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Here is an example from a textbook:

Suppose there is a unit charge or unit mass at the point $(x,y,z)=(-1,\sqrt{3},-2)$; then in rectangular coordinates, the density is $$\rho=\delta(x+1)\,\delta(y-\sqrt{3})\,\delta(z+2)\tag{1}$$ In cylindrical coordinates the point is $(r,\theta,z)=\left(2,\frac{2\pi}{3},-2\right)$, so in cylindrical coordinates the density is $$\rho=\frac{\delta(r-2)\,\delta\left(\theta -\frac{2\pi}{3}\right)\delta(z+2)}{r}\tag{2}$$ In spherical coordinates, the point is $(r,\theta,\phi)=\left(2\sqrt{2},\frac{3\pi}{4},\frac{2\pi}{3}\right)$, so in spherical coordinates the density is $$\rho=\frac{\delta(r-2\sqrt{2})\,\delta\left(\theta -\frac{3\pi}{4}\right)\delta\left(\phi-\frac{2\pi}{3}\right)}{{r^2}\sin\theta}\tag{3}$$

When I substitute the point $(x,y,z)=(-1,\sqrt{3},-2)$ into equation $(1)$ I get a density of unity as expected.

When I substitute the point $(r,\theta,z)=\left(2,\frac{2\pi}{3},-2\right)$ into equation $(2)$ I get a density of $$\color{red}{\frac{1}{2}\ne 1}$$

When I substitute the point $(r,\theta,\phi)=\left(2\sqrt{2},\frac{3\pi}{4},\frac{2\pi}{3}\right)$ into equation $(3)$ I get a density of $$\color{red}{\frac{1}{4\sqrt{2}}\ne 1}$$

I thought that choice of coordinate system is irrelevant since all coordinate systems you use will always lead you to the same result.

So why are the parts marked $\color{red}{\mathrm{red}}$ not equal to unity?

EDIT:

In response to the comments below, since we

"cannot evaluate the Dirac Delta as if it has a value at a point; it is not a function"

From $(1)$; Is it plausible to write $$\rho=\delta(0)\,\delta(0)\,\delta(0)$$ at the point $(x,y,z)=(-1,\sqrt{3},-2)$?

Notice that I did not evaluate the Dirac Delta, but I did substitute the coordinates in.

In other words; Is substitution allowed but evaluation not allowed?

BLAZE
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    You cannot evaluate the Dirac Delta as if it has a value at a point; it is not a function. See THIS ANSWER and THIS ONE for primers. – Mark Viola Mar 05 '16 at 00:12
  • So basically your question is why are there the factors in the denominators? – Korf Mar 05 '16 at 00:12
  • @Korf No, it's okay I understand why the denominators are present – BLAZE Mar 05 '16 at 00:14
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    @Blaze Ok. I'd say the answer more or less lies in what Dr. MV wrote. The important thing that is invariant under coordinate change is $\int \delta(x) f(x) \mathrm d x$. This is the invariant value, not the denominator. This also enforces the change of the denominator and your red inequalities, roughly speaking. – Korf Mar 05 '16 at 00:37
  • @Dr.MV Please see my edited post, thank you. – BLAZE Mar 05 '16 at 13:50
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    @BLAZE Since the Dirac Delta is not a function, it makes no sense to ask "what is the value of $\delta(x)$? for any $x$. And this does not exclude $x=0$. That is, $\delta (0)$ has no value. The notation $\delta(x)$ has meaning only as a distribution. If $f(x)$ is a sufficiently smooth function with compact support, then the functional $\langle f,\delta_0\rangle =f(0)$. Continued ... – Mark Viola Mar 05 '16 at 19:30
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    This is often written as $\langle f,\delta_0\rangle =\int_{-\infty}^\infty f(x)\delta(x),dx=f(0)$. But the integral notation represents neither an integral in the sense of Riemann nor an integral in the sense of Lebesgue. It is a functional that maps the test function $f(x)$ into the number $f(0)$. I hope this is clearer. – Mark Viola Mar 05 '16 at 19:30
  • @Dr.MV You have given an excellent explanation and for that I am very grateful. Just to be completely clear; It is not correct to write $\rho=\delta(0),\delta(0),\delta(0)$ since $\delta(0)$ is undefined; Is that correct? Thank you for your time. – BLAZE Mar 06 '16 at 09:01
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    Yes, you are correct. – Mark Viola Mar 06 '16 at 15:04

1 Answers1

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By definition $Q=\iiint \rho \, dV$

In Cartesian: $$Q=\iiint \rho(x,y,z) \, dx \, dy \, dz=1$$

In Cylindrical: $$Q=\iiint \rho(r,\theta,z) \, r\, dr \, d\theta \, dz=1$$

In Spherical: $$Q=\iiint \rho(r,\theta,\phi) \, r^2\sin \theta \, d\theta \, dr \, d\phi=1$$

Hence for point charge:

\begin{align*} \rho(x,y,z) &= \delta (x-x_{0}) \, \delta (y-y_{0}) \, \delta (z-z_{0}) \\ \rho(r,\theta,z) &= \frac{1}{r} \delta (r-r_{0}) \, \delta (\theta-\theta_{0}) \, \delta (z-z_{0}) \\ \rho(r,\theta,\phi) &= \frac{1}{r^2\sin \theta} \delta (r-r_{0}) \, \delta (\theta-\theta_{0}) \, \delta (\phi-\phi_{0}) \\ \end{align*}

Ng Chung Tak
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