Deriving electromagnetism from relativity and electrostatics.

Part 1: Liénard–Wiechert potential

The Liénard–Wiechert potential is given by-

$$\varphi(\mathbf{r}, t) = \frac{1}{4 \pi \epsilon_0} \left(\frac{q}{(1 - \mathbf{n} \cdot \mathbf{v_s})|\mathbf{r} - \mathbf{r}_s|} \right)_{t_r}$$

Here, $q$ is the source charge, $n$ is unit vector pointing towards the charge from location $\mathbf{r}$, $v_s$ is the source velocity at a retarded time, and $\mathbf{r}$ is the source location at a retarded time.

There is a similar potential given in Feynman Lecture Vol. 2 which is only an approximation and misses the $\frac{1}{1 - \mathbf{n} \cdot \mathbf{v_s}}$ factor. In spite of the extra factor in the LW potential, it is simpler because the local time component $\varphi$ of four vector field potential is invariant under Lorentz transformation. We will prove this now. The advantage of this invariance is that once we know the local field potential, we do not need to worry about change in the far away charge distribution after Lorentz transformation. The local four vector field potential contains the complete information about the far-away charges. The same is not true for the approximate potential given in the Feynman Lecture.

The equation given in Feynman lecture is different because it is inside an integral. When doing this integral, the required factor will turn up, due to apparent change in the size of a body approaching or receding from us, due to the fact that speed of light is finite for all observers. If speed of light was infinite, approaching bodies will not appear larger, and so would charge.

Let there be a source charge stationary at point $(x_0,y_0,0)$ with respect to an observer O at origin. Let there be another observer $O'$ at origin, moving with velocity $\mathbf{v}=-v \hat{i}$. Then locally, in the $O'$ frame, one would expect that $$\varphi '=\gamma \varphi$$, where $\gamma=\frac{1}{\sqrt{1-v^2}}$.

Let us see whether the same result follows from LW potential in the new frame.

In the frame O,

$$\varphi(0, t) = \frac{1}{4 \pi \epsilon_0} \left(\frac{q}{|\mathbf{r}|}\right)$$

In the new frame,

$$\varphi \prime(0, t) = \frac{1}{4 \pi \epsilon_0} \left(\frac{q}{(1 - \mathbf{n'} \cdot \mathbf{v})|\mathbf{r'}|}\right)$$

Note that in the new frame, $x_0 '=\gamma (x_0 + vr)$ where $r=\sqrt {x_0^2+y_0^2}$. Also, $y_0\prime=y_0$.

Also, Note that $$(1 - \mathbf{n'} \cdot \mathbf{v})=1-\frac{v x_0 '}{r'}$$, where $r'=\sqrt {x_0 '^2+y_0^2}$

$$=\frac{r\prime -v x_0\prime}{r\prime}$$

$$\implies \varphi \prime(0, t) = \frac{1}{4 \pi \epsilon_0} \left(\frac{q}{r\prime -v x_0\prime }\right)$$

Now, $r\prime -x_0\prime v$ turns out to be equal to $\frac {r}{\gamma}$, as can be checked. Hence, we obtain the expected equation.

Part 2: The actual Proof

Let there be a source charge stationary at point $(x_0,y_0,0)$ with respect to an observer O at origin. Let there be a charge $q$ at the origin moving with the velocity $\mathbf{u'}=u_x \hat{i}+u_y \hat{j}$ w.r.t O. Let there be another observer $O'$ at origin, moving with velocity $\mathbf{v}=v \hat{i}$, such that in the O' frame, the source appears to move with velocity $-v \hat{i}$.

What we are going to prove is, if we assume that for a stationary source, electrostatic force is as given by coulomb's law, then a Lorentz transformation of this force from frame O to O' is equivalent to the force arising from electromagnetic fields generated when solving the same result using the LW potential.

According to electromagnetic theory:-

$$\mathbf{E}=-\nabla \varphi - \frac{\partial \mathbf{A}}{\partial t}$$

$$\mathbf{B}=\nabla \times \mathbf{A}$$

$$ \mathbf{F}\propto \left( \mathbf{E} + \mathbf{B} \times \mathbf{v}\right)$$

Now, $\nabla \varphi = \varphi _x +\varphi _y + \varphi _z$. Also, 

$$\varphi_x= \frac{-x_0}{r^3}$$

$$\varphi_y= \frac{-y_0}{r^3}$$

$$\varphi_z=0$$

Then, in the O' frame,

$$\varphi '=\gamma \varphi$$

$$\mathbf{A}=-\gamma v \varphi$$

$$-\varphi _x' - \frac{\partial \mathbf{A'}}{\partial t} = \frac {\varphi \prime _x}{\gamma} = \varphi _x$$, because $\frac {\partial \mathbf {A'}}{\partial t}$ cancels the time component of $\varphi \prime _x'$ as seen in the O frame.
$$\implies \mathbf{E'}=\varphi _x \hat{i} + \gamma \varphi _y \hat{j}$$
Also, $$\mathbf{B'}=-v \gamma \varphi _y \hat{k}$$
$$\mathbf {F'}=q \left( \mathbf{E'} + \mathbf {B'}(u_y \prime \hat{i} - u_x \prime \hat{j})  \right)$$
Note that $u_y \prime$ and $u_x \prime$ are the velocity components of the second charge placed at origin, as seen in the O' frame.
Now in the frame O, $F=\varphi _x \hat{i} + \varphi _y \hat{j}= F_x \hat{i}+F_y \hat{j}$.
Let $ F'=F'_{x'} \hat{i}+ F'_y \hat{j}$.

Then $$F'_{x'}=F_x - \frac {F_y v u_y}{1- v u_x}$$.
Also, $$F'_y= \gamma F_y + v \gamma F_y \left( \frac {u_x -v}{1- v u_x} \right)$$
$$\implies F'_y= \frac {F_y}{\gamma (1-v u_x)}$$

This is the force that you would get when you apply Lorentz transformation from frame O to O'.

Concluding points

1. The LW potentials are completely part of classical electromagnetism. But it can be noted that classical electromagnetism is not consistent under Galilean transformation. The correct transformation that preserves invariance of experimental results between inertial frames is that given by Lorentz transformation, which was discovered before Einstein's result.
2. Hence magnetism is nothing more than relativistic effects on Force acting on a moving body as viewed from a frame of reference other than the one in which the object is stationary. It should be noted that this should apply for other forces of nature as well. For example gravito-magnetism is as real an effect.
3. What I did in this post was what was historically done in the reverse order. Relativity was deduced to make electromagnetism consistent. I, instead, assumed SR and derived EM as a consequence. It should be noted though that SR is more fundamental than EM.