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

[NFuller]

So, this has been bothering me for a few days and I'm having trouble understanding where the fault is. 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 $\nabla\cdot\mathbf{E}=0\ne\rho/\epsilon$. So where exactly is the argument breaking down? Is there something unusual about describing a vanishing divergence over infinite space?

 

[andrewkirk]

Interesting question. Upon reflection, my guess is that the integral form of Gauss's Law rests on an unstated assumption that the average charge density outside of the closed surface is zero. In practice, that will usually be close enough to correct to ignore any inaccuracy. A re-stated Gauss's Law that coped with this thought experiment might replace Q, the charge inside the surface, by $(Q-V\rho_0)$ where $V$ is the volume inside the surface and $\rho_0$ is the average charge density across all space. Similarly, for the differential case, one could replace $\rho$ by $\rho-\rho_0$.

In practice, one need not make these adjustments because presumably $\rho_0\approx. 0$.

 

[pixel]

What Gaussian surface are you choosing, and what symmetry assumptions about the direction of the E-field on that surface are you making?

 

[NFuller]

Interesting question. Upon reflection, my guess is that the integral form of Gauss's Law rests on an unstated assumption that the average charge density outside of the closed surface is zero. In practice, that will usually be close enough to correct to ignore any inaccuracy. A re-stated Gauss's Law that coped with this thought experiment might replace Q, the charge inside the surface, by $(Q-V\rho_0)$ where $V$ is the volume inside the surface and $\rho_0$ is the average charge density across all space. Similarly, for the differential case, one could replace $\rho$ by $\rho-\rho_0$.

In practice, one need not make these adjustments because presumably $\rho_0\approx. 0$.

Hey andrewkirk,
I am hesitant to think that Gauss's law is an approximation. I agree that what you have done seems to force out the correct solution but is the inclusion of $\rho_{0}$ justified? Basically, it seems that $\rho_{0}$ is highly artificial because it is zero until the charge distribution becomes infinite, then $\rho_{0}=\rho$ to fix the problem.

What Gaussian surface are you choosing, and what symmetry assumptions about the direction of the E-field on that surface are you making?

I was pointing out a problem with the differential form of Gauus's law so I did not specify a Gaussian surface. From what I can tell however, the integral form has the same problem. If you consider the Gaussian surface to be a spherical shell and take the radius out to infinity, it predicts a non-vanishing electric field.

 

[anorlunda]

Think of the integral form. Gauss' Law let's you calculate the field due to a charge within the surface. But that does not forbid other fields from charges outside the surface. For uniform density everywhere, I expect that the vector sums of all those forces to be zero because of the symmetry arguments the OP makes. So I agree with the OP and with @andrewkirk

 

[BruceW]

ooh, this is an interesting question. I believe the system you are talking about is called a "non-neutral Coulomb gas" in statistical physics. It is maybe quite a niche subject. I found a couple of papers talking about this problem in 2D http://www.sciencedirect.com/science/article/pii/092145349390271Q and https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.59.1001, which has application for superconducting films. From what I can work out, they include both ultraviolet and infrared cutoffs. In particular, the infrared cutoff means the electric potential obeys $$( \nabla^2 - {\lambda_c}^{-2}) V_{\lambda_c}(r) = -2 \pi \delta (r)$$ and this has something to do with the screening of the electric field. I guess the 2 cutoffs allows for calculations to be made, and then at the end of calculations, you can eliminate the cutoffs in a self-consistent way. Something along those lines, I don't know the detail really.

Edit : For clarity, $$\nabla^2 V(r) = -2 \pi \delta (r)$$ is the potential of point charge without using the cutoff.

 

[NFuller]

But that does not forbid other fields from charges outside the surface.

In the case of a spherical uniform charge distribution, you can use symmetry to argue that the electric field is radially symmetric at the Gaussian surface. Gauss's law then states that the Electric field at the surface is only a function of the charge enclosed by the surface. Now take the radius of the sphere to infinity. Apparently something about Gauss's law fails in this limit. My question is which part of the argument is failing here?

I believe the system you are talking about is called a "non-neutral Coulomb gas"

Not really. I'm just assuming all the charges are locked in place as if all space was filled with a good insulator with uniform charge density.

 

[andrewkirk]

it seems that $\rho_{0}$ is highly artificial because it is zero until the charge distribution becomes infinite, then $\rho_{0}=\rho$ to fix the problem.

Why do you think it would be zero? It will be negligible because charge only exists where there are particles and the density of particles in the universe in negligible - because of all the empty inter-galactic space. But there's no reason why it need be zero. Since it's so small, we can adopt an assumption that it is zero for the purpose of calculation, and that's what Gauss's law does. We only need to drop that assumption in the case of a thought experiment that (I expect) is unrealisable in practice.

 

[NFuller]

But there's no reason why it need be zero. Since it's so small, we can adopt an assumption that it is zero for the purpose of calculation, and that's what Gauss's law does.

This is a huge and unverified assumption. You are saying that there is some unidentified quantity which is so small it has never been detected but fixes the problems encountered when dealing with an infinite charge distribution.

 

[andrewkirk]

This is a huge and unverified assumption.

What exactly is the assumption that you are concerned about, and why do you believe it to be untenable?

 

[NFuller]

What exactly is the assumption that you are concerned about, and why do you believe it to be untenable?

It sounds like you are saying that Gauss's law is only an approximation and that to get the exact answer, you would need to invoke a new term $\rho_{0}$. The way you would calculate this term and when it becomes important is somewhat mysterious to me. As far as I know, Gauss's law is already exact without this corrective term.

 

[andrewkirk]

Yes that's right, the law is not exact. Quite apart from this issue, it is an approximation that only works when relativistic and quantum effects are small enough to be ignored.

The $\rho_0$ term never becomes important in this universe because, if we accept the cosmological principle that the universe is homogeneous and isotropic at the large scale (on which most cosmology is based), it follows that the term is negligible because the universe is almost entirely empty space - and hence chargeless.

 

[NFuller]

Yes that's right, the law is not exact. Quite apart from this issue, it is an approximation that only works when relativistic and quantum effects are small enough to be ignored.

This is a classical problem though and in classical ED, Gauss's law is exact.

The ρ0ρ0\rho_0 term never becomes important in this universe

What does this mean? I still don't know how you are defining $\rho_{0}$ so It's not clear to me if it is a physically or mathematically meaningful variable.

 

[andrewkirk]

I still don't know how you are defining $\rho_{0}$

See post #2.

 

[jasonRF]

I tend to take a pragmatic engineering approach, so I would just solve the equation. One solution to $\nabla \cdot \mathbf{E} = \rho / \epsilon_0$ for constant $\rho$ is $\mathbf{E} = \frac{\rho }{3 \epsilon_0}\mathbf{r}$. Since there is no preferred direction the fact that the field is radial seems reasonable by symmetry.

EDIT: in case you were wondering, I used no process to find this solution. When I first learned that $\nabla \cdot \mathbf{r} = 3$ I found it amusing for some reason and never forgot it. So when looking for a vector field that had a constant divergence it immediately came to mind. I also solved the PDE before attempting the integral form; the solution of the PDE was so simple and had such great symmetry that it was clear the integral form had to work.

Using the integral form of Gauss's law to solve for the field requires symmetry and a proper exploitation of that symmetry. I think there are very few problems that can be solved that way and I have goofed in this regards before, but this problem may be one . The charge density is uniform so is certainly spherically symmetric. If we assume the field is radial and only a function of $r$ and use a Gaussian surface that is a sphere of radius r, then the surface integral of the radial field yields $4 \pi r^2 E_r$ and the charge enclosed is $\frac{4}{3} \pi r^3 \rho$. Combining these results, I get that Gauss's integral law gives $E_r = \frac{\rho r}{3 \epsilon_0}$, in agreement with the solution to the PDE.

Does that make sense?

Jason

 

[jasonRF]

Now I see the problem with my post above. There is no way to tell where the origin is in the infinite charge density. If I move the origin, my solution above must change with it. This indicates that the uniqueness theorem may be violated. What are the exact mathematical requirements for uniqueness to hold? Is this simply the case of a mathematically ill-posed problem?

Jason

 

[NFuller]

Now I see the problem with my post above. There is no way to tell where the origin is in the infinite charge density. If I move the origin, my solution above must change with it. This indicates that the uniqueness theorem may be violated. What are the exact mathematical requirements for uniqueness to hold? Is this simply the case of a mathematically ill-posed problem?

Jason

Right! This is the issue I'm having.

 

[BruceW]

Not really. I'm just assuming all the charges are locked in place as if all space was filled with a good insulator with uniform charge density.

Oh, I see. I believe you should still be able to use the cut-offs to get an answer. I think that for electrostatics of charges that continue to infinity, you need to use the more complicated equations, rather than the usual Gauss' law. Sorry I don't know much about it, so I can't give a better answer. I guess it should be not so surprising that the usual method doesn't work, since trying to use the usual method, you get an infinite electric potential, which doesn't decay at infinity.

Edit: well, we should expect that it doesn't decay at infinity. The point I was meaning is that the usual derivation for uniqueness of the electric potential makes use of the fact that it decays sufficiently fast at infinity https://en.wikipedia.org/wiki/Uniqueness_theorem_for_Poisson's_equation

 

[tech99]

So, this has been bothering me for a few days and I'm having trouble understanding where the fault is. 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 $\nabla\cdot\mathbf{E}=0\ne\rho/\epsilon$. So where exactly is the argument breaking down? Is there something unusual about describing a vanishing divergence over infinite space?

E is defined as the force experienced by a unit charge. Force on an object can only be measured if we have a reference frame so we can measure acceleration, or alternatively, use the reaction of the force on another object. As your space is homogenous, we cannot measure force and we cannot measure an E. We cannot have a force if there is nothing for it to react against.

 

[rude man]

So, this has been bothering me for a few days and I'm having trouble understanding where the fault is. 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 $\nabla\cdot\mathbf{E}=0\ne\rho/\epsilon$. So where exactly is the argument breaking down? Is there something unusual about describing a vanishing divergence over infinite space?

This would seem another example of the age-old problem of infinities. I'd say the answer is there is no such thing as an infinite volume of charge (in all directions). For any finite volume the theorem gives the correct value of ∫∫D⋅dA = Q.

So similarly the potential of an infinite line of charge of finite charge density λ is also infinite. And so on.

 

[I like Serena]

I tend to take a pragmatic engineering approach, so I would just solve the equation. One solution to $\nabla \cdot \mathbf{E} = \rho / \epsilon_0$ for constant $\rho$ is $\mathbf{E} = \frac{\rho }{3 \epsilon_0}\mathbf{r}$. Since there is no preferred direction the fact that the field is radial seems reasonable by symmetry.

EDIT: in case you were wondering, I used no process to find this solution. When I first learned that $\nabla \cdot \mathbf{r} = 3$ I found it amusing for some reason and never forgot it. So when looking for a vector field that had a constant divergence it immediately came to mind. I also solved the PDE before attempting the integral form; the solution of the PDE was so simple and had such great symmetry that it was clear the integral form had to work.

Using the integral form of Gauss's law to solve for the field requires symmetry and a proper exploitation of that symmetry. I think there are very few problems that can be solved that way and I have goofed in this regards before, but this problem may be one . The charge density is uniform so is certainly spherically symmetric. If we assume the field is radial and only a function of $r$ and use a Gaussian surface that is a sphere of radius r, then the surface integral of the radial field yields $4 \pi r^2 E_r$ and the charge enclosed is $\frac{4}{3} \pi r^3 \rho$. Combining these results, I get that Gauss's integral law gives $E_r = \frac{\rho r}{3 \epsilon_0}$, in agreement with the solution to the PDE.

Does that make sense?

Jason

It seems to me that your solution is correct for a uniformly charged ball.
At the center of the ball the electric field is zero.
And the electric fields builds up radially to bigger and bigger values until we reach the surface of the ball.
This matches with the principle of super position, where we consider the electric field as a super position of the contributions from point charges.

If we take it to the limit, the electric field would become bigger and bigger ad infinitum, which is an impossibility.
This actually makes sense -- we have proven that an infinite universe with constant charge density is not possible.

 

[NFuller]

I'd say the answer is there is no such thing as an infinite volume of charge (in all directions).

we have proven that an infinite universe with constant charge density is not possible.

I think you are both missing the point. It's not about whether or not an infinite charge density exists, its about why the solution provided by Gauss's law fails. Remember that Gauss's law works for other infinite charge distributions such as an infinite rod or infinite sheet of charge even though these things don't actually exist.

 

[greypilgrim]

Here's a discussion about this question:
http://www.sbfisica.org.br/rbef/pdf/332701.pdf

 

[Charles Link]

In the differential equation, $\nabla \cdot E =\rho/\epsilon_o$, there is always a homogeneous solution to $\nabla \cdot E=0$ that may need to get added to it. $\\$ A similar thing occurs when you consider $B=\mu_o H+M$ and take $\nabla \cdot B=0$. (Take divergence of both sides of the equation). This gives $\nabla \cdot H=-\nabla \cdot M/\mu_o$ which gives $H(x)=-\int \frac{(\nabla' \cdot M(x')) (x-x')}{4 \pi \mu_o |x-x'|^3} \, d^3 x'$. The question is, where is the contribution to $B$ from any currents in conductors? And the answer is that it shows up in the homogeneous solution $\nabla \cdot H=0$.

 

[Vanadium 50]

I've spent a few days thinking about this, figuring that would be more productive than writing, and think I understand it.

The implicit assumption is that the electric field from a configuration of charges is unique. How do we know that? The proof involves something called a Helmholtz Decomposition. One of the conditions of this proof is that fields far away ("at infinity") vanish. This is true for a line of charge (symmetry reduces this to a 2-d problem) and sheet of charge (symmetry reduces this to a 1-d problem) but not a universe of charge.

 

[NFuller]

Here's a discussion about this question:
http://www.sbfisica.org.br/rbef/pdf/332701.pdf

This is what I'm looking for! So it looks like the problem is coming from the fact that Helmholtz's theorem doesn't guarantee that $\mathbf{E}$ can be described by its gradient and curl if $\mathbf{E}$ doesn't decay rapidly enough. Considering this, it's a bit surprising that Gauss's law works for the field above an infinite plane where $\mathbf{E}$ is constant out to infinity.

 

[Vanadium 50]

Considering this, it's a bit surprising that Gauss's law works for the field above an infinite plane where E\mathbf{E} is constant out to infinity.

The fact that fields vanish is a sufficient condition, not a necessary one. In the 1-d problem, all you really need to do is to toss out all the solutions that have the electric field blow up at infinity.

 

[NFuller]

Hey Vanadium 50, I saw your post just after I posted the one above. What are your thoughts for the infinite sheet of charge where the electric field is constant throughout space?

 

[Vanadium 50]

See message 27.

 

[Rap]

Take a sphere of radius R containing constant charge density ρ. Everything is radial, so we just need the radial coordinate r. The electric field E(r) is then radially directed, and proportional to r inside the sphere, because, as JasonRF said, ∇⋅r = 3, a constant. Specifically, the radial component is Er = r ρ /(3 εo), which is not a function of R. So you let R go to infinity and nothing changes, the E field is zero at the origin, its divergence is the density. But if you offset the origin of the sphere and let R go to infinity, you don't get the same answer.

Its kind of like the integral of x/(1+x^2) from -∞ to ∞. You can take the limit of the integral from -N to N as N goes to infinity and get one answer, then take the limit of the integral from -N to 2N as N goes to infinity and get another answer. The integral itself is undefined, and in the electric field problem, as posed, the electric field is undefined, because the idea of all space having a constant charge density is not a clearly defined concept, just as integrating x/(1+x^2) from -∞ to ∞ is not a clearly defined concept.

 

[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.