# Dirac delta function at singularities in spherical coordinates

Background Information

Let $$\delta^3(\vec x-\vec a)$$ represent a point density at $$\vec a$$. It satisfies $$f(\vec a)=\int \delta^3(\vec x-\vec a)f(\vec x)|J(\vec x)|\mathrm d^3\vec x,$$ where $$f(\vec x)$$ is an arbitrary function on the space, and $$|J(\vec x)|$$ is the Jacobian determinant.

On the order hand, the delta function of a vector can be decomposed into the product of several delta functions, of the vector's components, $$f(\vec a)=\int f(\vec x)\prod_i\delta^3(x_i-a_i)\mathrm d x_i,\tag{1}$$ where $$x_i,a_i$$ are components in the coordinate systems. Comparing the above two equations, we get $$\delta^3(\vec x-\vec a) = \frac{1}{|J(\vec a)|}\prod_i\delta(x_i-a_i).$$

For example, in Cartesian coordinates, $$|J|=1$$, so $$\delta^3(\vec x-\vec a)=\delta(x-x_a)\delta(y-y_a)\delta(z-z_a).$$ In spherical coodinates, $$|J|=r^2\sin\theta$$, and $$\delta^3(\vec x-\vec a)=\frac{1}{r^2\sin\theta}\delta(r-r_a)\delta(\theta-\theta_a) \delta(\varphi-\varphi_a).$$

Question

The derive above seems illegal when $$\vec a$$ is a singularity, where $$|J|=0$$. The coordinate of a singularity can be indefinite. For instance, points on $$z$$-axis (except the origin) in spherical coordinates can be endowed with arbitrary $$\varphi$$ component. At the origin, even the $$\theta$$ component is arbitrary, too.

Therefore, when decomposing $$\delta^3(\vec r-\vec a)$$, there's no corresponding $$\delta(\varphi)$$ or $$\delta(\theta)$$. We must turn back to the initial Equation (1).

For $$\vec a$$ on the positive half-axis of the $$z$$-axis, $$f(\vec a)=\int f(\vec x)\delta(r-r_a)\delta(\theta-0)\mathrm dr\mathrm d\theta \mathrm d\varphi =2\pi\int f(\vec x)\delta(r-r_a)\delta(\theta)\mathrm dr\mathrm d\theta,$$ so $$\delta^3(\vec r-\vec a)=\frac{1}{2\pi r^2\sin\theta}\delta(r-r_a)\delta(\theta).$$ If $$\vec a=0$$, then $$f(\vec a)=\int f(\vec x)\delta(r-0)\mathrm dr\mathrm d\theta \mathrm d\varphi =4\pi\int f(\vec x)\delta(r)\mathrm dr.$$ Thus $$\delta^3(\vec r-\vec a)=\frac{1}{4\pi r^2}\delta(r).$$ Is there anything wrong? I feel uncomfortable with this result.

This answer (though gained a downvote) states that symmetry results to the difference in the spherical coordinates. Is there any unified formula similar to Equation (1) that include singularities?

Or do I have to understand this in measure or distribution language? I'm not familiar with that.

• Would Mathematics be a better home for this question? Sep 1, 2022 at 15:24

## 1 Answer

The problem that you have is that when you calculate an integral over some volume $$\Omega$$ in spherical coordinates you are implicitly throwing out the integration region that is not covered by your spherical coordinates. This is no problem because that region is a null-set so it gives a zero contribution to your integral.
What you are using when writing your integral in spherical coordinates is the $$\int_{\Omega} f(y) dy = \int_{\Phi^{-1}(\Omega)} f(\Phi(x)) |\det(D\Phi(x))| dx,$$ where $$\Phi$$ is the coordinate transformation, that needs to be bijective(!, and differentiable), and $$\det(D\Phi(x)) = J(x)$$ is the corresponding Jacobi determinant.

Now look at spherical coordinates, where (naively) $$\Omega = \mathbb R^3$$ and $$\Phi((r,\theta,\varphi)) = (r \cos\varphi \sin \theta, r \sin \varphi \sin \theta, r \cos \theta).$$ Now if want $$\Phi$$ to be invertible we need to be careful with defining its domain. Naively we have $$\theta \in [0,\pi]$$, but there we run into problems when $$\theta = 0$$ or $$\pi$$, in which case $$\Phi^{-1}((0,0,z)) = \{ (z,0,\varphi) ~|~ \varphi \in[0,2\pi) \},$$ i.e. a point on the $$z$$-axis has infinite preimages... What we need to do is say that $$\Phi: [0,\infty) \times (0,\pi) \times [0,2\pi ) \rightarrow \mathbb R - \{ (0,0,z ) ~|~ z \in \mathbb R \}.$$ Now, as you can check, this is bijective, but now $$\Omega = \mathbb R - \{ (0,0,z ) ~|~ z \in \mathbb R \}$$. And the integral doesn't care that we removed the $$z$$-axis, since it is a null-set.

But now you see that you get into trouble when you integrate over $$f(\vec x) \delta^3(\vec x - \vec a),$$ where $$\vec a \in \{ (0,0,z ) ~|~ z \in \mathbb R \}$$. Since $$\vec a$$ is not in the integration region the integral must give zero $$\int_{\Omega} f(\vec x) \delta^3(\vec x - \vec a) d^3x = 0,$$ since the delta function vanishes on all of $$\Omega$$.
But actually it really just depends on how you define the delta function (which tells you that there is something wrong). I like to define it through a Dirac series, for example $$\delta_{\epsilon}(\vec x) = \frac{1}{2\pi \epsilon} \exp \Big ( - \frac{\vec x^2}{2 \epsilon} \Big ).$$ Then $$\int_{\Omega} d^3\vec x~ f(\vec x) \delta^3(\vec x - \vec a) := \lim_{\epsilon \rightarrow 0^+} \int_{\Omega} d^3\vec x~ \delta_{\epsilon}(\vec x) f(\vec x) = f(\vec a).$$ Now the delta function "stretches" into $$\Omega$$, so you actually get $$f(\vec a)$$.

If you want to use your formula $$\delta( \vec x - \vec a) = \frac{1}{|J(\vec a)|} \delta( \Phi^{-1}(\vec x - \vec a)),$$ you have to make sure that your cover $$\vec a$$ with your domain. For this you can just choose different spherical coordinates with respect to some other axis than $$z$$. Or you simply use symmetry to make $$\vec a$$ not point in the $$z$$-direction. In the end it is just an artificial problem that comes from an incorrect choice of coordinates.