# Electromagnetic "two-body" problem

Consider two electric charges $$q_1$$ and $$q_2$$ interacting by means of Lorentz forces:

$$\boldsymbol{F}_{i\to j} = q_j(\boldsymbol{E}_{i}+ \boldsymbol{v}_{j}\times\boldsymbol{B}_{i}), \qquad i,j\in\{1,2\},\text{and}\ i\neq j$$ using Coulomb's law and the Biot–Savart law, we get: $$\boldsymbol{F}_{1\to 2} = \frac{q_1q_2}{(4\pi \epsilon_0)d^2} \left( \boldsymbol{\hat{u}}_{12}+ \frac{\boldsymbol{v}_2\times(\boldsymbol{v}_1\times \boldsymbol{\hat{u}}_{12})}{c^2}\right)$$

$$\boldsymbol{F}_{2\to 1} = \frac{q_1q_2}{(4\pi \epsilon_0)d^2} \left( \boldsymbol{-\hat{u}}_{12}- \frac{\boldsymbol{v}_1\times(\boldsymbol{v}_2\times \boldsymbol{\hat{u}}_{12})}{c^2}\right)$$

Where we have used that $$\mu_0 = 1/(\epsilon_0 c^2)$$. Now, it is well known that in the presence of magnetic fields Newton's third law is not fulfilled (since the electromagnetic field is a "third body" that has linear momentum and energy). For this reason, $$\boldsymbol{F}_{1\to 2} \neq -\boldsymbol{F}_{2\to 1}$$, which can be verified by adding both forces. Using the vector identity $$\boldsymbol{a}\times(\boldsymbol{b}\times \boldsymbol{c}) = (\boldsymbol{a}\cdot\boldsymbol{c})\boldsymbol{b}-(\boldsymbol{a}\cdot\boldsymbol{b})\boldsymbol{c}$$, we get that:

$$\boldsymbol{F}_{1\to 2}+\boldsymbol{F}_{2\to 1} = \frac{q_1q_2}{4\pi \epsilon_0} \left(\frac{(\boldsymbol{v}_2\cdot\boldsymbol{\hat{u}}_{12})\boldsymbol{v}_1-(\boldsymbol{v}_1\cdot\boldsymbol{\hat{u}}_{12})\boldsymbol{v}_2}{d^2c^2}\right)$$

Now my questions are:

1. It is correct to say, assuming that the particle velocity is small relative to the speed of light, that the electromagnetic field undergoes a change of momentum in the form of radiation given by $$\text{d}p_{EM}/\text{d}t = \boldsymbol{F}_{1\to 2}+\boldsymbol{F}_{2\to 1} \neq 0$$? Or do we need to include radiative corrections such as those in the Larmor formula?
2. How to demonstrate in a simple way that if the charges move slowly (with respect to light) and are of opposite sign they would eventually approach each other? (it seems clear that if the velocities tend to align then $$\boldsymbol{F}_{1\to 2}+\boldsymbol{F}_{2\to 1}$$ tends to zero)
3. What about the angular momentum, is there a way to write a simple formula for the evolution of the angular momentum of the system? Does it happen that the electromagnetic field also drags some amount of angular momentum away from the charges?
• The formula you have described would not cause radiation anyway, the coulomb force and biot savart are not valid for moving point charges. You need to use the lienard-wichert potentials. Jul 24, 2022 at 21:24
• Momentum conservation can be expressed in a similar continuity equation as poyntings theorem en.m.wikipedia.org/wiki/Maxwell_stress_tensor however it must be noted this relation does not hold for point charges [99% sure] since its derivation assumes a vanishes small contribution to the force from a single $\rho$ dv element. Replace point charges with spherical balls and you're golden to use this relation. Jul 24, 2022 at 21:29
• This momentum conservation may or may not reduce to the form you've described, if $\nabla \cdot \sigma$ is small at low velocities for the fields described. However a modification of the lienard wichert potentials must be made to account for the fact we aren't using point charges Jul 24, 2022 at 21:32

1. No. EM field momentum in the vicinity of the particles changes in the opposite direction to momentum of the particles. So the approximate relation would be $$\frac{d\mathbf p_{EM}}{dt} = -\mathbf F_{12} - \mathbf F_{21}$$. Approximate because forces are not exactly given by the formulae you wrote. Indeed there are terms depending also on particle's accelerations, and all quantities are in the past, not present, due to retardation (assuming purely retarded fields without additional homogeneous solution are the correct description of EM field).