Physics

Physics · mechanics

Problem

B.4 Using the information provided above, relate the orbital angular velocity Ω\Omega with the GW frequency fGWf_{\mathrm{GW}}. Knowing that, for any smooth function F(t)F(t) and a≠1a \neq 1, dF(t)dt=χF(t)a⇒F(t)1−a=χ(1−a)(t−t0)(11)\frac{\mathrm{d} F(t)}{\mathrm{d} t}=\chi F(t)^{a} \quad \Rightarrow \quad F(t)^{1-a}=\chi(1-a)\left(t-t_{0}\right) \tag{11} where χ\chi is a constant and t0t_{0} is an integration constant, show that (10) implies that the GW frequency is fGW−8/3=8π8/3ξ(GMcc3)(2/3)+p(t0−t)2−p(12)f_{\mathrm{GW}}^{-8 / 3}=8 \pi^{8 / 3} \xi\left(\frac{G M_{c}}{c^{3}}\right)^{(2 / 3)+p}\left(t_{0}-t\right)^{2-p} \tag{12} and determine the constant pp. Given constants: 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 M1,M2M_{1}, M_{2}, at locations r⃗1,r⃗2\vec{r}_{1}, \vec{r}_{2}, respectively, with respect to the center-of-mass of the system, that is, M1r1→+M2r2→=0.(!)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 M1M_{1} can be expressed as d2r⃗1 dt2=−αr⃗1r1n(2)\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}} \tag{2} where r1=∣r⃗1∣,r2=∣r⃗2∣r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|. Find n∈Nn \in \mathbb{N} and α=α(G,M1,M2)\alpha=\alpha\left(G, M_{1}, M_{2}\right), where GG is Newton's constant [G≃6.67×10−11 N m2 kg−2]\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]. Context answer: n=3n=3 , α=GM23(M1+M2)2\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(μ,Ω,L)−GMμL,(3)E=A(\mu, \Omega, L)-G \frac{M \mu}{L}, \tag{3} where μ≡M1M2M1+M2,M≡M1+M2(4)\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, Ω\Omega is the angular velocity of each mass and LL is the total separation L=r1+r2L=r_{1}+r_{2}. Obtain the explicit form of the term A(μ,Ω,L)A(\mu, \Omega, L). Context answer: A(μ,Ω,L)=12μL2Ω2A(\mu, \Omega, L)=\frac{1}{2} \mu L^{2} \Omega^{2} Context question: A.3 Equation 3 can be simplified to E=βGMμLE=\beta G \frac{M \mu}{L}. Determine the number β\beta. Context answer: β=−12\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 P\mathcal{P}, which is dominated by Einstein's quadrupole formula, P=G5c5∑i=13∑j=13(d3Qij dt3)(d3Qij dt3)(5)\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, cc is the velocity of light c≃3×108 m/sc \simeq 3 \times 10^{8} \mathrm{~m} / \mathrm{s}. For a system of 2 pointlike particles orbiting on the x−yx-y plane, QijQ_{i j} is the following table ( i,ji, j label the row/column number) Q11=∑A=12MA3(2xA2−yA2),Q22=∑A=12MA3(2yA2−xA2),Q33=−∑A=12MA3(xA2+yA2),(6)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} Q12=Q21=∑A=12MAxAyA,(7)Q_{12} =Q_{21}=\sum_{A=1}^{2} M_{A} x_{A} y_{A}, \tag{7} and Qij=0Q_{i j}=0 for all other possibilities. Here, (xA,yA)\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 QijQ_{i j} can be expressed as a function of time tt as: Qii=μL22(ai+bicos⁡kt),Qij=i≠jμL22cijsin⁡kt.(8)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 kk in terms of Ω\Omega and the numerical values of the constants ai,bi,cija_{i}, b_{i}, c_{i j}. Context answer: k=2Ω,a1=a2=13,a3=−23,b1=1,b2=−1,b3=0,c12=c21=1,cij= otherwise 0k=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 P\mathcal{P} emitted in gravitational waves for that system, and obtain: P=ξGc5μ2L4Ω6(9)\mathcal{P}=\xi \frac{G}{c^{5}} \mu^{2} L^{4} \Omega^{6} \tag{9} What is the number ξ\xi ? [If you could not obtain ξ\xi, use ξ=6.4\xi=6.4 in the following.] Context answer: ξ=325\xi=\frac{32}{5} Context 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 dΩdt\frac{\mathrm{d} \Omega}{\mathrm{d} t} of the orbital angular velocity takes the form (dΩdt)3=(3ξ)3Ω11c15(GMc)5(10)\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 McM_{\mathrm{c}} is called the chirp mass. Obtain McM_{\mathrm{c}} as a function of MM and μ\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.] Context answer: Mc=(μ3M2)1/5M_{\mathrm{c}}=(\mu^{3} M^{2})^{1 / 5}

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