# Intuition behind Poynting's theorem: relative sign between the electric and magnetic contributions to the energy

The standard derivation of Poynting's theorem for EEs uses sinusoidal complex time dependence $$e^{\mathfrak{j} \omega t}$$, that is $$\mathbf{E}(t)=\Re [ \hat{\mathbf{E}}e^{\mathfrak{j} \omega t}]$$ and $$\mathbf{H}(t)=\Re [ \hat{\mathbf{H}}e^{\mathfrak{j} \omega t}]$$ as follows. Starting with Maxwell's equations

\begin {align} \operatorname{curl} \hat{\mathbf{H}} & = \hat{\mathbf{J}} + \mathfrak{j}\omega \epsilon \hat{\mathbf{E}}, \tag {1}\\ \operatorname{curl} \hat{\mathbf{E}} & = -\mathfrak{j}\omega \mu \hat{\mathbf{H}}, \tag {2} \end {align} first we take the complex conjugate of $$(1)$$: $$\operatorname{curl} \hat{\mathbf{H}}^*=\hat{\mathbf{J}}^* - \mathfrak{j}\omega \epsilon \hat{\mathbf{E}}^*, \tag{1a}$$ then multiply $$(\mathrm{1a})$$ and $$(2)$$ with $$\hat{\mathbf{E}}$$ and with $$\hat{\mathbf{H}}^*$$, respectively,

\begin {align} \hat{\mathbf{E}} \cdot \operatorname{curl} \hat{\mathbf{H}}^* & = \hat{\mathbf{E}} \cdot \hat{\mathbf{J}}^* - \mathfrak{j}\omega \epsilon \hat{\mathbf{E}} \cdot \hat{\mathbf{E}}^*, \tag {3}\\ \hat{\mathbf{H}}^*\cdot\operatorname{curl} \hat{\mathbf{E}} & = -\mathfrak{j}\omega \mu \hat{\mathbf{H}}^*\cdot\hat{\mathbf{H}}. \tag {4} \end {align}

Now subtract $$(3)$$ from $$(4)$$ and using the identity $$\operatorname{div}(\mathbf{a}\times \mathbf{b}) = \mathbf{b}\cdot\operatorname{curl}\mathbf{a} - \mathbf{a}\cdot \operatorname{curl}\mathbf{b}$$ get \begin{align} \operatorname{div} (\hat{\mathbf{E}}\times \hat{\mathbf{H}}^*) &= -\hat{\mathbf{E}} \cdot \hat{\mathbf{J}}^* - \mathfrak{j}\omega \epsilon \hat{\mathbf{E}} \cdot \hat{\mathbf{E}}^* + \mathfrak{j}\omega \mu \hat{\mathbf{H}}^*\cdot\hat{\mathbf{H}} \tag{5}\\ &=-\hat{\mathbf{E}} \cdot \hat{\mathbf{J}}^* +\mathfrak{j}\omega \big(\mu |\hat{\mathbf{H}}|^2 - \epsilon |\hat{\mathbf{E}}|^2 \big).\end{align}

The terms in parentheses represent reactive stored energy. The question is about the difference in sign between the stored magnetic and electric energies, how to explain in plain English the reason for the sign difference?

This is especially puzzling if you compare it with Poynting's theorem in real variables; the derivation is almost the same as above and the result is, see 1

$$\operatorname{div}(\mathbf{E}\times \mathbf{H}) = -\mathbf{E} \cdot \mathbf{J} - \big( \epsilon \mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} + \mu \mathbf{H} \cdot \frac{\partial \mathbf{H}}{\partial t} \big) \tag{6}\\$$

Notice that in this real form $$(6)$$ the time rates of the magnetic and electric stored energies have the same sign.

It is conventional to consider the magnetic and electric energies as being analogues of kinetic ($$T$$) and potential energies ($$V$$), resp., and then the EM Lagrangian density is $$T-V$$. Is this negative sign that shows up both in the Lagrangian density and the complex form $$(6)$$ of the Poynting's theorem a coincidence or does it have a deeper meaning?

• Just a thought... Maybe link the energy in electric field to 'potential energy' and in magnetic field to 'kinetic energy'. Now kinetic energy depends on velocity squared, which in your case, I guess, would produce a factor of $\left(i\omega\right)^2$, which would give the extra minus.
– Cryo
Commented Jan 26, 2020 at 22:45
• @Cryo I have added a few lines Commented Jan 26, 2020 at 23:01
• I've never understood these kinds of EE derivations. You can't just multiply together complexified versions of fields and expect to get a meaningful answer -- passing to complex fields, doing manipulations, and then taking the real part is only a correct procedure when everything you're doing is linear. Commented Jan 26, 2020 at 23:23
• I'm sure that $\mathbf{E} \times \mathbf{H}^*$ has some meaning, but whatever it is, it isn't the Poynting vector that every non-EE uses, even after taking its real part. Commented Jan 26, 2020 at 23:24
• @knzhou $\Re [\mathbf{E}\times \mathbf{H}^*]$ has well defined meaning that is the time average propagating power density of the real Poynting vector. You probably would not dismiss the derivation of Ehrenfest's theorem although it uses the same trick that got us eq. (6) and all EE EM books use. Commented Jan 26, 2020 at 23:49

I'll set $$\epsilon=1$$, $$\mu=1$$, and $$\mathbf{J}=0$$, because these simplifications don't affect the question about signs. The difference between the vacuum equations (using $$\mathbf{B}$$) and the in-material equations (using $$\mathbf{H}$$) is also peripheral to the question about signs, because the equations have the same structure either way. This answer considers the vacuum equations because this simplifies the wording.

The observables are the real-valued electric field $$\mathbf{E}(t)$$ and the real-valued magnetic field $$\mathbf{B}(t)$$, and the energy density is $$(\mathbf{E}^2+\mathbf{B}^2)/2$$. The question is about how this relates to a relative minus sign in the case of complex-valued fields. The answer may be more satisfying if we remember why complex-valued versions of $$\mathbf{E}$$ and $$\mathbf{B}$$ are sometimes used. We can always write $$\begin{gather*} \newcommand{\bfB}{\mathbf{B}} \newcommand{\bfH}{\mathbf{B}} \newcommand{\bfE}{\mathbf{E}} \newcommand{\real}{\text{real}} \bfE(t) = \bfE^+(t)+\bfE^-(t) \\ \bfB(t) = \bfB^+(t)+\bfB^-(t) \tag{1} \end{gather*}$$ where $$X^\pm(t)$$ are the positive- and negative-frequency parts of $$X(t)$$, which can each be obtained from the real-valued field $$X(t)$$ using a Hilbert transform. (In quantum field theory, they correspond to the photon-annihilating and photon-creating parts of the field operator, respectively.) The original real-valued fields $$\bfE(t)$$ and $$\bfB(t)$$ are the observables, but working with their complex-valued positive- and negative-frequency parts can sometimes be convenient. One application is highlighted below, one that relates directly to the question.

## A general mathematical result

Suppose that $$\bfE(t)$$ and $$\bfB(t)$$ are complex-valued fields satisfying the equations $$\begin{gather*} \nabla\times\bfH = \dot\bfE \\ \nabla\times\bfE = -\dot\bfH \tag{2} \end{gather*}$$ where $$\dot X$$ denotes the time-derivative of $$X$$. When the fields are real-valued, these are Maxwell's equations. Take the dot-product of $$\bfE$$ with the complex-conjugate of the first equation, take the dot-product of $$\bfH^*$$ with the second equation, subtract the two resulting equations, and use the identity $$\bfE\cdot(\nabla\times\bfH^*) - \bfH^*\cdot(\nabla\times\bfE) = \nabla\cdot(\bfH^*\times\bfE) \tag{3}$$ to get the general result $$\nabla\cdot(\bfH^*\times\bfE) = \bfE\cdot\dot\bfE^* + \bfH^*\cdot\dot\bfH. \tag{4}$$ This mathematical result holds independently of any physical interpretation of the complex-valued fields.

## Physical meaning in the real-valued case

When the fields are real-valued, equations (2) are Maxwell's equations, and the result (4) can be written $$\nabla\cdot(\bfH\times\bfE) = \frac{\partial}{\partial t}\frac{\bfE^2 +\bfH^2}{2}. \tag{5}$$ Equation (5) is a local conservation law: the left-hand side involves the spatial derivatives of the momentum density (Poynting vector), and the right-hand side involves the time-derivative of the energy density.

In a material medium, we can account for some of the material's properties by replacing $$\mathbf{B}$$ with $$\mathbf{H}$$ in all of these equations. Then the momentum and energy densities in (5) include effects of the material as well as the electromagnetic field itself, but the math is the same. I'm considering the vacuum case to avoid distracting caveats about the meaning of "energy density."

## Another conservation law

The positive- and negative-frequency parts of a field may be expressed in terms of the original real-valued field, and this relationship — the Hilbert transform — is linear. (It's non-local in time, but that's not an obstacle here.) Because it is linear, the fact that the real-valued fields satisfy Maxwell's equations implies that the positive- and negative-frequency parts do, too. Use this together with $$(X^+)^* = X^-$$ to see that equation (4) implies $$\nabla\cdot(\bfH^+\times\bfE^-) = \bfE^-\cdot\dot\bfE^+ + \bfH^+\cdot\dot\bfH^-. \tag{6}$$ Add equation (6) to its own complex-conjugate to get $$\nabla\cdot(\bfH^+\times\bfE^- + \bfH^-\times\bfE^+) = \frac{\partial}{\partial t} \big(\bfE^-\cdot\bfE^+ + \bfH^-\cdot\bfH^+\big). \tag{7}$$ This is another local conservation law, but it is not equivalent to (5). In particular, the right-hand side is not the time-derivative of the energy density. The energy density in (5) includes terms involving the squares of the positive- and negative-frequency parts. Those terms are missing in (7).

However, equation (7) can still be used as an approximation to (5) under special conditions. If the wave has a relatively narrow bandwidth, then the terms that are missing from (7) will be relatively rapidly oscillating functions of time, and we can eliminate these terms by averaging equation (5) in time — that is, by convolving equation (5) with a window function of sufficiently long duration. This is related to the fact that relationship between the original real-valued field $$X(t)$$ and its positive- and negative-frequency parts $$X^\pm(t)$$ is non-local in time.

By the way, in a quantum optics context, equations (5) and (7) become equivalent when restricted to states having a well-defined number of photons — but such states do not have well-defined values of the electric/magnetic fields.

Now suppose that we have a real-valued solution of Maxwell's equations involving only a single frequency. Then $$\bfE^+ = \bfE_0 e^{i\omega t} \hskip2cm \bfH^+ = \bfH_0 e^{i\omega t} \tag{8}$$ for some complex-valued coefficients $$\bfE_0$$ and $$\bfH_0$$. In this case, equation (6) becomes $$\nabla\cdot(\bfH^+\times\bfE^-) = i\omega\big(\bfE_0^*\cdot \bfE_0-\bfH_0^*\cdot\bfH_0\big). \tag{9}$$ Now we can clearly see the reason for the sign-difference noted in the question. It's simply because in equation (6), the time-derivative acts on $$\bfE^+$$ and $$\bfH^-$$, which in this example are proportional to $$e^{i\omega t}$$ and $$e^{-i\omega t}$$, respectively.
Another perspective: Equation (6) is only "half" of the conservation law (7). When we use (8) in equation (7), we see that the left- and right-hand sides are both equal to zero. In particular, the quantity $$(\bfE^-\cdot\bfE^+ + \bfB^-\cdot\bfB^+)/2$$ is indepenent of time. For a wave with narrow bandwidth (which is certainly true of the example (8)), this quantity is a time-smeared approximation to the full energy dentity in equation (5). The minus sign between the two terms on the right-hand side of (9) has no general physical significance, because (9) is just a special trick for dealing with single-frequency fields, and equations (5) and (7) show that even in this case the electric and magnetic terms always contribute to the energy density (time-averaged or not) with the same sign.
The sign difference in (9) also has nothing to do with the relationship between the lagrangian and the energy. However, the sign-difference between the energy $$\sim \bfE^2 + \bfB^2$$ and the lagrangian $$\sim \bfE^2 - \bfB^2$$ is directly related to the corresponding sign difference in single-particle mechanics, where the energy is $$\sim p^2+V$$ and the lagrangian is $$\sim p^2-V$$. The relationship becomes clear when the lagrangian is written in terms of a gauge field $$\mathbf{A}$$ in the Coulomb gauge.
• I meant by averaging $\frac{1}{T}\int_{(T)} a(t)b(t)dt = \frac{1}{2}Re[AB^*]$ where $(a,b)(t)=Re[(A,B)e^{j\omega t}]$. If in eq(7) you assume to enclose the fields into an ideal metal cavity, then $\mathbf{E} \parallel d\mathbf{A}$ over the surface of the cavity, and from Gauss' theorem the left side integral is zero and we get that in lossless matter inside the cavity $W_e=W_m$ (total time average energies Electric = Magnetic ). This is well known but I think is nontrivial and is the consequence of eq (7). My question is what does eq(7) mean when not in cavity, and possibly lossy medium. Commented Apr 24, 2020 at 12:26
• Thank you for your helpful review of my question. While I accepted your answer it really has not resolved my original problem that using your new equation numbering I repeat here with the hope that you can at least give me a hint. If in eq(9) you assume to enclose the fields into an ideal metal cavity, then $E∥dA$ over the surface of the cavity, and from Gauss' theorem the left side integral is zero and we get that in lossless matter inside the cavity $W_e=W_m$ (total time average energies Electric = Magnetic ). This is well known but I think is nontrivial and is the consequence of eq (9). Commented Apr 29, 2020 at 12:39
• My question: what does eq(9) mean when the fields are not in an ideal cavity but possibly also in a lossy medium. I do not believe that you cant derive the $W_e=W_m$ equality from the "real" version of Poynting's theorem, instead what you get from the "real" version is the energy conservation law. Commented Apr 29, 2020 at 12:41
• @hyportnex In the single-freq case, eqn (9) does show that if the time-averaged electric and magnetic contributions to the energy are equal, then $\nabla\cdot(B^+\times E^-)$ must be zero, which implies that $\nabla\cdot$ the (real) time-averaged Poynting vector is zero. You are right that eqn (9) gives a nice way to deduce this result. As far as I know, eqn (5) or (7) by itself is not sufficient for that. Eqn (9), being "half" of eqn (7), retains more info from (the single-freq version of) Maxwell's eqns: (9) by itself implies (7), but (7) by itself doesn't imply (9). Commented Apr 29, 2020 at 20:21