In cartesian coordinates since the metric components $g^{ab}$ are constant, we know that $\partial_c = \nabla_c$ where the RHS is the covariant derivative.

So we can write the laplacian in cartesian coordinates as $\nabla^2 \phi = g^{ab} \nabla_a \nabla_b \phi$. I am told that this is invariant under change of coordinates. How would I calculate this quantity in polar coordinates?

The best I've got is $\nabla^2 \phi = \frac{1}{r^2}\nabla_{\theta} \nabla_{\theta} \phi + \nabla_{r} \nabla_{r} \phi$.

I know what the laplacian should look like under polar, but I cannot see how to simplify further from here. I can find the christoffel symbols but how do I apply them to a scalar field?


1 Answer 1


Assuming a metric-compatible connection (which is what we use in GR), then we can write $$\nabla^2 \phi = g^{ab}\nabla_a \nabla_b \phi = g^{ab}\nabla_a (\partial_b\phi)$$ $$= g^{ab}\bigg(\partial_a\partial_b \phi - \Gamma^c_{ab}\partial_c \phi\bigg)$$ Where we have used the fact that $\phi$ is a scalar field and $\nabla\phi$ is a covector field. Given the Levi-Civita connection coefficients, and metric components, this can be expanded straightforwardly.

To see the conditions under which this formula is valid, note that the coordinate-independent action of the Laplace operator on a scalar field is to take the divergence of its gradient, i.e. $$\nabla^2 \phi \equiv \mathrm{div}\big(\mathrm{grad}(\phi)\big)$$ The (vector) gradient of a scalar field is a vector field with components $$\mathrm{grad}(\phi)^a \equiv g^{ab}\partial_b \phi$$ On the other hand, the (covariant) divergence of a vector field is given by $$\mathrm{div}(V) = \nabla_a V^a = \partial_a V^a + \Gamma^a_{a b}V^b$$

Putting those pieces together, we have that $$\nabla^2\phi = \mathrm{div}\big(\mathrm{grad}(\phi)\big) = \partial_a\big[\mathrm{grad}(\phi)^a\big] + \Gamma^a_{a b} \big[\mathrm{grad}(\phi)^b\big]$$ $$= \partial_a \big[g^{ab}\partial_b \phi\big] + \Gamma^a_{ab} \big[g^{bc} \partial_c \phi\big] = g^{ab}\partial_a\partial_b \phi + \big(\partial_a g^{ab}+ \Gamma^a_{ac} g^{cb}\big)\partial_b\phi$$ where we have sneakily relabeled the indices in the last line. If you know what the metric components and connection coefficients are, then you can just plug them in and go from there. In Cartesian coordinates, the entire second term vanishes (the $g$'s are constant and the $\Gamma$'s are zero), but in general this is of course not true.

If we're working with a metric-compatible connection, then $$\nabla_a g^{bc} = \partial_a g^{bc} + \Gamma^b_{ad} g^{dc} + \Gamma^c_{ad} g^{bd} = 0$$ for all $a,b,c$. In particular, $$\nabla_a g^{ab} = \partial_a g^{ab} + \Gamma^a_{ac} g^{cb} + \Gamma^b_{ac} g^{ac} = 0 \implies \partial_a g^{ab} + \Gamma^a_{ac} g^{cb} = - \Gamma^b_{ac} g^{ac}$$ and so our expression can be rewritten $$\nabla^2 \phi = g^{ab}\partial_a\partial_b \phi - \Gamma^b_{ac} g^{ac}\partial_b \phi = g^{ab} \nabla_a \partial_b \phi = g^{ab}\nabla_a \nabla_b \phi$$ where we've used the fact that $\nabla_b\phi = \partial_b \phi$ for a scalar field.

  • $\begingroup$ Ahh I completely forgot that by taking the first derivative you get a tensor! Thank you $\endgroup$
    – Governor
    Feb 23 at 18:24

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