Gravitational waves

The metric responds to matter, but it is not required to follow matter instantaneously. Einstein’s equation is a dynamical field equation. Once disturbed, geometry has propagating degrees of freedom of its own.

The key conceptual distinction is between a field’s source and the field already present. Maxwell’s equations permit light in a charge-free region. Einstein’s equation permits gravitational waves in a matter-free region. “The source is zero here” does not imply “the solution is zero here.”

18.1 Keeping the first-order gravitational field#

PREPARATION FOR THIS SECTION

Read a wave before reading its complex notation

How does an equation say that information travels at a definite speed?

Builds onPartial derivatives Differential equations and initial data

See the idea

Imagine a pulse with a shape $F$ moving toward increasing $z$ without changing shape. At time $t$, its profile is $F(z-ct)$. Keeping the pulse’s argument fixed gives $z=ct+\text{constant}$: the pulse moves at speed $c$. No complex numbers are needed.

Work it out

1. 01
   Check the wave equation with the chain rule

   For $h(t,z)=F(z-ct)$, differentiating twice in time gives $\partial_t^2h=c^2F''$, while differentiating twice in space gives $\partial_z^2h=F''$. Therefore every twice-differentiable shape of this form obeys the same wave equation. $G(z+ct)$ describes propagation in the opposite direction.

   $$-\frac1{c^2}\frac{\partial^2h}{\partial t^2}+\frac{\partial^2h}{\partial z^2}=0.$$

   Why this step works A partial differential equation relates changes at neighboring places and times. Its allowed solutions carry the propagation speed.

2. 02
   Turn crests into frequency and wavelength

   For a sinusoid $h=A\cos(kz-\omega t)$, increasing $z$ by $2\pi/k$ or time by $2\pi/\omega$ repeats the phase. Thus wavelength $\lambda=2\pi/k$, period $P=2\pi/\omega$, and frequency $f=1/P$. Substitution in the wave equation gives $\omega^2=c^2k^2$; choose positive $k,\omega$ for propagation toward increasing $z$.

   $$\omega=ck,\qquad f\lambda=c.$$

   Why this step works The coordinate z is a length, k has inverse-length units, and omega has inverse-time units. The phase is dimensionless.

3. 03
   Introduce the delayed source time

   A signal received at time $t$ from a source a distance $R$ away left at $t-R/c$. This is retarded time. In flat three-dimensional space, each source location $\mathbf x'$ has its own distance $R=|\mathbf x-\mathbf x'|$. A retarded field integral samples each source contribution at its corresponding departure time. The factor $1/R$ in the later solution describes spherical spreading; it is not part of the time delay.

   $$t_{\rm ret}=t-\frac{|\mathbf x-\mathbf x'|}{c}.$$

   Why this step works Selecting the retarded solution is a physical boundary condition: the measured response follows the source. The differential equation also permits added free waves.

4. 04
   Treat complex notation as an optional abbreviation

   Define $i^2=-1$ and use $e^{i\psi}=\cos\psi+i\sin\psi$. This identity packages two real functions together; taking the real part recovers the cosine. Differentiation multiplies $e^{i(kz-\omega t)}$ by $ik$ or $-i\omega$, making linear equations easier to solve. A complex amplitude $A=a+ib$ gives the real field $a\cos\psi-b\sin\psi$, so it also stores phase.

   $$h=\operatorname{Re}\!\left[Ae^{i(kz-\omega t)}\right].$$

   Why this step works The measured strain is real. Complex notation is a calculation tool; no imaginary displacement has been introduced.

Go deeper

In three spatial dimensions, $\nabla^2=\partial_x^2+\partial_y^2+\partial_z^2$ and $\Box=-c^{-2}\partial_t^2+\nabla^2$. The linearized Einstein equation applies this operator to each component of a suitable metric perturbation. A plane-wave phase can be written $k_\mu x^\mu$, with $x^0=ct$ and $k_\mu=(-\omega/c,k_x,k_y,k_z)$. Substitution gives $-\omega^2/c^2+|\mathbf k|^2=0$: the wave covector is null. This establishes the propagation law in the approximation. Gauge constraints and geodesic deviation are still needed to identify the two measurable polarizations.

Further readingMIT · The wave equation David Tong · When gravity is weak, §5.2

Test the idea

FIRST, PREDICT

Which profile moves toward increasing $z$ at speed $c>0$?

$F(z-ct)$$F(z+ct)$$F(z)\cos(ct)$

Compare the reasoning

$F(z-ct)$

A fixed feature has $z-ct=\text{constant}$, hence $dz/dt=c$.

$F(z+ct)$

A fixed feature now has $z+ct=\text{constant}$, so $dz/dt=-c$.

$F(z)\cos(ct)$

This keeps the spatial shape fixed while its amplitude changes. It does not generally describe a translating profile; moreover, ct alone has length units and cannot be the argument of a cosine.

A hint

Follow a particular crest by setting its argument equal to a constant.

NOW CHANGE THE EXAMPLE

A source is $9.0\times10^8\,\mathrm m$ away in a flat-space model with $c=3.0\times10^8\,\mathrm{m/s}$. A signal arrives at $t=12\,\mathrm s$. At what source time was it emitted?

A hint

Compute the travel time first, then subtract it from the arrival time.

Work through the solution

$R/c=3\,\mathrm s$, so $t_{\rm ret}=12-3=9\,\mathrm s$. In a curved spacetime the travel time must instead follow the appropriate null path.

Real wave profiles teach propagation; complex exponentials abbreviate the same real solutions.

Use Cartesian coordinates $x^0=ct$ on a Minkowski background, take $\Lambda=0$, and write

Smallness is asserted in a suitable background-adapted coordinate system; a wild coordinate transformation can make components large without creating strong physical gravity. We keep first-order terms in $h$. The inverse metric is

where perturbation indices are raised with $\eta$. Multiplying the two metrics verifies the minus sign: the first-order cross terms cancel.

The linearized connection is

The correction to the inverse metric is first order. Multiplying it by $\partial h$ would give a second-order term, which we discard here. For the same reason, the $\Gamma\Gamma$ terms do not enter first-order curvature: each connection is first order around this constant background.

Contracting the derivative terms in Riemann gives

with

The trace is

Two terms in Ricci involve divergences of $h$, one applies the wave operator to each component, and one differentiates the trace. Combining the trace with $h_{\mu\nu}$ will collect these terms into a simpler equation.

18.2 Trace reversal and Lorenz gauge#

Define the trace-reversed perturbation

In four dimensions its trace is $\bar h=-h$, since contracting $\eta_{\mu\nu}$ produces four. Consequently the inverse operation has the same form:

The name refers to the sign of the trace. It does not mean reversing every tensor component.

The linearized Einstein tensor becomes

Symmetrization includes its factor of $1/2$. The terms beyond the wave operator all contain the divergence of $\bar h$. A coordinate choice that makes this divergence zero will remove them together.

Under an infinitesimal coordinate change $x'^\mu=x^\mu+\xi^\mu$, the first-order perturbation changes by

This is a different coordinate description of the same metric, to the stated order. Its trace-reversed divergence transforms as

Locally, with suitable initial and boundary conditions, solve $\Box\xi_\nu=\partial^\mu\bar h_{\mu\nu}$. In the resulting coordinates,

This is usually called Lorenz, harmonic, or de Donder gauge in this context. It removes four coordinate-dependent combinations, not four physical forces.

The field equation collapses to

The coefficient follows directly from $G^{(1)}_{\mu\nu}=-\Box\bar h_{\mu\nu}/2$ and Einstein’s $8\pi G_N/c^4$. The extra factor of two comes from dividing by the coefficient $-1/2$ on the geometric side.

Taking a divergence requires $\partial^\mu T_{\mu\nu}=0$ at this order. The source cannot be chosen arbitrarily; its leading dynamics must be consistent with energy-momentum conservation. For a self-gravitating compact system, systematically including the gravitational contribution requires the appropriate perturbative expansion rather than inserting an inconsistent prescribed matter motion.

18.3 Why there are two polarizations, not ten#

In vacuum, $\Box\bar h_{\mu\nu}=0$. A plane wave has the form

Applying $\Box$ multiplies this by $-k_\alpha k^\alpha$, so a nontrivial wave requires

Its wave covector is null; for a wave propagating along $z$, the phase depends on $t-z/c$. Gravitational disturbances propagate on the background light cone at this order.

The gauge condition gives $k^\mu A_{\mu\nu}=0$, four restrictions on a symmetric tensor’s ten components. But this gauge is not fully fixed: transformations satisfying $\Box\xi^\mu=0$ preserve it. For a nonzero vacuum plane wave, four residual gauge choices remove four further amplitude combinations. What remains are two independent radiative polarizations.

This count applies to a nonzero vacuum plane wave satisfying both its field equation and gauge condition. The explicit reduction below shows which components disappear; counting alone would not establish their independence.

Further calculation: remove the four residual components

Write the unbarred amplitude as $H_{\mu\nu}$, so $h_{\mu\nu}=\operatorname{Re}[H_{\mu\nu}e^{ik_\alpha x^\alpha}]$. For propagation in the positive $z$ direction, take $k_\mu=(-q,0,0,q)$ with $q=\omega/c\ne0$. A residual coordinate change $\xi_\mu=\operatorname{Re}[B_\mu e^{ik_\alpha x^\alpha}]$ satisfies $\Box\xi_\mu=0$ because $k$ is null. It changes the amplitude by

$$H'_{\mu\nu}=H_{\mu\nu}-i(k_\mu B_\nu+k_\nu B_\mu).$$

The time-containing entries consequently obey

$$\begin{aligned} H'_{00}&=H_{00}+2iqB_0,\\ H'_{01}&=H_{01}+iqB_1,\qquad H'_{02}=H_{02}+iqB_2,\\ H'_{03}&=H_{03}+iq(B_3-B_0). \end{aligned}$$

Choose $B_0$ to set the first line to zero, $B_1,B_2$ to set the second line to zero, and then $B_3$ to set the third line to zero. Thus all $H'_{0\mu}$ vanish, while Lorenz gauge remains satisfied.

Since $k^\mu=(q,0,0,q)$, that gauge condition reads $\bar H'_{0\nu}+\bar H'_{3\nu}=0$. At $\nu=0$, the already zero time entries leave $H'/2=0$, where $H'$ is the trace. The remaining conditions then give $H'_{3\nu}=0$. Only $H'_{11}=-H'_{22}$ and $H'_{12}=H'_{21}$ are free. These are the two amplitudes displayed below.

A convenient representative is transverse-traceless, or TT, gauge. For propagation along $z$,

“Transverse” means the spatial perturbation has no component along the propagation direction. “Traceless” means its diagonal spatial entries sum to zero. Along the instantaneous principal axes, the two transverse eigenvalues are equal and opposite.

The plus polarization stretches an initially circular ring of free test particles along one axis and compresses it along the perpendicular axis. Half a cycle later the roles reverse. The cross polarization does the same with axes rotated by $45^\circ$. Superpositions produce elliptical or circular polarization.

Rotate the transverse basis by $\theta$, taking $e'_x=\cos\theta\,e_x+\sin\theta\,e_y$ and $e'_y=-\sin\theta\,e_x+\cos\theta\,e_y$. Applying the tensor transformation to its two inputs gives

The double angle follows from $\cos^2\theta-\sin^2\theta=\cos2\theta$ and $2\sin\theta\cos\theta=\sin2\theta$. In particular, a $45$-degree rotation exchanges the two patterns up to sign. This transformation is the classical meaning of the wave’s spin-2 angular response.

18.4 What a detector measures in TT coordinates#

For initially stationary free test particles in TT coordinates, $\Gamma^i{}_{00}=0$ at first order. Their spatial coordinates can remain constant while the metric changes the distance between them. Along a short arm directed along $x$,

Along $y$, the corresponding change is $-h_+/2$. This expression assumes an arm short compared with the wavelength, so the field is approximately uniform across it on the chosen time slice.

The same observable effect can be obtained from curvature, avoiding the impression that moving grid lines create physics. In a local inertial frame associated with the detector,

The geodesic-deviation equation therefore gives

To first order, with initial conditions corresponding to undisturbed separations before a passing wave,

In one coordinate description the masses stay at fixed coordinates and the metric changes their separation. In another their coordinates respond to tidal acceleration. They predict the same instrument response. Linearized Riemann is unchanged by a pure linearized gauge transformation about flat spacetime because the extra third derivatives cancel.

A laser interferometer compares the phases accumulated by light traversing differently oriented arms and returning to a common observer. In the ideal long-wavelength, favorably oriented case,

For $L_0=4\,\mathrm{km}$ and $h_+=10^{-21}$, the differential equivalent length is $4\times10^{-18}\,\mathrm m$. Each individual arm’s change in this idealized example is half that magnitude with opposite sign. Actual responses include source direction, polarization, optical configuration, and frequency-dependent light travel effects.

We can check the light signal explicitly in the same short-arm limit. During one round trip, treat $h_+$ as approximately constant. The null condition along the $x$ arm gives $c\,dt\simeq(1+h_+/2)|dx|$; the $y$ arm has the opposite sign. Their round-trip times are therefore

The clock at the beamsplitter measures this $t$ as proper time because $g_{00}=-1$ there. The arrival-time difference is $T_x-T_y=2L_0h_+/c$. A laser of local frequency $f_{\rm laser}$ converts it into a phase difference $2\pi f_{\rm laser}(T_x-T_y)$. Thus following the light gives a measurable change, consistent with the strain calculation. For longer arms relative to the wavelength, integrate the changing field along each outgoing and returning light path instead.

SPACETIME LAB / 07

One wave. Two ways to stretch.

Watch one detector ring. Compare plus and cross polarization, then separate what strain and frequency change.

ExploreDiagram

Interactive geometry is loading.

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Play

$+$Plus$\times$Cross

Phase at the middle detector

Start of cycleOne full cycle

Strain amplitude

Larger strain, larger deformation.

Angular frequency

$$\begin{gathered}h_+=h_0\sin(\omega z/c-\phi)\\x=x_0(1+h_+/2)\\y=y_0(1-h_+/2)\end{gathered}$$

Read the scene. The brighter middle ring is one detector plane; the quieter rings show different positions along the wave. Plus and cross are linear polarizations with stretching axes rotated by $45^\circ$. Guides mark fixed coordinates, not material arms. Amplitude $h_0$ is dimensionless strain and is exaggerated here; displacement is accurate only to first order in strain. Frequency is given in units $c/L_0$, where $L_0$ is the undeformed ring radius. Changing frequency changes wavelength as $\lambda=2\pi c/\omega$; the wave speed stays $c$. Animation advances $0.8L_0/c$ of model time per displayed second. These test particles do not form a material medium carrying sound.

18.5 Waves from a changing source#

PREPARATION FOR THIS SECTION

From an initial ripple to a retarded field

Which mathematical step chooses a wave arriving from the source, rather than one arriving from the future?

Builds onInitial data Point sources and Poisson Field energy

See the idea

A wave equation tells disturbances how to propagate, but does not decide which disturbance occurred. Initial data specify an initial displacement and velocity. A Green function instead builds a response from an elementary source. The retarded choice says that the response occurs after the source.

Work it out

1. 01
   Propagate initial data in one space dimension

   The equation $u_{tt}=c^2u_{xx}$ admits right-moving $F(x-ct)$ and left-moving $G(x+ct)$. With $u(x,0)=f(x)$ and $u_t(x,0)=g(x)$ on the whole line, solving for their sums and derivatives gives d’Alembert’s formula.

   $$u(x,t)=\frac{f(x-ct)+f(x+ct)}2+\frac1{2c}\int_{x-ct}^{x+ct}g(s)\,ds.$$

   Why this step works Differentiation checks the equation and both initial conditions; the integration interval exposes finite propagation speed.

2. 02
   Package oscillations with complex numbers

   Define $i^2=-1$ and $e^{i\theta}=\cos\theta+i\sin\theta$. The real part of $Ae^{i(kx-\omega t)}$ is a real wave. Substitution gives $\omega^2=c^2k^2$. Linear combinations also solve the linear equation, so a packet can be assembled from many wave numbers k.

   $$u=\operatorname{Re}\!\int a(k)e^{i[kx-\omega(k)t]}dk.$$

   Why this step works Complex notation packages phase; the physical field in this example is real.

3. 03
   Specify the kernel and its boundary condition

   In three flat spatial dimensions let $L=\nabla^2-c^{-2}\partial_t^2$ and $L\psi=-4\pi f$. Its retarded point-source kernel is $G_{\rm ret}=\delta(t-t'-R/c)/R$, with $R=|\mathbf x-\mathbf x'|$ and $LG_{\rm ret}=-4\pi\delta(t-t')\delta^{(3)}(\mathbf x-\mathbf x')$. Integrating it against the source collapses the time integral.

   $$\psi(t,\mathbf x)=\int\frac{f(t-R/c,\mathbf x')}{R}\,d^3x'.$$

   Why this step works The kernel normalization and retarded support specify which inverse of the differential operator is meant.

Go deeper

The kernel identity is understood distributionally: integrate it against a smooth test function, integrate derivatives by parts, and the shrinking sphere around $R=0$ supplies the same $-4\pi$ normalization as the Poisson kernel. Away from the source, radial propagation makes $R\psi$ obey a one-dimensional wave equation. A homogeneous solution can still be added; “retarded solution” here specifies the sourced response without an independently supplied incoming wave. In a dispersive medium $\omega(k)$ need not equal $c|k|$; phase velocity $\omega/k$ and narrow-packet group velocity $d\omega/dk$ differ. Neither alone universally determines front or signal speed. Curved-spacetime Green functions can also have support inside the light cone, so the flat-space formula must not be transplanted unchanged.

Further readingDavid Tong · Electromagnetism MIT · Differential equations and initial data

Test the idea

FIRST, PREDICT

Does the wave equation alone force every solution to be retarded?

Yes. The word wave means retarded.No. All wave solutions transmit information instantaneously.No. Initial or boundary conditions select the response.

Compare the reasoning

Yes. The word wave means retarded.

The equation also admits advanced kernels and incoming homogeneous solutions mathematically.

No. All wave solutions transmit information instantaneously.

Finite propagation and the choice of sourced response are separate issues.

No. Initial or boundary conditions select the response.

The differential operator allows homogeneous waves and other Green-function boundary conditions.

A hint

Try adding a free wave to a particular sourced solution.

NOW CHANGE THE EXAMPLE

In dimensionless units $c=1$, take $f(x)=x^2$ and $g(x)=0$. Use d’Alembert’s formula to find $u(0,2)$.

A hint

Average $f(-2)$ and $f(2)$.

Work through the solution

$u(0,2)=(4+4)/2=4$. The solution is $u=x^2+t^2$ and satisfies $u_{tt}=u_{xx}=2$.

The operator, source normalization, and initial or boundary conditions jointly define a wave prediction.

With no incoming radiation and an appropriate localized weak source, the retarded solution is

Every source element contributes at its own retarded time. The denominator gives the falloff with distance; the time argument samples the source when its signal had to leave to reach the observer now.

For a source much smaller than its characteristic gravitational wavelength, observed far away at distance $D$, approximate the denominator by $D$ and the leading source time by $t-D/c$. Then

The spatial stresses look inconvenient. Conservation converts them into a more intuitive changing mass distribution. Define the leading mass quadrupole moment before trace removal,

Here $\rho=\epsilon/c^2$ at leading nonrelativistic order. Spatial index positions in these Euclidean Cartesian formulas are interchangeable.

Use $\partial_tT^{00}=-c\partial_kT^{k0}$ and integrate by parts. Surface terms vanish for the localized source:

Differentiate again, use $\partial_tT_i{}^0=-c\partial_kT_i{}^k$, and integrate by parts again:

The factor of two comes from the two positions in the symmetric product $x_i x_j$. This manipulation is doing physical work: conservation relates momentum transport to changes in the shape of the mass distribution.

Define its trace-free part

The leading radiative field is

The unit vector $\mathbf n$ points toward the observer. The TT projector is

acting here on symmetric tensors. The first operation projects both indices into the observer’s transverse plane. The second removes the trace within that two-dimensional plane, hence $1/2$ rather than the $1/3$ used to remove a three-dimensional trace. That difference is a useful check that the geometry remains attached to the algebra.

18.6 Which source motions radiate#

The mass monopole is the total mass at the accuracy of this slow-motion calculation, $M=\int\rho\,d^3x$. The mass dipole is the vector $D_i=\int\rho x_i\,d^3x=M x_{{\rm CM},i}$, locating the center of mass. The quadrupole uses two position factors, as in $I_{ij}$ above.

Conservation gives $\dot M=0$, $\dot D_i=P_i$, and $\dot P_i=0$ for an isolated leading-order source, where $P_i$ is total momentum. Its dipole therefore has no second time derivative. The analogous first moment of the mass current describes total angular momentum, also conserved at this order. These lower moments cannot supply the varying radiative field. The quadrupole is the first available mass moment.

Two masses orbiting each other continually change their quadrupole even if their total mass and center of mass remain constant. A perfectly spherical body expanding and contracting has no trace-free mass quadrupole. Exact spherical symmetry also forbids tensor gravitational radiation beyond this approximation, consistent with Birkhoff’s theorem in the vacuum exterior.

The tempting rule “anything accelerating emits gravitational waves” is therefore too crude. Radiation depends on the collective multipole structure and conservation laws, not a tally of individual accelerations. Other theories with additional radiative fields can have different multipole channels.

18.7 Wave energy lives one order beyond the linear equation#

The field equation is nonlinear. Expand it schematically as

The linear term describes propagation. Quadratic terms describe how the waves themselves influence the slowly varying background. When the wavelength is short compared with the background curvature scale, one can average over several wavelengths while remaining local relative to that background. This produces an effective wave stress-energy tensor.

The scale separation is essential. This is not an exact, generally covariant pointwise gravitational-energy tensor for arbitrary geometries. Isaacson’s high-frequency treatment explicitly constructs the appropriate averaged effective description. See Isaacson, Gravitational Radiation in the Limit of High Frequency. II.

For a nearly planar wave in a locally flat wave zone, the average flux is

Angle brackets denote the averaging. The second equality uses the fact that the polarization matrix contributes twice each squared amplitude. Flux has units of power per area; $c^3/G_N$ times an inverse time squared has exactly those units.

The leading total radiated power from a slow isolated source is

The flux contains the square of a first time derivative of strain, whereas strain contains a second derivative of $Q$. That explains the third derivative in the power. The factor $1/5$ follows by integrating the transverse projection over all viewing directions, as we now calculate.

Write $A_{ij}=\dddot Q_{ij}$ at a fixed retarded time. Substituting the quadrupole strain into the flux and integrating over a large sphere gives

The distance cancels: wave amplitude falls as $1/D$, flux as $1/D^2$, and sphere area grows as $D^2$. The remaining calculation asks what fraction of a trace-free tensor survives transverse projection, averaged over viewing directions.

Define $S=A_{ij}A^{ij}$, $B_i=A_{ij}n^j$, and $s=A_{ij}n^i n^j$. Expanding the projector gives

For a uniform average over directions, isotropy requires

Why these forms? There is no preferred direction, so only Kronecker deltas can appear. Symmetry fixes the combinations. Contracting indices and using $n_i n^i=1$ fixes the denominators. Consequently $\langle B_iB^i\rangle_\Omega=S/3$ and, since $A$ is trace-free, $\langle s^2\rangle_\Omega=2S/15$. Therefore

Multiplying by the sphere’s solid angle $4\pi$ produces $8\pi S/5$. That cancels the flux prefactor’s $8\pi$ and leaves precisely $G_N/(5c^5)$. The time average and the angular average serve different purposes; their labels keep those operations distinct.

The $c^{-5}$ suppression makes ordinary laboratory gravitational radiation extremely weak. Large masses, rapid asymmetric motion, and compact configurations help overcome that suppression.

For compact bodies, the leading quadrupole law can describe their slow orbital dynamics even when gravity inside each body is strong; its derivation must then be embedded in a consistent approximation for the effective orbital source. It is not a demand that each black hole itself be a weak-field object.

For a circular binary with total mass $M$, reduced mass $\mu=m_1m_2/M$, and separation $r$, the leading result is

To see the binary coefficient, put the relative position at $\mathbf r=r(\cos\Omega t,\sin\Omega t,0)$ in its center-of-mass frame. The center-of-mass condition places the two bodies at $\mathbf r_1=(m_2/M)\mathbf r$ and $\mathbf r_2=-(m_1/M)\mathbf r$. Substituting these in $I_{ij}=m_1r_{1i}r_{1j}+m_2r_{2i}r_{2j}$ gives $I_{ij}=\mu r_i r_j$. The changing quadrupole components are

Three time derivatives remove the constant terms and give amplitudes $4\mu r^2\Omega^3$. Contracting the tensor counts both $xy$ and $yx$ and yields

Use $\Omega^2=G_NM/r^3$ in the quadrupole power law to obtain the displayed binary luminosity. This leading calculation treats the orbit as approximately circular and nearly unchanged over one cycle; the slow inspiral is then included through energy balance.

Its Newtonian binding energy is $E=-G_N\mu M/(2r)$. Because $dE/dt=-P_{\rm GW}<0$, the energy becomes more negative and $r$ decreases. Kepler’s relation $\Omega^2=G_NM/r^3$ then makes the orbital frequency increase. Its speed increases while its total energy decreases: the drop in gravitational potential energy is greater than the gain in kinetic energy.

The quadrupole components repeat at twice the orbital frequency, so the dominant wave frequency is $f=\Omega/\pi$. Kepler’s relation gives $r=(G_NM)^{1/3}(\pi f)^{-2/3}$. Define $\mathcal M=\mu^{3/5}M^{2/5}$, so $\mathcal M^{5/3}=\mu M^{2/3}$. Substitution into the energy and power gives

Differentiate $E$: $dE/df=-G_N^{2/3}\mathcal M^{5/3}\pi^{2/3}f^{-1/3}/3$. Energy balance, $(dE/df)\dot f=-P$, then yields

The combination $\mathcal M$ is the chirp mass. Its name is operational: the measured rate at which the signal’s pitch rises strongly constrains it. This is a leading inspiral formula, requiring slow enough orbital motion for the approximation. Near merger, higher-order analytic methods and numerical solutions of Einstein’s equation become necessary.

18.8 Comparing the prediction with GW150914#

On September 14, 2015, LIGO detected GW150914. The discovery report described a signal rising from approximately $35$ to $250$ Hz with peak strain about $10^{-21}$. Its inferred source was a merging binary black hole; the initial analysis estimated that roughly three solar masses of energy were radiated. These are findings of the original analysis, with model-dependent parameter estimates and uncertainties, rather than exact source properties. LIGO Scientific Collaboration and Virgo Collaboration, Observation of Gravitational Waves from a Binary Black Hole Merger.

The measured signal includes the increase in frequency described by the inspiral calculation. The final merger and settling require stronger-field predictions than the leading quadrupole formula. Comparing the complete predicted waveform with detector data tests this progression through different regimes.