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I know that in polar coordinates, it is $\frac{\partial \,{{\mathbf{e}}_{r}}}{\partial \theta }={{\mathbf{e}}_{\theta }}$ and $\frac{\partial \,{{\mathbf{e}}_{\theta }}}{\partial \theta }=-{{\mathbf{e}}_{r}}$ where ${{\mathbf{e}}_{r}}$ and ${{\mathbf{e}}_{\theta }}$ are the basis unit vectors.

Anyway, using the definition of the connection coefficients (Christoffel symbols) it should also be

$\frac{\partial \,{{\mathbf{e}}_{r}}}{\partial \theta }={{\Gamma }^{r}}_{r\theta }\,{{\mathbf{e}}_{r}}+{{\Gamma }^{\theta }}_{r\,\theta }\,{{\mathbf{e}}_{\theta }}$ and $\frac{\partial \,{{\mathbf{e}}_{\theta }}}{\partial \theta }={{\Gamma }^{r}}_{\theta \,\theta }\,{{\mathbf{e}}_{r}}+{{\Gamma }^{\theta }}_{\theta \,\theta }\,{{\mathbf{e}}_{\theta }}$

And since it is ${{\Gamma }^{r}}_{\theta \,\theta }=-r$ , ${{\Gamma }^{\theta }}_{r\,\theta }=\frac{1}{r}$ , ${{\Gamma }^{r}}_{r\,\theta }=0$, ${{\Gamma }^{\theta }}_{\theta \,\theta }=0$ (calculated with the metric) it should be

$\frac{\partial \,{{\mathbf{e}}_{r}}}{\partial \theta }=\frac{1}{r}{{\mathbf{e}}_{\theta }}$ and $\frac{\partial \,{{\mathbf{e}}_{\theta }}}{\partial \theta }=-r\,{{\mathbf{e}}_{r}}$

Where am I wrong?

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1 Answer 1

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You are using two different sets of basis vectors. $\frac{\partial\mathbf e_r}{\partial \theta} = \mathbf e_\theta$ and $\frac{\partial \mathbf e_\theta}{\partial \theta} = -\mathbf e_r$ hold for the orthonormal polar basis, in which the metric takes the form

$$g_{ij} = \pmatrix{1 & 0 \\ 0 & 1}$$

The connection coefficients you quote arise from the polar coordinate basis $\left\{\frac{\partial}{\partial r},\frac{\partial}{\partial \theta}\right\}$ which is not orthonormal, and in which the metric takes the form

$$g_{ij} = \pmatrix{1 & 0 \\ 0 & r^2}$$

The two bases are related via $\mathbf e_r = \frac{\partial}{\partial r}$ and $\mathbf e_\theta = r\frac{\partial }{\partial \theta}$.


An important thing to recognize is that the basis $\left\{\frac{\partial}{\partial r},\frac{\partial}{\partial \theta}\right\}$ arises naturally as the basis induced by the polar coordinates $(r,\theta)$. On the other hand, the orthonormal basis $\{e_r,e_\theta\}$ is not induced by a coordinate system; there is no set of coordinates $(u,v)$ such that $e_r = \frac{\partial}{\partial u}$ and $e_\theta =\frac{\partial}{\partial v}$. This is an example of a non-holonomic basis.

The reason that this is important is that in your first pass through GR, you will likely start off by using holonomic bases exclusively. Accidentally using a non-holonomic basis can lead to some apparent contradictions.

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    $\begingroup$ yes, and just to be clear: the vectors in the two sets have same directions but different sizes $\endgroup$ Sep 12, 2020 at 19:06
  • $\begingroup$ @AndrewSteane Yes, thanks. I've added to my answer to make that more clear. $\endgroup$
    – J. Murray
    Sep 12, 2020 at 20:29
  • $\begingroup$ j-murray, @AndrewSteane Thanks. It's all more clear now. I didn't expect such subtleties between the polar coordinate basis vectors (not unitary) and the polar unit vectors (not a coordinate basis). Now I know I must be careful...Many thanks. It's all more clear now. I didn't expect such subtleties in the difference between the polar coordinate basis vectors (not unitary) and the polar unit vectors (not a coordinate basis). Now I know I must be careful... $\endgroup$
    – Luca M
    Sep 13, 2020 at 11:54
  • $\begingroup$ This other related post can be useful: physics.stackexchange.com/questions/198280/… $\endgroup$
    – Luca M
    Sep 13, 2020 at 12:12

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