# Derivation of D-dimensional Laplacian in spherical coordinates

I'm trying to derive the D-dimensional laplacian over Euclidean space for a function $$f$$ that is invariant under D-dimensional Euclidean rotations.

Specifically i'm trying to go from the equation

$$\Delta_D \phi = U'(\phi)$$

to the equation

$$\frac{d^2\phi}{dr^2}+\frac{D-1}{r}\frac{d\phi}{dr} = U'(\phi)$$

where $$r = (x_1^2+x_2^2+x_3^2+\dots+x_D^2)^{1/2}$$. I'm a physicist and currently I don't have much knowledge about differential geometry and operators over manifolds, but still i wanted to know how, in a rigorous manner, to derive that equation under that change of coordinates.

Searching on the internet i found that the general form for the laplacian is given by the Laplace-Beltrami operator

$$\Delta_D \phi= \frac{1}{\sqrt{\det g}}\partial_i\left(\sqrt{\det g}g^{ij}\partial_j\phi\right)$$

but i don't know how the metric tensor is in the coordinates specified. I know that in Euclidian space is just a Kroneker delta, but what about spherical coordinates? On the Wikipedia article gives the result right away without any explanation.

I'm really curious to see how to derive it. I didn't find any explanation with computation in my google search.

Here is a longer derivation that requires mostly basic calculus.

The Laplacian is invariant under rotations. Specifically, if $$f:\mathbb R^n\rightarrow \mathbb R^n$$ and $$\nabla^2 f(x)\;=\alpha$$, then $$\nabla^2 g\; (R^{-1} x)=\alpha$$ where $$R$$ is any rotation matrix ($$R^TR=I$$ and $$\det(R)=1$$) and $$g(x) := f(Rx)$$.

Now suppose that $$f:\mathbb R^n\rightarrow \mathbb R^n$$ has the property that $$f(x)= h(||x||)$$ where $$h:\mathbb [0,\infty) \rightarrow \mathbb R$$, $$h$$ is twice differentiable, and $$||x||=\sqrt{\sum_i x_i^2}.$$

At $$x$$, we can form an orthonormal basis $$\{x/||x||, v_2, v_3, \ldots v_{n-1}\}$$. Now $$\nabla^2 f(x) = \sum_{i=1}^n \frac{\partial^2 f}{\partial x_i^2}(x)$$ and by invariance under rotation, we can use the basis $$\{x/||x||, v_2, v_3, \ldots v_{n-1}\}$$ to get $$$$(1)\quad\quad\nabla^2 f(x)= h''(||x||) + \sum_{i=2}^n p_i$$$$ where $$p_i:= k_i''(0)$$ and $$k_i(\epsilon) := f(x+\epsilon v_i)$$.

The fact that $$f$$ only depends on the radius implies that all of the $$k_i$$ are the same and for $$i=2, 3, \ldots n$$, $$f(x+\epsilon v_i) = k_i(\epsilon) = k(\epsilon) = f(x+ \epsilon v_2)=h(\sqrt{||x||^2+\epsilon^2}).$$

Let $$r=||x||$$. Using calculus, $$k'(\epsilon) = h'(\sqrt{r^2+\epsilon^2})\frac{\epsilon}{\sqrt{r^2+\epsilon^2}}, \quad \mathrm{and}$$ $$k''(\epsilon) = h''(\sqrt{r^2+\epsilon^2})\frac{\epsilon^2}{r^2+\epsilon^2} + h'(\sqrt{r^2+\epsilon^2})\frac{\sqrt{r^2+\epsilon^2} -\epsilon \frac{\epsilon}{\sqrt{r^2+\epsilon^2}} }{r^2+\epsilon^2}.$$ Thus $$p_i:= k''(0) = h''(r)\cdot 0 + h'(r)\frac{r}{r^2}=h'(r)/r$$. Applying this to equation (1) gives $$\nabla^2f(x)= h''(||x||) + \sum_{i=2}^n p_i = h''(r) + (n-1)\frac{h'(r)}{r}.$$

• Yes, even this i can understand! Very straight forward, thanks Aug 19, 2019 at 18:53

Since you're a physicist, I thought a variational derivation might be appealing. Consider the functional $$F[\phi]=\int \left[\frac12(\nabla \phi)^2+U(\phi)\right] dV.$$ Then the functional derivative is $$\frac{\delta}{\delta \phi}F[\rho]=\frac{\partial U}{\partial \phi}-\nabla^2 \phi =U'(\phi)-\Delta_D \phi,$$ so your first equation is equivalent to $$\delta F[\phi]/\delta \phi=0$$. But $$\phi$$ is a function of $$r$$ alone, so we can integrate out the irrelevant angular degrees of freedom to obtain $$F[\phi]=C_D\int \left[\frac12(\partial_r \phi)^2+U(\phi)\right] r^{D-1}dr$$ where $$C_D$$ is some $$D$$-dependent multiplicative constant. We may then compute the functional derivative once again as $$\dfrac{\delta}{\delta \phi}F[\phi]=C_n \left[U'(\phi)r^{D-1}-\partial_r(r^{D-1} \partial_r\phi)\right]=C_n r^{D-1}\left[U'(\phi)-\frac{D-1}{r}\frac{d\phi}{dr}-\frac{d^2\phi}{dr^2} \right].$$ For this variational derivative to vanish again, we must therefore have $$\Delta_D \phi = U'(\phi) = \frac{d^2\phi}{dr^2}+\frac{D-1}{r}\frac{d\phi}{dr}$$ which was the result to be derived.

• I love this! Thank you so much. This way is much clearer to me. Maybe in the future i'll go in depth in the differential geometry way. Aug 19, 2019 at 17:29