The formula for energy density of electromagnetic field in electrodynamics is $$\frac{1}{8\pi} (\vec E\cdot\vec D+\vec B\cdot\vec H).$$

This formula appears in all general physics courses I looked at. However Feynman writes in Section 27-4 of his well known course:

... we must say that we do not know for certain what is the actual location in space of the electromagnetic field energy.

This is due a non-unique way to get the above formula.

Feynman writes furthermore:

Anyway, everyone always accepts the simple expressions we have found for the location of electromagnetic energy and its flow. And although sometimes the results obtained from using them seem strange, nobody has ever found anything wrong with them—that is, no disagreement with experiment. So we will follow the rest of the world—besides, we believe that it is probably perfectly right.

I am wondering if nobody verified the above formula for energy density, what are the reasons to believe that “it is probably perfectly right”? Does this formula have any applications correctness of which can be verified (thus confirming indirectly the formula itself)? If there are no such applications, why this formula should be important at all?


4 Answers 4


It's certainly not the case that nobody has ever checked them. The issue is that by manipulating the Maxwell equations, one arrives at Poynting's theorem:

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

where $$u = \frac{\mathbf E \cdot \mathbf D + \mathbf B \cdot \mathbf H}{8\pi}\qquad \mathbf S = \frac{1}{4\pi} \mathbf E \times \mathbf H$$ are interpreted as the electromagnetic energy density and the electromagnetic energy flux, respectively. In words, Poynting's theorem says that in any fixed volume of space, the change of $u$ can be accounted for by (i) the flux of $\mathbf S$ through the boundary of the volume, and (ii) the physical work being done on moving charges.

Feynman's point is that if we define a new Poynting vector$^\dagger$, say, $$\mathbf S' = \mathbf S + \frac{1}{8\pi}\frac{\partial}{\partial t}\mathbf A \times \mathbf H$$ (where $\mathbf A$ is the vector potential) and a new electromagnetic energy density

$$u' = \frac{\mathbf E \cdot \mathbf D + \mathbf A \cdot \mathbf J}{8\pi}$$

then these quantities would also obey Poynting's theorem, insofar as $$\mathbf E \cdot \mathbf J = -\frac{\partial u'}{\partial t} -\nabla \cdot \mathbf S' $$

Defining $u'$ and $\mathbf S'$ amounts to adding something to $\mathbf S$ and subtracting the same quantity from $u$. How do we know that $u$ and $\mathbf S$ are the proper choices, rather than $u'$ and $\mathbf S'$? In the absence of additional inputs, there is no immediate answer.

Feynman is quick to note, of course, that there are perfectly reasonable conditions you may demand (i.e. gauge invariance, dependence on the fields but not their derivatives, correspondence with general relativity, etc) which would fix $u$ and $\mathbf S$ to be the unique choice. However, these demands are something we need to impose by hand, and do not emerge naturally from Maxwell's equations.

$^\dagger$ See H.M. Macdonald, Electric Waves(Cambridge U. Press, 1902), pp. 32, 72

  • 1
    $\begingroup$ I think this is precisely what Feynman claims: nobody checked this formula. If this is not the case, do you have a reference? Do you have a reference to your last comments that the energy density is uniquely characterized by symmetries and other conditions? Thank you. $\endgroup$
    – MKO
    Jun 29, 2020 at 7:28
  • $\begingroup$ @MKO What do you mean by checked? It's certainly fine mathematically, and follows from Maxwell's equations. The problem is that if you say $x+y=z$ and I say $(x+1)+(y-1)=z$, then we're saying the same thing in a slightly different way, and in the absence of additional inputs, there's no unambiguous way to say which of us is "right." $\endgroup$
    – J. Murray
    Jun 29, 2020 at 7:41
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    $\begingroup$ MKO, the energy density of the EM field is one component of the electromagnetic stress-energy tensor. The stress energy tensor has zero four-divergence, reflecting energy and momentum conservation. However, adding to the stress energy tensor another tensor field which has zero four-divergence also yields another viable candidate. Thus the stress energy tensor is only uniquely defined up to the addition of some arbitrary divergenceless function. Typically this freedom is used to impose conditions upon the SE tensor, such as rendering it symmetric in its indecies. $\endgroup$
    – AfterShave
    Jun 29, 2020 at 10:37
  • 1
    $\begingroup$ "Checked" of course means experimentally checked. Nobody has been able to do this. Your first sentence is incorrect. $\endgroup$
    – my2cts
    Jun 29, 2020 at 14:06
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    $\begingroup$ @my2cts At time of my writing, the OP read "if nobody has checked the formula..." which I interpreted to mean that the subject has not been subjected to scrutiny. As it currently reads, I agree that verified should be interpreted as experimental confirmation. $\endgroup$
    – J. Murray
    Jun 29, 2020 at 14:32

In classical electrodynamics there is no way to verify experimentally that the vacuum energy density is $$\varepsilon =\varepsilon _e +\varepsilon _m=\frac{1}{2}\mathbf{E}\cdot\mathbf{D}+\frac{1}{2}\mathbf{H}\cdot\mathbf{B} \label{1}\tag{1}$$, or more generally $$\delta \varepsilon = \mathbf{E}\cdot \textrm{d}\mathbf{D}+ \mathbf{H}\cdot \textrm{d}\mathbf{B} \label{2}\tag{2}$$ Instead we use this energy density in an integral to calculate the work to be done to change the electric and magnetic flux densities from their $\mathbf{D}=0$ and $\mathbf{B}=0$ neutral state to a final one while the temperature $T$ is kept constant, this will give us the free energy density:

$$\mathfrak{f}=\int_{0}^{D}\mathbf{E}\vert _{T=const}\cdot \textrm{d}{\mathbf{D}} + \int_{0}^{B}\mathbf{H}\vert _{T=const}\cdot \textrm{d}\mathbf{B} \label{3}\tag{3}$$

Let me quote here [1] (slightly rewritten for the equations)

In the past, the work and free energy have often been considered distributed in various ways. For instance, the currents, charges and field could be regarded as just the means through which all the work $$\delta w = \int \bf{E}\cdot \textrm{d}\bf{D}\textrm{dV} +\int \bf{H}\cdot \textrm{d}\bf{B}\textrm{dV} $$ is done on the body, so that it possesses all the Helmholtz free energy $\int_{\textrm{all space}} \mathfrak{f}dV$ .

Equally self-consistent is the Poynting-Heaviside viewpoint that the work is stored with an energy density $\eqref{2}$, and the free energy with a density \eqref{3} throughout the whole system including the free space. If, for instance, the field be purely electrostatic and the body be heated, changing its state of polarization without moving any charges, the net work $\delta w= \int \delta \varepsilon dV$ in the first interpretation would be zero, whereas in the Poynting-Heaviside view each element of volume would have to be considered as doing work (positive or negative) on the rest of the system. Alternatively, the energy has sometimes been divided, writing for instance $$\bf{H}\cdot \textrm{d}\bf{B}=\mu_0 \bf{H}\cdot \textrm{d}\bf{H} +\mu_0\bf{H}\cdot\textrm{d}\bf{M}$$ or $$\bf{H}\cdot \textrm{d}\bf{B}=\frac{1}{\mu_0} \bf{B}\cdot \textrm{d}\bf{B} - \bf{M}\cdot \textrm{d}\bf{B}$$

Then the first term is associated with the field and the second with the body. All such ways of distributing the energy are physically meaningless (cf. Guggenheim, p. 99 and Stratton, pp. 110, 133). Incidentally, each leads to a different value for 'the' free energy density in the body, and then to the correct entropy density!

Entropy density is calculated from the free energy, see Heine or Guggenheim:

$$\mathfrak{F}= \int_{\textrm{all space}}\mathfrak{f}dV =\int_{\textrm{all space}}dV\int_{0}^{D} \mathbf{E}\vert _{T=const}\cdot \textrm{d}\mathbf{D} +\int_{\textrm{all space}}dV\int_{0}^{B}\mathbf{H}\vert _{T=const}\cdot \textrm{d}\mathbf{B} \label{4}\tag{4}$$

and from $-SdT=d\mathfrak{F}-\delta w$ one can prove, see Guggenheim that:

$$S= \int_{V} dV \int_{0}^{D} \frac{1}{\epsilon_0} \left(\frac{\partial\mathbf{P}}{\partial T}\right)_{T,D}\cdot \textrm{d}\mathbf{D} +\int_{V}dV\int_{0}^{B}\left(\frac{\partial\mathbf{M}}{\partial T}\right)_{T,B}\cdot \textrm{d}\mathbf{B} \label{5}\tag{5}$$

The most interesting thing in this result is that while neither energy density nor free energy density is measurable as such, the total work at constant temperature equals total free energy change is measurable in the bias circuits and from that one can derive a unique entropy density that belongs directly to the ponderable matter, here enclosed in the volume $V$ over which the integration in $\eqref{5}$ is performed. Total energy change can be measured by the the total work done in the circuits that create the field to polarize the material, its local distribution is not measurable but the resulting entropy density is meaningful.

[1] Heine : THE THERMODYNAMICS OF BODIES IN STATIC ELECTROMAGNETIC FIELDS, Proceedings of the Cambridge Philosophical Society, 1956

[2] GUGGENHEIM, E. A. On magnetic and electrostatic energy and the thermodynamics of magnetization. Proc. Roy. Soc. A, 155 (1936), pp. 49 and 70.

[3] STRATTON, J. A. Electromagnetic theory (McGraw-Hill, 1941).


The only way to verify the EM energy formula would be to measure its gravitational field. Feynman states this in Part II, 27.4, 2nd paragraph. Such a measurement seems very much impossible even today. Feynman has confidence in the formula because it is consistent with the Maxwell equations and the Lorentz force, and describes a conserved quantity.

Feynman was, however, suspicious because of the many paradoxes involved with the EM conservation laws. This is clear from Part II ch. 27.4 and 27.5 from his choice of words: 'our “crazy” theory', "This mystic circulating flow of energy, which at first seemed so ridiculous, is absolutely necessary". And how can the energy of a capacitor not enter it via the wires?

In addition, there is the century old Abraham-Minkowski controversy over the EM energy density in media. Moreover, the EM formula and the Poynting vector do not follow from straightforward application of the Noether theorem to the generally accepted gauge invariant Lagrangian.

There is an alternative approach. I published a paper describing it, which can be found here. It is not generally accepted because it discards gauge invariance, which I consider not to be a principle of nature but rather a property of the Lorentz force.


I don't think it is the formula that is in question. As Feynman wrote, "What is the actual location in space of the electromagnetic field energy?" The question arises because there is another formula, easily derived from the one you wrote, that the EM energy is given by $W=\frac{1}{2}\int\rho\phi d^3r$. This would suggest that the energy resides in charged matter. A lesson from this is that 'energy density' cannot simply refer to where the energy is. Arguments can and have been made for either 'location' of the energy. This indicates that 'location' of EM energy is not a well defined concept. The arguments given tend to suggest matter for classical EM, but the fields for QED and for classical EM derived from QED, but I hold that EM energy density only has meaning when put into an integral.

  • $\begingroup$ your formula for $W$ has only electric energy, the magnetic energy would be $W_m= \frac{1}{2} \int \bf{J} \cdot \bf{A} d^3r$ $\endgroup$
    – hyportnex
    Jun 29, 2020 at 12:49
  • $\begingroup$ There can be no doubt that Feynman is talking about the energy distribution over space given by the formula. $\endgroup$
    – my2cts
    Jun 29, 2020 at 14:03

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