OII 2018 Teorica

Solutions to Theory Problem 1 LIGO-GW150914 (V. Cardoso, C. Herdeiro) July 15, 2018 v6.0

GW150914 (10 points)

Part A. Newtonian (conservative) orbits (3.0 points)

A.1 Apply Newton’s law to mass $M_1$:

$M_1\frac{d^2{\vec{r}}_1}{d{t}^2}=G\frac{M_1{M}_2}{|{\vec{r}}_2-{\vec{r}}_1{|}^2}\frac{{\vec{r}}_2-{\vec{r}}_1}{|{\vec{r}}_2-{\vec{r}}_1|}.(1)$

Use, from eq. (1) of the question sheet

${\vec{r}}_2=-\frac{M_1}{M_2}{\vec{r}}_1,(2)$

in eq. (1) above, to obtain

$\frac{d^2{\vec{r}}_1}{d{t}^2}=-\frac{G{M}_2^3}{(M_1+M_2{)}^2{r}_1^2}\frac{{\vec{r}}_1}{r_1}.(3)$

A.1 $n=3,\alpha =\frac{G{M}_2^3}{(M_1+M_2{)}^2}.$ 1.0pt

A.2 The total energy of the system is the sum of the two kinetic energies plus the gravitational potential energy. For circular motions, the linear velocity of each of the masses reads

$|{\vec{v}}_1|=r_1\Omega ,|{\vec{v}}_2|=r_2\Omega ,(4)$

Thus, the total energy is

$E=\frac{1}{2}(M_1{r}_1^2+M_2{r}_2^2){\Omega }^2-\frac{G{M}_1{M}_2}{L},(5)$

Now,

$(M_1{r}_1-M_2{r}_2{)}^2=0\Rightarrow M_1{r}_1^2+M_2{r}_2^2=\mu L^2.(6)$

Thus,

$E=\frac{1}{2}\mu L^2{\Omega }^2-\frac{GM\mu }{L}.(7)$

A.2 $A(\mu ,\Omega ,L)=\frac{1}{2}\mu L^2{\Omega }^2.$ 1.0pt

A.3 Energy (3) of the question sheet can be interpreted as describing a system of a mass $\mu$ in a circular orbit with angular velocity $\Omega$, radius $L$, around a mass $M$ (at rest). Equating the gravitational acceleration to the centripetal acceleration:

$G\frac{M}{L^2}={\Omega }^{2}L.(8)$

This is indeed Kepler’s third law (for circular orbits). Then, from (7),

$E=-\frac{1}{2}G\frac{M\mu }{L}.(9)$

A.3 $\beta =-\frac{1}{2}.$ 1.0pt

Part B - Introducing relativistic dissipation (7.0 points)

B.1 Some simple trigonometry for the $x,y$ motion of the masses (in an appropriate Cartesian system) yields:

$(x_1,y_1)=r_1(\cos (\Omega t),\sin (\Omega t)),(x_2,y_2)=-r_2(\cos (\Omega t),\sin (\Omega t)).(10)$

Then,

$Q_{ij}=\frac{M_1{r}_1^2+M_2{r}_2^2}{2}\left(\begin{array}{ccc}\frac{4}{3}{\cos }^2(\Omega t)-\frac{2}{3}{\sin }^2(\Omega t)& 2\sin (\Omega t)\cos (\Omega t)& 0\\ 2\sin (\Omega t)\cos (\Omega t)& \frac{4}{3}{\sin }^2(\Omega t)-\frac{2}{3}{\cos }^2(\Omega t)& 0\\ 0& 0& -\frac{2}{3}\end{array}\right),(11)$

or, using some simple trigonometry and (6),

$Q_{ij}=\frac{\mu L^2}{2}\left(\begin{array}{ccc}\frac{1}{3}+\cos 2\Omega t& \sin 2\Omega t& 0\\ \sin 2\Omega t& \frac{1}{3}-\cos 2\Omega t& 0\\ 0& 0& -\frac{2}{3}\end{array}\right).(12)$

B.1 $k=2\Omega ,a_1=a_2=\frac{1}{3},a_3=-\frac{2}{3},b_1=1,b_2=-1,b_3=0,c_{12}=c_{21}=1,c_{ij}^{\text{otherwise}}=0.$ 1.0pt

B.2 First take the derivatives:

$\frac{d^3{Q}_{ij}}{d{t}^3}=4{\Omega }^3\mu L^2\left(\begin{array}{ccc}\sin 2\Omega t& -\cos 2\Omega t& 0\\ -\cos 2\Omega t& -\sin 2\Omega t& 0\\ 0& 0& 0\end{array}\right).(13)$

Then perform the sum:

$\frac{dE}{dt}=\frac{G}{5c^5}(4{\Omega }^3\mu L^2{)}^2[2{\sin }^2(2\Omega t)+2{\cos }^2(2\Omega t)]=\frac{32}{5}\frac{G}{c^5}{\mu }^2{L}^4{\Omega }^6.(14)$

B.2 $\xi =\frac{32}{5}.$ 1.0pt

B.3 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,

$\frac{dE}{dt}=\frac{GM\mu }{2L^2}\frac{dL}{dt},(15)$

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

$\frac{GM\mu }{2L^2}\frac{dL}{dt}=-\frac{32}{5}\frac{G}{c^5}{\mu }^2{L}^4{\Omega }^6.(16)$

We can eliminate the $L$ and $dL/dt$ dependence in this equation in terms of $\Omega$ and $d\Omega /dt$, by using Kepler’s third law (8), which relates:

$L^3=G\frac{M}{{\Omega }^2},\frac{dL}{dt}=-\frac{2}{3}\frac{L}{\Omega }\frac{d\Omega }{dt}.(17)$

Substituting in (16), we obtain:

${\left(\frac{d\Omega }{dt}\right)}^3={\left(\frac{96}{5}\right)}^3\frac{{\Omega }^{11}}{c^{15}}{G}^5{\mu }^3{M}^2\equiv {\left(\frac{96}{5}\right)}^3\frac{{\Omega }^{11}}{c^{15}}(G{M}_c{)}^5.(18)$

B.3 $M_c=({\mu }^3{M}^2{)}^{1/5}.$ 1.0pt

B.4 Angular and cycle frequencies are related as $\Omega =2\pi f$. From the information provided above: GWs have a frequency which is twice as large as the orbital frequency, we have

$\frac{\Omega }{2\pi }=\frac{f_{\text{GW}}}{2}.(19)$

Formula (10) of the question sheet has the form

$\frac{d\Omega }{dt}=\chi {\Omega }^{11/3},\chi \equiv \frac{96}{5}\frac{(G{M}_c{)}^{5/3}}{c^5}.(20)$

Thus, from (11) of the question sheet

$\Omega (t{)}^{-8/3}=\frac{8}{3}\chi (t_0-t),(21)$

or, using (20) and the definition of $\chi$ gives

$f_{\text{GW}}^{-8/3}(t)=\frac{(8\pi {)}^{8/3}}{5}{\left(\frac{G{M}_c}{c^3}\right)}^{5/3}(t_0-t).(22)$

B.4 $p=1.$ 2.0pt

B.5 From the figure, we consider the two $\Delta t$‘s as half periods. Thus, the (cycle) GW frequency is $f_{\text{GW}}=1/(2\Delta t)$. Then, the four given points allow us to compute the frequency at the mean time of the two intervals as

	$t_{\text{AB}}$	$t_{\text{CD}}$
$t$ (s)	0.0045	0.037
$f_{\text{GW}}$ (Hz)	$(2\times 0.009{)}^{-1}$	$(2\times 0.006{)}^{-1}$

Now, using (22) we have two pairs of $(f_{\text{GW}},t)$ values for two unknowns $(t_0,M_c)$. Expressing (22) for both $t_{\text{AB}}$ and $t_{\text{CD}}$ and dividing the two equations we obtain:

$t_0=\frac{A{t}_{\text{CD}}-t_{\text{AB}}}{A-1},A\equiv {\left(\frac{f_{\text{GW}}(t_{\text{AB}})}{f_{\text{GW}}(t_{\text{CD}})}\right)}^{-8/3}.(23)$

Replacing by the numerical values, $A\simeq 2.95$ and $t_0\simeq 0.054$ s. Now we can use (22) for either of the two values $t_{\text{AB}}$ or $t_{\text{CD}}$ and determine $M_c$. One obtains for the chirp mass

$M_c\simeq 6\times 10^{31}\text{kg}\simeq 30\times M_{\odot }.(24)$

Thus, the total mass $M$ is

$M=4^{3/5}{M}_c\simeq 69\times M_{\odot }.(25)$

This result is actually remarkably close to the best estimates using the full theory of General Relativity! [Even though the actual objects do not have precisely equal masses and the theory we have just used is not valid very close to the collision.]

B.5 $M_c\simeq 30\times M_{\odot },M\simeq 69\times M_{\odot }.$ 1.0pt

B.6 From (8), Kepler’s law states that $L=(GM/{\Omega }^2{)}^{1/3}$. The second pair of points highlighted in the plot correspond to the cycle prior to merger. Thus, we use (19) to obtain the orbital angular velocity at $t_{\text{CD}}$:

${\Omega }_{t_{\text{CD}}}\sim 2.6\times 10^2\text{rad/s}.(26)$

Then, using the total mass (25) we find

$L\sim 5\times 10^2\text{km}.(27)$

Thus, these objects have a maximum radius of $R_{\text{max}}\sim 250$ km. Hence they have over 30 times more mass and,

$\frac{R_{\odot }}{R_{\text{max}}}\sim 3\times 10^3(28)$

they are 3000 times smaller than the Sun and!

Their linear velocity is

$v_{\text{col}}=\frac{L}{2}\Omega \simeq 7\times 10^4\text{km/s}.(29)$

They are moving at over 20% of the velocity of light!

B.6 $L_{\text{collision}}\sim 5\times 10^2\text{km},\frac{R_{\odot }}{R_{\text{max}}}\sim 3\times 10^3,\frac{v_{\text{col}}}{c}\sim \mathrm{0.2.}$ 1.0pt

Topic: Gravitation, Astrophysics, Oscillations & Waves Metodi: Newton’s Law of Gravitation, Kepler’s Laws, Energy Conservation Method, Calculus-Integration, Experimental Data Analysis, Differential Equations, Approximation & Series Expansion Competenze: Mathematical Modeling, Physical Reasoning, Experimental Data Analysis Objects: Black Hole Fonte: Testo (PDF) — p.1

Solutions to Theory Problem 1 The following information shall be provided: (V. Cardoso, C. Heir) July 15, 2018 v6.0

The Commission shall adopt implementing acts in accordance with Article 21 of Regulation (EU) No 1095/2012.

Part A. Newtonian (conservative) orbits (3.0 points)

A.1 Apply Newton’s law to mass $M_1$:

$M_1\frac{d^2{\vec{r}}_1}{d{t}^2}=G\frac{M_1{M}_2}{|{\vec{r}}_2-{\vec{r}}_1{|}^2}\frac{{\vec{r}}_2-{\vec{r}}_1}{|{\vec{r}}_2-{\vec{r}}_1|}.(1)$

Use, from eq. (1) of the question sheet

${\vec{r}}_2=-\frac{M_1}{M_2}{\vec{r}}_1,(2)$

in eq. (1) above, to obtain

$\frac{d^2{\vec{r}}_1}{d{t}^2}=-\frac{G{M}_2^3}{(M_1+M_2{)}^2{r}_1^2}\frac{{\vec{r}}_1}{r_1}.(3)$

A.1 $n=3,\alpha =\frac{G{M}_2^3}{(M_1+M_2{)}^2}.$ 1.0pt

A.2 The total energy of the system is the sum of the two kinetic energies plus the gravitational potential energy. For circular motions, the linear velocity of each of the masses reads

$|{\vec{v}}_1|=r_1\Omega ,|{\vec{v}}_2|=r_2\Omega ,(4)$

Thus, the total energy is

$E=\frac{1}{2}(M_1{r}_1^2+M_2{r}_2^2){\Omega }^2-\frac{G{M}_1{M}_2}{L},(5)$

Now,

$(M_1{r}_1-M_2{r}_2{)}^2=0\Rightarrow M_1{r}_1^2+M_2{r}_2^2=\mu L^2.(6)$

So,

$E=\frac{1}{2}\mu L^2{\Omega }^2-\frac{GM\mu }{L}.(7)$

A.2 $A(\mu ,\Omega ,L)=\frac{1}{2}\mu L^2{\Omega }^2.$ 1.0pt

A.3 Energy (3) of the question sheet can be interpreted as describing a system of a mass $\mu$ in a circular orbit with angular velocity $\Omega$, radius $L$, around a mass $M$ (at rest). Equating the gravitational acceleration to the centripetal acceleration:

$G\frac{M}{L^2}={\Omega }^{2}L.(8)$

This is indeed Kepler’s third law (for circular orbits). Then, from (7),

$E=-\frac{1}{2}G\frac{M\mu }{L}.(9)$

A.3 $\beta =-\frac{1}{2}.$ 1.0pt

Part B - Introducing relativistic dissipation (7.0 points)

B.1 Some simple trigonometry for the $x,y$ motion of the masses (in an appropriate Cartesian system) yields:

$(x_1,y_1)=r_1(\cos (\Omega t),\sin (\Omega t)),(x_2,y_2)=-r_2(\cos (\Omega t),\sin (\Omega t)).(10)$

Then,

$Q_{ij}=\frac{M_1{r}_1^2+M_2{r}_2^2}{2}\left(\begin{array}{ccc}\frac{4}{3}{\cos }^2(\Omega t)-\frac{2}{3}{\sin }^2(\Omega t)& 2\sin (\Omega t)\cos (\Omega t)& 0\\ 2\sin (\Omega t)\cos (\Omega t)& \frac{4}{3}{\sin }^2(\Omega t)-\frac{2}{3}{\cos }^2(\Omega t)& 0\\ 0& 0& -\frac{2}{3}\end{array}\right),(11)$

or, using some simple trigonometry and (6),

$Q_{ij}=\frac{\mu L^2}{2}\left(\begin{array}{ccc}\frac{1}{3}+\cos 2\Omega t& \sin 2\Omega t& 0\\ \sin 2\Omega t& \frac{1}{3}-\cos 2\Omega t& 0\\ 0& 0& -\frac{2}{3}\end{array}\right).(12)$

B.1 $k=2\Omega ,a_1=a_2=\frac{1}{3},a_3=-\frac{2}{3},b_1=1,b_2=-1,b_3=0,c_{12}=c_{21}=1,c_{ij}^{\text{otherwise}}=0.$ 1.0pt

B.2 First take the derivatives:

$\frac{d^3{Q}_{ij}}{d{t}^3}=4{\Omega }^3\mu L^2\left(\begin{array}{ccc}\sin 2\Omega t& -\cos 2\Omega t& 0\\ -\cos 2\Omega t& -\sin 2\Omega t& 0\\ 0& 0& 0\end{array}\right).(13)$

Then perform the sum:

$\frac{dE}{dt}=\frac{G}{5c^5}(4{\Omega }^3\mu L^2{)}^2[2{\sin }^2(2\Omega t)+2{\cos }^2(2\Omega t)]=\frac{32}{5}\frac{G}{c^5}{\mu }^2{L}^4{\Omega }^6.(14)$

B.2 $\xi =\frac{32}{5}.$ 1.0pt

B.3 Now we assume a sequence of Keplerian orbits, with decreasing energy, which is being taken from the system by the GWs.

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

$\frac{dE}{dt}=\frac{GM\mu }{2L^2}\frac{dL}{dt},(15)$

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

$\frac{GM\mu }{2L^2}\frac{dL}{dt}=-\frac{32}{5}\frac{G}{c^5}{\mu }^2{L}^4{\Omega }^6.(16)$

We can eliminate the $L$ and $dL/dt$ dependence in this equation in terms of $\Omega$ and $d\Omega /dt$, by using Kepler’s third law (8), which relates:

$L^3=G\frac{M}{{\Omega }^2},\frac{dL}{dt}=-\frac{2}{3}\frac{L}{\Omega }\frac{d\Omega }{dt}.(17)$

Substituting in (16), we obtain:

${\left(\frac{d\Omega }{dt}\right)}^3={\left(\frac{96}{5}\right)}^3\frac{{\Omega }^{11}}{c^{15}}{G}^5{\mu }^3{M}^2\equiv {\left(\frac{96}{5}\right)}^3\frac{{\Omega }^{11}}{c^{15}}(G{M}_c{)}^5.(18)$

B.3 $M_c=({\mu }^3{M}^2{)}^{1/5}.$ 1.0pt

B.4 Angular and cycle frequencies are related as $\Omega =2\pi f$. From the information provided above: GWs have a frequency which is twice as large as the orbital frequency, we have

$\frac{\Omega }{2\pi }=\frac{f_{\text{GW}}}{2}.(19)$

Formula (10) of the question sheet has the form

$\frac{d\Omega }{dt}=\chi {\Omega }^{11/3},\chi \equiv \frac{96}{5}\frac{(G{M}_c{)}^{5/3}}{c^5}.(20)$

Thus, from (11) of the question sheet

$\Omega (t{)}^{-8/3}=\frac{8}{3}\chi (t_0-t),(21)$

or, using (20) and the definition of $\chi$ gives

$f_{\text{GW}}^{-8/3}(t)=\frac{(8\pi {)}^{8/3}}{5}{\left(\frac{G{M}_c}{c^3}\right)}^{5/3}(t_0-t).(22)$

B.4 $p=1.$ 2.0pt

B.5 From the figure, we consider the two $\Delta t$‘s as half periods. Thus, the (cycle) GW frequency is $f_{\text{GW}}=1/(2\Delta t)$. Then, the four given points allow us to compute the frequency at the mean time of the two intervals as

	$t_{\text{AB}}$	$t_{\text{CD}}$
$t$ (s)	0.0045	0.037
$f_{\text{GW}}$ (Hz)	$(2\times 0.009{)}^{-1}$	$(2\times 0.006{)}^{-1}$

Now, using (22) we have two pairs of $(f_{\text{GW}},t)$ values for two unknowns $(t_0,M_c)$. Expressing (22) for both $t_{\text{AB}}$ and $t_{\text{CD}}$ and dividing the two equations we obtain:

$t_0=\frac{A{t}_{\text{CD}}-t_{\text{AB}}}{A-1},A\equiv {\left(\frac{f_{\text{GW}}(t_{\text{AB}})}{f_{\text{GW}}(t_{\text{CD}})}\right)}^{-8/3}.(23)$

Replacing by the numerical values, $A\simeq 2.95$ and $t_0\simeq 0.054$ s. Now we can use (22) for either of the two values $t_{\text{AB}}$ or $t_{\text{CD}}$ and determine $M_c$. One obtains for the chirp mass

$M_c\simeq 6\times 10^{31}\text{kg}\simeq 30\times M_{\odot }.(24)$

Thus, the total mass $M$ is

$M=4^{3/5}{M}_c\simeq 69\times M_{\odot }.(25)$

This result is actually remarkably close to the best estimates using the full theory of General Relativity! [Even though the actual objects don’t have exactly equal masses and the theory we just used is not valid very close to the collision.]

B.5 $M_c\simeq 30\times M_{\odot },M\simeq 69\times M_{\odot }.$ 1.0pt

B.6 From (8), Kepler’s law states that $L=(GM/{\Omega }^2{)}^{1/3}$. The second pair of points highlighted in the plot correspond to the cycle prior to merger. Thus, we use (19) to obtain the orbital angular velocity at $t_{\text{CD}}$:

${\Omega }_{t_{\text{CD}}}\sim 2.6\times 10^2\text{rad/s}.(26)$

Then, using the total mass (25) we find

$L\sim 5\times 10^2\text{km}.(27)$

Thus, these objects have a maximum radius of $R_{\text{max}}\sim 250$ km. Hence they have over 30 times more mass and,

$\frac{R_{\odot }}{R_{\text{max}}}\sim 3\times 10^3(28)$

They’re 3000 times smaller than the Sun and!

Their linear velocity is

$v_{\text{col}}=\frac{L}{2}\Omega \simeq 7\times 10^4\text{km/s}.(29)$

They’re moving at over 20% of the speed of light!

B.6 $L_{\text{collision}}\sim 5\times 10^2\text{km},\frac{R_{\odot }}{R_{\text{max}}}\sim 3\times 10^3,\frac{v_{\text{col}}}{c}\sim \mathrm{0.2.}$ 1.0pt

Topic: Gravitation, Astrophysics, Oscillations & Waves Metodi: Newton’s Law of Gravitation, Kepler’s Laws, Energy Conservation Method, Calculus-Integration, Experimental Data Analysis, Differential Equations, Approximation & Series Expansion Competenze: Mathematical Modeling, Physical Reasoning, Experimental Data Analysis Objects: Black Hole Fonte: Testo (PDF) — p.1