Hints and review to problem set 2

1. The electrostatic potential

By definition of the electrostatic potential for point charges, $\varphi = \frac{q}{4\pi\epsilon_0} \left[ -\frac{1}{\lvert (x, y, z - a/2)\rvert} + \frac{1}{\lvert (x, y, z + a/2)\rvert} \right]$

1.b) The limit of a thin shell

The key here is that we expect that as $a \to 0$, the potential is axi-symmetric. It would thus be wise to introduce spherical coordinates $x = r\cos\theta\sin\phi$, $y = r\sin\theta\sin\phi$, $z = r\cos\phi$. The two denominators become $\sqrt{x^2 + y^2 + (z \mp a/2)^2} = r \mp \frac{a\cos\phi}{2} + \ldots$

once expanded in a series expansion in $a$. Flip this expression and place in the potential expression to get $\varphi \sim -\frac{\cos\phi (qa)}{4\pi \epsilon_0 r^2}.$

To verify that $\varphi = \frac{1}{4\pi \epsilon_0} (\mb{p} \cdot \nabla) (1/r)$, compute $\nabla(1/r)$ and recall $\mb{r} = [x, y, r\cos\phi]$.

1.c) Equipotential lines

Examining the form of the potential, we see that equipotential lines are given by $r^2 = A\cos\phi$, for any constant, $A$. If we write this in Cartesian coordinates, we have $(x^2 + y^2 + z^2)^{3/2} = Az$. Set $y = 0$ and try plotting.

2. Gauss' Law and Poisson's equation

Gauss' Theorem states that $\iint_{\partial V} \nabla \phi \cdot \de{\mb{S}} = -4\pi G M(r).$

where $M(r)$ is the mass within a ball of radius $r$. Verify that $M(r) = \pi k r^4$.

2.a) Solve from Gauss' Law

Assume $\phi = \phi(r)$. Either evaluate the integral directly, or convert to a volume integral using the Divergence theorem. This gives $\iint_{\partial V} \nabla \phi \, \de{\mb{S}} = 4\pi^2 r^2 \dd{\phi}{r}.$

In any case, we are left to solve $4\pi r^2 \dd{\phi}{r} = -4\pi G M(r) = -4\pi G\begin{cases} \pi k r^4, & 0 \leq r < a \\ \pi k a^4, & r \geq a. \end{cases}$

Solving and applying boundary conditions gives $\phi = -\pi Gk \begin{cases} r^3/3 - 4a^3/3, & 0 \leq r < a \\ -a^4/r, & r > a \end{cases}$

2.b) Solve from Poisson's equation

Solving Poisson's equation is much the same process, but now we start with equating the integrands of $\iiint_{V(r)} \nabla^2 \phi \, \de{V} = -4\pi G \iiint_{V(r)} \rho_0 \, \de{V},$

giving $\nabla^2 \phi = \frac{1}{r^2} \dd{}{r} \left( r^2 \dd{\phi}{r}\right) = -4\pi G \begin{cases} kr, & 0 \leq r < a \\ 0, & r \geq a. \end{cases}$

The difference with evaluating the integrals directly, is that we now have four constants of integration. Two of the constants will be determined by imposing the continuity of $\mb{F} = \nabla \phi$ at $r = a$ and boundedness at $r = 0$. The other two constants are determined similarly to the same process above, with continuity at $r = a$ in $\phi$ and also the arbitrary choice of $\phi \to 0$ as $r \to \infty$.

3. Limits and jumps of Gauss' Law

a) Gravitational potential

First, we compute the mass of the ball: $M(r) = \begin{cases} 0, & 0 \leq r < a \\ \tfrac{4\pi}{3} \rho_0 (r^3 - a^3), & a \leq r < b \\ \tfrac{4\pi}{3} \rho_0 (b^3 - a^3), & b \leq r \end{cases}$

Then, seeking an axi-symmetric solution to Gauss' Law, we see to solve $4\pi r^2 \dd{\phi}{r} = -4\pi G M(r).$

This gives $\phi = \tfrac{4\pi}{3} G \rho_0 \begin{cases} \tfrac{3}{2}(b^2 - a^2), & 0 \leq r < a \\ -\tfrac{1}{2}r^2 - \tfrac{a^3}{r} + \tfrac{3}{2}b^2, & a \leq r < b \\ \tfrac{1}{r}(b^3 - a^3), & b \leq r \end{cases}$

once we have imposed the continuity of $\phi$ at $r = a$, $r = b$, and that $\phi \to 0$ as $r \to \infty$.

b) What is the limit

Note that in the limit that $b \to a$, \begin{align*} b^2 - a^2 &= 2a(b-a) + \ldots \\ b^3 - a^3 &= 3a^2(b-a) + \ldots \end{align*}

In this limit, the annular region $a \leq r \leq b$ shrinks down to a thin shell, and we will not worry about the behaviour of the potential within this region. Taking the limit of $b \to a$ for the potential gives $\phi = \tfrac{4\pi}{3} G \begin{cases} 3a (b-a)\rho_0 & 0 \leq r < a \\ 3a (b-a)\rho_0 \left(\tfrac{a}{r}\right) & 0 \leq r < a \end{cases}$

whose values agree on $r = a$, and we then set $\rho_0(b-a) = \sigma$.

c) Jump across the boundary

From the previous part, $\phi = \begin{cases} 4\pi G a\sigma & 0 \leq r < a \\ 4\pi G a\sigma \left(\tfrac{a}{r}\right) & 0 \leq r < a \end{cases}$

so that $\dd{\phi}{r} = \begin{cases} 0 & 0 \leq r < a \\ 4\pi G a^2 \sigma \left(\tfrac{1}{r^2}\right) & 0 \leq r < a \end{cases}$

and the jump in $-\nabla \phi/4\pi$ is given by $\left[-\frac{\nabla \phi}{4\pi}\right]_{a^-}^{a^+} = G\sigma \mb{n}.$

Notice that in general and from part a) the total mass is $M = \tfrac{4\pi}{3} \rho_0 (b^3 - a^3) \sim (4\pi a^2) \sigma,$

which is the surface area of a sphere, multiplied by the effective density, $\sigma$.