Part F · unit 38 · problems

Gravitational waves: ten components, two polarisations, one chirp

the slides · the curriculum

Checkpoint. From scratch, you derive the linearised wave equation \(\Box\bar h_{\mu\nu} = -16\pi GT_{\mu\nu}/c^4\) in Lorenz gauge, reduce ten components to two by explicit gauge fixing — naming the residual freedom at each stage — exhibit both polarisations' effect on a ring of free test masses, and reproduce the Hulse–Taylor orbital decay rate to within 5 per cent of the measured \(-2.423\times10^{-12}\).

Work each problem on paper first. Under it, Hints opens a ladder one rung at a time — each rung appears only once the one before it is open, and the last gives the first line of the working — and Worked solution has the whole thing: the idea, the working, then a check. Read the solution after your attempt, including when you got it right. F folds everything back; D switches dark and light.

Twelve problems, about six hours. Conventions as in the deck: signature \((-,+,+,+)\), \(x^0 = ct\), \(\Box = \eta^{\mu\nu}\partial_\mu\partial_\nu = -c^{-2}\partial_t^2 + \nabla^2\), indices on \(h\) moved with \(\eta\), gauge transformation \(h_{\mu\nu}\to h_{\mu\nu} + \partial_\mu\xi_\nu + \partial_\nu\xi_\mu\), and \(Q_{ij} = \int\rho\,(y^iy^j - \tfrac13\delta_{ij}|\vec y|^2)\,\mathrm d^3y\) — one third of unit 10's quadrupole moment.

Constants: \(G = 6.67430\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}\), \(c = 299\,792\,458\ \mathrm{m\,s^{-1}}\), \(GM_\odot = 1.32712440018\times10^{20}\ \mathrm{m^3\,s^{-2}}\) (so \(M_\odot = 1.9884\times10^{30}\) kg, \(GM_\odot/c^3 = 4.92549\ \mu\mathrm s\), \(GM_\odot/c^2 = 1476.6\) m), \(M_\oplus = 5.9722\times10^{24}\) kg, 1 au \(= 1.495978707\times10^{11}\) m, 1 Mpc \(= 3.0857\times10^{22}\) m, one year \(= 3.15576\times10^{7}\) s, \(L_\odot = 3.828\times10^{26}\) W.

Linearised gravity and its gauge

P1 Trace reversal, and why four dimensions routine

exercises: first-order expansion, index gymnastics with \(\eta\), a proof by direct computation

Work on \(\mathbb R^n\) with a Lorentzian \(\eta\) (so \(\eta^{\mu\nu}\eta_{\mu\nu} = n\)), and let \(g = \eta + h\).

  1. Prove that \(g^{\mu\nu} = \eta^{\mu\nu} - h^{\mu\nu} + O(h^2)\), where \(h^{\mu\nu} = \eta^{\mu\alpha}\eta^{\nu\beta}h_{\alpha\beta}\).
  2. Define \(\bar h_{\mu\nu} = h_{\mu\nu} - \tfrac12\eta_{\mu\nu}h\). Compute the trace \(\bar h\) and then \(\bar{\bar h}_{\mu\nu}\). Prove that trace reversal is an involution if and only if \(n = 4\).
  3. Find the constant \(\alpha_n\) for which \(\tilde h_{\mu\nu} = h_{\mu\nu} - \alpha_n\eta_{\mu\nu}h\) has trace \(-h\) in \(n\) dimensions, and check that \(\alpha_4 = \tfrac12\).
  4. Check the trace reversal in the Newtonian limit of unit 35: \(h_{00} = -2\Phi/c^2\), \(h_{ij} = -2\Phi\delta_{ij}/c^2\), \(h_{0i} = 0\), in four dimensions. Compute \(\bar h_{\mu\nu}\).
Hints · 3 rungs
Rung 1

Part 1 is a claim about a matrix inverse to first order; parts 2 and 3 are one-line trace computations; part 4 is a substitution. Everything reduces to one number: \(\eta^{\mu\nu}\eta_{\mu\nu}\).

Rung 2

For part 1, multiply your candidate by \(\eta_{\nu\lambda} + h_{\nu\lambda}\) and show the product is \(\delta^\mu_\lambda\) up to terms quadratic in \(h\). For part 2, take the trace with \(\eta^{\mu\nu}\) and use \(\eta^{\mu\nu}\eta_{\mu\nu} = n\).

Rung 3

\(\bar h = h - \tfrac n2h = \big(1 - \tfrac n2\big)h\). Now apply the bar again: \(\bar{\bar h}_{\mu\nu} = \bar h_{\mu\nu} - \tfrac12\eta_{\mu\nu}\bar h = \ldots\)

Worked solution

Idea. Everything is linear algebra at a point; the dimension enters only through the trace of the identity, \(\eta^{\mu\nu}\eta_{\mu\nu} = \delta^\mu_\mu = n\).

1. \((\eta^{\mu\nu} - h^{\mu\nu})(\eta_{\nu\lambda} + h_{\nu\lambda}) = \delta^\mu_\lambda + \eta^{\mu\nu}h_{\nu\lambda} - h^{\mu\nu}\eta_{\nu\lambda} - h^{\mu\nu}h_{\nu\lambda}\). The middle two terms are both \(h^\mu{}_\lambda\) and cancel; the last is \(O(h^2)\). The inverse of a matrix is unique, so \(g^{\mu\nu} = \eta^{\mu\nu} - h^{\mu\nu} + O(h^2)\). (For small enough \(h\) the inverse exists, because the invertible matrices form an open set.)

2. \(\bar h = \eta^{\mu\nu}\bar h_{\mu\nu} = h - \tfrac12nh = (1 - \tfrac n2)h\). Then \[\bar{\bar h}_{\mu\nu} = \bar h_{\mu\nu} - \tfrac12\eta_{\mu\nu}\bar h = h_{\mu\nu} - \tfrac12\eta_{\mu\nu}h - \tfrac12\big(1 - \tfrac n2\big)\eta_{\mu\nu}h = h_{\mu\nu} - \big(1 - \tfrac n4\big)\eta_{\mu\nu}h.\] This equals \(h_{\mu\nu}\) for every \(h\) if and only if \(1 - n/4 = 0\), i.e. \(n = 4\) (take \(h = \eta\), whose trace is \(n\ne0\), to see that "for every \(h\)" forces the coefficient to vanish). \(\square\)

3. \(\mathrm{tr}\,\tilde h = h - \alpha_nnh\). Setting this to \(-h\) gives \(\alpha_n = 2/n\), and \(\alpha_4 = \tfrac12\) ✓. In \(n\) dimensions the version of the linearised theory with trace \(-h\) uses \(2/n\); the one whose Einstein tensor has the four-term form uses \(\tfrac12\) regardless, and then the operation is not an involution. That is one of several places where \(n = 4\) is quietly special.

4. \(h = \eta^{\mu\nu}h_{\mu\nu} = -h_{00} + h_{ii} = 2\Phi/c^2 - 6\Phi/c^2 = -4\Phi/c^2\). Then \(\bar h_{00} = h_{00} - \tfrac12(-1)h = -2\Phi/c^2 - 2\Phi/c^2 = -4\Phi/c^2\) and \(\bar h_{ij} = -2\Phi\delta_{ij}/c^2 - \tfrac12\delta_{ij}(-4\Phi/c^2) = 0\).

Check. Part 4 reproduces unit 35: only \(\bar h_{00} = -4\Phi/c^2\) survives, which is the deck's static check of the retarded solution. And \(\bar h = -\bar h_{00} = 4\Phi/c^2 = -h\) ✓, as part 2 says for \(n = 4\). Dimensions: \(\Phi/c^2\) is dimensionless ✓.

P2 Proof: curvature does not see the gauge medium

exercises: gauge invariance by Clairaut, the linearised Riemann tensor, geodesic deviation in TT gauge

The linearised Riemann tensor, with all indices down, is

\[R_{\mu\nu\rho\sigma} = \tfrac12\big(\partial_\nu\partial_\rho h_{\mu\sigma} + \partial_\mu\partial_\sigma h_{\nu\rho} - \partial_\mu\partial_\rho h_{\nu\sigma} - \partial_\nu\partial_\sigma h_{\mu\rho}\big).\]
  1. Prove that it is unchanged by \(h_{\mu\nu}\to h_{\mu\nu} + \partial_\mu\xi_\nu + \partial_\nu\xi_\mu\) for any smooth \(\xi\).
  2. Compute \(R_{i0j0}\) for a wave in TT gauge, and show that unit 33's geodesic deviation equation, \(\ddot\xi^i = -c^2R^i{}_{0j0}\,\xi^j\) for slowly moving masses, becomes \(\ddot\xi^i = \tfrac12\ddot h_{ij}\xi^j\).
  3. Compute \(R_{i0j0}\) instead for the static Newtonian field \(h_{00} = -2\Phi/c^2\) (other components irrelevant for this component — say why), and recover unit 05's tidal equation.
  4. The electromagnetic analogue: prove that \(\vec E = -\nabla\varphi - \partial_t\vec A\) and \(\vec B = \nabla\times\vec A\) are unchanged by unit 24's gauge transformation. Name the one theorem both proofs use.
Hints · 3 rungs
Rung 1

Part 1 is a cancellation: the change in \(R\) is a sum of eight terms, each a third derivative of \(\xi\). Show they cancel in pairs. Parts 2 and 3 are substitutions; part 4 is the same argument with fewer indices.

Rung 2

Write \(\delta h_{\mu\sigma} = \partial_\mu\xi_\sigma + \partial_\sigma\xi_\mu\) and substitute into the first term: \(\partial_\nu\partial_\rho\partial_\mu\xi_\sigma + \partial_\nu\partial_\rho\partial_\sigma\xi_\mu\). Do the same for the other three and look for each third derivative with its opposite sign. For part 2, which components of \(h\) vanish in TT gauge?

Rung 3

In TT gauge \(h_{0\mu} = 0\), so in \(R_{i0j0} = \tfrac12(\partial_0\partial_jh_{i0} + \partial_i\partial_0h_{0j} - \partial_i\partial_jh_{00} - \partial_0\partial_0h_{ij})\) three terms die, leaving \(R_{i0j0} = -\tfrac12\partial_0^2h_{ij}\).

Worked solution

Idea. Gauge invariance of a field strength is always Clairaut: the gauge part of the potential is a derivative, and the field strength is an antisymmetrised derivative, so second (here third) derivatives meet with opposite signs.

1. The four terms of \(2\,\delta R_{\mu\nu\rho\sigma}\) are

\[\begin{aligned} &\partial_\nu\partial_\rho(\partial_\mu\xi_\sigma + \partial_\sigma\xi_\mu) + \partial_\mu\partial_\sigma(\partial_\nu\xi_\rho + \partial_\rho\xi_\nu)\\ &- \partial_\mu\partial_\rho(\partial_\nu\xi_\sigma + \partial_\sigma\xi_\nu) - \partial_\nu\partial_\sigma(\partial_\mu\xi_\rho + \partial_\rho\xi_\mu). \end{aligned}\]

Pair them by which component of \(\xi\) is differentiated: \(\partial_\nu\partial_\rho\partial_\mu\xi_\sigma\) against \(-\partial_\mu\partial_\rho\partial_\nu\xi_\sigma\); \(\partial_\nu\partial_\rho\partial_\sigma\xi_\mu\) against \(-\partial_\nu\partial_\sigma\partial_\rho\xi_\mu\); \(\partial_\mu\partial_\sigma\partial_\nu\xi_\rho\) against \(-\partial_\nu\partial_\sigma\partial_\mu\xi_\rho\); \(\partial_\mu\partial_\sigma\partial_\rho\xi_\nu\) against \(-\partial_\mu\partial_\rho\partial_\sigma\xi_\nu\). Each pair is the same third derivative with the order of differentiation changed, and cancels by Clairaut's theorem (unit 03) for \(C^3\) functions. \(\square\)

2. With \(h_{0\mu} = 0\), \(R_{i0j0} = -\tfrac12\partial_0^2h_{ij} = -\frac{1}{2c^2}\ddot h_{ij}\). Raising \(i\) with \(\delta\) changes nothing, so \(\ddot\xi^i = -c^2R^i{}_{0j0}\xi^j = \tfrac12\ddot h_{ij}\xi^j\). Integrating twice with \(\xi = \xi_0\) when \(h = 0\) and keeping first order, \(\xi^i = \xi_0^i + \tfrac12h_{ij}\xi_0^j\): the deck's proper-distance result, now from a gauge-invariant equation.

3. \(R_{i0j0}\) involves \(h_{i0}\), \(h_{0j}\) (differentiated in time), \(h_{00}\) and \(h_{ij}\) (differentiated twice in time). For a static field the time derivatives vanish, so only \(-\tfrac12\partial_i\partial_jh_{00}\) survives, whatever \(h_{ij}\) is. With \(h_{00} = -2\Phi/c^2\): \(R_{i0j0} = \partial_i\partial_j\Phi/c^2\), and the deviation equation reads \(\ddot\xi^i = -(\partial_i\partial_j\Phi)\xi^j\) — unit 05's tidal equation.

4. \(\varphi\to\varphi - \partial_t\chi\), \(\vec A\to\vec A + \nabla\chi\). Then \(\vec E\) changes by \(\nabla\partial_t\chi - \partial_t\nabla\chi = 0\) and \(\vec B\) by \(\nabla\times\nabla\chi\), whose components are \(\partial_j\partial_k\chi - \partial_k\partial_j\chi = 0\). Both proofs are Clairaut's theorem, and nothing else.

Check. Dimensions in part 2: \([R] = \mathrm{m^{-2}}\), and \(\ddot h/c^2\) is \(\mathrm{s^{-2}}/(\mathrm{m^2s^{-2}}) = \mathrm{m^{-2}}\) ✓. Sign in part 3: two masses separated along the line to a point mass (\(\partial_r^2\Phi = -2GM/r^3 \lt 0\)) accelerate apart, the tidal stretch ✓ — which fixes the sign in part 2 too, since both come from one formula.

P3 Count to two medium

exercises: the degree-of-freedom count done by hand, residual gauge, finding the TT representative

A vacuum plane wave along \(z\) in Lorenz gauge, \(\bar h_{\mu\nu} = e_{\mu\nu}\cos(k\cdot x)\), with units chosen so that \(\omega/c = 1\): \(k_\mu = (-1, 0, 0, 1)\).

  1. Write out the four Lorenz conditions \(k^\mu e_{\mu\nu} = 0\) and show they leave six free components. Say which six you choose as free.
  2. A residual gauge transformation \(\xi_\mu = \epsilon_\mu\sin(k\cdot x)\) changes \(e\) by \(p_{\mu\nu} = k_\mu\epsilon_\nu + k_\nu\epsilon_\mu - \eta_{\mu\nu}(k\cdot\epsilon)\). Show that \(\Box\xi = 0\), write out the ten components of \(p\), and identify the two combinations of \(e\) that no \(\epsilon\) can change.
  3. Take \[e = \begin{pmatrix}2&1&0&-2\\1&3&4&-1\\0&4&1&0\\-2&-1&0&2\end{pmatrix}.\] Check that it satisfies the Lorenz conditions, then find \(\epsilon\) bringing it to TT gauge, and read off \(h_+\), \(h_\times\).
Hints · 3 rungs
Rung 1

The answer to part 3 is a four-vector \(\epsilon\) and two numbers. The count in parts 1–2 is \(10 - 4 = 6\), then \(6 - 4 = 2\): make each stage explicit.

Rung 2

Raise the index on \(k\): \(k^\mu = (1, 0, 0, 1)\). So \(k^\mu e_{\mu\nu} = e_{0\nu} + e_{3\nu}\). And \(k\cdot\epsilon = k^\mu\epsilon_\mu = \epsilon_0 + \epsilon_3\). With these, \(p_{11} = p_{22} = -(\epsilon_0 + \epsilon_3)\) and \(p_{12} = 0\).

Rung 3

TT needs \(e_{11} + p_{11} = -(e_{22} + p_{22})\), i.e. \(3 - s = -(1 - s)\) with \(s = \epsilon_0 + \epsilon_3\): so \(s = 2\). Then set \(e_{01} + p_{01} = 0\) with \(p_{01} = k_0\epsilon_1 = -\epsilon_1\).

Worked solution

Idea. Do the deck's linear algebra in coordinates: the Lorenz conditions cut the ten components to six, the four residual gauge parameters move four of those six, and the two they cannot move are the polarisations.

1. \(k^\mu e_{\mu\nu} = e_{0\nu} + e_{3\nu} = 0\) for \(\nu = 0,1,2,3\): \(e_{00} + e_{30} = 0\), \(e_{01} + e_{31} = 0\), \(e_{02} + e_{32} = 0\), \(e_{03} + e_{33} = 0\). Four independent conditions (each involves a component of the third row the others do not), so six free components remain. Choose \(e_{00}, e_{01}, e_{02}, e_{11}, e_{12}, e_{22}\); then \(e_{30} = -e_{00}\), \(e_{31} = -e_{01}\), \(e_{32} = -e_{02}\) and \(e_{33} = -e_{03} = e_{00}\).

2. \(\Box\xi_\mu = \epsilon_\mu\Box\sin(k\cdot x) = -\epsilon_\mu(k^\alpha k_\alpha)\sin(k\cdot x) = 0\) since \(k^\alpha k_\alpha = -1 + 1 = 0\). With \(s = \epsilon_0 + \epsilon_3\) and \(\eta = \mathrm{diag}(-1,1,1,1)\):

\[\begin{aligned} p_{00} &= -2\epsilon_0 + s, & p_{01} &= -\epsilon_1, & p_{02} &= -\epsilon_2, & p_{03} &= \epsilon_0 - \epsilon_3,\\ p_{11} &= -s, & p_{22} &= -s, & p_{12} &= 0, & p_{33} &= 2\epsilon_3 - s,\\ p_{13} &= \epsilon_1, & p_{23} &= \epsilon_2 . \end{aligned}\]

On the six free components, \((p_{00}, p_{01}, p_{02}, p_{11})\) can be set to anything by choosing \((\epsilon_0,\epsilon_1,\epsilon_2,\epsilon_3)\) — four independent shifts. But \(p_{11} = p_{22}\) and \(p_{12} = 0\) always, so \(\tfrac12(e_{11} - e_{22})\) and \(e_{12}\) are unchanged by every residual transformation: these are \(h_+\) and \(h_\times\). Six minus four is two.

3. Lorenz: \(e_{00} + e_{30} = 2 - 2 = 0\), \(e_{01} + e_{31} = 1 - 1 = 0\), \(e_{02} + e_{32} = 0\), \(e_{03} + e_{33} = -2 + 2 = 0\) ✓. For TT:

  • traceless spatial block: \(3 - s = -(1 - s)\), so \(s = 2\);
  • \(e_{01} + p_{01} = 1 - \epsilon_1 = 0\), so \(\epsilon_1 = 1\); \(e_{02} + p_{02} = -\epsilon_2 = 0\), so \(\epsilon_2 = 0\);
  • \(e_{00} + p_{00} = 2 - 2\epsilon_0 + 2 = 0\), so \(\epsilon_0 = 2\), and \(\epsilon_3 = s - \epsilon_0 = 0\).

So \(\epsilon = (2, 1, 0, 0)\). The rest follow without further choice: \(e_{03} + p_{03} = -2 + 2 = 0\), \(e_{33} + p_{33} = 2 - 2 = 0\), \(e_{13} + p_{13} = -1 + 1 = 0\), \(e_{23} + p_{23} = 0\). The TT amplitude is

\[e^{\rm TT} = \begin{pmatrix}0&0&0&0\\0&1&4&0\\0&4&-1&0\\0&0&0&0\end{pmatrix},\qquad h_+ = 1,\quad h_\times = 4.\]

Check. The invariants of part 2 computed from the original \(e\): \(\tfrac12(3 - 1) = 1\) and \(e_{12} = 4\) ✓ — the TT representative was determined before any \(\epsilon\) was found, which is the theorem \(L = P\oplus TT\). And the TT projector applied to the original spatial block, \(\Lambda(S)\) for \(n = \hat z\), gives \(\tfrac12(S_{11} - S_{22}) = 1\) and \(S_{12} = 4\) ✓. The residual freedom named at each stage: four functions to reach Lorenz (not needed here, since \(e\) was given in Lorenz gauge), four constants \(\epsilon_\mu\) — one plane-wave \(\xi\) with \(\Box\xi = 0\) — to reach TT, and after that none: \(\epsilon\) was unique.

Polarisations

P4 A binary seen from any angle medium

exercises: the TT projector in a rotated basis, the quadrupole formula, inclination dependence

The binary of worked example 2 orbits in the \(xy\) plane, and is seen by an observer at distance \(D\) in the direction \(n = (0, \sin\iota, \cos\iota)\), where \(\iota\) is the inclination.

  1. Show that \(e_1 = (1,0,0)\) and \(e_2 = (0, \cos\iota, -\sin\iota)\) form an orthonormal basis of \(n^\perp\), and that for a symmetric \(S\) the TT projection has components \(h_+ = \tfrac12(e_1^{\mathsf T}Se_1 - e_2^{\mathsf T}Se_2)\) and \(h_\times = e_1^{\mathsf T}Se_2\) in that basis (times the prefactor).
  2. Using \(\ddot Q_{11} = -\ddot Q_{22} = -2\mu a^2\Omega^2\cos2\Omega t\) and \(\ddot Q_{12} = -2\mu a^2\Omega^2\sin2\Omega t\), show \[h_+ = -h_0\,\frac{1 + \cos^2\iota}2\cos2\Omega t_{\rm r},\qquad h_\times = -h_0\cos\iota\,\sin2\Omega t_{\rm r},\qquad h_0 = \frac{4G\mu a^2\Omega^2}{c^4D}.\]
  3. Describe the polarisation seen face-on (\(\iota = 0\)), edge-on (\(\iota = 90^\circ\)), and at \(\iota = 60^\circ\).
Hints · 3 rungs
Rung 1

Two functions of time with angle-dependent amplitudes. The projector is basis-free, so work in the basis \((e_1, e_2, n)\), where it looks exactly as it did for \(n = \hat z\).

Rung 2

\(Q\) has no \(z\) components (the orbit is planar). So \(e_2^{\mathsf T}\ddot Qe_2 = \cos^2\iota\,\ddot Q_{22}\) and \(e_1^{\mathsf T}\ddot Qe_2 = \cos\iota\,\ddot Q_{12}\).

Rung 3

\(h_+ = \frac{2G}{c^4D}\cdot\tfrac12\big(\ddot Q_{11} - \cos^2\iota\,\ddot Q_{22}\big) = \frac{G}{c^4D}\ddot Q_{11}(1 + \cos^2\iota)\), using \(\ddot Q_{22} = -\ddot Q_{11}\).

Worked solution

Idea. The projector \(\Lambda(n)\) is a linear map defined without a basis; choose the basis adapted to \(n\) and it takes the form found for \(n = \hat z\). Then only the components of \(\ddot Q\) in the plane of the sky matter.

1. \(e_1\cdot e_1 = 1\), \(e_2\cdot e_2 = \cos^2\iota + \sin^2\iota = 1\), \(e_1\cdot e_2 = 0\), \(e_1\cdot n = 0\), \(e_2\cdot n = \sin\iota\cos\iota - \cos\iota\sin\iota = 0\). So \((e_1, e_2, n)\) is orthonormal. In it \(P = 1 - nn^{\mathsf T}\) is \(\mathrm{diag}(1,1,0)\), and the deck's computation for \(n = \hat z\) applies verbatim: \(\Lambda(S)\) has \(\tfrac12(S_{e_1e_1} - S_{e_2e_2})\) and \(S_{e_1e_2}\) in the \(+\) and \(\times\) slots, where \(S_{uv} = u^{\mathsf T}Sv\).

2. \(e_1^{\mathsf T}\ddot Qe_1 = \ddot Q_{11}\); \(e_2^{\mathsf T}\ddot Qe_2 = \cos^2\iota\,\ddot Q_{22}\) (the \(\sin\iota\) parts multiply \(z\) components, which vanish); \(e_1^{\mathsf T}\ddot Qe_2 = \cos\iota\,\ddot Q_{12}\). With the prefactor \(2G/c^4D\):

\[h_+ = \frac{G}{c^4D}\big(\ddot Q_{11} - \cos^2\iota\,\ddot Q_{22}\big) = \frac{G}{c^4D}(1 + \cos^2\iota)\,\ddot Q_{11} = -\frac{4G\mu a^2\Omega^2}{c^4D}\cdot\frac{1 + \cos^2\iota}{2}\cos2\Omega t_{\rm r},\] \[h_\times = \frac{2G}{c^4D}\cos\iota\,\ddot Q_{12} = -\frac{4G\mu a^2\Omega^2}{c^4D}\cos\iota\,\sin2\Omega t_{\rm r}.\]

3. Face-on: \(h_+ = -h_0\cos2\Omega t\), \(h_\times = -h_0\sin2\Omega t\): equal amplitudes a quarter period apart, circular polarisation (P5). Edge-on: \(h_+ = -\tfrac12h_0\cos2\Omega t\), \(h_\times = 0\): linear polarisation with half the amplitude, the plus pattern aligned with the orbital plane. At \(\iota = 60^\circ\): amplitudes \(0.625\,h_0\) and \(0.500\,h_0\), elliptical polarisation.

Check. \(\iota = 0\) reproduces the deck's axial wave ✓. \(\iota\to180^\circ\) (seen from below) gives the same \(h_+\) and \(h_\times\to+h_0\sin2\Omega t\): the rotation sense reverses, as it must when the orbit is viewed from the other side ✓. Dimensions: \(G\mu a^2\Omega^2/c^4D\) is \(\mathrm J/(\mathrm N\cdot\mathrm m) = 1\) ✓. Measuring both amplitudes gives \(\iota\), which is how an observed binary's inclination is found — and why the inclination and the distance are hard to separate when \(\iota\) is near \(0\).

P5 A ring under circular polarisation routine

exercises: the ring of free masses, helicity 2, reading a wave as a rotating pattern

A ring of free masses of radius \(\rho_0\) lies in the \(xy\) plane, and a wave travels along \(z\) with \(h_+ = h\cos\omega t\), \(h_\times = h\sin\omega t\).

  1. Write the proper displacement \(\delta x^i = \tfrac12h_{ij}x_0^j\) of the mass at angle \(\phi\), and show that the ring is always an ellipse with semi-axes \(\rho_0(1 \pm \tfrac12h)\).
  2. Find the direction of the major axis as a function of time, and the angular speed at which it turns.
  3. The wave comes from a binary seen face-on, so \(\omega = 2\Omega\). Compare the ellipse's angular speed with the binary's. Which particle, if any, moves round the ring?
Hints · 3 rungs
Rung 1

The answer is a rotating ellipse: find its shape once and then its orientation angle as a function of \(t\). The spin-two slide's theorem does most of the work.

Rung 2

The matrix \(\begin{pmatrix}\cos\omega t&\sin\omega t\\\sin\omega t&-\cos\omega t\end{pmatrix}\) is symmetric with eigenvalues \(\pm1\). Find its eigenvector for \(+1\).

Rung 3

The eigenvector for \(+1\) is \((\cos\tfrac{\omega t}2, \sin\tfrac{\omega t}2)\): check with the double-angle formulas.

Worked solution

Idea. \(h_{ij}\) is \(h\) times a symmetric, traceless, orthogonal-like matrix with eigenvalues \(\pm1\); its eigenvectors turn at half the wave's angular frequency, because the wave has helicity 2.

1. \(h_{ij} = hH(t)\) with \(H = \begin{pmatrix}\cos\omega t&\sin\omega t\\\sin\omega t&-\cos\omega t\end{pmatrix}\). For \(x_0 = \rho_0(\cos\phi,\sin\phi)\): \(\delta x = \tfrac12h\rho_0\cos(\omega t - \phi)\), \(\delta y = \tfrac12h\rho_0\sin(\omega t - \phi)\), using the angle-difference formulas. \(H\) is symmetric with trace 0 and determinant \(-1\), so its eigenvalues are \(\pm1\), with orthonormal eigenvectors \(u_\pm\). In that basis \(x_0 + \delta x = (1 + \tfrac12hH)x_0\) scales the \(u_+\) component by \(1 + \tfrac12h\) and the \(u_-\) component by \(1 - \tfrac12h\): an ellipse with those semi-axes (times \(\rho_0\)).

2. \(Hu = u\) with \(u = (\cos\alpha,\sin\alpha)\): \(\cos\omega t\cos\alpha + \sin\omega t\sin\alpha = \cos(\omega t - \alpha)\) must equal \(\cos\alpha\), and \(\sin\omega t\cos\alpha - \cos\omega t\sin\alpha = \sin(\omega t - \alpha)\) must equal \(\sin\alpha\). Both hold for \(\alpha = \omega t/2\). So the major axis points along \(\omega t/2\) and turns at angular speed \(\omega/2\).

3. \(\omega/2 = \Omega\): the ellipse turns with the binary, its long axis following the line joining the stars (delayed by \(D/c\), and up to a sign convention in the phase). No mass moves round the ring: each mass moves on a small circle of radius \(\tfrac12h\rho_0\) about its rest position, at angular speed \(\omega\) — part 1's \(\delta x, \delta y\) — and the rotation is of the pattern, like a wave on a stadium crowd.

Check. At \(t = 0\) the wave is pure plus, and part 2 gives the major axis along \(x\) ✓ (the deck's figure at \(\omega t = 0\)). At \(\omega t = \pi/2\) it is pure cross, and the axis is at \(45^\circ\) ✓. The pattern returns to itself when \(\omega t/2\) advances by \(\pi\) — a half turn of the ellipse per wave period — which is helicity 2. Area: \(\pi\rho_0^2(1 - h^2/4)\), unchanged at first order ✓.

Generation and energy

P6 Proof: no dipole, and the virial lemma checked medium

exercises: conservation laws from \(\partial_\mu T^{\mu\nu} = 0\), the equivalence principle's role, the tensor virial theorem on an example

  1. For a slow source with \(T^{00} = \rho c^2\) and \(T^{0i} = c\rho v^i\), of bounded support and with \(\partial_\mu T^{\mu\nu} = 0\), prove \(\dot M = 0\), \(\dot D^i = P^i\) and \(\dot P^i = 0\), where \(M = \int\rho\), \(D^i = \int\rho y^i\), \(P^i = \int\rho v^i\).
  2. Two point charges \(q_1, q_2\) of masses \(m_1, m_2\) orbit their centre of mass. Show that the electric dipole moment is \(\vec p = \frac{q_1m_2 - q_2m_1}{M}\vec x\), with \(\vec x\) the separation, and that it is constant if and only if \(q_1/m_1 = q_2/m_2\). Which principle guarantees the gravitational analogue of that condition?
  3. Check the tensor virial theorem, component \(11\), for the circular binary. The source's total stress is the kinetic part \(\sum_am_av_a^iv_a^j\) plus the interaction part of the gravitational stress, which for two point masses is \(W_{ij} = -Gm_1m_2x^ix^j/a^3\) (cited: Chandrasekhar's potential-energy tensor, Ellipsoidal Figures of Equilibrium, ch. 2). Show that \(\sum m_av_a^1v_a^1 + W_{11} = \tfrac12\ddot I_{11}\).
Hints · 3 rungs
Rung 1

Part 1 is three integrations of the two conservation laws, one of them after multiplying by \(y^i\). Part 2 is the centre-of-mass reduction of worked example 2. Part 3 compares three explicit functions of \(t\).

Rung 2

The \(\nu = 0\) law is \(\partial_t\rho + \partial_k(\rho v^k) = 0\). The integral of a divergence of a field with bounded support is zero. For \(\dot D^i\), integrate \(y^i\partial_k(\rho v^k)\) by parts. For part 3, put \(\vec x = a(\cos\Omega t, \sin\Omega t, 0)\).

Rung 3

\(\sum m_av_a^1v_a^1 = \mu\dot x^2 = \mu a^2\Omega^2\sin^2\Omega t\), \(\ W_{11} = -\frac{Gm_1m_2}{a^3}a^2\cos^2\Omega t = -\mu a^2\Omega^2\cos^2\Omega t\), since \(Gm_1m_2/a^3 = \mu\,GM/a^3 = \mu\Omega^2\).

Worked solution

Idea. Gravity's dipole is weighted by the same mass that measures inertia, so its derivative is total momentum, which Newton's third law conserves. Electricity's dipole is weighted by charge, which does not have to be proportional to inertia.

1. \(\nu = 0\): \(\frac1c\partial_t(\rho c^2) + \partial_k(c\rho v^k) = 0\), i.e. \(\partial_t\rho + \partial_k(\rho v^k) = 0\). Integrate over a ball containing the source: \(\dot M = -\oint\rho v^kn_k = 0\). Multiply by \(y^i\) and integrate: \(\dot D^i = -\int y^i\partial_k(\rho v^k) = \int\delta^i_k\rho v^k = P^i\), since \(\partial_k(y^i\rho v^k) = \delta^i_k\rho v^k + y^i\partial_k(\rho v^k)\) and the divergence integrates to zero. \(\nu = i\): \(\partial_0T^{0i} + \partial_kT^{ki} = 0\), i.e. \(\partial_t(\rho v^i) = -\partial_kT^{ki}\); integrate: \(\dot P^i = 0\). Hence \(\ddot D = 0\). \(\square\)

2. With the centre of mass at rest at the origin, \(\vec y_1 = \frac{m_2}M\vec x\), \(\vec y_2 = -\frac{m_1}M\vec x\), so \(\vec p = q_1\vec y_1 + q_2\vec y_2 = \frac{q_1m_2 - q_2m_1}M\vec x\). On an orbit \(\vec x\) rotates, so \(\vec p\) is constant iff its coefficient vanishes: \(q_1m_2 = q_2m_1\), i.e. \(q_1/m_1 = q_2/m_2\). For gravity the "charge" is the gravitational mass, and the equivalence principle (unit 30) says it is proportional to the inertial mass for every body, with the same constant. So the gravitational dipole of any system is \(M\) times the centre of mass, and has no radiation.

3. \(I_{11} = \mu a^2\cos^2\Omega t = \tfrac12\mu a^2(1 + \cos2\Omega t)\), so \(\tfrac12\ddot I_{11} = -\mu a^2\Omega^2\cos2\Omega t\). Kinetic: \(\sum m_a(v_a^1)^2 = \mu(\dot x^1)^2 = \mu a^2\Omega^2\sin^2\Omega t\) (the same reduction as for \(I\)). Interaction: \(W_{11} = -\mu a^2\Omega^2\cos^2\Omega t\). Sum: \(\mu a^2\Omega^2(\sin^2\Omega t - \cos^2\Omega t) = -\mu a^2\Omega^2\cos2\Omega t\) ✓.

Check. The kinetic stress alone does not satisfy the lemma: it oscillates about \(+\tfrac12\mu a^2\Omega^2\), while \(\tfrac12\ddot I_{11}\) has mean zero. The gravitational "string" pulling the stars together is exactly what restores it — the deck's warning slide, verified. Trace: \(\sum mv^2 + W_{kk} = \mu a^2\Omega^2 - \mu a^2\Omega^2 = 0 = \tfrac12\ddot I_{kk}\): unit 08's virial theorem \(2T + V = 0\) for circular orbits ✓.

P7 Proof: the luminosity integral hard

exercises: isotropic tensors, angular integrals over \(S^2\), the angular pattern of a binary's radiation

  1. Compute \(\oint n_z^2\,\mathrm d\Omega\), \(\oint n_z^4\,\mathrm d\Omega\) and \(\oint n_x^2n_z^2\,\mathrm d\Omega\) directly in spherical coordinates.
  2. Given that \(\oint n_in_jn_kn_l\,\mathrm d\Omega = C(\delta_{ij}\delta_{kl} + \delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk})\), use part 1 to find \(C\) two ways, and check they agree.
  3. Prove that for symmetric traceless \(A\), \(\oint A_{ij}\Lambda_{ij,kl}(n)A_{kl}\,\mathrm d\Omega = \frac{8\pi}5A_{ij}A_{ij}\).
  4. From P4's waveform, the flux of a binary at inclination \(\iota\) is proportional to \(\big(\frac{1 + \cos^2\iota}{2}\big)^2 + \cos^2\iota\). Find the ratio of the flux along the axis to the flux in the orbital plane, and integrate the pattern over the sphere to recover worked example 2's \(P = \frac{32G\mu^2a^4\Omega^6}{5c^5}\).
Hints · 3 rungs
Rung 1

Parts 1–2 are numbers (multiples of \(\pi\)); part 3 is a proof that uses them; part 4 is one angular integral and some bookkeeping of constants.

Rung 2

Use \(u = \cos\theta\), \(\mathrm d\Omega = \mathrm d\phi\,\mathrm du\). For part 3, expand \(A_{ij}\Lambda_{ij,kl}A_{kl} = \mathrm{tr}(APAP) - \tfrac12(\mathrm{tr}\,AP)^2\) with \(P = 1 - nn^{\mathsf T}\), and use \(\mathrm{tr}\,A = 0\).

Rung 3

The integrand of part 3 becomes \(A_{ij}A_{ij} - 2(A^2)_{ij}n_in_j + \tfrac12(A_{ij}n_in_j)^2\). Integrate term by term with \(\oint n_in_j = \frac{4\pi}3\delta_{ij}\) and the rank-4 formula.

Worked solution

Idea. Rotation invariance reduces every angular integral of a polynomial in \(n\) to a few constants; two explicit integrals fix them, and then the power is algebra.

1. \(\oint n_z^2 = 2\pi\int_{-1}^1u^2\,\mathrm du = \frac{4\pi}3\). \(\oint n_z^4 = 2\pi\int_{-1}^1u^4\,\mathrm du = \frac{4\pi}5\). \(\oint n_x^2n_z^2 = \int_0^{2\pi}\cos^2\phi\,\mathrm d\phi\int_{-1}^1(1 - u^2)u^2\,\mathrm du = \pi\big(\frac23 - \frac25\big) = \frac{4\pi}{15}\).

2. All indices \(z\): \(C(1 + 1 + 1) = \frac{4\pi}5\), so \(C = \frac{4\pi}{15}\). Indices \(xxzz\): only \(\delta_{xx}\delta_{zz}\) survives, \(C = \frac{4\pi}{15}\) ✓. (A third check: contracting \(i = j\), \(k = l\) gives \(\oint1 = 4\pi = C(9 + 3 + 3)\) ✓.)

3. \(P = 1 - nn^{\mathsf T}\), so \(\mathrm{tr}(AP) = \mathrm{tr}A - n^{\mathsf T}An = -A_{ij}n_in_j\) and \(\mathrm{tr}(APAP) = \mathrm{tr}(A^2) - 2n^{\mathsf T}A^2n + (n^{\mathsf T}An)^2\). The integrand is \(A_{ij}A_{ij} - 2(A^2)_{ij}n_in_j + \tfrac12(A_{ij}n_in_j)^2\). Integrating:

\[4\pi A_{ij}A_{ij} - 2\cdot\frac{4\pi}3(A^2)_{ii} + \frac12\cdot\frac{4\pi}{15}\big((A_{ii})^2 + 2A_{ij}A_{ij}\big) = 4\pi A_{ij}A_{ij}\Big(1 - \frac23 + \frac1{15}\Big) = \frac{8\pi}5A_{ij}A_{ij},\]

using \((A^2)_{ii} = A_{ij}A_{ij}\) and \(A_{ii} = 0\). \(\square\)

4. Along the axis the bracket is \(1 + 1 = 2\); in the plane it is \(\tfrac14\). Ratio: \(8\). The integral: \(2\pi\int_{-1}^1\big[\tfrac14(1 + u^2)^2 + u^2\big]\mathrm du = 2\pi\big[\tfrac14\big(2 + \tfrac43 + \tfrac25\big) + \tfrac23\big] = 2\pi\cdot\tfrac{24}{15} = \frac{16\pi}5\). Now the flux is \(F = \frac{c^3}{16\pi G}\langle\dot h_+^2 + \dot h_\times^2\rangle\), and with P4's amplitudes at wave frequency \(2\Omega\), \(\langle\dot h_+^2 + \dot h_\times^2\rangle = (2\Omega)^2h_0^2\cdot\tfrac12\big[\big(\tfrac{1 + \cos^2\iota}2\big)^2 + \cos^2\iota\big]\). So

\[P = D^2\oint F\,\mathrm d\Omega = \frac{c^3}{16\pi G}\,4\Omega^2(h_0D)^2\cdot\frac12\cdot\frac{16\pi}5 = \frac{2c^3\Omega^2}{5G}\Big(\frac{4G\mu a^2\Omega^2}{c^4}\Big)^2 = \frac{32G\mu^2a^4\Omega^6}{5c^5}\ ✓.\]

Check. Two independent routes — the general \(\frac{G}{5c^5}\langle\overset{...}{Q}_{ij}\overset{...}{Q}_{ij}\rangle\) and the explicit angular pattern — give the same power, so both the \(\frac{8\pi}5\) and the \((1 + \cos^2\iota)/2\), \(\cos\iota\) amplitudes are right. Dimensions of the last line: \(c^3\Omega^2/G\) is \(\mathrm{kg\,s^{-3}}\) and \(h_0D\) is a length, so the product is \(\mathrm{kg\,m^2\,s^{-3}} = \mathrm W\) ✓. The pattern: strongest along the axis, weakest (but not zero) in the plane — unlike a dipole, which is silent along its axis.

P8 Why nobody built a laboratory source medium

exercises: the quadrupole luminosity for a rotating body, a strength-of-materials limit, estimation

A steel cylinder of length \(L = 20\) m and radius \(1\) m (density \(\rho = 7850\ \mathrm{kg\,m^{-3}}\), tensile strength \(\sigma = 5.0\times10^{8}\) Pa) is spun about a perpendicular axis through its centre.

  1. Treating it as a thin rod, show that the tension at the centre is \(\rho A\omega^2L^2/8\), and find the largest \(\omega\) before it breaks.
  2. Show that a rod rotating at \(\omega\) radiates \(P = \frac{2G}{45c^5}M^2L^4\omega^6\), and evaluate it.
  3. Estimate the strain it produces one wavelength away (roughly where the wave zone starts), using the axial amplitude \(h = 4G\mathcal I\omega^2/c^4r\) with \(\mathcal I = ML^2/12\). Compare with LIGO's \(10^{-21}\).
Hints · 3 rungs
Rung 1

Three numbers: a breaking speed, a power in watts, a strain. The key observation for part 2 is that the rod's second moment has exactly the shape of the binary's.

Rung 2

The tension at the centre supports the centripetal acceleration of everything on one side: \(\int_0^{L/2}\rho A\,\omega^2s\,\mathrm ds\). For part 2, with the rod along \(n(t) = (\cos\omega t, \sin\omega t, 0)\), \(I_{ij} = \int\lambda s^2\,\mathrm ds\;n_in_j = \mathcal I\,n_in_j\).

Rung 3

That is worked example 2 with \(\mu a^2\to\mathcal I = ML^2/12\) and \(\Omega\to\omega\): \(P = \frac{32G\mathcal I^2\omega^6}{5c^5}\).

Worked solution

Idea. A spinning rod is a binary with the masses smeared along a line. Its power is limited by \(\omega^6\), and \(\omega\) is limited by the steel. Nothing about the steel can compensate for \(G/c^5\).

1. \(T = \int_0^{L/2}\rho A\omega^2s\,\mathrm ds = \rho A\omega^2L^2/8\), so the stress is \(\rho\omega^2L^2/8\), and it breaks when this reaches \(\sigma\): \(\omega_{\max} = \frac1L\sqrt{8\sigma/\rho} = \frac1{20}\sqrt{8\times5.0\times10^8/7850} = 35.7\ \mathrm{rad\,s^{-1}}\) — 5.7 revolutions a second, tip speed 357 m/s.

2. \(M = \rho\pi R^2L = 4.93\times10^{5}\) kg. With \(\mathcal I = ML^2/12\), worked example 2's computation gives \(P = \frac{32G\mathcal I^2\omega^6}{5c^5} = \frac{32}{5\cdot144}\frac{GM^2L^4\omega^6}{c^5} = \frac{2G}{45c^5}M^2L^4\omega^6\). Numbers: \(\frac{2G}{45c^5} = 1.225\times10^{-54}\ \mathrm{s^3\,kg^{-1}\,m^{-2}}\), \(M^2 = 2.43\times10^{11}\), \(L^4 = 1.6\times10^{5}\), \(\omega^6 = 2.07\times10^{9}\):

\[P = 9.9\times10^{-29}\ \mathrm W.\]

At that rate it takes \(3\times10^{20}\) years to radiate one joule.

3. The wave frequency is \(2\omega\), so \(\lambda = 2\pi c/2\omega = 2.64\times10^{7}\) m — four Earth radii. There, \(h = \frac{4G\mathcal I\omega^2}{c^4\lambda}\) with \(\mathcal I = 1.64\times10^{7}\ \mathrm{kg\,m^2}\) gives \(h = 2.6\times10^{-41}\): twenty orders of magnitude below LIGO's \(10^{-21}\), and at \(11\) Hz, below LIGO's band besides. Closer than \(\lambda\) there is no wave, only the near-zone Newtonian field of a rotating rod.

Check. Dimensions of part 2: \(\frac{G}{c^5}M^2L^4\omega^6\) is \(\frac{\mathrm{m^3kg^{-1}s^{-2}}}{\mathrm{m^5s^{-5}}}\mathrm{kg^2m^4s^{-6}} = \mathrm{kg\,m^2s^{-3}} = \mathrm W\) ✓. Estimate check with the deck's \((c^5/G)(v/c)^{10}\)-type scaling: \(P\sim\frac{c^5}{G}\big(\frac{GM}{Lc^2}\big)^2\big(\frac vc\big)^6\) with \(GM/Lc^2 = 1.8\times10^{-23}\) and \(v/c = 1.2\times10^{-6}\) gives \(\sim10^{-29}\) W, the same order ✓. MTW's classic version of this problem (chapter 36, with \(\omega = 28\) rad/s) gets \(2\times10^{-29}\) W.

Binaries

P9 The Earth's orbit routine

exercises: worked examples 2 and 3 with numbers, sanity of scales

Treat the Earth–Sun system as a circular binary with \(a = 1\) au.

  1. Compute the gravitational-wave power, and compare with the Sun's luminosity.
  2. Compute \(\dot a\) in metres per year, and the coalescence time in years.
  3. Show that \(P = \frac{32}5\frac{c^5}G\big(\frac\mu M\big)^2\big(\frac vc\big)^{10}\), and evaluate \(v/c\) to confirm part 1.
Hints · 3 rungs
Rung 1

Four numbers and one identity. Everything is a substitution into the running example's formulas.

Rung 2

\(\mu = M_\odot M_\oplus/(M_\odot + M_\oplus)\) is \(M_\oplus\) to six figures. Keep \(G^4\mu^2M^3\) as \((GM)^3\cdot G\mu^2\) so that the accurately known \(GM_\odot\) does the work.

Rung 3

\(P = \frac{32}5\frac{(GM)^3G\mu^2}{c^5a^5}\) with \(GM = 1.32712\times10^{20}\cdot(1 + 3.0\times10^{-6})\ \mathrm{m^3s^{-2}}\).

Worked solution

Idea. The running example's formulas, with a very small \(\mu/M\) and a very small \(v/c\).

1. \(P = \frac{32}5\frac{G^4\mu^2M^3}{c^5a^5} = 196\) W — a small electric heater. The Sun radiates \(3.828\times10^{26}\) W of light, so the ratio is \(5.1\times10^{-25}\).

2. \(\dot a = -\frac{64}5\frac{G^3\mu M^2}{c^5a^3} = -1.11\times10^{-20}\ \mathrm{m\,s^{-1}} = -3.5\times10^{-13}\) m/yr, a few hundred proton diameters a year. \(\tau = \frac5{256}\frac{c^5a^4}{G^3\mu M^2} = 1.07\times10^{23}\) yr, about \(8\times10^{12}\) times the age of the universe. (The Sun will become a red giant in about \(5\times10^{9}\) years; gravitational radiation will not matter to the Earth.)

3. With \(v^2 = GM/a\), \(a^5 = (GM)^5/v^{10}\), so \(\frac{G^4\mu^2M^3}{c^5a^5} = \frac{G^4\mu^2M^3v^{10}}{c^5G^5M^5} = \frac{c^5}G\frac{\mu^2}{M^2}\frac{v^{10}}{c^{10}}\). Numbers: \(v = \sqrt{GM/a} = 2.978\times10^{4}\) m/s, \(v/c = 9.94\times10^{-5}\), \((v/c)^{10} = 9.37\times10^{-41}\), \((\mu/M)^2 = 9.02\times10^{-12}\), \(c^5/G = 3.628\times10^{52}\) W: \(P = 6.4\times3.628\times10^{52}\times9.02\times10^{-12}\times9.37\times10^{-41} = 196\) W ✓.

Check. \(\tau\) from \(a/(4|\dot a|)\): \(1.496\times10^{11}/(4\times3.5\times10^{-13}) = 1.07\times10^{23}\) yr ✓ (the \(4\) is the \(a^4\) law). Dimensions were checked in worked examples 2 and 3 ✓.

P10 The chirp, integrated hard

exercises: solving \(\dot f\propto f^{11/3}\), counting cycles, inverting for the chirp mass

A detector records a chirp whose frequency is \(f = 35\) Hz and rising at \(\dot f = 80\ \mathrm{Hz\,s^{-1}}\).

  1. Assuming the quadrupole chirp, find \(\mathcal M\) in solar masses.
  2. How long until coalescence? Show the answer is \(\frac38f/\dot f\), independent of \(\mathcal M\).
  3. Show that the number of wave cycles between \(f_1\) and \(f_2\) is \(N = \frac{1}{32\pi^{8/3}}\big(\frac{G\mathcal M}{c^3}\big)^{-5/3}\big(f_1^{-5/3} - f_2^{-5/3}\big)\), and evaluate it from 35 to 150 Hz, and the time taken.
  4. If the two masses are equal, find them, and the Newtonian separation at 150 Hz in units of \(GM/c^2\). Comment.
Hints · 3 rungs
Rung 1

Part 1 is worked example 4's boxed formula solved for \(\mathcal M\); part 2 is its step 4; part 3 is \(N = \int f\,\mathrm dt\), changed into an integral over \(f\).

Rung 2

\(\frac{G\mathcal M}{c^3} = \Big(\frac{5}{96}\pi^{-8/3}f^{-11/3}\dot f\Big)^{3/5}\). For part 3, \(\mathrm dt = \mathrm df/\dot f\).

Rung 3

\(N = \int_{f_1}^{f_2}\frac{f}{\dot f}\,\mathrm df = \frac{5}{96}\pi^{-8/3}\theta^{-5/3}\int_{f_1}^{f_2}f^{-8/3}\,\mathrm df\) with \(\theta = G\mathcal M/c^3\).

Worked solution

Idea. One differential equation, \(\dot f = Kf^{11/3}\) with \(K = \frac{96}5\pi^{8/3}\theta^{5/3}\), and one unknown, \(\theta = G\mathcal M/c^3\). Every observable of the inspiral follows from it.

1. \(\theta^{5/3} = \frac{5}{96}\pi^{-8/3}f^{-11/3}\dot f\). With \(f = 35\), \(\dot f = 80\): \(f^{-11/3} = 2.18\times10^{-6}\), \(\pi^{-8/3} = 0.0472\), so \(\theta^{5/3} = 0.05208\times0.0472\times2.18\times10^{-6}\times80 = 4.29\times10^{-7}\) and \(\theta = 1.512\times10^{-4}\) s. Divide by \(GM_\odot/c^3 = 4.9255\times10^{-6}\) s: \(\mathcal M = 30.7\,M_\odot\).

2. From \(\frac{\mathrm d}{\mathrm dt}f^{-8/3} = -\frac83K\): \(t_c - t = \frac{3}{8K}f^{-8/3} = \frac38\frac{f}{Kf^{11/3}} = \frac38\frac f{\dot f}\). Here \(\frac38\cdot\frac{35}{80} = 0.164\) s. It needs no \(\mathcal M\), because \(\dot f\) already contains it.

3. \(N = \int f\,\mathrm dt = \int_{f_1}^{f_2}\frac{f}{Kf^{11/3}}\mathrm df = \frac1K\cdot\frac35\big(f_1^{-5/3} - f_2^{-5/3}\big) = \frac{3}{5}\cdot\frac{5}{96}\pi^{-8/3}\theta^{-5/3}(\ldots) = \frac{1}{32\pi^{8/3}}\theta^{-5/3}(f_1^{-5/3} - f_2^{-5/3})\). Numbers: \(35^{-5/3} = 2.67\times10^{-3}\), \(150^{-5/3} = 2.36\times10^{-4}\), \(\theta^{-5/3} = 2.33\times10^{6}\): \(N = 8.4\) cycles. Time: \(t(150) - t(35) = 0.164 - 0.0034 = 0.161\) s.

4. Equal masses: \(\mathcal M = m\cdot\frac{m^{6/5}}{(2m)^{1/5}}\big/m = 2^{-1/5}m\), so \(m = 2^{1/5}\times30.7 = 35.3\,M_\odot\) each, \(M = 70.5\,M_\odot\). At \(f = 150\) Hz, \(a = (GM)^{1/3}(\pi f)^{-2/3} = 348\) km \(= 3.3\,GM/c^2\). Two objects of \(35\,M_\odot\) that close: no star is that compact, and the Newtonian formula is being used where \(v/c = (GM\pi f)^{1/3}/c\approx0.55\). The chirp measured the mass correctly (the early cycles are in the regime of validity); the late cycles are telling you it is time to stop using this unit.

Check. These are GW150914's numbers: 8 cycles from 35 to 150 Hz in about 0.2 s, detector-frame chirp mass about \(30\,M_\odot\) ✓. Dimensions: \(\theta\) is a time, so \(\theta^{-5/3}f^{-5/3}\) is dimensionless — a count ✓. Limit: \(f_2\to\infty\) gives a finite \(N\) — only finitely many cycles remain before coalescence, as \(t_c\) being finite requires ✓.

P11 Hulse–Taylor to five per cent hard

exercises: the checkpoint's number, end to end; the eccentricity factor; the orbital decay as a clock drift

PSR B1913+16: \(P_b = 0.322997448918\) d, \(e = 0.6171340\), \(m_p = 1.438\,M_\odot\), \(m_c = 1.390\,M_\odot\). Take as given (cited, Peters and Mathews 1963) that the orbit-averaged power of an eccentric orbit is the circular power at the same semi-major axis times \(f(e) = \big(1 + \frac{73}{24}e^2 + \frac{37}{96}e^4\big)(1 - e^2)^{-7/2}\).

  1. Derive \(\dot P_b = -\frac{192\pi}5\big(\frac{2\pi G\mathcal M}{c^3P_b}\big)^{5/3}f(e)\), explaining why the eccentric correction enters as a pure factor.
  2. Evaluate it and compare with the measured \(-2.423(1)\times10^{-12}\), and with \(-2.398(4)\times10^{-12}\) after the Galactic correction.
  3. Find the semi-major axis and \(\dot a\) in metres per year.
  4. By how many seconds has the time of periastron drifted after 30 years?
Hints · 3 rungs
Rung 1

The answer is one dimensionless number near \(10^{-12}\). The route is worked examples 2 to 4 rewritten in terms of \(P_b\), then one factor for the ellipse.

Rung 2

For an ellipse \(E = -G\mu M/2a\) depends only on \(a\), and \(P_b\propto a^{3/2}\). So \(\dot P_b/P_b = \tfrac32\dot a/a = -\tfrac32\dot E/E\), which is linear in the power.

Rung 3

\(\mathcal M = \mu^{3/5}M^{2/5}\) with \(M = 2.828\), \(\mu = 0.70680\) (solar masses): \(\mathcal M = 1.2307\,M_\odot\), so \(G\mathcal M/c^3 = 6.0620\ \mu\)s.

Worked solution

Idea. Energy balance turns the power into \(\dot E\), Kepler turns \(\dot E\) into \(\dot P_b\), and the ellipse only rescales the power. This is worked example 5.

1. \(E = -G\mu M/2a\) and \(P_b = 2\pi\sqrt{a^3/GM}\) hold for ellipses (unit 08). Then \(\frac{\dot P_b}{P_b} = \frac32\frac{\dot a}a = -\frac32\frac{\dot E}{E} = -\frac32\frac{\langle P\rangle}{|E|}\), linear in \(\langle P\rangle = P_{\rm circ}(a)f(e)\): so \(\dot P_b = \dot P_b^{\rm circ}f(e)\). For the circle, worked example 4 with \(f_{\rm gw} = 2/P_b\): \(\dot P_b = -\frac2{f^2}\dot f = -\frac{192}5\pi^{8/3}\theta^{5/3}f^{5/3} = -\frac{192\pi}{5}\big(\frac{2\pi\theta}{P_b}\big)^{5/3}\), since \(\pi^{8/3}f^{5/3} = \pi(\pi f)^{5/3}\) and \(\pi f = 2\pi/P_b\).

2. \(P_b = 27\,906.98\) s. \(\frac{2\pi\theta}{P_b} = \frac{2\pi\times6.0620\times10^{-6}}{27\,906.98} = 1.3648\times10^{-9}\); to the \(5/3\): \(1.6794\times10^{-15}\). Times \(\frac{192\pi}5 = 120.64\): \(\dot P_b^{\rm circ} = -2.026\times10^{-13}\). \(e^2 = 0.38085\), \(e^4 = 0.14505\): \(f(e) = \frac{2.2143}{0.18676} = 11.857\). So

\[\dot P_b = -2.402\times10^{-12}.\]

Against the raw measurement: \(2.423/2.402 = 1.0087\), within \(0.9\%\). Against the corrected one: \(2.398/2.402 = 0.9983\), within \(0.2\%\).

3. \(a = \big(GM(P_b/2\pi)^2\big)^{1/3}\) with \(GM = 2.828\times1.32712\times10^{20} = 3.7531\times10^{20}\): \(a = 1.949\times10^{9}\) m, 2.8 solar radii. \(\dot a = \frac23\frac a{P_b}\dot P_b = \frac23\cdot\frac{1.949\times10^{9}}{27\,906.98}\cdot(-2.402\times10^{-12}) = -1.118\times10^{-7}\) m/s \(= -3.53\) m/yr.

4. \(\Delta T = \frac12\frac{\dot P_b}{P_b}t^2\) with \(t = 30\) yr \(= 9.467\times10^{8}\) s: \(\frac12\times(-8.607\times10^{-17})\times8.963\times10^{17} = -38.6\) s.

Check. \(\dot P_b\) is seconds per second, dimensionless ✓. Limit \(e\to0\): \(f\to1\) and the circular formula returns ✓. The two classic slips give recognisable wrong answers: dropping \(f(e)\) gives \(-2.03\times10^{-13}\), off by 12; using the orbital frequency where the wave frequency belongs multiplies by \(2^{-5/3} = 0.315\). The checkpoint's \(5\%\) is met by a factor of five.

P12 LIGO's number medium

exercises: the chirp amplitude, the interferometer response, the Isaacson flux, estimation

GW150914: source-frame masses \(36\) and \(29\,M_\odot\), redshift \(z = 0.09\), luminosity distance \(D = 410\) Mpc. Take as given (cited; unit 39) that the detector sees the chirp mass multiplied by \(1 + z\) and that \(D\) is the distance to use in the amplitude.

  1. Compute the source-frame and detector-frame chirp masses.
  2. Estimate the strain for an optimally oriented source at \(f = 35\) Hz and \(150\) Hz, and compare with the measured peak of \(1.0\times10^{-21}\).
  3. With \(h = 1.0\times10^{-21}\), how far does each 4 km arm change length, and what is the differential change? Compare with a proton's radius, \(0.84\times10^{-15}\) m.
  4. The energy flux at Earth for a circularly polarised wave of that strain at 150 Hz, and the power implied if it were emitted isotropically from \(410\) Mpc. Compare with the reported peak of \(3.6\times10^{49}\) W.
Hints · 3 rungs
Rung 1

Four estimates, each a number with a comparison. The formulas are worked example 4's amplitude, the detector slide's round-trip times, and the Isaacson flux.

Rung 2

\(h = \frac4D\big(\frac{G\mathcal M}{c^2}\big)^{5/3}\big(\frac{\pi f}c\big)^{2/3}\), with \(G\mathcal M/c^2 = \mathcal M/M_\odot\times1476.6\) m. For part 4, \(F = \frac{c^3}{16\pi G}\langle\dot h_+^2 + \dot h_\times^2\rangle\) and circular polarisation makes the average \((2\pi fh)^2\).

Rung 3

\(\mathcal M_{\rm source} = (36\cdot29)^{3/5}/65^{1/5}\); \(G\mathcal M_{\rm det}/c^2 = 30.6\times1476.6\) m \(= 45.2\) km, and \(D = 1.265\times10^{25}\) m.

Worked solution

Idea. The quadrupole chirp gives the order of magnitude of everything LIGO measured; where it disagrees it is by the known factors — orientation, and the breakdown near merger.

1. \(36\times29 = 1044\), \(1044^{3/5} = 64.7\), \(65^{1/5} = 2.30\): \(\mathcal M = 28.1\,M_\odot\) at the source, and \(1.09\times28.1 = 30.6\,M_\odot\) at the detector.

2. \((G\mathcal M/c^2)^{5/3} = (4.518\times10^{4})^{5/3} = 5.73\times10^{7}\ \mathrm{m^{5/3}}\). At 35 Hz, \((\pi f/c)^{2/3} = (3.668\times10^{-7})^{2/3} = 5.12\times10^{-5}\ \mathrm{m^{-2/3}}\), so \(h = \frac{4}{1.265\times10^{25}}\times5.73\times10^{7}\times5.12\times10^{-5} = 0.93\times10^{-21}\). At 150 Hz the last factor is larger by \((150/35)^{2/3} = 2.64\): \(h = 2.5\times10^{-21}\). The measured peak, \(1.0\times10^{-21}\), sits between: the source was neither face-on nor overhead with the arms aligned (factors of order one each), and at 150 Hz the formula is past its validity anyway.

3. Each arm: \(\tfrac12hL = \tfrac12\times10^{-21}\times4000 = 2\times10^{-18}\) m; differential \(hL = 4\times10^{-18}\) m, which is \(4\times10^{-18}/0.84\times10^{-15} = 0.005\) of a proton radius.

4. \(F = \frac{c^3}{16\pi G}(2\pi fh)^2 = \frac{2.694\times10^{25}}{3.354\times10^{-9}}\times(9.42\times10^{-19})^2 = 7.1\times10^{-3}\ \mathrm{W\,m^{-2}}\). Times \(4\pi D^2 = 2.01\times10^{51}\ \mathrm{m^2}\): \(1.4\times10^{49}\) W. Reported peak: \(3.6\times10^{49}\) W. Same order; the difference is the orientation factor of part 2 (entering squared) and the fact that the emission is not isotropic (P7: the axis gets 8 times the plane).

Check. Dimensions: \(\frac{c^3}{G}\cdot\mathrm s^{-2}\) is \(\frac{\mathrm{m^3s^{-3}}}{\mathrm{m^3kg^{-1}s^{-2}}}\mathrm s^{-2} = \mathrm{kg\,s^{-3}} = \mathrm{W\,m^{-2}}\) ✓. The average power over the event, \(3.0\,M_\odot c^2/0.2\ \mathrm s = 2.7\times10^{48}\) W, is below the peak, as it must be ✓. And \(7\times10^{-3}\ \mathrm{W\,m^{-2}}\) — five millionths of sunlight — moved a mirror by \(2\times10^{-18}\) m: spacetime's stiffness \(c^4/G\) in one sentence.