# limits on a gauss box of light

Consider a wall defined by $w(x,y,z) = \Theta(x-L)$ which is nonzero in the infinite semi-space of $x \ge L$, as well as a coherent planar standing EM wave travelling in the $z$ plane given by its electric field:

$$E_x = \Theta(x - L) \sin(kz) \sin(\omega t), E_y = E_z = 0$$

Consider a complementary standing wave travelling in a opposite plane, in the region $x \le -L$:

$$E_x = - \Theta(L - x) \sin(kz) \sin(\omega t), E_y = E_z = 0$$

now, if i take a small closed box in the region $-L- \epsilon \le x \le L + \epsilon$ and z such that $0 \le kz \le \pi$, the net electric flux over this box at any given moment of time is:

$$\frac{ 8 \Delta y \sin(\omega t) }{k}$$

Naively, it would seem "conceivable" to setup standing waves in this way and produce a net electric flux in a given region of vacuum without any spatial charges anywhere. Now, since this apparent flux increases with the wavelength, i suspect there is some optical bound in the coherence that the walls can sustain due to dispersion, i.e: something like $\Delta k_z \Delta x \ge \hbar$, but i can figure out exactly what is the reason this won't work.

What i'm trying to figure out is how physical this solution is and what limits are imposed by optics in the realization of the charge-violating gauss box

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## 2 Answers

It is easy to see that your light cannot have such a well defined wall without any charges, i.e:

$$\nabla \cdot E = \lbrack \delta (x-L) + \delta (L-X) \rbrack \sin(kz) \sin(\omega t)$$

which means you need surface charges to sustain the wall.

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Oh I see--- the OP was taking the field to vanish inside the wall, and making the box straddle the wall so that part is in and part is out. I didn't understand. +1, you had the right interpretation. The reason I didn't understand is because he said "-(L+\epsilon)<x<L+\epsilon" where the flux is zero, and also -(L-\epsilon)<x<L-epsilon (where it's all plane wave) the flux is zero. He probably meant -L-\epsilon<x<L-\epsilon, where the flux is nonzero and it is what you said. – Ron Maimon Jul 23 '12 at 14:20

The net electric flux over any box in any configuration of electromagnetic waves without charges is zero. The electric field lines that enter the box also leave the box. In your case, it's obvious--- the E field is in the x-direction for both plane waves, and so the flux is only nonzero on the two opposite sides which are in the y-z plane. The flux entering one of the sides is equal point by point to the flux leaving the other side, because the plane waves are translationally invariant.

More generally, you can't violate Gauss's law by superposing plane-waves--- each one already satisfies Gauss's law by itself, and superposing two things that satisfy Gauss's law gives a third thing that also satisfies Gauss's law. The demonstration is by noting that for a plane wave

$$\nabla \cdot E \propto k\cdot E$$

that is, it is zero if E is perpendicular to k. This is just saying that the field lines in a plane wave go from one side of space to the other side of space at any one time, without being created or destroyed.

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