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.

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