Spherical symmetry math

Question

For spherical symmetry how the last four equations calculations is done? ccan you explain please?

For reference see the equations 44

Answer 1

We have $\rho(x) = \bigl(\sum_i x_i^2\bigr)^{1/2}$, hence \begin{align*} \partial_i \rho(x) &= \frac 1{2(\sum_j x_j^2)^{1/2}}\cdot 2x_i = \frac{x_i}{\rho(x)}\\ \partial_i^2\rho(x) &= \frac{\rho^2(x) - x_i^2}{\rho^3(x)} \end{align*} As $S$ is spherically symmetric, the values of $S$ only depend on $x$'s distance to the origin, that is we have $S(x) = \tilde S(\rho(x))$ for some function $\tilde S$. $\tilde S$ is denoted by $S$ again in your paper, but I will write $\tilde S$ here. By the chain rule, we have \begin{align*} \Delta S &= \sum_i \partial_i^2(\tilde S \circ \rho)\\ &= \sum_i \partial_i(\partial_\rho\tilde S \circ \rho \cdot \partial_i S)\\ &= \sum_i \partial_\rho^2 \tilde S \circ \rho \cdot (\partial_i \rho)^2 + \partial_\rho \tilde S \circ \rho \cdot \partial^2_i \rho\\ &= \partial_\rho^2 \tilde S \circ \rho \cdot \sum_i (\partial_i\rho)^2 + \partial_\rho\tilde S \circ \rho \cdot \sum_i \partial^2_i \rho \end{align*} Using the above, we have \begin{align*} \sum_i (\partial_i \rho)^2(x) &= \sum_i \frac{x_i^2}{\rho^2(x)}\\ &= 1\\ \sum_i \partial_i^2\rho(x) &= \sum_i \frac{\rho^2(x) - x_i^2}{\rho^3(x)}\\ &= \frac{d\rho^2(x) - \sum_i x_i^2}{\rho^3(x)}\\ &= \frac{d\rho^2(x) - \rho^2(x)}{\rho^3(x)}\\ &= \frac{d-1}{\rho(x)} \end{align*} Using this, we get $$ \Delta S(x) = \partial_\rho^2 \tilde S(\rho(x)) + \frac{d-1}{\rho(x)} \cdot \partial_\rho \tilde S(\rho(x)) $$ or in the notation of the paper $$ \Delta S = \partial_\rho^2 S + \frac{d-1}{\rho} \cdot \partial_\rho S $$