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Is this how one would express a sphere as a Dirac delta in Cartesian coordinates?

DiracDelta[Sqrt[R0^2-z^2-x^2],Sqrt[R0^2-z^2-y^2],Sqrt[R0^2-x^2-y^2]]

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.

RobPratt
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James Bowery
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  • @user64494 Yet, Barton et al. extract reasonable physical results from their use of Dirac delta. Your version of mathematics is not what we use in physics. –  Mar 20 '22 at 22:07
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    Unclear what you mean. A sphere and a Dirac delta are different mathematical objects. –  Mar 20 '22 at 14:42
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    It is not unclear @JohnDoty. He means to write a kernel $$\rho(x,y,z)$$ such that $$\int f \rho , \mathrm{d}V = \int_{S_{R_0}} f , \mathrm{d}S.$$ This is a common procedure. – Diffycue Mar 20 '22 at 15:17
  • @Diffycue I am glad it is clear to you, but it is definitely unclear to me as it was to John above (and, incidentally, your explanation did not clarify the problem to me either). This "common procedure" may be more discipline-specific than you think. –  Mar 20 '22 at 15:50
  • I've added some text clarifying in what sense I'm interpreting the 3 dimensional Dirac delta as a sphere. – James Bowery Mar 20 '22 at 17:20
  • The product $a(r)\delta(r)$ is defined only for smooth $a(r)$ (see Encyclopedia of Mathematics), but $\frac 1 {r^2}$ and $\frac 1 r$ have singularities at the origin. – user64494 Mar 20 '22 at 18:38
  • After the explanations, it is still unclear. Indeed, what is your question? If you ask: "Is this how one would express a sphere as a Dirac delta in Cartesian coordinates?" The answer is "No." It is not a sphere, but a delta-function localized on the sphere. That is, one can use it to integrate any function at the spherical surface, just as Diffycue has written. Only this, not more. However, what is the relation of this above question with the representation of the delta-function in spherical coordinates is still an enigma for me. – Alexei Boulbitch Mar 20 '22 at 19:28
  • What I had hoped to clarify by adding the Barton text book definitions was the fact that a model (as in model theory) for a "sphere" can use any "mathematical object" for which a truth function may map to a co-domain already accepted as representing a "sphere". For instance,
    user64494's statement that $\frac{1}{r^2}$ is a "singularity" at the origin is true for both $\delta(r^2)$ and for a "sphere" of radius 0, is, if we define "singularity" for a "sphere" as a "point".

    So I don't understand what people are confused about unless it is an unusual use to which I put $\delta$.

     – James Bowery Mar 20 '22 at 19:57
  • In any event, if people don't like my use of the word "sphere", or the way Barton seems to associate $\delta(\vec{r})$ with measures of corresponding dimensions (surface of sphere or length of a circle), perhaps I should avoid the word "sphere" and find some way to state the question in terms of Cartesian vs spherical coordinates as used in DiracDelta. Is that what people want? – James Bowery Mar 20 '22 at 20:02
  • @JohnDoty: I prefer formulas and references over ungrounded claims. This https://www.dropbox.com/s/w8tedgabl9me8sa/%D0%97%D0%BD%D1%96%D0%BC%D0%BE%D0%BA%20%D0%B5%D0%BA%D1%80%D0%B0%D0%BD%D0%B0%202022-03-21%2007.33.07.png?dl=0 is what I see in the link from the question. – user64494 Mar 21 '22 at 05:31
  • @user64494 It's you who have the ungrounded claims here. In mathematical physics, claims are grounded in experiments. If you doubt the math here, go into a laboratory and demonstrate that it gets the wrong answer. –  Mar 21 '22 at 11:17
  • @JohnDoty: Sorry, your claim "Barton et al. extract reasonable physical results from their use of Dirac delta" is empty words. – user64494 Mar 21 '22 at 12:05
  • @user64494 Show me the results of an experiment that demonstrates that they are "empty words". You are using technology designed using the mathematics you don't believe in to communicate your disbelief: don't you understand that by doing this you are demonstrating that it is your position that is ungrounded? –  Mar 21 '22 at 12:15

2 Answers2

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I assume you want an expression for the surface measure on a sphere having center at the origin and radius $R>0$. Let us call this measure $\sigma_R$.

Our sphere is the zero set of the function $g_R(\mathbf{x})=R^2-\lvert\mathbf x\rvert^2$. Now, according to this property of the Dirac delta distribution, we have the formula $$ \delta(g_R(\mathbf x)) = \frac{\sigma_R(\mathbf x)}{\lvert \nabla g_R(\mathbf x)\rvert} ,$$ and in our case, $\lvert \nabla g_R(\mathbf x)\rvert = 2 \lvert \mathbf x\rvert$, which reduces to $2R$ on the support of $\sigma_R$. We conclude that $$ \sigma_R (\mathbf x)= 2R\delta(R^2-\lvert \mathbf x\rvert^2).$$

In polar coordinates $(r, \theta, \phi)$, the right-hand side reads $2R\delta(R^2-r^2)$.

If we had used $f_R(\mathbf x)=R-\lvert \mathbf x\rvert$ instead of $g_R$, we would have obtained the alternative expression $$ \sigma_R(\mathbf x)=\delta(R-\lvert\mathbf x\rvert)=\delta(R-r).$$

Giuseppe Negro
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Is the thing you want something that yields the area of a spherical surface of radius r0, when integrated over all space? Then, in spherical coordinates it's r0^2 DiracDelta[r-r0], no? I believe that you may then substitute Sqrt[x^2+y^2+z^2] for r, and get what you want, formally. Unfortunately, Mathematica seems not to understand such a thing.