# What causes the electric field of a uniformly moving charge to update?

Note: I've edited my question to focus solely on the concept that I don't understand. If this isn't a physical situation, it's certainly a pedagogical one as it is exactly used in Purcell's E&M (3ed). And so, it should have some sort of explanation or cease to be used as a clarifying example (and I assume the former is more likely than the latter).

In frame F where a charge $$q$$ is moving with uniform velocity $$v$$, the electric field given by this charge depends on time. "It must be clearly understood that uniform velocity, as we have been using the term, implies a motion at constant speed in a straight line that has been going on forever" (Purcell E&M, 3ed).

To my understanding, it is radiation which perturbs existing electric fields and changes them. Radiation implies that an accelerating/accelerated charge must be present. There is no accelerating/accelerated charge in this situation.

Put simply, in the physical description I've described:

1. The electric field is changing with time $$\iff$$ Radiation is present

2. No accelerating/accelerated charge $$\implies$$ No radiation is present

To me, these two facts are inconsistent. How do I make sense of this apparent inconsistency?

• What you are describing is not what most people would call a "stationary point charge." Normally, we don't get to turn on or turn off the value of a point charge. Anyways, in your case, where you have a time-dependent electric charge distribution, you should be able to calculate the field and show whether or not it radiates. Did you attempt it?
– hft
Commented Aug 23, 2022 at 18:06
• @SillyGoose You are "turning on" the charge at $t=0$. It is time-dependent.
– hft
Commented Aug 23, 2022 at 18:13
• @SillyGoose You still are not locally conserving charge. But anyways, you can just solve the problem using the retarded potential.
– hft
Commented Aug 23, 2022 at 18:18
• @hft: It's worse than that. Local charge conservation is required for Maxwell's equations to make sense in the first place; there are no self-consistent solutions for the electromagnetic fields that do not obey local charge conservation. See my answer below. Commented Aug 23, 2022 at 19:00
• Your edited question is completely different from your initial question. You basically hollowed out the body of the question and replaced it with a different question. IMHO, that is not appropriate for edits directed towards re-opening. Just ask a new question.
– hft
Commented Aug 24, 2022 at 21:41

The situation you describe, with two equal and opposite charges popping into existence at a widely separated distance, obeys global charge conservation but not local charge conservation. Global charge conservation just means that the total amount of charge is the same at all times. Local charge conservation requires something stronger: it requires that the change in charge enclosed in any volume of space be exactly accounted for by the electric currents entering & leaving that volume. Hopefully you can see that the situation you describe does not satisfy this stronger condition. Mathematically, local charge conservation is expressed by the continuity equation for charge: $$\frac{\partial \rho}{\partial t} = - \vec{\nabla} \cdot \vec{J}.$$

But the problem is that Maxwell's equations require that charge be conserved locally, not just globally. This can be seen by noting that $$\frac{\partial \rho}{\partial t} = \frac{\partial}{\partial t} ( \epsilon_0 \nabla \cdot \vec{E}) = \vec{\nabla} \cdot \left( \epsilon_0\frac{\partial \vec{E}}{\partial t} \right) = \vec{\nabla} \cdot \left( \frac{1}{\mu_0} \vec{\nabla} \times \vec{B} - \vec{J} \right) = - \vec{\nabla} \cdot \vec{J},$$ which is the continuity equation. In other words, there is no solution to Maxwell's equations—the differential equations governing electromagnetic fields—in which positive and negative charges can "pop into existence" at a widely separated distance. It is mathematically inconsistent to discuss such a situation in the context of known physics.

• Okay, this is very helpful. The situation I described is physically impossible. Is it true in general (i.e. in theories beyond classical E&M) that any update in an existing electromagnetic field must be caused by radiation and thus by an accelerating charge? I am trying to build a visual in my head of how sources interact with existing electromagnetic fields. Commented Aug 23, 2022 at 19:02
• @SillyGoose: It's not even true in classical E&M. The field of a point charge moving at constant velocity, as measured at a fixed point in space, is constantly changing. But there is no radiation involved and no acceleration. Commented Aug 23, 2022 at 19:10
• You can satisfy continuity by introducing a delta current. The scalar and vector potentials can be solved for and a field can be found $\vec E = \theta(t-r/c)\frac{q\hat r}{4\pi \epsilon_0 r^2}$
– hft
Commented Aug 23, 2022 at 19:17
• Hm. (My knowledge is only based off what I've read recently in Purcell, so it may be very incomplete and wrong): Isn't a point charge moving with uniform velocity unphysical since it implies uniform travel since the beginning of time. At some point, the charge had to accelerate to its uniform velocity, creating the EM wave which propagates along and disturbs the "stationary field" into the "moving field"? Commented Aug 23, 2022 at 19:17
• @SillyGoose In both frames, there is no acceleration of the charge. There is no change in the field either in the usual limited context (i.e. a single point charge in a vacuum). If there is a change in the field, it must also involve an acceleration of one of the charges (i.e. as two electrons get closer together, they are accelerated away from one another, with a corresponding radiation). Most importantly, radiation isn't something separate from the fields - it's just a (rather general) wave in one of the fields. The wave equations describe all the changes in the fields. Commented Aug 24, 2022 at 6:05

Consider some vacuum in which $$\vec{E} = 0$$ everywhere. Then, place a charge $$q$$ in this space at $$t =0$$.

If you mean that you are suddenly "turning on" a point charge at $$t=0$$ then this is not how one typically thinks of a "stationary point charge."

In fact, it is pretty clear from the continuity equation (see below) that we can't be describing a point charge in the usual sense, since a "stationary point charge" has a current density equal to zero (because it is stationary).

Nevertheless, you can still figure out the fields and the potentials.

The charge density that you described is: $$\rho(\vec r, t) = q \delta(\vec r)\theta(t)$$

This leads to a retarded scalar potential, as usual, in the Lorentz gauge (and Gaussian units): $$\phi(\vec r, t) = \int d^3r'\frac{\rho(\vec r', t-|\vec r-\vec r'|/c)}{|\vec r - \vec r'|} = \frac{q}{r}\theta(t-r/c)\;. \qquad (1)$$

Although the problem did not specify a current density $$\vec J$$, we know there must be some current density in order to satisfy the continuity equation: $$-\frac{\partial \rho}{\partial t} = \vec{\nabla}\cdot\vec J$$ And so we introduce a current density $$\vec J$$: $$\vec J = -\delta(t)\frac{q\vec r}{4\pi r^3}\;,$$ which is clearly not the current density of a point charge. (The current density of a point charge is localized like $$\vec J(\vec r, t) = \vec v(t)\rho(\vec r, t)$$.)

The above current density leads to a retarded vector potential $$\vec A(\vec r, t) = \frac{1}{c}\int d^3r' \frac{\vec J(\vec r', t-|\vec r - \vec r'|/c)}{|\vec r - \vec r'|}\;.$$

The retarded scalar and vector potentials can be combined to find the electric field: $$\vec E = -\vec \nabla \phi - \frac{1}{c}\frac{\partial \vec A}{\partial t}\;.$$

The final result in SI units is: $$\vec E = \frac{1}{4\pi \epsilon_0}\int d^3r' \left(\frac{(\vec r - \vec r')}{|\vec r - \vec r'|^3}\rho(\vec r', t-|\vec r - \vec r'|/c) +\frac{(\vec r - \vec r')}{c|\vec r - \vec r'|^2}\frac{\partial \rho}{\partial t}(\vec r', t-|\vec r - \vec r|/c) -\frac{1}{c^2|\vec r - \vec r'|}\frac{\partial \vec J}{\partial t}(\vec r', t-|\vec r - \vec r|/c) \right)$$

• I am thinking: you must disturb an existing electric field to change it; thus, if any change occurs in an existing electric field, there must have been a disturbance. Commented Aug 23, 2022 at 18:22
• OK, I updated my answer.
– hft
Commented Aug 23, 2022 at 18:27
• I do feel like "ignoring the delta function portion of this" ignores the whole problem here. Commented Aug 23, 2022 at 18:37
• Also, that $\rho$ violates the continuity equation, so you'd need some sort of $\vec j$ that is proportional to a temporaral delta function and whose divergence is a spatial function. Commented Aug 23, 2022 at 18:39
• The point is that there IS a $1/r$ term in the field that comes from taking the gradient of the Heaviside function, because the information about the change of the field is communicated via electromagnetic waves. It's just in an infinitely thin wavefront Commented Aug 23, 2022 at 18:42

Your confusion comes from the fact that you identify radiation and time changing field. It can be shown that within Relativity, the electric field has two components: one longitudinal instantaneous electrostatic field and one transverse field generated by $$\frac{\partial \overrightarrow{B} }{\partial t}$$. Where longitudinal and transverse refers to the fields behavior in the Fourier Space. It can be shown that the instantaneous parts of the two fields cancel out in the real space. If the particles accelerate, there is a new component to the electric field that needs to be taken into account. It is the radiation field.

• "Note that ${\bf E}$ acts in line with the point which the charge occupies at the instant of measurement, despite the fact that, owing to the finite speed of propagation of all physical effects, the behaviour of the charge during a finite period before that instant can no longer affect the measurement." Is this the line you are referring to? If so, I am not sure how it reaffirms that this result does not violate Relativity since it doesn't provide an explanation as to why this is the case. Commented Aug 23, 2022 at 20:22
• You can find the full proof in: PHOTONS AND ATOMS INTRODUCTION TO QUANTUM ELECTRODYNAMICS by Claude Cohen-Tannoudji & al. The electric field has an instantaneous electrostatic component and a component generated by a varying magnetic field created by $\overrightarrow{j}( \overrightarrow{r},t) =q \overrightarrow{v}(t). \delta ( \overrightarrow{r}-\overrightarrow{r}(t))$, The second term cancels out among other things the non instantaneous part. Commented Aug 23, 2022 at 22:08

Electromagnetic radiation is when energy moves through space via electromagnetic fields. The fields have pressure, energy and momentum according to the Electromagnetic stress–energy tensor

$${\displaystyle T^{\mu \nu }={\begin{bmatrix}{\frac {1}{2}}\left(\epsilon _{0}E^{2}+{\frac {1}{\mu _{0}}}B^{2}\right)&{\frac {1}{c}}S_{\text{x}}&{\frac {1}{c}}S_{\text{y}}&{\frac {1}{c}}S_{\text{z}}\\{\frac {1}{c}}S_{\text{x}}&-\sigma _{\text{xx}}&-\sigma _{\text{xy}}&-\sigma _{\text{xz}}\\{\frac {1}{c}}S_{\text{y}}&-\sigma _{\text{yx}}&-\sigma _{\text{yy}}&-\sigma _{\text{yz}}\\{\frac {1}{c}}S_{\text{z}}&-\sigma _{\text{zx}}&-\sigma _{\text{zy}}&-\sigma _{\text{zz}}\end{bmatrix}},}$$ where $${\displaystyle \mathbf {S} ={\frac {1}{\mu _{0}}}\mathbf {E} \times \mathbf {B}}$$ is the Poynting vector, $$\sigma _{ij}$$ is the Maxwell stress tensor, and c is the speed of light.

As the electric and magnetic fields evolve over time according to Maxwell's Equations, the electromagnetic energy density $${\displaystyle u_{\mathrm {em} }={\frac {\epsilon _{0}}{2}}E^{2}+{\frac {1}{2\mu _{0}}}B^{2}\,}$$ changes over time satisfying the energy conservation law $${\displaystyle {\frac {\partial u_{\mathrm {em} }}{\partial t}}+\mathbf {\nabla } \cdot \mathbf {S} +\mathbf {J} \cdot \mathbf {E} =0}$$ where

• $$\mathbf {\nabla } \cdot \mathbf {S}$$ is the energy flowing out via radiation, given by the divergence of the Poynting vector
• $$\mathbf {J} \cdot \mathbf {E}$$ is the rate at which the fields do work on charges.

Radiation can be present without accelerating charges and can carry energy from one region to another.

No accelerating/accelerated charge $$\implies$$ Either no radiation is present, or radiation is present and is changing the electromagnetic energy density in the fields without doing work on charges.

In the example of a charge moving with uniform velocity, there is no accelerating charge and radiation is present: there is energy in the electromagnetic fields and this energy is moving through space (5 seconds ago most of the energy in the fields was concentrated around the region where the charge was 5 seconds ago, and now most the energy is concentrated around the region around where the charge is now).