I Momentum conservation for EM-Field/matter interaction

[euphoricrhino]

Hello,
I'm reading Feynman Lectures Vol II, and saw this "paradox" in section 26-2 (Figure 26-6), where two orthogonally moving charges can be shown to have unequal action and reactions. Later in Chapter 27, the explanation was given briefly citing field momentum.

I tried to prove this rigorously, but couldn't do so, below is my plan:
1. calculate the E and B field for each particle using formula for uniform-velocity charges and obtain E1,B1, E2, B2 as function of (x, y, z, t)
2. For any point in space, using superposition of fields, we can get E=E1+E2, and B=B1+B2
3. Obtain the Poynting vector ExB, thus the momentum density vector g as a function of (x, y, z, t)
4. Integrate g(x, y, z, t) over the whole space to get the total field momentum p at time t.
5. take the time derivative of p, and verify that this equals the difference between the action and reaction of the two charges.

The integration step is very complicated. I even simplified so one particle is taken to be at rest and the other particle moves straight to the first one. In this simplified case, the integration can be infinity since that the E field at distance-0 is infinity.

Is my plan at least reasonable (i.e., it's just a hard integration)? Another observation is that although E or B fields are additive from two charges, the momentum vector from two charges' fields is definitely not additive, am I understanding it correctly?

Thanks for any pointers.

 

[ergospherical]

In full generality it's helpful to consider a charge density $\rho(\mathbf{r},t)$. In the case of two particles you can of course specialise this to $\rho(\mathbf{r},t) = q_1\delta(\mathbf{r} - \mathbf{r}_1(t)) + q_2\delta(\mathbf{r} - \mathbf{r}_2(t))$. The Lorentz force per unit volume $f_i = \rho E_i + \rho \epsilon_{ijk} v_j B_k$ is related to the divergence $\partial_j \sigma_{ij}$ of the Maxwell stress tensor by the equation $f_i = \partial_j \sigma_{ij} - \frac{1}{c^2} \partial_t S_i$ (derivable from Maxwell's equations & the Lorentz force law). If you integrate this over some domain $V$ and invoke Gauss' theorem, you end up with\begin{align*}
-\frac{d}{dt} \int_V \frac{1}{c^2} S_i dV &= F_i - \int_{\partial V} \sigma_{ij} n_j dS \\
-\dot{p}_i^{\mathrm{field}} &= \dot{p}_i^{\mathrm{particles}} - \int_{\partial V} \sigma_{ij} n_j dS
\end{align*}and one identifies the rate of decrease of field momentum $p_i^{\mathrm{field}} = \frac{1}{c^2} S_i$ with the rate of increase of the momentum $p_i^{\mathrm{particles}}$ of the charge inside $V$ and the momentum flux $- \int_{\partial V} \sigma_{ij} n_j dS$ out of through the boundary.

 

[euphoricrhino]

Thanks so much, this is very clear. In fact, FLP has dropped the hint when talking about momentum density without giving the derivation. I was able to derive the exact form based on your hint.

I guess my plan would have worked but the integration is too complex to evaluate, and the fact I was planning to integrate over the whole space should account for the momentum flux term in your post above (which should evaluate to zero at infinity).

There is also this question of "how real" is the quantity $\mathbf{S}/c^2$ as momentum density. They are just algebraic terms being interpreted as field momentum density to make equations work. It's plausible to rearrange the terms to get another equivalent equation which can be interpreted differently. If we treat $\mathbf{S}/c^2$ as momentum density, this would mean that the momenta of the two fields cannot be additive point-wise, which seems counter intuitive.

 

[dRic2]

There is also this question of "how real" is the quantity S/c2 as momentum density. They are just algebraic terms being interpreted as field momentum density to make equations work.

Well, if you want to calculate the mechanical pressure on a surface due to the EM radiation, then S (its time average...) it's a pretty tangible thing... (https://en.wikipedia.org/wiki/Radiation_pressure)