Nonlinear tides in close binary systems

Weinberg, N. N., Arras, P., Quataert, E., & Burkart, J. (2012). Nonlinear tides in close binary systems. The Astrophysical Journal, 751(2), 136. https://doi.org/10.1088/0004-637X/751/2/136

Statement of the Problem

A primary star of mass MM and radius RR subject to a tidal acceleration from the secondary of mass MM'.
In a spherical coordinate system (r,θ,ϕ)(r,\theta,\phi) centered on the primary, take the orbit of the secondary to be (D(t),π/2,Φ(t))(D(t),\pi/2,\Phi(t)), where D(t)D(t) is the separation and Φ(t)\Phi(t) is the true anomaly corresponding to a Keplerian orbit with the semimajor axis aa, eccentricity ee, and orbital period Porb=2π[a3/G(M+M)]1/2P_\text{orb}=2\pi[a^3/G(M+M')]^{1/2}.

The tidal acceleration Uε(GM/R2)\boldsymbol{\nabla} U\propto\varepsilon(GM/R^2) with a small dimensionless strength of the tidal acceleration relative to internal gravity

ε=MM(Ra)3\varepsilon=\frac{M'}{M}\left(\frac{R}{a}\right)^{3}

Equations of Motion

x\boldsymbol{x}' is the position of a fluid element in the perturbed star
x\boldsymbol{x} is the position of the same fluid element in the background state
ξ\boldsymbol{\xi} is the Lagrangian displacement vector
x=x+ξ(x,t)\boldsymbol{x}'=\boldsymbol{x}+\boldsymbol{\xi}(\boldsymbol{x},t)

Write the internal forces due to pressure, buoyancy, and perturbed gravity
that are linear in ξ\boldsymbol{\xi} as f1[ξ]\boldsymbol{f}_1[\boldsymbol{\xi}]
and those due to leading-order nonlinear interactions as f2[ξ,ξ]\boldsymbol{f}_2[\boldsymbol{\xi},\boldsymbol{\xi}]

The external forcing terms due to the companion lead to the tidal acceleration

atide=U(ξ)U.\boldsymbol{a}_\text{tide}=-\boldsymbol{\nabla} U-(\boldsymbol{\xi}\cdot\boldsymbol{\nabla})\boldsymbol{\nabla} U.

Gathering terms, the second-order equation of motion including linear forces, three-wave nonlinear interactions and tidal forcing is

ρξ¨=f1[ξ]+f2[ξ,ξ]+ρatide.(4)\rho\ddot{\boldsymbol{\xi}}=\boldsymbol{f}_1[\boldsymbol{\xi}]+\boldsymbol{f}_2[\boldsymbol{\xi},\boldsymbol{\xi}]+\rho\boldsymbol{a}_\text{tide}.\tag{4}

There are two approaches to solve:
(1) expand quantities relative to the star’s unperturbed background state;
(2) expand quantities relative to the star’s linearly perturbed state.

Method 1

Expand using

ξ(x,t)=aqa(t)ξa(x)eiωat.\boldsymbol{\xi}(\boldsymbol{x},t)=\sum_a q_a(t)\boldsymbol{\xi}_a(\boldsymbol{x})e^{-i\omega_a t}.

The eigenmode labeled aa is specified by its frequency ωa\omega_a, eigenfunction ξa(x)\boldsymbol{\xi}_a(\boldsymbol{x}), and total amplitude qaq_a.

A set of coupled oscillator equations for each mode:

q˙a+iωaqa=γaqa+iωaUa(t)+iωabUab(t)qb+iωabcκabcqbqc.(6)\begin{aligned} \dot{q}_a+i\omega_a q_a=&-\gamma_a q_a + i\omega_a U_a(t) \\ &+i\omega_a\sum_b U_{ab}^*(t)q_b^* \\ &+i\omega_a\sum_{bc}\kappa_{abc}^*q_b^*q_c^*. \end{aligned}\tag{6}

The terms on the right-hand side represent linear damping ( γa\gamma_a ), the linear ( UaU_a ) and nonlinear ( UabU_{ab} ) tidal force, and three-wave coupling ( κabc\kappa_{abc} ).

Ua(t)=1E0d3xρξaU=kUa(k)eikΩt\begin{aligned} U_a(t)&=-\frac{1}{E_0}\int\mathrm{d}^3x\,\rho\boldsymbol{\xi}_a^*\cdot\boldsymbol{\nabla}U \\ &=\sum_k U_a^{(k)}e^{-ik\Omega t} \end{aligned}

Uab(t)=1E0d3xρξa(ξb)UU_{ab}(t)=-\frac{1}{E_0}\int\mathrm{d}^3x\,\rho\boldsymbol{\xi}_a\cdot(\boldsymbol{\xi}_b\cdot\boldsymbol{\nabla})\boldsymbol{\nabla}U

κabc=1E0d3xξaf2[ξb,ξc]\kappa_{abc}=\frac{1}{E_0}\int\mathrm{d}^3x\,\boldsymbol{\xi}_a\cdot\boldsymbol{f}_2[\boldsymbol{\xi}_b,\boldsymbol{\xi}_c]

Method 2

qa,linq_{a,\text{lin}} is the linear amplitude
raqaqa,linr_a\equiv q_a-q_{a,\text{lin}}
ξ=ξlin+ξnl\boldsymbol{\xi}=\boldsymbol{\xi}_\text{lin}+\boldsymbol{\xi}_\text{nl}

ρξ¨lin=f1[ξlin]ρU \rho\ddot{\boldsymbol{\xi}}_\text{lin}=\boldsymbol{f}_1[\boldsymbol{\xi}_\text{lin}]-\rho\boldsymbol{\nabla} U

ρξ¨nl=f1[ξnl]+f2[ξlin,ξlin]+2f2[ξlin,ξnl]+f2[ξnl,ξnl]ρ[(ξlin+ξnl)]U.\begin{aligned} \rho\ddot{\xi}_\text{nl}=&\boldsymbol{f}_1[\boldsymbol{\xi}_\text{nl}]+\boldsymbol{f}_2[\boldsymbol{\xi}_\text{lin},\boldsymbol{\xi}_\text{lin}]+2\boldsymbol{f}_2[\boldsymbol{\xi}_\text{lin},\boldsymbol{\xi}_\text{nl}] \\ &+\boldsymbol{f}_2[\boldsymbol{\xi}_\text{nl},\boldsymbol{\xi}_\text{nl}]-\rho[(\boldsymbol{\xi}_\text{lin}+\boldsymbol{\xi}_\text{nl})\cdot\boldsymbol{\nabla}]\boldsymbol{\nabla}U. \end{aligned}

Expand using

ξnl(x,t)=ara(t)ξa(x)eiωat.\boldsymbol{\xi}_\text{nl}(\boldsymbol{x},t)=\sum_a r_a(t)\boldsymbol{\xi}_{a}(\boldsymbol{x})e^{-i\omega_{a} t}.

The equation for each mode’s nonlinear amplitude

r˙a+(iωa+γa)ra=iωa(Va+Ka)+iωab(Uab+2Kab)rb+iωabcκabcrbrc.(20)\begin{aligned} \dot{r}_a+(i\omega_a+\gamma_a)r_a={}&i\omega_a(V_a^*+K_a^*) \\ &+i\omega_a\sum_b(U_{ab}^*+2K_{ab}^*)r_b^* \\ &+i\omega_a\sum_{bc}\kappa_{abc}^*r_b^*r_c^*. \end{aligned}\tag{20}

Va(t)1E0d3xρξa(ξlin)UV_a(t)\equiv-\frac{1}{E_0}\int\mathrm{d}^3x\,\rho\boldsymbol{\xi}_a\cdot(\boldsymbol{\xi}_\text{lin}\cdot\boldsymbol{\nabla})\boldsymbol{\nabla}U

Ka(t)1E0d3xξaf2[ξlin,ξlin]K_a(t)\equiv\frac{1}{E_0}\int\mathrm{d}^3x\,\boldsymbol{\xi}_a\cdot\boldsymbol{f}_2[\boldsymbol{\xi}_\text{lin},\boldsymbol{\xi}_\text{lin}]

Kab(t)1E0d3xξaf2[ξlin,ξb]K_{ab}(t)\equiv\frac{1}{E_0}\int\mathrm{d}^3x\,\boldsymbol{\xi}_a\cdot\boldsymbol{f}_2[\boldsymbol{\xi}_\text{lin},\boldsymbol{\xi}_b]

There are three types of three-wave coupling are
(1) linear-linear coupling (LLC)
(2) linear-nonlinear coupling (LNC)
(3) nonlinear-nonlinear coupling (NNC)

Linear Tide

The linear equation:

q˙a+iωaqa=γaqa+iωaUa(t),\dot{q}_a+i\omega_a q_a=-\gamma_a q_a+i\omega_a U_a(t),

whose steady-state solution is

qa,lin(t)=k=ωaUa(k)ωakΩiγaeikΩt.q_{a,\text{lin}}(t)=\sum_{k=-\infty}^\infty \frac{\omega_a U_a^{(k)}}{\omega_a-k\Omega-i\gamma_a}e^{-ik\Omega t}.

The linear response can be broken up into a zero frequency equilibrium tide and a dynamical tide, qa,lin(t)=qa,eq(t)+qa,dyn(t)q_{a,\text{lin}}(t)=q_{a,\text{eq}}(t)+q_{a,\text{dyn}}(t), where

qa,eq(t)k=Ua(k)eikΩtq_{a,\text{eq}}(t)\equiv\sum_{k=-\infty}^\infty U_a^{(k)}e^{-ik\Omega t}

and

qa,dyn(t)qa,lin(t)qa,eq(t)=k=(kΩ+iγaωakΩiγa)Ua(k)eikΩt.\begin{aligned} q_{a,\text{dyn}}(t)&\equiv q_{a,\text{lin}}(t)-q_{a,\text{eq}}(t) \\ &=\sum_{k=-\infty}^\infty \left(\frac{k\Omega+i\gamma_a}{\omega_a-k\Omega-i\gamma_a}\right)U_a^{(k)}e^{-ik\Omega t}. \end{aligned}

Stability Analysis

r˙b+(iωb+γb)rb=iωbc[Ubc(t)+2Kbc(t)]rc\dot{r}_b+(i\omega_b+\gamma_b)r_b=i\omega_b\sum_c[U_{bc}^*(t)+2K_{bc}^*(t)]r_c^*

Plug in rb(t)=Qb(t)eiωt/2r_b(t)=Q_b(t)e^{i\omega t/2} and similarly for rcr_c

Q˙b+(iΔb+γb)Qb=ic(ωbωc)1/2ΓbcQc\dot{Q}_b+(i\Delta_b+\gamma_b)Q_b=i\sum_c\left(\frac{\omega_b}{\omega_c}\right)^{1/2}\Gamma_{bc}^*Q_c^*

here Δk=ωk+ω/2\Delta_k=\omega_k+\omega/2 is the daughter detuning, define the daughter pair “driving rate”

Γbcωbωc(Ubc+2Kbc)\Gamma_{bc}\equiv\sqrt{\omega_b\omega_c}(U_{bc}+2K_{bc})

Writing these equations as Q˙=HQ\dot{\boldsymbol{Q}}=H\boldsymbol{Q}, the solutions are Qexp(st)\boldsymbol{Q}\propto\exp(st).
The system is unstable if there is an ss such that Re(s)>0\text{Re}(s)>0.


Nonlinear tides in close binary systems
http://jingliangwei.github.io/blog-hexo/2026/07/02/Nonlinear-tides-in-close-binary-systems/
Author
Arwell
Posted on
July 2, 2026
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