A Linearly Accelerating Charge: Radiation

The frequency or wavelength of light does not appear directly in Maxwell's equations. The frequency or wavelength of light also does not appear directly in the wave equation. This indicates there is nothing particularly unique about the wavelength of light, especially in vacuum. This seems to hint that we should look for some unified source of all electromagnetic radiation.

We find that the emission of radiation is a result of accelerating charge. This is somewhat expected because electromagnetic radiation is an oscillation in the electric and magnetic field, whereas the fields themselves emanate from charged particles.

If we take a look at Gauss's law for a point of charge q, then we get

$LaTeX: \oint_s\overset{\rightharpoonup}{E}\cdot d\overset{\rightharpoonup}{S}=\frac{q}{\epsilon_0}$ $\oint_s\overset{\rightharpoonup}{E}\cdot d\overset{\rightharpoonup}{S}=\frac{q}{\epsilon_0}$

for an enclosing surface of radius R, we have that the electric field is radial and constant in magnitude (at the distance R) and we integrate over the surface element $LaTeX: \left|d\overset{\rightharpoonup}{S} \right|=r^2\sin(\theta) d\theta d\varphi$ $\left|d\overset{\rightharpoonup}{S} \right|=r^2\sin(\theta) d\theta d\varphi$ , we get for the LHS

$LaTeX: E\int_0^{2\pi}\int_0^{\pi}r^2\sin(\theta)d\theta d\varphi=4\pi Er^2$ $E\int_0^{2\pi}\int_0^{\pi}r^2\sin(\theta)d\theta d\varphi=4\pi Er^2$

this is equal to the total charge over ε₀, so that

$LaTeX: E=\frac{q}{4\pi\epsilon_0r^2}$ $E=\frac{q}{4\pi\epsilon_0r^2}$

then the radial field drops off like 1/r². Now if that charge is moving at constant speed v the field lines are happy to move attached to the charge and do not change as a function of time. In fact, if an observer were to travel along side the charge at the same speed, based on their observation of the field lines, the charge would appear at rest as in (a) below. However, to an observer at rest, for speeds comparable to that of light, the field lines would appear compressed as in (b) below. If the charge were to be accelerating uniformly (for all time), then the field lines would appear as in (c) below

Moving Charge Length Contraction

It is only when the charge moves in a non-uniform motion (i.e., accelerates) that radiation occurs. The short movie below demonstrates the somewhat unusual effect when a charge, moving at constant velocity suddenly accelerates, and then returns to its initial velocity, by decelerating. A kink appears in the field lines. This is due to the fact that 1) the charge must move slower than the speed of light and 2) we have good reason to believe that the information transmitted by field lines moves exactly at the speed of light.

Accelerating Charge Electric Transverse Field

When the charge initially accelerates, the field lines nearest the charge must jolt forward, but the field lines a distance ct away from the charge cannot know at that instant that the charge had even moved, therefore they must stay in their initial orientation. The same argument applies to the case when the charge decelerates; the field lines nearest the charge suddenly return to nearly their original configuration, albeit they are now emanating from a charge that has translated. The result is that the field lines far enough away to not have noticed the charge originally accelerated, and the field lines nearest the charge after it returns to constant velocity are slightly displaced relative to one another. The only way for the field lines to stay connected is for a transverse component of the electric field to connect them.

The electric field from this accelerated charge is now composed of two components, one longitudinal and one transverse. The longitudinal component is the usual field emanating from a point charge and falls off as 1/r². We have derived this earlier using Gauss's law. The transverse component is a direct result of the acceleration and dies of as 1/r (the derivation can be found in the Feynman lectures, link below). Clearly, at a distance far from the point charge, the only field that remains is the transverse component and hence we get a traveling wave.

One more very important point to be made here is the fact that the transverse component of the electric field i.e., the radiation, is strongest perpendicular to the acceleration of the electron. In fact, looking at the diagram below provided by Yaron Hadad, we see that the field lines are not deformed at all in the direction of the acceleration.

Accelerating Charge Still Frame

(https://www.yaronhadad.com/the-geometry-of-electromagnetic-field-lines/)

In other words, the transverse field lines are most densely packed (which means strongest) in the direction rotated 90˚ from the direction of acceleration (where the rotation can be in any direction away from the acceleration vector) and have no strength whatsoever along the direction of acceleration. It turns out these facts lead to the radiation intensity pictured below

Accelerating Charge Intensity Pattern

and the electric field for the transverse component (i.e., the radiation component) when the charge is moving slowly compared to the speed of light (i.e., v/c ~ 0)

$LaTeX: E_{\text{radiation}}=\left(\frac{q}{4\pi\epsilon_0}\right)\frac{a}{c^2}\frac{\sin(\theta)}{r}$ $E_{\text{radiation}}=\left(\frac{q}{4\pi\epsilon_0}\right)\frac{a}{c^2}\frac{\sin(\theta)}{r}$

this expression was calculated using the expression derived by Feynman (LINK Links to an external site.) in Eqn. 26.2, Eqn. 26.3, and Eqn. 26.6. We can see from the equation here that the magnitude of the radiated field scales like sin(θ) and 1/r. Whenever the r-vector is in-line with the acceleration, the magnitude of the transverse component is zero. This again means that no "light" shines in the direction of the oscillation.