# Confusion on relativistic electromagnetism of two charge particles

When I was introduced to the relativistic nature of electromagnetism, the following scenario is introduced:

Two protons are fired such that they travel at the same velocity in parallel directions relative to the laboratory. They are mutually momentarily at rest in each others' frames of reference.

This and the following statements are understandable:

In a reference frame that sees the protons at rest, the protons experience a force in the opposite direction as a result of electrostatic repulsion.

However, the following is confusing:

In a reference frame that sees the protons moving, the same forces result from the magnetic force created by the protons' movement.

What I took away from this is that the moving protons are like current carrying-wires in that they create magnetic fields, and the magnetic fields acting on a moving charge create a magnetic force, but if that is true, protons in parallel motion should experience mutually attractive magnetic forces rather than repulsive forces. How does that work?

• This may help you. – Tofi Apr 7 '16 at 17:40

However, the following is confusing:

In a reference frame that sees the protons moving, the same forces result from the magnetic force created by the protons' movement.

I agree that it is confusing. In the instant that the two protons begin to move away from each other they are momentarily mutually at rest. At this instant, in a frame of reference in which the proton's are (momentarily) at rest, the force on each proton due to the other is purely electric in nature.

At this instant, in this frame, the coordinate acceleration of either proton is equal to its proper acceleration (the acceleration as measured by an 'accelerometer' on the proton)

However, in the lab frame, special relativity dictates that the coordinate acceleration of the protons away from each other $(\mathbf{a}'_\perp)$ is actually less than the proper acceleration away from each other $(\boldsymbol{\alpha}_\perp)$:

$$\mathbf{a}'_\perp = \frac{1}{\gamma^2}\boldsymbol{\alpha}_\perp$$

But we also have that

$$\mathbf{F}'_\perp = \gamma m_p \mathbf{a}'_\perp$$

$$\mathbf{F}_\perp = m_p \boldsymbol{\alpha}_\perp$$

Thus, it must be that

$$\mathbf{F}'_\perp = \frac{1}{\gamma}\mathbf{F}_\perp$$

That is, in the lab frame, there is less of a repulsive force on each proton.

Since charge is Lorentz invariant, the electric force of repulsion between the protons is unchanged from the (momentary) rest frame and so, if there were only an electric force, there would be an inconsistency.

However, there is also an attractive magnetic force component in the lab frame and so, the electromagnetic force of repulsion between the protons is less in the lab frame and is thus consistent with SR.

To be sure, let's check the calculation. The fields at the location of the 'upper' proton due to the 'lower' proton in the momentary rest frame transform to the lab frame as

\begin{align} & \mathbf {{E}_{\parallel}}' = \mathbf {{E}_{\parallel}}\\ & \mathbf {{B}_{\parallel}}' = \mathbf {{B}_{\parallel}} = 0\\ & \mathbf {{E}_{\bot}}'= \gamma \left( \mathbf {E}_{\bot} + \mathbf{ v} \times \mathbf {B} \right) = \gamma \mathbf {E}_{\bot} = \gamma \frac{e}{4\pi\epsilon_0d^2}\hat{\mathbf{z}}\\ & \mathbf {{B}_{\bot}}'= \gamma \left( \mathbf {B}_{\bot} -\frac{1}{c^2} \mathbf{ v} \times \mathbf {E} \right) = -\gamma \frac{1}{c^2} \mathbf{ v} \times \mathbf {E} = +\gamma \frac{v}{c^2}\frac{e}{4\pi\epsilon_0d^2}\hat{\mathbf{x}} \end{align}

where the lab frame has velocity $\mathbf{v} = -v\hat{\mathbf{y}}$ in the momentary rest frame.

Since the protons have zero velocity in the momentary rest frame, the Lorentz force on the 'upper' proton is

$$\mathbf{F} = e\left(\mathbf{E} + \mathbf{0} \times \mathbf{B} \right) = \frac{e^2}{4\pi\epsilon_0d^2}\hat{\mathbf{z}}$$

In the lab frame, the velocity of the protons is $\mathbf{u} = v\hat{\mathbf{y}}$ and the Lorentz force is

$$\mathbf{F}' = e\left(\mathbf{E}' + \mathbf{u} \times \mathbf{B}' \right)\frac{e^2}{4\pi\epsilon_0z^2} \gamma\left(1 - \frac{v^2}{c^2} \right)\hat{\mathbf{z}} = \frac{1}{\gamma}\mathbf{F}$$

as desired.

You are right that the two moving protons feel an attractive force and therefore have a smaller acceleration away from each other. There are two ways to get this result.

The first way, as you already know, is to see that the electrostatic force is partially canceled by a magnetic force.

The second way to see it is to find the acceleration by boosting (lorentz transforming) the protons from their reference frame. When the boost is done, the acceleration will be reduced due to time dilation, so you again get a smaller acceleration.

If you go through and do the math (which might be very complicated, at least for calculating the magentic fields of point charges) you should get that the answers of these two methods are the same.

See the relativistic phenomena should converge to a single reality. To an observer at rest relative to the moving proton, he sees a current. A current produces a magnetic field. So he sees a magnetic force on the charges.
But for an observer moving with the same velocity as the protons, he see both parallel layers of protons at rest. What he observes as force is the electrostatic repulsion between them. While you do the mathematics on the problem you will see that the magnetic force one observes in the rest frame is the electrostatic force the other observed on the moving frame. It has to be so. Relativity means symmetry of coordinate geometry. The physics should be the same on both frames. Both observers measure the same force but the forces themselves are different in their sources of origin. That is magnetism is a relativistic phenomenon. You observe magnetism on a charge only if the charge appears moving to you. If you move with the same velocity as the charge, you won't see the charge moving, which means what you observe is only just an electrostatic repulsive force between the two charges.
As an example let's take the case of length contraction in special relativity. Suppose you are a stationary observer on earth. You see a spacecraft travelling at a speed of 0.5c (0.5*3*10^8 m/s). You will see it's length contracted by a factor, say x meters. But to an observer (me) on the spaceship I cannot observe my ship contracted as me and everything inside the ship are moving with the same velocity, everything, including me, on the ship will be contracted by the same factor. So I can't feel that change. But what I observe is that I'am at rest and the rest of universe, including you that is moving with a velocity of 0.5c. So I measure the same factor of x meters by which the length of the universe contracted.
For a complete discussion see:
Introduction to Electrodynamics, 4th ed, by David.J.Griffith (chapter:Relativistic Electrodynamics)