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