How can two observers agree on the force between 2 moving charges?

Question

I'm having a hard time figuring this out. This question has been asked before, I saw the answers, but I'm still struggling with this. So I decided to ask.

I want to calculate the same force for both observer's given the following setup: (I hope everything is clear, don't hesitate asking for clarifications if something isn't clear)

In classical mechanics, the Lorentz force is dependent on the cross product between the velocity and the magnetic field.

$\vec{F_{12}} = Q_2(\vec{E_1} + \vec{v_2}\times \vec{B_1})$

$\vec{F_{21}} = Q_1(\vec{E_2} + \vec{v_1}\times \vec{B_2})$

For me at (0,0), the force felt by $Q_1$ would be: $\vec{F} = \vec{F_{21}} - \vec{F_{12}}$. Assuming I should add $-\vec{F_{12}}$ based on Newton's Third Law.

Both observers are going to disagree in this setup. The cross products for the second observer would be both = 0. Is it a correct assumption in the first place?

What's the correct way to mathematically calculate the same force for both observers. Specially the magnetic force which is the one which depends on velocities.

I suspect I should make use of the Electromagnetic tensor. (https://en.wikipedia.org/wiki/Electromagnetic_tensor).

Is it correct?, can you point me in the right direction? or even better for me, can you teach me how to calculate the same force?

EDIT: As @ProfRob pointed out, they may not agree on the forces, so what's the quantity they agree on?

Answer 1

Force is not an invariant in Special Relativity (as is the case for all 3-vectors). The observers will disagree on the force between the two charges.

Also, the two observers do not agree on what the electric and magnetic fields are.

To calculate the forces according to each observer you could use the Lorentz force calculated with the respective transformed electromagnetic fields and velocities.

https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Transformation_of_the_fields_between_inertial_frames

Edit: In terms of what they agree on:

• The size of the charges
• $\vec{E}\cdot \vec{B}$
• $E^2 - c^2 B^2$

Answer 2

You have to use the EM textbook equations, \begin{eqnarray} {\bf E}&=&\frac{q\gamma{\bf r}}{({\bf r_\perp}^2 +\gamma^2{\bf r_\parallel}^2)^{\frac{3}{2}}} = \frac{q{\bf r}} {\gamma^2[r^2-({\bf v\times r})^2]^{\frac{3}{2}}} \label{eq:erelv}\\ {\bf B}&=&{\bf v\times E} =\frac{q{\bf v\times r}} {\gamma^2[r^2-({\bf v\times r})^2]^{\frac{3}{2}}}, \label{eq:brelv} \end{eqnarray} for the electromagnetic fields of a charge moving at constant velocity. They are derived by Lorentz transforming the Coulomb field to the moving system.