Difference between revisions of "Binary systems"

Let the orbit lie in the $xy$-plane. Then the masses have trajectories \[\left( R\cos \omega t,\ R\sin \omega t,0 \right), \quad \left( -R\cos \omega t,\ -R\sin \omega t,0 \right)\] From the definition of the second mass moment \[I^{\alpha\beta}(t)=\int d^{3} x\; \rho (t,\mathbf{x})x^{\alpha}x^{\beta},\] ($\rho$ is the rest-mass density) we find for binary system \[I^{\alpha\beta}=MR^{2} \left( \begin{matrix} 1+\cos 2\omega t & \sin 2\omega t & 0 \\ \sin 2\omega t & 1-\cos 2\omega t & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right).\] As we have seen above in the long-wavelength approximation $\lambda \gg R$ and the distant-source approximation $r\gg R$ the strain tensor is \[ \bar{h}^{\alpha\beta}(t,\mathbf{x}) =\frac{2}{r}\ddot{I}^{\alpha\beta}(t-r)\] Therefore, \[\bar{h}^{\alpha\beta} =-\frac{8\omega^2 MR^{2}}{r} \left( \begin{matrix} \cos \left[ 2\omega (t-r) \right] & \sin \left[ 2\omega (t-r) \right] & 0 \\ \sin \left[ 2\omega (t-r) \right] & -\cos \left[ 2\omega (t-r) \right] & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right)\] Note that this representation of $\bar{h}^{\alpha\beta}$ is already in transverse traceless gauge for propagation in the $z$ direction. Also, the frequency of the emitted radiation is twice the orbital frequency by the symmetry of the problem.

The components of $Q^{\mu\nu}$ were calculated in the previous problem, so the quadrupole formula can be immediately applied to find the luminosity. The trace $Q_{\alpha}^{\alpha}=2MR^{2}$ is independent of time, so the time derivates of $I^{\alpha\beta}$ and $Q^{\alpha\beta}$ coincide. Averaging over the orbital period provides an additional factor of $1/2$ . The result for ${{L}_{GW}}$ is then \[L_{GW}=\frac{128}{5}M^{2}R^{4}\omega^{6}.\]

Let $R$ be the distance between the components. Then the quadrupole formula gives \[L_{GW}=\frac{32G}{5c^5} \Big(\frac{M_{1}M_{2}}{M_{1}+M_{2}}\Big)^{2} R^{4}\omega^6.\] Frequency $\omega$ is related to $R$ through \[\omega^{2}R^{3}=G(M_{1}+M_{2}),\] so for the gravitational luminosity we finally obtain \[L_{GW}=\frac{32}{5}\frac{G^{4}}{c^5} \frac{M_{1}^{2}M_{2}^{2}(M_{1}+M_{2})}{R^5}.\]

From the virial theorem $2\left\langle T \right\rangle =-\left\langle U \right\rangle$, the energy of the system is \[E=-G\frac{M_1 M_2}{2R},\] so \[\dot{R}=\frac{2R^{2}}{GM_{1}M_{2}}\frac{dE}{dt}.\] Then from $dE/dt=-L_{GW}$ and using the expression for the gravitational luminosity, we obtain \[\dot{R}=-\frac{64G^{3}}{5c^{2}} \frac{M_{1}M_{2}(M_{1}+M_{2})}{R^3}.\] Integrating from present time $t$ to the future time of coalescence ${{t}_{coal}}$, we have $R(t)$: \[ R(t)=2\;\Big[\frac{16}{5} \frac{G^{3}M_{1}M_{2}(M_1 +M_2)}{c^5} (t_{coal}-t)\Big]^{1/4}.\]

For a binary with $M_1 =M_2 =M/2$ a rough estimate would be \[t_{chirp}\sim\frac{MV^{2}}{L_{GW}} =\frac{5M}{2}\Big(\frac{R}{M}\Big)^{4}.\] More accurate evaluation would use the time evolution of $R(t)$: for $M_1 =M_2 =M/2$ we get \[t_{chirp}=\frac{5}{16}\frac{c^5 R^4}{G^3 M^3}.\] This is of the order of $45\times 10^9$ years.

From the previous problem for gravitational wave frequency for a binary with components of equal masses we get \begin{align*} f_{GW}&=\frac{\omega }{\pi } =\frac{1}{\pi }\Big(\frac{GM}{4R^3}\Big)^{1/2} =\frac{2^{1/8}5^{3/8}}{8\pi} \Big(\frac{c^3}{GM}\Big)^{5/8}(t_{coal}-t)^{-3/8}=\\ &=(1.9Hz)\Big(\frac{1.4M_{\odot}}{M}\Big)^{5/8} \Big(\frac{days}{\tau}\Big)^{3/8}, \quad \tau \equiv t_{coal}-t. \end{align*} Hence the mass of the compact binary system can be derived from the frequency evolution.