The Larmor formula is used to calculate the total power radiated by a nonrelativistic point charge as it accelerates. When accelerating, any charged particle (such as an electron) radiates away energy in the form of EM waves. When the velocity is small compared to the speed of light, the total power radiated is given by the Larmor formula:
where a’ is the acceleration, q is the charge, and c is the speed of light. A relativistic generalization is given by the Liénard-Wiechert Potentials, which are the most complete form of electromagnetic potential functions generated by charges. This formula has important applications in the theory of scattering and radiative resistance.
We’ll start off with something familiar, Coulomb’s Law:
where er is the unit vector that points from the field point (the point where the field in measured/calculated) to the source point (the point where the charge is). It is important to note that the electric field generated by a point charge is not entire given by this formula. Firstly, we can observe a few peculiarities simply by observing the formula. Let’s say I suddenly place a charge at the other end of the universe (or anywhere VERY far away), will I feel the field generated by that charge immediately? According to this formula, yes, since there is no dependence on time at all. But of course we know that is not the case, since nothing can travel faster than the speed of light, including electric effects such as the electric field. It turns out that a few more terms are required for us to have a complete formula. Not only that, all the positions, velocities and accelerations of the point charge in question but be written in terms of its retarded positions, velocities, and accelerations. These quantities must all be evaluated at the retarded time, t’:
Where r’ is the retarded position. You can see that this formula makes sense because the retarded time is the time at the present moment, minus the time it takes for electric influences (which travels at the speed of light, c) to reach the field point. Now we have fixed the retardation problem, but the formula is still not complete. It turns that in general, the electric field also depends on the acceleration and velocity of the charge (as you might expect, since if not an accelerating charge would not generate EM waves!):
The first term in the bracket is just the familiar Coulomb’s law term. Now two additional terms have appeared. A quick observation reveals that the first two terms fall off with 1/r’2. It may not be immediately obvious, but the third term, which is responsible for radiation, falls off with 1/r’. (Energy has to be conserved. The power radiated from a source cannot disappear. Imagine a source radiating energy in a spherically symmetric manner. The surface area of the sphere increases with r’2, which means the intensity of radiation falls with 1/r’2. The square of the electric field is proportional to the intensity of radiation, thus the electric field falls with 1/r’). Over large distances, the 1/r’2 terms quickly become insignificant as compared to the 1/r’ term. Thus when the field point is far away from the source point, as often is the case when it comes to radiation, the formula is just:
This is the starting formula from which all radiative effects can be worked out. Even relativistic effects are included in it.
Now we turn our attention to the term. What does it mean? Suppose we are standing at the field point, looking at a point charge that is rather far away. The charge is wiggling about, and because it is wiggling it is emitting radiation that we detect at the field point. er’ is the unit vector that points from our position to the wiggling charge, and is the acceleration of this unit vector. Now we know that the end of er’ is restricted to move on a sphere of radius one centered on us. Its acceleration will thus have two components: a radial component (the centripetal acceleration) and a transverse, or tangential, component. As the distance between us and the charge increases, the wiggling gets smaller and smaller, just like how anything appears smaller when they are further away. It is easy to see that the transverse component of the acceleration will drop with 1/r’. The radial component of the acceleration is:
Since the radial acceleration varies with the inverse square of the retarded distance, we can disregard it in comparison to the tangential acceleration. The tangential acceleration is simply
To avoid confusion, please note that the tangential acceleration at on the left hand side of the equation refers to the tangential acceleration of the unit vector er’, while the tangential acceleration at’ refers to the retarded tangential acceleration of the charge. That is, the acceleration of the charge that is perpendicular to our line of sight. This also means that if a charge is oscillating along your line of sight, you won’t detect any radiation from it. You might be expecting some “transverse” EM wave from this motion, but wait, you do know that there is no such thing as “transverse” EM waves right? XD
Anyway, making these substitutions into our equation, we get,
For simplicity, I have dropped the vectors from the equation. Now to proceed with any further calculations, we have to know how our point charge is actually moving. We will take the simplest case. The charge is simply going up and down (the exact motion doesn’t matter that much), as shown in the figure below.
We would expect the at’ term in our present formula to be replaced by a’ of the point charge, multiplied by a correction factor (since when we are looking at the charge at an angle, only a component of the acceleration is tangential). From geometry, that factor is sin(θ), where θ is defined in the diagram above. So,
There is one more formula to be introduced before we can get the result. I’m not going to prove it, but the intensity of the radiation, S, (that is, the energy of the radiation that is flowing per unit area per unit time) is related the electric field E by:
Substituting our formula for the electric field into this equation, we obtain:
The final step is just to integrate this over the whole surface of the sphere. The sin(θ) guarantees cylindrical symmetry, so the area element we pick is a ring, as shown below,
And from geometry, the area of this element is equal to
Thus our integral is:
If you actually want to see the maths…
And we arrive at the final equation for Larmor formula,