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

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.

Answer 4

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).