Period Shift in Hulse-Taylor Binaries

Classical Mechanics Level 2

Suppose that an inspiraling black hole binary system is at radius $R_0$ at time $t=0$ and that the radius of the binary changes as

$\frac{dR}{dt} = -\frac{\kappa}{R^3}$

for some constant $\kappa$ . What is the period shift in the rotation of the binary as a function of time (taking the shift to be zero at $t=0$ ), noting that the frequency of rotation in a Hulse-Taylor binary is

$\Omega = \left(\frac{GM}{4R^3}\right)^{1/2}?$

\frac{2\pi}{\sqrt{GM}} (R_0^2 - 2\kappa t)^{3/8}-2\pi \sqrt{\frac{R_0^3}{GM}}

\frac{4\pi}{\sqrt{GM}} (R_0^4 - 4\kappa t)^{3/8}-4\pi \sqrt{\frac{R_0^3}{GM}}

\frac{2\pi}{\sqrt{2GM}} (R_0^4 - \kappa t)^{1/2}-2\pi \sqrt{\frac{R_0^3}{2GM}}

\frac{4\pi}{\sqrt{3GM}} (4\kappa t)^{1/4}-4\pi \sqrt{\frac{R_0^3}{3GM}}

1 solution

Matt DeCross
Feb 12, 2016

Solving the differential equation for the radius as a function of time by the separate-and-integrate method: $R^4 - R_0^4 = -4\kappa t.$ This rearranges to: $R(t) = (R_0^4 - 4\kappa t)^{1/4}$ Plugging into the frequency of rotation $\Omega$ and using the fact that: $T = \frac{2\pi}{\Omega}$ one obtains the period as a function of time, $T(t) = \frac{4\pi}{\sqrt{GM}} (R_0^4 - 4\kappa t)^{3/8}.$ Subtracting from this $T(0)$ gives the result.

@Matt DeCross , shouldn't the exponent be $3/8$ ?

A Former Brilliant Member - 3 years, 5 months ago

Ack, good catch. I will report the error.

Matt DeCross - 3 years, 3 months ago

Period Shift in Hulse-Taylor Binaries

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