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I Varying Charge Densities with Constant Current Density

  1. Sep 11, 2016 #1
    1) If I vary charge densities, but keep current density constant, do I get any sort of electromagnetic wave?

    2) If the answer to question 1 is no, then if I vary charge densities, but keep current density constant, could I conceivably have a two isolated "open circuit" current elements of finite length, one consisting of charge flowing from point A to point B, and another consisting of charge flowing from point Y to point Z?

    3) If the answer to question 2 is yes, then wouldn't there be possible case where the Lorentz force on one of these "open circuit" current elements is not in the exact opposite direction to the Lorentz force on the other "open circuit" current element?

    4) If the answer to question 3 is yes, then what bears the difference between these two Lorentz forces so that conservation of momentum is obeyed, if not for the emission of electromagnetic waves (as per answering "no" to question 1)?

    Sincerely,

    Kevin M.
     
  2. jcsd
  3. Sep 11, 2016 #2

    vanhees71

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    I don't know, what specifically you mean in 1). Note that electric charge must be conserved, because otherwise Maxwell's equations become unsolvable. Your changes of the charge-current density vector must fulfill the continuity equation
    $$\partial_{\mu} j^{\mu} \equiv \partial_t \rho + \vec{\nabla} \cdot \vec{j}=0.$$
    If this is fulfilled there's a solution of Maxwell's equations (e.g., by the retarded solutions).
     
  4. Sep 11, 2016 #3
    Yes, the current densities are constant, but there is also simultaneously a divergence of them which gives rise to a changing charge density.

    1) is basically asking if this changing charge density will give rise to an electromagnetic wave that could conceivably give off momentum.
     
    Last edited: Sep 11, 2016
  5. Sep 12, 2016 #4

    vanhees71

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    It should. As long as the continuity equation (and conditions for the charge-current density at infinity) are fulfilled, the retarded solutions (Jefimenko equations) exist.
     
  6. Sep 12, 2016 #5
    Upon looking at the Jefimenko equations, there is term in the electric field for changing charge densities, however the magnetic field does not appear to be affected by changes in charge densities. The result is that the magnetic field would not be the same as it is in an electromagnetic wave, where it varies as 1/r. The magnetic field still drops as 1/r^2 (for a finite current element) or faster as 1/r^3 (for a magnetic dipole element or opposed current elements). Furthermore, if we are taking about changes of charge densities while still conserving charge, the result would be two additional electric fields, per Jefimenko's equations, which are effectively that of an electric dipole, and so the sum of those fields should drop with 1/r^2 (because the fields due to their changes in charge density would, individually, drop with 1/r). So I don't see how the momentum would radiate. It seems like the near field would absorb all the momentum. Can that be right?
     
    Last edited: Sep 12, 2016
  7. Sep 13, 2016 #6

    vanhees71

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    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" ;-)).
     
  8. Sep 14, 2016 #7
    From Wikipedia's entry on the Jefimenko's equation, we have:

    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):

    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:

    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?

    Sincerely,

    Kevin M.
     
    Last edited: Sep 14, 2016
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