In an answer to this Physics SE question, @ChiralAnomaly demonstrated that, indeed, there is a minimum field energy density observable at any point in an EM field. With a bit more calculation, it's easy to show that if we define the usual field invariants as $k_1$ and $k_2$ where $$k_1 = (|E_p|^2 -|B_p|^2)$$ and $$k_2 = |E_p| |B_p|$$ and $E_p$ and $B_p$ are respectively the values of $\vec {E}$ and $\vec{B}$ in any inertial frame in which $\vec {E}$ and $\vec{B}$ are parallel, then the minimum field energy $H_0$ is: $$H_0^2 = (k_1)^2 + 4(k_2)^2.$$ Because $H_0$ is composed of field invariants, it is also a field invariant. Just for fun, I'll call $\frac{H_0}{c^2}$ the "rest mass density" of the field.

Here's my question: Is the integral of $\frac{H_0}{c^2}$ over all space conserved?

$H_0$ is invariant with respect to Lorentz transformations, but Lorentz invariance of a quantity that has the dimensions of energy (eg, a Lagrangian) does not imply that the quantity is conserved. Such a quantity might be conserved, but it might not. I think $\frac{M_0}{c^2}$ is an example of such a quantity that is conserved ($M_0$ is rest mass density of, e.g., a fluid).

What I would like to know is whether or not there is something resembling a continuity equation for $H_0$: if $H_0$ decreases in one place, does it increase in another place, more or less the way $\frac{M_0}{c^2}$ does?

Edit #2: I have tried taking the 4-gradient of H, but run into terms like $(E \cdot \nabla) E$ and $(E \cdot \nabla )B$, and don't know what to do with them.

I'm hoping to find an equation that shows "where changes in $H_0$ go to or come from", more or less the way that this equation shows where changes in the field energy density "go to or come from":

$$ \vec E\cdot\vec j=\nabla{(\epsilon_o c^2\vec B\times\vec E)}-\frac {\partial}{\partial t}(\frac{\epsilon_o c^2}{2}\,\vec B\cdot\vec B+ \frac{\epsilon_o}{2}\,\vec E\cdot\vec E). $$

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    $\begingroup$ I think you intended to have a minus sign in $k_1$. In that case, the scalar you construct in $H_0$ is not the energy density because it involves a different transformation to the parallel frame at each point. $\endgroup$ – octonion Mar 24 '19 at 15:21
  • $\begingroup$ Thanks, I fixed the sign typo. $\endgroup$ – S. McGrew Mar 24 '19 at 15:49
  • $\begingroup$ @Octonion, In the referenced answer, it was shown that the field energy density is the same in all frames where E is parallel to B, and that it is a minimum over all frames. $\endgroup$ – S. McGrew Mar 24 '19 at 15:53
  • $\begingroup$ Again, you constructed a valid scalar ('invariant' in your language), and you can interpret it as the minimum energy density over all frames. But that particular frame will be different for each point, so you are not integrating over an energy density, and it is not conserved in time. $\endgroup$ – octonion Mar 24 '19 at 17:33
  • $\begingroup$ The "comoving" or "rest" frame of a fluid would be different at each point; but the rest mass is conserved. $\endgroup$ – S. McGrew Mar 24 '19 at 19:13

Is the integral of $\frac{H_0}{c^2}$ over all space conserved?

No. As a counterexample, consider an oscillating LC circuit. (If you like, you can have the inductor and the capacitor overlap in space so that their fields are parallel.) At a time when the capacitor has zero field, we have $k_1<0$ and $k_2=0$ everywhere, so the integral is negative. At a time when the inductor has zero field, $k_1>0$ and $k_2=0$ everywhere, so the integral is positive.

In general, we can have systems that include EM fields and act like they have a rest mass, but in those systems the fields' contribution to the rest mass isn't given by the integral of $H_0$. For example, a box full of photons (a photon gas) does act like it has a rest mass equal to the energy of the photons, but $H_0$ will average to zero for this configuration. In relativistic terms, perfect fluids do have a comoving frame, but in the comoving frame, their stress-energy tensor doesn't usually look like the stress-energy of matter at rest (a pure energy density with no pressure).

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  • $\begingroup$ It's clear enough that the integral of $H_0$ won't be conserved if there are sources that are being driven, as in the LC circuit. However, I don't think the situation is quite that simple. The changing electric field in a charging capacitor will produce a magnetic field very similar to that of a current-carrying wire, which crossed with the E field results in a Poynting vector pointing radially in or out from the axis of the capacitor. So there is energy flow in and out of the space between the capacitor plates. I'd like to know the relationship of that energy flow to changes in $H_0$. $\endgroup$ – S. McGrew Mar 25 '19 at 5:05
  • $\begingroup$ The energy flow described by the Poynting vector is the flow of the electromagnetic energy density $U\propto E^2+B^2$, not the flow of your quantity $H_0$. $\endgroup$ – user4552 Mar 25 '19 at 16:38
  • $\begingroup$ I'm aware that the Poynting vector is the flow of $E^2 + B^2$; but $H_0$ is a component of $E^2 + B^2$ . The induced B field that, crossed with E, produces the Poynting vector varies with position, so the value of $H_0$ varies with respect to both time and space while E is changing. This suggests that there is a flow associated with $H_0$. If the "equation of motion" for$ H_0$ looks like a continuity equation that would be fun; but I'll be satisfied just to see the equation, whatever it is. $\endgroup$ – S. McGrew Mar 25 '19 at 17:02

Here's another counterexample. $$\vec{E}=\hat{x}E_0 \sin \omega t \cos k z$$ $$\vec{B}=-\hat{y}E_0 \cos \omega t \sin k z$$ It is just a superposition of two transverse EM waves moving in opposite directions on the z-axis. I'm using $c=1$ units.

Here the scalar $k_2=\vec{E}\cdot\vec{B}=0$. So your scalar field that you are hoping has a conserved spatial integral is $$H_0=|\vec{E}|^2-|\vec{B}|^2=E_0^2 \left(\sin^2\omega t\cos^2 kz-\cos^2\omega t\sin^2 kz\right)$$ or actually you may prefer the absolute value of this, but it won't matter.

Integrate over a wavelength in the z direction, and any fixed length in the x and y directions. Whatever is happening in this volume $V$ that you are integrating over is happening identically in any other adjacent volume involving a wavelength so you can't claim the $H_0$ is flowing somewhere else.

After integration you get $$\int H_0 dV=\frac{1}{2}E_0^2 V\left(\sin^2\omega t-\cos^2\omega t\right)$$ This oscillates sinusoidally in time so it is definitely not conserved.

If you instead considered the energy density, which is not a scalar, there would be a plus sign, so this integral would be a constant which reflects the fact energy is conserved.

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  • $\begingroup$ The problem with this counterexample is that it does not meet the criterion of the question: $H_0$ is defined by finding an inertial frame in which $E$ and $B$ are parallel; but if $E \cdot B = 0$ in any frame, they can never be parallel. $\endgroup$ – S. McGrew Mar 28 '19 at 2:28
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    $\begingroup$ @S.McGrew, Good, I was waiting for that response. Simply superimpose the solution on a background electric field in the y-direction, and you will get $E\dot B\neq 0$ for almost all times (except when $B=0$). You still see non-conservation. $\endgroup$ – octonion Mar 28 '19 at 2:31
  • $\begingroup$ The expression for $H_0^2$ gets a bit messy ($H_0$ is even worse). However, I see your point that its integral remains time dependent and is therefore not conserved. Still, it would be nice to see a compact formula for the 4-gradient of $H_0$. $\endgroup$ – S. McGrew Mar 28 '19 at 4:00
  • $\begingroup$ I have edited the question to provide an example of an equation that has a structure similar to what I'm looking for. $\endgroup$ – S. McGrew Mar 30 '19 at 4:12

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