\omega_{32} = -cos\theta d\phi## (for work, look at: https://www.physicsforums.com/threa...ctor-in-spherical-coords.903231/#post-5687781)
We can then use the fact that ##\omega^i_j = \Gamma^i_{jk} \sigma^k## to compute the Christoffel symbols.
Notice that if ##i = j## all the Christoffel symbols will vanish due to ##\omega^i_i = 0## thus, right off the bat we know that ##\Gamma^1_{11} = \Gamma^1_{12} = \Gamma^1_{13} = \Gamma^2_{21} = \Gamma^2_{22} = \Gamma^2_{23} = \Gamma^3_{31} = \Gamma^3_{32} = \Gamma^3_{33} = 0##
With metric compabitibility, we know that ##\Gamma^i_{jk} = -\Gamma^i_{kj}## thus... ## \Gamma^1_{21} = \Gamma^1_{31} = \Gamma^2_{12} = \Gamma^2_{32} = \Gamma^3_{13} = \Gamma^3_{23} = 0##
So onto the ones we can calculate, ##\omega^1_2 = -d \theta = \Gamma^1_{21} \sigma^1 + \Gamma^1_{22} \sigma^2 + \Gamma^1_{23} \sigma^3 = 0 + \Gamma^1_{22} r d\theta + \Gamma^1_{23} r\sin\theta d\phi##
The only way for this equation to be true is for##\Gamma^1_{22} = \frac{-1}{r} ## and ##\Gamma^1_{23} = 0 ## We then can set ##\Gamma^1_{32} = 0 ##
Next, ##\omega^1_3 = -sin\theta d \phi = \Gamma^1_{31} \sigma^1 + \Gamma^1_{32} \sigma^2 + \Gamma^1_{33} \sigma^3 = 0 + 0 + \Gamma^1_{33} r\sin\theta d\phi## therefore, we see that ##\Gamma^1_{33} = \frac{-1}{r}##Next, ##\omega^2_3 = cos\theta d\phi = \Gamma^2_{31} \sigma^1 + \Gamma^2_{32} \sigma^2 + \Gamma^2_{33} \sigma^3 = \Gamma^2_{31} dr + \Gamma^2_{33} r\sin\theta d\phi## To make this true, we then see that ##\Gamma^2_{31} = 0 ## and ##\Gamma^2_{33} = \frac{\cos\theta}{r\sin\theta} = \frac{\cot\theta}{r} ##
Which raises a red flag! I've never see a Christoffel symbol where we have that r in the denominator for ##\Gamma^{\theta}_{\phi \phi}## so if anyone can point out my error, i'd appreciate it.