# Delta function of the Euclidean norm $\delta(|\mathbf x|)$ / in polar coordinates at origin $\delta(r)$

Several posts discuss the representation of the delta function in polar coordinates in 2D or 3D, e.g. Dirac delta in polar coordinates or Delta function at the origin in polar coordinates

Does anyone have a reference for a representation of $$\delta(|\mathbf x|)$$ or $$\delta(r)$$ in general dimension $$n \geq 2$$? I am guessing the following: $$\delta(r)/r^{n-1} = s_{n-1} \delta(\mathbf x), \qquad s_{n-1} := {2\pi^{n/2} \over \Gamma(n/2)} \text{ (area of unit sphere)}$$ or equivalently (?) $$\delta(|\mathbf x|) = s_{n-1} |\mathbf x|^{n-1} \delta(\mathbf x)$$

My proof: For $$r_0 \neq 0$$ the general curvilinear correspondence is $$\delta(\mathbf {x - x}_0) \equiv \delta(r - r_0) \delta(\mathbf u - \mathbf u_0) / r^{n-1}, \qquad \mathbf {x, x}_0\in\mathbb R^n\quad r,r_0>0\quad \mathbf{u, u}_0\in\mathbb S^{n-1} \text{ (unit sphere)}$$ The case $$\mathbf x_0 = \mathbf 0 \Leftrightarrow (r_0 = 0, \mathbf u_0\in\mathbb S^{n-1})$$ corresponds to a singular Jacobian of the transformation and we may "integrate out" all ignorable spherical coordinates $$\mathbf u$$: \begin{aligned}\delta(\mathbf x) \int_{\mathbb S^{n-1}} d\mathbf u & = \delta(r) / r^{n-1} \int_{\mathbb S^{n-1}} \delta(\mathbf u - \mathbf u_0) d\mathbf u, \qquad\text{i.e.} \\ s_{n-1} \delta(\mathbf x) & = \delta(r) / r^{n-1} \end{aligned} as required.

I will write $$\omega_d := s_{d-1} = \frac{2\,\pi^{d/2}}{\Gamma(d/2)}$$.
If $$f(x) = \tilde{f}(|x|)$$ with $$\tilde{f}\in L^1_{\mathrm{loc}}$$, then for any $$\varphi\in C^\infty(\mathbb{R})$$, $$\langle \tilde{f},\varphi\rangle = \int_{\mathbb{R}_+} \tilde{f}(r)\,\varphi(r)\,\mathrm{d}r = \frac{1}{\omega_d}\int_{\mathbb{R}^d} f(x)\frac{\varphi(|x|)}{|x|^{d-1}}\,\mathrm{d}x = \frac{1}{\omega_d}\left\langle f, \frac{\varphi(|x|)}{|x|^{d-1}}\right\rangle$$ so if we want to define a generalization of the notion of radial change of variable, we might want to define $$\tilde{f}(r)$$ by setting $$\langle \tilde{f},\varphi\rangle := \frac{1}{\omega_d}\left\langle f, \frac{\varphi(|x|)}{|x|^{d-1}}\right\rangle$$ in the general case. In particular, we will have the relation $$\omega_d\,|x|^{d-1} \tilde{f}(|x|) = f(x)$$. Remark however that the above definition does not make sense if we replace $$f$$ by $$\delta_0$$. $$\frac{\delta_0(x)}{|x|^{d-1}} \text{ is not a priori a well defined distribution}$$ Defining $$g(r) = \tilde{f}(|r|)$$ for $$r\in\mathbb{R}$$, we can however try to solve the equation $$\omega_d\,r^{d-1} g(r) = \delta_0$$. Taking the Fourier transform ($$\hat{g}(y) = \int_{\mathbb{R}} g(x)\,e^{-2i\pi x\cdot y}\,\mathrm{d}x$$) we get $$\frac{\omega_d}{(-2i\pi)^{d-1}} \hat{g}^{(d-1)} = 1$$ so by integrating $$d-1$$ times, we get $$\hat{g}(y) = \frac{(-2i\pi)^{d-1}}{\omega_d} \left(\frac{y^{d-1}}{(d-1)!} + \sum_{k=1}^{d-1} a_k\,x^{k-1}\right)$$ so that (if I did not messed up with the power on the $$-1$$) $$g(r) = \frac{(-1)^{d-1}}{\omega_d} \left(\frac{\delta_0^{(d-1)}}{(d-1)!} + \sum_{k=1}^{d-1} C_k\,\delta_0^{(k-1)}\right)$$ (where $$\delta_0^{(n)}$$ is the $$n$$-th derivative of the Dirac delta centered in $$0$$) and we can see that there is no uniqueness of $$\delta_0(r)/r^{d-1}$$ (which is natural since $$x^n \delta_0^{(n-1)} = 0$$). However, the Dirac is homogeneous, and thus so is its radial representation. Therefore $$g = \frac{(-1)^{d-1}}{\omega_d\,(d-1)!} \,\delta_0^{(d-1)}$$ which can be written in a more informal way with $$r=|x|$$ $$\boxed{ \delta_0(x) = \frac{(-1)^{d-1}}{\omega_d\,(d-1)!} \,\delta_0^{(d-1)}(r) }$$
Remark: Of course, we can now choose to define $$\frac{\delta_0(r)}{r^{d-1}} := \frac{(-1)^{d-1}}{\,(d-1)!} \,\delta_0^{(d-1)}(r)$$ as the homogeneous solution of the equation $$r^{d-1} g(r) = \delta_0$$, and so we find as you was writing $$\omega_d\,\delta_0(x) = \frac{\delta_0(r)}{r^{d-1}}$$, but now we know the true meaning of this notation.
• I do not understand your $|λx| = λ^d|x|$. It should be $|λx| = λ\,|x|$. It seems to me it is the same homogeneity on both sides. Each derivative change your homogeneity of one, and so $$\lambda^{-1-(d-1)} = \lambda^{-d}$$ Commented Aug 12, 2020 at 16:06
• My mistake, but I think there's still a mismatch: Taking x=λx homogeneity gives us $λ^{-d} δ_0(x)$ on the LHS, while RHS gives $\delta_0^{(d−1)}0(|λx|)=\delta_0^{(d−1)}(λ|x|)=λ^{−(d-1)} \delta_0^{(d−1)}(|x|)$ ??? Commented Aug 12, 2020 at 17:01
• My mistake again I got it $\delta_0^{(n)}$ is homogeneous of degree $-n-1$ and for $n=d-1$ the correct RHS is $\lambda^{-(d-1)-1} = \lambda^{-d}$ as you wrote. Commented Aug 12, 2020 at 17:21
• Yes, exact. And the advantage of this representation is that it is a well defined distribution (i.e. we know its action on test functions) while this is not the case of $\delta_0(r)/r^{d-1}$. In distributions theory, one usually only defines the multiplication of elements that are in dual space of each other. As we see here, the multiplication of $\delta_0(r)$ with $1/r^{d-1}$ has no unique clear meaning. Commented Aug 13, 2020 at 8:42