Part F · Einstein · branch

Gravitational waves: ten components, two polarisations, one chirp

Linearise the field equations about flat spacetime and they become unit 24's wave equation. Count what the gauge leaves and two polarisations remain; solve with the retarded Green's function and a binary star's orbital period shrinks by 2.40×10⁻¹² seconds every second, which is what the sky does.

needs 35, 21, 24, 20~3 h slides + 6 h problemsproblems: ./mr problems 38

Unit 35 ended by linearising the vacuum equations for one slide and finding a wave with two polarisations. This unit does that properly, then asks the two questions the slide could not: what makes such waves, and how would anyone know? The answers are the quadrupole formula and a binary pulsar whose orbit has been shrinking, on schedule, for as long as it has been watched.

Each prerequisite does one job. Unit 35 supplies the field equations, the coupling $8\pi G/c^4$ and the first look at the linearised theory. Unit 24 supplies the retarded Green's function — computed there by unit 20's contour integrals — and the electromagnetic gauge freedom that gravity's turns out to mirror. Unit 21 supplies the method of multiple scales, which is what licenses treating a slowly decaying orbit as a sequence of circles. Unit 20 is there because unit 24's Green's function is its theorem.

How to read this deck

One example runs through everything: a binary star on a circular orbit. Section 1 linearises the field equations and meets the gauge freedom. Section 2 counts what the gauge leaves — two polarisations — and shows what each does to a ring of free masses. Section 3 solves the sourced equation and derives the quadrupole formula for the field the binary makes. Section 4 asks what energy the wave carries, and why that question is subtler than it looks. Section 5 is the binary's inspiral: power, decay, coalescence time, chirp. Section 6 evaluates it twice, for the Hulse–Taylor pulsar and for LIGO's first event, and ends at the interferometer.

Conventions

Signature $(-,+,+,+)$, $x^0 = ct$, indices raised and lowered with $\eta$ at first order, $\Box = \eta^{\mu\nu}\partial_\mu\partial_\nu$ as in unit 35. Latin indices run over space. Summation over repeated spatial indices regardless of height, since the spatial metric is $\delta_{ij}$ at this order. Constants: $G = 6.67430\times10^{-11}\ \mathrm{m^3\,kg^{-1}\,s^{-2}}$, $c = 299\,792\,458$ m/s, $GM_\odot = 1.32712440018\times10^{20}\ \mathrm{m^3\,s^{-2}}$, so $GM_\odot/c^3 = 4.9255\ \mu$s and $GM_\odot/c^2 = 1476.6$ m. Every number in the deck was computed from these.

38 · overview

One binary, two polarisations, one number

the running example

Two masses on a circular orbit. Newton says it goes round for ever. General relativity says it radiates, the orbit shrinks, the frequency rises — a chirp — and the stars collide in a finite time that depends on the masses through one combination only.

Evaluated twice: for PSR B1913+16, whose period falls by $2.402\times10^{-12}$ s per second against $2.423\times10^{-12}$ measured; and for GW150914, which moved each of LIGO's 4 km mirrors by $2\times10^{-18}$ m.

section	what it does to the binary	and gets
1 linearise	the equation its field obeys	$\Box\bar h_{\mu\nu} = -\frac{16\pi G}{c^4}T_{\mu\nu}$
2 polarisations	what that field does to a ring	$10 - 4 - 4 = 2$
3 quadrupole	the field it actually makes	$h\sim\frac{2G}{c^4r}\ddot Q$
4 energy	what the field carries off	$P = \frac{G}{5c^5}\langle\overset{...}{Q}\overset{...}{Q}\rangle$
5 the chirp	how the orbit decays	$\dot f\propto\mathcal M^{5/3}f^{11/3}$
6 measured	two binaries, two numbers	$\dot P_b = -2.402\times10^{-12}$

$Q_{ij}$: the source's trace-free second mass moment. $\mathcal M = \mu^{3/5}M^{2/5}$: the chirp mass.

getting around

A dotted underline is a glossary term: click it. G lists them all; N opens the notes, where the long proofs are, under Full proof. A solid underline, like the binary, shows the running example on hover.

Why a binary. It is the simplest source that radiates at all — section 3 proves that anything simpler cannot — and it is the source nature actually provides. Every gravitational wave detected so far has come from a binary, and the only other precision test of the radiation, the binary pulsars, is a binary too.

The route, in the table's last column. Each entry is the result its section boxes, and each feeds the next: the wave equation (section 1) is solved with unit 24's retarded Green's function to give the field as the second time derivative of the trace-free mass moment $Q$ (section 3); the field's rate of change, squared and integrated over a sphere, is the power (section 4); setting that power equal to the loss of orbital energy $-G\mu M/2a$ gives the chirp, with the masses entering only through the chirp mass $\mathcal M$ (section 5); and the chirp written in terms of the orbital period, times a cited factor for the ellipse, is the pulsar's $\dot P_b$ (section 6). Section 2 is the side road that says what the field is: two numbers per direction, and what they do to free masses.

The one number. The checkpoint asks you to reproduce the Hulse–Taylor period derivative to within five per cent. The deck gets it to within one per cent of the raw measurement and 0.2 per cent of the corrected one. The only input beyond this unit is a cited factor for the orbit's eccentricity, and the deck says exactly where it comes from.

What is not here. The last few orbits and the merger of two black holes are not computable by the methods of this unit: the linear theory, the slow-motion expansion and the adiabatic approximation all fail there, and the deck shows where with numbers. Black holes themselves are units 36 and 40; this unit never uses their metric.

section 1

Linearised gravity

A derivative of the field equations, and a redundancy met for the second time.

38 · linearise

A metric near Minkowski's

definition · linearisation

Let $g(\varepsilon)$ be a family of metrics on $\mathbb R^4$, smooth in $\varepsilon$, with $g(0) = \eta$. The metric perturbation is $$h = \frac{\mathrm d}{\mathrm d\varepsilon}\Big|_{\varepsilon=0} g(\varepsilon),$$ a symmetric $(0,2)$-tensor field. The linearisation of any quantity $F[g]$ is $\frac{\mathrm d}{\mathrm d\varepsilon}F[g(\varepsilon)]\big|_0$ — its derivative at $\eta$ in the direction $h$.

Physics. Unit 35's weak field: $g = \eta + h$ with $|h_{\mu\nu}|\ll1$ in some chart, keeping terms linear in $h$ and dropping the rest. The definition says precisely what "dropping" means: it is taking a derivative.

lemma · the linear pieces

To first order, $g^{\mu\nu} = \eta^{\mu\nu} - h^{\mu\nu}$, and $$\Gamma^\lambda_{\mu\nu} = \tfrac12\eta^{\lambda\rho}\big(\partial_\mu h_{\rho\nu} + \partial_\nu h_{\rho\mu} - \partial_\rho h_{\mu\nu}\big).$$

proof

$(\eta^{\mu\lambda} - h^{\mu\lambda})(\eta_{\lambda\nu} + h_{\lambda\nu}) = \delta^\mu_\nu + O(h^2)$. In unit 32's formula for $\Gamma$ every term is $g^{-1}\partial g$; $\partial\eta = 0$, so only $\eta^{-1}\partial h$ survives at first order. $\square$

Check. $h$ is dimensionless, $[\Gamma] = \mathrm{m^{-1}}$ ✓. Static, with $h_{00} = -2\Phi/c^2$: $\Gamma^i_{00} = -\tfrac12\partial_ih_{00} = \partial_i\Phi/c^2$, unit 32's Newtonian limit ✓.

Why define it as a derivative. "Small $h$" is not a coordinate-free statement: the Minkowski metric in rotating coordinates has components that grow without bound, and a perfectly curved metric can be made to look like $\eta$ at a point. What is invariant is the family $g(\varepsilon)$ and its derivative at $\varepsilon = 0$. Everything this unit calls "first order" is a derivative in $\varepsilon$, and "$h$ is small" is shorthand for "we are computing that derivative". This is also why the linear theory is exact as a statement about tangent vectors to the space of solutions, even when no actual solution is exactly linear.

Indices. At first order an index on $h$ is raised with $\eta$, not $g$: the difference is $O(h^2)$. So $h^{\mu\nu} = \eta^{\mu\alpha}\eta^{\nu\beta}h_{\alpha\beta}$ and $h = \eta^{\mu\nu}h_{\mu\nu}$ is the trace. With the signature $(-,+,+,+)$, $h = -h_{00} + h_{11} + h_{22} + h_{33}$.

The $\Gamma\Gamma$ terms. Each Christoffel symbol is $O(\varepsilon)$, so any product of two is $O(\varepsilon^2)$ and has zero derivative at $\varepsilon = 0$. That is what makes the curvature linear in $h$ on the next slide: of unit 33's $R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho_{\nu\sigma} - \partial_\nu\Gamma^\rho_{\mu\sigma} + \Gamma\Gamma - \Gamma\Gamma$ only the two derivative terms survive.

38 · linearise

The linearised field equations

definition · the trace-reversed perturbation

$\bar h_{\mu\nu} = h_{\mu\nu} - \tfrac12\eta_{\mu\nu}h$. Its trace is $\bar h = -h$, and in four dimensions the operation is an involution: $\bar{\bar h}_{\mu\nu} = h_{\mu\nu}$.

theorem · the linearised Einstein tensor $$\begin{aligned}G^{(1)}_{\mu\nu} = -\tfrac12\Big(&\Box\bar h_{\mu\nu} + \eta_{\mu\nu}\partial^\alpha\partial^\beta\bar h_{\alpha\beta}\\ &- \partial_\mu\partial^\alpha\bar h_{\nu\alpha} - \partial_\nu\partial^\alpha\bar h_{\mu\alpha}\Big)\end{aligned}$$

with $\Box = \eta^{\mu\nu}\partial_\mu\partial_\nu = -c^{-2}\partial_t^2 + \nabla^2$. The field equations at first order are $G^{(1)}_{\mu\nu} = \dfrac{8\pi G}{c^4}T_{\mu\nu}$.

proof: linearise Ricci from the previous slide's $\Gamma$, contract, substitute — Full proof in the notes (N)

Why trace-reverse. In terms of $h$ the operator has six terms; in terms of $\bar h$ it has four, and three of them carry a divergence $\partial^\alpha\bar h_{\alpha\beta}$. The next section's gauge choice makes that divergence zero.

checks

Dimensions. $[\partial\partial\bar h] = \mathrm{m^{-2}} = [8\pi GT/c^4]$ ✓ (unit 35).

Static limit. $\bar h_{00} = -4\Phi/c^2$, others zero: the $00$ equation is $-\tfrac12\nabla^2\bar h_{00} = 8\pi G\rho/c^2$, i.e. $\nabla^2\Phi = 4\pi G\rho$ — Poisson ✓.

Full proof

Step 1, Ricci. With unit 33's $R_{\mu\nu} = R^\lambda{}_{\mu\lambda\nu}$ and the $\Gamma\Gamma$ terms gone, $R^{(1)}_{\mu\nu} = \partial_\lambda\Gamma^\lambda_{\nu\mu} - \partial_\nu\Gamma^\lambda_{\lambda\mu}$. Insert the previous slide's $\Gamma$. The first term gives $\tfrac12(\partial_\lambda\partial_\nu h^\lambda{}_\mu + \partial_\lambda\partial_\mu h^\lambda{}_\nu - \Box h_{\mu\nu})$; for the second, $\Gamma^\lambda_{\lambda\mu} = \tfrac12\partial_\mu h$ because the other two terms cancel under the contraction. So $$R^{(1)}_{\mu\nu} = \tfrac12\big(\partial_\lambda\partial_\mu h^\lambda{}_\nu + \partial_\lambda\partial_\nu h^\lambda{}_\mu - \Box h_{\mu\nu} - \partial_\mu\partial_\nu h\big).$$ Step 2, the scalar. $R^{(1)} = \eta^{\mu\nu}R^{(1)}_{\mu\nu} = \partial_\lambda\partial_\mu h^{\lambda\mu} - \Box h$. Step 3, Einstein. $G^{(1)}_{\mu\nu} = R^{(1)}_{\mu\nu} - \tfrac12\eta_{\mu\nu}R^{(1)}$ (the correction from $g_{\mu\nu}R$ versus $\eta_{\mu\nu}R$ is second order, since $R$ itself is first order). Step 4, substitute $h_{\mu\nu} = \bar h_{\mu\nu} - \tfrac12\eta_{\mu\nu}\bar h$ (the involution) and $h = -\bar h$. The terms in $\bar h$ with no divergence cancel: $-\tfrac12\Box h_{\mu\nu} = -\tfrac12\Box\bar h_{\mu\nu} + \tfrac14\eta_{\mu\nu}\Box\bar h$, the $-\tfrac12\partial_\mu\partial_\nu h$ term is $+\tfrac12\partial_\mu\partial_\nu\bar h$ and cancels against the $-\tfrac12\partial_\mu\partial_\nu\bar h$ coming from the two $\partial_\lambda\partial h^\lambda$ terms, and the $\eta_{\mu\nu}\Box\bar h$ pieces cancel between Ricci and $-\tfrac12\eta R$. What is left is the slide's four terms. $\square$ The author checked the identity in full generality, symbolically, for arbitrary $h_{\mu\nu}(x)$; do it by hand once, with every index written out, because the cancellations are the content — problem P1 does the trace-reversal algebra they turn on.

The involution. $\bar h = \eta^{\mu\nu}(h_{\mu\nu} - \tfrac12\eta_{\mu\nu}h) = h - 2h = -h$, using $\eta^{\mu\nu}\eta_{\mu\nu} = 4$. Then $\bar{\bar h}_{\mu\nu} = \bar h_{\mu\nu} - \tfrac12\eta_{\mu\nu}\bar h = h_{\mu\nu} - \tfrac12\eta_{\mu\nu}h + \tfrac12\eta_{\mu\nu}h = h_{\mu\nu}$. In $n$ dimensions the trace is multiplied by $1 - n/2$ and the involution fails unless $n = 4$ (problem P1).

The name. $\bar h$ is related to $h$ exactly as $G_{\mu\nu}$ is related to $R_{\mu\nu}$: subtract half the trace times the metric. Unit 35's trace-reversed field equation is the same move.

38 · linearise

Gauge: a diffeomorphism, infinitesimally

theorem · linearised gauge transformation

Let $\varphi_\varepsilon$ be the flow of a vector field $\xi$. Then $$\varphi_\varepsilon^*\big(g(\varepsilon)\big) = \eta + \varepsilon\big(h + \mathcal L_\xi\eta\big) + O(\varepsilon^2),$$ $$(\mathcal L_\xi\eta)_{\mu\nu} = \partial_\mu\xi_\nu + \partial_\nu\xi_\mu .$$ So $h$ and $h + \partial_\mu\xi_\nu + \partial_\nu\xi_\mu$ are two descriptions of one spacetime.

proof

Differentiate $\varphi_\varepsilon^*g(\varepsilon)$ at $0$ by the product rule: $\frac{\mathrm d}{\mathrm d\varepsilon}\varphi^*_\varepsilon\eta + \varphi_0^*h = \mathcal L_\xi\eta + h$ (unit 31's definition of $\mathcal L$). Unit 31's coordinate formula, with $\partial\eta = 0$, gives $\eta_{\lambda\nu}\partial_\mu\xi^\lambda + \eta_{\mu\lambda}\partial_\nu\xi^\lambda$. $\square$

Why it is the same spacetime. $\varphi_\varepsilon$ is an isometry from $(\mathbb R^4, \varphi_\varepsilon^*g)$ to $(\mathbb R^4, g)$. Every length, angle, geodesic and curvature invariant of one is the same number for the other. Unit 35 counted these four functions as the four that "change no physics".

For the trace-reversed field, with $\partial\cdot\xi = \partial^\lambda\xi_\lambda$:

$$\bar h_{\mu\nu}\ \to\ \bar h_{\mu\nu} + \partial_\mu\xi_\nu + \partial_\nu\xi_\mu - \eta_{\mu\nu}\,\partial\cdot\xi,$$ $$\partial^\mu\bar h_{\mu\nu}\ \to\ \partial^\mu\bar h_{\mu\nu} + \Box\xi_\nu .$$

a sign you will meet

Unit 35 wrote $h\to h - \partial\xi - \partial\xi$: a coordinate change $x\to x + \xi$ rather than a flow. It is the same set of transformations, with $\xi\to-\xi$.

The product rule, carefully. $F(\varepsilon) = \varphi_\varepsilon^*(g(\varepsilon))$ is a function of $\varepsilon$ in which $\varepsilon$ appears twice. Write $F(\varepsilon) = \varphi_\varepsilon^*\eta + \varphi_\varepsilon^*(g(\varepsilon) - \eta)$. The first term's derivative at $0$ is $\mathcal L_\xi\eta$ by unit 31's definition. The second is $\varphi_\varepsilon^*$ applied to something $O(\varepsilon)$; its derivative at $0$ is $\varphi_0^*h = h$, since $\varphi_0$ is the identity and any correction is $O(\varepsilon^2)$.

The trace-reversed version. The change in $h$ is $\delta h_{\mu\nu} = \partial_\mu\xi_\nu + \partial_\nu\xi_\mu$, whose trace is $2\partial\cdot\xi$. So $\delta\bar h_{\mu\nu} = \delta h_{\mu\nu} - \tfrac12\eta_{\mu\nu}\cdot2\partial\cdot\xi$. Take $\partial^\mu$: $\Box\xi_\nu + \partial_\nu\partial\cdot\xi - \partial_\nu\partial\cdot\xi = \Box\xi_\nu$.

History. Einstein found the waves in 1916 and made an error in the radiated power; his 1918 paper corrected it and gave the quadrupole formula of section 3 (itself off by a factor of 2, fixed later by Eddington). In 1922 Eddington also showed that some of the "waves" in Einstein's 1916 treatment were pure coordinate effects that could be made to travel at any speed — they "propagate with the speed of thought", in his phrase. The theorem on this slide is how to tell the difference: a perturbation of the form $\mathcal L_\xi\eta$ is a coordinate effect, whatever it looks like. Section 2 finds what is left when all of those are removed.

38 · linearise

The same redundancy, met twice

	electromagnetism (unit 24)	linearised gravity
field variable	$(\varphi, \vec A)$: 4 functions	$h_{\mu\nu}$: 10 functions
gauge parameter	$\chi$: 1 function	$\xi^\mu$: 4 functions
transformation	$\varphi\to\varphi - \partial_t\chi,\ \vec A\to\vec A + \nabla\chi$	$h_{\mu\nu}\to h_{\mu\nu} + \partial_\mu\xi_\nu + \partial_\nu\xi_\mu$
invariant, measurable	$\vec E,\ \vec B$	linearised Riemann $R^{(1)}_{\mu\nu\rho\sigma}$
Lorenz gauge	$\nabla\cdot\vec A + c^{-2}\partial_t\varphi = 0$	$\partial^\mu\bar h_{\mu\nu} = 0$
reached by solving	a wave equation for $\chi$	a wave equation for each $\xi_\nu$
residual freedom	$\chi$ with $\Box\chi = 0$	$\xi$ with $\Box\xi_\nu = 0$
source must obey	$\partial_t\rho + \nabla\cdot\vec J = 0$	$\partial^\mu T_{\mu\nu} = 0$

theorem · curvature is gauge invariant

$R^{(1)}_{\mu\nu\rho\sigma} = \tfrac12\big(\partial_\nu\partial_\rho h_{\mu\sigma} + \partial_\mu\partial_\sigma h_{\nu\rho} - \partial_\mu\partial_\rho h_{\nu\sigma} - \partial_\nu\partial_\sigma h_{\mu\rho}\big)$ is unchanged by $h\to h + \partial\xi + \partial\xi$: the eight new terms, each a third derivative of $\xi$, cancel in pairs once partials commute (Clairaut). Problem P2.

one difference

Unit 24's transformation is exact. This one is the first-order shadow of the full diffeomorphism group, whose composition does not commute; linearising makes it abelian, like electromagnetism's.

Why meet it twice. Unit 24 met gauge freedom as an embarrassment of riches: the potentials are not determined by the fields, and one chooses a gauge to make the equations pleasant. Gravity has the same structure for a deeper reason — the gauge parameter is a coordinate change, and coordinates were never physical. The table is the dictionary. Reading it row by row is most of section 1 and section 2: once you know how unit 24 went, you know how this goes.

The electromagnetic check. Under unit 24's transformation, $\vec E = -\nabla\varphi - \partial_t\vec A$ changes by $\nabla\partial_t\chi - \partial_t\nabla\chi = 0$, and $\vec B = \nabla\times\vec A$ by $\nabla\times\nabla\chi = 0$ — both by Clairaut. The gravitational theorem is the same argument with more indices.

The source condition. In both theories the Lorenz-gauge equation is a wave equation whose divergence must vanish, and the divergence of the left side vanishes only if the source's does. For electromagnetism that is charge conservation. For gravity it is $\partial^\mu T_{\mu\nu} = 0$ in flat space — a condition section 3 will find a binary star does not satisfy, with consequences.

The Riemann formula. It is the linearisation of unit 33's Riemann tensor with an index lowered, obtained exactly as Ricci was on the previous slide; the author checked it symbolically against unit 33's definition. Its $i0j0$ component in section 2 reproduces unit 33's tidal limit $R_{i0j0} = \partial_i\partial_j\Phi/c^2$ when $h_{00} = -2\Phi/c^2$, which fixes every sign.

38 · linearise

Lorenz gauge, and the wave equation

lemma · Lorenz gauge can always be reached

Given $h$, solve $\Box\xi_\nu = -\partial^\mu\bar h_{\mu\nu}$ — four wave equations, solvable by unit 24's retarded solution. The new $\bar h$ has $\partial^\mu\bar h_{\mu\nu} = 0$, the Lorenz gauge.

theorem · the linearised field equation

In Lorenz gauge,

$$\Box\bar h_{\mu\nu} = -\frac{16\pi G}{c^4}\,T_{\mu\nu}.$$

Proof. Three of the four terms of $G^{(1)}_{\mu\nu}$ carry $\partial^\alpha\bar h_{\alpha\beta} = 0$, leaving $-\tfrac12\Box\bar h_{\mu\nu} = 8\pi GT_{\mu\nu}/c^4$. $\square$

unit 24's box has the other sign

Unit 24 wrote $\Box_{24} = c^{-2}\partial_t^2 - \nabla^2 = -\Box$, so that its retarded solution solves $\Box_{24}u = f$. In that convention the equation above is

$$\Box_{24}\,\bar h_{\mu\nu} = \frac{16\pi G}{c^4}\,T_{\mu\nu}.$$

Section 3 uses this form.

checks

Consistency. $\partial^\mu$ of the left side is $\Box\,\partial^\mu\bar h_{\mu\nu} = 0$, so the source must obey $\partial^\mu T_{\mu\nu} = 0$ — the table's last row ✓.

Vacuum. $\Box\bar h_{\mu\nu} = 0$: ten copies of unit 35's wave equation, every one travelling at $c$ ✓.

Why the gauge exists. The lemma asks for a solution of $\Box\xi = s$ with a given source $s = -\partial^\mu\bar h_{\mu\nu}$. Unit 24's retarded-solution theorem supplies one whenever $s$ is $C^2$ and vanishes before some time and outside a bounded set; for radiation from an isolated source that is the case, and for more general $h$ one adds a homogeneous solution. Unit 35 had to cite this step; this unit's prerequisites prove it.

Why "Lorenz". Ludvig Lorenz, Danish, 1867 — not Hendrik Lorentz, whose name is on the transformations. The gauge is often misattributed; both are in this course.

The running example, so far. The binary's field obeys this equation with $T_{\mu\nu}$ the stars' energy and momentum. Section 3 solves it. Before that, section 2 asks what the vacuum solutions are — what arrives at a detector far from the binary.

section 2

Two polarisations

Ten components, four conditions, four more gauge functions. What is left, and what it does.

38 · polarisations

Counting to two, honestly

A vacuum plane wave in Lorenz gauge: $\bar h_{\mu\nu} = e_{\mu\nu}\cos(k_\lambda x^\lambda)$ with $e$ a constant symmetric matrix. $\Box\bar h = 0$ forces $k$ null (unit 35). Work in $V = \mathrm{Sym}^2(\mathbb R^4)^*$, $\dim V = 10$.

theorem · the degree-of-freedom count

Fix a null $k\ne0$. Let

$L = \{e\in V : k^\mu e_{\mu\nu} = 0\}$, the Lorenz-gauge amplitudes;
$P = \{k_\mu\epsilon_\nu + k_\nu\epsilon_\mu - \eta_{\mu\nu}(k\cdot\epsilon) : \epsilon\in\mathbb R^4\}$, the pure-gauge ones.

Then $P\subset L$, $\ \dim L = 6$, $\ \dim P = 4$, and the physical amplitudes $L/P$ form a space of dimension $\mathbf 2$.

proof, in outline

$\dim L = 6$: $e\mapsto k^\mu e_{\mu\nu}$ maps $V$ onto $\mathbb R^4$; $L$ is its kernel.
$P\subset L$: $k^\mu(k_\mu\epsilon_\nu + k_\nu\epsilon_\mu - \eta_{\mu\nu}k\cdot\epsilon) = 0 + k_\nu(k\cdot\epsilon) - k_\nu(k\cdot\epsilon) = 0$, using $k\cdot k = 0$.
$\dim P = 4$: $\epsilon\mapsto$ that tensor is injective. Details in the notes.

the count, in words

$10$ components $-\ 4$ Lorenz conditions $-\ 4$ residual gauge functions $= 2$. The residual four are the $\xi$ with $\Box\xi = 0$: a plane-wave $\xi = \epsilon\sin(k\cdot x)$ shifts $e$ by exactly an element of $P$.

Full proof

Surjectivity. Since $k\ne0$ pick a vector $a$ with $k\cdot a = 1$. Given any $v\in\mathbb R^4$ put $S_{\mu\nu} = a_\mu v_\nu + v_\mu a_\nu - (k\cdot v)\,a_\mu a_\nu$, which is symmetric. Then $k^\mu S_{\mu\nu} = v_\nu + (k\cdot v)a_\nu - (k\cdot v)a_\nu = v_\nu$. So the map is onto, and by rank–nullity $\dim L = 10 - 4 = 6$.

Injectivity of $\epsilon\mapsto p(\epsilon)$. Suppose $k_\mu\epsilon_\nu + k_\nu\epsilon_\mu - \eta_{\mu\nu}(k\cdot\epsilon) = 0$. Its trace is $2k\cdot\epsilon - 4k\cdot\epsilon = -2k\cdot\epsilon$, so $k\cdot\epsilon = 0$ and $k_\mu\epsilon_\nu + k_\nu\epsilon_\mu = 0$ for all $\mu,\nu$. Choose $\mu$ with $k_\mu\ne0$. Setting $\nu = \mu$ gives $\epsilon_\mu = 0$; then for any $\nu$, $k_\mu\epsilon_\nu = -k_\nu\epsilon_\mu = 0$ gives $\epsilon_\nu = 0$. So $\epsilon = 0$, the map is injective, and $\dim P = 4$.

The residual gauge, concretely. With $\xi_\mu = \epsilon_\mu\sin(k\cdot x)$, $\partial_\mu\xi_\nu = k_\mu\epsilon_\nu\cos(k\cdot x)$, so the transformation of the previous section shifts $\bar h$ by $p(\epsilon)\cos(k\cdot x)$. And $\Box\xi = -(k\cdot k)\,\xi = 0$, so it preserves the Lorenz condition, as it must, since $P\subset L$.

Against unit 35's count. Unit 35 wrote $10 - 4 - 4$ with the second four being constraint equations. Here the second four are residual gauge and the first four are the Lorenz conditions. These are two bookkeepings of one fact: in Lorenz gauge the four Lorenz conditions are what the constraints become, and the gauge has been spent in two instalments — four functions to reach Lorenz gauge, four more to exhaust it. Either way two remain. The electromagnetic count is $4 - 1 - 1 = 2$ by the same two routes.

38 · polarisations

One representative per class: TT gauge

Take the wave along $z$: $k_\mu = \frac\omega c(-1,0,0,1)$. Call $e\in L$ transverse-traceless gauge (TT) if $e_{0\mu} = 0$ and $e^\mu{}_\mu = 0$. Lorenz then forces $e_{3\mu} = 0$, and

$$e = \begin{pmatrix}0&0&0&0\\0&h_+&h_\times&0\\0&h_\times&-h_+&0\\0&0&0&0\end{pmatrix}.$$

theorem

$L = P\oplus TT$. Every Lorenz-gauge wave is gauge-equivalent to exactly one TT wave, and $$h_+ = \tfrac12(e_{11} - e_{22}),\qquad h_\times = e_{12}$$ are gauge invariant on $L$.

Schematic: $L$ drawn as a plane, $P$ and $TT$ as lines through $0$. A gauge class is a translate $e + P$, parallel to $P$. $L = P\oplus TT$ says each class crosses $TT$ exactly once, at the class's TT representative; its coordinates there are $(h_+, h_\times)$.

Reading the figure. The previous slide's quotient $L/P$ is the set of parallel translates of $P$ inside $L$ — the gauge classes. A quotient is abstract; a complement of $P$ is concrete, and $TT$ is one: a subspace meeting $P$ only at $0$ whose dimension makes up the rest of $L$. Then each translate meets it in exactly one point, and that point is the class's canonical representative. The real dimensions are $6 = 4 + 2$; the picture draws them as $2 = 1 + 1$, which is enough for the geometry.

Lorenz forces $e_{3\mu} = 0$. With $k^\mu = \frac\omega c(1,0,0,1)$ (index raised with $\eta$), the Lorenz condition $k^\mu e_{\mu\nu} = 0$ reads $e_{0\nu} + e_{3\nu} = 0$. If $e_{0\nu} = 0$ then $e_{3\nu} = 0$. The remaining spatial block is $e_{11}, e_{12}, e_{22}$, and tracelessness $e_{11} + e_{22} = 0$ leaves two.

The general function. By linearity and Fourier superposition (unit 20), a vacuum wave travelling along $z$ with any profile is $h_+(t - z/c)$ and $h_\times(t - z/c)$ in the two slots: two arbitrary functions of one variable, which is unit 35's "two free functions".

38 · polarisations

TT gauge: the proof, and the two polarisations

proof of the theorem

For $p(\epsilon)\in P$: $p_{11} = p_{22} = -k\cdot\epsilon$ and $p_{12} = 0$, so $h_\pm$ are unchanged by gauge. If $p(\epsilon)\in TT$ then $p_{11} = -p_{22}$ forces $k\cdot\epsilon = 0$, and $p_{0\mu} = k_0\epsilon_\mu + k_\mu\epsilon_0 = 0$ forces $\epsilon = 0$. So $P\cap TT = 0$; dimensions $4 + 2 = 6 = \dim L$. $\square$

$h_+$ and $h_\times$ are the plus and cross polarisations. In TT gauge $h = 0$, so $\bar h_{\mu\nu} = h_{\mu\nu}$ and the bar is dropped.

check

Two free numbers, as counted ✓. Both components lie in the plane $\perp$ to the direction of travel: the wave is transverse, like light.

The proof, in the figure's terms. "$h_\pm$ unchanged by gauge" says $h_\pm$ are constant along every translate of $P$, so they are functions on the classes. "$P\cap TT = 0$" says the two lines cross only at the origin, and the dimension count says they span $L$.

The proof's middle step, written out. $p_{0\mu} = k_0\epsilon_\mu + k_\mu\epsilon_0 - \eta_{0\mu}(k\cdot\epsilon)$ and $k\cdot\epsilon = 0$ by then. For $\mu = 0$: $2k_0\epsilon_0 = 0$, so $\epsilon_0 = 0$. For $\mu = 1, 2$: $k_0\epsilon_\mu = 0$. For $\mu = 3$: $k_0\epsilon_3 + k_3\epsilon_0 = k_0\epsilon_3 = 0$. Since $k_0 = -\omega/c\ne0$, all four vanish.

How to find the TT representative in practice. Problem P3 does it for a specific Lorenz-gauge amplitude, solving for $\epsilon$ component by component. The answer depends only on the gauge-invariant pair $(h_+, h_\times)$, which is the theorem.

38 · polarisations

The transverse-traceless projector

definition · the TT projector

For a unit vector $n\in\mathbb R^3$ let $P = 1 - nn^{\mathsf T}$, the orthogonal projector onto $n^\perp$. On symmetric $3\times3$ matrices define $$\Lambda(S) = PSP - \tfrac12P\,\mathrm{tr}(PS),$$ in components $\Lambda_{ij,kl}S_{kl}$ with $\Lambda_{ij,kl} = P_{ik}P_{jl} - \tfrac12P_{ij}P_{kl}$.

lemma

$\Lambda(S)$ is symmetric, transverse ($\Lambda(S)n = 0$) and traceless; $\Lambda^2 = \Lambda$; $\Lambda$ is self-adjoint for $\langle A,B\rangle = \mathrm{tr}(AB)$; and $\Lambda(\mathbb 1) = 0$.

proof

$Pn = 0$ gives transversality. $\mathrm{tr}P = 2$ and $P^2 = P$ give $\mathrm{tr}\,\Lambda(S) = \mathrm{tr}(PS) - \mathrm{tr}(PS) = 0$. For a transverse traceless $A$, $PAP = A$ and $\mathrm{tr}(PA) = 0$, so $\Lambda(A) = A$: hence $\Lambda^2 = \Lambda$. $\Lambda(\mathbb 1) = P - \tfrac12P\cdot2 = 0$. Self-adjointness: $\mathrm{tr}(\Lambda(S)T)$ is symmetric in $S, T$ by cyclicity. $\square$

For $n = \hat z$: $\Lambda(S)$ has $\tfrac12(S_{11} - S_{22})$ and $S_{12}$ in the $h_+$, $h_\times$ slots — the gauge invariants of the TT-gauge theorem. So for a wave along $n$, $$h^{\rm TT}_{ij} = \Lambda_{ij,kl}\,\bar h_{kl} = \Lambda_{ij,kl}\,h_{kl}.$$

What it is for. Section 3 computes $\bar h_{ij}$ in Lorenz gauge from a source. That is not the TT field, but the TT field is its gauge-invariant part, and the projector extracts it without solving for any $\epsilon$. Far from the source the wave is locally a plane wave along $n = \vec x/r$, which is where the projector is applied.

Why $\Lambda h = \Lambda\bar h$. They differ by a multiple of $\delta_{ij}$ in the spatial block, and $\Lambda(\mathbb 1) = 0$. So it does not matter which one is projected.

Self-adjointness, in full. $\mathrm{tr}(\Lambda(S)T) = \mathrm{tr}(PSPT) - \tfrac12\mathrm{tr}(PS)\mathrm{tr}(PT)$. The first term equals $\mathrm{tr}(PTPS)$ by cyclicity and $P^{\mathsf T} = P$; the second is visibly symmetric. So $\Lambda$ is an orthogonal projector, and $\langle\Lambda S, \Lambda S\rangle = \langle S,\Lambda S\rangle$ — the identity that section 4 uses to compute the power.

Its image has dimension 2. In a basis with $n = e_3$, a transverse traceless symmetric matrix has its third row and column zero and a traceless symmetric $2\times2$ block: two parameters. The linear algebra of the counting and TT-gauge slides and the projector agree.

The mathematics first, as ever. $\Lambda$ is a linear map on $\mathrm{Sym}^2(\mathbb R^3)$, and $\Lambda_{ij,kl}$ are its matrix entries in the standard basis. That it is a projector is a fact about the map; that its entries have the form above is a fact about the basis.

38 · polarisations

Free masses sit still, and distances change

lemma · rest is a geodesic in TT gauge

In TT gauge the worldline $x^i = \text{const}$, $t = \tau$ is a geodesic.

proof

With $u = (c,0,0,0)$ the geodesic equation needs $\Gamma^\lambda_{00} = 0$. By the first slide's lemma, $\Gamma^i_{00} = \tfrac12(2\partial_0h_{i0} - \partial_ih_{00})$ and $\Gamma^0_{00} = -\tfrac12\partial_0h_{00}$; both vanish because $h_{0\mu} = 0$. $\square$

So TT coordinates are carried by the free masses. The coordinates do not move; the wave shows up in proper distance.

proper separation

Masses at the origin and at $L_0n$, with $L_0\ll\lambda$: $$L = \int_0^{L_0}\sqrt{1 + h_{ij}n^in^j}\,\mathrm ds = L_0\big(1 + \tfrac12h_{ij}n^in^j\big).$$ Along $x$ under $h_+$: $L_0(1 + \tfrac12h_+)$. Along $y$: $L_0(1 - \tfrac12h_+)$.

checks

$h\to0$: $L = L_0$ ✓. Along $z$, the direction of travel: $h_{3j} = 0$, no change — transverse ✓. The strain $\Delta L/L_0 = \tfrac12h$ is dimensionless ✓.

Why $L_0\ll\lambda$. The integral takes $h$ constant along the segment, which needs the separation to be small compared with the wavelength. For LIGO, $L_0 = 4$ km and at 100 Hz $\lambda = 3000$ km. For the full expression without that assumption, integrate $h(t - s\,n_z/c)$ along the path; the problem sheet does not need it.

The gauge-invariant version. Proper distance is invariant, but the lemma was proved in a gauge. To see that nothing depends on the gauge, use unit 33's geodesic deviation equation, which involves only curvature. In TT gauge, the slide "The same redundancy, met twice" gives $R_{i0j0} = \tfrac12(\partial_0\partial_jh_{i0} + \partial_i\partial_0h_{0j} - \partial_i\partial_jh_{00} - \partial_0^2h_{ij}) = -\tfrac12\partial_0^2h_{ij}$. For slowly moving masses with $u\approx(c,\vec0)$ and $\tau\approx t$, the deviation equation is $$\ddot\xi^i = -c^2R^i{}_{0j0}\,\xi^j = \tfrac12\ddot h_{ij}\,\xi^j,$$ with dots $\mathrm d/\mathrm dt$ (the $c^2$ cancels $\partial_0^2 = c^{-2}\partial_t^2$). Integrating twice with $\xi = \xi_0$ when $h = 0$ gives $\xi^i = \xi_0^i + \tfrac12h_{ij}\xi_0^j$ to first order — exactly the proper-distance result. The curvature is gauge invariant, so this is the answer in every gauge. Newtonian check of the sign: with $h_{00} = -2\Phi/c^2$ the same formula gives $R_{i0j0} = -\tfrac12\partial_i\partial_jh_{00} = \partial_i\partial_j\Phi/c^2$, unit 33's tidal limit.

So is the wave a force? In TT gauge, no mass accelerates: each is in free fall and stays at its coordinates. In a freely falling lab frame centred on one mass (Fermi normal coordinates, the local version of unit 32's normal coordinates), the others feel a tidal acceleration $\tfrac12\ddot h_{ij}\xi^j$. Same physics, different bookkeeping, which is the lesson of section 1.

38 · polarisations

A ring of free masses

Twelve free masses on a ring in the $xy$ plane, the wave travelling along $z$ (out of the page), $h = 0.5$ for visibility. Top: $h_+$. Bottom: $h_\times$, the same pattern turned by $45^\circ$. The marked mass only moves to and fro along a line: nothing goes round.

worked example 1 · the ring

Step 1. Mass at $x_0 = \rho_0(\cos\phi, \sin\phi)$. By the last slide its proper position is $x^i = x_0^i + \tfrac12h_{ij}x_0^j$.

Step 2, plus. $h_+ = h\cos\omega t$: $\ \delta x = \tfrac12h\rho_0\cos\phi\cos\omega t$, $\ \delta y = -\tfrac12h\rho_0\sin\phi\cos\omega t$. An ellipse with axes along $x$ and $y$, swapping every half period.

Step 3, cross. $h_\times = h\cos\omega t$: $\ \delta x = \tfrac12h\rho_0\sin\phi\cos\omega t$, $\ \delta y = \tfrac12h\rho_0\cos\phi\cos\omega t$. The same ellipse, axes along the diagonals.

Step 4, the number. $\rho_0 = 4$ km and GW150914's $h = 10^{-21}$: largest displacement $\tfrac12h\rho_0 = 2\times10^{-18}$ m.

Step 2 in full. With only $h_{11} = -h_{22} = h\cos\omega t$ nonzero, $\delta x = \tfrac12h_{11}x_0 = \tfrac12h\rho_0\cos\phi\cos\omega t$ and $\delta y = \tfrac12h_{22}y_0 = -\tfrac12h\rho_0\sin\phi\cos\omega t$. At $\omega t = 0$ the ring is stretched along $x$ by the factor $1 + h/2$ and squeezed along $y$ by $1 - h/2$; at $\omega t = \pi$ the reverse; at $\pi/2$ and $3\pi/2$ it is round.

Step 3 in full. With $h_{12} = h_{21} = h\cos\omega t$, $\delta x = \tfrac12h_{12}y_0$ and $\delta y = \tfrac12h_{21}x_0$. In the rotated basis $u = (x + y)/\sqrt2$, $v = (x - y)/\sqrt2$ the matrix $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ is $\mathrm{diag}(1,-1)$: a plus pattern along the diagonals. That is the content of the name.

Checks. Area: to first order the ellipse's area is $\pi\rho_0^2(1 + \tfrac12h)(1 - \tfrac12h) = \pi\rho_0^2(1 + O(h^2))$ — the wave preserves area at first order, because $h_{ij}$ is traceless. A breathing ring would need a trace, which TT gauge has proved absent. Dimensions of step 4: metres ✓.

The number, honestly. Step 4 uses $\rho_0 = 4$ km because LIGO's arms are 4 km, but LIGO is not a ring: its two arms measure the stretch along $x$ and along $y$, and section 6 computes the actual measured quantity. The order of magnitude, $10^{-18}$ m, is right — about a thousandth of a proton's diameter.

38 · polarisations

Spin two

theorem · the helicity of the wave

Rotate the transverse axes by $\psi$ about the direction of travel. The amplitudes transform as $$\begin{pmatrix}h_+'\\h_\times'\end{pmatrix} = \begin{pmatrix}\cos2\psi&\sin2\psi\\-\sin2\psi&\cos2\psi\end{pmatrix}\begin{pmatrix}h_+\\h_\times\end{pmatrix},$$ a rotation by $2\psi$.

proof

A bilinear form's matrix changes to $R^{\mathsf T}HR$ (unit 16). With $H = \begin{pmatrix}h_+&h_\times\\h_\times&-h_+\end{pmatrix}$ and $R$ the rotation by $\psi$, multiply out and use the double-angle formulas. $\square$

One pattern turned through $\psi$. A vector (top) is back to itself only after $360^\circ$. The ring's ellipse (bottom) is pure $h_\times$ at $45^\circ$, the negative of where it started at $90^\circ$, and back at $180^\circ$: its amplitudes $(h_+, h_\times)$ go round twice as fast as the pattern.

Reading the figure. The numbers under each drawing are the components in the fixed axes after the pattern has been turned through $\psi$: for the vector, $(\cos\psi, \sin\psi)$; for the ellipse, $(h_+, h_\times) = (\cos2\psi, \sin2\psi)$, starting from pure plus. Turning the pattern by $\psi$ is the same as turning the axes by $-\psi$, which is why the theorem's matrix has the opposite sign of $\sin2\psi$.

The multiplication. $R = \begin{pmatrix}c&-s\\s&c\end{pmatrix}$ with $c = \cos\psi$, $s = \sin\psi$. Then $(R^{\mathsf T}HR)_{11} = h_+(c^2 - s^2) + 2h_\times sc = h_+\cos2\psi + h_\times\sin2\psi$ and $(R^{\mathsf T}HR)_{12} = -2h_+sc + h_\times(c^2 - s^2) = -h_+\sin2\psi + h_\times\cos2\psi$. The $22$ entry is minus the $11$ entry, so the result is again of the TT form. The direction of the rotation (active or passive) only changes the sign of $\psi$.

38 · polarisations

Helicity two, read off

$\psi = 45^\circ$ swaps $h_+$ and $h_\times$ (up to sign): the two polarisations are one pattern, turned by $45^\circ$ — the two rows of the ring figure.
$\psi = 180^\circ$ changes nothing: the pattern is an ellipse, symmetric under a half turn.
Light's polarisation vector rotates by $\psi$, returning only after $360^\circ$ (unit 24). An angle multiplied by $s$ is helicity $s$: light has 1, gravity 2.

check

$h_+^2 + h_\times^2$ is invariant, as a rotation must leave it ✓. It is the combination that the energy flux of section 4 turns out to contain.

Why the angle doubles. A vector changes sign under a half turn; a symmetric bilinear form is quadratic in the vector it eats, so the sign cancels. Anything built from $s$ copies of a vector picks up $s$ factors of $e^{i\psi}$, and the traceless transverse part of a symmetric pair of indices picks up $e^{\pm2i\psi}$.

Circular polarisation. Put $h_+ = h\cos\omega t$ and $h_\times = h\sin\omega t$. By the theorem this is the pure-plus pattern rotated by the angle $\psi$ with $2\psi = \omega t$: the ellipse keeps its shape and turns, at angular speed $\omega/2$. Section 5 shows that a binary seen face-on emits exactly this wave with $\omega = 2\Omega$ — so a ring of masses above the binary's pole has an ellipse that rotates with the binary. Problem P5.

The word "spin". In quantum theory helicity $s$ means the quanta carry angular momentum $s\hbar$ along their motion; the graviton, if it exists as a particle, has $s = 2$. This course is classical and makes no use of that. The classical content is the theorem: the transformation under rotations by $e^{\mp2i\psi}$ on $h_+\pm ih_\times$.

section 3

The quadrupole formula

Unit 24's Green's function, a slow source, and two conservation laws that silence the first two multipoles.

38 · quadrupole

The retarded solution

theorem · the field of a source

Let $T_{\mu\nu}$ be $C^2$, of bounded spatial support, zero before some time, with $\partial^\mu T_{\mu\nu} = 0$. Then

$$\bar h_{\mu\nu}(t,\vec x) = \frac{4G}{c^4}\int\frac{T_{\mu\nu}(t_{\rm r},\vec y)}{|\vec x - \vec y|}\,\mathrm d^3y,$$

$t_{\rm r} = t - |\vec x - \vec y|/c$, solves the linearised equation in Lorenz gauge.

proof

Component by component, $\Box_{24}\bar h_{\mu\nu} = f$ with $f = 16\pi GT_{\mu\nu}/c^4$. Unit 24's retarded solution is $\int f(t_{\rm r})/(4\pi|\vec x - \vec y|)$, and $16\pi/4\pi = 4$. The Lorenz condition holds because $\partial^\mu T_{\mu\nu} = 0$, as unit 24's retarded potentials satisfy theirs because charge is conserved. $\square$

Spacetime, one space direction drawn. The field at the event $(t, \vec x)$ is built only from where the source's worldtube meets the past light cone of that event (highlighted): each source point $\vec y$ enters as it was at its own retarded time $t_{\rm r}$. What the source does at time $t$ has not arrived yet.

Reading the figure. Light moves at $45^\circ$ with $ct$ up the page. The integral in the theorem runs over every source point $\vec y$, but at each $\vec y$ it samples $T_{\mu\nu}$ at one instant only, $t_{\rm r} = t - |\vec x - \vec y|/c$, which is where that point's worldline crosses the cone. Points of the source nearer to $\vec x$ are sampled later. For a slow source the spread of those instants, about $d/c$, is short compared with the time the source takes to change, and the slide "Far away and slow" turns that into the slow-motion approximation.

The Lorenz condition, in full. Write $\bar h_{\mu\nu} = \frac{4G}{c^4}\int T_{\mu\nu}(t - |\vec x - \vec y|/c, \vec y)\,|\vec x - \vec y|^{-1}\mathrm d^3y$ and change variables to $\vec z = \vec x - \vec y$. Then $\partial/\partial x^j$ acts only on the second argument of $T$ (and on nothing else), and $\partial_t$ on the first, so $\partial^\mu\bar h_{\mu\nu} = \frac{4G}{c^4}\int(\partial^\mu T_{\mu\nu})(t - |\vec z|/c, \vec x - \vec z)\,|\vec z|^{-1}\mathrm d^3z = 0$. This is unit 24's proof for its potentials, word for word, with $\partial_t\rho + \nabla\cdot\vec J = 0$ replaced by $\partial^\mu T_{\mu\nu} = 0$.

38 · quadrupole

The kernel, causality, and two checks

The kernel is unit 24's retarded Green's function $\delta(t - r/c)/4\pi r$, found there from unit 20's contour integral with the poles pushed below the axis. That choice — the field at $\vec x$ now reports the source as it was at the retarded time — is causality, and it is the only input here that is not the field equations.

checks

Static dust. $T_{00} = \rho c^2$: $\bar h_{00} = \frac{4G}{c^2}\int\frac{\rho}{|\vec x - \vec y|} = -\frac{4\Phi}{c^2}$, unit 35's result ✓.

Dimensions. $\frac{G}{c^4}\cdot\mathrm{J\,m^{-3}}\cdot\mathrm{m^2} = \frac{\mathrm N^{-1}\cdot\mathrm J}{\mathrm m} = 1$ ✓.

38 · quadrupole

Far away and slow

Three lengths: the source's size $d$, the distance $r = |\vec x|$, and the reduced wavelength $\lambda/2\pi = c/\omega$. Put $n = \vec x/r$.

the wave zone and the slow-motion approximation

Far: $r\gg d$ and $r\gg\lambda/2\pi$. Then $|\vec x - \vec y| = r - n\cdot\vec y + O(d^2/r)$, and in the denominator $1/|\vec x - \vec y| = 1/r$ up to a relative $O(d/r)$.

Slow: $d\ll\lambda/2\pi$, which for a source of speed $v\sim\omega d$ is $v\ll c$. Then in the retarded time $$T\big(t - \tfrac rc + \tfrac{n\cdot\vec y}{c}\big) = \sum_{m\ge0}\frac1{m!}\Big(\frac{n\cdot\vec y}{c}\Big)^m\partial_t^mT\big(t - \tfrac rc\big),$$ and the $m$-th term is $O\big((v/c)^m\big)$ relative to the first.

Not to scale. Seen from far away, a source point $\vec y$ is nearer than the centre by its projection $n\cdot\vec y$ on the line of sight (highlighted), up to $O(d^2/r)$. In $1/|\vec x - \vec y|$ that is negligible; in the retarded time it is a delay of $n\cdot\vec y/c$ across the source, small only if the source is slow.

Reading the figure. The two lines to the observer, from the centre and from $\vec y$, become parallel as $r\to\infty$, and their lengths then differ by exactly the projection of $\vec y$ on $n$: the dashed perpendicular is the wavefront through $\vec y$. The arcs are wave crests a wavelength apart, drawn crowded so they fit; the regime of this section has $d\ll\lambda/2\pi\ll r$, so that the whole source sits well inside one crest-to-crest gap.

The expansion of the distance. $|\vec x - \vec y|^2 = r^2 - 2r\,n\cdot\vec y + |\vec y|^2$, so $|\vec x - \vec y| = r\sqrt{1 - 2n\cdot\vec y/r + |\vec y|^2/r^2} = r - n\cdot\vec y + O(d^2/r)$ by the binomial series. In the denominator the correction is relative $O(d/r)$ and is dropped: it gives terms falling faster than $1/r$, which carry no energy to infinity (unit 24's three zones). In the time argument it is multiplied by $\omega/c$ and becomes $\omega d/c$, which is not small because $r$ is large but small because the source is slow. That is why two different conditions are needed.

The Taylor series in time. It is Taylor's theorem in the first argument of $T$ with step $n\cdot\vec y/c$, valid when $T$ is smooth in time; the ratio of consecutive terms is $|n\cdot\vec y|\,|\partial_tT|/(c|T|)\sim d\omega/c$. Truncating after $m = 0$ is unit 21's order notation made concrete: the error is $O(v/c)$.

38 · quadrupole

The leading term, and the multipole ladder

leading order $$\bar h_{\mu\nu}(t,\vec x) \approx \frac{4G}{c^4r}\int T_{\mu\nu}\big(t - r/c,\ \vec y\big)\,\mathrm d^3y .$$

Every component falls as $1/r$ and depends on $t - r/c$: an outgoing wave.

The multipole ladder. The $m$-th term involves the $m$-th moment $\int T\,y^{i_1}\cdots y^{i_m}$ — unit 10's multipole expansion of $1/|\vec x - \vec y|$, with time added and each rung costing a factor $v/c$.

For the binary: $d = a$, $\omega = 2\Omega$, so $\dfrac{d}{\lambda/2\pi} = \dfrac{2\Omega a}{c} = \dfrac{2v}{c}$. For Hulse–Taylor, $v/c = 1.5\times10^{-3}$.

38 · quadrupole

Why gravity has no dipole radiation

For a slow source $T^{00} = \rho c^2$ and $T^{0i} = c\rho v^i$ at leading order. Then

$\bar h_{00}\approx\dfrac{4GM}{c^2r}$ with $M = \displaystyle\int\rho$, $\quad\bar h_{0i}\approx-\dfrac{4GP^i}{c^3r}$ with $P^i = \displaystyle\int\rho v^i$.

theorem · two conservation laws

$\partial_\mu T^{\mu\nu} = 0$ gives $\dot M = 0$ and $\dot P^i = 0$; and the mass dipole $D^i = \int\rho\,y^i$ has $\dot D^i = P^i$, so $\ddot D^i = 0$.

proof

$\nu = 0$: $\partial_t\rho + \partial_k(\rho v^k) = 0$. Integrate: $\dot M = 0$. Multiply by $y^i$ and integrate by parts: $\dot D^i = \int\rho v^i = P^i$. $\nu = i$: $\partial_t T^{0i}/c = -\partial_kT^{ki}$ integrates to $\dot P^i = 0$. $\square$

The same orbit, two theories. Left: the mass dipole $\vec D = m_1\vec y_1 + m_2\vec y_2$ is the total mass times the centre of mass, which sits still — nothing to radiate. Right: two opposite charges on the same orbit have $\vec p = q(\vec y_1 - \vec y_2)$, which turns with the orbit, and $\ddot{\vec p}\ne0$ radiates.

Reading the figure. With the centre of mass at the origin, $m_1\vec y_1 + m_2\vec y_2 = 0$ at every instant, so the heavier body runs on the smaller circle, radii in the ratio $m_2 : m_1$ (drawn for $m_1 = 2m_2$). For charges the weights in the dipole are $+q$ and $-q$ rather than the masses, and no choice of origin makes the sum vanish: $\vec p$ is $q$ times the separation, a vector of fixed length turning at $\Omega$.

The monopole and dipole fields are static. $\bar h_{00} = 4GM/c^2r$ with constant $M$ is the Newtonian potential of the whole system, $-4\Phi/c^2$ with $\Phi = -GM/r$. $\bar h_{0i}$ with constant $P$ is the same potential seen from a frame moving with the centre of mass; in the centre-of-mass frame $P = 0$ and it vanishes. The next term in the slow-motion series gives $\bar h_{00}$ a piece proportional to $n\cdot\dot{\vec D}$, constant again. None of these fields changes with $t - r/c$, so none carries energy away.

The integration by parts. $\int y^i\partial_k(\rho v^k)\,\mathrm d^3y = -\int\delta^i_k\rho v^k\,\mathrm d^3y = -P^i$, the boundary term vanishing because the source has bounded support. Hence $\dot D^i = \int y^i\partial_t\rho = -\int y^i\partial_k(\rho v^k) = P^i$.

38 · quadrupole

Mass is the charge: the structural reason

electromagnetism is different

Unit 24's dipole $\vec p = \sum q_a\vec x_a$ has $\dot{\vec p} = \sum q_a\vec v_a$, which is not conserved unless every particle has the same $q/m$. Larmor radiation is $\ddot{\vec p}\ne0$.

the structural reason

Gravity's charge is inertial mass — the equivalence principle of unit 30. So its dipole's derivative is total momentum, conserved by Newton's third law (unit 13's Noether, for translations). Mass conservation kills the monopole; momentum conservation kills the dipole. The first multipole that can radiate is the quadrupole, and each rung up the ladder costs $(v/c)^2$ in power.

The magnetic dipole too. Electromagnetism has a second dipole, the magnetic one, $\tfrac12\sum q_a\vec x_a\times\vec v_a$. Its gravitational analogue is $\tfrac12\sum m_a\vec x_a\times\vec v_a$ — half the total angular momentum, conserved. So gravity's "current dipole" is silenced as well. What survives at the next order is the mass quadrupole, and at the order after it the current quadrupole and mass octupole.

Self-gravitating sources. For a binary, $\partial_\mu T^{\mu\nu} = 0$ in flat space is false — the stars accelerate. But the two conservation laws used here hold anyway, for any isolated Newtonian system: $M$ is conserved because mass is, and $P$ because of Newton's third law. The theorem survives; the next slides need more care, and the warning after the quadrupole formula says how much.

38 · quadrupole

Lemma: the stresses are the second moment

The wave lives in $\bar h_{ij}$, which by the leading-order formula needs $\int T^{ij}$ — the source's stresses, including the kinetic $\rho v^iv^j$. Those are hard to know. The lemma trades them for the mass distribution.

lemma · tensor virial theorem

If $\partial_\mu T^{\mu\nu} = 0$ and $T$ has bounded spatial support, then $$\int T^{ij}\,\mathrm d^3y = \frac12\frac{\mathrm d^2}{\mathrm dt^2}\int\rho\,y^iy^j\,\mathrm d^3y,\qquad\rho = T^{00}/c^2.$$

proof

1. From $\nu = 0$ and $\nu = k$, with $x^0 = ct$: $\partial_0^2T^{00} = -\partial_0\partial_kT^{k0} = \partial_k\partial_lT^{kl}$.

2. Multiply by $y^iy^j$ and integrate. Integrating by parts twice, with no boundary terms, $$\int y^iy^j\partial_k\partial_lT^{kl} = \int\partial_k\partial_l(y^iy^j)\,T^{kl} = 2\int T^{ij},$$ since $\partial_k\partial_l(y^iy^j) = \delta^i_k\delta^j_l + \delta^i_l\delta^j_k$ and $T$ is symmetric.

3. So $\frac1{c^2}\frac{\mathrm d^2}{\mathrm dt^2}\int T^{00}y^iy^j = 2\int T^{ij}$. $\square$

Check. Both sides are $\mathrm{kg\,m^2\,s^{-2}} = \mathrm J$ ✓.

Step 1 in full. The $\nu = 0$ law is $\partial_0T^{00} + \partial_kT^{k0} = 0$ and the $\nu = l$ law is $\partial_0T^{0l} + \partial_kT^{kl} = 0$. Differentiate the first by $\partial_0$ and use Clairaut and the symmetry $T^{k0} = T^{0k}$: $\partial_0^2T^{00} = -\partial_k\partial_0T^{0k} = \partial_k\partial_lT^{lk}$.

Step 2's derivatives. $\partial_l(y^iy^j) = \delta^i_ly^j + y^i\delta^j_l$, and $\partial_k$ of that is $\delta^i_l\delta^j_k + \delta^i_k\delta^j_l$. Contracting with symmetric $T^{kl}$ gives $T^{ji} + T^{ij} = 2T^{ij}$.

The name. Its trace is the scalar virial theorem: $\int T^{kk} = \tfrac12\ddot I$ with $I = \int\rho|\vec y|^2$. For a bound system in a steady state $\ddot I$ averages to zero, and $\int T^{kk}$ — kinetic energy twice plus the gravitational stresses — averages to zero too: unit 08's $2\langle T\rangle + \langle V\rangle = 0$, if you include in $T^{ij}$ the gravitational stresses of the slide "The binary breaks the hypothesis".

An example. Two masses on a circle about their centre of mass: the left side is the kinetic stress $\sum m v^iv^j$ plus the tension of the gravitational "string" between them; the right side is half the second derivative of $\sum my^iy^j$. Problem P6 checks one component.

38 · quadrupole

The quadrupole formula

theorem · the quadrupole formula

With $I_{ij}(t) = \int\rho\,y^iy^j\,\mathrm d^3y$ and its trace-free part $Q_{ij} = I_{ij} - \tfrac13\delta_{ij}I_{kk}$, far from a slow source

$$h^{\rm TT}_{ij}(t,\vec x) = \frac{2G}{c^4r}\,\Lambda_{ij,kl}(n)\,\ddot Q_{kl}\big(t - r/c\big).$$

proof

Leading order and the lemma give $\bar h_{ij} = \frac{4G}{c^4r}\cdot\frac12\ddot I_{ij}$. Project with $\Lambda(n)$. Since $\Lambda(\mathbb 1) = 0$, replacing $I$ by $Q$ changes nothing. $\square$

a factor of three

Unit 10's quadrupole moment was $\int\rho(3y^iy^j - |\vec y|^2\delta_{ij}) = 3Q_{ij}$. Both are the same symmetric bilinear form up to scale; the gravitational-wave literature uses $Q$. And $I_{ij}$ here is the second mass moment, not unit 16's inertia tensor $\int\rho(|\vec y|^2\delta_{ij} - y^iy^j)$.

checks

Dimensions. $\frac{G}{c^4r}\cdot\mathrm{kg\,m^2\,s^{-2}} = \frac{\mathrm J}{\mathrm N\cdot\mathrm m} = 1$ ✓.

A pulsating sphere. Spherical symmetry makes $I_{ij}\propto\delta_{ij}$, so $Q = 0$: no quadrupole radiation, however violently it pulsates ✓.

A rotating axisymmetric body about its axis has constant $I_{ij}$: silent ✓. A spinning neutron star radiates only through its asymmetry.

Einstein, 1918, with the factor fixed later. Every gravitational-wave amplitude in this unit comes from this line.

Why the TT part is the whole story. Far away, the $\bar h_{00}$ and $\bar h_{0i}$ components are the static fields of the slide "Why gravity has no dipole radiation", and the gauge-dependent parts of $\bar h_{ij}$ (anything of the form $p(\epsilon)$ in section 2's language) carry nothing physical. Locally the outgoing wave is a plane wave along $n$, and section 2 showed that the physical content of such a wave is its TT part. The projector extracts it.

The mathematics first. $I$ is a symmetric bilinear form on $\mathbb R^3$, $I(u,v) = \int\rho\,(u\cdot y)(v\cdot y)$, and $Q$ is its trace-free part; $\Lambda(n)$ is a linear map on symmetric forms. The formula is a statement about maps, and the components in any orthonormal basis follow from unit 16's change-of-basis theorem. That $Q$ is unit 10's moment divided by 3 is visible from the definitions.

History. Einstein's 1916 paper had an error; his 1918 paper "Über Gravitationswellen" derived the quadrupole formula, with an overall factor of 2 wrong, found by Eddington in 1922. From the 1950s to the early 1980s a "quadrupole formula controversy" asked whether the formula applies to self-gravitating systems at all; the next slide is that controversy in one box.

38 · quadrupole

The binary breaks the hypothesis

the honest difficulty

The linear theory needs $\partial^\mu T_{\mu\nu} = 0$ in flat spacetime. For dust that means $\partial_t(\rho v^i) + \partial_k(\rho v^iv^k) = 0$: every particle moves in a straight line. A binary held together by gravity violates it. Strictly, the quadrupole formula has been derived for a source that cannot orbit.

What saves it is that the lemma only needs some conserved symmetric $T^{\mu\nu}$ whose $T^{00}$ is the mass density.

proposition · Newtonian gravity has a stress

With $\nabla^2\Phi = 4\pi G\rho$, the gravitational stress $$t^{ij} = \frac1{4\pi G}\Big(\partial_i\Phi\,\partial_j\Phi - \tfrac12\delta_{ij}|\nabla\Phi|^2\Big)$$ has $\partial_jt^{ij} = \rho\,\partial_i\Phi$. So $\partial_t(\rho v^i) + \partial_j(\rho v^iv^j + t^{ij}) = 0$ holds for self-gravitating dust.

Cited. That the full nonlinear theory reproduces the quadrupole formula with this source, at leading post-Newtonian order, is a theorem of the post-Newtonian and post-Minkowskian programme: Blanchet, Living Reviews in Relativity 17 (2014) 2, §§1–3; Maggiore, Gravitational Waves vol. 1, ch. 5.

Proof of the proposition. $\partial_j(\partial_i\Phi\,\partial_j\Phi) = \partial_j\partial_i\Phi\,\partial_j\Phi + \partial_i\Phi\,\nabla^2\Phi$ and $\partial_j(\tfrac12\delta_{ij}|\nabla\Phi|^2) = \partial_j\Phi\,\partial_i\partial_j\Phi$. Subtract: the first terms cancel by Clairaut, leaving $\partial_i\Phi\,\nabla^2\Phi = 4\pi G\rho\,\partial_i\Phi$. Divide by $4\pi G$. $\square$ Then Newton's law for dust, $\rho(\partial_t + v\cdot\nabla)v^i = -\rho\partial_i\Phi$, combined with continuity, is $\partial_t(\rho v^i) + \partial_j(\rho v^iv^j) = -\rho\partial_i\Phi = -\partial_jt^{ij}$. $\square$ It is the gravitational analogue of unit 24's Maxwell stress tensor, with the opposite sign because like charges attract.

So what is proved here, and what is cited. Proved: the virial lemma applies to the effective source $T^{ij} = \rho v^iv^j + t^{ij}$, and gives $\int T^{ij} = \tfrac12\ddot I_{ij}$ with the Newtonian $\rho$ only. So if the linear formula is applied with the effective source, the answer depends only on the mass distribution and is the quadrupole formula. Cited: that doing this is legitimate — that the nonlinear terms of the field equations act, at this order, as exactly this source and nothing more. It took until the early 1980s (Damour's 1983 treatment of binary motion, and the Walker–Will analysis) to settle, and the Hulse–Taylor measurement of section 6 was the experimental half of the settlement.

Where it certainly fails. When $v/c$ is not small, when the bodies are close enough that their own fields are strong, and at merger. Section 6 shows all three with numbers for GW150914.

section 4

Energy

Why there is no energy density at a point, what survives an average, and how much a quadrupole radiates.

38 · energy

No energy density at a point

proposition

Let $t_{\mu\nu}$ be a tensor built at each point from $g$ and $\partial g$ alone, vanishing whenever $\partial g = 0$ in some chart at that point. Then $t_{\mu\nu} = 0$ everywhere.

proof

At any point $p$, unit 32's normal coordinates give $\partial_\lambda g_{\mu\nu}(p) = 0$. So $t(p) = 0$ in that chart; a tensor zero in one basis is zero in every basis. $\square$

Physics. The equivalence principle (unit 30): a freely falling observer feels no gravitational field at a point. Whatever gravitational energy is, it cannot be a local density built from the field's first derivatives — the way $\tfrac12(\varepsilon_0E^2 + B^2/\mu_0)$ is for electromagnetism.

the phrase to be careful with

"The energy of the gravitational field" has no local meaning. Unit 35 said this about pseudotensors. The proposition is why.

what does survive

Curvature cannot be transformed away — it involves $\partial^2g$. A wave has curvature $R_{i0j0} = -\tfrac12\partial_0^2h_{ij}$, and a region several wavelengths across cannot be put in normal coordinates all at once.

So an energy for waves must be an average over a region, not a value at a point. The next slide is the average.

The hypothesis "vanishing whenever $\partial g = 0$". It is what "energy of the field" ought to satisfy: in flat spacetime in Cartesian coordinates, where the field is trivially absent, the energy is zero. A tensor with that property and no second derivatives is identically zero. This does not forbid quantities built from curvature (which involve $\partial^2g$), and it does not forbid non-tensorial quantities — the pseudotensors of unit 35 — whose values depend on the chart and which therefore do not answer the question "how much energy is here".

What the average buys. Normal coordinates remove $\partial g$ at one point, and in a neighbourhood whose size is small compared with the curvature radius. A wave of wavelength $\lambda$ has curvature radius comparable to $\lambda/\sqrt h$, and no chart flattens it over a region containing several wavelengths. Averaging over such a region produces a quantity that no choice of coordinates can remove.

Global energy. For an isolated system, the total mass measured at infinity — the ADM mass, and the Bondi mass at null infinity — is well defined, and the Bondi mass decreases by exactly the energy carried off by waves (cited: Bondi, van der Burg and Metzner 1962; Sachs 1962; see Wald, General Relativity, §11.2). That is the theorem behind the energy balance of section 5.

38 · energy

The Isaacson average

theorem · the Isaacson stress-energy tensor (cited)

For waves whose wavelength is short compared with the curvature scale of the background, the average $\langle\cdot\rangle$ over a region several wavelengths across gives

$$t_{\mu\nu} = \frac{c^4}{32\pi G}\big\langle\partial_\mu h^{\rm TT}_{ij}\,\partial_\nu h^{\rm TT}_{ij}\big\rangle,$$

a tensor, gauge invariant, that sources the background's curvature as matter does. Isaacson 1968; MTW ch. 35, §35.13–35.15; Maggiore vol. 1, §1.4.

Energy density and flux ($\partial_0 = c^{-1}\partial_t$, and $h_{ij}h_{ij} = 2(h_+^2 + h_\times^2)$):

$$u = t_{00} = \frac{c^2}{16\pi G}\big\langle\dot h_+^2 + \dot h_\times^2\big\rangle,\qquad F = c\,u .$$

the coefficient, checked against unit 18

Per polarisation, the second-order Einstein–Hilbert action (cited, same sources) is a stretched string with density $\mu_{\rm s} = c^2/16\pi G$ and tension $\mathcal T_{\rm s} = c^4/16\pi G$. Unit 18's energy density $\tfrac12\mu_{\rm s}\dot u^2 + \tfrac12\mathcal T_{\rm s}u'^2$ on an outgoing wave is $\mu_{\rm s}\dot u^2$ — the formula above ✓. Wave speed $\sqrt{\mathcal T_{\rm s}/\mu_{\rm s}} = c$ ✓.

Number. GW150914 at Earth, $h = 10^{-21}$ at 150 Hz, circularly polarised: $F = 7.1\times10^{-3}\ \mathrm{W\,m^{-2}}$, five millionths of sunlight.

What is cited and why. Isaacson's derivation expands the Einstein tensor to second order in $h$, averages, and shows that the averaged second-order piece acts as a source for the background: $G_{\mu\nu}[\bar g] = \kappa\,(T_{\mu\nu} + t_{\mu\nu})$. It needs two scales, a second-order expansion of the curvature, and care with what the average means. None of it is beyond this course's tools, but it is long, and the coefficient can be cross-checked in two lines, which is what the box does. Isaacson, Phys. Rev. 166 (1968) 1263 and 1272.

The unit 18 check, in full. The second-order expansion of the Einstein–Hilbert action about flat space, in TT gauge, is $S^{(2)} = -\frac{c^4}{64\pi G}\int\partial_\alpha h_{ij}\,\partial^\alpha h_{ij}\,\mathrm dt\,\mathrm d^3x$ (cited). With $\partial_\alpha h\,\partial^\alpha h = -c^{-2}\dot h^2 + |\nabla h|^2$ and $h_{ij}h_{ij} = 2(h_+^2 + h_\times^2)$, each polarisation has Lagrangian density $\frac{c^2}{32\pi G}\dot h_+^2 - \frac{c^4}{32\pi G}|\nabla h_+|^2$, which is unit 18's string, $\tfrac12\mu_{\rm s}\dot u^2 - \tfrac12\mathcal T_{\rm s}u'^2$, with $\mu_{\rm s} = c^2/16\pi G$ and $\mathcal T_{\rm s} = c^4/16\pi G$. On $u = F(t - z/c)$, $u' = -\dot u/c$, so $\tfrac12\mathcal T_{\rm s}u'^2 = \tfrac12\mu_{\rm s}\dot u^2$ and the energy density is $\mu_{\rm s}\dot u^2$. Summing the two polarisations reproduces $u = \frac{c^2}{16\pi G}(\dot h_+^2 + \dot h_\times^2)$. The tension $c^4/16\pi G$ is half of unit 35's stiffness $c^4/8\pi G = 4.8\times10^{42}$ N.

The number. Circular polarisation, $h_+ = h\cos\omega t$, $h_\times = h\sin\omega t$, has $\dot h_+^2 + \dot h_\times^2 = \omega^2h^2$ exactly, with $\omega = 2\pi\times150\ \mathrm{s^{-1}}$. Then $F = \frac{c^3}{16\pi G}\omega^2h^2 = 7.13\times10^{-3}\ \mathrm{W\,m^{-2}}$, against $1361\ \mathrm{W\,m^{-2}}$ of sunlight: a ratio of $5.2\times10^{-6}$. A large flux carried by a tiny strain: the tension $c^4/G$ again.

Gauge invariance of the average. Under a residual gauge transformation, the change in $\langle\partial_\mu h\,\partial_\nu h\rangle$ is a sum of terms each of which, after moving one derivative across inside the average, contains $\Box\xi = 0$ or $\partial^\alpha\bar h_{\alpha\beta} = 0$, plus total derivatives. The average of a total derivative over a region of size $\ell$ is smaller by $\lambda/\ell$ and is discarded. That is the precise sense in which averaging buys gauge invariance.

38 · energy

The quadrupole luminosity

theorem · the quadrupole luminosity $$P = \frac{G}{5c^5}\,\big\langle\overset{...}{Q}_{ij}\,\overset{...}{Q}_{ij}\big\rangle .$$

proof

1. $P = r^2\oint F\,\mathrm d\Omega$, and $F = \frac{c^3}{32\pi G}\langle\dot h^{\rm TT}_{ij}\dot h^{\rm TT}_{ij}\rangle$.

2. $\dot h^{\rm TT} = \frac{2G}{c^4r}\Lambda\overset{...}{Q}$, and $\Lambda$ is an orthogonal projector: $\dot h^{\rm TT}_{ij}\dot h^{\rm TT}_{ij} = \frac{4G^2}{c^8r^2}\overset{...}{Q}_{ij}\Lambda_{ij,kl}\overset{...}{Q}_{kl}$.

3. For symmetric traceless $A$: $\oint A_{ij}\Lambda_{ij,kl}A_{kl}\,\mathrm d\Omega = \frac{8\pi}5A_{ij}A_{ij}$ (notes).

4. $\frac{c^3}{32\pi G}\cdot\frac{4G^2}{c^8}\cdot\frac{8\pi}5 = \frac G{5c^5}$. $\square$

checks

Dimensions. $\frac{\mathrm{m^3\,kg^{-1}\,s^{-2}}}{\mathrm{m^5\,s^{-5}}}\cdot(\mathrm{kg\,m^2\,s^{-3}})^2 = \mathrm{kg\,m^2\,s^{-3}} = \mathrm W$ ✓.

Static source: $\overset{...}{Q} = 0$, $P = 0$ ✓.

beside Larmor

Unit 24: $P = \dfrac{\mu_0}{6\pi c}|\ddot{\vec p}|^2$. The same shape — a constant, a derivative of the lowest radiating moment, squared — with one more time derivative because the moment has one more power of length.

Full proof of step 3

Two isotropic integrals. $\oint n_in_j\,\mathrm d\Omega = \frac{4\pi}3\delta_{ij}$ and $\oint n_in_jn_kn_l\,\mathrm d\Omega = \frac{4\pi}{15}(\delta_{ij}\delta_{kl} + \delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk})$. Proof: each integral is invariant under rotations, so it is an isotropic tensor, and the only isotropic symmetric tensors of rank 2 and 4 are the ones written, with unknown constants. Fix the constants by contracting: $\oint n_in_i = 4\pi = 3C$, so $C = 4\pi/3$; and $\oint(n_in_i)^2 = 4\pi = C'(9 + 3 + 3) = 15C'$. That rank-4 isotropic tensors are spanned by those three is a fact of linear algebra (cited: Weyl, The Classical Groups, ch. II; or check it directly: problem P7 does the needed case by computing two components).

The contraction. For symmetric traceless $A$, with $P = 1 - nn^{\mathsf T}$, $A_{ij}\Lambda_{ij,kl}A_{kl} = \mathrm{tr}(APAP) - \tfrac12(\mathrm{tr}AP)^2$. Expanding, $\mathrm{tr}(APAP) = A_{ij}A_{ij} - 2(A^2)_{ij}n_in_j + (A_{ij}n_in_j)^2$ and $\mathrm{tr}(AP) = -A_{ij}n_in_j$. So the integrand is $A_{ij}A_{ij} - 2(A^2)_{ij}n_in_j + \tfrac12(A_{ij}n_in_j)^2$. Integrate: $4\pi A_{ij}A_{ij} - 2\cdot\frac{4\pi}3A_{ij}A_{ij} + \frac12\cdot\frac{4\pi}{15}\cdot2A_{ij}A_{ij}$ (the rank-4 integral against $A_{ij}A_{kl}$ gives $\frac{4\pi}{15}(0 + 2A_{ij}A_{ij})$ since $A$ is traceless). Total: $4\pi A_{ij}A_{ij}(1 - \tfrac23 + \tfrac1{15}) = \frac{8\pi}5A_{ij}A_{ij}$. $\square$

Step 2. $\Lambda$ is self-adjoint and idempotent, so $(\Lambda A)_{ij}(\Lambda A)_{ij} = \langle\Lambda A,\Lambda A\rangle = \langle A,\Lambda A\rangle = A_{ij}\Lambda_{ij,kl}A_{kl}$.

The average. $\langle\cdot\rangle$ is over several periods; for periodic sources it is the time average over one period, which is all section 5 needs.

38 · energy

Ten powers of $v/c$

Estimate. A source of mass $M$, size $d$, frequency $\omega$ has $\overset{...}{Q}\sim Md^2\omega^3 = Mv^3/d$. So

$$P\sim\frac{G}{c^5}\frac{M^2v^6}{d^2} = \frac{c^5}{G}\Big(\frac{GM}{dc^2}\Big)^2\Big(\frac vc\Big)^6 .$$

For a gravitationally bound source $v^2\sim GM/d$, and $P\sim(c^5/G)(v/c)^{10}$. Worked example 2 finds the exact version for the binary: $$P = \frac{32}5\,\frac{c^5}G\Big(\frac\mu M\Big)^2\Big(\frac vc\Big)^{10}.$$

the luminosity scale

$c^5/G = 3.63\times10^{52}$ W, about $10^{26}$ Suns, and independent of any source.

binary	$v/c$	$P$
Earth–Sun	$9.9\times10^{-5}$	$196$ W
PSR B1913+16	$1.5\times10^{-3}$	$7.8\times10^{24}$ W
GW150914, 150 Hz	$0.55$	$\sim4\times10^{49}$ W

The Earth radiates less than a kettle. A laboratory source is hopeless (problem P8: $10^{-28}$ W). Only compact bodies near $c$ make detectable waves — and that is exactly where the slow-motion expansion is failing.

The exact form. Worked example 2 gives $P = \frac{32G^4\mu^2M^3}{5c^5a^5}$. With $v^2 = GM/a$, $a = GM/v^2$ and $a^5 = G^5M^5/v^{10}$, so $P = \frac{32G^4\mu^2M^3v^{10}}{5c^5G^5M^5} = \frac{32}5\frac{c^5}G\big(\frac\mu M\big)^2\big(\frac vc\big)^{10}$.

The rows. Earth–Sun: $a = 1$ au, $M = M_\odot + M_\oplus$, $\mu\approx M_\oplus = 5.972\times10^{24}$ kg: $v/c = 9.94\times10^{-5}$, $P = 196$ W (problem P9). PSR B1913+16: the circular formula at $a = 1.949\times10^{9}$ m gives $6.55\times10^{23}$ W, and the cited eccentricity factor $11.86$ of section 6 raises it to $7.76\times10^{24}$ W, about 2 per cent of the Sun's light. GW150914: $\mu/M = 0.247$ and the formula at $v/c = 0.55$ gives $3.6\times10^{49}$ W, the same as the measured peak — but at $v/c = 0.55$ the formula has no business being right, and the agreement to two figures is luck; at $v/c = 0.5$ it gives $1.4\times10^{49}$ W. The row is marked "$\sim$" for that reason.

Why $c^5/G$. It is the only combination of $G$ and $c$ with the dimensions of power. It is also, up to a number, the largest luminosity anything can have in general relativity — a conjecture usually credited to Dyson and Thorne, not a theorem. The peak of GW150914 was $10^{-3}c^5/G$.

section 5

The chirp

The running example, end to end: power, decay, coalescence, and the one mass that shows.

38 · the chirp

The binary's quadrupole

worked example 2 · the power radiated by a circular binary

Step 1, reduce. Centre of mass at the origin: $\vec y_1 = \frac{m_2}M\vec x$, $\vec y_2 = -\frac{m_1}M\vec x$ with $\vec x$ the separation. Then $I_{ij} = m_1y_1^iy_1^j + m_2y_2^iy_2^j = \mu\,x^ix^j$.

Step 2, the orbit. $\vec x = a(\cos\Omega t, \sin\Omega t, 0)$. With double angles, $$I_{11,22} = \tfrac{\mu a^2}2(1 \pm \cos2\Omega t),\quad I_{12} = \tfrac{\mu a^2}2\sin2\Omega t.$$ The trace is constant, so $\overset{...}{Q} = \overset{...}{I}$.

Step 3, differentiate. $\overset{...}{I}_{11} = -\overset{...}{I}_{22} = 4\mu a^2\Omega^3\sin2\Omega t$, $\ \overset{...}{I}_{12} = -4\mu a^2\Omega^3\cos2\Omega t$.

Steps 1–2 over one orbit. The second moment $I_{ij} = \mu x^ix^j$ is even in $\vec x$, so half an orbit — the stars swapped, the same line through the centre (top) — leaves it unchanged. Both components repeat twice per orbit: the source, and so the wave, runs at $2\Omega$.

Reading the figure. The curves are step 2's $I_{11}/\mu a^2 = \cos^2\Omega t$ and $I_{12}/\mu a^2 = \tfrac12\sin2\Omega t$; $I_{22}$, not drawn, is $\sin^2\Omega t$, the mirror image of $I_{11}$, so that the trace stays $1$. The icons along the top are the separation vector at each quarter orbit. An electric dipole $q\vec x$ would follow $\vec x$ itself and repeat only once per orbit; the mass dipole is frozen (the slide "Why gravity has no dipole radiation"), and the quadrupole is the first moment that moves.

Step 1. $\vec y_1 - \vec y_2 = \vec x$ and $m_1\vec y_1 + m_2\vec y_2 = 0$ give the two formulas. Then $m_1(m_2/M)^2 + m_2(m_1/M)^2 = m_1m_2(m_1 + m_2)/M^2 = \mu$. Unit 08 reduced the two-body problem the same way.

Step 3. $\frac{\mathrm d^3}{\mathrm dt^3}\cos2\Omega t = 8\Omega^3\sin2\Omega t$ and $\frac{\mathrm d^3}{\mathrm dt^3}\sin2\Omega t = -8\Omega^3\cos2\Omega t$; times $\mu a^2/2$ gives the slide. The $33$ and $i3$ components vanish because the orbit is planar.

38 · the chirp

The binary's power

worked example 2, continued

Step 4, contract. $\overset{...}{Q}_{ij}\overset{...}{Q}_{ij} = \overset{...}{I}_{11}^2 + \overset{...}{I}_{22}^2 + 2\overset{...}{I}_{12}^2 = 16\mu^2a^4\Omega^6(2\sin^2 + 2\cos^2) = 32\mu^2a^4\Omega^6$, constant: no average needed.

Step 5, the power, then Kepler ($\Omega^2 = GM/a^3$, unit 08):

$$P = \frac{32G\mu^2a^4\Omega^6}{5c^5} = \frac{32}{5}\frac{G^4\mu^2M^3}{c^5a^5}.$$

Checks. Dimensions: $\frac{G^4\mu^2M^3}{c^5a^5}$ is $\frac{\mathrm{m^{12}kg^{-4}s^{-8}}\,\mathrm{kg^5}}{\mathrm{m^5s^{-5}}\,\mathrm{m^5}} = \mathrm{kg\,m^2s^{-3}}$ ✓. Test mass, $m_2\to0$: $P\propto\mu^2\to0$ ✓. The wave frequency is $2\Omega$ ✓ — the ring's ellipse repeats every half orbit.

Which way it shines. Problem P4's amplitudes $h_+ = h_0\frac{1 + \cos^2\iota}2$ and $h_\times = h_0\cos\iota$, with $h_0 = 4G\mu a^2\Omega^2/c^4D$, at angle $\iota$ from the orbit's axis (vertical; the orbit, edge-on, is the horizontal bar). Along the axis they are equal: circular polarisation. In the orbital plane only $h_+$ is left, at half strength.

Reading the figure. Each curve is a polar plot: the distance from the centre in direction $\iota$ is the amplitude an observer in that direction sees, in units of the on-axis $h_0$. Rotating the picture about the vertical axis gives the full three-dimensional pattern. The flux goes as $h_+^2 + h_\times^2$, so the axis receives $1 + 1 = 2$ and the orbital plane $\tfrac14$: eight times less. Step 5's power is this flux integrated over the sphere, the integral that step 3 of the quadrupole-luminosity proof did once for every source.

The wave itself. Along the orbit's axis, $n = \hat z$, the projector keeps $\tfrac12(\ddot Q_{11} - \ddot Q_{22})$ and $\ddot Q_{12}$. With $\ddot I_{11} = -2\mu a^2\Omega^2\cos2\Omega t$ and $\ddot I_{12} = -2\mu a^2\Omega^2\sin2\Omega t$, $$h_+ = -\frac{4G\mu a^2\Omega^2}{c^4D}\cos2\Omega t_{\rm r},\qquad h_\times = -\frac{4G\mu a^2\Omega^2}{c^4D}\sin2\Omega t_{\rm r},$$ with $D$ the distance: circularly polarised, the case worked out in the notes of "Helicity two, read off". In the orbital plane the amplitude halves and only $h_+$ survives: linearly polarised. Problem P4 does every inclination.

Kepler. $\Omega^6 = G^3M^3/a^9$, so $a^4\Omega^6 = G^3M^3/a^5$.

38 · the chirp

A slow drift on a fast orbit: two times

The orbit loses energy $P$ while its energy is $E = -G\mu M/2a$. The fraction lost per radian of orbit is

$$\epsilon = \frac{P}{|E|\,\Omega}\ll1 .$$

the adiabatic inspiral, as unit 21's two-timing

Fast time $t$ (the orbital phase), slow time $T = \epsilon t$ (the radius). Ansatz: $\vec x = a(T)\big(\cos\phi, \sin\phi\big)$ with $\dot\phi = \Omega(a(T))$. At leading order this is a Kepler circle for every $T$; at the next order, the solvability condition — no secular growth in the fast time — is the energy balance

$$\frac{\mathrm dE}{\mathrm dt} = -\langle P\rangle .$$

The leading-order inspiral, $a^4$ falling linearly in time, with the loss per orbit exaggerated about $4\times10^{10}$ times over Hulse–Taylor's. Each turn is almost a Kepler circle (dashed, the first turn's); only over many turns does the radius drift — the fast time and the slow one.

Reading the figure. The curve is $\vec x = a(\cos\phi, \sin\phi)$ with the radius from the next worked example, $a^4 = a_0^4 - 4\beta t$, and the phase from $\dot\phi = \Omega(a) = \sqrt{GM/a^3}$, integrated exactly: $\phi = \frac{2}{5}\frac{\sqrt{GM}}{\beta}(a_0^{5/2} - a^{5/2})$. The constant $\beta$ is chosen so that the radius falls from $a_0$ to $0.35\,a_0$ in six turns. For the real Hulse–Taylor orbit the radius changes by $1.6$ parts in $10^{12}$ per turn ($3.5$ m a year out of $1.95\times10^{9}$ m, over 1131 orbits a year): a million turns, nearly nine centuries, change it by two parts in a million.

Why this is unit 21's method. In unit 21 the Duffing oscillator had a fast phase and a slowly drifting amplitude and phase, and the slow evolution came from the solvability condition at order $\varepsilon$: the coefficient of the resonant term had to vanish. Here the small parameter is $\epsilon$, the leading-order solution is a Kepler circle of radius $a$, and the radiation-reaction force is the perturbation. Projecting it on the orbit's velocity and averaging over one fast period is exactly the solvability condition, and the result is $\dot E = -\langle P\rangle$ — the damped-oscillator amplitude equation $a' = -\mu a$ of unit 21 in a different costume. The ansatz is justified on time scales $O(1/\epsilon)$, and then repeatedly, by unit 21's cited Grönwall estimate; the inspiral takes many such scales, and the adiabatic approximation is justified throughout as long as $\epsilon$ stays small.

The energy. $E = \tfrac12\mu v^2 - G\mu M/a$ with $v^2 = GM/a$ gives $-G\mu M/2a$: unit 08's virial relation for circles. Decreasing $E$ means decreasing $a$ and increasing speed: radiation makes the stars go faster. Friction on a satellite does the same.

38 · the chirp

How adiabatic is it?

system	$\dot f/f^2$	verdict
PSR B1913+16	$\sim10^{-12}$	adiabatic, superbly
GW150914, 35 Hz	$0.065$	adiabatic, just
GW150914, 150 Hz	$0.73$	not adiabatic

$\dot f/f^2$ is the relative change of frequency per cycle, the observable version of $\epsilon$.

cited, not derived

The radiation-reaction force, whose average the solvability condition takes (MTW §36.11; Maggiore vol. 1, ch. 3). Energy balance itself rests on the Bondi mass-loss theorem of the energy section. Radiation reaction circularises orbits (Peters 1964), so circles stay circles.

38 · the chirp

How fast it falls in

worked example 3 · decay and coalescence

Step 1, balance. $E = -\frac{G\mu M}{2a}$ gives $\dot E = \frac{G\mu M}{2a^2}\dot a$. Set equal to $-P$ from worked example 2: $$\dot a = -\frac{2a^2}{G\mu M}\cdot\frac{32G^4\mu^2M^3}{5c^5a^5} = -\frac{64}{5}\frac{G^3\mu M^2}{c^5a^3}.$$

Step 2, separate. $a^3\,\mathrm da = -\beta\,\mathrm dt$ with $\beta = \frac{64G^3\mu M^2}{5c^5}$. Integrate from $a_0$ at $t = 0$: $$a(t)^4 = a_0^4 - 4\beta t .$$

Step 3, coalescence ($a = 0$): $$\tau = \frac{a_0^4}{4\beta} = \frac{5}{256}\,\frac{c^5a_0^4}{G^3\mu M^2}.$$

Step 2's $a(t) = a_0(1 - t/\tau)^{1/4}$, the same curve for every binary once time is measured in its own $\tau$. The radius barely moves for most of the life: half of it is still there at $t = \tfrac{15}{16}\tau$, and it goes to zero with a vertical tangent.

Reading the figure. $a^4 = a_0^4(1 - t/\tau)$ by steps 2–3, since $4\beta = a_0^4/\tau$. At $a = a_0/2$, $a^4/a_0^4 = 1/16$, so $t = \tfrac{15}{16}\tau$. The slope $\dot a = -\beta/a^3$ grows without bound as $a\to0$: the plunge at the end is the same power law, not a new effect.

Step 1's algebra. $\frac{2a^2}{G\mu M}\cdot\frac{32G^4\mu^2M^3}{5c^5a^5} = \frac{64G^3\mu M^2}{5c^5a^3}$: the $G$'s go $4 - 1 = 3$, the $\mu$'s $2 - 1 = 1$, the $M$'s $3 - 1 = 2$, the $a$'s $2 - 5 = -3$.

Step 2. $\frac{\mathrm d}{\mathrm dt}\frac{a^4}4 = a^3\dot a = -\beta$, so $a^4/4$ decreases linearly. This is unit 00's separable case, and the same computation as unit 24's classical atom, which collapses as $a^2\dot a = -\frac43r_e^2c$: the radiating orbit's radius obeys a power law and reaches zero in finite time. There it was a failure of classical physics. Here it is a prediction.

38 · the chirp

Coalescence times, from the Earth to Hulse–Taylor

worked example 3, continued

Checks. $[\beta] = \frac{\mathrm{m^9kg^{-3}s^{-6}}\cdot\mathrm{kg^3}}{\mathrm{m^5s^{-5}}} = \mathrm{m^4\,s^{-1}}$, so $[\tau] = \mathrm s$ ✓. $\mu\to0$: $\tau\to\infty$, a test mass never falls in ✓. $a_0\to0$: $\tau\to0$ ✓.

Numbers.

Earth–Sun	$\tau = 1.1\times10^{23}$ yr
B1913+16, as if circular	$1.6\times10^{9}$ yr
B1913+16, eccentric (cited eqs.)	$3.0\times10^{8}$ yr
B1913+16, $\dot a$ with $e$	$-3.5$ m/yr

The Earth's orbit outlives the universe by a factor of $10^{13}$. Hulse–Taylor's does not.

The numbers. Earth–Sun: $a_0 = 1.496\times10^{11}$ m gives $\tau = 1.07\times10^{23}$ yr, and $\dot a = -3.5\times10^{-13}$ m/yr. Hulse–Taylor as if circular at $a = 1.949\times10^{9}$ m: $1.64\times10^{9}$ yr. The actual orbit is eccentric; integrating Peters' coupled equations for $\dot a$ and $\dot e$ (cited: Peters, Phys. Rev. 136 (1964) B1224) numerically from today's $a$ and $e = 0.617$ gives $3.0\times10^{8}$ yr, shorter because the periastron passages radiate so strongly. The rate $\dot a = \tfrac23(a/P_b)\dot P_b = -3.53$ m/yr uses the measured-and-predicted $\dot P_b$ of section 6 and $P_b\propto a^{3/2}$.

What "coalescence" means here. $a\to0$ is where the formula says the stars meet. Long before then the stars touch or, for black holes, the slow-motion approximation fails; $\tau$ is still an excellent estimate of the remaining lifetime because almost all of it is spent at large $a$ — $a^4$ is linear in time, so the last half of the radius goes in the last $1/16$ of the time.

38 · the chirp

The chirp, and the one mass that shows

worked example 4 · $\dot f$ from $\dot a$

Step 1, Kepler in terms of $f$. The wave frequency is $f = \Omega/\pi$, and $\Omega^2 = GM/a^3$ gives $a = (GM)^{1/3}(\pi f)^{-2/3}$. Differentiating the log, $\dfrac{\dot f}f = -\dfrac32\dfrac{\dot a}a$.

Step 2, insert $\dot a$. $$\frac{\dot f}f = \frac{96}5\frac{G^3\mu M^2}{c^5a^4} = \frac{96}5\frac{G^{5/3}\mu M^{2/3}}{c^5}(\pi f)^{8/3}.$$

Step 3, one combination. $\mu M^{2/3} = \mathcal M^{5/3}$ with the chirp mass $\mathcal M = \mu^{3/5}M^{2/5} = \dfrac{(m_1m_2)^{3/5}}{(m_1 + m_2)^{1/5}}$. So

$$\boxed{\ \dot f = \frac{96}{5}\,\pi^{8/3}\Big(\frac{G\mathcal M}{c^3}\Big)^{5/3}f^{11/3}\ }$$

Step 4, integrate. $\frac{\mathrm d}{\mathrm dt}f^{-8/3} = -\frac83f^{-11/3}\dot f$ is constant, so $f$ blows up at a finite $t_c$: $$t_c - t = \frac{5}{256}\Big(\frac{G\mathcal M}{c^3}\Big)^{-5/3}(\pi f)^{-8/3} = \frac38\frac f{\dot f}.$$

Step 5, the amplitude (notes, from the wave of worked example 2): $h = \dfrac4D\Big(\dfrac{G\mathcal M}{c^2}\Big)^{5/3}\Big(\dfrac{\pi f}c\Big)^{2/3}$ on the axis.

Read it. Frequency, its rate, and the amplitude up to distance all contain the masses only through $\mathcal M$. Measure $f$ and $\dot f$: get $\mathcal M$. Add $h$: get $D$.

Checks. $\theta = G\mathcal M/c^3$ is a time, so $[\dot f] = \mathrm s^{5/3}\,\mathrm s^{-11/3} = \mathrm s^{-2}$ ✓. $\mathcal M\to0$: $\dot f\to0$ ✓. Equal masses $m$: $\mathcal M = 2^{-1/5}m = 0.87m$.

Step 1. $\ln a = \tfrac13\ln GM - \tfrac23\ln(\pi f)$, so $\dot a/a = -\tfrac23\dot f/f$.

Step 2. $-\tfrac32\cdot\big(-\tfrac{64}5\big) = \tfrac{96}5$, and $a^4 = (GM)^{4/3}(\pi f)^{-8/3}$ turns $G^3\mu M^2/a^4$ into $G^{5/3}\mu M^{2/3}(\pi f)^{8/3}$.

Step 3. $\mathcal M^{5/3} = \mu M^{2/3}$ by definition, and $\frac{G^{5/3}\mathcal M^{5/3}}{c^5} = \big(\frac{G\mathcal M}{c^3}\big)^{5/3}$. Then $\frac{\dot f}f = \frac{96}5\big(\frac{G\mathcal M}{c^3}\big)^{5/3}\pi^{8/3}f^{8/3}$, and multiply by $f$.

Step 4. With $K = \frac{96}5\pi^{8/3}(G\mathcal M/c^3)^{5/3}$, $\frac{\mathrm d}{\mathrm dt}f^{-8/3} = -\frac83K$, so $f^{-8/3} = \frac83K(t_c - t)$. Then $t_c - t = \frac{3}{8K}f^{-8/3} = \frac38\frac{f}{\dot f}$, and $\frac3{8}\cdot\frac5{96}\pi^{-8/3} = \frac5{256}\pi^{-8/3}$.

The check line, cleanly. Write $\theta = G\mathcal M/c^3$, a time. Then $[\dot f] = [\theta]^{5/3}[f]^{11/3} = \mathrm s^{5/3}\,\mathrm s^{-11/3} = \mathrm s^{-2}$ ✓.

Step 5. Worked example 2's axial wave has amplitude $\frac{4G\mu a^2\Omega^2}{c^4D}$. With $a^2\Omega^2 = (GM)^{2/3}\Omega^{2/3}$ and $\Omega = \pi f$, this is $\frac{4G^{5/3}\mu M^{2/3}(\pi f)^{2/3}}{c^4D} = \frac4D\big(\frac{G\mathcal M}{c^2}\big)^{5/3}\big(\frac{\pi f}c\big)^{2/3}$.

Distance, and redshift. That one waveform gives both $\mathcal M$ and $D$ makes a binary a "standard siren" (Schutz 1986). For a source at cosmological distance the observed $\mathcal M$ is $(1 + z)$ times the source's, and $D$ is the luminosity distance of unit 39; that is cited here and used only in the numbers of section 6.

38 · the chirp

What the chirp looks like

$h(t)\propto f(t)^{2/3}\cos\Phi(t)$ from worked example 4, with $\mathcal M = 30.6\,M_\odot$ — GW150914's — from $0.2$ s before coalescence, where $f = 33$ Hz, until $f$ reaches $250$ Hz, where the drawing stops because the approximations have long since failed. Dashed: 35 and 150 Hz, the band in which GW150914 was seen.

What rises. The frequency, as $(t_c - t)^{-3/8}$, and the amplitude, as $f^{2/3}\propto(t_c - t)^{-1/4}$. Both diverge at $t_c$, which the real signal does not: it saturates at merger and then rings down, and none of that is in this formula.

Heard as sound, the band from 35 to 250 Hz is audible: a rising "whoop" lasting a fifth of a second. The name chirp is literal.

How the figure was drawn. $f(t)$ from step 4 of worked example 4, inverted: $f = \frac1\pi\big(\frac{5}{256\,(t_c - t)}\big)^{3/8}\big(\frac{G\mathcal M}{c^3}\big)^{-5/8}$, with $G\mathcal M/c^3 = 1.507\times10^{-4}$ s. The phase is $\Phi = 2\pi\int f\,\mathrm dt$, integrated numerically; the amplitude is $f^{2/3}$ normalised to its final value. Between $0.2$ s before coalescence and the $250$ Hz cut there are about ten cycles. At $0.2$ s the formula gives $f = 32.6$ Hz.

Why the frequency is cut at 250 Hz. At a wave frequency $f$ the separation is $a = (GM)^{1/3}(\pi f)^{-2/3}$. With $M = 70.9\,M_\odot$ (the detector-frame total), $a = 920$ km at 35 Hz and $349$ km at 150 Hz — that is $8.8$ and $3.3$ times $GM/c^2$. By 250 Hz the "stars" would be closer than any black holes of that mass can be without merging, and the formula is describing nothing.

The exponents. $f\propto(t_c - t)^{-3/8}$ from step 4. $h\propto f^{2/3}$ from step 5. So $h\propto(t_c - t)^{-1/4}$: slower growth than the frequency, which is visible in the drawing as oscillations bunching up faster than they grow.

section 6

Two binaries, measured

A pulsar's clock, forty years of it; and a mirror that moved two thousandths of a proton.

38 · measured

PSR B1913+16

Found by Hulse and Taylor at Arecibo in 1974: a pulsar spinning every 59 ms in a 7.75-hour orbit with an unseen neutron star. A pulsar is a clock, and timing its pulses maps its orbit.

orbital period $P_b$	$27\,906.98$ s
eccentricity $e$	$0.6171340$
pulsar, companion	$1.438,\ 1.390\,M_\odot$
chirp mass $\mathcal M$	$1.2307\,M_\odot$
semi-major axis $a$	$1.949\times10^{9}$ m

Weisberg and Huang, ApJ 829 (2016) 55. The masses come from two other relativistic effects in the timing (periastron advance, clock rate), not from $\dot P_b$ — so $\dot P_b$ is a genuine test.

the eccentric orbit (cited)

Averaged over an orbit of semi-major axis $a$ and eccentricity $e$, the quadrupole luminosity is $P_{\rm circ}(a)\,f(e)$ — $f(e)$ a function of the eccentricity, not the wave frequency — with $$f(e) = \frac{1 + \frac{73}{24}e^2 + \frac{37}{96}e^4}{(1 - e^2)^{7/2}}.$$ Peters and Mathews, Phys. Rev. 131 (1963) 435; Maggiore vol. 1, ch. 4.

$f(0) = 1$ ✓. At $e = 0.6171$: $f(e) = 2.2143/0.18676 = 11.857$. The periastron passages, where the speed peaks, dominate — $v^{10}$ again.

What is cited and what is not. $f(e)$ comes from worked example 2's computation done on a Kepler ellipse instead of a circle: $I_{ij} = \mu x^ix^j$ along the ellipse, three time derivatives, contract, and average over one orbit using unit 08's orbit equation. Every step uses tools this course has; the averaging integrals are long, which is why it is cited. You can check the limit $e\to0$ and the leading correction $1 + \frac{157}{24}e^2 + O(e^4)$ — the expansion of $f$ — if you want a test of your own algebra.

Why energy loss sets $\dot P_b$ even for an ellipse. The orbital energy of a Kepler ellipse depends only on $a$: $E = -G\mu M/2a$ (unit 08). And $P_b\propto a^{3/2}$ by Kepler's third law. So $\dot P_b/P_b = \tfrac32\dot a/a = -\tfrac32\dot E/E$, exactly as for a circle, and the eccentric $\dot P_b$ is the circular one at the same $a$ multiplied by $f(e)$.

The masses. Fitting the timing model to the pulse arrival times measures the periastron advance, $4.2266^\circ$ per year — unit 37's effect, in a far stronger field — and the combined gravitational redshift and time dilation of the pulsar clock around the orbit. Each gives one equation in $m_p$ and $m_c$; together they fix both. The period derivative is then a third measurement with no free parameter left.

The numbers. $e^2 = 0.380854$, $e^4 = 0.145050$; the numerator $1 + 3.041667\times0.380854 + 0.385417\times0.145050 = 2.21434$; $(1 - e^2)^{7/2} = 0.186756$.

38 · measured

The period derivative

worked example 5 · $\dot P_b$ for PSR B1913+16

Step 1, circular, in terms of $P_b$. $P_b = 2\pi/\Omega = 2/f$, so $\dot P_b = -2\dot f/f^2$. With worked example 4 and $f = 2/P_b$: $$\dot P_b = -\frac{192\pi}{5}\Big(\frac{2\pi G\mathcal M}{c^3P_b}\Big)^{5/3}.$$

Step 2, the numbers. $G\mathcal M/c^3 = 1.2307\times4.9255\ \mu\mathrm s = 6.0620\ \mu$s. So $\frac{2\pi G\mathcal M}{c^3P_b} = 1.3648\times10^{-9}$, and its $5/3$ power is $1.6794\times10^{-15}$.

Step 3, circular answer. $-120.637\times1.6794\times10^{-15} = -2.026\times10^{-13}$.

Step 4, eccentric. Multiply by $f(e) = 11.857$:

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

	$\dot P_b/10^{-12}$
this example	$-2.402$
observed	$-2.423 \pm 0.001$
Galactic acceleration	$-0.025 \pm 0.004$
observed, corrected	$-2.398 \pm 0.004$

checks

$\dot P_b$ is dimensionless (seconds per second) ✓. Within $0.9\%$ of the raw value; corrected/predicted $= 0.9983 \pm 0.0016$ ✓. Circular orbit alone: wrong by $12\times$ — the eccentricity is not optional.

Step 1 in full. $\dot P_b = -\frac2{f^2}\cdot\frac{96}5\pi^{8/3}\theta^{5/3}f^{11/3} = -\frac{192}5\pi^{8/3}\theta^{5/3}f^{5/3}$ with $\theta = G\mathcal M/c^3$. Now $\pi^{8/3}f^{5/3} = \pi\cdot(\pi f)^{5/3} = \pi\,(2\pi/P_b)^{5/3}$, since $\pi f = \Omega = 2\pi/P_b$. So $\dot P_b = -\frac{192\pi}5\big(\frac{2\pi\theta}{P_b}\big)^{5/3}$.

Step 2 in full. $\mathcal M = \mu^{3/5}M^{2/5}$ with $M = 2.828\,M_\odot$ and $\mu = 1.438\times1.390/2.828 = 0.70680\,M_\odot$: $\mathcal M = 1.23075\,M_\odot$. $GM_\odot/c^3 = 4.92549\ \mu$s. $2\pi\times6.06203\times10^{-6}/27\,906.98 = 1.36485\times10^{-9}$. Its $5/3$ power: $\log_{10}$ is $-8.86491$, times $5/3$ is $-14.77485$, so $1.67935\times10^{-15}$.

The corrections. The observed $\dot P_b$ includes a kinematic part that is not gravitational radiation: the pulsar and the Sun accelerate differently in the Galaxy's potential, and the pulsar's proper motion across the sky changes its distance (the Shklovskii effect). Both change the apparent period. Damour and Taylor (1991) computed them; the uncertainty in the pulsar's distance limits the test to about $0.2\%$, not the timing. Cited: Weisberg and Huang 2016.

The checkpoint. It asks for $5\%$. Worked example 5 without the Galactic correction is $0.9\%$ from the raw measurement. The only way to miss by more than $5\%$ with correct algebra is to forget $f(e)$, which misses by a factor of 12.

History. Taylor, Fowler and McCulloch announced the decay in 1979, after four years of data, agreeing with the quadrupole formula to about 20 per cent. Hulse and Taylor received the 1993 Nobel Prize. It was the first evidence that gravitational waves exist, and it came twenty-one years before any detector saw one.

38 · measured

Forty years of a clock running down

The cumulative shift in the time of periastron, $\Delta T(t) = \tfrac12(\dot P_b/P_b)\,t^2$, predicted from worked example 5; marked, the slide's two numbers. Weisberg and Huang's measured points lie on this curve (see their paper).

Why a parabola. Each orbit is shorter than the last by $\dot P_bP_b$. After $N$ orbits the accumulated lag is $\sum_{k\le N}k\,\dot P_bP_b\approx\tfrac12\dot P_bP_bN^2$, and $N = t/P_b$.

The numbers. $-38.6$ s after 30 years, $-68.6$ s after 40 — a drift of about a minute, read from pulse arrival times good to tens of microseconds.

the running example, instance 1, closed

Quadrupole formula, energy balance, Kepler, one cited factor for the ellipse: $-2.402\times10^{-12}$. The sky: $-2.398\times10^{-12}$ after correction. No parameter was adjusted.

The sum. If the $k$-th orbit has period $P_b + k\,\delta$ with $\delta = \dot P_bP_b$, then after $N$ orbits the elapsed time differs from $NP_b$ by $\delta\sum_{k=1}^N k = \delta N(N + 1)/2\approx\tfrac12\delta N^2$. With $N = t/P_b$ this is $\tfrac12(\dot P_b/P_b)t^2$.

The numbers. $\dot P_b/P_b = -2.402\times10^{-12}/27\,906.98\ \mathrm s = -8.607\times10^{-17}\ \mathrm{s^{-1}}$. After 30 years ($9.467\times10^{8}$ s): $-38.6$ s. After 40: $-68.6$ s. The curve is drawn from 1975; the measurements start in late 1974.

Why the figure has no data points. The course states only numbers it has computed. The measured points are in Weisberg and Huang (2016) and Weisberg, Nice and Taylor, ApJ 722 (2010) 1030. Look them up: they sit on the curve to within their error bars, across four decades.

38 · measured

GW150914

14 September 2015, both LIGO detectors, 7 ms apart (less than the 10 ms light time between them). The frequency rose from 35 to 150 Hz in about 0.2 s and 8 cycles.

Worked example 4, backwards. Measuring $f$ and $\dot f$ gave $\mathcal M\approx30\,M_\odot$ as seen at the detector. With $\mathcal M = 30.6\,M_\odot$:

$\dot f$ at 35 Hz	$79.6$ Hz/s
time to $t_c$ from 35 Hz	$0.165$ s
cycles, 35 to 150 Hz	$8.4$
$h$, optimal, 35 Hz	$0.93\times10^{-21}$
$h$, optimal, 150 Hz	$2.5\times10^{-21}$

Measured peak strain $1.0\times10^{-21}$: the source was not optimally placed or oriented for the detectors.

the energy

$3.0 \pm 0.5\,M_\odot c^2 = 5.4\times10^{47}$ J in about 0.2 s. Peak power $3.6\times10^{49}$ W, about $200\,M_\odot c^2$ per second and $10^{-3}c^5/G$.

where this unit stops

Newtonian energy between 35 and 150 Hz: $1.5\,M_\odot c^2$. The rest is merger and ringdown, at $v/c\approx0.5$, separation $3GM/c^2$, $\dot f/f^2\approx0.7$. Linear theory, slow motion and adiabaticity have all failed. What replaces them is numerical relativity, and the black holes are units 36 and 40.

Abbott et al. (LIGO and Virgo), PRL 116 (2016) 061102.

The numbers. With $\theta = G\mathcal M/c^3 = 30.6\times4.9255\ \mu\mathrm s = 1.5072\times10^{-4}$ s: $\dot f(35) = \frac{96}5\pi^{8/3}\theta^{5/3}35^{11/3} = 79.6$ Hz/s. $t_c - t = \frac5{256}\theta^{-5/3}(35\pi)^{-8/3} = 0.165$ s. The number of cycles between two frequencies is $\int f\,\mathrm dt = \int f/\dot f\,\mathrm df = \frac{1}{32\pi^{8/3}}\theta^{-5/3}(f_1^{-5/3} - f_2^{-5/3}) = 8.4$ (problem P10). Amplitude from step 5 with $D = 410$ Mpc $= 1.265\times10^{25}$ m and $G\mathcal M/c^2 = 45.2$ km: $0.93\times10^{-21}$ at 35 Hz, $2.45\times10^{-21}$ at 150 Hz.

Redshift, cited. The source's masses are $36$ and $29\,M_\odot$, chirp mass $28.1\,M_\odot$; at redshift $z = 0.09$ the detector sees every time scale stretched by $1 + z$, which is indistinguishable from masses larger by $1 + z$: $30.6\,M_\odot$. That, and the luminosity distance, are unit 39's business; the numbers here only use the detector-frame chirp mass.

Why "optimal". Step 5's amplitude is for a detector directly above the orbit's axis with its arms aligned to the polarisation. A real detector responds to a combination $F_+h_+ + F_\times h_\times$ with antenna factors $|F|\le1$ depending on the sky position, and the orbit's inclination reduces $h$ further (problem P4). The factor between $2.5$ and $1.0$ is of that kind.

The Newtonian energy. $E = -\tfrac12\mu(GM\Omega)^{2/3}$ with source-frame $M = 65\,M_\odot$, $\mu = 16.06\,M_\odot$, and source-frame frequencies $(1 + z)\times35$ and $(1 + z)\times150$ Hz gives $\Delta E = 1.50\,M_\odot c^2$. The measured $3.0\,M_\odot c^2$ is twice that; the difference was radiated in the last orbit, the merger and the ringdown of the final black hole.

Why black holes. At 150 Hz the Newtonian separation is 349 km for a total mass of $71\,M_\odot$: two objects of $30\,M_\odot$ each within a few hundred kilometres. No star is that compact. The identification, and the ringdown frequency of the remnant, need the Kerr metric of unit 40.

38 · measured

The detector is Michelson's

A beam splitter sends light down two perpendicular arms of length $L = 4$ km to mirrors hung as free masses, and recombines the returns. Any difference in round-trip time shows as a change in the interference.

round-trip times in TT gauge

Wave along $z$, arms along $x$ and $y$, $L\ll\lambda$. Light along $x$ has $0 = -c^2\mathrm dt^2 + (1 + h_+)\,\mathrm dx^2$, so $c\,\mathrm dt = (1 + \tfrac12h_+)\,\mathrm dx$, and by the lemma the mirrors stay at $x = 0, L$:

$$T_{x,y} = \frac{2L}c\big(1 \pm \tfrac12h_+\big).$$

Schematic; the arm changes are exaggerated about $10^{20}$ times. With the wave along $z$ and $h_+ > 0$, light takes longer on the $x$ arm and less long on the $y$ arm, which a lab frame reads as the end mirrors moving by $\pm\tfrac12h_+L$. Half a wave period later the roles swap.

Reading the figure. The solid mirrors are where the free masses sit in TT coordinates, which the lemma of section 2 says do not change; the dashed ones are where they sit in proper distance from the beam splitter, which is what a lab frame calls their position and what the interference measures. The two descriptions agree on the one thing the beam splitter can compare, the difference of the two round-trip times. The ring of free masses is this picture with every direction at once; the interferometer keeps its centre and two of its masses, a quarter-turn apart.

The round trip. Along the $x$ arm $h_+$ is effectively constant during one round trip ($2L/c = 27\ \mu$s, against a wave period of $7$ ms at 150 Hz), so $T_x = 2\int_0^L(1 + \tfrac12h_+)\,\mathrm dx/c = \frac{2L}c(1 + \tfrac12h_+)$. Along $y$, $g_{yy} = 1 - h_+$. The mirrors do not move in TT coordinates (the free-masses lemma); the light's coordinate speed changes instead. In a local lab frame one would say the mirrors moved and light's speed is $c$; the time difference is the same, being a comparison of two proper times at the beam splitter.

What the real instrument adds (cited: Saulson, Fundamentals of Interferometric Gravitational Wave Detectors). Each arm is a Fabry–Pérot cavity in which light makes hundreds of round trips, multiplying the phase shift; the mirrors are 40 kg fused-silica masses on multi-stage pendulums, free along the arm above a few hertz; and the laser power is recycled to reduce the photon counting noise. None of that changes the principle on the slide.

38 · measured

What the two arms measure

The measured quantity. $T_x - T_y = \dfrac{2Lh_+}c$: the arms behave as if lengthened and shortened by $\tfrac12h_+L$ each, a differential length change $h_+L$.

For GW150914, $h = 1.0\times10^{-21}$: each arm $\pm2\times10^{-18}$ m, differential $4\times10^{-18}$ m, about $1/200$ of a proton radius.

the same experiment, twice

Michelson and Morley (1887) compared round-trip times along perpendicular arms, looking for the ether, and found no difference. LIGO compared round-trip times along perpendicular arms, and in 2015 found one — made by a wave that had travelled for over a billion years.

38 · summary

What the running example bought

result	formula	from
field equation	$\Box\bar h_{\mu\nu} = -\frac{16\pi G}{c^4}T_{\mu\nu}$	35, Lorenz gauge
polarisations	$10 - 4 - 4 = 2$	linear algebra
field	$h^{\rm TT} = \frac{2G}{c^4r}\Lambda\ddot Q$	24's $G_{\rm ret}$, virial lemma
power	$\frac G{5c^5}\langle\overset{...}{Q}\overset{...}{Q}\rangle$	Isaacson (cited)
binary	$\frac{32G^4\mu^2M^3}{5c^5a^5}$	worked ex. 2
decay	$\tau = \frac5{256}\frac{c^5a_0^4}{G^3\mu M^2}$	21's two-timing
chirp	$\dot f\propto\mathcal M^{5/3}f^{11/3}$	worked ex. 4
B1913+16	$\dot P_b = -2.402\times10^{-12}$	$+f(e)$ (cited)

the binary, closed

Linearised, its field obeys unit 24's wave equation. Gauge leaves two polarisations, each squeezing a ring along one pair of axes. Two conservation laws silence its monopole and dipole, so it radiates through its quadrupole, weakly — ten powers of $v/c$. Energy balance on a slowly shrinking circle gives a chirp governed by one mass. At $v/c = 10^{-3}$ that predicts a pulsar's clock to 0.2 per cent. At $v/c = 0.5$ it predicts the shape of what LIGO heard, and then stops being true.

Next. Unit 39 applies the field equations to the whole universe; unit 40 to rotating holes, and says where this course's methods end.

The citations, collected. What is used without proof is labelled on its own slide: the Isaacson stress-energy tensor (and the second-order action used to check its coefficient); the Peters–Mathews eccentricity factor; and the post-Newtonian justification of the quadrupole formula for self-gravitating sources, with its companions, the radiation reaction force whose average is the energy balance and the Bondi mass-loss theorem that balance rests on. Three smaller ones go the same way: the span of the rank-four isotropic tensors, the numerical integration of Peters' eccentric equations for the merger time, and unit 39's redshifted chirp mass and luminosity distance. Everything else in the unit is proved on the slides or in these notes from units 08, 10, 13, 16, 20, 21, 24, 30, 31, 32, 33 and 35.

Reading. Maggiore, Gravitational Waves, vol. 1 (OUP, 2008): chapter 1 (the linearised theory, TT gauge, the Isaacson tensor) and chapters 3–4 (the quadrupole formula and compact binaries) at exactly this level; skip §3.4 onwards on first reading, which uses spherical tensor harmonics this course has not built. Schutz, A First Course in General Relativity, chapter 9, for a shorter route. Hartle, Gravity, chapter 23, for the physics with little formalism.

38 · checkpoint

Checkpoint

you can now do this, or you cannot

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 $-2.423\times10^{-12}$.

Failing it is unambiguous. If your count ends at 6, you stopped after Lorenz gauge and forgot the residual $\Box\xi = 0$. If your ring breathes, you kept a trace. If your $\dot P_b$ is $-2.0\times10^{-13}$, you forgot the eccentricity factor; if it is off by a power of 2, you used the orbital frequency where the wave frequency belongs.

Twelve problems, about six hours: ./mr problems 38. P3 is the count; P5 the ring; P11 the checkpoint's number; P8 and P12 the two estimates. When the checkpoint is true: ./mr done 38.