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What is the general way in which radiation density for a given surface is expressed using Dirac deltas?

Consider this surface expressed in cylindrical coordinates (for any $\phi$ and $r_0$ an oblateness parameter):

$\sqrt{r^2+z^2}-\sqrt{r_0^2-z^2}==0$

It isn't a cylinder, so, were it not for the fact that it is symmetric about the origin (inviting degenerate solutions), and were it not for the fact that it is an oblate spheroid (inviting coordinate transforms involving oblate spheroid coordinates) it would be a generic example as to how to express radiation flux density as a Dirac delta in cylindrical coordinates.

Taking a stab in the dark at this, is this correct?

$Q\frac{\delta(\sqrt{r^2+z^2}-\sqrt{r_0^2-z^2})}{4\pi(r^2+z^2)}$?

Note that all I did was to place the surface equation to 0 in the Dirac delta, multiply it by the quantity $Q$ and normalize it to the area of a sphere of the same radius as the norm of the vector to the point at which the radiation flux density is measured.

Is it that simple?

Before I get into a potentially misleading description of how I (probably mistakenly) came up with that example, here is the general challenge:

Pick any 3D surface expressed in any coordinate system except one that matches the surface, and express that surface's radiation flux density as a $\delta$. Examples of what I mean by "a coordinate system that matches the surface" would be spherical coordinates matching a sphere, Cartesian coordinates matching a box, cylindrical coordinates matching a cylinder, oblate spheroid coordinate system matching an oblate spheroid, etc.

Nor am I interested in using coordinate transforms to go from a shape that matches a coordinate system to a coordinate system that doesn't match the shape.

If necessary to avoid falling into the trap of thinking of this as a coordinate transform question, pick a surface that has no corresponding (in the above sense) orthogonal coordinate system. (If, for example, the surface $x+y+\sqrt{x^2+y^2+z^2}==0$ has no matching orthogonal coordinate system, it might do.)

So, now to get to the meat of my question about "radiation flux density through a surface" and its relationship to Dirac delta:

p 32 Barton et al gives the strong definition of the 3D (spherical coordinate) radial Dirac delta as:

$$\delta^3(\vec{r}) = \frac{\delta(r)}{4\pi r^2}\tag{1}$$

and, correspondingly in 2D (polar coordinate):

$$\delta^2(\vec{r}) = \frac{\delta(r)}{2\pi r}.\tag{2}$$

Since $4\pi r^2$ and $2\pi r$ measure the area and length of sphere and circle respectively, and the surface and line integrals of these two Dirac deltas are 1 (by definition) it seems natural to, in appropriate circumstances, use the radial Dirac delta in modeling density distributions, normalized to 1, over radial surfaces and radial lines respectively.

However, I'm not looking for trivial examples. A trivial example would be a spherical surface, $Q\frac{\delta(r)}{4\pi r^2}$ for a given total radiated quantity Q, in spherical coordinates.

At first I thought of using an oblate spheroid centered at the origin, in either Cartesian or spherical coordinates, with radiation coming from the origin but it's symmetric about the origin (inviting degenerate solutions for this case) and there is such a thing as "oblate spheroid coordinates" (inviting degenerate solutions for this case).

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    $\begingroup$ Note that eqs. (1) & (2) are mathematically ill-defined. $\endgroup$
    – Qmechanic
    Mar 18, 2022 at 15:21
  • $\begingroup$ Are you asking what the delta function looks like in an oblate spheroidal coordinate system? Or something else? $\endgroup$
    – Kyle Kanos
    Mar 18, 2022 at 15:25
  • $\begingroup$ @Qmechanic, I changed the markup to, I believe, match exactly Barton et al. Is it still mathematically ill defined? $\endgroup$ Mar 18, 2022 at 16:27
  • $\begingroup$ You're dividing by zero. $\endgroup$ Mar 18, 2022 at 16:31
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    $\begingroup$ Pay no attention to @Qmechanic. Mathematicians constantly complain about how expressions involving delta functions are ill-defined. Physicists and engineers use them all the time without serious difficulty. Physics is founded on experiment, not on definitions. Whether the application of a mathematical concept leads to agreement with experiment is what matters to physics. $\endgroup$
    – John Doty
    Mar 18, 2022 at 16:40

2 Answers 2

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The trick to switching coordinates with Dirac deltas is to equate two definite integrals (whose limits I won't show). You were at least subconsciously familiar with this in inferring (1) for spherically symmetric Schwartz functions $f$, viz.$$f(O)=\int f(x)\delta^{(3)}(x)\underbrace{dxdydz}_{4\pi r^2dr}=\int\delta(r)f(x)dr\implies\delta^{(3)}(x)=\frac{\delta(r)}{4\pi r^2},$$where $O$ denote the origin.

Now let's consider cylindrical coordinates. Since $dxdy=rdrd\theta$ in the $z=0$ plane that passes through $O$, $\theta$-symmetric $f$ satisfy$$\delta^{(2)}(x)=\frac{\delta(r)}{2\pi r}.$$That looks awfully $2$-dimensional. If you want to make the fact the space is $3$-dimensional explicit, multiply the above by $\delta(z)$ to get$$\delta^{(3)}(x)=\frac{\delta(r)\delta(z)}{2\pi r}.$$This follows in particular from $dxdydz=2\pi r drdz$.

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  • $\begingroup$ I'm still not clear on how one expresses a surface density except in the trivial case of a sphere or circle which I presume would be $\frac{Q\delta(r)}{4\pi r^2}$ for a sphere given emission of quantity Q, or, $\frac{Q\delta(r)}{2\pi r}$ for a circle in a nonphysical flatland. When you say $f$ is a symmetric Schwarz function, WP says of the 2D (elipse) case: $S(z) = \frac{(a^2+b^2)z-2a b \sqrt{z^2+b^2-a^2}}{a^2-b^2}$ where $a$ and $b$ are the axes (I presume). But that's just 2D and it isn't clear how $f(O)$ relates to $S(z)$ nor how the integral provides the density at a point. $\endgroup$ Mar 19, 2022 at 3:10
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For a general orthogonal coordinate system $\xi_i$, the delta function takes the form, $$ \delta\left(\mathbf r-\mathbf r_0\right)=\frac{\delta\left(\xi_1-\xi_{10}\right)}{h_1}\,\frac{\delta\left(\xi_2-\xi_{20}\right)}{h_2}\,\frac{\delta\left(\xi_3-\xi_{30}\right)}{h_3} \tag{1}$$ where $h_i$ is the scale factor, $$ h_i^2=\sum_j\left(\frac{\partial r_j}{\partial \xi_i}\right)^2.\tag{2}$$ Thus, you should be able to compute the delta function for any orthogonal coordinate system by using (1) and (2).

For oblate spheroids, the coordinate transformations $\{x,\,y,\,z\}\to\{\mu,\,\nu,\,\phi\}$ follow, \begin{align} x &= a\cosh\mu\cos\nu\cos\phi \\ y &= a\cosh\mu\cos\nu\sin\phi \\ z &= a\sinh\mu\sin\nu \end{align} The scale factors and delta function will follow from this.

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  • $\begingroup$ In my prior response to you I said "Perhaps" but I see that when I said "spherical coordinate system" it didn't override your interpretation of my "oblate spheroid" as meaning "oblate spheroid coordinate system". What I meant by "oblate spheroid" is restricting to spherical coordinates, and then expressing a radial Dirac delta in those coordinates that describes not just an oblate spheroid but, by integrating across all $r$, at a given $\theta$ and a given $\phi$, to provide a number: the density at that point on the surface. $\endgroup$ Mar 19, 2022 at 3:26
  • $\begingroup$ Okay, how does expressing the Dirac delta function in the coordinates best suited for oblate spheroids not do what you want? Because it's not the familiar $\{r,\,\theta,\,\phi\}$? $\endgroup$
    – Kyle Kanos
    Mar 19, 2022 at 15:02
  • $\begingroup$ I used the oblate spheroid as an example of density surface that would not be trivially expressed as $Q\frac{\delta(r)}{4\pi r}$ in spherical coordinates. I chose that example not because I wanted another coordinate system in which to trivially express the density surface, but rather so that I could figure out how to express and/or interpret nontrivial delta expressions. For example, in cylindrical coordinates, would a density surface of an oblate spheroid be expressed as $Q\frac{\delta(r-\sqrt{r_0^2-z^2})}{4\pi(r^2+z^2)}$? $\endgroup$ Mar 19, 2022 at 18:10
  • $\begingroup$ I added to the original question to try and clarify what I'm asking. $\endgroup$ Mar 19, 2022 at 18:27
  • $\begingroup$ I've read your remarks several times now, and I'm not at all sure how "express and/or interpret nontrivial delta expressions" isn't what I answered. You are expressing a want for a coordinate transform of $Q\left(\mathbf r\right)\sim f\left(\mathbf r\right)\delta\left(\mathbf r\right)$ from one system to another. I gave you the troublesome part of the delta function, do you need help also with $f\left(\mathbf r\right)$ part as well? $\endgroup$
    – Kyle Kanos
    Mar 19, 2022 at 19:26

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