# Conceptual question about special relativity in electrodynamics

The Lorentz force is *not* a Lorentz invariant, so if you get the same total force in S' as S, then you are doing something wrong.

In the stationary frame of the electrons, the force on one electron due the other is given by a pure Coulomb force $$ {\bf F'} = -e{\bf E'} = \frac{e^2}{4\pi \epsilon_0z'^2}\ {\bf \hat{z}} ,$$ where $z'$ is their separation along the z-axis.

In your laboratory frame (in which the electrons are moving) then the electric field is transformed according to the usual special relativistic transformations of the electromagnetic fields for a frame velocity difference of ${\bf v} = v{\bf \hat{x}}$ (i.e. perpendicular to a line joining the charges) $$ {\bf E} = \gamma E'\ {\bf \hat{z}}$$ and there is now a magnetic field $$ {\bf B} = -\gamma \frac{vE'}{c^2}\ {\bf \hat{y}}\ ,$$ and $z=z'$.

The total Lorentz force is then $${\bf F} = -e \left( {\bf E} + {\bf v} \times {\bf B}\right) = -\gamma eE' \left(1 - \frac{v^2}{c^2}\right)\ {\bf \hat{z}} = \gamma^{-1}\ {\bf F'}$$

Thus in the laboratory frame the force between the electrons diminishes as they get faster.

**Edit:** I just noticed that Rob's much more succinct answer basically says the same thing. This answer is mainly for people who'd like to derive how the electric and magnetic fields transform under a boost.

The short answer is that when you have a charge moving in space, there *is* a charge density as well as a current density, though it's not quite as simple to work with as in the case of an infinite line of charges, since -- as you're dealing with point object -- these densities are *singular*. The density of a point charge at rest at some point $\vec{r}_0$ can be given by $$\rho(\vec{r}) = q \delta^3(\vec{r}-\vec{r}_0),$$

and if that point is moving then the charge density is given by $$\rho(\vec{r},t) = q \delta^3(\vec{r}-\vec{v}t),$$

which is a bit of a bother to work with. If you're interested in finding the electric and magnetic fields of a point charge without explicitly using the charge and current densities, read on. My arguments will follow the Feynman Lectures on Physics (see here).

**General Transformations of Fields:**

I'm going to assume that you know what four-vectors are, and how they transform. I'm also going to assume that you know that the electrostatic potential and the magnetic vector potential together form a four-vector $A^\mu$. It is possible to do this entire analysis without using these assumptions, but they make it quite straightforward.

$$\mathcal{A} = \begin{pmatrix}\phi/c\\A_x\\A_y\\A_z\end{pmatrix}$$

Let's consider first that we have a charge $q$ that is at rest in the frame $S^\prime$ which is moving with respect to $S$ at a velocity $v$. An observer in $S$ would thus see the charge moving with a velocity $v$.

Since the four-potential $A^\mu$ is a four-vector, we can relate the potentials in $S^\prime$ with the potentials in $S$ using the (inverse) Lorentz Transformations:

\begin{equation*} \begin{aligned} \phi/c &= \gamma \left( \frac{\phi^\prime}{c} + \beta A^\prime_x \right)\\ A_x &= \gamma \left( A_x^\prime + \beta \frac{\phi^\prime}{c}\right)\\ A_y &= A_y^\prime\\ A_z &= A_z^\prime \end{aligned} \end{equation*}

Now, in $S^\prime$ the charge is at rest, and we simply have $$\phi^\prime = \frac{1}{4 \pi \epsilon_0} \frac{q}{\left({x^\prime}^2 + {y^\prime}^2 + {z^\prime}^2 \right)^{1/2}}, \quad \quad \vec{A}^\prime = 0 \text{ (No magnetic field)}$$

We can then find $\phi$ and $A_x$ as measured in $S$:

\begin{equation*} \begin{aligned} \phi/c &= \frac{\gamma}{4\pi \epsilon_0 c}\frac{q}{\left({x^\prime}^2 + {y^\prime}^2 + {z^\prime}^2 \right)^{1/2}} = \frac{\gamma}{4\pi\epsilon_0 c} \frac{q}{\left(\gamma^2 ( x - v t)^2 + y^2 + z^2 \right)^{1/2}}\\ A_x &= \frac{\gamma}{4\pi\epsilon_0 c^2} \frac{q v}{\left(\gamma^2 ( x - v t)^2 + y^2 + z^2 \right)^{1/2}}\\ A_y &= 0\\ A_z &= 0 \end{aligned} \end{equation*}

which may look complicated but is basically a simple substitution. The only other thing I've done is to write the RHS in terms of $x,y,z$ as measured in $S$, using the fact that $x^\prime = \gamma( x- v t)$ and so on.

We can get the Electric and Magnetic fields in $S$ from the potentials using: $$\vec{E} = -\vec{\nabla}{\phi} - \frac{\partial \vec{A}}{\partial t} \quad \quad \vec{B} = \vec{\nabla}\times \vec{A},$$ and you can show that $$\vec{E} = \frac{\gamma q}{4 \pi \epsilon_0} \frac{(x-vt)\hat{x} + y \hat{y} + z\hat{z}}{\left(\gamma^2 ( x - v t)^2 + y^2 + z^2 \right)^{3/2}} \quad \text{ and } \quad \vec{B} = \frac{\vec{v}}{c^2} \times \vec{E} $$

**Force on a second charge:**

Suppose now you had a second point charge $Q$ at rest with respect to $q$ in $S^\prime$. Imagine that the coordinates of $q$ and $Q$ are $(0,0,0)$ and $(0,y,0)$ respectively. The force on $Q$ due to $q$ would just be $$F_Q^\prime = \frac{1}{4\pi \epsilon_0} \frac{qQ}{y^2} = Q \vec{E}_q(0,y,0). \text{ (Since there is no magnetic field)}$$

Now what about the force observed by someone in $S$? According to this observer, the coordinates of the charges are $(vt,0,0)$ and $(vt,y,0)$, and the force is

$$F_Q = Q \left(E_q(vt,y,0) + \vec{v}\times \vec{B}(vt,y,0) \right) = \gamma Q \left( 1 - \frac{v^2}{c^2}\right) \vec{E}^\prime = \frac{q \vec{E}^\prime}{\gamma} = \frac{F_Q^\prime}{\gamma}.$$

Thus, the force is *not* the same in both frames, it is a component of a four-vector itself (the four-force) which I feel is not mentioned enough in most courses on Special Relativity.