Adiabatic Tidal Dynamics

Adiabatic tidal dynamics is obtained by eliminating a body’s induced multipoles at essentially zero forcing frequency and inserting the resulting local interaction into the orbital action. The result is conservative: it shifts the binding energy, the circular frequency–radius relation, and periastron advance. It also changes the radiative multipoles, so a waveform phase cannot be computed by altering the conservative energy alone. This page derives the leading quadrupolar effects in a fixed convention and marks where the static approximation must give way to frequency-dependent response.

Helpful background. The Einstein Equation supplies the curvature dynamics; Weak Metric Perturbations and Gauge Freedom supplies the gauge logic behind invariant energy–frequency relations; Presymplectic Potentials and Currents supplies action variation; Relativistic Matter and Compact Objects supplies the stellar response coefficients; Relativistic Two-Body Dynamics supplies orbital Hamiltonian methods; and Black-Hole Perturbation Theory supplies black-hole response boundary conditions.

Static response in the binary action

Consider two nonspinning bodies with masses $m_1,m_2$, total mass $M=m_1+m_2$, reduced mass $\mu=m_1m_2/M$, and separation $r$. We retain only electric quadrupoles and use the convention established on the tidal moments and multipoles page:

$$L_{Q_A} =-\frac12Q_A^{ij}\mathcal E^A_{ij} -\frac{1}{4\lambda_A}Q_A^{ij}Q^A_{ij}, \qquad Q_A^{ij}=-\lambda_A\mathcal E_A^{ij}.$$

Here $\lambda_A$ is real and evaluated at zero frequency. Eliminating $Q_A^{ij}$ gives

$$L_{\rm tidal} =\sum_{A=1}^2\frac{\lambda_A}{4} \mathcal E^A_{ij}\mathcal E_A^{ij}.$$

The companion’s Newtonian tidal tensor at body 1 is

$$\mathcal E^1_{ij} =\frac{Gm_2}{r^3} \left(\delta_{ij}-3n_i n_j\right), \qquad \mathcal E^1_{ij}\mathcal E_1^{ij} =6\left(\frac{Gm_2}{r^3}\right)^2,$$

and body labels are exchanged for body 2. The relative Hamiltonian through leading static-tidal order is therefore

$$H =\frac{p_r^2}{2\mu} +\frac{L^2}{2\mu r^2} -\frac{GM\mu}{r} -\frac{A}{r^6}, \qquad A=\frac32G^2 \left(\lambda_1m_2^2+\lambda_2m_1^2\right).$$

The deformation lowers the potential energy at fixed separation. Flanagan and Hinderer derive the same induced-quadrupole action and its orbital equations in Flanagan and Hinderer 2008, Eqs. (3)–(8). Numerical factors elsewhere may look different because some authors define $\mathcal E_{ij}$ or the quadrupole coupling with the opposite sign; the invariant comparison is the final energy or phasing after translating all conventions.

This local action assumes more than $R_A/r\ll1$. If $\omega_{\rm tide}$ denotes a forcing frequency in the body’s frame and $\omega_{\alpha}$ its lowest appreciably coupled mode frequency, the static elimination requires

$$\epsilon_{\rm ad} =\max_\alpha\left| \frac{\omega_{\rm tide}}{\omega_\alpha} \right|\ll1, \qquad \epsilon_{\rm lin} =\frac{|Q_A|}{m_AR_A^2}\ll1.$$

For a circular, nonrotating quadrupolar tide, the dominant body-frame forcing has $|\omega_{\rm tide}|=2\Omega$. Static response is not synonymous with absence of orbital evolution: it means the internal configuration tracks the slowly changing field with corrections ordered by $\epsilon_{\rm ad}^2$ and by the radiation-reaction rate. A real constant $\lambda_A$ contains neither a resonant pole nor material damping.

Circular energy and a local orbital check

At fixed angular momentum, a circular orbit obeys $\partial H/\partial r=0$. The Hamiltonian above gives

$$\frac{L^2}{\mu} =GM\mu r+\frac{6A}{r^4},$$

and Hamilton’s equation $\Omega=\partial H/\partial L=L/(\mu r^2)$ gives

$$\Omega^2 =\frac{GM}{r^3} +\frac{6A}{\mu r^8}.$$

The tide therefore raises the circular frequency at fixed coordinate separation in this Newtonian gauge. Substitution of the circularity condition into $H$ gives

$$E_{\rm circ}(r) =-\frac{GM\mu}{2r} +\frac{2A}{r^6}.$$

The positive correction here does not contradict the negative tidal potential. Maintaining a circular orbit at the same $r$ requires additional kinetic energy. To compare at fixed invariant frequency, define $x=(GM\Omega)^{2/3}$ and the untidally perturbed radius $r_0=GM/x$. Expanding the frequency equation to first order in $\epsilon_A=A/(GM\mu r_0^5)$ gives

$$r=r_0\left(1+2\epsilon_A\right)+O(\epsilon_A^2r_0), \qquad E_{\rm circ}(x) =-\frac{\mu x}{2}+\frac{3A x^6}{(GM)^6} +O(\epsilon_A^2\mu x).$$

One third of the tidal energy correction at fixed frequency comes from shifting the Newtonian radius. This is why inserting $r=GM/x$ directly into $E_{\rm circ}(r)$ would miss a contribution. Vines, Flanagan, and Hinderer extend this elimination through first relative PN tidal order in Vines, Flanagan, and Hinderer 2011, § II, Eqs. (2.3)–(2.12).

The same central perturbation produces prograde apsidal advance. Write $p=L^2/(GM\mu^2)$ for the unperturbed semilatus rectum and assume $A/(GM\mu p^5)\ll1$. The Binet equation for $u=1/r$ is

$$u''+u =\frac{GM\mu^2}{L^2} +\frac{6\mu A}{L^2}u^5.$$

Linearizing about a slightly eccentric orbit gives the tidal advance per radial period

$$\Delta\varpi_{\rm tidal} =\frac{30\pi A}{GM\mu p^5} +O(A^2,e^2).$$

This expression is a useful leading-order sign and scaling check, not a relativistic prediction for strong-field periastron advance. In a post-Newtonian or effective-one-body description, coordinate radii and individual Hamiltonian terms change under canonical transformations. One should compare a gauge-invariant relation such as $E(\Omega)$, periastron advance as a function of invariant frequencies, or a scattering observable.

From conservative dynamics to waveform phase

For a slowly evolving quasicircular inspiral, orbital balance reads

$$\frac{dE(\Omega)}{dt} =-\mathcal F_\infty(\Omega) -\mathcal F_{\rm body}(\Omega).$$

In the strictly adiabatic, nondissipative approximation used here, $\mathcal F_{\rm body}=0$. Nevertheless, $\lambda_A$ modifies both $E(\Omega)$ and the gravitational flux $\mathcal F_\infty$, because the induced quadrupoles contribute to the binary’s radiative multipole moments. Keeping the tidal correction in only one side of the balance equation gives an inconsistent phase. The 1PN-accurate calculation of both energy and radiation is displayed in Vines, Flanagan, and Hinderer 2011, §§ II–III, Eqs. (2.11), (3.7), and (3.9)–(3.13).

Keep $G$ explicit and define $v=(\pi GMf)^{1/3}$, where $f=\Omega/\pi$ is the dominant quadrupolar gravitational-wave frequency in the source frame. The dimensionless deformability is

$$\Lambda_A=\frac{G\lambda_A}{(Gm_A)^5}.$$

It reduces to $\lambda_A/m_A^5$ in geometrized units. The mass-weighted combination entering the leading phase is

$$\widetilde\Lambda =\frac{16}{13M^5} \left[ (m_1+12m_2)m_1^4\Lambda_1 +(m_2+12m_1)m_2^4\Lambda_2 \right].$$

For the positive-frequency quadrupole with time dependence $e^{-2i\phi(t)}$, use $\widetilde h(f)=\int h(t)e^{2\pi ift}dt=\mathcal A(f)e^{i\Psi(f)}$. Stationary phase then gives the leading static quadrupolar contribution

$$\Psi(f) =\Psi_{\rm pp}(f) +\frac{3}{128\nu v^5} \left[ -\frac{39}{2}\widetilde\Lambda v^{10} +O(v^{12}) \right], \qquad \nu=\frac{\mu}{M}.$$

Thus electric quadrupolar tides enter at relative 5PN order. This counting says that their leading scaling is $v^{10}$ relative to the Newtonian point-mass phase; it does not say that all point-mass terms at the same formal order are numerically interchangeable with tides. The tidal coefficient carries the large material scale $\Lambda_A\sim C_A^{-5}$. The original low-frequency derivation and its domain are given in Flanagan and Hinderer 2008, Eqs. (8)–(11).

For equal masses and equal equations of state, $\widetilde\Lambda=\Lambda_1=\Lambda_2$. For a black-hole–neutron-star binary in four-dimensional general relativity with the usual stationary black-hole boundary condition, the static black-hole Love coefficient is zero, but the black hole can still absorb flux. Setting its $\Lambda$ to zero therefore removes a conservative static response, not every finite-size effect.

PN, EFT, and EOB representations

The same low-frequency physics can be organized in three common ways. A post-Newtonian Lagrangian expands the tidal interaction and orbital field simultaneously in $v$ and $R/r$. A worldline effective action collects local operators such as $\mathcal E_{\mu\nu}\mathcal E^{\mu\nu}$, with coefficients fixed by matching. An effective-one-body model places equivalent information into deformations of its radial potentials and waveform multipoles. Damour and Nagar give an explicit EOB realization and its tidal coupling constants in Damour and Nagar 2010, §§ II–IV.

These are representations, not different physical tides. A canonical transformation can move a term between kinetic and potential sectors; a resummation can alter the apparent strong-field behavior without changing the known series; and an effective Love number can fold frequency dependence into a convenient model. Agreement with the static expansion must be checked before such a model is called adiabatic. Conversely, agreement at $\omega=0$ does not validate extrapolation through a mode resonance.

The controlled truncation should identify four different omissions: higher tidal multipoles; additional PN corrections within each multipole; frequency dependence, beginning at relative $O[(\omega_{\rm tide}/\omega_\alpha)^2]$ for a regular nondissipative mode; and nonlinear response in $Q/(mR^2)$. For natural static deformabilities, an electric multipole of degree $\ell$ contributes relative orbital-energy scaling $(R/r)^{2\ell+1}$, with mass-ratio and response coefficients still required. This is power counting, not a numerical error bound. Near contact or disruption, the spatial and amplitude expansions may cease to be hierarchical even if more PN terms remain algebraically available.

Exercises

1. Circular-orbit consistency. Starting from the Hamiltonian above, verify both $\Omega^2$ and $E_{\rm circ}(r)$.

Solution

At $p_r=0$,

$$\frac{\partial H}{\partial r} =-\frac{L^2}{\mu r^3} +\frac{GM\mu}{r^2} +\frac{6A}{r^7}=0,$$

which gives $L^2/\mu=GM\mu r+6A/r^4$. Then

$$\Omega^2 =\frac{L^2}{\mu^2r^4} =\frac{GM}{r^3}+\frac{6A}{\mu r^8}.$$

The circular kinetic energy is $L^2/(2\mu r^2)=GM\mu/(2r)+3A/r^6$. Adding $-GM\mu/r-A/r^6$ yields $-GM\mu/(2r)+2A/r^6$.

2. Why the phase needs two tidal inputs. Suppose $E=E_0+\delta E$ and $\mathcal F=\mathcal F_0+\delta\mathcal F$. Expand $\dot\Omega=-\mathcal F/(dE/d\Omega)$ to first order.

Solution

Using $(E_0'+\delta E')^{-1}=E_0'^{-1}(1-\delta E'/E_0')+O(\delta^2)$,

$$\dot\Omega =-\frac{\mathcal F_0}{E_0'} \left[ 1+\frac{\delta\mathcal F}{\mathcal F_0} -\frac{\delta E'}{E_0'} \right] +O(\delta^2).$$

Equivalently, the first-order correction is

$$\delta\dot\Omega =-\frac{\delta\mathcal F}{E_0'} +\frac{\mathcal F_0\delta E'}{(E_0')^2}.$$

Both $\delta E$ and $\delta\mathcal F$ contribute at the same perturbative order. Dropping either produces an incomplete tidal chirp.

References

• Damour, Thibault, and Alessandro Nagar. “Effective One Body Description of Tidal Effects in Inspiralling Compact Binaries.” Physical Review D 81, 084016 (2010). DOI; Open PDF.
• Flanagan, Éanna É., and Tanja Hinderer. “Constraining Neutron-Star Tidal Love Numbers with Gravitational-Wave Detectors.” Physical Review D 77, 021502(R) (2008). DOI; Open PDF.
• Vines, Justin, Éanna É. Flanagan, and Tanja Hinderer. “Post-1-Newtonian Tidal Effects in the Gravitational Waveform from Binary Inspirals.” Physical Review D 83, 084051 (2011). DOI; Open PDF.