Transmitted Power and Poynting's theorem contradiction?

Question

I was reading Chapter 12.1 in Hayt & Buck "Engineering Electromagnetics" 8-th edition. Here they discuss the reflection of uniform plane waves at normal incidence.

They derived the following expressions for the reflection and transmission coefficients:

$\Gamma = \frac{E_{x10}^{-}}{E_{x10}^{+}} = \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}$

$\tau = \frac{E_{x20}^{+}}{E_{x10}^{+}} = \frac{2\eta_2}{\eta_1+\eta_2} = 1 + \Gamma$

where $\eta_1, \eta_2$ are the intrinsic impedances of the two materials (which may be complex), and the electric field is uniform in the x direction, parallel to the interface.

Then they consider the power reflected and the power transmitted.

They use Poynting's theorem in phasor form: $\left<S\right> = \left|\frac{1}{2}\Re\left\{\mathbf{E}_s \times \mathbf{H}_s\right\}\right|$. From here, they use the reflection and transmission coefficients and the intrinsic impedances to conclude that

$\left<S_{1r}\right> = \left|\Gamma\right|^2\left<S_{1i}\right>$

$\left<S_{2}\right> = \frac{\Re\left\{1/\eta_2^*\right\}}{\Re\left\{1/\eta_1^*\right\}}\left|\tau\right|^2\left<S_{1i}\right>$

On the other hand, conservation of energy implies that the transmitted power must be the incident power minus the reflected power, so another expression is

$\left<S_{2}\right> = \left(1 - \left|\Gamma\right|^2\right)\left<S_{1i}\right>$

But the two forms of the coefficients obtained for the transmitted power are not in general equal. From my calculations, they would be equal if and only if the two intrinsic impedances have a ratio which is a real number. But I don't see any reason for this to be the case in general, and yet none of the steps seem to make that assumption.

So, where does the contradiction arise?

Answer 1

If the impedances are complex then it means you have dissipative terms. For example, if the problem was normal incidence from vacuum into a conductor, then energy conservation is not as simple as saying the (magnitude of the) Poynting vector of the incident wave equals the sum of the Poynting vectors in the transmitted and reflected waves.

In a conductor, then there is an ${\bf E} \cdot {\bf J}$ term that must be included in any conservation of energy calculation, because the transmitted E-field does work on the conduction charges (not the case if the impedances are real).

For example, in a conducting material with a complex wavevector, the electric field could have the form $$ \vec{E} = E_0 \exp(-\alpha x) \exp[i(kx-\omega t)]\ \hat{j}$$ and the time-averaged Poynting vector will be proportional to $\exp(-2\alpha x)$, but is a vector along the x-axis.

The divergence of this is non-zero, which tells you that the Poynting vector is not a conserved quantity, so you cannot just equate the sums of the Poynting vectors.

Answer 2

I came back to this problem a few years later, and I finally figured out what was wrong.

First of all, the Poynting vectors are indeed conserved across the boundary (placed at $z = 0$), but you cannot use the $\vec{S}_{1i}$ and $\vec{S}_{1r}$ vectors independently in the energy conservation equation. This is because there is interference between the incident and reflected waves, which causes a standing wave pattern in medium 1, whose average intensity is position-dependent.

In general, interference effects cause the intensity of the sum of two waves to be different from the sum of the intensities of the two waves. The interference means that you cannot simply say $\langle S_{1i}\rangle = \langle S_{1r}\rangle + \langle S_2\rangle$ or $(1 - |\Gamma|^2)\langle S_{1i}\rangle = \langle S_2\rangle$

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Let $\vec{S}_1$ be the total Poynting vector in medium 1, and $\vec{S}_2$ the same for medium 2.

By definition, $\vec{S}_1 = \vec{E}_1 \times \vec{H}_1$ where $\vec{E}_1$, $\vec{H}_1$ are the total electric and magnetic fields in medium 1. Assuming the interface between medium 1 and 2 has no current sheets ($\vec{K}_s = \vec{0}$), then the boundary conditions imply that $\vec{E}_1 = \vec{E}_2$ and $\vec{H}_1 = \vec{H}_2$. Therefore we clearly must also have $\vec{S}_2 = \vec{S}_1$, and intensity is conserved across the interface (at $z = 0$).

This means we should still be able to say that $\langle S_1\rangle = \langle S_2\rangle$ (at $z = 0$, again). So how do we correctly find $\langle S_1\rangle$ given $\langle S_{1i}\rangle$?

The incident wave in medium 1 propagates in the $\hat{z}$ direction, with $\vec{E}_{1i}$ in $\hat{x}$ and $\vec{H}_{1i}$ in $\hat{y}$. Moreover, in phasor form we know that $\mathbf{H}_{1i} = \frac{\mathbf{E}_{1i}}{\eta_1}$

The reflected wave in medium 1 propagates in the $-\hat{z}$ direction, with $\vec{E}_{1r}$ again in $\hat{x}$. This implies $\vec{H}_{1r}$ is in $-\hat{y}$, because the direction of the wave has been flipped. In phasor form, we thus have $\mathbf{H}_{1r} = -\frac{\mathbf{E}_{1r}}{\eta_1}$. In addition, we know that $\mathbf{E}_{1r} = \Gamma\mathbf{E}_{1i}$, and hence $\mathbf{H}_{1r} = -\Gamma\mathbf{H}_{1i}$.

The total phasors in medium 1 are then $\mathbf{E}_1 = \mathbf{E}_{1i} + \mathbf{E}_{1r} = (1 + \Gamma)\mathbf{E}_{1i}$ and $\mathbf{H}_1 = \mathbf{H}_{1i} + \mathbf{H}_{1r} = (1 - \Gamma)\mathbf{H}_{1i}$.

Therefore the Poynting vector in phasor form is $\mathbf{S}_1 = \frac{1}{2}\mathbf{E}_1\times\mathbf{H}_1^* = \frac{1}{2}(1 + \Gamma)(1 - \Gamma^*)\mathbf{E}_{1i}\times\mathbf{H}_{1i} = (1 + \Gamma)(1 - \Gamma^*)\mathbf{S}_{1i}$

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Note that $\mathbf{S}_1 = (1 - |\Gamma|^2)\mathbf{S}_{1i} + 2j\Im\{\Gamma\}\mathbf{S}_{1i}$, so there is an additional term besides the $(1 - |\Gamma|^2)$ factor that comes from the interference effect. When $\Gamma$ and $\mathbf{S}_{1i}$ have imaginary components (i.e. when $\frac{\eta_2}{\eta_1}$ is not real and $\eta_1$ is not real) we get a nonzero contribution to the real part of the Poynting phasor. So we can only use $\langle S_{1i}\rangle = \langle S_{1r}\rangle + \langle S_2\rangle$ when $\eta_1$ is real or when $\frac{\eta_2}{\eta_1}$ is real.

We can substitute $\Gamma = \frac{\eta_2 - \eta_1}{\eta_2 + \eta_1}$ and $\mathbf{S}_{1i} = \frac{|\mathbf{E}_{1i}|^2}{2\eta_1}$ to obtain $$\mathbf{S}_1 = \frac{2\eta_2}{\eta_1 + \eta_2}\frac{2\eta_1^*}{\eta_1^* + \eta_2^*}\frac{|\mathbf{E}_{1i}|^2}{2\eta_1} = \frac{2\eta_2}{|\eta_1 + \eta_2|^2}|\mathbf{E}_{1i}|^2 = |\tau|^2\frac{|\mathbf{E}_{1i}|^2}{2\eta_2^*} = \frac{|\mathbf{E}_2|^2}{2\eta_2^*} = \mathbf{S}_2$$

which confirms that $\vec{S}_1 = \vec{S}_2$.

Moreover, $\langle S_1\rangle = \langle S_2\rangle = \Re\left\{|\tau|^2\frac{|\mathbf{E}_{1i}|^2}{2\eta_2^*}\right\} = \frac{|\tau|^2|\mathbf{E}_{1i}|^2}{2}\Re\{1/\eta_2^*\} = \frac{\Re\{1/\eta_2^*\}}{\Re\{1/\eta_1^*\}}|\tau|^2\langle S_{1i}\rangle$ and so we confirmed the formula for $\langle S_2\rangle$.

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The comments about power dissipation due to $\vec{E}\cdot\vec{J}$ terms are valid, but they do not imply that $\vec{S}$ is discontinuous across the interface. The dissipative terms will produce an absorption coefficient which causes $\vec{S}$ to decay as the distance increases. But right across an interface, the enclosed volume is zero, so the volume integral of $\vec{E}\cdot\vec{J}$ will be nil (assuming the current densities are distributed over a volume, and there are no current sheets), and hence the inward flux equals the outward flux across the interface (i.e. the Poynting vector is continuous).

Additionally, absorption coefficients can arise even if the medium is non-conductive, i.e. even if $\vec{J}_f = \vec{0}$. This is because power can also be dissipated by bound currents and movement of bound charges which do not get accounted for explicitly in Poynting's theorem. Instead they show up by causing complex electric and magnetic permittivities.

Another way to understand this is that although $\vec{S}$ has generally nonzero divergence, the divergence is still finite, so fluxes of $\vec{S}$ can only vary continuously across a volume, and they cannot just change right across an interface. For this to happen you would need infinite volume current densities, or basically something like current sheets or current wires.

Answer 3

You only expect transmitted power to be equal to the incident power minus the reflected power when no energy is supplied to charges (which can then lose energy to heat).

The Poynting vector does not have a divergence free energy flow, it can gain or lose energy by getting or supplying energy to charges. And the charges can give or get energy from the fields but can also gain or lose energy to heat.

An energy conservation argument tells you nothing unless you know there is nothing else the energy is going to.