For my course in General Relativity I am given the problem to find the expressions for the gradient, laplacian, divergence and curl in spherical coordinates using covariant derivatives.

I have tried multiple ways to arrive at the answer but I have yet to find an answer that I can fully understand. Here is what I tried, let's start of with the gradient of an arbitrary scalar function $f$. \begin{equation} \vec{\nabla}f = \nabla^if \hat{e}_i = \partial^i f\hat{e}_i = g^{ij} \partial_j f\hat{e}_i \end{equation} In this equation $\hat{e}_i$ is a basis vector of the tangent space. Since $f$ is a scalar function, the covariant derivative reduces to just the partial derivative. Working this out reveals we lack a factor $\sqrt{g_{ij}}$. An answer to a question similar to mine, Gradient, divergence and curl with covariant derivatives, revealed to me that the basis vectors in the tangent space are not orthonormal, $\hat{e}_i \hat{e}_j = g_{ij}$. Is this a definition I can just assume or is there some proof behind it?

To get the normalized unit vectors in polar coordinates we will thus have to change these unit vectors to: $\hat{e'}_i = \frac{\hat{e}_i}{\sqrt{g_{ii}}}$. Substituting this in our equation for the gradient and we find the missing factor and get the correct result. This I can understand.

Continuing to the divergence (Laplacian is just the divergence of the gradient, so I'll cover that later) of an arbitrary vector $\vec{V}$: \begin{equation} \vec\nabla\cdot\vec V = \nabla_i V^i \end{equation} This equation is correct, but the basis of $\vec V$ here is also the unnormalized basis $\hat{e}_i$, $\vec V=V^i \hat{e}_i$ while the divergence in spherical coordinates has derivatives of the vector w.r.t. the normalized basis. We can thus again substitute the normalized basis to get $\vec V = V^i \sqrt{g_{ii}} \hat{e'}_i$. The normalized vector components are thus equal to $\bar V^i = V^i \sqrt{g_{ii}}$. We can now rewrite this and substitute in the equation for the diverence and get: \begin{equation} \vec\nabla\cdot\vec V = \nabla_i \frac{\bar V^i}{\sqrt{g_{ii}}} \end{equation} Which yield the desired equation for spherical coordinates.

Applying the divergence on the gradient to get the Laplacian is quite straightforward and yields the correct equation.

Now comes the curl. We can write the curl as: \begin{equation} \vec\nabla\times\vec V = (\vec\nabla\times\vec V)^i \hat e_i = g^{ik}(\vec\nabla\times\vec V)_k \hat e_i = g^{ik} \epsilon_{klm} \nabla^lV^m \hat e_i = g^{ik} g^{ln} \epsilon_{klm} \nabla_nV^m \hat e_i \end{equation}

It seems logical to me to now both apply the normalization on the unit vector as we did with the gradient as well as change to the normalized basis of our vector: \begin{equation} g^{ik} g^{ln} \epsilon_{klm} \nabla_nV^m \hat e_i = g^{ik} g^{ln} \sqrt{\frac{g_{ii}}{g_{mm}}} \epsilon_{klm} \nabla_n \bar V^m \hat e'_i = g^{ii} g^{nn} \sqrt{\frac{g_{ii}}{g_{mm}}} \epsilon_{inm} \nabla_n \bar V^m \hat e'_i \end{equation} In the last step we used that the metric is diagonal.

Let us now compute the result for $i=1$

\begin{equation} g^{rr} g^{nn} \sqrt{\frac{g_{rr}}{g_{mm}}} \epsilon_{rnm} \nabla_n \bar V^m \hat e'_i = \frac{1}{r^2}\frac{1}{r\sin(\theta)} \times (\partial_\theta \bar V^\phi + \cot(\theta)\bar V^\phi) - \frac{1}{r^2\sin^2(\theta)}\frac{1}{r} \times(\partial_\phi \bar V^\theta - \sin(\theta)\cos(\theta)\bar V^\phi) \end{equation}

Two things are wrong here, one being a missing factor of $r^2\sin(\theta)$ and the other being the last term which follows from the connection $\Gamma^\theta_{\phi\phi} = -\sin(\theta)\cos(\theta)$. If this term is absent then the only difference would be the factor.

For $i=2$ we have: \begin{equation} g^{\theta\theta} g^{nn} \sqrt{\frac{g_{\theta\theta}}{g_{mm}}} \epsilon_{\theta nm} \nabla_n \bar V^m \hat e'_i = \frac{1}{r^2}\frac{1}{r^2\sin^2(\theta)} r \times (\partial_\phi \bar V^r - r\sin^2(\theta) \bar V^\phi) - \frac{1}{r^2}\frac{1}{\sin(\theta)} \times (\partial_r \bar V^\phi + \frac{1}{r}\bar V^\phi) \end{equation}

This again almost works out except we are again missing the same factor $r^2\sin(\theta)$ as well as the connection $\Gamma^r_{\phi\phi}$ yielding an unwanted term.

Lastly we have $i=3$ which yields:

\begin{equation} g^{\phi\phi} g^{nn} \sqrt{\frac{g_{\phi\phi}}{g_{mm}}} \epsilon_{\phi nm} \nabla_n \bar V^m \hat e'_i = \frac{1}{r^2\sin^2(\theta)} \sin(\theta) \times (\partial_r \bar V^\theta + \frac{1}{r} \bar V^\theta) - \frac{1}{r^2\sin^2(\theta)}\frac{1}{r^2} r\sin(\theta) \times (\partial_\theta \bar V^r - r \bar V^\theta) \end{equation}

We are again missing the factor $r\sin^2(\theta)$ and again have an unwanted term $\Gamma^r_{\theta\theta}$.

I notice the unwanted connection terms all contain the two equal indices downstairs. Can someone explain to me both where the factor $r^2\sin(\theta)$ I am missing should come from, and tell me where my mistake is in the connection terms.


1 Answer 1


I doubt that you are still worrying about this, but I ran into the same issues when attempting this problem, so I wanted to post my answer for any future readers.

Firstly, the missing factor of $r^2 \sin\theta$ comes from the Levi-Civita tensor $\epsilon$, which is related to the Levi-Civita symbol $\tilde{\epsilon}$ by $$ \epsilon_{\mu_{1}\cdots\mu_{n}}=\sqrt{|\det g|}\tilde{\epsilon}_{\mu_{1}\cdots\mu_{n}} $$ In this case, $\sqrt{|\det g|}=r^{2}\sin\theta$.

Secondly, the unwanted appearance of various connection terms is due to the tempting but incorrect statement $$ \nabla_{n}\bar{V}^{m}=\partial_{n}\bar{V}^{m}+\Gamma^{m}_{nk}\bar{V}^{k} $$ This isn't right because $\bar{V}$ is not a (rank 1 contravariant) tensor. Instead, we must use $$ \begin{aligned} \nabla_{n}\bar{V}^{m} &=\sqrt{|g_{mm}|}\nabla_{n}V^{m}\\ &=\sqrt{|g_{mm}|}(\partial_{n}V^{m}+\Gamma^{m}_{nk}V^{k})\\ &=\sqrt{|g_{mm}|}\partial_{n}\left(\frac{\bar{V}^{m}}{\sqrt{|g_{mm}|}}\right)+\sqrt{\left\lvert\frac{g_{mm}}{g_{kk}}\right\rvert}\Gamma^{m}_{nk}\bar{V}^{k} \end{aligned} $$

  • $\begingroup$ Thank you for providing the answer! It is indeed long past the due date of this question, but I am amazed it got solved this long after posting the problem. I've marked your answer so hopefully you can help people in the future with it :) $\endgroup$
    Mar 24, 2022 at 8:18

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