A rotation-powered pulsar has period and period derivative (dimensionless)

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Astrophysics & Cosmology · Stellar death and compact objects

Stars from the Inside, Ch Neutron stars and pulsars: the densest matter, the maximum mass and the spinning magnet

Problem

A rotation-powered pulsar has period P=0.75 sP = 0.75\,\mathrm{s} and period derivative P˙=8.0×10−15\dot P = 8.0\times10^{-15} (dimensionless). Assume a moment of inertia I=1.0×1038 kg m2I = 1.0\times10^{38}\,\mathrm{kg\,m^2} and an orthogonal rotating dipole (α=90∘\alpha = 90^\circ). ((a)) Compute the spin-down luminosity E˙=4π2IP˙/P3\dot E = 4\pi^2 I\dot P/P^3. ((b)) Compute the equatorial surface magnetic field B=3.2×1015PP˙ TB = 3.2\times10^{15}\sqrt{P\dot P}\,\mathrm{T} (with PP in seconds). ((c)) Compute the characteristic age τc=P/(2P˙)\tau_c = P/(2\dot P), in years.

Browse the problem library