# Conceptual question about special relativity in electrodynamics

I‘m currently taking Physics II and I have a conceptual question about SR in edyn.

Suppose I have two electron beams with linear charge density $$\lambda$$ paralell two each other in a resting system $$S$$ and each electron moves with a velocity $$v$$. Now I can calculate the electric and magnetic force between them, add them up, and get the total force between the two beams.

Now I move into the system $$S‘$$ which is moving with a velocity $$v$$. Here, because the electrons seem to be at rest, the magnetic force should be zero and the total force should be due to the electric force alone. Applying the lorentz factor $$\gamma$$ to the linear charge density $$\lambda$$ I get the same total force as in system $$S$$. Everything‘s fine.

Now, I don‘t have beams but two single electrons with charge q moving paralell with velocity $$v$$ in a resting system $$S$$. Here, I too can calculate $$F_{total} = F_E + F_B$$.

Again I‘m moving into a non resting system $$S‘$$ with velocity $$v$$. Here, the magnetic force is also zero and I‘m left with the electric force between the two particles.

And this is where my understanding fails. Since I have no charge density anywhere and q is invariant, I can‘t apply the lorentz transformstion and suddenly the total force in system $$S‘$$ is not equal to that in system $$S$$.

My understanding of the whole thing is, that the total force of both systems should not change and be equal to eachother. Can this only be explained by quantum mechanics? I don‘t know.

I’m sorry for this long question but this is really puzzling me and it would be great if someone could help me out with it

• The electric field of a moving charge is not the same as that of a stationary charge. If you take this into account it will all add up. May 28, 2020 at 18:34

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.

A point charge $$q$$ can be thought of as the charge density $$\rho(x^\mu) = q\ \delta^4(x^\mu)$$ if you want. But in the case of point charges you could apply the transformation directly to the Lorentz force

$$f = \frac q{m_0}\iota_p F,$$

where $$p$$ is the 4-momentum and $$F$$ is the electromagnetic tensor. Indeed, this last expression alone, which in coordinates reads

$$f^\mu = \frac q{m_0} F^{\mu\nu}p_\nu,$$

is enough to conclude that the force, which is a 4-vector, is covariant. Its magnitude $$g(f,f) = g_{\mu\nu}f^\mu f^\nu$$ is clearly a scalar, hence invariant under any Lorentz transformation.

• Note that this "four-force", as it is often called, is a different object than the usual force vector (which in these contexts is often called the "three-force"). May 28, 2020 at 20:28