Linear charge density of a path on a surface 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.
 A: 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 $\gamma$.
