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I'm trying to find sufficient additional conditions to derive Coulomb equation for the electric field generated by a steady point charge in free space from Maxwell equations in said conditions.

I know that a way to do this is assuming that the solution of Maxwell equations must have spherical symmetry due to the disposition of charges (a single point charge infact) having such a degree of symmetry.

The thing is, how can I prove that the solution of maxwell equations has to be symmetrical if I don't know in advance the equation of the electric field? And if the answer is "because the space is isotropic", how can I mathematically write such a condition?

Infact, if I consider Maxwell's equations (I write just the first two for the case we are considering) I get

\begin{align} \int_{\partial \Omega} E \cdot dS &= \delta(\Omega)q \\ \int_\gamma E \cdot ds =& 0 \quad \text{if} \quad \partial\gamma=\emptyset \end{align}

($\delta$ is $1$ if the carge is inside of $\Omega$, $0$ if it is outside and $q$ is the value of the point charge) so if $E$ solves them, also $E+K$ does for every constant field $K$.

I get that space should not "choose" any direction for $K$, but how to mathematically state it? There is clearly something missing here.

I thought of adding as a condition that

\begin{equation} \lim_{|x|\to +\infty} E = 0 \end{equation}

and this infact eliminates the possibility of adding a costant field, but I think I'm far from proving uniqueness of the solution.

Also another thing that bugged me, is the fact that it is not always the case that $\lim_{|x|\to +\infty} E = 0$, since if we impose this condition for the solution in the case of a homogeneus charged infinite plane, we don't get a solution at all.

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  • $\begingroup$ Do you like this? At the very top of the page, starting with the charge density $\rho(\mathbf{r}) = q_i \delta(\mathbf{r}-\mathbf{r}_i)$ basics.altervista.org/test/Physics/EM/… $\endgroup$
    – basics
    Feb 20 at 17:58
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    $\begingroup$ @RogerV. I know, and I know how to do it, my question is really different. $\endgroup$ Feb 21 at 13:22
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    $\begingroup$ Regarding why spherical symmetry is justified, the answer is that it follows by the symmetry of the equations (including the delta source, which is ‘clearly’ symmetric) and the fact that the initial and boundary conditions are symmetric. There are various well-posedness theorems for Maxwell’s equations (which are really a coupled system of evolution equations) which you’ll find once you start studying PDEs proper. Anyway, see Formalizing the arguments of symmetry for some of these details. $\endgroup$
    – peek-a-boo
    Feb 21 at 18:28
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    $\begingroup$ ^ there I pretty much restricted myself to talk about smooth stuff, but the Dirac delta is obviously not smooth; it’s a distribution, but similar arguments work (once you have precise definitions in place for what distributions, pullback of distributions etc are). You’re worried about infinite charge distributions not giving $E\to 0$ at infinity, and you’re right; but the thing is that is not the right boundary condition. The ultimate answer of whether a collection of initial and boundary conditions are ‘good’/‘correct’ is whether the resulting system is well-posed. $\endgroup$
    – peek-a-boo
    Feb 21 at 18:29
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    $\begingroup$ It is not a “symmetry of the space” that is required, it is a symmetry of the charge distribution, which implies a symmetry of the electric field it creates. See my answer below where no fancy functions are used, just that essential argument. $\endgroup$
    – Ben H
    Feb 21 at 21:19

6 Answers 6

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I'd like to point out that Maxwell's equations permit static divergence-free and curl-free background solutions that are more general than just a constant E-field, namely: $$\vec E=a_x(b_y+a_yy)(b_z+a_zz)\hat x + a_y(b_x+a_xx)(b_z+a_zz)\hat y + a_z(b_x+a_xx)(b_y+a_yy)\hat z $$ where $a_i$ and $b_i$ are constants. We cannot mathematically show that these solutions aren't "real." A common feature of such solutions though is that $\lVert \vec E\rVert$ does not decay to zero as $\vec r \to \infty$. This is a bit disconcerting (worse, some of the solutions blow up at infinity), and we do not observe such background fields, so it is safe to assume they don't exist.

I would also note that infinite charge distributions are rather pathological. In dealing with these, we can either "zoom out" far enough that they do not extend to infinity, or impose reasonable assumptions of symmetry (e.g. $\lVert\vec E\rVert$ being the same on either side of an infinite charged sheet).

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  • $\begingroup$ So is there an additional condition to enshure uniqueness of the solution and rule out all these "unreal" solutions? Also, I get that infinite distributions of charges are pathological from a physical prospective, but they are just fine from a mathematical point of view, so I would expect a mathematical theory capable of handling these things, am I wrong here? $\endgroup$ Feb 20 at 19:28
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    $\begingroup$ Keep in mind that Maxwell's equations are a mathematical model of electromagnetism, motivated by experiments and with the aim of accurately predicting their results. We can only rule out these background solutions using the fact that we don't detect them in experiments, or perhaps by arguing that a successful model ought to produce solutions that have certain symmetry properties. Mathematically valid solutions may not be physical, nor may hypothetical infinite charge distributions. $\endgroup$
    – Puk
    Feb 20 at 19:38
  • $\begingroup$ well indeed, but to say that we have to manually rule out these solutions, is basically to say that the model we are using is not sufficient to provide unique solutions, and from the mathematical point of view is quite catastrofic. That's why I was hoping to find some conditions to enshure uniqueness. $\endgroup$ Feb 20 at 20:00
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    $\begingroup$ Most physicists would consider $\lVert \vec E \rVert \to 0$ as $\vec r \to 0$ a simple and reasonable condition to impose on the solution for a finite charge distribution, and it does ensure uniqueness. Infinite distributions can be treated as I explain above. There is no catastrophe in terms of the ability of Maxwell's equations to explain and predict physical phenomena. If you are interested in the problem from a purely mathematical point of view, Math SE might be a more suitable place for your question. $\endgroup$
    – Puk
    Feb 20 at 20:07
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    $\begingroup$ @LorenzoVanni The result you seek follows from Helmholtz's theorem, you can start there. $\endgroup$
    – Puk
    Feb 20 at 20:27
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From empirical observations, we have reasons to believe that our universe obeys the symmetries of spacetime being homogeneous and isotropic and parity symmetric. The Maxwell's equations are also homogeneous, isotropic and parity symmetric.

However, these are not at all sufficient to show that the solutions to Maxwell's equations are symmetric in any way. This is because you can easily have a non-symmetric distribution of charges, and immediately the solutions will also fail to have symmetry.

Thus, if you want to obtain Coulomb's Law, you have to consider a spherically symmetric charge distribution. Only in that case can you put in some mathematics to prove that the solutions are also symmetric. To that end, you use polar coördinates and show that the derivative of the charge distribution is zero along the angular coördinates. Then you can express the vectors generally, and then derive that the solution also has to have zero components in those angular directions.

You were also confused as to how to deal with, say, uniformly charged infinite plate. For ease of derivation, you should apply Helmholtz theorem in the general case of having non-trivial boundary conditions at infinity. It says that a vector field can always be decomposed to a purely gradient part, a purely curl part, and a part that satisfies the boundary conditions at infinity. You should also use this for the solution in general, because this means to work with potentials, and that is always much less work to do than to solve for vectors directly.

Finally, we should give an example of something failing the symmetries in a totally different way. It is known that the weak interaction happens to be maximally violating parity symmetry. That means that, if we want to derive the fields corresponding to the weak interaction, then we cannot use parity symmetry to assert that the fields have to be spherically symmetric. That is, consider a perfectly spherical conducting shell with charge $Q$ on it, we can derive that the electric field outside of it is spherically symmetric by first considering a randomly chosen point to have a randomly pointing electric vector. With one rotation, we can move this considered point to hover above the North pole, again that defines a new direction in which it is pointing towards. Usually, at this step, we would invoke mirror symmetry about a vertical plane and conclude that if there was any component that is not radial, then the mirror symmetry will cancel it out. This is how we can assert that the electric field has to be purely spherically symmetric, and that means that only the radial component is left.

However, if in some weird way, we manage to get an equivalent thing for weak charges, then we cannot invoke the mirror symmetry step. That is still ok, because the mirror symmetry step is not necessary. We can simply apply a rotation around the North-South axis and also get the same result.

It is thus interesting to note that a single point charge electron will fail this. This is because there is a preferred z-axis for the electron because the electron has spin, which thus picks out a direction along which its spin has an eigenvalue of either along or against that direction. Because of that, the weak interaction will obey only cylindrical symmetry, and then the mirror symmetry cannot be circumvented by the other rotation. Hence, the weak interaction will not obey an equivalent of Coulomb's Law.

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    $\begingroup$ Excellent point! When one chooses $\rho(\vec r)=q\delta(\vec r)$ for a charge distribution, then they have already assumed spherical symmetry in the form of the charge's distribution in space. $\endgroup$ Feb 20 at 18:23
  • $\begingroup$ I'm not saying that space is not isotropic, I'm rather asking of a condition to put it in mathematically. You however say that using poolar coordinates I can show that the solution is symmetric, but I feel we still miss something, since I indeed considered a symmetric distribution of charges, but I already found solutions that don't have such degree of symmetry, so my question was, what is the mathematical condition I'm missing that enables us to obtain uniqueness of the solution? $\endgroup$ Feb 20 at 19:22
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    $\begingroup$ @LorenzoVanni then you have to specifically ask a question on your particular derivation. Because we cannot know what it is you "found" as solutions that should have been discarded. If you did it correctly, you should have the unique solution. $\endgroup$ Feb 20 at 20:04
  • $\begingroup$ I'm sorry, I wasn't clear, I'm referring to what i wrote in my question. As I said, clearly given a point charge (that is a symmetric distribution), the field $E$ given by Coulomb's law solves Maxwell's equations, but also does every field $E+K$ where $K$ is a constant field. This problem goes away if we assume spherical symmetry of the solution (not of the distribution), so I was asking how to prove that such a configuration produce a symmetric field, and if the answer is "by space isotropy" how to put space isotropy in mathematical therms. $\endgroup$ Feb 20 at 20:13
  • $\begingroup$ 1) constant field K breaks space isotropy because, while it respects $\varphi$ symmetry if we align z axis with K, it will fail to have $\vartheta$ symmetry. 2) Constant field breaks the boundary condition at infinity if you wanted Coulomb's Law, which is that the field should vanish at infinity (and at the correct rate, no faster, no slower). $\endgroup$ Feb 20 at 20:32
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Coulomb's law is in fact, equivalent to one of Maxwell's equations, viz.: $$\nabla\cdot \vec E(\vec r)=\rho(\vec r)/\epsilon_0.$$ Now if the case under quesiton concerns a point charge, then: $$\nabla\cdot \vec E(\vec r)=q\delta(\vec r)/\epsilon_0.$$ One can also write that: $$-\nabla^2\phi(\vec r)=q\delta(\vec r)/\epsilon_0,$$ utilizing the fact that: $\vec E(\vec r)=-\nabla\phi(\vec r)$. Thus, we have arrived at Poisson's equation. Now it is well known that: $$\nabla^2({1\over r}\hat r)=4\pi\delta(\vec r)$$ Thus, we must have that: $$\phi(\vec r)=-{q\over 4\pi\epsilon_0r},$$ which is the potential for a point charge, taking the gradient: $$\vec E(\vec r)={q\over 4\pi\epsilon_0r^2}\hat r.$$ It is quite evident then, that for any charge $q^\prime$ there is a force: $$\vec F=q^\prime \vec E(\vec r)={q^\prime q\over 4\pi\epsilon_0r^2}\hat r,$$ which is Coulomb's law for a point charge $q$ at the origin. The only assumptions needed was the validity of Maxwell's equations as a description of empirical phenomena. For the most part the differential operators of mathematical physics will admit many more solutions than the ones commonly used for the description of physical phenomena, however, it is the observations that refine the solution set down. In this sense much of physics cannot be put on a mathematical footing alone.

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    $\begingroup$ Shouldn't your fourth equation be $$\nabla^2 \left( \frac 1r \right)=4\pi \delta (\vec r)\ ?$$ $\endgroup$ Feb 21 at 19:39
  • $\begingroup$ @PhilipWood not even that, there should by a minus sign: $\nabla^2(1/r)=-4\pi\delta(\vec{r})$. The correct potential is then $\phi=\frac{q}{4\pi\epsilon_0r}$, no minus sign (there’s already the minus sign in $\vec{E}=-\nabla\phi$, so we get the correct electric field $\vec{E}=\frac{q}{4\pi\epsilon_0}\frac{\hat{\mathbf{r}}}{r^2}$). Also, @ Albertus, your conclusion that “Hence $\phi(r)=…$” is wrong. All your previous line shows is that if $\phi(r)=\frac{q}{4\pi\epsilon_0r}$, then $-\Delta\phi=\frac{q}{\epsilon_0}\delta$. $\endgroup$
    – peek-a-boo
    Feb 21 at 20:19
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    $\begingroup$ But you claimed the converse, which is actually incorrect based merely on what you wrote. There are MANY more solutions to $-\Delta\phi=\frac{q}{\epsilon_0}\delta$ than the coulomb potential. Namely, we can add to $\phi(r)=\frac{q}{4\pi\epsilon_0r}$, any solution of Laplace’s equation $\Delta\psi=0$ (which actually forms an infinite-dimensional vector space), so it goes back to naturallyInconsistent’s comment in the beginning, that your reasoning is incomplete. $\endgroup$
    – peek-a-boo
    Feb 21 at 20:24
  • $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Feb 22 at 19:31
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I'm answering my own question because I think that maybe I've found a way to put space isotropy in mathematical therms, and infact (in this case) it solves all my problems, so I'd really like an opinion on this.

I think this could be extended with little effort to moving distributions, when I will have studied the subject in greater detail I think I will be able to update this answer.

I'm therefore assuming constitutive relation of the form:

\begin{equation} D = D(x, E, H), \; B=B(x, E ,H), \; J = 0 \end{equation} and the measure with sign $Q$ independent of time.

The condition I think I'm searcing is:

With $A$ being the domain (open connected) we are solving the problem in, let $f: A \to A$ of the form $f(x) = x_0 + \Lambda(x-x_0)$ bijective, where $\Lambda$ is linear and $\det(\Lambda) = \pm1$. If $Q(\Omega) = Q(f(\Omega))$ for every $Q$-measurable set, and we have connected closed bounded 2 and 3 manifolds $\Omega$ and $S$ rispectively for wich integrals are defined, then the solution $(E,H)$ to Maxwell's equations

\begin{align} &\oint_{\partial \Omega} D \cdot dS = Q(\Omega) \\ \oint_{\partial S}& E \cdot ds = - \frac{d}{dt} \int_S D \cdot dS \\ \frac{d}{dt}&\int_S D \cdot dS = \oint_{\partial S} H \cdot ds \\ &\oint_S D \cdot dS = 0 \end{align}

is such that $(E, H)(t,f(x)) = (\Lambda E, \Lambda B)(t,x).$

EDIT: I think I have to require also $D(f(x), \Lambda E, \Lambda H)= \Lambda D(x, E,H), \; B(f(x), \Lambda E, \Lambda H)= \Lambda B(x, E,H)$. END EDIT

For example in the case of a point charge in free space fixed in the origin, we get constitutive relations

\begin{equation} D = \varepsilon_0 E, \; B=\mu_0H, \; J = 0 \end{equation}

and $Q(\Omega) = q\delta_0(\Omega) = q\delta_0(f(\Omega))$ whenever $f$ is a reflection or a rotation, hence with little effort we get spherical symmetry of both $E$ and $B$, and then $E(t,x)=\frac{q}{4\pi\varepsilon_0|x|^3}x, B(t,x)=0$, as desired.

Let me know if you think this could be good.

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  • $\begingroup$ Mathematically, your reasoning is as circular as was my example, e.g. when you write $Q(\Omega)=q\delta_0(\Omega)$ you are assuming spherical symmetry, for how else do you know that $\oint_{\partial (f(\Omega))} \vec D\cdot ds=Q(f(\Omega))$? $\endgroup$ Feb 21 at 16:12
  • $\begingroup$ @AlbertusMagnus Doing what you say I assumed spherical symmetry of the distribution, not of the solution. What I postulated is really different, infact I'm saying that if the distribution has some degree of symmetry then the solution has it too, that is I'm postulating space isotropy. $\endgroup$ Feb 21 at 16:16
  • $\begingroup$ Where is your proof that a spherically symmetric distribution implies that the Field is also spherically symmetric? I did it basically by thinking about $\lim_\limits{\partial S\rightarrow 0}\oint \vec E\cdot \hat n d(\partial S)=\delta(\vec r)/\epsilon_0$, and how the spherically symmetric distribution must yield only radial fields. $\endgroup$ Feb 21 at 16:26
  • $\begingroup$ I didn't prove it, I postulated it, since it is not provable without other assumption as I showed in my original question. You are assuming that "spherically symmetric distribution must yield only radial fields" that comes from the thing you are triyng to prove. Also it is not obvious that the limit of solutions as the distribution tends to the point charge is the solution in the case for the point charge (if i'm understeanding correctly you are taking a limit on a radius of a ball, and on the dirac distribution). $\endgroup$ Feb 21 at 16:32
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This question illustrates a more general issue that often arises with partial differential equations. There are similar issues in General Relativity for example.

The first thing to note is that a given source distribution is almost never enough to settle uniquely a solution to a set of partial differential equations, because you can add in whatever functions satisfy the source-free equations and you still have a solution. Physically, if you have a point charge then around it you may find a Coulomb-law field, or you may find a Coulomb-law field plus electromagnetic waves. (For linear equations this 'adding in' is exactly an addition operation; for non-linear equations it is not simply addition but a similar conclusion holds.)

It follows that to handle this issue in general you need either a Cauchy surface on which the fields have been given, or else further conditions in addition to the source distribution. Symmetry is often brought in to play that role.

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how can I prove that the solution of maxwell equations has to be symmetrical if I don't know in advance the equation of the electric field?

The answer is: if the charged source has some symmetry (precisely: a particular transformation that leaves it unchanged), then the electric field produced by that source must have the same symmetry. The source and the field are tied together, so a transformation acts on both of them. If the source is unchanged by a transformation but the field is different, then you would have the same source producing two different fields.

If you want to derive Coulomb’s law from Gauss’s law without reference to Coulomb’s law (e.g., "we know the electric field points away from a positive charge"), then you need to make a rigorous symmetry argument to prove a couple facts about the electric field before using Gauss’s law.

If we have, e.g., a spherical ball of charge (includes the limit of a point charge), then we must prove first that:

  1. The electric field at any point in space points toward or away from the center of the ball (or has zero magnitude).
  2. The magnitude of the electric field is uniform on a sphere of radius $d$.

As you will see below, these two facts must be assumed to perform the Gauss’s law calculation.

Both of those facts follow from this symmetry of the ball of charge:

The ball of charge is symmetric (unchanged) with respect to a rotation, by any angle, about an axis through the center of the ball.

Because the charged ball produces the electric field, the electric field must also be unchanged by such rotations.

Now we prove the above facts:

  1. For any point in space I can find such a rotational axis that passes through that point (and, by definition of the transformation, through the center of the ball). The electric field at that point must lie along that axis, because if it pointed off axis the electric field vector would be changed by the rotation. Thus, the electric field points “radially” (in/out) at any point.
  2. For any two points in space that are the same distance $d$ from the center of the ball, we can find such an axis “between them” such that a 180 degree rotation takes one point into the other. The electric field vectors at the two points must have the same magnitude, because if they did not the transformed field would be different (the shorter magnitude vector replacing the longer and vice versa).

Now that we have proven these facts (using only geometric arguments --- the only reference to physics is that the ball “produces” the field), we can do Gauss’s law: $$ \begin{align} \int_S \vec{E} \cdot \hat{n} \, dA &= \frac{Q\text{enc by S}}{\epsilon_0}\\ \int_S E |\hat{n}| \cos(0) \, dA &= \frac{Q\text{ball}}{\epsilon_0}\\ E \int_S \, dA &= \frac{Q}{\epsilon_0}\\ E (4\pi r^2) & = \frac{Q}{\epsilon_0}\\ & \quad \rightarrow \quad E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} \end{align} $$ We have chosen the Gaussian surface, $S$, to be a sphere of radius $r$ (with $r$ greater than the radius of the ball). In going from the first to second line, we have used the fact that the normal to that surface is the same direction as the electric field vector, at any point on the surface (point 1, above). In going from the second to the third line, we have used the fact that the magnitude of the electric field is the same at every point on the Gaussian surface (point 2, above).

We've implicitly assumed that the direction of the electric field was outward, but this was not necessary. The signs on the two sides of Gauss's Law must agree, so, either $Q>0$ and $\vec{E}$ is parallel to $\hat{n}$, or $Q < 0$ and $\vec{E}$ is antiparallel to $\hat{n}$. Thus the direction of the field (in or out) is determined by Gauss's Law, even though its radial nature, and its uniformity at equal distances from the center, were conclusions from symmetry alone.

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  • $\begingroup$ Thanks for the answer, but as i said, I know this derivation, my question was infact how to mathematically state that the solutions to Maxwell's equation should have some degree of symmetry if the distribution has it. I was searching for a mathematical condition similar to the one (maybe wrong) I tried to formulate in my own answer. $\endgroup$ Feb 21 at 23:53
  • $\begingroup$ Ah. That is contained in my answer: "Because the charged ball produces the electric field, the electric field must also be unchanged by such rotations." This conclusion is not specific to Maxwell's equations, because it is a more general idea. If the source of a field is unchanged by some transformation, then the field itself must be unchanged by that transformation. Why? Because the transformation is acting on both the source and the field. And if the source is unchanged by the transformation, but the field is different, then you have two different fields from the same source. $\endgroup$
    – Ben H
    Feb 22 at 0:10
  • $\begingroup$ Transformation are often physically realizable. A rotation, for instance, is just the same as an observer moving to a different position to view the object. Imagine viewing a uniformly-charged sphere producing an electric field. Now walk around to the other side of that sphere. If it is producing a different looking electric field (despite looking exactly the same from this perspective), then you would be very confused. $\endgroup$
    – Ben H
    Feb 22 at 0:15
  • $\begingroup$ I intuitively get that solutions should be symmetric if the distribution is, I was asking of a way to axiomatize this fact, and not leaving it to intuition as a general idea. $\endgroup$ Feb 22 at 9:14
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    $\begingroup$ Yes i guess I’m just not being clear enough (or precise enough); I’ll think some more about it. I’m not claiming it is intuition, I’m trying to say it is a rigorous step in the argument: given a source and it’s field, we can transform them; if the source is unchanged by the transformation, then the field must also be. If that were not the case, the solution would not be unique. $\endgroup$
    – Ben H
    Feb 22 at 10:09

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