# 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.

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. – Davide Morgante Aug 19 at 17:29

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 – Davide Morgante Aug 19 at 18:53