5th of November, 2008
The equilibrium is stable if when the charge is displaced from the equilibrium position by a small distance, the forces acting on it push it back towards the equilibrium point (and not away from it).
Suppose that there is a stable equilibrium point in a static electric field. The criterion of stability requires that the electric field point inwards on every point of a surface enclosing the equilibrium point. This would mean that the surface integral of the electric field in non-zero on this surface—a result which contradicts that the equilibrium point is in vacuum, i.e. there are no charges inside the surface. Therefore, the original assumption, that there exists a stable equilibrium point, must be false.
But it can be proven that the Laplcaian of any component of the magnetic field is 0 in vaccum. Taking the curl of both sides of the equation $\nabla \times \overrightarrow{B}=0$, and using that $\nabla \cdot \overrightarrow{B}=0$, we get
$$\begin{array}{cc}\begin{array}{rl}\nabla \times (\nabla \times \overrightarrow{B})& =0\\ \nabla (\nabla \cdot \overrightarrow{B})-{\nabla}^{2}\overrightarrow{B}& =0\\ ({\nabla}^{2}{B}_{x},{\nabla}^{2}{B}_{y},{\nabla}^{2}{B}_{z})& =0.\end{array}& \end{array}$$ |
The divergence of $\overrightarrow{F}$ is zero, therefore, in analogy with the case of the electric field, a small magnet (with fixed orientation) cannot have a stable equilibrium point in a magnetic field.
Let us calculate the divergence of the force acting on a piece of magnetizable material:
But ${\nabla}^{2}{B}^{2}$ can be shown to be positive or zero:
$$\begin{array}{cc}\begin{array}{rl}{\nabla}^{2}{B}^{2}& ={\nabla}^{2}{B}_{x}^{2}+{\nabla}^{2}{B}_{y}^{2}+{\nabla}^{2}{B}_{z}^{2}\\ & =\nabla (2{B}_{x}\nabla {B}_{x})+\cdots \\ & =2|\nabla {B}_{x}{|}^{2}+2{B}_{x}{\nabla}^{2}{B}_{x}+\cdots \\ & =2|\nabla {B}_{x}{|}^{2}+\cdots \ge 0,\end{array}& \end{array}$$ |
and therefore $\nabla \cdot {\overrightarrow{F}}_{m}\ge 0$.
With a reasoning similar to the one used in the previous point it can be shown that a force field having $\nabla \cdot \overrightarrow{F}\ge 0$ only has unstable equilibrium points. (The integral over the Gaussian surface drawn around the equilibrium point is positive or zero, therefore there must exist at least some points on it where $\overrightarrow{F}$ is pointing outwards, or $\overrightarrow{F}$ must be zero everywhere.)
Therefore neihter a small piece of magnetic material with $k>0$, nor a little magnet has stable equilibrium points in a static magnetic field. But for a diamagnet $k<0$, i.e. the magnetic moment is always oriented antiparallel to $\overrightarrow{B}$, so a diamagnet does have stable equilibrium points.
Let ${n}_{i}$ denote the index of refraction of the $i$th layer from the ground, and ${\phi}_{i}$ denote the altitude of the star when viewed from that layer. Writing Snell’s law of refraction for the consecutive layers,
$$\begin{array}{llll}\hfill {n}_{1}sin(\pi \u22152-{\phi}_{1})& ={n}_{2}sin(\pi \u22152-{\phi}_{2})\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {n}_{2}sin(\pi \u22152-{\phi}_{2})& ={n}_{3}sin(\pi \u22152-{\phi}_{3})\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \dots \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {n}_{i}sin(\pi \u22152-{\phi}_{i})& ={n}_{i+1}sin(\pi \u22152-{\phi}_{i+1})\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \dots \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$we see that ${n}_{1}sin(\pi \u22152-{\phi}_{1})={n}_{i}sin(\pi \u22152-{\phi}_{i})$ for any $i$. As we go very high up, the atmosphere gets thinner, so $n$ approaches 1 and $\phi $ approaches ${\phi}_{r}$. Thus the real altitude of the star can be obtained from the equation
$${n}_{\mathrm{\text{ground}}}sin(\pi \u22152-{\phi}_{v})=sin(\pi \u22152-{\phi}_{r}$$ |
$$\begin{array}{cc}\begin{array}{rl}{W}_{\mathrm{\text{ion}}}& =-\frac{1}{4\pi {\epsilon}_{0}}\frac{2{q}^{2}}{a}+\frac{1}{4\pi {\epsilon}_{0}}\frac{2{q}^{2}}{2a}-\frac{1}{4\pi {\epsilon}_{0}}\frac{2{q}^{2}}{3a}+\frac{1}{4\pi {\epsilon}_{0}}\frac{2{q}^{2}}{4a}-\cdots \\ & =-\frac{1}{4\pi {\epsilon}_{0}}\frac{2{q}^{2}}{a}\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots \phantom{\rule{0em}{0ex}}\right).\end{array}& \end{array}$$ |
The Taylor expansion of $ln(1+x)$ is
Substituting $x=1$ we find that
and therefore
Notice that if we add the energy of all ions in the crystal, we get double the energy of the complete crystal. (If we have only two ions, the energy of the complete system is the energy of one ion in the electric field of the other. Adding the energy of both gives twice this value.) So the electrostatic energy of the crystal per ion is
Now let us examine the problem from the reference frame moving together with the ball. The electrostatic analogy still holds in this reference frame. In this reference frame the ball is not moving, but the water is flowing past the ball. Since the water cannot flow into or out of a solid ball, on the surface of the ball it must be true that
$${v}_{\perp}=-\partial \psi \u2215\partial r=0.$$ | (A) |
($r$ is the radial coordinate measured from the centre of the ball.) Very far from the ball the flow is constant, $\overrightarrow{v}={\overrightarrow{v}}_{0}$ (where $-{v}_{0}$ is the velocity of the ball in the original reference frame.)
How should the homogeneous flow field ${\overrightarrow{v}}_{0}$ be changed so that condition (A) will be satisfied? We need to find a field ${\overrightarrow{v}}^{\prime}$ which vanishes at infinity, cancels the normal component of ${\overrightarrow{v}}_{0}$ on the surface of the sphere, and satisfies $\nabla \cdot {\overrightarrow{v}}^{\prime}=\nabla \times {\overrightarrow{v}}^{\prime}=0$. The sum ${\overrightarrow{v}}_{0}+{\overrightarrow{v}}^{\prime}$ will be the solution.
We have seen a similar problem in electrostatics: a conducting sphere placed in a homogeneous electric field. The difference is that there the tangential (and not normal) component of the electric field needed to be cancelled on the surface of the sphere. There we found that the distortion in the field caused by the sphere is a dipole field.
It turns out that a suitable chosen dipole field can cancel not only the tangential component of a homogeneous field on the surface of a sphere, but the normal component too. A dipole potential has the form
where $\overrightarrow{p}$ is a constant vector, and the dipole field is
Let us choose $\overrightarrow{p}$ so that it is parallel to ${\overrightarrow{v}}_{0}$ and calculate the normal components of ${\overrightarrow{v}}_{0}$ and ${\overrightarrow{v}}^{\prime}$ at the surface of the sphere. We shall use spherical coordinates, with $\theta $ being the angle between $\overrightarrow{p}$ and $\overrightarrow{r}$. Let $R$ be the radius of the sphere.
$$\begin{array}{llll}\hfill {v}_{0\perp}& ={\overrightarrow{v}}_{0}\cdot \widehat{r}={v}_{0}cos\theta \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {v}_{\perp}^{\prime}& ={\overrightarrow{v}}^{\prime}\cdot \widehat{r}=\frac{3pcos\theta -pcos\theta}{{R}^{3}}=\frac{2p}{{R}^{3}}cos\theta \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$$If $p$ is chosen so that $2p\u2215{R}^{3}=-{v}_{0}$ then the ${\overrightarrow{v}}^{\prime}$ will cancel the normal component of ${\overrightarrow{v}}_{0}$ on the surface of the sphere. The solution is
in the reference frame moving together with the ball.