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My problem is somewhat general. I do not think it has been posted before, however, I am new to Physics Stack Exchange so please, if I'm wrong, feel free to let me know.
I will give an example problem and then talk about the general case I'm interested in.

Given the outer surface of a cylinder with height $l$ with a surface density $$\sigma(\theta,z)$$ how do I get the linear charge density of a path $$\theta(z)$$ on the surface? I realize that if $\sigma$ is constant on the surface and the path is perpendicular to the symmetry axis of the cylinder it should be $$\lambda = \frac{\sigma}{l}$$ However, this does not make sense looking at the dimensions. Also, I am searching for a more general insight.
Given a volume charge density $$\rho(x,y,z)$$ (if the charge density can be expressed as a surface, $\rho$ would just be a surface charge density with a $\delta$-distribution) how do I get the linear/surface charge density of a path/surface (could be sphere, cylinder, plane etc.) which lies in the same volume? I would be very happy if you could direct me to a book/website where this is explained or, even better, explain it here. This problem has been bugging me a lot.

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    $\begingroup$ You can’t talk about the linear charge density of a path on a surface. It would depend on the “width” of the path, and a 1D path doesn’t have a width. $\endgroup$
    – G. Smith
    Aug 12, 2020 at 20:15
  • $\begingroup$ @vreithinger Philip has already pointed out that your equation $\lambda=\sigma/l$ is nonsense on dimensional grounds, but you have not edited your question. $\endgroup$
    – G. Smith
    Aug 13, 2020 at 5:38
  • $\begingroup$ @Philip Hopefully you agree that in SI a surface charge density is some number of coulombs per square meter and a linear charge density is some number of coulombs per meter. If so, tell me: If a surface has a charge density of one coulomb per square meter, then how many coulombs per meter does a path on it have? (The answer is supposed to be obvious: Zero.) Hopefully thinking in this way will make it clear that the concept makes no sense unless one says that the path is, say, 1 mm wide. Yes, we use delta functions as you say. My argument cannot be used in the way you claim. $\endgroup$
    – G. Smith
    Aug 13, 2020 at 5:52
  • $\begingroup$ @G.Smith Well I agree with you that the density can't be used naively as a finite-valued function in this case. The point I was trying to make is that despite this fact, we can use the $\delta-$function to define a charge density. This density will be infinite on the surface, and zero away from it. I believe this is the sort of density that the OP is asking about. I'm not sure I understand the problem you have with it, your argument above would seem to suggest that you think that a point charge has either zero charge density, or finite size, both of which I believe are not true... $\endgroup$
    – Philip
    Aug 13, 2020 at 6:01
  • $\begingroup$ @Philip A point charge has a finite charge, zero size, and an infinite charge density. It seems that you and I understand what the OP means by “linear charge density of a path on a surface” in completely different ways. I’m sorry that I haven’t been able to make my argument understandable. I can’t explain it any better than I already have, so I won’t try further. $\endgroup$
    – G. Smith
    Aug 13, 2020 at 16:32

1 Answer 1

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Let's take a region having charge density $\rho (x,y,z)$. Now on I will only be dealing in cartesian coordinates, however, you can easily switch to any other coordinate system if needed. Aleso, in the following answer, I am assuming charge distributions having finite characteristic parameters (volume charge density or surface charge density or linear charge density).

Mathematical derivation

Surface charge density

Let's choose a surface $S(x,y,z)$ having an infinitesimal thickness $\mathrm d t$. Now let's choose an infinitesimal area element on the surface, at the point $(x_0,y_0,z_0)$, having an area $\mathrm d A$. Thus the charge contained in that infinitesimal volume formed by $\mathrm dA$ and $\mathrm dt$ is

$$\mathrm dq =\rho(x_0,y_0,z_0)\:\mathrm dA\:\mathrm dt\tag{1}$$

Now the surface charge density is defined as $\sigma =\mathrm d q/\mathrm dA$. Using this, and equation$(1)$, we get

$$\sigma(x_0,y_0,z_0)=\frac{\rho(x_0,y_0,z_0)\:\mathrm dA\:\mathrm dt}{\mathrm dA}=\rho(x_0,y_0,z_0)\:\mathrm dt$$

However, since we are talking about a surface, thus the thickness being infinitesimally small, the surface charge density ($\sigma$) must vanish.

Linear charge density

Applying the above process to linear charge density, we get (here, our infinitesimal volume element is a cuboid):

$$\mathrm d q=\rho(x_0,y_0,z_0)\:\mathrm dl \:\mathrm dh \:\mathrm dw$$

where $\mathrm dl$ is the infinitesimal length element of the curve, $\mathrm dh$ is the thickness of the line and $\mathrm dw$ is the depth of the line. Now using the definition of linear charge density ($\lambda=\mathrm dq/\mathrm dl$), we get

$$\lambda(x_0,y_0,z_0)=\frac{\rho(x_0,y_0,z_0):\mathrm dl :\mathrm dh :\mathrm dw}{\mathrm dl}=\rho(x_0,y_0,z_0) :\mathrm dh :\mathrm dwdd

which again gives us a zero linear charge density.

Let's, instead, try finding the linear charge density of a curve located on a surface having surface charge density $\sigma(x,y,z)$. Applying the above process, we see that we can now drop the depth term ($\mathrm dw$), since there is no depth to a 2D surface. Thus we get

$$\lambda(x_0,y_0,z_0) = \sigma (x_0, y_0,z_0)\:\mathrm dh$$

Againg, the linear charge density vanishes.

This implies that you cannot have a surface of a $N-1$ dimensions, with a finite (relevant) charge density inside an $N$ dimensional space having finite (relevant) charge density everywhere.

Physical explanation

There's a nice and intuitive way of why this isn't possible. Imagine a finite $N$-dimensional space. Now let's, for the sake of argument, assume that all the hypersurfaces inside that $N$-dimensional space have a non zero finite charge density everywhere. If this is true, then we can find the charge contained by that surface, which would be finite. Now, infinitely many such surfaces exist, and to make up the finige $N$-dimensional space, you would need infinite of such $N-1$ dimensional hypeesurfaces. This implies that the final charge contained in our space, is equal to the sum of the charges contained in each of the infinitely many hypersurfaces. But this implies that the charge contained in our space is infinite, since we are adding a finite non-zero charge (for each surface), infinitely many times. But we already assumed that the charge density of our finite $N$-dimensional space is finite everywhere, so the charge contained in that ginite space, must be finite as well. This shows we have a contradiction, implying that both of our initial assumptions

  • Finite space having finite charge density

  • Hypersurface having finite non-zero charge density

cannot be simultaneously true. Hence, we have reached the same conclusion, the one which the math suggested.

Charge distributions involving Dirac delta functions

In the following part, I am only considering a specific example, where I will be trying to convert surface charge density, to linear charge density. It won't be hard to generalise this to other scenarios as well.

Let's say the surface charge density is of the form

$$\sigma(\mathbf r)=q(\mathbf r) \delta (\mathbf s)$$

where $\delta$ is the Dirac delta function, $q:V\to \mathbb R$ is a function from vector space to real numbers, and $\mathbf s=f(\mathbf r)$, where $f:V\to V$ is function mapping vactors to vectors in the vector space. Let the solution of the equation $\mathbf s=f(\mathbf r)=\boldsymbol{0}$ be the curve $\gamma$. Now, let's find the linear charge density at a point $\mathbf r_0$ lying on the curve $\gamma$. To do that, we need to determine the thickness of our curve.

Notice that the magnitude of a first order infinitesimal change in $\mathbf s$, corresponds to translating the curve $\gamma$, forming a new curve $\gamma '$, which doesn't intersect $\gamma$. The collection of such neighbouring curves, make up a "thick" curve, say $\Gamma$. So $\Gamma$ is essentially an area, which, at any point has a thickness $\mathrm d \mathbf r$ (i.e. the change in the position vector of that point, which was initially on the curve). Thus, writing the change in $f$ upto the first linear term, we get

$$ f(\mathbf r)+\frac{\mathrm df(\mathbf r)}{\mathrm dr}\mathrm d r=\mathbf s + \mathrm d\mathbf s$$

But we know, that initially $\mathbf r$ lay on the curve $\gamma$, so $\mathbf s=f(\mathbf r)=0$. Applying this to above equation, we get

$$\frac{\mathrm d f(\mathbf r)}{\mathrm dr}\mathrm d r= \mathrm d\mathbf s$$

Taking the magnitude of both the sides, we get

$$\left|\frac{\mathrm d f(\mathbf r)}{\mathrm dr}\right | \mathrm dr = \mathrm ds$$

Rearranging the above expreesion, we get the thickness $\mathrm d r$ as

$$\mathrm dr =\frac{\mathrm ds}{\left|\frac{\mathrm d f(\mathbf r)}{\mathrm dr}\right |}$$

Now, we have gotten the thickness at every point. Let's take a small element at $\mathbf r_0$ of length $\mathrm dl$. This the charge of that element would be

\begin{align} \mathrm dq &=\left(\int \frac{q(\mathbf r) \delta (\mathbf s)}{ \left|\frac{\mathrm d f(\mathbf r)}{\mathrm dr}\right |} \mathrm ds \right) \mathrm dl\\ \mathrm dq&=\frac{\mathbf r_0}{\left|\frac{\mathrm d f(\mathbf r)}{\mathrm dr}\right |_{\mathbf r=\mathbf r_0}}\mathrm dl \end{align}

Using the definition of linear charge density, $\lambda=\mathrm dq/\mathrm dl$, we get

$$\lambda(\mathbf r_0)=\frac{\mathbf r_0}{\left|\frac{\mathrm d f(\mathbf r)}{\mathrm dr}\right |_{\mathbf r=\mathbf r_0}}$$

This is the final expression. However, you might see that the function we given in the start should be such that $\left|\frac{\mathrm d f(\mathbf r)}{\mathrm dr}\right |\neq 0$, for all $\mathbf r$ on the curve $\gamma$.

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  • $\begingroup$ This is not correct. You can certainly have infinite charge densities, but finite total charges. Consider the point charge, whose density is $\rho(\mathbf{r}) = q \delta^3(\mathbf{r})$. This is clearly infinite at $\mathbf{r}=0$ because of the 3D Dirac-delta, but nevertheless gives a finite total charge when integrated over space. The idea is that charge densities are distributions, not functions. A similar thing can be done for the infinite plate of charge. $\endgroup$
    – Philip
    Aug 13, 2020 at 5:00
  • $\begingroup$ @Philip In my answer, I have stated that I am only talking about finite charge densities. If you somehow include the Dirac delta function, you are essentially no longer expressing the charge density, instead you are talking about a charge distribution in a lower dimension (in your given example, the charge distribution is equivalen to a point charge, 0-dimensions). So defining charge densities using Dirac delta functions and then finding the charge density of a lower dimensional charge density is useless, since you could directly use the properties of Dirac delta function to do the same. $\endgroup$
    – user258881
    Aug 13, 2020 at 5:47
  • $\begingroup$ @Philip Also, to define charge density on every possible hypersurface, you need to have Dirac delta-ish charge density everywhere, which definitely implies that the finite $N$-dimensional space is having infinite charge. $\endgroup$
    – user258881
    Aug 13, 2020 at 5:51
  • $\begingroup$ I'm afraid I don't understand what you've said at all. I agree that you were talking about finite densities. I also agree that this problem cannot admit a finite density everywhere (I believe that the OP knows this, as they have mentioned the $\delta-$function in their question). You seem to agree that an object like a point-charge or a sheet of charge can be defined to have a charge density using a $\delta-$function. While certain configurations (like the sheet of charge with constant $\sigma$) lead to infinite charge, others (like the point charge) don't. $\endgroup$
    – Philip
    Aug 13, 2020 at 6:10
  • $\begingroup$ @Philip Let's suppose that you are given the surface charge density $\sigma(\mathbf r)= q(\mathbf r) \delta (\mathbf s)$, where $\mathbf s=f(\mathbf r)$ where $f:V\to V$ is a function, and $q:V\to\mathbb R$. The solution of the equation $f(\mathbf r)=0$ is the curve $\gamma$. Now, it is trivial to see that only the curve $\gamma$ has non-zero linear charge density throughout its complete length. The relevant linear charge density is simply $\frac{\mathrm d q(\mathbf r)}{\mathrm dr}$. Is this what you were looking for? $\endgroup$
    – user258881
    Aug 13, 2020 at 7:29

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