Subfield: Mechanics

LIGO-GW150914

In 2015, the gravitational-wave observatory LIGO detected, for the first time, the passing of gravitational waves (GW) through Earth. This event, named GW150914, was triggered by waves produced by two black holes that were orbiting on quasi-circular orbits. This problem will make you estimate some physical parameters of the system, from the properties of the detected signal.

Part A: Newtonian (conservative) orbits
Context question:
A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is,

$$
M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 .
\tag{!}
$$

The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as

$$
\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}}
\tag{2}
$$

where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$.
Context answer:
\boxed{$n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$}


Context question:
A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as:

$$
E=A(\mu, \Omega, L)-G \frac{M \mu}{L},
\tag{3}
$$

where

$$
\mu \equiv \frac{M_{1} M_{2}}{M_{1}+M_{2}}, \quad M \equiv M_{1}+M_{2}
\tag{4}
$$

are the reduced mass and total mass of the system, $\Omega$ is the angular velocity of each mass and $L$ is the total separation $L=r_{1}+r_{2}$. Obtain the explicit form of the term $A(\mu, \Omega, L)$.
Context answer:
\boxed{$A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2}$}


Context question:
A.3 Equation 3 can be simplified to $E=\beta G \frac{M \mu}{L}$. Determine the number $\beta$.
Context answer:
\boxed{$\beta=-\frac{1}{2}$}


Extra Supplementary Reading Materials:

Part B: Introducing relativistic dissipation

The correct theory of gravity, General Relativity, was formulated by Einstein in 1915, and predicts that gravity travels with the speed of light. The messengers carrying information about the interaction are called GWs. GWs are emitted whenever masses are accelerated, making the system of masses lose energy.

Consider a system of two point-like particles, isolated from the rest of the Universe. Einstein proved that for small enough velocities the emitted GWs: 1) have a frequency which is twice as large as the orbital frequency; 2 ) can be characterized by a luminosity, i.e. emitted power $\mathcal{P}$, which is dominated by Einstein's quadrupole formula,

$$
\mathcal{P}=\frac{G}{5 c^{5}} \sum_{i=1}^{3} \sum_{j=1}^{3}\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right)\left(\frac{\mathrm{d}^{3} Q_{i j}}{\mathrm{~d} t^{3}}\right)
\tag{5}
$$

Here, $c$ is the velocity of light $c \simeq 3 \times 10^{8} \mathrm{~m} / \mathrm{s}$. For a system of 2 pointlike particles orbiting on the $x-y$ plane, $Q_{i j}$ is the following table ( $i, j$ label the row/column number)

$$
Q_{11}=\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 x_{A}^{2}-y_{A}^{2}\right), \quad Q_{22} =\sum_{A=1}^{2} \frac{M_{A}}{3}\left(2 y_{A}^{2}-x_{A}^{2}\right), \quad Q_{33}=-\sum_{A=1}^{2} \frac{M_{A}}{3}\left(x_{A}^{2}+y_{A}^{2}\right),
\tag{6}
$$
$$
Q_{12} =Q_{21}=\sum_{A=1}^{2} M_{A} x_{A} y_{A},
\tag{7}
$$

and $Q_{i j}=0$ for all other possibilities. Here, $\left(x_{A}, y_{A}\right)$ is the position of mass A in the center-of-mass frame.
Context question:
B.1 For the circular orbits described in A.2 the components of $Q_{i j}$ can be expressed as a function of time $t$ as:

$$
Q_{i i}=\frac{\mu L^{2}}{2}\left(a_{i}+b_{i} \cos k t\right), \quad Q_{i j} \stackrel{i \neq j}{=} \frac{\mu L^{2}}{2} c_{i j} \sin k t .
\tag{8}
$$

Determine $k$ in terms of $\Omega$ and the numerical values of the constants $a_{i}, b_{i}, c_{i j}$.
Context answer:
$$
k=2 \Omega, \quad a_{1}=a_{2}=\frac{1}{3}, a_{3}=-\frac{2}{3}, \quad b_{1}=1, b_{2}=-1, b_{3}=0, c_{12}=c_{21}=1, c_{i j} \stackrel{\text { otherwise }}{=} 0
$$


Context question:
B.2 Compute the power $\mathcal{P}$ emitted in gravitational waves for that system, and obtain:

$$
\mathcal{P}=\xi \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6}
\tag{9}
$$

What is the number $\xi$ ? [If you could not obtain $\xi$, use $\xi=6.4$ in the following.]
Context answer:
\boxed{$\xi=\frac{32}{5}$}


Question: B.3 In the absence of GW emission the two masses will orbit on a fixed circular orbit indefinitely. However, the emission of GWs causes the system to lose energy and to slowly evolve towards smaller circular orbits. Obtain that the rate of change $\frac{\mathrm{d} \Omega}{\mathrm{d} t}$ of the orbital angular velocity takes the form

$$
\left(\frac{\mathrm{d} \Omega}{\mathrm{d} t}\right)^{3}=(3 \xi)^{3} \frac{\Omega^{11}}{c^{15}}\left(G M_{\mathrm{c}}\right)^{5}
\tag{10}
$$

where $M_{\mathrm{c}}$ is called the chirp mass. Obtain $M_{\mathrm{c}}$ as a function of $M$ and $\mu$. This mass determines the increase in frequency during the orbital decay. [The name "chirp" is inspired by the high pitch sound (increasing frequency) produced by small birds.]

Solution:
Now we assume a sequency of Keplerian orbits, with decreasing energy, which is being taken from the system by the GWs.

First, from (9), differentiating with respect to time,

$$
\frac{\mathrm{d} E}{\mathrm{~d} t}=\frac{G M \mu}{2 L^{2}} \frac{\mathrm{d} L}{\mathrm{~d} t}
\tag{15}
$$

Since this loss of energy is due to GWs, we equate it with (minus) the luminosity of GWs, given by (14)

$$
\frac{G M \mu}{2 L^{2}} \frac{\mathrm{d} L}{\mathrm{~d} t}=-\frac{32}{5} \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6}
\tag{16}
$$

We can eliminate the $L$ and $\mathrm{d} L / \mathrm{d} t$ dependence in this equation in terms of $\Omega$ and $\mathrm{d} \Omega / \mathrm{d} t$, by using Kepler's third law (8), which relates:

$$
L^{3}=G \frac{M}{\Omega^{2}}, \quad \frac{\mathrm{d} L}{\mathrm{~d} t}=-\frac{2}{3} \frac{L}{\Omega} \frac{\mathrm{d} \Omega}{\mathrm{d} t}
\tag{17}
$$



Substituting in (16), we obtain:

$$
\left(\frac{\mathrm{d} \Omega}{\mathrm{d} t}\right)^{3}=\left(\frac{96}{5}\right)^{3} \frac{\Omega^{11}}{c^{15}} G^{5} \mu^{3} M^{2} \equiv\left(\frac{96}{5}\right)^{3} \frac{\Omega^{11}}{c^{15}}\left(G M_{\mathrm{c}}\right)^{5}
\tag{18}
$$

Final answer: $M_{\mathrm{c}}=(\mu^{3} M^{2})^{1 / 5}$