Binary Black Hole Inspiral

Classical Mechanics Level 4

Two black holes in a mutually orbiting binary system both have mass $M$ and orbit at radius $R$ to their center of rotation.

What is the magnitude of the rate of change of their radius (the rate at which they inspiral due to energy loss to gravitational radiation)?

Assumptions and Details:

The power radiated in gravitational waves by an inspiraling binary is $P = -\dfrac25 \dfrac{G^4 M^5}{c^5 R^5}.$
Apart from the expression for gravitational wave radiation, assume Newtonian mechanics applies.

\frac{2}{5} \frac{G^3 M^3}{c^5 R^3}

\frac{8}{5} \frac{G^3 M^3}{c^5 R^3}

\frac{16}{5} \frac{G^3 M^3}{c^5 R^3}

\frac{128}{5} \frac{G^3 M^3}{c^5 R^3}

1 solution

Matt DeCross
Feb 14, 2016

First, compute the orbital velocity $\Omega$ of the black holes. Equating the gravitational force with the centripetal force in the Newtonian approximation, we have:

$MR \Omega^2 = \frac{GM^2}{(2R)^2}.$

Note the factor of two multiplying the radius because the distance between masses is twice the distance to the center of rotation. Solving for $\Omega$ , find:

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

Given $\Omega$ , the total energy of the binary system is:

$E = 2\left(\frac12 MR^2 \Omega^2\right) - \frac{GM^2}{(2R)} = -\frac14 \frac{GM^2}{R}.$

Note again the factors of two: the first because both masses in the binary system are rotating, the second again because the distance to the center of rotation is half the distance between masses.

The power dissipated in gravitational waves is equal to the rate of orbital energy loss, and we have $P = \frac{dE}{dt}$ . Taking a time derivative of the orbital energy and equating the power lost in orbit and the power of the gravitational waves, have:

$-\frac25 \frac{G^4 M^5}{c^5 R^5} = \frac14 \frac{GM^2}{R^2} \frac{dR}{dt} .$

Solving for $\frac{dR}{dt}$ obtains the answer.

This is a much better version, now that you've given the means for readers to attempt a solution.

Michael Mendrin - 5 years, 4 months ago

You don't need to calculate the total energy, because P=dE/dt=dE/dR * dR/dt=F * dR/dT (where F is the gravitational force between them). Hence, dR/dt=P/F and that's it.

Alexis Warnier - 3 years, 6 months ago

You should write it up as a solution. It seems the better solution.

Desmond Campbell - 3 years, 5 months ago

I'm not a huge fan of this type of approach because it's very easy to make an error when symbol-pushing that clearly explaining in English would avoid. My main beef with the equations above is the replacement "dE/dR = F". Sure, we know generally that forces are (negative!) gradients of potentials, but E is the total mechanical energy and not the gravitational potential alone, and that sign has disappeared.

In fact if you do the computation you see that the "F" you get is the force from one black hole to the other (i.e. with corresp. distance 2R as in F=GM^2/(2R)^2), as opposed to say the force due to gravity on one black hole directed towards the center of mass of the binary (with corresp. distance R), providing the instantaneous centripetal acceleration. Especially since the parameter in the problem is R and not 2R, it isn't completely obvious a priori which "F" you would plug in without having computed "E". Maybe you already knew that dE/dR for a binary system gives the "F" directed from one body to the other, but this is not a reasonable assumption to me without independently deriving it - if I came across a claim of that order in a paper I would definitely verify it by hand myself. Moral of the story, brevity might be at the expense of clarity (or trustworthiness - see my problem https://brilliant.org/problems/attractive-force-of-spherical-capacitors-2/ where a similar sleight of hand actually gets you the wrong answer by a factor of 2).

Matt DeCross - 3 years ago

Binary Black Hole Inspiral

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