Question about intensity of EM waves

For electromagnetic wave if it's reflected from a perfect conductor standing wave can be form. I wonder why Poynting vector can be used to describe the intensity of standing EM wave. (see p.19 of http://web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide13.pdf ).

From some textbooks the Poynting vector is used to describe the intensity of traveling wave. It is derived from wave without reflection and since Maxwell's equations are obeyed and therefore cB=E. However in standing wave since $\vec{E}=2\vec{E}_0 \cos(\vec{k}\cdot\vec{x})\cos(\omega t)$ and $\vec{B}=2\vec{B}_0 \sin(\vec{k}\cdot\vec{x})\sin(\omega t)$, the $\vec{E}$ and $\vec{B}$ do not obey $c|\vec{B}|=|\vec{E}|$ at every point and the overall wave is not traveling in any direction.

Another question is, the Poynting vector of standing wave is pointing perpendicular to the $\vec{E}$ and $\vec{B}$ field and sometime has positive and negative value. It means that energy is transmitting to either directions in different times. However for standing wave $c|\vec{B}|=|\vec{E}|$ is not obeyed, does it contradict Maxwell's equation?

The Poynting vector is useful not because we say so, but because of Poynting's theorem, which in essence states that the Poynting vector can usefully model how electromagnetic energy is moved around a system of changing electromagnetic fields.

More precisely, you can define a quantity $$u=\frac12\left(\varepsilon_0\mathbf E^2+\frac1{\mu_0}\mathbf B^2\right)$$ and a vector $$\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B$$ in terms of which the total power delivered per unit volume by the electromagnetic fields to the charges and currents, $P_\mathrm{mech}=\mathbf J·\mathbf E$ obeys $$P_\mathrm{mech}+\frac{\partial u}{\partial t}+\nabla·\mathbf S=0.$$ This is exactly the form of a continuity equation for energy, so you can interpret it as saying that the electromagnetic fields hold energy with a density $u$ and which flows with a flux $\mathbf S$, potentially being lost to the field as it delivers work to charges and currents. In fact, you can go beyond 'interpreting': it looks like a duck and it quacks like a duck, which is really the only things that we require from mathematical models of ducks.

The Poynting vector, to emphasize, is not 'derived from wave without reflection'; it is part of the fundamental theorem of energy conservation in classical electrodynamics. That is why it is useful.

That said, the Poynting vector can indeed be a bit confusing when standing waves are involved. Standing waves are formed when you superpose one wave travelling in one direction, $$\mathbf E_1=E_0\mathbf i \cos(kz-\omega t),\quad \mathbf B_1=B_0\mathbf j \cos(kz-\omega t),$$ with an equivalent wave going in the opposite direction: $$\mathbf E_2=E_0\mathbf i \cos(kz+\omega t),\quad \mathbf B_2=-B_0\mathbf j \cos(kz+\omega t).$$ Each wave has a Poynting vector which points along its propagation direction, $$\mathbf S_1 =\frac1{\mu_0}\mathbf E_1\times\mathbf B_1 =\frac{E_0B_0}{\mu_0}\mathbf k\cos^2(kz-\omega t)$$ and $$\mathbf S_2 =\frac1{\mu_0}\mathbf E_2\times\mathbf B_2 =-\frac{E_0B_0}{\mu_0}\mathbf k\cos^2(kz+\omega t),$$ so you'd sort of expect the combined energy flow to be zero. On average, this is indeed the case, but when you do the full calculation you get something different: $$\mathbf S_ =\frac1{\mu_0}(\mathbf E_1+\mathbf E_2)\times(\mathbf B_1+\mathbf B_2) =\mathbf S_1+\mathbf S_2 +\frac1{\mu_0}\mathbf E_1\times\mathbf B_2 +\frac1{\mu_0}\mathbf E_2\times\mathbf B_1 .$$ In addition to the two single-wave terms, you get two cross-terms which muddy the picture a bit. If you do the math, you get what your worksheet gives you: a standing-wave like term that oscillates at half the wavelength, $$\mathbf S=\frac{E_0B_0}{\mu_0}\mathbf k\sin(2kz)\sin(2\omega t).$$ What's going on here? As it turns out, the interference between the two travelling waves that makes the standing wave works differently for the electric and the magnetic fields. In particular, the $-z$-propagating wave has a minus sign on the magnetic field, which means that the magnetic field will make a standing wave which is displaced both in space and in time from the electric field: $$\mathbf E=2E_0\mathbf i \cos(kz)\cos(\omega t) \quad\text{while}\quad \mathbf B=2B_0\mathbf j \sin(kz)\sin(\omega t).$$ As time moves forward, the electric and magnetic fields oscillate out of step and in different regions of space.

In terms of energy, this means that there is a constant flux of energy from the regions where the electric field has maxima to the regions where it has a node but the magnetic field has a maximum.

As you can see, the energy is being passed back and forth between such regions, with no net flow in either direction. This is reflected by the fact that if you time-average the Poynting vector, you get zero, which is exactly what you'd expect from the fact that each travelling wave is carrying energy at the same rate, but they do so in different directions.

To go back to the specifics of your question,

the $\mathbf E$ and $\mathbf B$ do not obey $c|\mathbf B |=|\mathbf E |$ at every point,

is of course true, and it is perfectly valid because the relation $c|\mathbf B |=|\mathbf E |$ is valid but it is valid only for plane waves of a single frequency and direction. If you superpose them you'll get different things there you will always have solutions as long as you superpose solutions of Maxwell's equations. If in doubt differentiate! Plug the fields you're unsure about into the equations and see if they satisfy them.

the overall wave is not traveling in any direction.

This is in general the case for arbitrary solutions of the Maxwell equations. Some special solutions (which we use a lot) can be seen to travel in a specific direction, but in general this won't be the case. This does not prevent the fields from having a Poynting vector, which is always defined as $\mathbf S=\tfrac1{\mu_0}\mathbf E\times\mathbf B$, though in general it will be a complicated space- and time-dependent field.

The standing wave solution you quote is a solution to Maxwell's equations - as is any linear superposition of travelling waves.

The relationship that $E = cB$ is really only applicable to travelling transverse electromagnetic waves in vacuum, however I note that in this case $E_0 = cB_0$.

The Poynting vector (in vacuum) is defined by $$S = \frac{\vec{E} \times \vec{B}}{\mu_0}$$

For your standing wave, the E-field and B-field are perpendicular and so the Poynting vector is in the mutually perpendicular direction (parallel to the surface vector of your reflector, which is the $x$ direction according to you E- and B-field definitions) and had a magnitude of $$S = 4E_0B_0 \sin kx \cos kx \sin \omega t \cos \omega t = E_0 B_0 \sin 2kx \sin 2\omega t$$

Thus as you say, the Poynting vector oscillates backwards and forwards with time at any given $x$, but it has a time-averaged value of zero. Intuitively this is what you expect since the standing wave is the superposition of two travelling waves of equal amplitude but in opposite directions.

• for the standing wave since there is a factor of sine at every point, even Eo=cBo, the actual amplitude at a specific point x is Eo sin(kx) and E=cB is not obeyed. Did I get anything wrong? – Kelvin S Jul 29 '15 at 15:36
• $E_0 = cB_0$. I make no further claim. The property that $E=cB$ is not a general property of electromagnetic waves. – Rob Jeffries Jul 29 '15 at 15:39

I wonder why Poynting vector can be used to describe the intensity of standing EM wave.

It's because a standing wave isn't really standing. Hence the photon in the cavity is off like a shot when you "open the box". It didn't accelerate from zero to c in an instant, it was always propagating at c. The Poynting vector denotes the wave motion even when it looks like there isn't any motion.

From some textbooks the Poynting vector is used to describe the intensity of traveling wave.

Because the Poynting vector denotes the wave motion, as per as per this Blaze labs picture:

It is derived from wave without reflection

Perhaps the point you're missing is that you don't need a reflection for a standing wave. Take a look at the static fields section of the Wikipedia Poynting vector article. The Poynting vector is orthogonal to E and H and is circular:

Public domain image by Michael Lenz, see Wikipedia

Imagine you've got a single electromagnetic wave wrapped round twice round a circle such that the minima coincides with the maxima. You go round twice for one wavelength. You go round 720 degrees, then you've got a static field: "While the circulating energy flow may seem nonsensical or paradoxical, it is necessary to maintain conservation of momentum. Momentum density is proportional to energy flow density, so the circulating flow of energy contains an angular momentum". This angular momentum is demonstrated via the Einstein-de Haas effect.

Another question is, the Poynting vector of standing wave is pointing perpendicular to the E and B field and sometime has positive and negative value. It means that energy is transmitting to either directions in different times. However for standing wave cB=E is not obeyed, does it contradict Maxwell's equation?

I'm sorry, I'm not clear what you mean here. But can I volunteer that a standing wave isn't travelling in any direction provided it's symmetrical, and there's no contradiction of Maxwell's equations that I know of.