Confusion in Poynting's theorem

Question

The total force on a charge is equal to $\mathbf {F}=q\mathbf {(E+ v×B)}$ where everything have their usual meanings . We can say that:

$$dW= \mathbf {F\cdot dl} = {\mathbf{F}\cdot \frac{d \mathbf{l} }{dt}} dt= \mathbf {(F\cdot v)}dt = q(\mathbf {E\cdot v})dt = \mathbf {(E\cdot j)} dt ...... $$

where $\mathbf {j}$ is the current density vector created by the charge and it is non zero only at the point of point charge. Now from Maxwell's laws:

$$\mathbf {j}= \frac{1}{\mu}\mathbf {\nabla×B} -\epsilon \frac{d\mathbf {\tilde E}}{dt}$$ where $\epsilon$ and $\mu$ are electric and magnetic permittivities. We have to note here that $\mathbf {\tilde E}$ is the total electric field, not only the electric field which acts on the charge . Thus $\mathbf {E}$ and $\mathbf {\tilde E}$ are not same. But in Poynting's theorem both are treated same and thus dot produced with $\mathbf {j}$. So is the theorem wrong?

Answer 1

Whenever you have a conflict between some established principle and the concept of a classical point charge, the issue is the classical point charge. They lead to all sorts of oddities like infinite energy, weird self-forces, and other such things.

Poynting's theorem follows directly from Maxwell's equations. So it can be used any time that Maxwell's equations apply. To resolve the issue you mention, simply, use continuous charge and current distributions, $\rho$ and $\vec J$. These are the variables that appear in Maxwell's equations, so applying them makes direct sense. In terms of those variables the Lorentz force density is $\vec f = \rho \vec E + \vec J \times \vec B$ and it can be applied directly.

Answer 2

The total force on a charge is equal to $\mathbf {F}=q\mathbf {(E+ v×B)}$ where everything have their usual meanings . We can say that:

$$dW= \mathbf {F\cdot dl} = {\mathbf{F}\cdot \frac{d \mathbf{l} }{dt}} dt= \mathbf {(F\cdot v)}dt = q(\mathbf {E\cdot v})dt = \mathbf {(E\cdot j)} dt ...... $$

where $\mathbf {j}$ is the current density vector created by the charge and it is non zero only at the point of point charge. [...]

You're assuming above that $\mathbf E$ above is not the total field, but only that part of the total field that acts on the charged particle. This is correct in point particle theories a la Tetrode, Fokker, Frenkel, Feynman-Wheeler, but it is not the usual meaning of the symbol $\mathbf E$; the usual meaning is that this is total field, and that all charged particles are spatially extended bodies, with finite (or regular enough) charge density everywhere, so one total field $\mathbf E$ can be used in all equations, and then density expressions in the Poynting theorem are all finite and can be interpreted in terms of energy.

Now from Maxwell's laws:

$$\mathbf {j}= \frac{1}{\mu}\mathbf {\nabla×B} -\epsilon \frac{d\mathbf {\tilde E}}{dt}$$ where $\epsilon$ and $\mu$ are electric and magnetic permittivities. We have to note here that $\mathbf {\tilde E}$ is the total electric field, not only the electric field which acts on the charge .

Actually, this is not necessary, because Maxwell's equations (with appropriate source terms) hold for each and every component of EM field defined by its source, and also for total field and total (sum of ) sources. So it is quite possible to write down the above equation only for that part of total electric field which acts on the point particle.

But let's assume $\tilde{\mathbf E}$ in the above equation is the total field.

Thus $\mathbf {E}$ and $\mathbf {\tilde E}$ are not same. But in Poynting's theorem both are treated same and thus dot produced with $\mathbf {j}$. So is the theorem wrong?

In the Poynting theorem (for charges in vacuum), total fields $\tilde{\mathbf E},\tilde{\mathbf B}$ are used only. Thus the term $\tilde{\mathbf E}\cdot\mathbf j$ in the Poynting theorem involves total fields $\tilde{\mathbf E}$ and $\mathbf j$.

This is mathematically invalid in case of point particles, because the field is too singular at the particle, so the product $\tilde{\mathbf E}\cdot \mathbf j$ is undefined (meaning, its integral is undefined). So for point particles, the Poynting theorem has singular points, and cannot be interpreted in terms of work/energy. One has to formulate different (but mathematically similar) local conservation law, without singular points. This is possible using adjunct fields (one distinct field for each particle) and then that theorem, with defined expressions that have finite integrals, can be interpreted in terms of work and energy.

Answer 3

There are not two different $E$s. The $E$ in the first equation is the total electric field at the given point, including the $E$ produced by the charge itself; an accelerating charge actually can have work done on it by the field that it itself produces, which results in a transfer of energy away from the charge and into the surrounding field. The $E$ in the second equation is also the total electric field at the given point.