Solution

For a localized charge and current distribution, take the retarded vector potential

A (x, t) = \frac{μ _{0}}{4 π} \int \frac{J ( x ^{'} , t - ∣ x - x ^{'} ∣/ c )}{∣ x - x ^{'} ∣} d^{3} x^{'} .

Let

R = ∣ x ∣

n = x / R

, and

t_{R} = t - R / c

. If the source size is much smaller than both

R

and the characteristic radiation wavelength, the leading dipole term is

A (x, t) ≃ \frac{μ _{0}}{4 π R} \int J (x^{'}, t_{R}) d^{3} x^{'} = \frac{μ _{0}}{4 π R} \overset{p}{˙} (t_{R}),

where the last identity follows from charge conservation.

Keeping only the

1/ R

terms in the radiation zone gives the transverse fields

E_{rad} = \frac{μ _{0}}{4 π R} n \times (n \times \overset{p}{¨} (t_{R})), B_{rad} = \frac{1}{c} n \times E_{rad} .

The radial Poynting vector therefore yields

\frac{d P}{d Ω} = \frac{μ _{0}}{16 π ^{2} c} ∣ n \times \overset{p}{¨} (t_{R}) ∣^{2} .

Using

\int n_{i} n_{j} d Ω = \frac{4 π}{3} δ_{ij}

or equivalently

\int sin^{2} θ d Ω = 8 π /3

P ≃ \frac{μ _{0}}{6 π c} ∣ \overset{p}{¨} (t - R / c) ∣^{2} .

This requires a localized, nonrelativistic source of size

d ≪ λ

, observation distance

R ≫ d

and

R ≫ λ

, and neglect of higher multipoles and faster-decaying near fields.

For the pulsar, write its magnetic dipole moment as

m (t)

to distinguish it from the electric dipole above. The magnetic dipole radiation formula differs by a factor

c^{- 2}

P = \frac{μ _{0}}{6 π c ^{3}} ∣ \overset{m}{¨} ∣^{2} .

Here

\overset{m}{¨} (t) = - p_{0} Ω^{2} sin α [cos (Ω t) x + sin (Ω t) y],

and hence

P = C Ω^{4}, C = \frac{μ _{0} p _{0}^{2} sin ^{2} α}{6 π c ^{3}} .

On the slow spin-down timescale, energy conservation gives the magnetic-dipole spin-down equation

I Ω \dot{Ω} = - C Ω^{4},

\frac{1}{Ω ( t ) ^{2}} = \frac{1}{Ω _{0}^{2}} + \frac{2 C}{I} t .

Half of the initial rotational energy remains when

Ω = Ω_{0} / 2

. Therefore

t_{1/2} = \frac{I}{2 C Ω _{0}^{2}} = \frac{6 π M R ^{2} c ^{3}}{5 μ _{0} p _{0}^{2} Ω _{0}^{2} sin ^{2} α},

where

I = 2 M R^{2} /5

. Since

E_{0} = I Ω_{0}^{2} /2

, the equivalent expression is

t_{1/2} = \frac{6 π M ^{2} R ^{4} c ^{3}}{25 μ _{0} p _{0}^{2} E _{0} sin ^{2} α} .

Solved by gpt-5.6-sol high.

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