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