Laplacian in n dimensions

Question

There is a ready formula for Laplacian in hyper spherical coordinates, but I want to know how to get the radial derivative from this form $$\frac{1}{r^{n-1}} \frac {\partial} {\partial r} (r^{n-1} \frac{\partial}{\partial r}) \tag{1}$$ to the form $$\frac{(n-1)}{r^2}\sum_{k=1}^n x_k \frac {\partial}{\partial x_k}+\sum_{j,k=1}^n \frac{x_jx_k}{r^2} \frac{\partial^2}{\partial x_j\partial x_k}, \tag{2}$$ and also the derivation of the angular derivative?

Answer 1

I will show (1) can be written as (2).

Since $\frac{\partial}{\partial r}=\frac{x^i}{r}\frac{\partial}{\partial x^i}$, we can replace $\partial/\partial r$ in (1) with $x^i$ derivatives. Differentiating a smooth function $u=u(x)$ gives three terms: \begin{align} \frac{x^j}{r^{n}}\frac{\partial}{\partial x^j}(r^{n-2}\cdot x^k\cdot u_k)&=\frac{x^j}{r^n}((n-2)r^{n-3}\frac{x^j}{r}\cdot x^k\cdot u_k+r^{n-2}\cdot \delta_j^k\cdot u_k+r^{n-2}\cdot x^k\cdot u_{jk})\\ &=(n-2)\frac{x^k}{r^2}u_k+\frac{x^j}{r^2}u_j+\frac{x^jx^k}{r^2}u_{jk}\\ &=(n-1)\frac{x^k}{r^2}u_k+\frac{x^jx^k}{r^2}u_{jk}. \end{align} Here, $u_k=\partial u/\partial x^k,u_{jk}=\partial^2u/\partial x^j\partial x^k$, and $\delta^j_k$ is the Kronecker Delta, and we implicitly sum over repeated indicices $j,j$ and $k,k$. This is the desired result.

EDIT:

To verify $\frac{\partial}{\partial r}=\frac{x^i}{r}\frac{\partial}{\partial x^i}$, we will use the chain rule in spherical coordinates: $$ \frac{\partial }{\partial r}=\frac{\partial x^i}{\partial r}\frac{\partial}{\partial x^i}, $$ so it suffices to show $\partial x^i/\partial r=x^i/r$. But this is clear from how spherical coordinates are defined. In case these expressions are not explicitly available, we could also argue as follows. The function $x^i/r$ represents the position vector restricted to the unit sphere, so it can only depend on the angles, i.e. $x^i/r=f^i(\Theta)$. This means $x^i=rf^i(\Theta)$.