# No magnetic field from a static charge - Is there a simple physical argument to show why?

For a charge moving in an electric field $\vec E$, its equation of motion is given by the electric part of the Lorentz force $$\frac d {dt}\gamma m \vec v = e\vec E$$This comes from the conservation of relativistic energy in a static electric field. But a magnetic field would still make this conservation law true since the magnetic force is always orthogonal to the velocity of the charge and therefore doesn't change its energy.

Is there a simply physical argument that shows why a static charge doesn't create a magnetic field?

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I don't really get what the preliminary discussion has to do with the question at the end. If you're interested in a static charge, why are we talking about accelerating charges? –  Mark Eichenlaub Jan 15 '13 at 23:56
Whatever arguments people put forth will hopefully not be inconsistent with the fact that an electron at rest DOES have a magnetic field. :-D –  Steve B Jan 16 '13 at 0:20
@MarkEichenlaub you're right that it doesn't really need the preamble. But I thought it worth putting in to emphasise that the conservation of relativistic energy partly explains the form of the electric Lorentz force created by a static electric change, and perhaps give a clue to the additional physical argument that is needed to show why it mustn't create a magnetic field. –  John McVirgo Jan 16 '13 at 0:24
@SteveB well this question is about a static charge, i.e. an ideal electric monopole, not an electron. –  David Z Jan 16 '13 at 1:00
@Chris -- It is not true that "magnetic fields can only be caused by time varying electric fields". One counterexample is a loop of wire carrying a DC current. Another counterexample is an electron at rest (since it's at rest, its electric field is not changing over time). –  Steve B Feb 1 '13 at 16:08
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The electric and magnetic fields form one object, the electromagnetic field tensor. This tensor represents an oriented plane at each point in space(time). An easy way to visualize this is to think of the magnetic field vector. Instead of the vector, think about the plane to which that vector is normal. This is the fundamental nature of the EM field.

The electric field is like this, too, except the planes have one direction along the time axis and one direction along a spatial axis.

When a point charge moves through space and time, it traces out a plane. One of the plane's directions is the direction it moves--a stationary charge "moves" through time. The other direction is based on the direction between it and an observer.

Since a charge at rest only moves through time, it sweeps out planes that have at least one timelike direction. This means its EM field contains only electric--not magnetic--components.

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The fact is, magnetism is nothing more than electrostatics combined with relativistic motion. Other definitions fall into two classes:

1. Really advanced speculation in the form of theory-of-everything-ness that I won't go into here.

2. Empirical observations from circa the year 1900, where people thought "huh, isn't it nifty that electromagnetism turns out to be relativistically invariant?" If you really believe that the magnetic field is some magical, independent entity that has to be tested in every conceivable configuration, then sure, it's possible that a static charge (or a flux capacitor or a unicorn) will produce a magnetic field.

But, these days we understand that there is only electrostatic attraction/repulsion between charges. A particle's acceleration can always be seen to be due solely to electrostatic effects if you transform into its rest frame. Thus you shouldn't be looking at any energy-conserving rule to derive the Lorentz force. You should start with relativistic mechanics and $F \propto Qq/r^2$, and all of Maxwell's/Lorentz's laws will follow.

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As a guess, I'd say for a static charge, the conservation of momentum requires the direction of the Lorentz force to be independent of the direction of the other moving charge's velocity. This then implies a magnetic field can't be created by a static electric charge. I haven't got a clue how to prove this though lol.

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A static charge only produces an electric field; cf. Maxwell's equations, Gauss' law.

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Edit. As Ben Crowell points out in his comments below (and in the subsequent chat), my argument below is incomplete. In particular, I used a specific integral expression for the electric field (written in terms of the charge density) that is not generally valid. In my own mind, that integral expression was justified by writing the electric field as $\mathbf E = -\nabla\Phi$, but as Ben pointed out in the chat, this cannot be done in general; one must also include a vector potential term. As a result, my claim that the time-independence of the charge density leads to the time-independence of the electric field does not hold with the assumptions stated. To make something like my argument work, one really needs to ensure, either by assumption, or perhaps by some other formulation of the problem, that the time derivative of the electric field vanishes. Thanks to Ben for his time!

There is a somewhat simple, rigorous argument coming from applying Maxwell's equations. User sybtc refers to Maxwell's equations in his answer, but he provides nothing in the way of detail, and I don't think the result is just an obvious application of Gauss's law. Gauss's law shows that a static charge distribution will lead to a static electric field, but one needs to analyze the other equations to show that such a charge distribution also does not create a magnetic field. One can actually prove the following claim which I think is what you are looking for (although I'm not sure it's a "simple physical" result).

Claim. For any localized charge distribution $\rho$ for which the corresponding current $\mathbf J$ vanishes, the magnetic field is everywhere zero.

Proof. First, recall the continuity equation: \begin{align} \frac{\partial \rho}{\partial t} +\nabla\cdot\mathbf J = 0 \end{align} The hypothesis $\mathbf J = \mathbf 0$ therefore implies that the charge distribution is static; $\partial\rho/\partial t = 0$. By Gauss's Law \begin{align} \nabla\cdot \mathbf E = \frac{\rho}{\epsilon_0}, \end{align} the electric field of a localized charge distribution has the following integral expression: \begin{align} \mathbf E(t,\mathbf x) = \int_{\mathbb R^3}d^3x'\,\rho(t,\mathbf x')\frac{\mathbf x - \mathbf x'}{|\mathbf x - \mathbf x'|^3} \end{align} It follows that since $\rho$ is time-independent, then so is $\mathbf E$: \begin{align} \frac{\partial\mathbf E}{\partial t} = \mathbf 0 \end{align} That fact, the fact that the current vanishes, and Ampere's Law \begin{align} \nabla\times \mathbf B = \mu_0\mathbf J +\mu_0\epsilon_0 \frac{\partial \mathbf E}{\partial t} \end{align} combine to require the magnetic field to have zero curl; \begin{align} \nabla\times\mathbf B = 0 \end{align} But now recall that one of Maxwell's equations tells us that the divergence of the magnetic field is zero; \begin{align} \nabla\cdot \mathbf B = 0 \end{align} It follows from the Helmholtz Decomposition, that the magnetic field must vanish everywhere provided it falls off sufficiently rapidly at infinity, a reasonable property that should be true of physical charge distributions.

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Thanks, I guess Maxwell's equations says so and that's it really. I still think there's a possible simple argument though: boosting to the COM frame of two charges moving at identical velocities towards one another, and looking at the forces which should be identical. –  John McVirgo Aug 29 '13 at 16:55
@JohnMcVirgo Sure thing. I'm not sure I follow your argument. Would you elaborate? Are you attempting to demonstrate that there is no magnetic field produced in that simple situation of the two charges moving towards each other? –  joshphysics Aug 29 '13 at 19:15
The forces are assumed to be general (electric + magnetic + other) and symmetrical to one another in the COM frame, and this limits the type of force experienced by a charge in the rest frame of the other. As a guess, it means it can't be velocity dependent and therefore can only be electric. Maybe the direction of the acceleration is limited in some way? –  John McVirgo Aug 29 '13 at 20:39
This isn't quite right. A plane wave in vacuum is a counterexample to the claim as currently stated. You really need a separately stated assumption that the electric field has no time variation. –  Ben Crowell Sep 2 '13 at 22:17
@BenCrowell I'm not convinced. I probably should have stated this in the claim, but I assumed that the fields vanish rapidly at $\infty$ (I explicitly stated this at the end for the magnetic field). The plane wave doesn't satisfy this assumption. Correct me if I'm wrong, but once one assumes this, and once you assume that the current vanishes everywhere in space, you're ok. –  joshphysics Sep 2 '13 at 23:49
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I'm construing the question as follows. Two of Maxwell's equations are

$$(1) \qquad \nabla\cdot\textbf{B}=0 \qquad \text{and}$$

$$(2) \qquad \nabla\times \textbf{B}=(\ldots)\frac{\partial\textbf{E}}{\partial t}+(\ldots)\textbf{J} \qquad .$$

What is a simple physical argument to prove we can't possibly have an alternative form in which charge appears on the right-hand-side of one of these equations as an additional source term?

We can't do it with equation 2, because it's a vector equation, and charge is a scalar. We can't do it with equation 1, because charge is even under time-reversal and parity, but the magnetic field is odd.

As a side remark, it's possible to find examples of systems hypothetically behaving in a way that we know is inconsistent with the equations of motion, but that is not obviously ruled out by conservation laws. An example would be a rock hovering in mid-air, which conserves energy and momentum just fine.

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