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Derivation

Quadrupole Formula for Radiated Power

Statement

For a slowly moving, weakly self-gravitating source, the total power carried away by gravitational waves equals the time-averaged square of the third time derivative of the source's reduced (traceless) mass quadrupole moment \(Q_{ij}=\int\rho\left(x_i x_j-\tfrac{1}{3}\delta_{ij}r^2\right)d^3x\), namely \(\;P=\dfrac{G}{5c^5}\left\langle \dddot{Q}_{ij}\,\dddot{Q}_{ij}\right\rangle\).

Why it matters

This is the master formula of gravitational-wave astrophysics. It converts a purely mechanical property of a source — how its mass is distributed and how fast that distribution changes — into a radiated luminosity, without ever solving the full nonlinear field equations. It is what tells us that a spinning steel bar radiates nothing measurable while a coalescing neutron-star binary can briefly outshine the entire electromagnetic universe.

Historically it delivered the first quantitative confirmation of gravitational radiation: the observed orbital decay of the Hulse–Taylor binary pulsar matches this formula to better than a percent, and it sets the leading-order template that LIGO/Virgo waveforms are built upon.

Assumptions
Weak field (linearized gravity).If dropped, \(g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}\) with \(|h|\ll1\) fails; the field equations become nonlinear, waves gravitate and scatter off curvature (tails), and one must use numerical relativity as near a merger.
Slow motion / long wavelength.The multipole expansion parameter is \(\varepsilon\sim \omega d/c=v/c\), with \(d\) the source size. If \(\varepsilon\not\ll1\), retardation across the source and higher multipoles (mass octupole, current quadrupole) and post-Newtonian corrections all enter at the same order and cannot be dropped.
Far (wave) zone observation.Flux is evaluated at \(r\gg\lambda\gg d\). Closer in, near-zone \(1/r^2,\,1/r^3\) pieces of the field are not pure radiation and contaminate any naive energy integral.
Flat-space conservation \(\partial_\mu T^{\mu\nu}=0\) holds for the matter.The tensor virial identity that produces \(\ddot M_{ij}\) uses special-relativistic conservation. For a source held together by its own strong gravity this is only approximately true; the honest justification requires the post-Newtonian effective stress tensor (the "quadrupole-formula controversy", resolved in favour of the formula).
Energy defined by the Isaacson average.Gravitational energy has no gauge-invariant local density; only the wave-zone stress tensor \(t^{\rm GW}_{\mu\nu}\), averaged over several wavelengths (the angle brackets), is meaningful. Without averaging, "energy flux" is gauge.
Derivation
1
\[ \Box\,\bar h_{\mu\nu}=-\frac{16\pi G}{c^4}\,T_{\mu\nu},\qquad \bar h_{\mu\nu}\equiv h_{\mu\nu}-\tfrac12\eta_{\mu\nu}h,\quad \partial^\mu \bar h_{\mu\nu}=0. \]
Linearized Einstein equations in the trace-reversed variable and Lorenz (harmonic) gauge — imported from linearized-gravity-and-wave-equation. A
2
\[ \bar h_{\mu\nu}(t,\vec x)=\frac{4G}{c^4}\int \frac{T_{\mu\nu}\!\left(t-\tfrac{|\vec x-\vec x'|}{c},\,\vec x'\right)}{|\vec x-\vec x'|}\,d^3x'. \]
Retarded Green's function of \(\Box\); the source is isolated so the advanced/homogeneous parts vanish. A
3
\[ \bar h_{ij}(t,\vec x)\simeq\frac{4G}{c^4 r}\int T_{ij}\!\left(t-\tfrac{r}{c},\,\vec x'\right)d^3x',\qquad r\equiv|\vec x|. \]
Far zone \(r\gg d\): set \(|\vec x-\vec x'|\to r\) in the denominator, and (slow motion) evaluate the retarded time at the source centre \(t-r/c\), dropping \(O(\vec x'\!\cdot\hat n/c)\) retardation corrections inside the integral. Only the spatial components \(ij\) are needed; gauge fixes the rest. B
4
\[ \int T^{ij}\,d^3x=\frac12\frac{d^2}{dt^2}\int T^{00}x^i x^j\,d^3x=\frac12\,\ddot M^{ij},\qquad M^{ij}\equiv\frac{1}{c^2}\int T^{00}x^i x^j\,d^3x. \]
Tensor virial identity. From \(\partial_\mu T^{\mu\nu}=0\): \(\partial_0^2 T^{00}=\partial_i\partial_j T^{ij}\); multiply by \(x^k x^l\), integrate over all space, and discard surface terms (compact source). At leading order \(T^{00}=\rho c^2\), so \(M^{ij}=\int\rho\,x^i x^j\,d^3x\) is the Newtonian second mass moment. C
5
\[ \bar h_{ij}(t,\vec x)=\frac{2G}{c^4 r}\,\ddot M_{ij}\!\left(t-\tfrac{r}{c}\right). \]
Insert step 4 into step 3. This is the quadrupole formula for the metric perturbation. A
6
\[ h^{\rm TT}_{ij}=\Lambda_{ij,kl}(\hat n)\,\bar h_{kl}=\frac{2G}{c^4 r}\,\Lambda_{ij,kl}\,\ddot Q_{kl},\qquad \Lambda_{ij,kl}=P_{ik}P_{jl}-\tfrac12 P_{ij}P_{kl},\ \ P_{ij}=\delta_{ij}-n_i n_j. \]
Project onto the physical transverse–traceless part along the propagation direction \(\hat n=\vec x/r\). Because \(\Lambda\) is traceless, its trace part kills the \(\delta_{ij}r^2\) piece, so \(M_{kl}\) may be replaced by the reduced quadrupole \(Q_{kl}=M_{kl}-\tfrac13\delta_{kl}M_{mm}\) for free. B
7
\[ \frac{dE}{dt\,dA}=c\,t^{\rm GW}_{0r}=\frac{c^3}{32\pi G}\left\langle \dot h^{\rm TT}_{ij}\,\dot h^{\rm TT}_{ij}\right\rangle. \]
Radial energy flux from the Isaacson wave-zone stress–energy tensor \(t^{\rm GW}_{\mu\nu}=\frac{c^4}{32\pi G}\langle\partial_\mu h^{\rm TT}_{ij}\partial_\nu h^{\rm TT}_{ij}\rangle\) — imported from stress-energy-tensor-from-noether-currents. B
8
\[ P=\int r^2\,d\Omega\,\frac{dE}{dt\,dA}=\frac{G}{8\pi c^5}\left\langle\dddot Q_{kl}\,\dddot Q_{mn}\right\rangle\int d\Omega\;\Lambda_{kl,mn}. \]
Insert step 6 into step 7 (each \(\dot h\) brings one extra time derivative, so \(\dddot Q\)), cancel \(r^2\), and use the projector idempotency \(\Lambda_{ij,kl}\Lambda_{ij,mn}=\Lambda_{kl,mn}\). A
9
\[ \int\frac{d\Omega}{4\pi}\,\Lambda_{kl,mn}=\frac{11}{30}\delta_{km}\delta_{ln}+\frac{1}{30}\delta_{kn}\delta_{lm}-\frac{2}{15}\delta_{kl}\delta_{mn}. \]
Angular average, using \(\int\frac{d\Omega}{4\pi}n_i n_j=\tfrac13\delta_{ij}\) and \(\int\frac{d\Omega}{4\pi}n_i n_j n_k n_l=\tfrac1{15}(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk})\). C
10
\[ \int\frac{d\Omega}{4\pi}\,\Lambda_{kl,mn}\,\dddot Q_{kl}\dddot Q_{mn}=\left(\tfrac{11}{30}+\tfrac{1}{30}\right)\dddot Q_{ij}\dddot Q_{ij}=\frac{2}{5}\,\dddot Q_{ij}\dddot Q_{ij}. \]
Contract with the symmetric traceless tensor \(\dddot Q\): the two double-Kronecker terms each give \(\dddot Q_{ij}\dddot Q_{ij}\) (symmetry), while \(\delta_{kl}\delta_{mn}\dddot Q_{kl}\dddot Q_{mn}=(\dddot Q_{kk})^2=0\) (tracelessness). B
11
\[ P=\frac{G}{8\pi c^5}\cdot 4\pi\cdot\frac{2}{5}\left\langle\dddot Q_{ij}\dddot Q_{ij}\right\rangle=\frac{G}{5c^5}\left\langle\dddot Q_{ij}\,\dddot Q_{ij}\right\rangle. \]
Assemble step 8 with step 10 (\(\int d\Omega=4\pi\)). A
Result
\[ \boxed{\,P=\frac{G}{5c^5}\left\langle \dddot{Q}_{ij}\,\dddot{Q}_{ij}\right\rangle\,},\qquad Q_{ij}=\int\rho\left(x_i x_j-\tfrac13\delta_{ij}r^2\right)d^3x. \]

Reading. Gravitational luminosity is set by the third time derivative of the traceless mass quadrupole, squared and summed over the two free indices, then averaged over several wave periods (the brackets). There is no monopole term (total mass is conserved) and no dipole term (its second derivative is the conserved total momentum's rate of change, which vanishes) — the quadrupole is the leading radiating multipole, and the ferocious \(c^{-5}\) is why gravity radiates so feebly.

Units check. \([Q]=\mathrm{kg\,m^2}\Rightarrow[\dddot Q]=\mathrm{kg\,m^2\,s^{-3}}\), so \([\dddot Q^2]=\mathrm{kg^2\,m^4\,s^{-6}}\). With \([G/c^5]=\dfrac{\mathrm{m^3\,kg^{-1}\,s^{-2}}}{\mathrm{m^5\,s^{-5}}}=\mathrm{kg^{-1}\,m^{-2}\,s^{3}}\), the product is \(\mathrm{kg\,m^2\,s^{-3}}=\mathrm{J\,s^{-1}}=\mathrm{W}\). ✓

Limiting cases
  • Exact spherical symmetry: \(Q_{ij}\equiv0\) at all times, so \(P=0\) — the gravitational analogue of Birkhoff's theorem; even a violently pulsating spherical star radiates nothing.
  • Axisymmetric body spinning about its symmetry axis: the quadrupole is constant in time, \(\dddot Q=0\), so \(P=0\); only a triaxial (or wobbling) rotor radiates.
  • Rigid rotation at angular frequency \(\omega\): quadrupole components oscillate at \(2\omega\); hence \(P\propto\omega^6\) and the wave frequency is twice the mechanical one.
  • Newtonian point-mass binary, circular orbit: \(P=\dfrac{32}{5}\dfrac{G}{c^5}\mu^2 a^4\omega^6=\dfrac{32}{5}\dfrac{G^4}{c^5}\dfrac{m_1^2 m_2^2 (m_1+m_2)}{a^5}\).
  • \(v/c\to0\): the formula is the leading (Newtonian, "2.5 PN in the flux count") term of the systematic post-Newtonian multipole series.
Breaks when
  • Strong field / near merger. When the source scale approaches its Schwarzschild radius (\(Gm/c^2 a\sim1\)), linearization fails: nonlinear tail and memory terms, higher PN orders, and horizon dynamics all contribute at leading order — only numerical relativity is reliable.
  • Relativistic internal motion. If \(v/c\) is not small, the slow-motion truncation in step 3 is illegitimate: retardation across the source and the mass octupole, current quadrupole, and beyond enter at the same order as the mass quadrupole.
  • Strongly self-gravitating matter. The virial identity in step 4 used flat-space \(\partial_\mu T^{\mu\nu}=0\); for a body bound by its own strong gravity this must be replaced by the (approximately conserved) effective post-Newtonian stress tensor, or the naive derivation is invalid.
  • Static or steady-state configuration. If nothing time-varies (or only the monopole/dipole do), \(\dddot Q=0\) and the formula returns zero — it says nothing about, e.g., a boosted static mass.
Failure modes
  • Plugging the full second moment \(M_{ij}\) into the boxed formula. \(\frac{G}{5c^5}\dddot M_{ij}\dddot M_{ij}\) contains spurious trace terms and is numerically wrong; you must use the traceless \(Q_{ij}\). (The replacement was only free inside the \(\Lambda\)-projection of step 6.)
  • Reporting the wave frequency as \(\omega\) instead of \(2\omega\). Quadrupole components of a rotor go as \(\cos 2\omega t\); forgetting the doubling wrecks any spectral estimate.
  • Omitting the time average. The angle brackets are essential; an instantaneous "\(\dddot Q^2\)" is gauge-dependent and can even be negative-looking after projection.
  • Using \(\ddot Q\) (two dots) rather than \(\dddot Q\). Two derivatives give the strain amplitude \(h\); the power needs three.
  • Losing the factor \(\tfrac12\) in \(\int T^{ij}=\tfrac12\ddot M^{ij}\). Propagates to a factor-4 error in \(P\).
  • Expecting dipole radiation by analogy with electromagnetism. Mass-energy has one sign and momentum is conserved, so both mass monopole and dipole are non-radiating; there is no gravitational analogue of \(\ddot{\vec d}\)-radiation.
Discussion

The absence of the lower multipoles is the physically deepest feature. In electromagnetism the leading radiator is the electric dipole \(\vec d=\sum q\vec r\), because charge can be positive or negative and separated. Gravitationally, the "charge" is mass-energy, which is single-signed and, more decisively, conserved: the mass monopole is the constant total mass (no radiation), and the mass dipole's first derivative is the total momentum, so its second derivative — what would source dipole waves — is the rate of change of a conserved quantity and vanishes. Radiation therefore begins at the quadrupole, one full order higher than electromagnetism, which alone accounts for much of gravity's radiative feebleness.

The remaining feebleness lives in the prefactor \(G/c^5\). Numerically \(c^5/G\approx3.6\times10^{52}\ \mathrm{W}\) is the "Planck luminosity", the natural unit of gravitational power; every real source radiates a dimensionless fraction of it equal to \(\tfrac15\langle(\dddot Q/{\rm this\ scale})^2\rangle\). Only astrophysical masses in relativistic motion make that fraction appreciable — which is exactly why the first direct detections were black-hole mergers, where \(Q\sim M a^2\) is enormous and \(\omega\) approaches \(c^3/GM\).

The formula is also the engine of inspiral. Radiated energy is drawn from the orbit, \(dE_{\rm orb}/dt=-P\); for a circular binary \(E_{\rm orb}=-Gm_1m_2/2a\), so the separation shrinks, \(\omega\) rises, and \(P\propto\omega^6\) accelerates — the runaway "chirp" whose rising tone LIGO records. The same balance, applied to the Hulse–Taylor pulsar, predicted an orbital-period decay confirmed to sub-percent accuracy, earning the 1993 Nobel Prize and giving gravitational waves their first hard evidence.

Rigorously, the boxed result is only the leading term of the post-Newtonian, multipolar expansion of the field equations (Thorne's multipole formalism; the Blanchet–Damour–Iyer matched-asymptotic and MPM–PN schemes). The 1970s "quadrupole-formula controversy" questioned whether flat-space conservation — and hence step 4 — could legitimately be used for a self-gravitating source; careful PN analysis showed the effective stress tensor restores conservation to the required order, vindicating the formula. Modern waveform models push this series to 4PN and beyond, adding current-multipole, tail (wave–curvature scattering), spin, and nonlinear memory contributions that the leading formula cannot see.

Common misconceptions. "Any accelerating mass radiates gravitationally" — false: a spherically or axially symmetric acceleration has \(\dddot Q=0\). "The wave carries away the source's rotational energy at the rotation frequency" — the waves come out at \(2\omega\). "A single mass moving inertially radiates" — no, uniform motion is a boost of a static field, \(\dddot Q=0\).

Worked examples

Example 1 — a spinning laboratory dumbbell. Two equal point masses \(m\), each at radius \(\ell\), on a rigid massless rod rotating in the \(xy\)-plane at angular frequency \(\omega\). How much gravitational power does it emit?

1
\[ M_{xx}=2m\ell^2\cos^2\omega t=m\ell^2(1+\cos2\omega t),\quad M_{yy}=m\ell^2(1-\cos2\omega t),\quad M_{xy}=m\ell^2\sin2\omega t. \]
Second mass moment of the two masses at \(\pm\ell(\cos\omega t,\sin\omega t,0)\); both contribute equally. A
2
\[ \dddot Q_{xx}=8m\ell^2\omega^3\sin2\omega t,\quad \dddot Q_{yy}=-8m\ell^2\omega^3\sin2\omega t,\quad \dddot Q_{xy}=-8m\ell^2\omega^3\cos2\omega t. \]
The constant trace subtracts out; three derivatives of the \(2\omega\) oscillation bring \((2\omega)^3=8\omega^3\). B
3
\[ \left\langle\dddot Q_{ij}\dddot Q_{ij}\right\rangle=\left\langle\dddot Q_{xx}^2+\dddot Q_{yy}^2+2\dddot Q_{xy}^2\right\rangle=(8m\ell^2\omega^3)^2(1+1+2)\!\cdot\!\tfrac{1}{1}=128\,m^2\ell^4\omega^6. \]
Here \(\sin^2+\sin^2+2\cos^2=2\); the result is already time-independent, so the average is trivial. B
4
\[ P=\frac{G}{5c^5}\,(128\,m^2\ell^4\omega^6)=\frac{128}{5}\,\frac{G}{c^5}\,m^2\ell^4\omega^6. \]
Insert the boxed formula. Symbolic answer before numbers. A
5
\[ m=1000\,\mathrm{kg},\ \ell=1\,\mathrm{m},\ f=100\,\mathrm{Hz}\Rightarrow\omega=2\pi f=628\ \mathrm{s^{-1}},\ \omega^6=6.1\times10^{16}\,\mathrm{s^{-6}}. \]
Choose plausible (if optimistic) laboratory numbers. A
\[ P=\frac{128}{5}\,(2.76\times10^{-53})\,(10^{6})(1)(6.1\times10^{16})\ \mathrm{W}\approx 4\times10^{-29}\ \mathrm{W}. \]

Reading. A tonne-scale bar spun to its bursting point radiates a few \(10^{-29}\ \mathrm W\) — about one graviton every \(10^{5}\) years. This is why gravitational waves cannot be made in the lab; only astronomical sources qualify.

Units check. \(\mathrm{(kg^{-1}m^{-2}s^3)(kg^2)(m^4)(s^{-6})=kg\,m^2\,s^{-3}=W}\). ✓

Example 2 — gravitational luminosity of the Hulse–Taylor binary pulsar (PSR B1913+16). Two neutron stars, \(m_1=1.441\,M_\odot\), \(m_2=1.387\,M_\odot\), orbital period \(P_b=7.75\ \mathrm{hr}\). Estimate the emitted power.

1
\[ M=m_1+m_2=2.828\,M_\odot=5.62\times10^{30}\,\mathrm{kg},\qquad \mu=\frac{m_1 m_2}{M}=0.707\,M_\odot=1.41\times10^{30}\,\mathrm{kg}. \]
Total and reduced mass; \(M_\odot=1.989\times10^{30}\,\mathrm{kg}\). A
2
\[ \omega=\frac{2\pi}{P_b}=2.25\times10^{-4}\,\mathrm{s^{-1}},\qquad a=\left(\frac{GM}{\omega^2}\right)^{1/3}=1.95\times10^{9}\,\mathrm{m}. \]
Circular-orbit approximation with Kepler's third law \(\omega^2=GM/a^3\). A
3
\[ P_{\rm circ}=\frac{32}{5}\,\frac{G}{c^5}\,\mu^2 a^4\omega^6. \]
Circular-binary specialization of the boxed formula (single reduced mass \(\mu\) at separation \(a\); the quadrupole oscillates at \(2\omega\)). Symbols first. B
4
\[ P_{\rm circ}=6.4\,(2.76\times10^{-53})(1.98\times10^{60})(1.44\times10^{37})(1.30\times10^{-22})\ \mathrm{W}\approx 6.6\times10^{23}\ \mathrm{W}. \]
Numbers: \(\mu^2=1.98\times10^{60}\), \(a^4=1.44\times10^{37}\), \(\omega^6=1.30\times10^{-22}\). A
5
\[ f(e)=\frac{1+\tfrac{73}{24}e^2+\tfrac{37}{96}e^4}{(1-e^2)^{7/2}}\Big|_{e=0.617}=11.9,\qquad P=f(e)\,P_{\rm circ}\approx 7.9\times10^{24}\ \mathrm{W}. \]
The real orbit is eccentric; orbit-averaging multiplies the circular result by the Peters enhancement factor \(f(e)\). C
\[ P\approx 7.9\times10^{24}\ \mathrm{W}\quad(\sim 2\times10^{-2}\,L_\odot). \]

Reading. The system radiates gravitationally at a few percent of the Sun's total electromagnetic luminosity — invisibly, yet enough to shrink the orbit measurably. Fed back through \(dE/dt=-P\), this predicts the observed period decay \(\dot P_b\approx-2.4\times10^{-12}\), matching timing data to \(\lesssim1\%\).

Units check. Same as Example 1; \(G\mu^2 a^4\omega^6/c^5\) has units of watts. ✓

Problems
  1. (A) Frequency doubling. A rigid body rotates about the \(z\)-axis at angular frequency \(\omega\). Show that its gravitational radiation emerges at \(2\omega\), and identify the one configuration for which it vanishes entirely.
    Solution Under rotation by angle \(\phi=\omega t\), the in-plane quadrupole components transform as \(Q_{xx},Q_{yy},Q_{xy}\propto\cos2\phi,\ \sin2\phi\) (a rank-2 tensor picks up \(e^{\pm2i\phi}\) under a rotation, i.e. \(2\phi=2\omega t\)). Three time derivatives preserve this frequency, so \(P\) oscillates at \(2\times2\omega\) but the wave field \(h\propto\ddot Q\) itself carries frequency \(2\omega\). It vanishes iff the body is axisymmetric about \(z\) (\(Q_{xx}=Q_{yy}\), \(Q_{xy}=0\), all constant), giving \(\dddot Q=0\) and \(P=0\).
  2. (B) Bigger dumbbell. Two \(500\,\mathrm{kg}\) masses are separated by \(2\,\mathrm{m}\) (so \(\ell=1\,\mathrm{m}\)) and spun at \(f=60\,\mathrm{Hz}\). Using \(P=\tfrac{128}{5}\tfrac{G}{c^5}m^2\ell^4\omega^6\), find the radiated power.
    Solution \(\omega=2\pi(60)=377\,\mathrm{s^{-1}}\), \(\omega^6=(377)^6=2.87\times10^{15}\,\mathrm{s^{-6}}\). \(m^2=2.5\times10^{5}\), \(\ell^4=1\). With \(G/c^5=2.76\times10^{-53}\) and \(128/5=25.6\): \(P=25.6\times2.76\times10^{-53}\times2.5\times10^{5}\times2.87\times10^{15}\ \mathrm{W}\). Multiply: \(25.6\times2.76=70.7\); \(\times2.5\times10^{5}=1.77\times10^{7}\); \(\times2.87\times10^{15}=5.07\times10^{22}\); \(\times10^{-53}\Rightarrow P\approx5\times10^{-31}\ \mathrm{W}\). Utterly undetectable — the point of the exercise.
  3. (B) Orbital decay law. For a circular Newtonian binary with \(E_{\rm orb}=-\tfrac{Gm_1m_2}{2a}\), impose \(dE_{\rm orb}/dt=-P\) and derive \(\dot a\), then the merger time \(\tau\) from initial separation \(a_0\).
    Solution \(\dfrac{dE}{dt}=\dfrac{Gm_1m_2}{2a^2}\dot a=-P=-\dfrac{32}{5}\dfrac{G^4}{c^5}\dfrac{m_1^2m_2^2 M}{a^5}\Rightarrow \boxed{\dot a=-\dfrac{64}{5}\dfrac{G^3 m_1 m_2 M}{c^5 a^3}}\). Then \(a^3\,da=-\tfrac{64}{5}\tfrac{G^3 m_1m_2M}{c^5}\,dt\), i.e. \(a^4(t)=a_0^4-\tfrac{256}{5}\tfrac{G^3 m_1m_2M}{c^5}t\). Setting \(a=0\): \(\boxed{\tau=\dfrac{5}{256}\dfrac{c^5 a_0^4}{G^3 m_1 m_2 M}}\).
  4. (C) Hulse–Taylor period decay. Using \(\dot P_b/P_b=\tfrac32\,\dot a/a\) and the result of Problem 3, compute the circular-orbit \(\dot P_b\) for PSR B1913+16 (data from Example 2), then apply the eccentricity factor \(f(0.617)=11.9\) and compare with the observed \(-2.40\times10^{-12}\).
    Solution From Problem 3, \(\dfrac{\dot a}{a}=-\dfrac{64}{5}\dfrac{G^3 m_1 m_2 M}{c^5 a^4}\). With \(m_1m_2=7.91\times10^{60}\), \(M=5.62\times10^{30}\), \(a^4=1.44\times10^{37}\), \(G^3=2.97\times10^{-31}\), \(c^5=2.42\times10^{42}\): numerator \(G^3 m_1m_2M=1.32\times10^{61}\); denominator \(c^5a^4=3.49\times10^{79}\); ratio \(3.78\times10^{-19}\); \(\times(64/5)=4.84\times10^{-18}\,\mathrm{s^{-1}}\). So \(\dot P_b^{\rm circ}=\tfrac32(-4.84\times10^{-18})(27900\,\mathrm{s})=-2.0\times10^{-13}\). Multiply by \(f(e)=11.9\): \(\dot P_b\approx-2.4\times10^{-12}\), in agreement with observation — the classic confirmation of the quadrupole formula.
  5. (B) A symmetric non-radiator. A uniform-density oblate spheroid rotates about its (short) symmetry axis at \(\omega\). Show it emits no gravitational waves, and state what physical feature a neutron star must have to radiate continuously.
    Solution Choose body axes along the principal axes; the inertia/quadrupole tensor is diagonal with \(Q_{xx}=Q_{yy}\ne Q_{zz}\). Rotation about \(z\) leaves this tensor invariant (it is a rotation about a symmetry axis), so in the lab frame \(Q_{ij}\) is time-independent, \(\dddot Q=0\), and \(P=0\). Continuous emission requires a time-varying quadrupole about the rotation axis — a non-axisymmetric "mountain" or ellipticity \(\varepsilon=(I_{xx}-I_{yy})/I_{zz}\ne0\), or free precession/wobble, so that \(Q_{xy}\) etc. oscillate at \(2\omega\).