Connection 1-forms to Christoffel symbols

  • #1
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.
 

Answers and Replies

  • #2
Could it be that I'm so use to seeing people set r=const.=1 that I usually do not see the r?
 
  • #3
Orodruin
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Since your connection one-form is dimensionless (basically as a consequence of using an orthonormal basis), the physical dimension of your Christoffel symbols must be the inverse of the physical dimension of the corresponding ##\sigma^i##. For both ##\sigma^2## and ##\sigma^3##, the physical dimension is ##\mathsf L## and hence the corresponding Christoffel symbols must have physical dimension ##\mathsf L^{-1}##. This means the Christoffel symbols have to be proportional to ##1/r## as ##r## is the only dimensionful parameter you have.

Also note that what you are computing are the Christoffel symbols of the orthonormal basis you have chosen. I would reserve ##\Gamma^\theta_{\phi\phi}## etc for the Christoffel symbols in holonomic basis.
 
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  • #4
lavinia
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Since your connection one-form is dimensionless (basically as a consequence of using an orthonormal basis), the physical dimension of your Christoffel symbols must be the inverse of the physical dimension of the corresponding ##\sigma^i##. For both ##\sigma^2## and ##\sigma^3##, the physical dimension is ##\mathsf L## and hence the corresponding Christoffel symbols must have physical dimension ##\mathsf L^{-1}##. This means the Christoffel symbols have to be proportional to ##1/r## as ##r## is the only dimensionful parameter you have.

Also note that what you are computing are the Christoffel symbols of the orthonormal basis you have chosen. I would reserve ##\Gamma^\theta_{\phi\phi}## etc for the Christoffel symbols in holonomic basis.
Could you explain this more?
 
  • #5
Orodruin
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Could you explain this more?
I assume you mean the first part? I guess dimensional analysis is a concept more used in physics. Mathematicians would probably prefer to refer to it as a scaling property when ##r \to kr##. Under this transformation the connection 1-forms do not change, but the expressions for the dual basis vectors ##\sigma^2## and ##\sigma^3## do. In order to maintain the relation between the connection 1-forms, the dual basis vectors, and the Christoffel symbols, the Christoffel symbols ##\Gamma^i_{j2}## and ##\Gamma^i_{j3}## must scale as ##1/r##.
 
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  • #6
lavinia
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Thanks. I see now. Nice argument.
 

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