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General setup

Consider the following thought experiment: A capacitor $C$ and an inductor $L$ are mounted with fixed joints on a plate in outer space at say $1$ meter apart. They are connected in the $LC$ harmonic oscillator configuration. When powered the energy is exchanged between one to the other one. The sum of the energies on both remains constant over time.

One can eventually even drop the existence of the plate and just assume that the capacitor and the inductor are at a fixed $1$ meter distance apart.

Question

Using the mass energy equivalence formula $E = m\cdot c^2$ follows that the mass of $C$ varies with the energy of the capacitor as well as the mass of L varies with the energy of the inductor. Is therefore the center of gravity of the assembly moving in time (oscillating) without any external force being acted on it? Does this violate the law of the conservation of linear momentum?

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  • $\begingroup$ What a nice paradox! Put simply, mass shuffles backwards and forwards at a finite speed from C to L and vice versa, implying momentum first in one direction then in the other and so on. I'm wondering about whether we should consider the momentum of the photons emitted by accelerating charges in the wires, but there's probably something much simpler than that going on! $\endgroup$ Nov 27 '21 at 9:56
  • $\begingroup$ @PhilipWood I am not quite sure if this is or not sarcastic, but assuming is not, I do not think that photons/anything are/is being emitted because in such a case the movement would need to eventually stop (or "something" should be also received), right?! This is not the case in the presented experiment, is it not ?! $\endgroup$
    – C Marius
    Nov 27 '21 at 10:01
  • $\begingroup$ No sarcasm intended – desperation, perhaps. There certainly will be photons emitted, because we have accelerating charges, though the effect will be very small (because the accelerations are so small). But then the mass transfer is also very small. But as I've said, I expect that your paradox has a simpler resolution! $\endgroup$ Nov 27 '21 at 10:33
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    $\begingroup$ @m4r35n357 For the little it's worth I make the maximum momentum in the system due to the mass transfer $$p_{max} =\frac{l\ Q_0^2}{2 c^2 \sqrt{LC^3}}$$ in which $l$ is the distance between L and C. $\endgroup$ Nov 27 '21 at 11:11
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    $\begingroup$ @m4r35n357 A neater and more instructive version of my earlier formula is simply $$p_{\text{max}}=\frac{l \omega}{c^2} U_\text{max}$$ in which $l$ is the distance between L and C and $U_\text{max}$ is the maximum energy stored in the inductor or capacitor. $\endgroup$ Nov 27 '21 at 12:39
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Is therefore the center of gravity of the assembly moving in time (oscillating) without any external force being acted on it? Does this violate the law of the conservation of linear momentum?

Not even remotely. Circuit theory is an approximation of Maxwell’s equations. It is intrinsically non-relativistic, so using circuit theory in the context of relativity is inherently problematic. On the one hand you are ignoring relativity while on the other you are not. You are essentially guaranteed to get nonsensical results.

To do this properly you would need to use Maxwell’s equations, which are fully relativistic. If you analyze the circuit using Maxwell’s equations then you will find momentum is conserved. This has been proven in general, not merely for your specific case, by Poynting’s theorem.

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  • $\begingroup$ So, the logical conclusion is that the physical circuit is moving left and right such that the COM is stationary. This movement I imagine consumes the energy in the LC oscillator which will eventually stop oscillating. Is this what happens ? The LC tank circuit can never be undamped !? $\endgroup$
    – C Marius
    Nov 27 '21 at 22:07
  • $\begingroup$ Does relativity theory allow undamped LC oscillators? $\endgroup$
    – C Marius
    Nov 27 '21 at 22:17
  • $\begingroup$ I don’t think that special relativity would forbid undamped oscillations, but Maxwell’s equations do. The circuit will radiate $\endgroup$
    – Dale
    Nov 28 '21 at 1:48
  • $\begingroup$ I am having difficulties in understanding for this specific case how the Poynting vector shows that the C component should experience a linear moment when discharging/ charging. Looking at the $S = \frac{1}{\mu_0} E \times H$ this will be symmetrically orthogonal on the wires and C. So, yes, it might radiate, but why does not radiate equally in all directions? $\endgroup$
    – C Marius
    Nov 28 '21 at 13:39
  • $\begingroup$ The easiest way to see this is to write Poynting’s theorem in fully relativistic form: $\partial_\nu T^{\mu \nu}+f^\mu =0$, where $T$ is the EM field stress energy tensor, which includes energy density, Poynting vector, and EM stress, and $f$ is the Lorentz power and force density on matter. So any energy and momentum that leaves the field goes to matter and vice versa. $\endgroup$
    – Dale
    Nov 28 '21 at 14:25
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It is a theorem in relativity that the centre of "mass" defined by $$ {\mathcal E}X^\mu_{\rm cofm} = \int T^{00}(x) x^\mu \sqrt{g}d^3x $$ does not move for isolated system with total momentum zero. Here $T^{00}$ is the energy density and ${\mathcal E}$ the total energy $$ {\mathcal E}= \int T^{00}(x) \sqrt{g}d^3x $$

Despite this theorem there are many appararent paradoxes that come from combining non-relativistic mechanics with the fully relativistic Maxwell equations. They are resolved when you realize the the field momentum density
$$ {\bf p}=\frac 1{c^2} {\bf E}\times {\bf H} $$ having $1/c^2$ factor means that one has to keep $1/c^2$ relativistic corrections to the mechanics. The general heading for these effects is hidden momentum.

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  • $\begingroup$ By looking at the Poynting vector $S=\frac{1}{\mu_0} E \times H$ (collinear with $p$) we get that the energy always flow into the wires symmetrically. Moreover. for instance, for the case when the capacitor is discharging, it loses energy to inductor (so it should move to left, assuming L is to the right, to keep the COM in place), but computing the Poynting vector the energy flows radially/symmetrically out of and orthogonally on the capacitor. Therefore, I still do not get practically why the capacitor is moving! Those moments should be symmetrical! $\endgroup$
    – C Marius
    Nov 28 '21 at 13:32
  • $\begingroup$ Energy is moving about, some of it in the energy of the electromagnetic field and so of it in the $mc^2$'s of the coil and capacitor. The center of energy stays put, so the tank cicuit will jiggle (impercepably) from side to side. $\endgroup$
    – mike stone
    Nov 28 '21 at 16:58
  • $\begingroup$ Or it will radiate energy in one side then to the other ... the center of energy includes the energy lost to the radiation right? $\endgroup$
    – C Marius
    Nov 28 '21 at 17:33
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When capacitor shoots energy to the right, then capacitor recoils to the left.

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  • $\begingroup$ How can I observe that? By looking at the Poynting vector $S = \frac{1}{\mu+0}E \times B$ we get that the energy always flow into the wires. Moreover, the energy flow and the current flow can never be in the same direction. The energy flow is always perpendicular on the electric field. For instance, for the case when the capacitor is discharging, it loses energy, but computing the Poynting vector the energy flows radially/symmetrically out of and orthogonally on the capacitor. Am I wrong? $\endgroup$
    – C Marius
    Nov 28 '21 at 13:26
  • $\begingroup$ @CMarius OK. If capacitor emits energy to all directions, then it's the wires that direct that energy to where the wires lead. And the wires experience the recoil. Hmm yes, if we have a plate capacitor and we start discharging it from the middle of the plates, then the the plates do not recoil. $\endgroup$
    – stuffu
    Nov 28 '21 at 15:20
  • $\begingroup$ As I said, the energy flows into the wires and, from what I can imagine, there is no net momentum applied to them because of this. $\endgroup$
    – C Marius
    Nov 28 '21 at 16:02
  • $\begingroup$ @C Marius Capacitor shoots energy to two wires that are left to the capacitor. Capacitor recoils to the right. Capacitor shoots energy to two wires, that are left to the capacitor and right to the capacitor. Capacitor does not recoil, energy has no net momentum. If we want to connect those wires to a inductor, we must bend the wires, the energy pushes the wire where the bend is, centrifugal force. $\endgroup$
    – stuffu
    Nov 28 '21 at 16:29
  • $\begingroup$ @CMarius If flowing energy manages to move the wire then the wire gains energy and energy loses energy. If the motion is transmissed to another wire then that wire may do work on the energy flowing inside of it. This is some kind of answer to your energy problem that I spotted in the comments. $\endgroup$
    – stuffu
    Nov 28 '21 at 20:53

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