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How do we know that $$u = \frac{1}{2}\left(\epsilon_0E^2 + \frac{1}{\mu_0}B^2\right)$$ gives the energy density of electromagnetic fields? Is it a postulate of classical electrodynamics? Griffith seems to think that Poynting's theorem, $$\frac{\mathrm dW}{\mathrm dt} = -\frac{\mathrm d}{\mathrm dt}\int_\mathcal{V} u \,\mathrm d\tau - \oint_\mathcal{S} \mathbf{S} \cdot \mathrm d\mathbf{a},$$ where $\mathbf{S} = \dfrac{1}{\mu_0}\mathbf{E} \times \mathbf{B}$ is the pointing vector, implies that it is so, but I don't see how.

Related to this, is the Poynting vector without question the energy flux density? Given Poynting's theorem and the fact that $u$ is the energy density, Griffiths seems to argue, the natural interpretation of $\mathbf{S}$ is that it's the energy flux density, but can this be proven definitively?

Edit: Avoid using the Lagrangian or Hamiltonian formulation if possible.

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    $\begingroup$ What would you like energy to mean? Certainly that equation states that $\vec{S}$ is the flux of $u$. If by 'energy' you mean something like 'quantity conserved due to time translation invariance' then I think you are (mostly) stuck with $u$ and $\vec{S}$ as they are. $\endgroup$ – jacob1729 Feb 27 at 10:29
  • $\begingroup$ @jacob1729 Replying to your statement that $\mathbf{S}$ is the flux of $u$, I think you're incorrect for $\oint \mathbf{S} \cdot d\mathbf{a}$ may represent the power passing through a closed surface, but that doesn't mean that $\int \mathbf{S} \cdot d\mathbf{a}$ is the power passing through an open surface. See The footnote on page 358 of Griffiths's Electrodynamics (Fourth Edition) if you have a copy. $\endgroup$ – PiKindOfGuy Feb 27 at 10:36
  • $\begingroup$ See the answer I gave in physics.stackexchange.com/questions/253870/… $\endgroup$ – Iván Mauricio Burbano Feb 28 at 18:22
  • $\begingroup$ I strongly advice you to read this related chapter of the Feynman lectures: feynmanlectures.caltech.edu/II_27.html $\endgroup$ – Run like hell Apr 10 at 8:31
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As already stated by @jacob1729, a good definition of energy is something that is conserved in time if the system (Lagrangian density$\mathcal{L}$) has no explicit time dependence ($\partial_t \mathcal{L}=0$ if $t$ is time).

The way to arrive at this quantity is to start with the Lagrangian for the electromagnetic field $\mathcal{L}=\mathcal{L}\left(t, r^i, \nabla_j \phi, \nabla_j A^i,\dot{A}^i\right)$, where $r^i$ is the position coordinates, $\phi$ is scalar potential and $A$ is vector potential (I will stick to Lorentz gauge).

we can then construct

$\mathcal{H}=\partial_{\dot{A}^i}\mathcal{L}\,\dot{A}^i -\mathcal{L}$

One can then proove that

$- \frac{d\mathcal{H}}{dt}= \partial_t \mathcal{L} + \nabla_j\left(\dot{A}^i \partial_{\nabla_j A^i}\mathcal{L} + \dot{\phi}\partial_{\nabla_j\phi}\mathcal{L}\right)$

Now integrate over some closed surface and apply Gauss' theorem:

$\int_V d^3 r \left(- \frac{d\mathcal{H}}{dt}\right) = -\frac{d}{dt}\int_V d^3 r \mathcal{H} = \int_V d^3 r \,\partial_t \mathcal{L} + \oint_{\partial V} \mathbf{\hat{n}}^j.\left(\dot{A}^i \partial_{\nabla_j A^i}\mathcal{L} + \dot{\phi}\partial_{\nabla_j\phi}\mathcal{L}\right)$

If your field vanishes outside of the volume, $V$, the second term on the right is not important. Then, provided the Lagrangian density has no explicit time dependence you see that:

$\frac{d}{dt}\int_V d^3 r \mathcal{H} = 0$

i.e. the quantity, $\int_V d^3 r \mathcal{H}$ is conserved. Finally, you should be able to prove that $\mathcal{H}$ is the energy density as you know it.


If you don't like Lagrangian formulations go from simple Maxwell's equations

Let there be some volume $V$. Using Maxwell's equations in vacuum, with no charges and currents, evaluate

$\frac{d}{dt}\int_V d^3 r \left( \frac{\epsilon_0 E^2}{2} + \frac{1 B^2}{2\mu_0}\right)=\int_V d^3 r \left( \epsilon_0\dot{\mathbf{E}}.\mathbf{E} + \mu_0^{-1} \dot{\mathbf{B}}.\mathbf{B}\right) = \mu_0^{-1}\int_V d^3 r \left( \mathbf{\nabla}\times\mathbf{B}.\mathbf{E} - \mathbf{\nabla}\times\mathbf{E}.\mathbf{B}\right)=-\mu_0^{-1}\int_V d^3 r \mathbf{\nabla}.\left(\mathbf{E}\times\mathbf{B}\right)$

Now apply Gauss theorem

$\frac{d}{dt}\int_V d^3 r \left( \frac{\epsilon_0 E^2}{2} + \frac{1 B^2}{2\mu_0}\right)=-\mu_0^{-1}\oint_{\partial V} d^2 r \:\mathbf{\hat{n}}.\left(\mathbf{E}\times\mathbf{B}\right)$

And you have a quantity that is conserved in time provided the stuff on the right-hand side vanishes. What Lagrangian formulation allows you to do is to justify why this quantity is energy

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I pointed out in comments that this really can't be settled unless you have a definition of energy you're willing to work with. Let's assume we don't know about Lagrangian mechanics and try to come up with heuristics on what the EM field energy should be:

1) Energy must be conserved. In particular we want the total energy of all charges and fields to be constant. Local conservation of field energy can be represented as:

$$ \frac{d}{dt} \int_V u \ dV = -\int_{\partial V}\vec{S}\cdot d \vec{S}$$

The left hand side represents the change of energy (whose density is $u$) in a volume $V$ whilst the right hand side is a flux through the bounding surface $\partial V$. The quantity $\vec{S}$ is to be interpreted as the flux of $u$ and at the moment has no interpretation other than that. However there is also the work done on charges to consider. Magnetic fields do no work, so the rate at which work is done on charges is simply:

$$ \frac{dW}{dt} = \vec{F}\cdot \vec{v} = q\vec{E}\cdot\vec{v} = \vec{E}\cdot(q\vec{v}) $$

the field should lose this amount of energy per unit time, so conservation of energy must be represented by an equation (passing to the differential form using Gauss' theorem and recognising $q\vec{v}$ as a vector current):

$$ \frac{\partial u}{\partial t}+\vec{\nabla}\cdot\vec{S} = -\vec{E}\cdot\vec{J}$$

and our job is to find a candidate $u$ and $\vec{S}$ that satisfy such a relation. Any pair have some right to be called an energy and an energy flux.

2) We can also use the fact that we know what the work is to construct the energy in some simple situations. These are the cases of a uniform electric field in a fixed volume and a uniform magnetic field. Specifically, these can be realised by ideal parallel plate capacitors and solenoidal inductors and finding the average energy density is a standard (high school or first year undergraduate) problem. The results are:

$u_{C}=\frac{1}{2} \epsilon_0 |E|^2 , u_{L}= \frac{1}{2\mu_0} |B|^2 $

(the subscripts C,L refer to the device in question as a capacitor or an inductor). So whatever $u$ we pick satisfying the conservation equation, we want it to reduce in the limits of constant fields to these cases.

With both restriction (1) and (2) it seems unlikely that we can come up with candidates for the field energy other than the usual ones. I believe restriction (2) is actually important though, as there are other conserved quantities (satisfying some form of $\partial_t u' + \text{div}\left(\vec{S'}\right)=0$) that we could add to get a new $u, \vec{S}$ that would still have the same loss term. However it would no longer give the required result for a constant field.

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  • $\begingroup$ Actually, the strategy of your point 2, for static and quasi-static ields, can be implemented in much more generality than the case of uniform electric and magnetic fields. See Landau&Lifshitz Electrodynamics of continuous media. The only delicate point in L&L derivation is related to a question I asked without answer in this site ( physics.stackexchange.com/questions/444539/… ) $\endgroup$ – GiorgioP Apr 11 at 7:14
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Here's a simple (and simple-minded) approach. We postulate that energy is stored in electric and magnetic fields, and that if both types of field are present we can sum the the energies in each, as if the other weren't there.

We now consider cases in which the fields are uniform: the parallel plate capacitor (with plate separation, t << square root of plate area, A), and the toroidally-wound inductor (with square root coil cross-section area, A' << toroid axial circumference, $\ell$ ). It is easy to derive the equations:, $$U_{cap}=\frac12 C V^2,\ \ C=\frac{\epsilon_0 A}{t},\ E=\frac Vt$$ and $$U_{ind}=\frac12 L I^2,\ \ L=\frac{\mu_0 n^2 A'}{\ell},\ B=\frac{\mu_0 n I}{\ell}$$ We find the energies stored per unit volume to be $$u_E=\frac{U_{cap}}{At}=\frac12 \epsilon_0 E^2\ \ \ \text{and}\ \ \ \ u_B=\frac{U_{ind}}{A' \ell}=\frac{1}{2 \mu_0} B^2$$

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    $\begingroup$ This approach would get the wrong energy if the correct answer happened to include spatial derivatives of $E,B$ or terms like $E\cdot B$. I do however think its important to keep these limiting cases in mind. $\endgroup$ – jacob1729 Feb 28 at 17:44
  • $\begingroup$ Agreed. I've rewritten the postulates (at the beginning) to acknowledge your point about terms like $\vec E.\vec B.$ I suspect that the delivery of formula for $u_E$ and $u_B$ independent of actual volume between capacitor plates or inside toroid suggests that spatial derivatives of $E$ and $B$ are not involved, but I don't have a rigorous argument for this assertion. I'd claim that my simple-minded approach produces plausible formulae for $u_E$ and $u_B,$ but does not establish them rigorously. $\endgroup$ – Philip Wood Feb 28 at 18:33
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I just noticed I'm using different units then you're using. I'll leave it to you as an exercise to correct the units by comparing Maxwell's equations.

The speed of light is $c=\frac {1}{(\epsilon_{0}\mu_{0})^{1/2}}$ regardless of the units.

In Gaussian units - the units which I'm using, $\epsilon_{0}=1$ and $\mu_{0}=1$.

The total rate of doing work assuming a continuous distribution of charges and currents is

$$ \int_\tau \vec J\cdot \vec E{d\tau}$$

Using the Ampere-Maxwell law

$$\nabla \times \vec B -\frac {1}{c}\frac {\partial \vec E}{\partial t}=\frac {4\pi}{c} \vec J$$

and solving for the current density

$$\vec J=\frac {c}{4\pi}(\nabla \times \vec B -\frac {1}{c}\frac {\partial \vec E}{\partial t})$$

yields

$$ \int_\tau \vec J\cdot \vec E d\tau=\frac{1}{4\pi}\int_\tau \lbrack c\vec E\cdot(\nabla \times \vec B -\frac {1}{c}\frac {\partial \vec E}{\partial t})\rbrack d\tau \tag{1} $$

Using the vector identity $$\nabla\cdot(\vec E \times \vec B)=\vec B\cdot(\nabla \times \vec E)-\vec E\cdot(\nabla \times \vec B)$$

rearranging

$$\vec E\cdot(\nabla \times \vec B)=-\nabla\cdot(\vec E \times \vec B)+\vec B\cdot(\nabla \times \vec E) \tag{2}$$

inserting Faraday's law equation $(3)$ $$ \nabla \times \vec E=-\frac {1}{c}\frac{\partial \vec B}{\partial t} \tag{3} $$ into equation $(2)$ yields

$$\vec E\cdot(\nabla \times \vec B)=-\nabla\cdot(\vec E \times \vec B)-\vec B\cdot\frac{\partial \vec B}{\partial t} \tag{4} $$

Substituting equation $(4)$ into $(1)$ $$ \frac {dW}{dt}=\frac{-1}{4\pi}\int_\tau \lbrack c\vec \nabla \cdot (\vec E \times \vec B) +\vec E\cdot \frac {\partial \vec E}{\partial t}+\vec B \cdot\frac{\partial B}{\partial t}\rbrack d\tau $$

where $$ \frac {dW}{dt}=\int_\tau \vec J\cdot \vec Ed\tau $$

Define $$u=\frac {1}{8\pi}(\vec E\cdot\vec E + \vec B \cdot \vec B) $$ $$\vec S= \frac{c}{4\pi}(\vec E \times \vec B) $$

then $$ \frac {dW}{dt}=-\int_\tau \lbrack \frac {\partial u}{\partial t}+\nabla \cdot \vec S\rbrack d\tau $$ Noting, that

$$\int_{\tau}\nabla\cdot\vec S d\tau= \oint_A \vec S\cdot d\vec A $$

Hence $$ \frac {dW}{dt}=-\frac{d}{dt}\int_\tau u d\tau -\oint_A \vec S\cdot d\vec A $$

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Energy and momentum are the source of gravitational fields. Therefore in principle one can decide your question by experiment. In practice the gravitational effect are so small that at present such an experiment is highly unfeasible. Therefore we have to rely on theory. If you require energy and momentum to be conserved and to be expressed in terms of E and B, then the present expressions are your only option. Nevertheless, read my peer reviewed paper for an alternative: https://arxiv.org/abs/physics/0106078.

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