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

[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.

 

[Orodruin]

that the interior of the upper half would have this boundary condition

That would not be much of a boundary condition now would it?

However, there would a priori not be anything necessarily horrendous about if the potential looked like that. The charge distribution leading to it would be the idealisation of two equal and opposite surface charges approaching zero distance between each other. The field would be described by something like $\vec E = E_0 \delta(z) \vec e_z$ (again, considering the field as a distribution rather than a function). Clearly this is different from a constant charge distribution though.

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

I agree, but, as you pointed out earlier, that was not the original question in this thread. The original question stated that there should be a uniform non-zero charge throughout all of space.

 

[Metmann]

That would not be much of a boundary condition now would it?

Not in the strict sense ^^ But one could for example consider the situation: You have a some charge density in some domain outside of a sphere and as a "boundary condition" you require the potential to be constant inside the whole sphere. Of course this would immidiately imply vanishing density inside of the sphere, but in this sense one could also use the term "boundary condition" for a condition that is applied to a volume instead of a surface, even though the mathematical concept of boundary condition applies of course only to boundaries, i.e. $d-1$-dimensional domains.

However, there would a priori not be anything necessarily horrendous about if the potential looked like that [...]. Clearly this is different from a constant charge distribution though.

I don't have a problem with the potential ;) It just wouldn't be consistent with a constant charge density, that's the point, just as you say.

The original question stated that there should be a uniform non-zero charge throughout all of space.

Well, even a uniform non-zero charge density is a priori consistent with maximal symmetry (before you solve Gauß's equation). Boundary conditions are something you impose from the exterior. If there exists nothing else in the universe (infinite) than the background charge and small enough test charges, than there is no physical boundary condition and only maximal symmetry restricts solutions to the problems. I think this approach is still an answer to the original question. And within this approach the answer is, that there is no solution with non-vanishing density.

 

[tom.stoer]

I agree, but, as you pointed out earlier, that was not the original question in this thread. The original question stated that there should be a uniform non-zero charge throughout all of space.

I fully agree.

Starting with uniform non-zero charge $\rho$ yields a class of non-trivial solutions for $E \neq 0$.
Starting with maximum symmetry = homogeneity and isotropy yields $\rho = 0$ and $E = 0$.

 

[Orodruin]

If there exists nothing else in the universe (infinite) than the background charge and small enough test charges, than there is no physical boundary condition and only maximal symmetry restricts solutions to the problems.

I disagree with this. A behaviour at infinity is a boundary condition even if it is not posed on a physical boundary. The underlying assumption is that one is making when ignoring infinity is that the contribution from ever more distant shells vanishes. As we have already noted, this is not the case and the potential formally diverges. The boundary condition you put is equivalent to a prescription on how to renormalise the potential. One such way is to introduce a radial cutoff in the integral (formally, subtracting all contributions from radii larger than $r_0$). However, this will naturally break your translational invariance.

 

[Metmann]

A behaviour at infinity is a boundary condition even if it is not posed on a physical boundary.

Sure, no doubt. But in the situation I describe, there is (in my opinion) no physical way to introduce such a condition without yielding $\rho=0$. I mean a meaningful condition would definitely be, that the total energy should be finite (what you do with the cutoff), hence the potential should vanish at infinity. Can you solve $\Delta \Phi = \rho = \text{const} \neq 0$ with $\lim_{|\vec{x}|\rightarrow \infty} \Phi(\vec{x}) = 0$ without a cutoff? Spontaneously I cannot think of any other meaningful boundary condition within this setting.

The boundary condition you put is equivalent to a prescription on how to renormalise the potential. One such way is to introduce a radial cutoff in the integral (formally, subtracting all contributions from radii larger than r0r_0).

But where does this cutoff come frome? In the situation I described, there is no natural cutoff. Of course you can randomly introduce it, but why should that be physical if the charge density is filling the whole universe? And your results depend on the chosen cutoff so its highly subjective. Of course, if you also consider special relativity, you can use as a cutoff the maximal distance a photon could have traveled since BB, but in classical physics? (Yes I know, ED incorporates Lorentz-invariance and hence SR, but let us for know consider we wouldn't know about it. Hmm, but maybe that's also the problem. One probably shouldn't think about electrostatics in purely classical ways.).

 

[Orodruin]

Sure, no doubt. But in the situation I describe, there is (in my opinion) no physical way to introduce such a condition without yielding $\rho=0$. I mean a meaningful condition would definitely be, that the total energy should be finite (what you do with the cutoff), hence the potential should vanish at infinity. Can you solve $\Delta \Phi = \rho = \text{const} \neq 0$ with $\lim_{|\vec{x}|\rightarrow \infty} \Phi(\vec{x}) = 0$ without a cutoff? Spontaneously I cannot think of any other meaningful boundary condition within this setting.

With your assumptions, I agree that they imply $\rho = 0$. However, that was not the original question of the thread. The original question of the thread was stating that you have a non-zero uniform charge density. Your assumptions, while reasonable at face value, clearly violate the statement of non-zero uniform charge density. That does not mean that solutions that include non-zero uniform charge density do not exist.

But where does this cutoff come frome? In the situation I described, there is no natural cutoff. Of course you can randomly introduce it, but why should that be physical if the charge density is filling the whole universe?

You have to introduce the limit of several cutoffs at increasing radius. Of course, you can take the limit in several ways, which will give you different results. It is a matter of definition of how the integral with $r \to \infty$ is considered. Since the integral is not convergent, different prescriptions will give you different results, but none will be translationally invariant. It is the type of prescription that you make that will give you the behaviour at infinity and therefore the boundary condition.

What you assume about your limiting behaviour is physical because it affects the field configuration.

Of course, if you also consider special relativity, you can use as a cutoff the maximal distance a photon could have traveled since BB, but in classical physics?

This is not at all applicable to this thread. This is about electrostatics. The bottom line is that there exist solutions to the field equations that do satisfy the condition of having a uniform non-zero charge density. However, they do not display the same symmetries as the charge distribution. Since this symmetry breaking does not come from the differential equation, nor from the charge distribution, the only other place it can come from is the boundary conditions, i.e., the behaviour of the fields at infinity. In order to have a solution with uniform charge distribution, these boundary conditions must therefore break the symmetry. This was the original question: "If you have uniform non-zero charge distribution, what is the field?" The answer is that it depends on the specified behaviour at infinity and that that behaviour breaks the symmetry of the charge distribution. Of course, the question could have been "What kind of solutions exist with maximal symmetry?" In that case the answer would have been that the only such solution is $\vec E = 0$ and that this solution has zero charge density, but it would have been a different question.

 

[Metmann]

The original question of the thread was stating that you have a non-zero uniform charge density. Your assumptions, while reasonable at face value, clearly violate the statement of non-zero uniform charge density. That does not mean that solutions that include non-zero uniform charge density do not exist.

I've not claimed that non-zero charge density solutions do not exist, I'm just trying to elaborate if physically meaningful solutions exist.

It is the type of prescription that you make that will give you the behaviour at infinity and therefore the boundary condition.

Ok, and which prescription would be physical if they give you different answers? Surely the physics should only depend on the physics, not on the mathematics.

The bottom line is that there exist solutions to the field equations that do satisfy the condition of having a uniform non-zero charge density

That's trivial and I thought there would be more to the question. You can directly write down a bunch of solutions (also when I was misstaken and thought that Gauß' law wouldn't be applicable I did not state that there wouldn't be solutions to the differential equation), but since we are dealing with physics, I'm wondering which of these solutions can really be regarded as physical.

the original question: "If you have uniform non-zero charge distribution, what is the field?"

If this question does not incorporate physical meaning, than you're right, the question has already been answered. But then I want to extend the question ;)

 

[Orodruin]

Ok, and which prescription would be physical if they give you different answers? Surely the physics should only depend on the physics, not on the mathematics.

That depends on what you measure. I think you are seeing the boundary conditions as something to be imposed apart from the physical situation. They are not. Without them, you have not fully specified the physical situation. You can have different boundary conditions, leading to different fields, but they correspond to different physical situations, regardless of whether the charge density happens to be the same or not.

I'm wondering which of these solutions can really be regarded as physical.

Sure they can. But the limiting behaviour is part of the physics and you need to take it into account. The limiting behaviour is physical precisely because it affects the fields.

 

[Metmann]

Sure they can. But the limiting behaviour is part of the physics and you need to take it into account. The limiting behaviour is physical precisely because it affects the fields.

Yes sure, if you assume that there are different things in the Universe than background charge and test charges. But you have to assume this additionally. All the time I have understood the question in the way I wrote before: infinite Universe filled with background, that's it. And then I wondered if there exists a non-trivial physical solution. The answer is of course: no.

I think you are seeing the boundary conditions as something to be imposed apart from the physical situation. They are not. Without them, you have not fully specified the physical situation. You can have different boundary conditions, leading to different fields, but they correspond to different physical situations, regardless of whether the charge density happens to be the same or not.

It depends: If we consider the Universe to be infinite and do not know anything about its "boundary" and if we assume that apart from test charges there is only background charge, then any boundary condition would be something additional! You can still solve the problem by the way I did, using symmetries. That's exactly what is done in Cosmology. Other boundary conditions assume implicitly that there is more physics apart from the specific situation considered.
But of course if you know e.g. that you can "earthen" the potential at some plane (due to aspects not incorporated in the "theory", namely Gauß's law + charge distribution (symmetry) + infinite universe, without any additional information), but I would call this "extrinsic" boundary conditions. An "intrinsic" "boundary" condition would be given be the symmetry.

I have to admit, that maybe I failed to make my point clear earlier, my language was probably not clear enough.

Finally we can conclude: The topic is now exhausted, everything has been said. Sorry for dragging it out.

 

[Orodruin]

All the time I have understood the question in the way I wrote before: infinite Universe filled with background, that's it.

But this is the point. It is not it. Without a specification of the boundary behaviour, you have not specified the physical situation.

If we consider the Universe to be infinite and do not know anything about its "boundary" and if we assume that apart from test charges there is only background charge, then any boundary condition would be something additional!

If you do not know anything about the boundary, then you do not know the full physical situation and clearly in this case you cannot compute the field.

You can still solve the problem by the way I did, using symmetries.

Not if none of the boundary conditions that are compatible with the differential equation breaks those symmetries. Then there is no physical situation with your assumed symmetric charge distribution that has those symmetries.

An "intrinsic" "boundary" condition would be given be the symmetry.

But this is exactly the point. Given the charge distribution, there is no such boundary condition.

Other boundary conditions assume implicitly that there is more physics apart from the specific situation considered.

Yet, those boundary conditions have to be compatible with the rest of the theory.

Finally we can conclude: The topic is now exhausted, everything has been said. Sorry for dragging it out.

I really think that we actually agree on most points, the main ones being:

• There are a multitude of solutions to Gauss's law with uniform non-zero charge density.
• Those solutions are separated based on their behaviour at infinity
• None of those solutions display the maximal symmetry of the charge distribution itself except in the case $\rho = 0$, which is trivial.

 

[Likith D]

This is why the boundary behaviour breaks translational invariance.

Well, I remember integrating painstakingly to find E field of a planar disc and letting the disc tend to infinity... to actually get E field value that agrees with gauss' law... (E field of infinite uniformly charged sheet)
I have done the same to a uniformly charged wire segment and letting length tend to infinity also to get an answer that agrees with gauss' law... (E field of infinite linear charge distribution)
No "broken transnational invarience" here...
But once we step into the case of 3d spheres with radius tending to infinity and we suddenly have "transnational invarience"? How do you justify that?

 

[Stephen Tashi]

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

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.

To talk about the electric field (as opposed to the electric flux) at the surface aren't you assuming the spherical uniform charge distribution contains all the charges in the problem? A charge outside the sphere contributes no net flux through the surface of the sphere. It can still make a contribution to the electric field.

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?

If the (net) electric field is zero on the surface of a sphere of finite radius (existing within the supposed uniform distribution of charge throughout space) then it is no surprise that it stays zero as the radius is increased. One amusing thought is to represent what takes place using two spheres. Use a big sphere with a uniform charge density to represent "the universe". Use a much smaller sphere inside it to represent "the sphere we are worried about". Then let the radius of both spheres approach infinity with the big sphere always containing the smaller one.

 

[Orodruin]

No "broken transnational invarience" here...

This is not true. Both a wire and a plane break translational invariance and rotational invariance in many of the directions and you cannot let the field go to zero at infinity in some of the directions. (Namely, along the wire/plane)

I also cannot see why you are surprised that different charge distributions lead to different requirements on the boundary conditions that are compatible with the charge distribution.

and we suddenly have "transnational invarience"?

The charge distribution has translational invariance in all directions, just as the line charge has translational invariance in the direction of the line. However, you always have to look at what boundary conditions are compatible with the differential equation and you cannot impose boundary conditions that aren't. In the case of a uniform charge distribution, a boundary condition that preserves translational and rotational invariance is incompatible with Gauss' law.

 

[yosimba2000]

I'm not an expert and it's a dead thead but I want to state my thoughts anyway.

Isn't Gauss's Law responsible for finding the E-fields at the boundaries of the enclosed volume due to the ENCLOSED charge? So due to the almost infinite amount of charge in our volume (let's say a sphere), we can assume E-field to be non-zero, radial, and uniform strength everywhere along the volume boundary.

But once you take into account the E-field from the infinite charges outside the boundary, they will counteract the E-field along the volume boundary, which were from the charges in the enclosed volume, giving us the expected effect of net 0 E-field everywhere.

 

[Orodruin]

Isn't Gauss's Law responsible for finding the E-fields at the boundaries of the enclosed volume due to the ENCLOSED charge?

Gauss’ law is a tool, it is not responsible for anything. All it does is to tell you about the relation between the enclosed charge and the flux integral over a closed surface. In the case of a spherically symmetric charge distribution and spherically symmetric boundary conditions, you can use symmetry to argue that the field must be radial and therefore directly use Gauss’ law to relate the field to the enclosed charge. In the case of boundary conditions that are not spherically symmetric you cannot do this. The point is that, as already discussed in this thread, boundary conditions compatible with the differential equation for a constant charge distribution and spherically symmetric relative to one point will not be spherically symmetric relative to other points. Thus, there is only one point where you can use Gauss’ law to directly relate the field (rather than its integral) to the enclosed charge.

But once you take into account the E-field from the infinite charges outside the boundary, they will counteract the E-field along the volume boundary, which were from the charges in the enclosed volume, giving us the expected effect of net 0 E-field everywhere.

No. The shell theorem tells you that the effect from any shell outside your sphere is zero. The external effect can also not have a divergence inside your sphere and thus not affect Gauss’ law. A constant zero field would clearly violate Gauss’ law. To have zero E everywhere is therefore certainly not expected. What you are missing is the contribution from the boundary at infinity, which can only be spherically symmetric relative to a single point (and that assumes no higher multipoles, see above).

 

[rude man]

I'm not an expert and it's a dead thead but I want to state my thoughts anyway.

Isn't Gauss's Law responsible for finding the E-fields at the boundaries of the enclosed volume due to the ENCLOSED charge? So due to the almost infinite amount of charge in our volume (let's say a sphere), we can assume E-field to be non-zero, radial, and uniform strength everywhere along the volume boundary.
.

You're thinking of the total freely-MOVING charge in a sphere (if it's a conductor), but Gauss's law specifies that only NET free charge causes the surface integral enclosing said free charge to be non-zero. There are just as many protons as electrons in an uncharged object!