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

[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.

 

[Ken G]

This question is actually of significant historical importance, but in the context of Newton's gravity, rather than Maxwell's equations. The basic issue for static electric fields is of course the same.

Newton assumed the universe should be infinite and homogeneous, and he came to the same resolution as NFuller, that the gravitational field should be zero by symmetry arguments-- if it doesn't know how to point, it must be zero. He regarded that as a deeper principle than the equations he was solving, so although the issues about the problems with applying symmetry arguments to Gauss' law that have been raised are entirely valid, Newton, who knew the necessary version of Gauss' law (i.e. that a spherically symmetric mass distribution had to produce a field toward the origin that depended only on the enclosed charge), felt the symmetry argument supercedes the mathematics of forming a proper solution. Perhaps it could be said he felt the boundary condition must be whatever is necessary to produce a zero field, because a field that doesn't know which way to point must be zero. The point is, Newton felt his gravity was consistent with a static universe, and this mistake caused him to miss that the universe should be dynamical.

Of course general relativity reaches a different conclusion about an infinite homogeneous mass distribution, but that's a different theory, so it's easy to overlook that Newton could also have realized the universe should be dynamical. I have always felt that Newton was basically wrong, in the sense that even if Newton's own theory of gravity were applicable, the gravitational field should not be taken to be zero, and the universe could not be static. I think the problem is in the implicit assumption that the field must be a "thing," handed to us without our input or participation, so must be unique. But this is never actually required in empirical sciences-- we only need the observations to come out as predicted, including whatever input or participation we need to supply. The observation is that a homogeneous mass distribution that is bounded and stationary undergoes contraction, and the timescale for density increase is given by 1/sqrt(rho*G). A similar expression would hold for the expansion of a homogeneous charge distribution with fixed charge-to-mass ratio. Notice this timescale depends on the local density, so the global scale is irrelevant and should exhibit no problems even when extended to infinity. The contraction/expansion timescale is what must be unique, not the fields, because observing fields requires a somewhat arbitrary calibration procedure, which may need to dovetail with the frame of reference or perspective of the observer.

So I think the physical problem here is in the implicit assumption that the field must be unique. It seems odd that the field you calculate could depend on where you set the origin, but that's not odd if the behavior you are trying to understand (contraction or expansion toward or away from some origin) also depends on where you set the origin-- the timescale for the contraction or expansion does not depend on where you set the origin. Also, the OP was bothered by the increasing magnitude of the field as you go away from the origin, but this can depend on the calibration procedure you use to measure fields, and there is no rule that says the field cannot depend on the calibration, nor that it cannot grow without limit if that's how you choose to calibrate it. All that has to be satisfied in an empirical science is that once you choose your origin (i.e., local reference frame), and once you have a self-consistent procedure for obtaining correct results to all your measurements from that perspective, then all observers must get the results they predicted, even if those results depend on the procedure used.

In short, I feel we should never have assumed the field must be unique, as that assumption is more than we get to assume when using empirical science. I agree with Newton that the physical outcome should always trump the formal mathematical issues, but I would add that a physical outcome is a description by an observer using a given procedure, and is not in and of itself something that needs to be unique. This is very much the perspective of relativity, so I feel that relativity has already solved the issue even without replacing Newton's gravity with Einstein's, which is reasonable because the solution can hold even at times when the speeds are much less than the speed of light. Hence I'm saying not only that I agree the solution is allowed to depend on implicit boundary conditions, but also that different boundary conditions are allowed within the same physical problem, if they dovetail with different procedures by different observers. Relativity basically says that it is only necessary that the observers be right, not that they agree!

 

[TeethWhitener]

If we consider a uniform charge density ρρ\rho extending through all space

Other people have basically mentioned this, but the divergence theorem (and Stokes' theorem in general) only holds when $V$ is a compact subset of $\mathbb{R}^n$. $\mathbb{R}^n$ itself is not compact. Thus you aren't guaranteed that
$$\int_{\mathbb{R}^n} \nabla\cdot{\bf F}dV=\int_{\partial \mathbb{R}^n} {\bf F}\cdot {\bf dA}$$
So you can't make the claim that $\nabla\cdot\bf E=\rho/\varepsilon_0$. I'm not even clear if it's possible to formulate the meaning of $\partial \mathbb{R}^n$ in a coherent way.

 

[tom.stoer]

So I think the physical problem here is in the implicit assumption that the field must be unique ...

... and is not in and of itself something that needs to be unique. This is very much the perspective of relativity, so I feel that relativity has already solved the issue

Field equations for the electric field do have unique solutions if boundary conditions are included.

Hence I'm saying not only that I agree the solution is allowed to depend on implicit boundary conditions, but also that different boundary conditions are allowed within the same physical problem

Changing the boundary or boundary condition usually changes the physical problem.

 

[Ken G]

Changing the boundary or boundary condition usually changes the physical problem.

Which is my point-- it is appropriate to change the physical problem when one changes the observer's perspective/procedure. So it's not a mathematics problem, or a symmetry problem. It's a physics problem, so requires a physics solution, by which I mean, a solution that merely connects the observer to their own predictions, with no need to use an absolute language about what is happening.

 

[tom.stoer]

Ok, sorry, then somehow I got you wrong

 

[Ken G]

There could be an important difference between Newton's gravity and Coulomb's electric force, in that all matter has the same gravitational charge-to-mass ratio, whereas we can build instruments with various different charge-to-mass ratios. Hence, one does not say that the Coulomb force admits to an equivalence principle. Still, that issue doesn't seem to be the crux of the problem with using Gauss' law in the gravitational context, so it seems a similar solution as relativity finds in the gravitational case should be possible in the Coulomb case, even in the limit of taking c to infinity so no new theory would be needed. We would only need to relax the requirement that physics regard itself as enabled to make absolute statements like "what is the actual field here."

ETA: Let me express this point in terms of a trivial but interesting theorem that applies to any sphere of constant charge density. Let the origin be at the center, and then the usual electric field is equal to -x, where x is the displacement vector from the origin (and the charge density is suitably scaled). Similarly, the field at y is -y. But now let us regard y as a new origin for our coordinates, such that the coordinates at x become x-y. If an observer at y simply adds y to every field we calculated before, then the field at x is -x+y, which again obeys the same rule-- it is negative the displacement from the new origin! What this means is, if we have a spherical homogeneous charge distribution, we can regard any point within it as the origin where the field is zero, and use Gauss' law around that point, ignoring all charges outside the Gaussian sphere, as long as we are willing to have a field that differs from the original by a fixed vector y. All we then need is a way to regard that fixed field difference as an ignorable difference (stemming from the different perspectives of the observers), and we allow all observers to correctly infer what they will observe, without any need for a concept of an "absolute field," or any machinery other than the integral form of Gauss' law.

Notice this is the 3D homogeneous analog to the points made above in 1D involving an arbitrary constant of integration, but my point is that this constant of integration is not a different boundary condition in the sense of a different physical situation, it is a different constraint reflecting a different point of reference within the same physical situation. Hence we should not say the nonuniqueness of the field stems from insufficiently describing the physical situation, we should say it is nonunique because we have insufficiently described the point of reference of the observer, and what aspects of that observer's reality that we are allowed to regard as ignorable differences between observers. That solves Newton's problem and gives rise to the equivalence principle, but for electric forces, the problem seems perhaps trickier because an equivalence principle may not exist. (Yet people do seek ways to unify gravity with electric forces, so perhaps it is indeed a similar situation)

 

[NFuller]

I said that you have to put boundary conditions that are compatible with the differential equation.

This is what is not making sense to me. Generally speaking, you can set the boundary condition to whatever you want: a constant, a sine function, a spherical harmonic, etc. I don't need to pick a boundary condition that makes a particular solution work. I can pick anything, and the solution must conform to that boundary condition.

That is not how infinity works. Look at the (spatially) one-dimensional case E′=κE′=κE' = \kappa, where you obtain E=κx+AE=κx+AE = \kappa x + A. A behaviour at infinity that fixes AAA will be of the type limx→∞[E(x)+E(−x)]=E0limx→∞[E(x)+E(−x)]=E0\lim_{x\to \infty} [E(x) + E(-x)] = E_0 and set A=E0/2A=E0/2A = E_0/2 and therefore single out x=−E0/2κx=−E0/2κx = - E_0/2\kappa as the point with E=0E=0E = 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.

Again, it is not clear how the origin is set here. Am I at the origin, or is two meters to my right the origin? I see that you have found where $E=0$ with respect to the origin, but you have not defined where that origin actually is.

 

[NFuller]

Other people have basically mentioned this, but the divergence theorem (and Stokes' theorem in general) only holds when $V$ is a compact subset of $\mathbb{R}^n$. $\mathbb{R}^n$ itself is not compact. Thus you aren't guaranteed that
$$\int_{\mathbb{R}^n} \nabla\cdot{\bf F}dV=\int_{\partial \mathbb{R}^n} {\bf F}\cdot {\bf dA}$$
So you can't make the claim that $\nabla\cdot\bf E=\rho/\varepsilon_0$. I'm not even clear if it's possible to formulate the meaning of $\partial \mathbb{R}^n$ in a coherent way.

I've read about this before but did not really understand it. Could you explain more about what a compact subset is? I'm not very familiar with this terminology.

 

[Metmann]

A compact subset in $\mathbb{R}^n$ is a closed and limited subset, meaning it has a boundary and is 'finite' in the sense that it does not stretch out to infinity. There is a more general topological definition, but that's not the matter here.
The divergence theorem holds in general only for compactly supported $\nabla \cdot \mathbf{F}$ or equivalently only on compact subsets of manifolds, because this (plus continuity) ensures existence of the integral. $\mathbb{R}^n$ does not have a natural boundary a priori.(and of course it is not compact, because it stretches out to infinity).

 

[Orodruin]

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)?

The 1/r^2 Green's function pre-assumes that the field of the point charge vanishes at infinity. The integrals you will obtain from this will clearly not be convergent (working in the potential now, which is usually much more convenient than working with the fields). This is just the point you noticed about the total potential energy.

Generally speaking, you can set the boundary condition to whatever you want

This is not generally true. Take f’(x) = 1/x. Having a boundary condition f(0) = 1 is clearly incompatible with this differential equation. Just an example.

Again, it is not clear how the origin is set here.

The origin is arbitrary. The natural choice would be the point where E=0. If not you will end up with a different boundary condition. The boundary condition is not invariant under translations.

Other people have basically mentioned this, but the divergence theorem (and Stokes' theorem in general) only holds when $V$ is a compact subset of $\mathbb{R}^n$. $\mathbb{R}^n$ itself is not compact. Thus you aren't guaranteed that
$$\int_{\mathbb{R}^n} \nabla\cdot{\bf F}dV=\int_{\partial \mathbb{R}^n} {\bf F}\cdot {\bf dA}$$
So you can't make the claim that $\nabla\cdot\bf E=\rho/\varepsilon_0$. I'm not even clear if it's possible to formulate the meaning of $\partial \mathbb{R}^n$ in a coherent way.

As already stated, this is not what is done when going from the integral form of Gauss’s law to the differential form. You only need the divergence theorem for compact volumes to do that.

 

[TeethWhitener]

As already stated, this is not what is done when going from the integral form of Gauss’s law to the differential form. You only need the divergence theorem for compact volumes to do that.

I'm not sure I understand. I thought this was the divergence theorem for compact volumes.

 

[Metmann]

The 1/r^2 Green's function pre-assumes that the field of the point charge vanishes at infinity. The integrals you will obtain from this will clearly not be convergent (working in the potential now, which is usually much more convenient than working with the fields). This is just the point you noticed about the total potential energy.

But then my statement was right. Gauß's law in differential form is not valid for an infinitely stretched constant charge density.

As already stated, this is not what is done when going from the integral form of Gauss’s law to the differential form. You only need the divergence theorem for compact volumes to do that.

But you can perform this transition only when the integrals are well-defined. For the infinitely stretched constant charge density this is not the case. Resp. in other words: In my opinion you can use Gauß's differential form only in those cases, where the integral formulation is also well-defined. The differential formulation is a local formulation which holds within certain domains only, not globally! It holds, where the integral formulation is defined.

 

[Orodruin]

I'm not sure I understand. I thought this was the divergence theorem form compact volumes.

The argument roughly goes as follows: Assume that Gauss's law holds for any compact volume $V$, i.e.,
$$
\frac{Q}{\epsilon_0} = \frac{1}{\epsilon_0} \int_V \rho\, dV = \oint_S \vec E \cdot d\vec S = \int_V \nabla \cdot \vec E dV
$$
where we have only applied the divergence theorem for a compact volume $V$. Now, take a series of smaller and smaller compact volumes $V_i$ containing the point $\vec x_0$, spheres centred at $\vec x_0$ with decreasing radii will do. Essentially by the multi-dimensional equivalent of the mean value theorem for integrals, you find that
$$
\rho(\vec x_i) V_i/\epsilon_0 = (\nabla \cdot \vec E)|_{\vec x = \vec x'_i} V_i
$$
for some points $\vec x_i$ and $\vec x'_i$ in $V_i$. Cancelling the volumes on both sides and noting that both $\vec x_i$ and $\vec x'_i$ must approach $\vec x$ as $i$ increases gives you $\rho(\vec x_0)/\epsilon_0 = (\nabla \cdot \vec E)|_{\vec x = \vec x_0}$. Nowhere is the divergence theorem for the entire space necessary.

Edit: Addition. In essence, the differential version of Gauss's law essentially follows from the infinitesimal version of the divergence theorem, not the extended one.

 

[Metmann]

Nowhere is the divergence theorem for the entire space necessary.

As stated above, the differential formulation holds only locally! It is valid within domains given by the integral formulation, in general not outside of them!

Edit: Addition. In essence, the differential version of Gauss's law essentially follows from the infinitesimal version of the divergence theorem, not the extended one.

What's the infinitesimal version of the divergence theorem? Divergence theorem = Stoke's Theorem = Integrals, at least to my knowledge. Maybe I am missing something.

 

[TeethWhitener]

Nowhere is the divergence theorem for the entire space necessary.

I was saying that the divergence theorem doesn't apply over the whole space. Are you saying it can be done?

 

[Orodruin]

As stated above, the differential formulation holds only locally! It is valid within domains given by the integral formulation, in general not outside of them!

I do not see where you think that I have claimed anything else. Nowhere have I used the global properties. The point $\vec x_0$ in #73 is arbitrary.

 

[Ken G]

Yet we should not lose sight of the fact that Gauss' law in an infinite homogeneous charge distribution with fixed charge-to-mass ratio allows you to pick any origin you like, and correctly deduce the differential equation for the time dependence of the vector displacement r between any two points, which will be d²r/dt² = 4*pi*rho*r/3 for unit charge-to-mass ratio and charge density rho. That is true if you keep all speeds much less than c, and assume an arbitrarily large charge distribution-- it need not be infinite, merely so large that any boundaries are far away compared to the r values of interest. So it would seem that the concern about the mathematics of infinity cannot be part of the physics problem-- physics problems should give reasonable results even when we are agnostic about the possibility of an infinite distribution of charge. So Gauss' law is doing something right, even if the observer cannot know where the "true center" of the charge distribution is, or even if there is any meaning to such a true center (as there isn't when dealing with gravity in a homogeneous universe).

The physics question that I see as remaining unsolved is what will happen to clumps whose charge-to-mass ratio deviates from unity in a truly infinite charge distribution. These clumps (or test charges) could be used to determine the "true field" in a way not possible with gravity, and perhaps one would need to know the boundary conditions to understand what happens to such clumps. One could then say that simply saying you have an infinite homogeneous charge distribution leaves something unspecified, but in a way that only appears if you allow test charges to deviate from the prevailing charge-to-mass ratio. It thus seems that unlike with gravity, here Gauss' law is only sufficient to give r(t) in the homogeneous dynamics, but further information is needed to know what instruments that are equipped with test charges of large charge-to-mass ratio will detect for the "actual field."

 

[Orodruin]

What's the infinitesimal version of the divergence theorem? Divergence theorem = Stoke's Theorem = Integrals, at least to my knowledge. Maybe I am missing something.

By this I mean what follows when you take region around a given point and let its volume approach zero and how this relates to the surface integral. In essence, for a small enough volume $V$
$$
\oint_S \vec F \cdot d\vec S = (\nabla \cdot \vec F) V,
$$
where $S$ is the bounding surface of $V$.

 

[Metmann]

I do not see where you think that I have claimed anything else. Nowhere have I used the global properties. The point ⃗x0\vec x_0 in #73 is arbitrary.

Ah yes, I get your point, sorry, I messed something up. I've got to sort myself.

By this I mean what follows

Ok, didn't know it under this name.

 

[Fishytwofeathers]

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?

MMMM, I see your quandary? Perhaps it's the lens from which you are seeing, don't forget you are in the here and now looking back in retrospect? Or perhaps you are not allowing yourself to see it through the lens of Gauss? Perhaps he was the seed, why else would you be asking the question?

 

[NFuller]

Assume that Gauss's law holds for any compact volume VVV,

Isn't the validity of this assumption part of the argument though? I will admit that I am not familiar with some of the mathematical terms you and others are discussing so forgive me if I'm overlooking the obvious.

The origin is arbitrary.

Is the field not unique then?

Newton assumed the universe should be infinite and homogeneous, and he came to the same resolution as NFuller,

Well, if my initial assumption was wrong, I guess it's comforting that I did no worse than Newton.

 

[Orodruin]

Isn't the validity of this assumption part of the argument though?

The question that post answered was ”can you go from Gauss’s law on integral form to the differential form only from its validity for compact volumes?”

Is the field not unique then?

Given an origin and a corresponding boundary behaviour the field is unique. However, if you change the origin you will not have the same boundary behaviour. This is why the boundary behaviour breaks translational invariance.

Well, if my initial assumption was wrong, I guess it's comforting that I did no worse than Newton.

Anyone who has an advanced physics education and says this problem has never puzzled or bothered them is lying or did not think about it enough.

 

[Demystifier]

Even for $\rho=0$, the equation $\nabla {\bf E}=0$ has an infinite number of nontrivial solutions
$${\bf E}=E_x{\bf e}_x+E_y{\bf e}_y+E_z{\bf e}_z$$
where $E_x$, $E_y$ and $E_z$ are arbitrary constants.

 

[Orodruin]

Even for $\rho=0$, the equation $\nabla {\bf E}=0$ has an infinite number of nontrivial solutions
$${\bf E}=E_x{\bf e}_x+E_y{\bf e}_y+E_z{\bf e}_z$$
where $E_x$, $E_y$ and $E_z$ are arbitrary constants.

There are many more solutions than that even. As discussed in posts #35 to #38.

 

[NFuller]

Given an origin and a corresponding boundary behaviour the field is unique. However, if you change the origin you will not have the same boundary behaviour. This is why the boundary behaviour breaks translational invariance.

So if you change the origin the boundary conditions change?

Working off of post #58, let's say that the boundary condition is $\lim_{x\rightarrow\infty}[\mathbf{E}(\mathbf{x})+\mathbf{E}(-\mathbf{x})]=E_{0}\hat{r}$. Then $\mathbf{E}=\rho\mathbf{x}/\epsilon_{0}+E_{0}/2\hat{x}$. The thing I still don't get is where is $\mathbf{x}=0$? Metaphorically speaking, if I had a meter that measured the electric field, and I was somewhere in this space, what would the meter read?

 

[Orodruin]

So if you change the origin the boundary conditions change?

Yes. This would be the very essence of the boundary conditions not being translationally invariant.

Working off of post #58, let's say that the boundary condition is $\lim_{x\rightarrow\infty}[\mathbf{E}(\mathbf{x})+\mathbf{E}(-\mathbf{x})]=E_{0}\hat{r}$. Then $\mathbf{E}=\rho\mathbf{x}/\epsilon_{0}+E_{0}/2\hat{x}$. The thing I still don't get is where is $\mathbf{x}=0$? Metaphorically speaking, if I had a meter that measured the electric field, and I was somewhere in this space, what would the meter read?

This is not the appropriate generalisation of the boundary condition (the correct generalisation would involve integrals of spherical harmonics). Your origin does not have to be anywhere in particular. You are free to pick whatever point you want as the origin and whatever compatible boundary conditions that you prefer. However, there are going to be different choices where only some correspond to the same physical situation.

Regardless, the question in itself is rather unphysical and you will not find a physical situation where it can be applied - even as an approximation.

 

[pierce15]

Forgive me for not having read the past 5 pages of debate, but I have a few thoughts.

First, we can stop arguing about whether this is a problem of the differential form of Gauss's law. The derivation of the differential form from the integral form uses the divergence theorem which holds in this case. In fact you can show that the "charge" has to be 0 everywhere from the integral form alone. Take a cube Gaussian pillbox with two faces perpendicular to our constant E, then Gauss's law in integral form says that $E \cdot A - E \cdot A = Q_\text{in_cube} = 0$, the cube was arbitrary so the charge is zero everywhere.

Anyway, it seems to me that the issue we are having is "how is there a nonzero electric field when there is no charge?" I don't see a problem. You picked an aphysical field and got an aphysical result. The next question is "then why do our toy models of infinite plates and line charges with finite charge density not give aphysical results?" The answer is that they are completely aphysical but model certain situations that can actually occur (i.e. are the limit of other situations), such as the electric field along the axis of a large charged circular laminar surface. On the other hand, the situation in question of a uniform electric field throughout space does not model any realistic scenario.

 

[Metmann]

First, we can stop arguing about whether this is a problem of the differential form of Gauss's law. The derivation of the differential form from the integral form uses the divergence theorem which holds in this case.

I agree, I was totally wrong.

Take a cube Gaussian pillbox with two faces perpendicular to our constant E

E doesn't have to be constant.I think, we should discard every solution that leads to infinite (total) field energy and which breaks the symmetry of the problem (yes, spontaneous symmetry [here I just mean: solution doesn't share the symmetry of the problem] breaking is a well-known concept, but maybe if possible we should try without in classical physics). With symmetry I mean here: Isotropy around every point. If we assume these two conditions, the only solution in both cases ($\rho=0$ and $\rho=\text{const}$) is $\vec{E}=0$, and hence automatically $\rho=0$.
What do you think about this?

 

[tom.stoer]

I think, we should discard every solution that leads to infinite field energy and which breaks the symmetry of the problem (yes, spontaneous symmetry breaking is a well-known concept, but maybe if possible we should try without in classical physics). If we assume these two conditions, the only solution in both cases ($\rho=0$ and $\rho=\text{const}$) is $\vec{E}=0$.
What do you think about this?

$E=0$ does not solve $\nabla E = \rho = \text{const} \neq 0$

The symmetry of the problem is not completely specified w/o boundary condition. Given $\rho = \text{const} \neq 0$ there are two choices:
1) a boundary condition compatible with $\rho = \text{const} \neq 0$; then the allowed solutions obey $E \neq 0$
2) a boundary condition compatible with $E = \text{const} = 0$ following your idea; but then $\nabla E = 0 \neq \rho$, so there is no solution

 

[Metmann]

=0E=0 does not solve ∇E=ρ=const≠0

No, but it would show, that the constant has to be zero, because else the both conditions would not be satisfied.

The symmetry of the problem is not completely specified w/o boundary condition

Well, the symmetry given by constant $\rho$, is: isotropy and homogeneity or equivalently: isotropy around every point.

 

[tom.stoer]

No, but it would show, that the constant has to be zero, because else the both conditions would not be satisfied.

It depends on your starting point. I said "Given ..." so I started with non-zero charge density which was the original question; this rules out maximal symmetry. If you start with maximal symmetry then this rules out non-zero charge.

Well, the symmetry given by constant $\rho$, is: isotropy and homogeneity or equivalently: isotropy around every point.

Obviously not - only in terms of the charge, not in terms of the electromagnetic field. So there is some essential ingredient missing to rule out the mathematical solution for non-zero E; could be a boundary condition, could be finite energy, could be maximal symmetry including E (which is trivial b/c any non-zero E breaks this symmetry :-)

 

[Metmann]

Obviously not - only in terms of the charge, not in terms of the electromagnetic field.

That's what I stated as the second assumption. The E-field is the solution of the equation, so I stated that it should follow the symmetry of the given physical situation, which is represented by $\rho$.

so I started with non-zero charge density

Yes, but if you start with arbitrary constant charge density, my discussion would result in: charge density has to be zero. So with both my assumptions, non-zero charge density would lead to a contradiction.

 

[Orodruin]

Well, the symmetry given by constant ρρ\rho, is: isotropy and homogeneity or equivalently: isotropy around every point.

As has been stated several times over in this thread, the symmetry of the charge distribution is not sufficient to conclude that the problem displays symmetry. In particular, the symmetries of the boundary conditions also need to be checked. There is no way around it. In the case of the constant non-zero charge, there is no boundary condition that is compatible with the differential equation and shows the same symmetries as the charge distribution.

Compare to the finite case of a constant charge inside a sphere of finite radius where the upper half of the sphere is held at potential V0 and the lower half is grounded. The charge distribution clearly displays full rotational symmetry, but the boundary condition does not and therefore your solution will not have full rotational symmetry. You simply cannot conclude symmetry based on the differential equation alone.

yes, spontaneous symmetry [here I just mean: solution doesn't share the symmetry of the problem] breaking is a well-known concept, but maybe if possible we should try without in classical physics

This is a bit off-topic, but why do you think spontaneous symmetry breaking does not exist in classical physics (it does). All it requires is a system with degenerate potential minima and a symmetry of the potential that transform these into each other.

Yes, but if you start with arbitrary constant charge density, my discussion would result in: charge density has to be zero. So with both my assumptions, non-zero charge density would lead to a contradiction.

This only means that one of the assumptions cannot be satisfied. In other words, it rules out solutions to the equation that have constsnt charge density and satisfy your requirements. It does not rule out solutions that do not, but it shows that those solutions cannot satisfy those requirements.

 

[Metmann]

I start from the end:

This only means that one of the assumptions cannot be satisfied. In other words, it rules out solutions to the equation that have constsnt charge density and satisfy your requirements. It does not rule out solutions that do not, but it shows that those solutions cannot satisfy those requirements.

That's exactly what I've written. Constant charge density + requirements = contradiction. Solutions with constant charge density are physically ruled out by the requirements (if one assumes the requirements to be viable, that's another question)

This is a bit off-topic, but why do you think spontaneous symmetry breaking does not exist in classical physics (it does). All it requires is a system with degenerate potential minima and a symmetry of the potential that transform these into each other.

Indeed.

Compare to the finite case of a constant charge inside a sphere of finite radius where the upper half of the sphere is held at potential V0 and the lower half is grounded. The charge distribution clearly displays full rotational symmetry, but the boundary condition does not and therefore your solution will not have full rotational symmetry. You simply cannot conclude symmetry based on the differential equation alone.

In this case, in the interior of the sphere these boundary conditions are also not compatible with the charge distribution and the differential equation. Constant potential implies zero field which implies zero charge density according to Gauß' law, or am I mistaken? Furthermore at the equator of the sphere you would have a discontinuity and hence infinite field. Therefore these boundary conditions in fact cannot be imposed. So in my opinion this situation is not compatible to the present problem, where we want to find a solution within the domain of the charge distribution.

But I see the problem. There are just no meaningful boundary conditions in the present case.

 

[tom.stoer]

That's what I stated as the second assumption. The E-field is the solution of the equation, so I stated that it should follow the symmetry of the given physical situation, which is represented by $\rho$.

The symmetry is encoded in $\rho$ and in the boundary conditions for $E$. $\rho$ alone does not specify the symmetry completely.

Look at the 1-dim toy model. They system and its symmetry can be specified by
1) $\partial_x E = \rho$
2) $E(0) = 0$

This breaks translational invariance and results in $E = x\rho$.

What's wrong with that?

 

[Metmann]

This breaks translational invariance and results in E=xρE = x\rho.

Yes, but if we require that the solution should follow the symmetry of $\rho$ alone, than this implies $\rho = 0$ (or if $\rho \neq 0$ is fixed, there wouldn't be a solution fulfilling the requirement. If the requirement makes sense is another question.)

My way of thinking about the current problem is: Suppose the whole universe (assuming it is infinite) would be filled with a constant charge density, as a background, and no a priori boundary condition is imposed from the exterior of the system (in finite cases humans can always impose a boundary condition, but in the infinite case this is more subtle). Furthermore all further charges in the Universe could be treated as test charges and do not modulate the background distribution in a measurable sense. Is this possible? Approaching this, I would assume that the solution should follow the symmetries of the background. Then the only situation in which this would be possible, would be a vanishing background density.

But now I understand where I lacked understanding all the time. I did not think about boundary conditions, that are imposed within the system due to aspects, not taken into account by the equation alone. All the time I was thinking about the following: The background distribution is the only thing we have and we do not modulate it from without or within the system.

 

[Orodruin]

In this case, in the interior of the sphere these boundary conditions are also not compatible with the charge distribution and the differential equation.

I am sorry, but this is just wrong. It is perfectly possible to find a solution with this boundary condition and charge distribution. It is the superposition of a radial field that grows with the radius (from the charge distribution) and the gradient of the potential field that solves Laplace's equation with the inhomogeneous boundary conditions.

Constant potential implies zero field which implies zero charge density according to Gauß' law, or am I mistaken?

I said constant potential on the sphere. Not inside the sphere. The potential inside the sphere is not a step function, it is a superposition of spherical harmonics multiplied by $r^\ell$.

Furthermore at the equator of the sphere you would have a discontinuity and hence infinite field.

No, this is not the case. The potential step is on the boundary of the sphere. Anywhere inside the sphere you will have a finite field. The field will grow towards infinity as you approach it, but there is no problem with that (the point charge field also does this inside the volume). In general, you should consider these configurations as distributions, not as functions. Besides, the step function is just an example so this is completely besides the point. You can put any function on the sphere that is not constant, $V_0 \cos(\theta)$ works perfectly fine as well and will single out the $Y_{10}$ harmonic in the solution (in fact, the contribution from the boundary condition will be a constant field in the $z$-direction).

Yes, but if we require that the solution should follow the symmetry of $\rho$ alone, than this implies $\rho = 0$ (or if $\rho \neq 0$ is fixed, there wouldn't be a solution fulfilling the requirement. If the requirement makes sense is another question.)

This is the point, this is what is inconsistent. You simply cannot conclude that the problem displays a symmetry unless you take the boundary conditions into account.

 

[Metmann]

I am sorry, but this is just wrong. It is perfectly possible to find a solution with this boundary condition and charge distribution. It is the superposition of a radial field that grows with the radius (from the charge distribution) and the gradient of the potential field that solves Laplace's equation with the inhomogeneous boundary conditions.

I am sorry, too. I missunderstood your boundary conditions. You said "constant charge density inside of the sphere" and then "upper half of the sphere held at constant potential", so I was thinking you would mean, that the interior of the upper half would have this boundary condition, but of course you just mean, that the boundary condition is applied to the sphere itself, not the interior. My fault, sorry.

This is the point, this is what is inconsistent. You simply cannot conclude that the problem displays a symmetry unless you take the boundary conditions into account.

Concerning this, see my post above.

 

[tom.stoer]

You simply cannot conclude that the problem displays a symmetry unless you take the boundary conditions into account.

Of course one could follow a different approach by requiring maximal symmetry instead of specifying boundary conditions.

[We do that in general relativity quite frequently. As an example: we do not require asymptotic flatness but homogenity plus isotropy, resulting in FRW cosmologies; the main difference is that in GR we then find solutions with non-zero density, whereas in electrostatics this is not possible; the main difference is the non-linearity of GR]

 

[Metmann]

We do that in general relativity quite frequently

Probably that's also why my approach here was different. It has been a long time since I've last done classical E-dynamics, while cosmology is all around me.