Connection 1-forms to Christoffel symbols

[WendysRules]

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.

 

[WendysRules]

Could it be that I'm so use to seeing people set r=const.=1 that I usually do not see the r?

 

[Orodruin]

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.

 

[lavinia]

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?

 

[Orodruin]

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 $1/r$.

 

[lavinia]

Thanks. I see now. Nice argument.