IPhO 2018 · Theory problem 1

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 (3.0 points)

A.1 1.0pt

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,

\[ M_1\vec r_1+M_2\vec r_2=0\,. \tag{1} \]

The stars are isolated from the rest of the Universe and moving at non-relativistic velocities. Using Newton's laws, the acceleration vector of mass \(M_1\) can be expressed as

\[ \frac{d^2\vec r_1}{dt^2} =-\alpha\frac{\vec r_1}{r_1^n}\,, \tag{2} \]

where \(r_1=|\vec r_1|,r_2=|\vec r_2|\). Find \(n\in\mathbb N\) and \(\alpha=\alpha(G,M_1,M_2)\), where \(G\) is Newton's constant \([G\simeq6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}]\).

A.2 1.0pt

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

\[ \mu\equiv\frac{M_1M_2}{M_1+M_2}\,, \qquad M\equiv M_1+M_2\,, \tag{4} \]

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 \(A(\mu,\Omega,L)\).

A.3 1.0pt

Equation 3 can be simplified to \(E=\beta G\frac{M\mu}{L}\). Determine the number \(\beta\).

Part B: Introducing relativistic dissipation (7.0 points)

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,

\[ \mathcal P=\frac{G}{5c^5} \sum_{i=1}^{3}\sum_{j=1}^{3} \left(\frac{d^3Q_{ij}}{dt^3}\right) \left(\frac{d^3Q_{ij}}{dt^3}\right)\,. \tag{5} \]

Here, \(c\) is the velocity of light \(c\simeq3\times10^8\,\mathrm{m/s}\). For a system of 2 pointlike particles orbiting on the \(x-y\) plane, \(Q_{ij}\) is the following table (\(i,j\) label the row/column number)

\[ \begin{aligned} Q_{11}&=\sum_{A=1}^{2}\frac{M_A}{3}(2x_A^2-y_A^2)\,,\\ Q_{22}&=\sum_{A=1}^{2}\frac{M_A}{3}(2y_A^2-x_A^2)\,,\\ Q_{33}&=-\sum_{A=1}^{2}\frac{M_A}{3}(x_A^2+y_A^2)\,, \end{aligned} \tag{6} \]

\[ Q_{12}=Q_{21}=\sum_{A=1}^{2}M_Ax_Ay_A\,, \tag{7} \]

and \(Q_{ij}=0\) for all other possibilities. Here, \((x_A,y_A)\) is the position of mass A in the center-of-mass frame.

B.1 1.0pt

For the circular orbits described in A.2 the components of \(Q_{ij}\) can be expressed as a function of time \(t\) as:

\[ Q_{ii}=\frac{\mu L^2}{2}(a_i+b_i\cos kt)\,, \qquad Q_{ij}\overset{i\ne j}{=}\frac{\mu L^2}{2}c_{ij}\sin kt\,. \tag{8} \]

Determine \(k\) in terms of \(\Omega\) and the numerical values of the constants \(a_i,b_i,c_{ij}\).

B.2 1.0pt

Compute the power \(\mathcal P\) emitted in gravitational waves for that system, and obtain:

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

What is the number \(\xi\)? [If you could not obtain \(\xi\), use \(\xi=6.4\) in the following.]

B.3 1.0pt

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 \(\frac{d\Omega}{dt}\) of the orbital angular velocity takes the form

\[ \left(\frac{d\Omega}{dt}\right)^3 =(3\xi)^3\frac{\Omega^{11}}{c^{15}}(GM_c)^5\,, \tag{10} \]

where \(M_c\) is called the chirp mass. Obtain \(M_c\) 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.]

B.4 2.0pt

Using the information provided above, relate the orbital angular velocity \(\Omega\) with the GW frequency \(f_{\mathrm{GW}}\). Knowing that, for any smooth function \(F(t)\) and \(a\ne1\),

\[ \frac{dF(t)}{dt}=\chi F(t)^a \quad\Rightarrow\quad F(t)^{1-a}=\chi(1-a)(t-t_0)\,, \tag{11} \]

where \(\chi\) is a constant and \(t_0\) is an integration constant, show that (10) implies that the GW frequency is

\[ f_{\mathrm{GW}}^{-8/3} =8\pi^{8/3}\xi \left(\frac{GM_c}{c^3}\right)^{(2/3)+p} (t_0-t)^{2-p} \tag{12} \]

and determine the constant \(p\).

On September 14, 2015 GW150914 was registered by the LIGO detectors, consisting of two L-shaped arms, each 4 km long. These arms changed by a relative length according to Fig. 1. The arms of the detector respond linearly to a passing gravitational wave, and the response pattern mimics the wave. This wave was created by two black holes on quasi-circular orbits; the loss of energy through gravitational radiation caused the orbit to shrink and the black holes to eventually collide. The collision point corresponds, roughly, to the peak of the signal after point D, in Fig. 1.

Figure 1

Figure 1. Strain, i.e. relative variation of the size of each arm, at the LIGO detector H1. The horizontal axis is time, and the points A, B, C, D correspond to \(t=0.000,0.009,0.034,0.040\) seconds, respectively.

B.5 1.0pt

From the figure, estimate \(f_{\mathrm{GW}}(t)\) at

\[ t_{\overline{\mathrm{AB}}}=\frac{t_{\mathrm B}+t_{\mathrm A}}{2} \qquad\text{and}\qquad t_{\overline{\mathrm{CD}}}=\frac{t_{\mathrm D}+t_{\mathrm C}}{2}\,. \tag{13} \]

Assuming that (12) is valid all the way until the collision (which strictly speaking is not true) and that the two objects have equal mass, estimate the chirp mass, \(M_c\), and total mass of the system, in terms of solar masses \(M_\odot\simeq2\times10^{30}\,\mathrm{kg}\).

B.6 1.0pt

Estimate the minimal orbital separation between the two objects at \(t_{\overline{\mathrm{CD}}}\). Hence estimate a maximum size for each object, \(R_{\max}\). Obtain \(R_\odot/R_{\max}\) to compare this size with the radius of our Sun, \(R_\odot\simeq7\times10^5\,\mathrm{km}\). Estimate also their orbital linear velocity at the same instant, \(v_{\mathrm{col}}\), comparing it with the speed of light, \(v_{\mathrm{col}}/c\).

Conclude that these are extremely fast moving, extremely compact objects indeed!

Official English text of the problem.

Answers

  • A.1: \(\displaystyle n=3,\qquad \alpha=\frac{GM_2^3}{(M_1+M_2)^2}\).
  • A.2: \(\displaystyle A(\mu,\Omega,L)=\frac12\mu\Omega^2L^2\).
  • A.3: \(\displaystyle \beta=-\frac12\).
  • B.1: With the masses on the \(x\)-axis at \(t=0\) and counterclockwise orbital motion, \[ k=2\Omega,\qquad (a_1,a_2,a_3)=\left(\frac13,\frac13,-\frac23\right),\qquad (b_1,b_2,b_3)=(1,-1,0). \] Also, \(c_{12}=c_{21}=1\); all other off-diagonal \(c_{ij}\) vanish.
  • B.2: \(\displaystyle \xi=\frac{32}{5}=6.4,\qquad \mathcal P=\frac{32G}{5c^5}\mu^2L^4\Omega^6\).
  • B.3: \(\displaystyle M_c=\mu^{3/5}M^{2/5}\), with \[ \frac{d\Omega}{dt} =3\xi\frac{(GM_c)^{5/3}}{c^5}\Omega^{11/3}. \]
  • B.4: \(\displaystyle f_{\mathrm{GW}}=\frac{\Omega}{\pi},\qquad p=1\), and \[ f_{\mathrm{GW}}^{-8/3} =8\pi^{8/3}\xi \left(\frac{GM_c}{c^3}\right)^{5/3}(t_0-t). \]
  • B.5: \(\displaystyle f_{\mathrm{GW}}(0.0045\,\mathrm s)\simeq56\,\mathrm{Hz}\), \(\displaystyle f_{\mathrm{GW}}(0.037\,\mathrm s)\simeq83\,\mathrm{Hz}\); \(\displaystyle M_c\simeq30M_\odot,\quad M\simeq69M_\odot\).
  • B.6: \(\displaystyle L_{\min}\simeq5.1\times10^2\,\mathrm{km}\), \(\displaystyle R_{\max}\simeq2.6\times10^2\,\mathrm{km}\), \(\displaystyle R_\odot/R_{\max}\simeq2.7\times10^3\). Each object's orbital speed is \(\displaystyle v_{\mathrm{col}}\simeq6.7\times10^7\,\mathrm{m/s}\), so \(\displaystyle v_{\mathrm{col}}/c\simeq0.22\).

Solution

A.1

Newton's law of gravitation gives the acceleration of the first star:

\[ \frac{d^2\vec r_1}{dt^2} =-GM_2\frac{\vec r_1-\vec r_2}{|\vec r_1-\vec r_2|^3}. \]

The center-of-mass condition implies

\[ \vec r_2=-\frac{M_1}{M_2}\vec r_1, \qquad \vec r_1-\vec r_2 =\frac{M_1+M_2}{M_2}\vec r_1. \]

Consequently,

\[ \frac{d^2\vec r_1}{dt^2} =-\frac{GM_2^3}{(M_1+M_2)^2}\frac{\vec r_1}{r_1^3}. \]

Comparison with (2) gives

\[ \boxed{n=3,\qquad \alpha=\frac{GM_2^3}{(M_1+M_2)^2}}. \]

A.2

The center-of-mass condition and \(L=r_1+r_2\) give

\[ r_1=\frac{M_2}{M}L, \qquad r_2=\frac{M_1}{M}L. \]

The stars have orbital speeds \(v_1=\Omega r_1\) and \(v_2=\Omega r_2\). Their total kinetic energy is therefore

\[ \begin{aligned} T &=\frac12 M_1\Omega^2r_1^2+\frac12 M_2\Omega^2r_2^2\\ &=\frac12\Omega^2L^2 \left(\frac{M_1M_2^2+M_2M_1^2}{M^2}\right)\\ &=\frac12\mu\Omega^2L^2. \end{aligned} \]

The gravitational potential energy, taking zero at infinite separation, is

\[ U=-\frac{GM_1M_2}{L}=-\frac{GM\mu}{L}. \]

Thus the term \(A\) is the kinetic energy:

\[ \boxed{A(\mu,\Omega,L)=\frac12\mu\Omega^2L^2}. \]

A.3

For the first star, gravity supplies the centripetal force:

\[ M_1\Omega^2r_1=\frac{GM_1M_2}{L^2}. \]

Using \(r_1=M_2L/M\), we obtain the circular-orbit relation

\[ \boxed{\Omega^2=\frac{GM}{L^3}}. \]

Hence

\[ T=\frac12\mu\Omega^2L^2=\frac{GM\mu}{2L}, \]

and

\[ E=T+U =\frac{GM\mu}{2L}-\frac{GM\mu}{L} =-\frac{GM\mu}{2L}. \]

Therefore,

\[ \boxed{\beta=-\frac12}. \]

B.1

Choose the time origin so that the first mass lies on the positive \(x\)-axis at \(t=0\), with counterclockwise motion. Then

\[ \begin{aligned} (x_1,y_1)&=(r_1\cos\Omega t,\ r_1\sin\Omega t),\\ (x_2,y_2)&=(-r_2\cos\Omega t,\ -r_2\sin\Omega t). \end{aligned} \]

From A.2,

\[ M_1r_1^2+M_2r_2^2=\mu L^2. \]

Substitution into the quadrupole components gives

\[ \begin{aligned} Q_{11} &=\frac{\mu L^2}{3}(2\cos^2\Omega t-\sin^2\Omega t) =\frac{\mu L^2}{2}\left(\frac13+\cos2\Omega t\right),\\ Q_{22} &=\frac{\mu L^2}{3}(2\sin^2\Omega t-\cos^2\Omega t) =\frac{\mu L^2}{2}\left(\frac13-\cos2\Omega t\right),\\ Q_{33} &=-\frac{\mu L^2}{3} =\frac{\mu L^2}{2}\left(-\frac23\right),\\ Q_{12}=Q_{21} &=\mu L^2\cos\Omega t\sin\Omega t =\frac{\mu L^2}{2}\sin2\Omega t. \end{aligned} \]

Thus

\[ \boxed{k=2\Omega}. \]

The diagonal coefficients are

\(i\) \(a_i\) \(b_i\)
\(1\) \(1/3\) \(1\)
\(2\) \(1/3\) \(-1\)
\(3\) \(-2/3\) \(0\)

The off-diagonal coefficients are

\[ \boxed{c_{12}=c_{21}=1,\qquad c_{13}=c_{31}=c_{23}=c_{32}=0}. \]

B.2

Write \(q=\mu L^2/2\) and \(k=2\Omega\). For the circular orbit, the nonzero third derivatives are

\[ \begin{aligned} \dddot Q_{11}&=qk^3\sin kt,\\ \dddot Q_{22}&=-qk^3\sin kt,\\ \dddot Q_{12}=\dddot Q_{21}&=-qk^3\cos kt. \end{aligned} \]

Both off-diagonal entries contribute to the double sum. Therefore,

\[ \begin{aligned} \sum_{i,j}(\dddot Q_{ij})^2 &=2q^2k^6\sin^2kt+2q^2k^6\cos^2kt\\ &=2q^2k^6\\ &=2\left(\frac{\mu L^2}{2}\right)^2(2\Omega)^6\\ &=32\mu^2L^4\Omega^6. \end{aligned} \]

The luminosity is consequently

\[ \boxed{\mathcal P =\frac{32G}{5c^5}\mu^2L^4\Omega^6}, \qquad \boxed{\xi=\frac{32}{5}=6.4}. \]

The result is independent of orbital phase.

B.3

We approximate the inspiral as a slowly evolving sequence of circular orbits. The masses remain constant, and the energy carried away by the waves obeys

\[ \frac{dE}{dt}=-\mathcal P. \]

Eliminating \(L\) with the circular-orbit relation,

\[ L=(GM)^{1/3}\Omega^{-2/3}, \]

gives

\[ E=-\frac12\mu(GM)^{2/3}\Omega^{2/3}. \]

Its time derivative is

\[ \frac{dE}{dt} =-\frac13\mu(GM)^{2/3}\Omega^{-1/3}\frac{d\Omega}{dt}. \]

The luminosity, expressed in terms of \(\Omega\), is

\[ \mathcal P =\xi\frac{G}{c^5}\mu^2(GM)^{4/3}\Omega^{10/3}. \]

Equating the energy-loss rate to \(-\mathcal P\) yields

\[ \frac{d\Omega}{dt} =3\xi\frac{G^{5/3}\mu M^{2/3}}{c^5}\Omega^{11/3}. \]

Cubing,

\[ \left(\frac{d\Omega}{dt}\right)^3 =(3\xi)^3\frac{\Omega^{11}}{c^{15}}G^5\mu^3M^2. \]

This has the required form if

\[ M_c^5=\mu^3M^2. \]

Thus

\[ \boxed{M_c=\mu^{3/5}M^{2/5} =\frac{(M_1M_2)^{3/5}}{(M_1+M_2)^{1/5}}}. \]

In particular, \(d\Omega/dt>0\): as the system loses energy, its orbit shrinks and its orbital frequency increases.

B.4

The orbital frequency is \(f_{\mathrm{orb}}=\Omega/(2\pi)\). The given factor of two between GW and orbital frequencies therefore implies

\[ \boxed{f_{\mathrm{GW}}=2f_{\mathrm{orb}}=\frac{\Omega}{\pi}}. \]

Taking the positive cube root of (10),

\[ \frac{d\Omega}{dt} =3\xi\frac{(GM_c)^{5/3}}{c^5}\Omega^{11/3}. \]

With \(\Omega=\pi f_{\mathrm{GW}}\), this becomes

\[ \frac{df_{\mathrm{GW}}}{dt} =3\xi\pi^{8/3} \left(\frac{GM_c}{c^3}\right)^{5/3} f_{\mathrm{GW}}^{11/3}. \]

Using (11) with \(a=11/3\), or directly differentiating \(f_{\mathrm{GW}}^{-8/3}\), gives

\[ \frac{d}{dt}\left(f_{\mathrm{GW}}^{-8/3}\right) =-8\xi\pi^{8/3} \left(\frac{GM_c}{c^3}\right)^{5/3}. \]

Integration yields

\[ \boxed{ f_{\mathrm{GW}}^{-8/3} =8\pi^{8/3}\xi \left(\frac{GM_c}{c^3}\right)^{5/3}(t_0-t) }. \]

Comparing the exponents with (12),

\[ \frac23+p=\frac53,\qquad 2-p=1, \]

so

\[ \boxed{p=1}. \]

Here \(t_0\) is the formal time at which this approximate inspiral model predicts a divergent frequency.

B.5

A and B mark consecutive zero crossings on either side of one positive crest; C and D do likewise for a later crest. Each interval therefore spans approximately half a GW cycle.

Approximating the frequency over each short interval by its midpoint value,

\[ f_{\mathrm{GW}}(t_{\overline{\mathrm{AB}}}) \simeq\frac{1}{2(t_{\mathrm B}-t_{\mathrm A})} =\frac{1}{2(0.009\,\mathrm s)} \simeq55.6\,\mathrm{Hz}, \]

\[ f_{\mathrm{GW}}(t_{\overline{\mathrm{CD}}}) \simeq\frac{1}{2(t_{\mathrm D}-t_{\mathrm C})} =\frac{1}{2(0.006\,\mathrm s)} \simeq83.3\,\mathrm{Hz}. \]

The corresponding times are

\[ t_{\overline{\mathrm{AB}}}=0.0045\,\mathrm s, \qquad t_{\overline{\mathrm{CD}}}=0.037\,\mathrm s, \]

so their difference is \(0.0325\,\mathrm s\).

Let \(f_{\mathrm{AB}}\) and \(f_{\mathrm{CD}}\) denote these frequency estimates. Subtracting the B.4 relation evaluated at the two times eliminates \(t_0\):

\[ f_{\mathrm{AB}}^{-8/3}-f_{\mathrm{CD}}^{-8/3} = 8\pi^{8/3}\xi \left(\frac{GM_c}{c^3}\right)^{5/3} \left(t_{\overline{\mathrm{CD}}}-t_{\overline{\mathrm{AB}}}\right). \]

Solving for the chirp mass,

\[ M_c=\frac{c^3}{G} \left[ \frac{f_{\mathrm{AB}}^{-8/3}-f_{\mathrm{CD}}^{-8/3}} {8\pi^{8/3}\xi \left(t_{\overline{\mathrm{CD}}}-t_{\overline{\mathrm{AB}}}\right)} \right]^{3/5}. \]

Using the supplied constants and \(\xi=6.4\),

\[ M_c\simeq6.02\times10^{31}\,\mathrm{kg}, \]

or, to appropriate precision,

\[ \boxed{M_c\simeq30M_\odot}. \]

For equal component masses, \(M_1=M_2=M/2\), so

\[ \mu=\frac{M}{4}, \qquad M_c=\left(\frac M4\right)^{3/5}M^{2/5} =\frac{M}{4^{3/5}}. \]

Thus

\[ M=4^{3/5}M_c \simeq1.38\times10^{32}\,\mathrm{kg}, \]

and

\[ \boxed{M\simeq69M_\odot}. \]

Each component therefore has a mass of approximately \(35M_\odot\).

B.6

At the later measured time,

\[ \Omega_{\overline{\mathrm{CD}}} =\pi f_{\mathrm{CD}} \simeq2.62\times10^2\,\mathrm{rad/s}. \]

The circular-orbit relation gives

\[ L=\left(\frac{GM}{\Omega_{\overline{\mathrm{CD}}}^2}\right)^{1/3}. \]

The equal-mass choice also gives the minimum total mass, and hence the minimum separation, compatible with the inferred chirp mass and frequency. Indeed,

\[ \frac{\mu}{M}=\frac{M_1M_2}{(M_1+M_2)^2}\le\frac14, \]

so

\[ M_c=M\left(\frac{\mu}{M}\right)^{3/5} \le\frac{M}{4^{3/5}}. \]

Equality holds for equal masses. Using the mass found in B.5,

\[ L_{\min} \simeq \left[ \frac{(6.67\times10^{-11})(1.383\times10^{32})} {(2.618\times10^2)^2} \right]^{1/3}\mathrm m, \]

which gives

\[ \boxed{L_{\min}\simeq5.1\times10^5\,\mathrm m =5.1\times10^2\,\mathrm{km}}. \]

Using a common radius for the equal-mass objects, their remaining distinct requires approximately \(2R\lesssim L\). Hence

\[ \boxed{R_{\max}\simeq\frac{L_{\min}}2 \simeq2.6\times10^2\,\mathrm{km}}. \]

The comparison with the Sun is

\[ \frac{R_\odot}{R_{\max}} \simeq\frac{7\times10^5\,\mathrm{km}} {2.56\times10^2\,\mathrm{km}}, \]

so

\[ \boxed{\frac{R_\odot}{R_{\max}}\simeq2.7\times10^3}. \]

Each object orbits the center of mass at radius \(L/2\). Its orbital speed at this instant is therefore

\[ v_{\mathrm{col}} =\Omega_{\overline{\mathrm{CD}}}\frac{L}{2} \simeq(2.618\times10^2)(2.562\times10^5)\,\mathrm{m/s}, \]

giving

\[ \boxed{v_{\mathrm{col}}\simeq6.7\times10^7\,\mathrm{m/s}}, \qquad \boxed{\frac{v_{\mathrm{col}}}{c}\simeq0.22}. \]

Thus each object contains roughly \(35\) solar masses within a radius no greater than about \(260\,\mathrm{km}\), while moving at about one fifth of the speed of light. These are extremely compact and fast-moving objects. The numerical estimates use the stipulated Newtonian circular-orbit and quadrupole approximations, whose accuracy deteriorates near merger.

The solution was checked against the official solution.