I Understanding Gauss's Law: Where Does the Argument Break Down?

[greypilgrim]

I think the symmetry argument is flawed from the start. Here's why: Let's look at the electric field at $(0,0,0)$. An infinite sheet parallel to the $y$-$z$-plane at a distance $x>0$ creates an electric field
$$\left(\begin{array}{c}
-\frac{d\sigma}{2\varepsilon_0}\\
0\\
0\end{array}\right)
=
\left(\begin{array}{c}
-\frac{\rho}{2\varepsilon_0}dx\\
0\\
0\end{array}\right)
$$
So the half-space $x>0$ creates a field $\int_0 ^\infty -\frac{\rho}{2\varepsilon_0}dx$ which does not converge. If you're trying to argue that this should cancel with the integral over the other half-space, you'd need such integrals to converge.

It's like saying the integral $\int_{-\infty} ^\infty \text{sign}(x)dx$ has value 0, but it actually doesn't exist except in some Cauchy principal value sense.

Remember that in the usual case of a charged infinite sheet or wire the solution is only defined away from the charge, which obviously isn't possible in the 3D case. I guess the electric field of an uniformly charged space can just not be well-defined.

 

[Meir Achuz]

Gauss's law states that the integral of the normal component of electric field around a closed surface equals the charge inside the surface.
It says nothing about the angular distribution of the electric field at the surface. In order to find the electric field, some symmetry must be used to make the surface integral trivial. Otherwise, applying Gauss's law is more complicated than applying Coulomb's law.
In the case of a uniform charge density throughout 'all space' the electric field depends on how a large enclosing surface approaches infinity. Infinity is not a number. A limit must be taken as the surface grows larger and larger. If the bounding surface is a sphere about a single point, then the electric field is equal to \epsilon_0\rho{\bf r}/3, and spherically symmetric about that single point. I know of no other simple symmetry as the bounding surface approaches infinity. In that case, the electric field is indeterminate unless the Coulomb integral is performed. The electric field cannot be zero. Gauss did have something to tell us.

 

[tom.stoer]

If we consider a uniform charge density $\rho$ extending through all space, then by symmetry, I would argue that $\mathbf{E}=0$ in all space. However, this does not agree with what a naive application of Gauss's Law would predict since $\nabla\cdot\mathbf{E}=0\ne\rho/\epsilon$.

The electric field must not vanish. The very first assumption is already not correct.

In 1+1 dimensions the Gauss-law reads

$$\partial_x E_x = \rho$$

Therefore

$$E_x = \rho x + \text{const.}$$

It's more complicated in 3+1 dimensions, but it should be clear that there's no good reason why the electric field should vanish.

However the total charge is infinite which may force us to exclude this as an unphysical situation.

The solution is not unique. Uniqueness should follow from boundary conditions which haven't been discussed yet. From the simple solution in 1+1 dimensions it's obvious that they are hidden in the "+ const." term.

 

[Orodruin]

Gauss's law then states that the Electric field at the surface is only a function of the charge enclosed by the surface.

This is not what Gauss's law states, it is in essence a statement about the divergence of the electric field and the surface integral can never let you uniquely compute the electric field. As @Vanadium 50 has already pointed out, you also need additional information about the boundary conditions, which is what is missing from the original statement. A constant charge spread through all of space is incompatible with the usual boundary condition that the field goes to zero at infinity. Any boundary condition that you can impose on the solution is going to break the translational symmetry and thereby uniquely single out one solution.

You can also see this in the one-dimensional analogue of the problem $f'(x) = \kappa$, which has the solutions $f(x) = \kappa x + A$ with $A$ being an arbitrary constant. Unless you specify the behaviour at infinity, e.g., $\lim_{x\to\infty}[f(x) + f(-x)] = 2A = 0$, your solution will have undetermined constants. In the 3D case, you would generally find $\vec E = \rho_0 \vec x / 3\epsilon_0 + \vec k$, where $\vec k$ is a constant vector.

 

[tom.stoer]

In the 3D case, you would generally find $\vec E = \rho_0 \vec x / 3\epsilon_0 + \vec k$, where $\vec k$ is a constant vector.

The solution can be much more complex b/c you can add any $\vec{E}_0$ with $\nabla\vec{E}_0 = 0$, e.g.

$$\vec{E}_0 = \left(\begin{array}{c} 0 \\ y \\ -z \end{array}\right)\,f(x)$$

 

[Orodruin]

The solution can be much more complex b/c you can add any $\vec{E}_0$ with $\nabla\vec{E}_0 = 0$, e.g.

$$\vec{E}_0 = (0,y\,f(x),-z\,f(x))$$

You also need to satisfy $\nabla \times \vec E = 0$ and with that $\vec E_0$ generally
$$
\nabla \times \vec E = f'(x) [z \vec e_y + y \vec e_z] \neq 0.
$$

Edit: I will give you that you can add any divergence and curl free field.

 

[tom.stoer]

You also need to satisfy $\nabla \times \vec E = 0$ and with that $\vec E_0$ generally
$$
\nabla \times \vec E = f'(x) [z \vec e_y + y \vec e_z] \neq 0.
$$

you are right; sorry for being imprecise

Edit: I will give you that you can add any divergence and curl free field.

yes, that's basically the result; we may also discuss this in terms of the potential and harmonic functions

 

[Orodruin]

yes, that's basically the result; we may also discuss this in terms of the potential and harmonic functions

Indeed, in fact this is how I would prefer to approach this. Start with the particular solution $\vec E = \vec x \rho/3\epsilon_0$ and then solve $\nabla^2 V = 0$.

The general solution to the homogeneous equation for the potential is on the form
$$
V = \sum_{\ell = 0}^\infty \sum_{|m| \leq \ell} f_{\ell m} Y_{\ell m}(\theta,\varphi) r^\ell
$$
in spherical coordinates, where $f_{\ell m}$ must be chosen to satisfy whatever behaviour is imposed at infinity.

Edit: The $\ell = 0$ term is a constant that does not change the field at all. The $\ell = 1$ terms essentially correspond to the addition of a constant field. The higher $\ell$ terms all correspond to fields that are not constant, but still both divergence and curl free. The field $y\vec e_y - z \vec e_z$ should be among the $\ell = 2$ terms.

 

[NFuller]

The electric field must not vanish. The very first assumption is already not correct.

but it should be clear that there's no good reason why the electric field should vanish.

Again, you can argue that the electric field vanishes by symmetry.

it is in essence a statement about the divergence of the electric field

Yes, that is how I originally posed the problem.

You can also see this in the one-dimensional analogue of the problem f′(x)=κf′(x)=κf'(x) = \kappa, which has the solutions f(x)=κx+Af(x)=κx+Af(x) = \kappa x + A with AAA being an arbitrary constant. Unless you specify the behaviour at infinity, e.g., limx→∞[f(x)+f(−x)]=2A=0limx→∞[f(x)+f(−x)]=2A=0\lim_{x\to\infty}[f(x) + f(-x)] = 2A = 0, your solution will have undetermined constants. In the 3D case, you would generally find ⃗E=ρ0⃗x/3ϵ0+⃗kE→=ρ0x→/3ϵ0+k→\vec E = \rho_0 \vec x / 3\epsilon_0 + \vec k, where ⃗kk→\vec k is a constant vector.

So you are arguing that $\mathbf{E}\neq0$?

 

[Orodruin]

Again, you can argue that the electric field vanishes by symmetry.

No, you really cannot. To make a symmetry argument also the boundary conditions (in this case posed at infinity) must also be symmetric under whatever transformation you refer to. In the "standard" case of the point charge, you can assume the field to be vanish at infinity and this condition is perfectly symmetric under rotations. However, the case with a constant charge density is incompatible with a field that goes to zero at infinity and any behaviour at infinity is going to break your symmetry.

 

[tom.stoer]

Again, you can argue that the electric field vanishes by symmetry.

No, you can't.

If you have a charge density $\rho$ and an equation which allows you to solve for $E$ in terms of $\rho$, and if your symmetry argument results in an $E$ that does not solve the equation, then the argument violates the equation and is therefore wrong (or incompatible with the equation).

The Gauß law is a local equation. To construct a global solution you can start with a local one and extend it globally. But you must start with a solution. You must not start with some $E$ that does not solve the equation locally but hope that it does so globally.

So you are arguing that $\mathbf{E}\neq0$?

Yes.

(and if you don't like that you are free to call this situation unphysical)

 

[NFuller]

No, you really cannot. To make a symmetry argument also the boundary conditions (in this case posed at infinity) must also be symmetric under whatever transformation you refer to. In the "standard" case of the point charge, you can assume the field to be vanish at infinity and this condition is perfectly symmetric under rotations. However, the case with a constant charge density is incompatible with a field that goes to zero at infinity and any behaviour at infinity is going to break your symmetry.

So why are these same symmetry arguments used when finding the electric field due to infinite charge distributions in 1 and 2 dimensions? I have a hard time believing that $\mathbf{E}$ is non zero. If the universe looks the same in all directions out to infinity, then why would an electric field have any directional preference?

 

[NFuller]

If you have a charge density ρρ\rho and an equation which allows you to solve for EEE in terms of ρρ\rho, and if your symmetry argument results in an EEE that does not solve the equation, then the argument violates the equation and is therefore wrong (or incompatible with the equation).

Or the equation is not valid in the system given. This point has already been eluded to in the paper reference by greypilgrim, where the field produced does not satisfy the field equations.

 

[I like Serena]

Suppose the entire universe on one side of a plane has constant charge density. Then the electric field at the plane is infinite, isn't it?
Symmetry or not, this is a problem.
Now make the superposition with the charge on the other side of the plane.
The result is infinity minus infinity.
Gauss or not, something is going wrong there.

 

[Orodruin]

So why are these same symmetry arguments used when finding the electric field due to infinite charge distributions in 1 and 2 dimensions?

I assume here that you are talking about 1- and 2-dimensional charge distributions in 3-dimensional space, because the corresponding problem in 1-dimensional space does have the same problem. The point is that, unlike for the constant distribution in all of space, you can find boundary conditions at infinity that does have the same symmetries as the charge distribution itself. In particular, for the infinite line charge, the boundary condition that the field goes to zero as you go far away from the line as well as the translational symmetry along the line is satisfied by a $\vec e_\rho/\rho$ field (where $\rho$ is the radial polar coordinate) and for the infinite surface charge you can find a boundary condition such that the field goes to a constant field far away from the surface and is translationally invariant for translations within the surface. For the infinite volume charge, this is no longer possible. You cannot find a condition that is both rotationally and translationally symmetric (which are the symmetries of the charge distribution) and so you must impose boundary conditions that break these symmetries.

I have a hard time believing that $\mathbf{E}$ is non zero. If the universe looks the same in all directions out to infinity, then why would an electric field have any directional preference?

The point is that the universe does not look the same in all directions and/or does not display translational invariance. You are thinking only of the charge distribution but the boundary conditions must form a part of that statement. It is simply inconsistent to assume that. You really should not have a hard time believing that $\vec E$ is non-zero. In order for $\vec E$ to be an electric field at all it must satisfy Maxwell's equations and your assertion that it is zero is a direct violation of Maxwell's equations.

Note that the derivation of the field from a point charge (for example) is directly dependent on the implicit assumption that the boundary conditions satisfy the same symmetry properties as the charge distribution. If you put different boundary conditions, you will add a divergence and curl free field to the solution and it still satisfies Maxwell's equations.

 

[Orodruin]

Or the equation is not valid in the system given. This point has already been eluded to in the paper reference by greypilgrim, where the field produced does not satisfy the field equations.

This is what the paper says:

One way out of it is to consider Maxwell’s two equations for electrostatics ∇ · E = ρ/ε₀ , (11) and ∇ × E = 0, (12) and realize that the first one is not compatible with the answer dictated by the underlying symmetry of the distribution, that is, E = 0 everywhere is not a valid solution of Maxwell’s equations when we take ρ as a continuous uniform charge distribution fulfilling all the space.

This is not saying that Maxwell's equations are not valid. It is saying that Maxwell's equations are incompatible with the underlying symmetry of the charge distribution. The question then becomes: "Why is it incompatible?" The correct reply to that question is that it must be broken by the boundary conditions, just as indicated above, it is the only thing you have left that can break the symmetries. And not only can it break the symmetries, it has to break the symmetries when you take that charge distribution. If you are not talking about a solution to Maxwell's equations, you are really not talking about an electric field.

 

[tom.stoer]

Or the equation is not valid in the system given. This point has already been eluded to in the paper reference by greypilgrim, where the field produced does not satisfy the field equations.

Your intended „solution” is extremely strange; it solves a pseudo-problem, namely your (incomplete) symmetry argument, by introducing other problems.

I can‘t see any reason why this (incomplete) symmetry argument is stronger or better than well-defined solutions of Maxwell‘s equations.

A charge distribution causing problems with established physics is to be ruled out, instead of ruling out established physics.

 

[NFuller]

This is not saying that Maxwell's equations are not valid. It is saying that Maxwell's equations are incompatible with the underlying symmetry of the charge distribution. The question then becomes: "Why is it incompatible?" The correct reply to that question is that it must be broken by the boundary conditions, just as indicated above, it is the only thing you have left that can break the symmetries. And not only can it break the symmetries, it has to break the symmetries when you take that charge distribution. If you are not talking about a solution to Maxwell's equations, you are really not talking about an electric field.

So what boundary conditions are you assuming? What would the field look like under that assumption?

 

[Orodruin]

So what boundary conditions are you assuming? What would the field look like under that assumption?

It does not matter as long as they are compatible with the differential equation, and if they are, then they will break translational symmetry. What the field will look like has been discussed already. The general expression for the potential of the homogeneous problem in spherical coordinates is given in post #38. Regardless of the arbitrary constants, a field $\vec E = - \nabla V$ for that $V$ is going to satisfy $\nabla \cdot \vec E = 0$. To this you add the particular solution given earlier in the same post and you will have the general solution. The arbitrary coefficients have to be fixed based on the behaviour at infinity. Note that none of the terms (except the constant $\ell = 0$ term) leads to a field that is zero at infinity, which is why you can make the assumption that the field vanishes at infinity in the case of a point charge. However, in the case of a constant charge density, you cannot cancel the growth of the particular solution in all directions simultaneously and therefore also not assume that the field vanishes at infinity.

Edit: To elaborate on this. In the same spirit as putting bounds such that the solution is zero at infinity in the case of a point charge, you can impose boundary conditions such that the field grows at most as $r$ as $r \to \infty$. This rules out all of the modes with $\ell \geq 1$ from the expansion of the homogeneous potential and leaves the solution on the form $\vec E = \rho \vec x/3\epsilon_0 + \vec k$.

 

[Orodruin]

then the electric field at the plane is infinite, isn't it?

Again, this depends on you giving appropriate boundary conditions to your region of interest.

 

[I like Serena]

Again, this depends on you giving appropriate boundary conditions to your region of interest.

Doesn't it follow directly from the principle of superposition?
Infinite force from one side (on a test charge) and no force from the other side?

 

[Orodruin]

Doesn't it follow directly from the principle of superposition?

Sure, you can superpose solutions. But if you superpose solutions with inhomogeneous boundary conditions the superposition will satisfy boundary conditions that are the linear combination of the boundary conditions you superposed.

Also, this is not what I was replying to. The solution being infinite at the surface is not well defined. Instead, you have to impose proper boundary conditions at $r\to \infty$. If you do so, your solution will be finite.

 

[I like Serena]

Sure, you can superpose solutions. But if you superpose solutions with inhomogeneous boundary conditions the superposition will satisfy boundary conditions that are the linear combination of the boundary conditions you superposed.

Also, this is not what I was replying to. The solution being infinite at the surface is not well defined. Instead, you have to impose proper boundary conditions at $r\to \infty$. If you do so, your solution will be finite.

As yet I'm not talking about solutions to Gauss's law.
@NFuller uses symmetry in an infinitely charged universe to conclude that the superposition of all forces on a test charge must be zero.
My point is that if we have a half charged universe, that same superposition would lead to an infinite force on a test charge, so a fully charged universe would lead to subtracting infinity from infinity, defeating the argument that it is supposed to be zero - the result is not-a-number (NaN), or just 'undefined'.
As such there is no contradiction to Gauss's law, since the result is undefined.
It seems to fit your conclusion that the solution to Gauss's law has an unknown constant vector (or some such) that cannot be determined.

 

[Orodruin]

As yet I'm not talking about solutions to Gauss's law.

Then you are not talking about an electric field. Electric fields satisfy Gauss’s law. If you want to make your own model of electrostatics you are welcome to do so (elsewhere, personal theories are against forum rules).

@NFuller uses symmetry in an infinitely charged universe to conclude that the superposition of all forces on a test charge must be zero.

Where do you think the superposition principle comes from? It comes from the equation of motion, ie, Gauss’s law, being a linear differential equation. To use the superposition principle, you are implicitly assuming Gauss’s law. Now, it is generally not presented like this in high-school because most people have an easier time accepting heuristic arguments and ”field from separate particles”, but that does not make it any less true that superposition being possible relies on the field satisfying Gauss’s law. In the typical superposition of point charges argumentation, the second implicit assumption is that the field vanishes at infinity. This assumption can no longer hold in the case of an infinitely extended charge and in general the translational symmetry is broken by the boundary condition you have to impose.

Let me repeat the main message again, because it is important: You simply cannot ignore the symmetry (or lack thereof) of the boundary conditions if you want to apply symmetry arguments. It does not matter if the charge distribution displays a symmetry - if the boundary condition does not display the same symmetry - then it is not a symmetry of the system.

My point is that if we have a half charged universe, that same superposition would lead to an infinite force on a test charge, so a fully charged universe would lead to subtracting infinity from infinity, defeating the argument that it is supposed to be zero - the result is not-a-number (NaN), or just 'undefined'.
As such there is no contradiction to Gauss's law, since the result is undefined.
It seems to fit your conclusion that the solution to Gauss's law has an unknown constant vector (or some such) that cannot be determined.

You are making the same mistake as the OP here. You fail to account for the fact that with a charge distribution that extends to infinity, you need non-zero boundary conditions on the limiting behaviour. Undefined functions do not solve differential equations, functions with well defined behaviour do.

 

[I like Serena]

Then you are not talking about an electric field. Electric fields satisfy Gauss’s law. If you want to make your own model of electrostatics you are welcome to do so (elsewhere, personal theories are against forum rules).

Sure, electric fields satisfy Gauss's law.
That doesn't change the fact that the superposition principle - also observed empirically - is supposed to coincide with it.
Isn't that the whole point of this thread?

 

[Metmann]

If you start from the (Lorentz-invariant) Lagrangian of electrodynamics you arrive at the Maxwell Equations if and only if charge and current vanish at infinity, else you get boundary terms. So if we consider the extremization of the action as the basic physical principal, than the answer is:

Maxwell's equations (and hence also Gauß's law) are just not valid for a constant charge density without spatial cutoff.

I think this answer is quite reasonable, since an infinitely stretched constant charge density is just unphysical. Also: If you think about how one usually derives the Gauß law in non-relativistic electrodynamics, you notice, that you always start with the integral formulation which necessarily incorporates compact volumes. The transition to the differential formulation is only valid within these volumes. The extension to whole Euclidean space can be done only if the charge density is integrable, i.e. if there exists a finite total charge.

You could also argue, that a constant charge density on the whole space is equivalent to empty space with shifted ground state energy. Since except for GR the total energy is meaningless in physics (only energy differences matter), constant charge density (infinitely stretched) and empty space are totally equivalent in non-gravitational electrodynamics, if you neglect possible further quantum influences. Hence the solution is trivially a constant E-field.

 

[NFuller]

It does not matter as long as they are compatible with the differential equation

The solution doesn't set the boundary condition, the boundary condition sets the solution. Even with some boundary condition at infinity, how do you know where to set the origin in space to make the electric field you suggested valid? Infinity is the same distance away from every point in space, so the boundary condition doesn't really set an origin.

I have been looking into some more generalized constructions of Maxwell's equations. In particular, that the action becomes zero if the fields satisfy the correct equations of motion, which are of course the Maxwell Equations. However, the Lagrangian formalism again relies on the fields decaying at infinity. I'm wondering if the principle of least action would lead to a different solution, or any solution, without assuming the fields vanish at infinity.

 

[Orodruin]

The solution doesn't set the boundary condition, the boundary condition sets the solution.

I never said that. I said that you have to put boundary conditions that are compatible with the differential equation.

Even with some boundary condition at infinity, how do you know where to set the origin in space to make the electric field you suggested valid? Infinity is the same distance away from every point in space, so the boundary condition doesn't really set an origin.

That is not how infinity works. Look at the (spatially) one-dimensional case $E' = \kappa$, where you obtain $E = \kappa x + A$. A behaviour at infinity that fixes $A$ will be of the type $\lim_{x\to \infty} [E(x) + E(-x)] = E_0$ and set $A = E_0/2$ and therefore single out $x = - E_0/2\kappa$ as the point with $E = 0$. Your constructions in the three-dimensional case will look similar and instead of the sum of two points involve integrals over the spherical harmonics.

In particular, that the action becomes zero if the fields satisfy the correct equations of motion, which are of course the Maxwell Equations.

Yes, the action may be zero, but that is not what you want to find when you do Lagrangian mechanics. You want to find out for which field configurations the variation of the action is zero, i.e., when the action is stationary.

Edit: Also, another pet peeve of mine. "Principle of least action" is a confusing misnomer, it should really be called the "principle of stationary action".

If you start from the (Lorentz-invariant) Lagrangian of electrodynamics you arrive at the Maxwell Equations if and only if charge and current vanish at infinity, else you get boundary terms.

The action would formally be infinite, but that does not really stop you from having variations of the fields that are localised, which will give you finite variations without boundary terms and those variations will give you Maxwell's equations. Even if you allow variations that do not vanish at infinity, the variation of the action must be zero also for those variations that do vanish. This is the underlying reason why, when dealing with variations on spaces with a boundary, the Euler-Lagrange equations always hold and the effect of allowing variations on the boundary is to impose natural boundary conditions. A good question might be what those natural boundary conditions would be in this scenario. I suspect that you will get inconsistencies there.

If you think about how one usually derives the Gauß law in non-relativistic electrodynamics, you notice, that you always start with the integral formulation which necessarily incorporates compact volumes. The transition to the differential formulation is only valid within these volumes. The extension to whole Euclidean space can be done only if the charge density is integrable, i.e. if there exists a finite total charge.

You do not need the integral form of Gauss's law to hold for the entire space to derive the infinitesimal form. You only need it to hold for any compact sub-volume. In this context I would also point out that fundamental laws are not derived, but introduced based on observation. The "derivation" of Gauss's law in electrodynamics is essentially a collection of arguments starting from observed properties of electromagnetism that just so happen to be the properties of solutions to the Laplace equation. In fact, this was actually already set in stone once you observed and assumed the $1/r^2$ behaviour of the potential of a point charge and the superposition of fields, as the Green's function of the Laplace equation just goes as $1/r^2$ and the superposition principle requires a linear differential equation.

You could also argue, that a constant charge density on the whole space is equivalent to empty space with shifted ground state energy.

To be honest, this sounds very fishy and ad hoc. Can you back up how it would be equivalent to empty space?

 

[tom.stoer]

You may want to study this equation in one spatial dimension. There are essentially four different topologies: $R$ = the real line, $R^+$ which allows you to add a static charge at $x = 0$, $[x,L]$ which allows for two charges, and compact $S^1$ w/o boundary. You may derive the Maxwell equations including boundary terms and investigate whether a topology with some appropriate boundary condition allows for constant $\rho$ and vanishing $E = 0$.

For $S^1$ constant charge always implies $\rho = 0$.

 

[Metmann]

when dealing with variations on spaces with a boundary, the Euler-Lagrange equations always hold and the effect of allowing variations on the boundary is to impose natural boundary conditions. A good question might be what those natural boundary conditions would be in this scenario. I suspect that you will get inconsistencies there.

That's what I mean with 'boundary terms'. I also don't think there is a natural and consistent way to choose these.

The "derivation" of Gauss's law in electrodynamics is essentially a collection of arguments starting from observed properties of electromagnetism that just so happen to be the properties of solutions to the Laplace equation.

You can also 'derive' it using the classical concept of charge conservation and that uses the integral formulation of Gauß's law.

In fact, this was actually already set in stone once you observed and assumed the 1/r21/r^2 behaviour of the potential of a point charge and the superposition of fields, as the Green's function of the Laplace equation just goes as 1/r21/r^2 and the superposition principle requires a linear differential equation.

That's indeed true. I am curious: Does this 'derivation' break down somewhere during the transition from the local statement (localized point charge) to the global statement (uncountably infinite number of point charges spread to infinity)?

To be honest, this sounds very fishy and ad hoc. Can you back up how it would be equivalent to empty space?

If you consider just electrostatics, a constant infinitely spread charge density $\rho$ amounts to an infinite total energy
$$ E \sim \rho^2 \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{x} \, \mathrm{d}^3\vec{x}'}{|\vec{x}-\vec{x}'|} = \infty. $$
Energy density would be
$$ e(\vec{x}) \sim \rho^2 \int_{\mathbb{R}^3} \frac{\mathrm{d}^3\vec{x}'}{|\vec{x}-\vec{x}'|}, $$
which is also infinite everywhere, but the integrand is definitely translationally invariant, hence energy density is of course homogeneous.
Since there is no difference between any pair of points, I would suggest to treat the infinite $E$ as a vacuum energy, hence subtracting it from the physical energy by setting $\rho=0$.
In my opinion, only spatial and temporal deviations of physical quantities have physical meaning in classical physics. Hence, the subtraction of the infinite background energy does not change the physics.
Of course here I neglected possible varying mass density or some temporal behaviour.

For S1S^1 constant charge always implies ρ=0\rho = 0.

Interesting. What about $S^n$ in general? My first guess would be, that for $n>=2$ the situation is different due to simply connectedness. Solving the problem for $S^3$ instead of $\mathbb{R}^3$ would probably help.