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Imagine a particle with charge $q$ at rest at the origin.

It is surrounded by a concentric spherical insulating shell, also at rest, with charge $Q$ and radius $R$.

At time $t=0$ I apply a constant horizontal acceleration $\mathbf{a}$ to the particle.

The electromagnetic field spreads out radially in all directions from the charge at the speed of light (Appendix 1).

The momentum density in the field is directed radially and also spreads out from the charge at the speed of light (Appendix 2).

As the momentum density at opposite field points cancel the total momentum in the field is always zero (Appendix 2).

At time $t=R/c$ the field reaches the charged shell and imparts a horizontal force on it (Appendix 3).

What balances this force if the rate of change of horizontal momentum in the field is always zero?

I think it has to be balanced by an opposite force on the particle that develops due to an advanced electromagnetic interaction from the spherical shell at time $t$ backwards in time to the particle at time $t_r$.

Appendix 1: EM field of an accelerated charge with velocity zero

Specializing the Lienard-Wiechert fields for an accelerating charge with zero velocity we find that the electric field is given by:

$$\mathbf{E}(\mathbf{r},t)=\frac{q}{4\pi\epsilon_0}\left(\frac{\mathbf{\hat r}}{r^2}+\frac{\mathbf{\hat r}\times(\mathbf{\hat r}\times \mathbf{a})}{c^2r}\right)_{t_r}$$

and the magnetic field is given by:

$$\mathbf{B}(\mathbf{r},t)=\frac{\mathbf{\hat r}(t_r)}{c}\times\mathbf{E}(\mathbf{r},t) $$

where $\mathbf{\hat r}$ is the unit vector in the direction of the field point and the retarded time $t_r$ at the charge $q$ is given by:

$$t_r=t-\frac{r(t_r)}{c}.$$

Appendix 2: Zero total momentum in EM field from an accelerated charge

The momentum density $\mathbf{g}(\mathbf{r},t)$ in the electromagnetic field is given by:

$$\mathbf{g}(\mathbf{r},t)=\epsilon_0\mathbf{E}(\mathbf{r},t)\times\mathbf{B}(\mathbf{r},t).$$

The momentum density is directed radially so that:

$$\mathbf{g}(\mathbf{r},t)=\frac{\epsilon_0E^2}{c}\mathbf{\hat r}$$

where the magnitude of the electric field $E$ is given by

$$E = -\frac{q a \sin \theta}{4\pi\epsilon_0c^2r}$$

and $\theta$ is the angle between acceleration $\mathbf{a}$ and radial direction $\mathbf{\hat r}$.

As $\sin(\pi-\theta)=\sin(\theta)$ then we have:

$$\mathbf{g}(-\mathbf{r},t)=-\mathbf{g}(\mathbf{r},t)$$

so that the momentum density at opposite field points cancel.

Thus the total momentum in the fields is always zero.

Appendix 3: Total force on charged spherical shell from accelerated charge

The electric field at the spherical shell consists of a static radial part and an acceleration part. The total force from the static radial part will cancel out in opposite pairs so that we only have to worry about the acceleration part.

The total horizontal force on the spherical shell with charge $Q$ and radius $R$ is given by:

$$F = \int_{sphere} E\sin \theta\ \sigma\ dA$$

where

$$E = -\frac{q a \sin \theta}{4\pi\epsilon_0c^2R}$$

$$\sigma=\frac{Q}{4\pi R^2}$$

$$dA = 2 \pi R^2 \sin \theta\ d\theta$$

Performing this integration over the sphere using:

$$\int_0^\pi \sin^3 \theta\ d\theta=\frac{4}{3}$$

We find that the total horizontal force is given by:

$$F = -\frac{2}{3}\frac{qQa}{4\pi\epsilon_0c^2R}.$$

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  • $\begingroup$ Your conclusion of zero net field momentum is based on the unphysical premise that only an inifitesimal moment of source charge acceleration is adequate for evaluation. With acceleration comes velocity and displacement. Try doing a finite time integral that evaluates the Poynting momentum flow through some fixed spherical surface, over the relevant interval of present time. $\endgroup$ – user50679 Jun 24 '14 at 10:20
  • $\begingroup$ Force can only be balanced by a time-rate change in momentum, due to conservation of momentum. $\endgroup$ – jjack Aug 16 '15 at 11:21
  • $\begingroup$ what force is accelerating the small charge? How do you apply it? it will be another field? rocket ? The acceleration will generate radiation which will take away momentum.. $\endgroup$ – anna v Sep 11 '17 at 7:45
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The EM theory allows for energy and momentum conservation, quite irrespective of which particular solutions, retarded, advanced or other, we choose to work with.

In the picture where only retarded fields are present, which is the most natural one, the answer to your question is as follows.

In your reasoning you use the particular EM field $\mathbf E_q, \mathbf B_q$ due to the charge $q$ inside the sphere in the role of the total EM field (in the expression $\mathbf E\times \mathbf B/\mu_0$ for momentum density). That is not correct, because the total field $\mathbf E,\mathbf B$ differs substantially from the charge field $\mathbf E_q$ where the sphere is; there are other charges there. When these charges are acted upon the radiation field of the charge $q$, the fields of the sphere charges and the field of the charge $q$ combine to form the total field, which can carry momentum.

If the sphere charges are heavy so they do not move much, their fields are almost static and intense only in the vicinity of the sphere (the Coulomb electric field gets very strong near the point-like charge). Then the EM field acquires appreciable momentum only in the vicinity of the charges of the sphere. The matter of the sphere get momentum in one direction, the field inside and near the sphere gets momentum in the opposite direction.

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  • $\begingroup$ But what if the charges in the sphere are fixed to the sphere so that their acceleration in response to the $\mathbf{E_q}$ field is small. In that case the EM fields and the field momentum that the sphere charges generate will be negligible. $\endgroup$ – John Eastmond Oct 19 '13 at 23:29

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