vanhees71 said:
Can you give a specific example for your case of charge-current densities? It's hard to discuss this in "many words" instead of "a few formulae" ;-)).
From Wikipedia's entry on the Jefimenko's equation, we have:
Jefimenko's equations - Wikipedia said:
$$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{\rho(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t}\right)(\mathbf{r}-\mathbf{r}') - \frac{1}{|\mathbf{r}-\mathbf{r}'| c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \mathrm{d}^3 \mathbf{r}'$$
$$\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \times (\mathbf{r}-\mathbf{r}') \mathrm{d}^3 \mathbf{r}'$$
With
constant current densities, several terms drop. The result is:
$$\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{\rho(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r}'$$
$$\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} \right] \times (\mathbf{r}-\mathbf{r}') \mathrm{d}^3 \mathbf{r}'$$
B(
r,t) gives us a time-delayed version of the Biot-Savart Law. In the event that the current density distribution is constant, then this should yield the same field as when you don't take the time-delay into account. However, if you look at the formula for
E(r,t), then you can see that there is an extra term associated with the changing charge density. This term drops off inversely with the distance. This is made evident by analyzing the term itself:
$$\frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r}'$$
We can see that the term contains the following ratio whose magnitude varies inversely with the "retarded-distance"
r-
r':
$$\frac{(\mathbf{r}-\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|^2}$$
So it would be correct to say that this term drops inversely to |
r-
r'|, or roughly speaking, inversely to the distance. So if we had charge piling up in one location, we would see a radial electric field emanating from that location.
In the case of negative charge piling up, the result is an additional "electric field term" pointing towards the negative charge which drops inversely with the distance. If that location is taken to be at rest with respect to a conveniently chosen inertial observer, then the "retarded-distance"
r-
r' is no different than just a distance between two points in that inertial observer's frame of reference. However, taking the divergence of that term would yield a negatively-valued charge distribution surrounding the charge per Gauss' law (from Maxwell's equations):
Maxwell's equations - Wikipedia said:
$$\nabla \cdot \mathbf{E} = \frac {\rho} {\varepsilon_0}$$
This would be an obvious violation of the conservation of charge if it wasn't for the fact that
we must also add to the picture another changing charge density, which fixes the problem. This is because as the other changing charge density must change in the opposite direction in order to conserve charge, the radial electric field associated with its changing charge density would point outward in contrast to inwards. However, now when we include the other changing charge density, we can note that the spatially-separated sources effectively generate a dipole-like field, and so the sum of these field-terms of type
$$\frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r}'$$
for two source field locations, where the changes of charge density are equal and opposite, drops not inversely to the distance, but
inverse to the square of the distance. Now, couple this with a static magnetic field that drops inversely with the square of the distance, as per the Biot-Savart law. The "time-delayed" version of this is:
$$\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} \right] \times (\mathbf{r}-\mathbf{r}') \mathrm{d}^3 \mathbf{r}'$$
The result is a Poynting vector which does not radiate into infinity:
Poynting vector - Wikipedia said:
$$\mathbf{S} = \frac{1}{\mu_0} \mathbf{E} \times \mathbf{B}$$
Where both
E and
B drop inversely with the square of the distance.
At distances sufficiently close the the wire, the magnetic field
B of a wire element of finite length drops near-inversely with the distance. The most extreme example of this is the magnetic field around infinitely long wire current element, where the distance from the wire is arbitrarily smaller than the length of wire. Also, we should note that at positions very close to the increasing charge density and far from the decreasing charge density (or vice-versa), the sum of extra electric field terms will drop near-inversely to the distance. So one could say that very close to the ends of these "open" finite wire elements (which carry constant currents) that the field "radiates" near the electric poles of each wire.
However, by analyzing the terms for just one straight wire element, we can see that the radial
E field crossed with the cylindrical
B field, results in a Poynting vector along the path of the conductor, in the space surrounding the conductor. On the other end, we see that such a Poynting vector will point in the opposite direction because the radial direction of the extra
E term reverses, but not the
B term. So this near-field "radiation" produced by each wire element taken separately does not radiate in a direction which can account for the thrust. Of course, we will need to know how the
E field of one wire element interacts with the
B field of the wire element, but it is clear that these Poynting vector components also do not radiate, because both the
E and the
B fields of each will vary with the inverse square of the distance, causing the Poynting vector to drop inversely to the fourth power of the distance (ergo, no radiation).
What then should we say of the overall forces in the wire elements? These wire elements do not appear to radiate electromagnetic waves to the far-field. So energy is not said to leave the system. Per the center of energy theorem, we should then expect the forces on the wire elements to cancel. Do we know what the necessary forces are?
The field from the "time-delayed" Biot-Savart law (applying for constant current densities), as per above, is:
$$\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} \right] \times (\mathbf{r}-\mathbf{r}') \mathrm{d}^3 \mathbf{r}'$$
This may only generate a force at right angles to the wire current elements. So what happens if I have:
1) A wire current element from point (0,0) to (1,0)
2) A wire current element from point (0,0) to (0,1)
3) Increasing N electrons at (1,0) and (0,1)
4) Decreasing N electrons at (0,0)
The magnetic fields of 1 and 2 would exert Lorentz forces on each other's current elements. The result would be a total force on the wire elements aligned with a vector from (0,0) to (0+1,1+0) or from (0,0) to (1,1).
What we need next is an accounting for the forces due to electric fields. In the case that the embedded positive charges in the wire do not redistribute, we see that the changing charge densities are entirely due to time-varying concentrations of electrons at the ends of each wire element. Thus, if we imagine the wire elements to be neutral, then the ends of each wire element must bear opposite charges. So we have the following charges at each end at time t:
5) A charge of -q*N(t) at point (1,0)
6) A charge of -q*N(t) at point (0,1)
7) A charge of 2q*N(t) at point (0,0)
It is safe to say that any electric forces between the charge at (1,0) and (0,1) will be equal and opposite because they bear the same charge at time (t), and they subject each other to equal and opposite electric fields at time (t) as a result of the simple bilateral symmetry along the line y=x. The same concept should, by extension, apply to forces between the points (0,0) and (1,0) as well as between the points (0,0) and (0,1), but in this case there is a charge ratio between the ends of -2 (instead of 1) and the ratio between the electric fields received between each is 2 (instead of -1), resulting once again in equal and opposite electric forces along y=0 and x=0, respectively.
What am I missing here?
Kevin M.
