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.


2 Answers 2


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 \begin{equation} (1)\quad\quad\nabla^2 f(x)= h''(||x||) + \sum_{i=2}^n p_i \end{equation} 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 $$ \nabla^2f(x)= h''(||x||) + \sum_{i=2}^n p_i = h''(r) + (n-1)\frac{h'(r)}{r}. $$

  • $\begingroup$ Yes, even this i can understand! Very straight forward, thanks $\endgroup$ 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.

  • $\begingroup$ 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. $\endgroup$ Aug 19, 2019 at 17:29

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