# Lorentz Force law for charge density

I was going through chapter 8 of Griffiths' "Introduction to Electrodynamics", where he introduces the Maxwell Stress Tensor, by deriving it from the Lorentz law for some charge density $$\rho$$,

$$\vec{F}=\int_\mathcal V(\rho\vec E+\vec J\times\vec B) \ d\tau$$

Then, he uses Maxwell's equations to rewrite this in terms of the fields only, by using:

$$\epsilon_0(\nabla\cdot\vec E)\vec E=\rho\vec E$$

and:

$$\vec J\times\vec B=(\frac{1}{\mu_0}\nabla\times\vec B-\epsilon_0\frac{\partial\vec E}{\partial t})\times\vec B$$

However, if the original fields are external (not produced by $$\rho$$), then the charge density should not be equal to the divergence of the field since it isn't it's source. If on the other hand the charge density is the source of these fields, then isn't this expression calculating the force the fields do on their own source? Thanks.

• "external" to what? Aren't you are integrating over all space?
– hft
Jun 10, 2022 at 18:51
• You already introduced self action/self force/self energy/etc when you admitted the field concept in the first place.
– hft
Jun 10, 2022 at 18:52

The fields in this expression is the TOTAL E field, not just "external" fields that aren't produced by the charge your currently looking at calculating the force for.

You may say that "if we include the FULL E field, you are finding the force experienced by that charge, due to its own field aswell" which is technically true.

Doing so for point charges for example, you only focus on the field produced by the OTHER charge, and not the field produced by its own charge.

However since each $$\rho dv$$ element is basically zero, if we keep this contribution to the field, the answer doesn't change.

I believe this analysis only works if $$\rho$$ is finite.

Using this expression for point charges where $$\rho = Q \delta^3(r)$$ would yield incorrect results since the contribution of this single element is not negligible and must be ignored. Meaning poyntings theorem etc is not valid for point charges.

Maxwell’s equations make no distinction between external and internal fields, not between self forces and external forces. The $$\rho$$ is just whatever charge density is at the given location and the $$F$$ is the force density of whatever happens to be there due to all the fields at that location.

Griffiths' intention is to show how the Maxwell stress tensor and the equation of local conservation of momentum is derived in EM theory. The standard way is to start with net EM force on the material system, both due to external EM field, and due to EM field generated by the system. So his $$\mathbf E,\mathbf B$$ are net fields, not merely external or partial fields.

One could re-do the procedure while starting with force due to external fields $$\mathbf E_{ext},\mathbf B_{ext}$$ only. But then the contribution of internal forces to total EM force (which can be non-zero in EM theory unlike in classical mechanics) would not be accounted for. Also, as you point out, $$\rho$$ and $$\mathbf j$$ in the system could not be expressed using the external fields $$\mathbf E_{ext},\mathbf B_{ext}$$. So one can't formulate general law of local momentum conservation using only external fields, or only incomplete set of components of total EM field. One needs to use all EM field components, because they can all potentially cause forces. In the simplest case when all distributions are regular enough (no point charges, no line charges), one can do this with net fields alone and there is no need to distinguish between its components, and this is the scenario where the method Griffiths is explaining is applicable.

If there are point charges, this method fails at their positions. However, the method can be adapted to point charges, if net force and EM momentum is expressed in a consistent way using the partial fields due to individual particles $$\mathbf E_a,\mathbf B_a$$; however the expressions for EM momentum and energy are then different from the standard Maxwell and Poynting expressions.