# Is there any way to calculate Christoffel symbols of the second kind for spherical polar coordinates directly using metric tensor?

Determine the Christoffel symbols of the second kind for spherical polar coordinates.

From, $$ds^2=dr^2+r^2d\theta^2+r^2\sin\theta d\phi^2$$ We can easily calculate the metric tensor $$g_{ij}$$ and get,

$$g_{i j}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & r^{2} & 0 \\ 0 & 0 & r^{2} \sin ^{2} \theta \end{array}\right]$$ and $$g^{i j}=\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & r^{-2} & 0 \\ 0 & 0 & \left(r^{2} \sin ^{2} \theta\right)^{-1} \end{array}\right]$$

I knew the relation between Christoffel symbols of the second kind and metric tensor, $$\Gamma_{\alpha \beta}^{\mu}=\frac{1}{2} g^{\mu \lambda}\left(\frac{\partial g_{\lambda \alpha}}{\partial x^{\beta}}+\frac{\partial g_{\lambda \beta}}{\partial x^{\alpha}}-\frac{\partial g_{\alpha \beta}}{\partial x^{\lambda}}\right)$$

To calculate those symbols, we need to choose $$\alpha,\beta$$ and $$\mu$$ carefully. But is there any way to obtain those directly by using the matrix $$(g_{ij})$$? Because $$\Gamma_{i j}^{1},\Gamma_{i j}^{2}$$ and $$\Gamma_{i j}^{3}$$ seems to be related with the matrix.

\begin{aligned} &\Gamma_{i j}^{1}=\left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & -r & 0 \\ 0 & 0 & -r \sin ^{2} \theta \end{array}\right]\\ &\Gamma_{i j}^{2}=\left[\begin{array}{ccc} 0 & \frac{1}{r} & 0 \\ \frac{1}{r} & 0 & 0 \\ 0 & 0 & -\sin \theta \cos \theta \end{array}\right] \end{aligned} $$\Gamma_{i j}^{3}=\left[\begin{array}{ccc} 0 & 0 & \frac{1}{r} \\ 0 & 0 & \cot \theta \\ \frac{1}{r} & \cot \theta & 0 \end{array}\right]$$

• Not sure how you did it but you can compare the results with this. Commented Aug 19, 2022 at 7:37
• Aha, nice blog post. Thanks, @KurtG. Commented Aug 19, 2022 at 8:46
• That link is about a 2-dimensional spherical surface embedded in $\mathbb{R}^3$. Not to be confused with what is done here. Commented Aug 20, 2022 at 6:06
• @falamiw: typo metric Commented Aug 20, 2022 at 8:07

Yes. Smart observation, but also not quite.

What you can do is consider the dot product of basis and implicitly differentiate that.

$$\vec e_r.\vec e_r=1$$

Differentiate the above with $$r, \theta$$ and $$\phi$$ for the symbol relations.

You can get further ones by considering the dot product of radial vector with the other basis vectors like azimuth or polar and differentiating that too.

# Calculating $$\Gamma_{r \theta}^{\theta}$$:

WE have:

$$e_{\theta} \cdot e_{\theta} = r^2$$

Differentiating with $$r$$:

$$(\partial_r e_{\theta}) \cdot e_{\theta} =r \tag{2}$$

Now to actually get the Christoffels from the above, a small observation must be done. We have that:

$$\partial_r e_{\theta} = \Gamma_{r \theta}^r e_r + \Gamma_{r \theta}^{\theta} e_{\theta} + \Gamma_{r \theta}^{\phi} e_{\phi}$$

Dot both side with $$e_{\theta}$$:

$$\partial_r e_{\theta} \cdot e_{\theta} = \Gamma_{r \theta}^{\theta} e_{\theta} \cdot e_{\theta}$$

Or,

$$\frac{(\partial_r e_{\theta}) \cdot e_{\theta} }{e_{\theta} \cdot e_{\theta}} = \Gamma_{r \theta}^{\theta}$$

Now , put $$2$$ in this and use that $$e_{\theta} \cdot e_{\theta} = r^2$$. Done!

I use an identity from page-275 of Pavel Grinfeld's Tensor Calculus book,

$$Z_i^{\alpha}$$ is the basis on the surface, $$S^{\gamma}$$ are the parameters of the surface ( the chart coordinates).

• If someone could format this that'd be nice ( I wrote on phone) Commented Aug 19, 2022 at 7:10
• I think adding an example would be a great help to grasp the answer @Beautifullyirrational. Thanks for your response. Commented Aug 19, 2022 at 8:50
• @falamiw here u go Commented Aug 19, 2022 at 9:04
• You need to use the contravariant basis as well, like this $$\Gamma^j_{hk}=\mathbf{e}^j\cdot \frac{\partial \mathbf{e}_h}{\partial x^k}$$ Commented Aug 19, 2022 at 9:22
• Why tho? Afaik christoffel is just the components of the basis with the coordinates, so directly reading off here should be okie. @ContraKinta Commented Aug 19, 2022 at 9:24