Towards explaining the forces between accelerated charges – 2

In my last post, I presented a mathematical representation according to classical theory of the forces, resulting from acceleration, between two charges q₁ and q₂, where q₂ moves at the constant velocity v and q₁ moved at the velocity [u] and with the acceleration [a] at the retarded time. I also presented some terms of that representation that resulted from my own calculations, notably without any reference to magnetism or special relativity. Here, I will begin to explain how I arrived at those terms. Let me first repeat the basic image that describes the situation:

Just as in the case of constant velocities (copyright: Physics Essays Publications – also consider this document for more detailed calculations), the force due to acceleration is influenced by a change in the ‘frequency’ with which [a] is applied to q₂, from a situation in which both q₁ and q₂ are static to one in which q₁ moved at u at the retarded time, when it had the acceleration [a], and q₂ moves at v. To arrive at this frequency factor, similar to the case of constant velocities, I will first establish the projection of [a] onto a plane that is perpendicular to n – u/c (or to r₁₂*, which amounts to the same thing). In other words, I need to find x so that:

In other words

From this we obtain

Hence

where

To calculate the change in frequency, we now need to consider an extremely small ball of radius r₀ around q₂, which moves at v and in which q₂ is displaced by vr₀/c in the direction of the x-axis:

We need to calculate the length of the vector in the direction of where  cuts through the circle of radius r₀. We can do this be calculating k in

with

It follows that

The vector  now needs to be projected, in the plane defined by r₁₂* and , onto the plane that is vertical to n (and thus to [r₁₂]). This is because it is only in such a plane that the acceleration leads to a change in the effective charge:

To calculate this projection, we first calculate  by stipulating that, for a suitable x:

Bearing in mind that

the result is

This means that

We can now calculate d from

Hence

What does this d_1,2 and  enable us to do? Well, we can now write down two plane equations for planes that are vertical to  and that are at a distance d₁ and d₂, respectively, from zero (i.e. the current location of q₂), and at opposite sides of zero. We can then make those planes move at c towards zero and determine the times they take to meet q₂, which at the same time moves at v in the direction of the x-axis. We can then find out how those two times differ from the time that applies when a=v=u=0.

Just as a write this down, I wonder whether instead I need to look at just a=0, and v and u being any velocities. I shall look into that later. For now, let me persist with a=v=u=0.

In my next post.