Earnshaw's theorem for electrostatics rests only on two facts: the potential energy…

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Relativity & Gravitation · Newtonian gravitation as a field theory

The Boundary-Value Problem, Ch The field as a differential equation: Poisson, Laplace and the well-posed problem

Problem

Earnshaw's theorem for electrostatics rests only on two facts: the potential energy of the test object is proportional to a potential obeying Laplace's equation in the region free of sources, and nothing about the sign of the charge enters the argument. The Newtonian gravitational potential Φ(r)\Phi(\mathbf{r}) is defined so that the gravitational field is g=−∇Φ\mathbf{g} = -\nabla\Phi and obeys ∇2Φ=4πGρm\nabla^2\Phi = 4\pi G\rho_m, with ρm\rho_m the mass density and G=6.674×10−11 N m2 kg−2G = 6.674\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}; in vacuum ∇2Φ=0\nabla^2\Phi = 0. The potential energy of a small test mass mq>0m_q > 0 placed at r\mathbf{r} is U(r)=mqΦ(r)U(\mathbf{r}) = m_q\Phi(\mathbf{r}). Two point masses, each M=5.0×105 kgM = 5.0\times10^{5}\,\mathrm{kg}, are fixed at z=+bz = +b and z=−bz = -b on the zz axis, with b=2.0 mb = 2.0\,\mathrm{m}, in otherwise empty space; gravity is the only force acting. A small test mass sits at the origin, midway between them. ((a)) Write Φ(z)\Phi(z) on the axis for −b<z<b-b < z < b and show the origin is an equilibrium point, dΦ/dz=0d\Phi/dz = 0 there. ((b)) Find d2Φ/dz2d^2\Phi/dz^2 at the origin and say whether a small axial displacement grows or is restored. ((c)) Using ∇2Φ=0\nabla^2\Phi = 0 at the origin (mass-free there) together with the rotational symmetry of the arrangement about the zz axis, find ∂2Φ/∂x2\partial^2\Phi/\partial x^2 and ∂2Φ/∂y2\partial^2\Phi/\partial y^2 at the origin, and say whether a small sideways displacement grows or is restored. ((d)) State whether any fixed arrangement of point masses could make the equilibrium of a test mass stable against a displacement in every direction at once, using the same mean-value argument as Earnshaw's theorem, and say whether the positive sign of mqm_q changes that conclusion.

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