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  • Non-axisymmetric potentials
  • Weak Bars
  1. 天体物理动力学

Week7: Orbits

Non-axisymmetric potentials

Weak Bars

We choose a polar coordinate $(R,\phi)$ in which the line $\phi=0$ coincides with the long axis of the potential. Since we assume the bar is weak, we may write

Φ(R,ϕ)=Φ0(R)+Φ1(R,ϕ)\Phi(R,\phi)=\Phi_0(R)+\Phi_1(R,\phi)Φ(R,ϕ)=Φ0​(R)+Φ1​(R,ϕ)

where $\Phi_1/\Phi_0\ll1$ (Pertubation).

  • Lindblad resonances (with a steady pattern speed $\Omega_b$)

    We seek to represent a general loop orbit as a superposition of the circular motion of a guiding center and small oscillations around this guiding center.

    Then the Lagrangian is

    L=12R˙2+12[R(ϕ˙+Ωb)]2−Φ(R,ϕ)\mathcal{L}=\frac12\dot R^2+\frac12\left[R(\dot\phi+\Omega_b)\right]^2-\Phi(R,\phi)L=21​R˙2+21​[R(ϕ˙​+Ωb​)]2−Φ(R,ϕ)

    So the equations of motion are

    R¨=R(ϕ˙+Ωb)2−∂Φ(R,ϕ)∂Rddt[R2(ϕ˙+Ωb)]=−∂Φ(R,ϕ)∂ϕ\ddot R=R(\dot\phi+\Omega_b)^2-\frac{\partial\Phi(R,\phi)}{\partial R}\\ \frac{\text d}{\text dt}\left[R^2(\dot\phi+\Omega_b)\right]=-\frac{\partial\Phi(R,\phi)}{\partial \phi}R¨=R(ϕ˙​+Ωb​)2−∂R∂Φ(R,ϕ)​dtd​[R2(ϕ˙​+Ωb​)]=−∂ϕ∂Φ(R,ϕ)​

    since we regard the weak bar as perturbation, we also divide $R$ and $\phi$ into zeroth- and first-order parts

    R(t)=R0+R1(t),ϕ(t)=ϕ0(t)+ϕ1(t)R(t)=R_0+R_1(t),\quad \phi(t)=\phi_0(t)+\phi_1(t)R(t)=R0​+R1​(t),ϕ(t)=ϕ0​(t)+ϕ1​(t)

    For the zeroth-order

    R0(ϕ˙0+Ωb)2=(dΦ0dR)R0,ϕ¨0=0R_0(\dot\phi_0+\Omega_b)^2=\left(\frac{\text d\Phi_0}{\text dR}\right)_{R_0},\quad \ddot\phi_0=0R0​(ϕ˙​0​+Ωb​)2=(dRdΦ0​​)R0​​,ϕ¨​0​=0

    which is the usual equation for centrifugal equilibrium at $R_0$. We further define $\Omega_0\equiv\Omega(R_0)$, where

    Ω(R)=±1RdΦ0dR\Omega(R)=\pm\sqrt{\frac1R\frac{\text d\Phi_0}{\text dR}}Ω(R)=±R1​dRdΦ0​​​

    is the circular frequency at $R$ in the potential $\Phi_0$. The angular speed for the circular motion then becomes

    ϕ˙0=Ω0−Ωb\dot\phi_0=\Omega_0-\Omega_bϕ˙​0​=Ω0​−Ωb​

    we choose the origin of time so that

    ϕ0(t)=(Ω0−Ωb)t\phi_0(t)=\left(\Omega_0-\Omega_b\right)tϕ0​(t)=(Ω0​−Ωb​)t

The equations of motion yield

R¨1=(R0+R1)(ϕ˙0+ϕ˙1+Ωb)2−[∂∂R(Φ0+Φ1)]R0+R1\ddot R_1=\left(R_0+R_1\right)\left(\dot \phi_0+\dot \phi_1+\Omega_b\right)^2-\left[\frac{\partial}{\partial R}\left(\Phi_0+\Phi_1\right)\right]_{R_0+R_1}R¨1​=(R0​+R1​)(ϕ˙​0​+ϕ˙​1​+Ωb​)2−[∂R∂​(Φ0​+Φ1​)]R0​+R1​​
ddt[(R0+R1)(ϕ˙0+ϕ˙1+Ωb)]=−[∂∂ϕ(Φ0+Φ1)]ϕ0+ϕ1\frac{\text d}{\text dt}\left[(R_0+R_1)(\dot\phi_0+\dot\phi_1+\Omega_b)\right]=-\left[\frac{\partial}{\partial \phi}\left(\Phi_0+\Phi_1\right)\right]_{\phi_0+\phi_1}dtd​[(R0​+R1​)(ϕ˙​0​+ϕ˙​1​+Ωb​)]=−[∂ϕ∂​(Φ0​+Φ1​)]ϕ0​+ϕ1​​

Since

[∂∂R(Φ0+Φ1)]R0+R1=[∂∂R(Φ0+Φ1)]R0+[∂2∂R2(Φ0+Φ1)]R0R1=(∂Φ0∂R)R0+[∂Φ1∂R+d2Φ0dR2R1]R0+O(R12)\begin{align*} \left[\frac{\partial}{\partial R}\left(\Phi_0+\Phi_1\right)\right]_{R_0+R_1} &=\left[\frac{\partial}{\partial R}\left(\Phi_0+\Phi_1\right)\right]_{R_0}+\left[\frac{\partial^2}{\partial R^2}\left(\Phi_0+\Phi_1\right)\right]_{R_0}R_1\\ &=\left(\frac{\partial\Phi_0}{\partial R}\right)_{R_0}+\left[\frac{\partial\Phi_1}{\partial R}+\frac{\text d^2\Phi_0}{\text d R^2}R_1\right]_{R_0}+\mathcal{O(R_1^2)}\\ \end{align*}[∂R∂​(Φ0​+Φ1​)]R0​+R1​​​=[∂R∂​(Φ0​+Φ1​)]R0​​+[∂R2∂2​(Φ0​+Φ1​)]R0​​R1​=(∂R∂Φ0​​)R0​​+[∂R∂Φ1​​+dR2d2Φ0​​R1​]R0​​+O(R12​)​
[∂∂ϕ(Φ0+Φ1)]ϕ0+ϕ1=(∂Φ0∂ϕ)ϕ0+[∂Φ1∂ϕ+∂2Φ0∂ϕ2ϕ1]R0+O(ϕ12)=(∂Φ1∂ϕ)R0+O(ϕ12)\begin{align*} \left[\frac{\partial}{\partial \phi}\left(\Phi_0+\Phi_1\right)\right]_{\phi_0+\phi_1} &=\left(\frac{\partial\Phi_0}{\partial \phi}\right)_{\phi_0}+\left[\frac{\partial\Phi_1}{\partial \phi}+\frac{\partial^2\Phi_0}{\partial \phi^2}\phi_1\right]_{R_0}+\mathcal{O(\phi_1^2)}\\ &=\left(\frac{\partial\Phi_1}{\partial \phi}\right)_{R_0}+\mathcal{O(\phi_1^2)} \end{align*}[∂ϕ∂​(Φ0​+Φ1​)]ϕ0​+ϕ1​​​=(∂ϕ∂Φ0​​)ϕ0​​+[∂ϕ∂Φ1​​+∂ϕ2∂2Φ0​​ϕ1​]R0​​+O(ϕ12​)=(∂ϕ∂Φ1​​)R0​​+O(ϕ12​)​

for the first-order, we have

R¨1=(R0+R1)(ϕ˙0+ϕ˙1+Ωb)2−(∂Φ0∂R)R0−[∂Φ1∂R+d2Φ0dR2R1]R0=2R0ϕ˙1(ϕ˙0+Ωb)+R1(ϕ˙0+ϕ˙1+Ωb)2−[∂Φ1∂R+d2Φ0dR2R1]R0≡2R0Ω0ϕ˙1+R1(Ω2−d2Φ0dR2)R0−(∂Φ1∂R)R0\begin{align*} \ddot R_1 &=\left(R_0+R_1\right)\left(\dot \phi_0+\dot \phi_1+\Omega_b\right)^2-\left(\frac{\partial\Phi_0}{\partial R}\right)_{R_0}-\left[\frac{\partial\Phi_1}{\partial R}+\frac{\text d^2\Phi_0}{\text d R^2}R_1\right]_{R_0}\\ &=2R_0\dot\phi_1\left(\dot\phi_0+\Omega_b\right)+R_1\left(\dot \phi_0+\dot \phi_1+\Omega_b\right)^2-\left[\frac{\partial\Phi_1}{\partial R}+\frac{\text d^2\Phi_0}{\text d R^2}R_1\right]_{R_0}\\ &\equiv2R_0\Omega_0\dot\phi_1+R_1\left(\Omega^2-\frac{\text d^2\Phi_0}{\text d R^2}\right)_{R_0}-\left(\frac{\partial\Phi_1}{\partial R}\right)_{R_0} \end{align*}R¨1​​=(R0​+R1​)(ϕ˙​0​+ϕ˙​1​+Ωb​)2−(∂R∂Φ0​​)R0​​−[∂R∂Φ1​​+dR2d2Φ0​​R1​]R0​​=2R0​ϕ˙​1​(ϕ˙​0​+Ωb​)+R1​(ϕ˙​0​+ϕ˙​1​+Ωb​)2−[∂R∂Φ1​​+dR2d2Φ0​​R1​]R0​​≡2R0​Ω0​ϕ˙​1​+R1​(Ω2−dR2d2Φ0​​)R0​​−(∂R∂Φ1​​)R0​​​
  ⟺  R¨1−2R0Ω0ϕ˙1+R1(d2Φ0dR2−Ω2)R0=−(∂Φ1∂R)R0\iff \ddot R_1-2R_0\Omega_0\dot\phi_1+R_1\left(\frac{\text d^2\Phi_0}{\text d R^2}-\Omega^2\right)_{R_0}=-\left(\frac{\partial\Phi_1}{\partial R}\right)_{R_0}⟺R¨1​−2R0​Ω0​ϕ˙​1​+R1​(dR2d2Φ0​​−Ω2)R0​​=−(∂R∂Φ1​​)R0​​

and

0=ddt[(R0+R1)2(ϕ˙0+ϕ˙1+Ωb)]+[∂∂ϕ(Φ0+Φ1)]ϕ0+ϕ1=2(R0+R1)R˙1Ω0+(R0+R1)2ϕ¨1+(∂Φ1∂ϕ)R0=2R0R˙1Ω0+R02ϕ¨1+(∂Φ1∂ϕ)R0\begin{align*} 0 &=\frac{\text d}{\text dt}\left[(R_0+R_1)^2(\dot\phi_0+\dot\phi_1+\Omega_b)\right]+\left[\frac{\partial}{\partial \phi}\left(\Phi_0+\Phi_1\right)\right]_{\phi_0+\phi_1}\\ &=2(R_0+R_1)\dot R_1\Omega_0+(R_0+R_1)^2\ddot\phi_1 +\left(\frac{\partial\Phi_1}{\partial \phi}\right)_{R_0}\\ &=2R_0\dot R_1\Omega_0+R_0^2\ddot\phi_1 +\left(\frac{\partial\Phi_1}{\partial \phi}\right)_{R_0} \end{align*}0​=dtd​[(R0​+R1​)2(ϕ˙​0​+ϕ˙​1​+Ωb​)]+[∂ϕ∂​(Φ0​+Φ1​)]ϕ0​+ϕ1​​=2(R0​+R1​)R˙1​Ω0​+(R0​+R1​)2ϕ¨​1​+(∂ϕ∂Φ1​​)R0​​=2R0​R˙1​Ω0​+R02​ϕ¨​1​+(∂ϕ∂Φ1​​)R0​​​
  ⟺  ϕ¨1+2Ω0R˙1R0=−1R02(∂Φ1∂ϕ)R0\iff \ddot\phi_1+2\Omega_0\frac{\dot R_1}{R_0}=-\frac1{R_0^2}\left(\frac{\partial\Phi_1}{\partial \phi}\right)_{R_0}⟺ϕ¨​1​+2Ω0​R0​R˙1​​=−R02​1​(∂ϕ∂Φ1​​)R0​​

To proceed further we must select a specific form of $\Phi_1$

Φ1(R,ϕ)=Φb(R)cos⁡(mϕ),Φb(R)<0\Phi_1(R,\phi)=\Phi_b(R)\cos(m\phi),\quad \Phi_b(R)<0Φ1​(R,ϕ)=Φb​(R)cos(mϕ),Φb​(R)<0

where $m\in N^$, since any potential that is an even function of $\phi$ can be expanded as a Fourier series. If $m=2$ the potential is *barred.

At first we assume that $\phi_1\ll1$ and hence $\phi(t)$ always remains close to $(\Omega_0-\Omega_b)t$. With this assumption we have

R¨1−2R0Ω0ϕ˙1+R1(d2Φ0dR2−Ω2)R0≡−(dΦbdR)R0cos⁡[m(Ω0−Ωb)t]\ddot R_1-2R_0\Omega_0\dot\phi_1+R_1\left(\frac{\text d^2\Phi_0}{\text d R^2}-\Omega^2\right)_{R_0}\equiv-\left(\frac{\text d\Phi_b}{\text d R}\right)_{R_0}\cos\left[m\left(\Omega_0-\Omega_b\right)t\right]R¨1​−2R0​Ω0​ϕ˙​1​+R1​(dR2d2Φ0​​−Ω2)R0​​≡−(dRdΦb​​)R0​​cos[m(Ω0​−Ωb​)t]
ϕ¨1+2Ω0R˙1R0=mΦb(R0)R02sin⁡[m(Ω0−Ωb)t]\ddot\phi_1+2\Omega_0\frac{\dot R_1}{R_0}=\frac{m\Phi_b(R_0)}{R_0^2}\sin\left[m\left(\Omega_0-\Omega_b\right)t\right]ϕ¨​1​+2Ω0​R0​R˙1​​=R02​mΦb​(R0​)​sin[m(Ω0​−Ωb​)t]
⇒ϕ˙1=−2Ω0R1R0−Φb(R0)R02(Ω0−Ωb)cos⁡[m(Ω0−Ωb)t]+Const\Rightarrow \dot \phi_1=-2\Omega_0\frac{R_1}{R_0}-\frac{\Phi_b(R_0)}{R_0^2\left(\Omega_0-\Omega_b\right)}\cos\left[m\left(\Omega_0-\Omega_b\right)t\right]+Const⇒ϕ˙​1​=−2Ω0​R0​R1​​−R02​(Ω0​−Ωb​)Φb​(R0​)​cos[m(Ω0​−Ωb​)t]+Const

Then we can eliminate $\dot\phi_1$ from the equations

R¨1+4R1Ω02+2Ω0Φb(R0)R0(Ω0−Ωb)cos⁡[m(Ω0−Ωb)t]+R1(d2Φ0dR2−Ω2)R0≡−(dΦbdR)R0cos⁡[m(Ω0−Ωb)t]+Const\ddot R_1+4R_1\Omega_0^2+\frac{2\Omega_0\Phi_b(R_0)}{R_0\left(\Omega_0-\Omega_b\right)}\cos\left[m\left(\Omega_0-\Omega_b\right)t\right]+R_1\left(\frac{\text d^2\Phi_0}{\text d R^2}-\Omega^2\right)_{R_0}\equiv-\left(\frac{\text d\Phi_b}{\text d R}\right)_{R_0}\cos\left[m\left(\Omega_0-\Omega_b\right)t\right]+ConstR¨1​+4R1​Ω02​+R0​(Ω0​−Ωb​)2Ω0​Φb​(R0​)​cos[m(Ω0​−Ωb​)t]+R1​(dR2d2Φ0​​−Ω2)R0​​≡−(dRdΦb​​)R0​​cos[m(Ω0​−Ωb​)t]+Const
  ⟺  R¨1+(d2Φ0dR2+3Ω2)R0R1=−[dΦbdR+2ΩΦbR0(Ω−Ωb)]R0+Const\iff \ddot R_1+\left(\frac{\text d^2\Phi_0}{\text d R^2}+3\Omega^2\right)_{R_0}R_1=-\left[\frac{\text d\Phi_b}{\text d R}+\frac{2\Omega\Phi_b}{R_0\left(\Omega-\Omega_b\right)}\right]_{R_0}+Const⟺R¨1​+(dR2d2Φ0​​+3Ω2)R0​​R1​=−[dRdΦb​​+R0​(Ω−Ωb​)2ΩΦb​​]R0​​+Const

We define

κ2=d2Φ0dR2+3Ω2\kappa^2=\frac{\text d^2\Phi_0}{\text d R^2}+3\Omega^2κ2=dR2d2Φ0​​+3Ω2

and

κ02=(d2Φ0dR2+3Ω2)R0=(RdΩ2dR+4Ω2)R0(κ02<4Ω02)\kappa_0^2=\left(\frac{\text d^2\Phi_0}{\text d R^2}+3\Omega^2\right)_{R_0}=\left(R\frac{\text d\Omega^2}{\text d R}+4\Omega^2\right)_{R_0}\quad\left(\kappa_0^2<4\Omega_0^2\right)κ02​=(dR2d2Φ0​​+3Ω2)R0​​=(RdRdΩ2​+4Ω2)R0​​(κ02​<4Ω02​)

since

d2Φ0dR2=d(RΩ2)dR=RdΩ2dR+Ω2\frac{\text d^2\Phi_0}{\text d R^2} =\frac{\text d(R\Omega^2)}{\text d R}=R\frac{\text d\Omega^2}{\text d R}+\Omega^2dR2d2Φ0​​=dRd(RΩ2)​=RdRdΩ2​+Ω2

$\kappa_0$ is just the usual epicycle frequency. The constant in the equation is not important as it can be absorbed into $R_1$. Then we have got an equation of motion of a harmonic oscillator of natural frequency of $\kappa_0$ that is driven at a frequency of $m\left(\Omega_0-\Omega_b\right)$. The general solution is

R1(t)=C1cos⁡(κ0t+α)−[dΦbdR+2ΩΦbR0(Ω−Ωb)]R0cos⁡[m(Ω0−Ωb)t]ΔR_1(t)=C_1\cos(\kappa_0t+\alpha)-\left[\frac{\text d\Phi_b}{\text d R}+\frac{2\Omega\Phi_b}{R_0\left(\Omega-\Omega_b\right)}\right]_{R_0}\frac{\cos\left[m\left(\Omega_0-\Omega_b\right)t\right]}{\Delta}R1​(t)=C1​cos(κ0​t+α)−[dRdΦb​​+R0​(Ω−Ωb​)2ΩΦb​​]R0​​Δcos[m(Ω0​−Ωb​)t]​

where $C_1$ and $\alpha$ are arbitrary constants, and

Δ≡κ02−m2(Ω0−Ωb)2\Delta\equiv \kappa_0^2-m^2\left(\Omega_0-\Omega_b\right)^2Δ≡κ02​−m2(Ω0​−Ωb​)2

We may also eliminate $t$ from this solution with the equation $\phi_0=\left(\Omega_0-\Omega_b\right)t$

R1(ϕ0)=C1cos⁡(κ0tΩ0−Ωb+α)+C2cos⁡(mϕ0)R_1(\phi_0)=C_1\cos\left(\frac{\kappa_0t}{\Omega_0-\Omega_b}+\alpha\right)+C_2\cos(m\phi_0)R1​(ϕ0​)=C1​cos(Ω0​−Ωb​κ0​t​+α)+C2​cos(mϕ0​)

where

C2=−1Δ[dΦbdR+2ΩΦbR0(Ω−Ωb)]R0C_2=-\frac{1}{\Delta}\left[\frac{\text d\Phi_b}{\text d R}+\frac{2\Omega\Phi_b}{R_0\left(\Omega-\Omega_b\right)}\right]_{R_0}C2​=−Δ1​[dRdΦb​​+R0​(Ω−Ωb​)2ΩΦb​​]R0​​

If $C_1=0$, $R_1(\phi_0)$ becomes periodic in $\phi_0$ with period $2\pi/m$, so the orbit is a closed loop one. Orbits with $C_1\neq0$ are non-closed loop ones parented by this closed loop.

For the $C_1=0$ case, there are two sorts of singular points

  • $\Omega_0-\Omega_b=0$

  • $\Delta=0$

    The former, of which $\Omega_0=\Omega_b$, corresponds to the case of corotation resonance. The guiding center simply corotates with the potential.

    The latter, of which $m\left(\Omega_0-\Omega_b\right)=\pm\kappa_0$, is known as Lindblad resonances. Radii at which such resonances occur are called Lindblad radii.

  • Inner Lindblad resonance - plus sign - larger $\Omega_0$

  • Outer Lindblad resonance - minus sign - smaller $\Omega_0$

  • Orbits trapped at resonance

    When $R_0$ approches the radius of either a Linblad resonance or the corotation resonance, the value of $R_1$ predicted with linear approximation diverges, thus we should modify the analysis.

    In the previous analysis we have assumed $\phi_1\ll1$, which is not neccessarily satisfied. If the bar strength $\Phi_1$ is proportional to some small parameter $\epsilon$, we assume $\phi_1$ is of order unity, $R_1$ is of order $\epsilon^{1/2}$, and the time derivative of any quantity is smaller than that quantity by of order $\epsilon^{1/2}$. When we further put the guiding center at $L_5$, the equations of motion go like

    R¨1+(κ02−4Ω02)R1−2R0Ω0ϕ˙1=−(∂Φ1∂R)R0\ddot R_1+\left(\kappa_0^2-4\Omega_0^2\right)R_1-2R_0\Omega_0\dot\phi_1=-\left(\frac{\partial\Phi_1}{\partial R}\right)_{R_0}R¨1​+(κ02​−4Ω02​)R1​−2R0​Ω0​ϕ˙​1​=−(∂R∂Φ1​​)R0​​
    ϕ¨1+2Ω0R˙1R0=−1R02(∂Φ1∂ϕ)R0\ddot\phi_1+2\Omega_0\frac{\dot R_1}{R_0}=-\frac1{R_0^2}\left(\frac{\partial\Phi_1}{\partial \phi}\right)_{R_0}ϕ¨​1​+2Ω0​R0​R˙1​​=−R02​1​(∂ϕ∂Φ1​​)R0​​

    According to our ordering, in the first equation, $\ddot R_1$ is of order $\epsilon^{3/2}$, $\Phi_1$ is of order unity, while other terms are of order $\epsilon^{1/2}$. In the second equation, each term is of order $\epsilon^{1/2}$. Thus

    (κ02−4Ω02)R1−2R0Ω0ϕ˙1=0,ϕ¨1+2Ω0R˙1R0=−1R02(∂Φ1∂ϕ)R0\left(\kappa_0^2-4\Omega_0^2\right)R_1-2R_0\Omega_0\dot\phi_1=0,\quad \ddot\phi_1+2\Omega_0\frac{\dot R_1}{R_0}=-\frac1{R_0^2}\left(\frac{\partial\Phi_1}{\partial \phi}\right)_{R_0}(κ02​−4Ω02​)R1​−2R0​Ω0​ϕ˙​1​=0,ϕ¨​1​+2Ω0​R0​R˙1​​=−R02​1​(∂ϕ∂Φ1​​)R0​​
    ⇒ϕ¨1(κ02κ02−4Ω02)=−1R02(∂Φ1∂ϕ)(R0,ϕ0+ϕ1)\Rightarrow\ddot\phi_1\left(\frac{\kappa_0^2}{\kappa_0^2-4\Omega_0^2}\right)=-\frac1{R_0^2}\left(\frac{\partial\Phi_1}{\partial \phi}\right)_{\left(R_0,\phi_0+\phi_1\right)}⇒ϕ¨​1​(κ02​−4Ω02​κ02​​)=−R02​1​(∂ϕ∂Φ1​​)(R0​,ϕ0​+ϕ1​)​

    Recall $\Phi_1=\Phi_1(R,\phi)=\Phi_b(R)\cos(m\phi)$ and let $m=2$

    ϕ¨1=−2ΦbR02(4Ω02−κ02κ02)sin⁡[2(ϕ0+ϕ1)]\ddot\phi_1=-\frac{2\Phi_b}{R_0^2}\left(\frac{4\Omega_0^2-\kappa_0^2}{\kappa_0^2}\right)\sin\left[2\left(\phi_0+\phi_1\right)\right]ϕ¨​1​=−R02​2Φb​​(κ02​4Ω02​−κ02​​)sin[2(ϕ0​+ϕ1​)]

    where $\Phi_b<0$. At $L_5$ we have $\phi_0=\pi/2$, so

    ϕ¨1=2ΦbR02(4Ω02−κ02κ02)sin⁡(2ϕ1)\ddot\phi_1=\frac{2\Phi_b}{R_0^2}\left(\frac{4\Omega_0^2-\kappa_0^2}{\kappa_0^2}\right)\sin\left(2\phi_1\right)ϕ¨​1​=R02​2Φb​​(κ02​4Ω02​−κ02​​)sin(2ϕ1​)
      ⟺  ψ¨=−p2sin⁡ψ\iff \ddot\psi=-p^2\sin\psi⟺ψ¨​=−p2sinψ

    where $\psi=2\phi_1$, and

    p2=−4∣Φb∣R02(4Ω02−κ02κ02)p^2=-\frac{4\left|\Phi_b\right|}{R_0^2}\left(\frac{4\Omega_0^2-\kappa_0^2}{\kappa_0^2}\right)p2=−R02​4∣Φb​∣​(κ02​4Ω02​−κ02​​)

    This is simply the equation of a pendulum. Here the singularity appeared at corotation ($\Omega_0=\Omega_b$) no longer exists in this more careful analysis. Also, the stable equilibrium point of the pendulum, $\phi_1 = 0$, is at the maximum, not the minimum, of the potential $\Phi_1$ (the donkey effect).

    • The donkey effect - in azimuth stars behave like donkeys, slowing down when pulled forwards and speeding up when held back - Lynden–Bell & Kalnajs (1972)

    If the integral of motion

    Ep=12ψ˙2−p2cos⁡ψE_p=\frac12\dot\psi^2-p^2\cos\psiEp​=21​ψ˙​2−p2cosψ

    is less than $p^2$, the star oscillates slowly or librates about the Lagragian point, whereas if $E_p>p^2$, the star is not trapped by the bar but circulates about the center of the galaxy.

    For small oscillations, the libration frequency is $p$.

    We may also obtain the shape of the orbit

    R1=R0Ω0κ02−4Ω02ψ˙=±R0Ω0κ02−4Ω022[Ep+p2cos⁡(2ϕ1)]R_1=\frac{R_0\Omega_0}{\kappa_0^2-4\Omega_0^2}\dot\psi=\pm \frac{R_0\Omega_0}{\kappa_0^2-4\Omega_0^2}\sqrt{2\left[E_p+p^2\cos\left(2\phi_1\right)\right]}R1​=κ02​−4Ω02​R0​Ω0​​ψ˙​=±κ02​−4Ω02​R0​Ω0​​2[Ep​+p2cos(2ϕ1​)]​
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