B.4 Using the information provided above, relate the orbital angular velocity Ω with the GW frequency fGW. Knowing that, for any smooth function F(t) and a=1,
dtdF(t)=χF(t)a⇒F(t)1−a=χ(1−a)(t−t0)(11)
where χ is a constant and t0 is an integration constant, show that (10) implies that the GW frequency is
fGW−8/3=8π8/3ξ(c3GMc)(2/3)+p(t0−t)2−p(12)
and determine the constant p.
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,M2, at locations r1,r2, respectively, with respect to the center-of-mass of the system, that is,
M1r1+M2r2=0.(!)
The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass M1 can be expressed as
dt2d2r1=−αr1nr1(2)
where r1=∣r1∣,r2=∣r2∣. Find n∈N and α=α(G,M1,M2), where G is Newton's constant [G≃6.67×10−11Nm2kg−2].
Context answer:
n=3 , α=(M1+M2)2GM23
Context question:
A.2 The total energy of the 2-mass system, in circular orbits, can be expressed as:
E=A(μ,Ω,L)−GLMμ,(3)
where
μ≡M1+M2M1M2,M≡M1+M2(4)
are the reduced mass and total mass of the system, Ω is the angular velocity of each mass and L is the total separation L=r1+r2. Obtain the explicit form of the term A(μ,Ω,L).
Context answer:
A(μ,Ω,L)=21μL2Ω2
Context question:
A.3 Equation 3 can be simplified to E=βGLMμ. Determine the number β.
Context answer:
β=−21
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, which is dominated by Einstein's quadrupole formula,
P=5c5Gi=1∑3j=1∑3(dt3d3Qij)(dt3d3Qij)(5)
Here, c is the velocity of light c≃3×108m/s. For a system of 2 pointlike particles orbiting on the x−y plane, Qij is the following table ( i,j label the row/column number)
Q11=A=1∑23MA(2xA2−yA2),Q22=A=1∑23MA(2yA2−xA2),Q33=−A=1∑23MA(xA2+yA2),(6)Q12=Q21=A=1∑2MAxAyA,(7)
and Qij=0 for all other possibilities. Here, (xA,yA) 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 Qij can be expressed as a function of time t as:
Qii=2μL2(ai+bicoskt),Qij=i=j2μL2cijsinkt.(8)
Determine k in terms of Ω and the numerical values of the constants ai,bi,cij.
Context answer:
k=2Ω,a1=a2=31,a3=−32,b1=1,b2=−1,b3=0,c12=c21=1,cij= otherwise 0
Context question:
B.2 Compute the power P emitted in gravitational waves for that system, and obtain:
P=ξc5Gμ2L4Ω6(9)
What is the number ξ ? [If you could not obtain ξ, use ξ=6.4 in the following.]
Context answer:
ξ=532
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 dtdΩ of the orbital angular velocity takes the form
(dtdΩ)3=(3ξ)3c15Ω11(GMc)5(10)
where Mc is called the chirp mass. Obtain Mc as a function of M and μ. 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)1/5