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Why is the Poynting Vector symmetric in E and H? I always thought that E and B were the analogous fields, so I would think that any equation using magnetic and electric fields should be symmetric in E/B or D/H. I've seen the derivation, but I have yet to find the point where my want for symmetry is being broken.

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4 Answers 4

The discrete symmetry in Maxwell equations in vacuum is not precisely that. What Maxwell equation say is that $$\mathrm d * F=0 \qquad \mbox{and } \qquad \mathrm d F=0.$$

The theory is invariant under

$$\mathbf B \mapsto \mathbf {+E},$$ $$\mathbf E \mapsto \mathbf {-B},$$

which is implemented by the Hodge dual on $F$, $F \mapsto *F$, and which leaves the Poynting vector invariant.

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Could you elaborate on how this addresses the question of why $\mathbf H$ appears in the Poynting vector as opposed to $\mathbf B$? Perhaps I'm just dense, but it's not clear to me how Hodge duality sheds light on this when we're dealing with electrodynamics in a medium. –  joshphysics Jun 5 at 5:50
@joshphysics Sorry it took so long to answer. You're right, my answer doesn't address the question: I'm not used to electrodynamics out of vacuum. So I'll erase it. –  c.p. Nov 11 at 9:12

When you say

...so I would think that any equation using magnetic and electric fields should be symmetric in E/B or D/H

You're taking the "correspondence" between these pairs too far. It is easy to construct all sorts of electrodynamical identities in which, $\mathbf E$ and $\mathbf H$ appear together. For example, using Ampere's Law in a medium \begin{align} \nabla\times\mathbf H - \frac{\partial \mathbf D}{\partial t} = \mathbf J \end{align} and some vector calculus, it is possible to show that \begin{align} \frac{\partial u}{\partial t} + \nabla\cdot (\mathbf E\times\mathbf H) = -\mathbf J \cdot \mathbf E \end{align} where $u$ is the energy density in the fields, and $\mathbf J\cdot \mathbf E$ is the work per unit volume performed by the fields on whatever charges are present. This equation is essentially a statement of the conservation of energy for electrodynamics. This motivates the definition \begin{align} \mathbf S = \mathbf E\times\mathbf H \end{align} which has the interpretation of the energy per unit area per unit time being carried by the electromagnetic fields. Of course, homogeneous, isotropic, linear dielectric, for example, $\mathbf H = \mathbf B/\mu$ so that \begin{align} \mathbf S = \frac{1}{\mu}\mathbf E\times\mathbf B \end{align} in which case you have your wish that $\mathbf E$ and $\mathbf B$ appear together. In particular, in vacuum this is the case.

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But now why do we only have a $\mu$ and no $\epsilon$? –  Anthony Jun 5 at 7:43
@Anthony The presence, or lack thereof, of $\epsilon$ in the last expression for $\mathbf S$ in a linear dielectric depends on whether you decide to write it in terms of $\mathbf E$ or $\mathbf D = \epsilon \mathbf E$. In the latter choice, we would have $\mathbf S = \frac{1}{\epsilon\mu}\mathbf D\times \mathbf B$. Also, there is no rule of electrodynamics that says "in linear media, any equation written in terms of $\mathbf E$ and $\mathbf B$ must contain both $\epsilon$ and $\mu$." In short, your expectations about how equations "should" look are too high and not supported by the math. –  joshphysics Jun 5 at 8:23

Since Poynting vector S

$$\vec{S}=\vec{E}\times \vec{H}$$ and $$\vec{H}=\frac{\vec{k}\times \vec{E}}{\mu\omega}$$

$\vec{S}$ is symmetric in $\vec{E}$ and $\vec{H}$ because

If $\vec{E} \rightarrow -\vec{E}$ then $\vec{H}$ also change its direction and $\vec{H} \rightarrow -\vec{H}$ and vice-versa.

So $\vec{S}$ is symmetric about $\vec{E}$ as for $\vec{E} \rightarrow -\vec{E} \Rightarrow \vec{H} \rightarrow -\vec{H} $

so $$\vec{S}=\vec({-E})\times \vec({-H})$$ $$\vec{S}=\vec{S}$$

and similarly for $\vec{H}$.

And another point is that poynting vector is always in the direction of $\vec{k}$ in vacuum.

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I don't think expression you've written for $\vec H$ in terms of $\vec E$ is general; one can in principle have all sorts of strange constitutive relations. –  joshphysics Jun 5 at 5:52

The electromagnetic stress-energy tensor is $$T^{\mu\nu} = \frac{1}{\mu_0} \left[ F^{\mu \alpha}F^\nu{}_{\alpha} - \frac{1}{4} \eta^{\mu\nu}F_{\alpha\beta} F^{\alpha\beta}\right]\text{.}$$ Its terms are antisymmetric, e.g.,: $$\begin{eqnarray*} T_0{}^0 &=& -\frac{1}{2}\left[\epsilon_0 E^2 + \frac{1}{\mu_0}B^2\right]\text{,}\\ T_a{}^0 &=& \epsilon_0\left[\mathbf{E}\times\mathbf{B}\right]\text{,}\\ T_0{}^a &=& \frac{1}{\mu_0}\left[\mathbf{B}\times\mathbf{E}\right]\text{.} \end{eqnarray*}$$ However, if one applies the following definitions of the auxilary fields $\mathbf{D}$, $\mathbf{H}$ (thus there is no need to assume a linear dielectric, etc.): $$\begin{eqnarray*} \epsilon_0\mathbf{E} = \mathbf{D}-\mathbf{P},&\quad&\frac{1}{\mu_0}\mathbf{B} = \mathbf{H} + \mathbf{M} \end{eqnarray*}$$ to separate free and bound charges, the stress-energy tensor decomposes into (the first and second square-bracketed terms correspond to the free and bound terms, respectively): $$\begin{eqnarray*} -T_0{}^0 &=& \frac{1}{2}\left[\mathbf{E}\cdot\mathbf{D} + \mathbf{B}\cdot\mathbf{H}\right] + \frac{1}{2}\left[-\mathbf{E}\cdot\mathbf{P}+\mathbf{B}\cdot\mathbf{M}\right]\text{,}\\ +T_a{}^0 &=& \left[\mathbf{D}\times\mathbf{B}\right] + \left[-\mathbf{P}\times\mathbf{B}\right]\text{,}\\ -T_0{}^a &=& \left[\mathbf{E}\times\mathbf{H}\right] + \left[\mathbf{E}\times\mathbf{M}\right]\text{,} \end{eqnarray*}$$ as well as a more complicated "Maxwell stress" terms in the space-space components.

Thus, fundamentally situation is (anti)symmetric, but the apparent break is due to a conventional decomposition to free and bound charges. In terms of the full stress-tensor tensor, $\mathbf{D}\times\mathbf{B}$ is part of the field momentum density, while $\mathbf{E}\times\mathbf{H}$ is part of energy flux density. The latter is the standard Poynting vector, but because of antisymmetry, the choice of which to use is actually arbitrary, so long as it's consistently applied.

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