Short Stories about integral theorems


December 9. 2000, Math 21a
Oliver Knill

The dynamo

The mathematics of the dynamo is tied to the theorem of Stokes. Experiments show that if one moves a closed wire through a magnetic field, then a current flows in the wire. The voltage is proportional to the change of the magnetic flux through a surface spanned by the wire.
Maxwell derived from such knowledge the equation ${\rm curl}(E) = - \frac{1}{c} \frac{\partial}{\partial t} B$. To see that this law explains the experiments, we take a surface $S$ which has the wire as a boundary. Compute the flux integral of both sides of this Maxwell equation and get $\int\int_S {\rm curl}(E) \; \cdot dS = \frac{1}{c} \frac{\partial}{\partial t} \int\int_S B \; \cdot dS$. The left hand side is by Stokes theorem the line integral $\int_{\gamma} E \; ds$ which is the voltage. The right hand side is the change of the magnetic flux through the surface.

Digging into the moon

We recently bought some estate on the moon at the Lunar Embassy. The size of Manhatten can currently be obtained for about 30 Dollars. We wonder whether we should use this to build our private telescope... In one of the dreams, we plan therefore to to dig a hole to the center of the moon. The question arrizes: what is the magnitude of the gravity force at radius $r$ from the center of the moon? This question can be answered quite nicely using the divergence theorem. The law of gravity can be reformulated by saying that the gravitational field $F$ satisfies the law ${\rm div}(F) = 4 \pi \rho$, where $\rho$ is the density of the mass distribution. We want to see whether this is reasonable. Consider the case of the moon, a ball of radius $R$, then by rotational symmetry, the gravitational force is rotationally symmetric and normal to the surface of the moon. Therefore, we can write $F(x)=F(r) x/\vert\vert x\vert\vert$ for the force at a point $x$. We have $\int_{S_r} F(x) \; dS = 4 \pi r^2 F(r)$. By the divergence theorem, the left hand side is $\int \int \int_{B_r} \rho(x) \; dV$ which is the mass of the material inside the sphere of radius $r$. This is $4\pi r^3/3$ for $r<R$ and $4 \pi R^3/3$ for $r \geq R$. We see that inside the moon the gravitational force decreases linearly when we dig in while decreases like $1/r^2$ when going away from the moon.

A bugs struggle

Below one of the bridges on the Charles river, several vortices appear. You observe that the water flows around some of the vortices counterclockwise, and clockwise around others. A model for the velocity field of a single vortex is the vector field $F_{\pm}(x,y) = \pm (-y/(x^2+y^2),x/(x^2+y^2))$ which is conservative away from the origin: a potential function is $h(x,y) = \pm {\rm arctan}(y/x)$. The potential of a bunch of vortices located at points $(x_i,y_i)$ is the sum $\sum_i a_i {\rm arctan}( (y-y_i)/(x-x_i))$, where $a_i=\pm1$. Assume we move around a closed path on the water surface we experience at every point a force which is equal to the water velocity at that point. A fly which fell into the water swims around a closed path $\gamma$. It surrounds two positive vortices and one negative vortex. What is the work, it did? You have seen in one of the homework problems that a path around a vortex leads to a line integral value $2 \pi$ and that a path surrounding no vortex produces a zero line integral. Without doing any calculation, we see that the fly's work is $4\pi - 2\pi = 2 \pi$. More generally, the work needed to go around $n$ vortices with positive vorticity and $m$ vortices with negative vorticity is $(n-m) 2\pi$.

We build a Perpetuum mobile

A perpetuum mobile is a machine which produces more energy then what is put into it. A handy thing! No electric bills anymore.
The quest is find a force field which is not conservative. When moving along a force field along a path $\gamma$, the line integral $\int_{\gamma} F \; ds$ is the work done. With a nonconservative field, one would be able to go around some path such that the line integral (=work) is positive (if negative, we could reverse the direction).
Many proposals for perpetuum mobiles exist. One of the ideas is to fill a toral tube on one side with water. A piece of wood falls in the air on one side and dragged in the water to the top on the other side.

And there shall be light

The miraculous Maxwell equations describe the electric field $E$ and magnetic field $B$:
\fbox{ \parbox{10cm}{ \begin{center}\begin{tabular}{lr} ${\rm curl}(E) = -\frac{... ...\rm div}(B) = 0$, & ${\rm div}(E) = 4 \pi \rho$\ \\ \end{tabular}\end{center}}}
where $\rho$ is the charge density a scalar field, $i$ is the electric current vector field and $c$ is a constant, the speed of light.
In vacuum $\rho=0, i=0$ the electric field satisfies $\frac{\partial^2 E}{(\partial t)^2} E = c \frac{\partial \; {\rm curl}(B)}{\pa... ...; {\rm curl} \frac{\partial B}{\partial t} = c \; {\rm curl} \; {\rm curl} (E)$. We see for example that $u=E_1$ satisfies the wave equation $u_{tt} = u_{xx}$. Actually, all components $E_i$ of the electric field and all components $B_i$ of the magnetic field satisfy the wave equation.

Magnetic Monopoles

Physicists still search for magnetic monopoles, the magnetic analogue of electric charges. If they would exist, the Maxwell were more symmetric. Experiments however indicate that one can not cut away the positive pole of the magnet and separate it from the negative pole.
The nonexistence of magnetic monopoles is mathematically expressed by the equation ${\rm div}(B)=0$. This means that there are no places, where the magnetic field is created. If we take a closed surface $S$ enclosing a region $R$ and look at the flux of the magnetic field through it then, by the divergence theorem, $\int \int_S B \; dS = \int \int \int_R {\rm div}(B) \; dV = 0$. This means that no magnetic field is created inside $R$ nor distroyed. The law ${\rm div}(B)=0$ indeed means: "There are no magnetic monopoles".

Spilling a bottle of soda

The continuity equation is $\frac{\partial \rho}{\partial t} + {\rm div}( \rho v ) = 0$, where $\rho$ is the density of a fluid and where $v$ is the velocity of the fluid. In order to see what it means, we introduce the current $i=\rho v $ which is a vector field.
Take a closed surface $R$ which encloses a region $R$. The divergence theorem tells us that the flux of $i$ through the surface is equal to the integral $\int \int \int_R {\rm div}(i) \; dV$ which is by the continuity equation equal to $-\int\int\int_R \frac{\partial \rho}{\partial t} \; dV$. This is the negative of the time derivative of the total mass $\int\int\int_R \rho \; dV$. In other words, the flux through the surface $S$ is minus the change of the mass in the interior of the surface. Therefore, if a positive amount of fluid escapes through the surface, then the mass inside decreases.
The continuity equation applies in other contexts also, for example, when $\rho$ is the electric charge and $i$ is the electric current, or when $\rho$ is the energy density of the heat and $i$ is the heat current.

Why does a balloon not start to move spontaneously?

Let us reformulate this problem in multivariable calculus language: we have a closed surface $S$, the skin of a balloon. The air pressure inside the balloon produces a force on $S$ which is normal to the surface and proportional to the area. The total force on the balloon in the $x$ direction is the flux of the vector field $(1,0,0)$ through the boundary. (Why? Think it maybe first through at the example of a box $0 \leq x \leq 1, 0 \leq y \leq 1, 0 \leq z \leq z$. The x component of the force at the wall $x=1$ is positive, the force at the wall $x=0$ is negative and zero at the other walls. )
By the divergence theorem, the flux integral is equal to the average of the divergence of the vector field $(1,0,0)$ in the interior of the ball. It is zero by the divergence theorem.

What is the magnetic field produced by a wire?

A researcher investigating the relation between electromagnetic waves and cancer is interested in the magnetic field strength away from a wire carrying a current $J$. Assume that in a cylindrical wire pointing in the $z$ direction flows a current $i$, The magnetic field lines go circular around the wire and the strength depends only on the distance from the $z$ axes. The magnetic field is therfore of the form $B(x,y,z) = B(r) ( -y,x,0)$ where $B(r)$ is a still unknown function of $r$. If we compute the line integral of $B$ around a circle $(\cos(t),\sin(t),z)$, we get $2 \pi r B(r)$ which is what we want to know. The line integral is by Stokes theorem equal to the flux of ${\rm curl}(B)$ through the surface. From Maxwell equations, we know that this is $4 \pi/c$ times the flux of $j$ thought the wire. We simply call this $J$, the current. We see from $2 \pi r B(r) = 4 J \pi/c$ that the magnetic field strength decays up to a multiplicative constant like $1/r$, where $r$ is the distance from the wire.