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Let $f: \mathbb{R}^3 \rightarrow \mathbb{R}$ be a real valued scalar field and $\mathbf{r}\in\mathbb{R}^3$ a vector with $r = \sqrt{\mathbf{r}\cdot\mathbf{r} }$ its norm. Let's say that $f$ is invariant under rotations, such that \begin{equation} f(R\mathbf{r}) = f(\mathbf{r}) \ , \end{equation} for all $R$, with $RR^{T} = \mathbb{I}$. Is there a straightforward way to prove that there must exist a function $h: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(\mathbf{r}) = h(r)$ for all $\mathbf{r}\in\mathbb{R}^3$?

I understand that this is almost self-evident. If a function is invariant under rotations then it should only depend on the norm. However, I am not able to prove it more formally without relying on that intuition.

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    $\begingroup$ Have you tried writing down a general function $f(x,y,z)$ in terms of the components and seeing what the consequences of the invariance are? $\endgroup$
    – Buzz ♦
    Commented Dec 15, 2023 at 4:35
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    $\begingroup$ In addition to Buzz's suggestion, try finding $g(r,\vartheta,\varphi)=f(r\sin\vartheta\cos\varphi,r\sin\vartheta\sin\varphi,r\cos\vartheta)$ and consider what a rotation does to those; you should be able to derive that $\frac{\partial g}{\partial\varphi}=0$ fairly easily. $\endgroup$
    – naturallyInconsistent
    Commented Dec 15, 2023 at 5:12

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If it is invariant under rotation, you can always rotate the vector from $(x,y,z) \rightarrow (0,0,r)$. Done.

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  • $\begingroup$ Thanks! I didn't see it. That does it. $\endgroup$
    – Pere Rosselló
    Commented Dec 15, 2023 at 18:05

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