Gauss’ Law

Since my blog claims to talk about physics sometimes and I just finished teaching multivariable calculus, I thought I’d do a post on one form of Gauss’ law. As a teacher of the course, I found this to be an astonishingly beautiful “application” of the divergence theorem. It turned out to be a touch too difficult for my students (and I vaguely recall being extremely confused about this when I took the class myself).

First, I’ll remind you what some of this stuff is if you haven’t thought about these concepts for awhile. Let’s work in {\mathbb{R}^3} for simplicity. Consider some subset {U\subset \mathbb{R}^3}. Let {F: U\rightarrow \mathbb{R}^3} be a vector field. Mathematically this is just assigning a vector to each point of {U}. For calculus we usually put some fairly restrictive conditions on {F}, such as all partial derivatives exist and are continuous.

The above situation is ubiquitous in classical physics. The vector field could be the gravitational field or the electric field or it could describe velocity of a flowing fluid or … One key quantity you might want to know about your field is what is the flux of the field through a given surface {S}? This measures the net change of the field flowing through the surface. If {S} is just a sphere, then it is easy to visualize the flux as the amount leaving the sphere minus the amount flowing in.

Let’s suppose {S} is a smooth surface bounding a solid volume {E} (e.g. the sphere bounding the solid ball). In this case we have a well-defined “outward normal” direction. Define {\mathbf{n}} to be the unit vector field in this direction at all points of {S}. Just by definition the flux of {F} through {S} must be “adding up” the values of {F\cdot \mathbf{n}} over {S}, because this dot product just tells us how much {F} is pointing in the outward direction.

Thus we define the flux (using Stewart’s notation) to be:

\displaystyle \iint_S F\cdot d\mathbf{S} := \iint_S F\cdot \mathbf{n} \,dS

Note the second integral is integrating a scalar valued function with respect to surface area “dS.” Now recall that the divergence theorem says that in our situation (given that {F} extends to a vector field on an open set containing {E}) we can calculate this rather tedious surface integral by converting it to a usual triple integral:

\displaystyle \iint_S F\cdot d\mathbf{S} = \iiint_E div(F) \,dV

If you’re advanced, then of course you could just work this out as a special case of Stoke’s theorem using the musical isomorphisms and so on. Let’s now return to our original problem. Suppose I have a charge {Q} inside some surface {S} and I want to compute the flux of the associated electric field through {S}.

From my given information this would seem absolutely impossible. If {S} can be anything, and {Q} can be located anywhere inside, then of course there are just way too many variables to come up with a reasonably succinct answer. Surprisingly, Gauss’ law tells us that no matter what {S} is and where {Q} is located, the answer is always the same, and it is just a quick application of the divergence theorem to prove it.

First, let’s translate everything so that {Q} is located at the origin. Since flux is translation invariant, this will not change our answer. We first need to know what the electric field is, and this is essentially a direct consequence of Coloumb’s law:

\displaystyle F(x,y,z)=\frac{kQ}{(x^2+y^2+z^2)^{3/2}}\langle x, y, z\rangle

If we care about higher dimensions, then we might want to note that the value only depends on the radial distance from the origin and write it in the more succinct way {\displaystyle F(r)=\frac{kQ}{|r|^3}r}, where {k} is just some constant that depends on the textbook/units you are working in. Let’s first compute the partial of the first coordinate with respect to {x} (ignoring the constant factor for now):

\displaystyle \frac{\partial}{\partial x}\left(\frac{x}{(x^2+y^2+z^2)^{3/2}}\right) = \frac{-2x^2+y^2+z^2}{(x^2+y^2+z^2)^2}

You get similar things for taking the other derivatives involved in the divergence except the minus sign moves to {-2y^2} and {-2z^2} respectively. When you add all these together you get in the numerator {-2x^2-2y^2-2z^2+2x^2+2y^2+2z^2=0}. Thus the divergence is {0} everywhere and hence by the divergence theorem the flux must be {0} too, right? Wrong! And that’s where I lost most of my students.

Recall that pesky hypothesis that {F} can be extended to a vector field on an open neighborhood of {E}. Our {F} can’t even be defined at all to extend continuously across the origin. Thus we must do something different. Here’s the idea, we just change our region {E}. Since {E} is open and contains the origin, we can find a small sphere of radius {\varepsilon>0} and centered at {(0,0,0)} whose interior is properly contained in {E}, say {S_\varepsilon}.

Let {\Omega} be the region between these two surfaces. Effectively this “cuts out” the bad point of {F} and now we are allowed to apply the divergence theorem to {\Omega} where our new boundary is {S} oriented outwards and {S_\varepsilon} oriented inward (negatively). We already calculated that {div F=0}, thus one side of the equation is {0}. This gives us

\displaystyle \iint_S F\cdot d\mathbf{S} = \iint_{S_\varepsilon} F\cdot d\mathbf{S}

This is odd, because it says that no matter how bizarre or gigantic {S} was we can just compute the flux through a small sphere and get the same answer. At this point we’ve converted the problem to something we can do because the unit normal is just {\mathbf{n}=\frac{1}{\sqrt{x^2+y^2+z^2}}\langle x, y, z\rangle}. Direct computation gives us

\displaystyle F\cdot \mathbf{n} = \frac{kQ (x^2+y^2+z^2)}{(x^2+y^2+z^2)^3}=\frac{kQ}{(x^2+y^2+z^2)^2}

Plugging this all in we get that the flux through {S} is

\displaystyle \iint_{S_\varepsilon} \frac{kQ}{\varepsilon^2} \,dS = \frac{kQ}{\varepsilon^2}Area(S_\varepsilon) = 4\pi k Q.

That’s Gauss’ Law. It says that no matter the shape of {S} or the location of the charge inside {S}, you can always compute the flux of the electric field produced by {Q} through {S} as a constant multiple of the amount of charge! In fact, most books use k=1/(4\pi \varepsilon_0) where $\varepsilon_0$ is the “permittivity of free space” which kills off practically all extraneous symbols in the answer.