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Jefimenko's Equations are:

$$ \begin{align} &\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{\rho(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t}\right)(\mathbf{r}-\mathbf{r}') - \frac{1}{|\mathbf{r}-\mathbf{r}'| c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \mathrm{d}^3 \mathbf{r} \\ &\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \times (\mathbf{r}-\mathbf{r}') \,\mathrm{d}^3 \mathbf{r} \\ & \mbox{where the retarded time is: }t_r = t - \frac{|\mathbf{r}-\mathbf{r}'|}{c} \end{align} $$

I noticed the following: In the E-field, the second and third terms fall at $1/r$. Meaning, they are radiative terms. That is, electromagnetic radiation. Analogously, In the B-field, the second term falls at $1/r$, thus radiative. However, in the B-Field, there is no radiative term depending on $\partial\rho / \partial t$.

With that in mind, consider the following situation: $$ \frac{\partial\mathbf J}{\partial t} = 0 \quad\quad\mbox{and}\quad\quad \frac{\partial\rho}{\partial t} \neq 0 $$

Then, the magnetic field will not be time varying, and won't be radiative. In this situation, the only radiative fields would be electric. Meaning, an electromagnetic radiation without the magnetic part! How is this possible? Is there a mistake somewhere? What prevents it from happening?


An approach was to allow time-varying $\rho$ with charge conservation by means of constant not-null-everywhere $\mathbf J(\mathbf r)$. By continuity equation, one can have a time varying $\rho$ one wants, by simply choosing $\mathbf J$ cleverly. I haven't made further progress with this. $$ \nabla\cdot\mathbf J + \frac{\partial\rho}{\partial t} = 0 \quad\implies\quad \frac{\partial\rho}{\partial t} = -\nabla\cdot\mathbf J $$

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  • $\begingroup$ I realized I made a mistake in my original answer and edited it. The correct answer is an unambiguous "no": the $1/r$ parts of the expression for ${\bf E}$ cancel out in the situation that you're considering. $\endgroup$ – tparker Jul 18 '17 at 5:44
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Arguably, it might be, but the sources you've provided cannot be sustained for long: since the current doesn't change in time, you have $$ \frac{\partial^2\rho}{\partial t^2} = -\nabla \cdot\frac{\partial \mathbf J}{\partial t} = 0, $$ i.e. if the charge density is increasing with time, then it is doing so linearly with time and without any way to stop, so you will necessarily end up with regions of arbitrarily high charges over sufficiently long times, and that is going to take an unbounded amount of energy.

If you're OK with that, then yeah, you can go on and find out whatever weird properties the emitted field has, but it's not really something that's broadly considered to be a physically allowed situation.

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  • $\begingroup$ Ohhh!!! Well Thought! So I have $\rho(\mathbf r) = a(\mathbf r) + t b(\mathbf r)$. At principle, one could engineer a machine to do this. Granted, one couldn't maintain this forever, but, one can maintain it for arbitrarily long (right?) (especially if $b$ is small). So, if it is not "considered to be a physically allowed situation", what could prevent one from doing this? $\endgroup$ – Physicist137 Jul 13 '17 at 21:35
  • $\begingroup$ The point is partly that for a field to be called radiative you normally require that it can be sustained for arbitrarily long periods of time, and this is not the case here. Given a specific $b(\mathbf r)$ (where you can essentially ignore the background $a(\mathbf r)$), you have a set time before your sources can't hold it and you need to change the current, which bumps you back to the standard case. If you really don't care about the $t\to\infty$ limit, then yeah, you can run wild with whatever you get, but stuff that requires infinite energy is not considered to be too interesting. $\endgroup$ – Emilio Pisanty Jul 13 '17 at 21:43
  • $\begingroup$ Actually, even if you could build up regions of arbitrary strong charge forever, you still wouldn't get $1/r$ fields if $\partial {\bf J}/\partial t = {\bf 0}$; see my updated answer. $\endgroup$ – tparker Jul 18 '17 at 5:46
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To add to Emilio Pisanty's answer: for the situation you are considering, Jefimenko's equations simplify to

$$ \begin{align} \mathbf{E}(\mathbf{r}, t) &= \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{\rho(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r} \\ \mathbf{B}(\mathbf{r}, t) &= \frac{\mu_0}{4 \pi} \int \frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} \times (\mathbf{r}-\mathbf{r}') \,\mathrm{d}^3 \mathbf{r} \\ \end{align} $$

As Emilio points out, if ${\bf J}$ does not depend on time then the continuity equation requires that $\rho$ depends linearly on time. Writing it as $\rho({\bf r'}, t) = a({\bf r'}) + b({\bf r'})\, t$ gives $$ \begin{align*} \mathbf{E}(\mathbf{r}, t) &= \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{a({\bf r'}) + b({\bf r'})\, t_r}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{b({\bf r'})}{|\mathbf{r}-\mathbf{r}'|^2 c}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r} \\ &= \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{a({\bf r'}) + b({\bf r'})\, \left(t - \frac{|{\bf r} - {\bf r'}|}{c} \right)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{b({\bf r'})}{|\mathbf{r}-\mathbf{r}'|^2 c}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r} \\ &= \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{a({\bf r'}) + b({\bf r'})\, t}{|\mathbf{r}-\mathbf{r}'|^3}\right)(\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r} \\ &= \frac{1}{4 \pi \epsilon_0} \int \left[ \frac{\rho({\bf r', t})}{|\mathbf{r}-\mathbf{r}'|^3} (\mathbf{r}-\mathbf{r}') \right] \mathrm{d}^3 \mathbf{r}. \end{align*} $$ All retardation effects cancel out, and the electromagnetic fields are given by the usual Coulomb and Biot-Savart laws applied to the instantaneous rather than retarded sources! In particular, they fall of like $1/r^2$, and are not radiative.

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  • $\begingroup$ Huh, that's weird. And the same happens for $\mathbf B$? (would upvote again if I could.) $\endgroup$ – Emilio Pisanty Jul 18 '17 at 9:30
  • $\begingroup$ @EmilioPisanty Yes, the $\bf{B}$ fields are only sourced by the currents ${\bf J}$, which don't depend on time, so you get the same answer whether or not you retard the Biot-Savart law. $\endgroup$ – tparker Jul 18 '17 at 16:26
  • $\begingroup$ Ah yes, of course. Weird little bundle, this one. $\endgroup$ – Emilio Pisanty Jul 18 '17 at 16:46
  • $\begingroup$ @EmilioPisanty There aren't any magnetic monopoles, so technically it's a trivial bundle ;) $\endgroup$ – tparker Jul 18 '17 at 20:39
  • $\begingroup$ @EmilioPisanty This situation is discussed at the beginning of Section III of aapt.scitation.org/doi/abs/10.1119/1.16589. $\endgroup$ – tparker Dec 29 '17 at 5:52

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