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Tidal Torque for Inclined and Eccentric Orbits {#InclinationEccentricity_torque}
==============================================
Starting from eq. 20 of Lai (2012):
\f{eqnarray*}{
    \mathbf{T}&=&-\int d^3 x \delta\rho(\mathbf{r}, t) \mathbf{r}\times
        \nabla U*(\mathbf{r}, t)\\
        &=&-\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \int d^3 x
            \left\{
                \sum_{m,m'}\mathcal{U}_{m,m'}
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    \exp(-im'\Omega t + i\Delta_{m,m'})
            \right\}
            \mathbf{r}\times
            \left\{
                \sum_{\mu,\mu'}\mathcal{U}_{\mu,\mu'}
                \exp(i \mu'\Omega t)
                \nabla \left[r^2 Y_{2,\mu}^*(\theta, \phi)\right]
            \right\}
\f}
Since \f$\mathbf{r}\times\nabla r^2=0\f$:
\f{eqnarray*}{
    \mathbf{T}&=&-\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \int d^3 x
            \left\{
                \sum_{m,m'}\mathcal{U}_{m,m'}
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    \exp(-im'\Omega t + i\Delta_{m,m'})
            \right\}
            \mathbf{r}\times
            \left\{
                \sum_{\mu,\mu'}\mathcal{U}_{\mu,\mu'}
                \exp(i \mu'\Omega t) r^2
                \nabla Y_{2,\mu}^*(\theta, \phi)
            \right\}\\
        &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m',\mu,\mu'} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,\mu'}
                \exp(i(\mu'-m')\Omega t + i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2 \mathbf{r}\times
                    \nabla Y_{2,\mu}^*(\theta, \phi)
\f}

The torque in the z direction
-----------------------------

Taking the dot product of the torque with
\f$\hat{z}=\cos\theta \hat{r} - \sin\theta \hat{\theta}\f$, and noting that
the \f$\hat{r}\f$ part of \f$\hat{z}\f$ does not contribute, since at any
point it is orthogonal to
\f$\mathbf{r}\times \nabla Y_{2,\mu}^*(\theta, \phi)\f$, we get:
\f{eqnarray*}{
    T_z&=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m',\mu,\mu'} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,\mu'}
                \exp(i(\mu'-m')\Omega t + i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2 \sin\theta \hat{\theta}\cdot \left[
                        \mathbf{r}\times
                        \nabla Y_{2,\mu}^*(\theta, \phi)
                    \right]\\
    &=& \left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m',\mu,\mu'} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,\mu'}
                \exp(i(\mu'-m')\Omega t + i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2 \sin \theta (\mathbf{r}\times\hat{\theta})\cdot
                    \nabla Y_{2,\mu}^*(\theta, \phi)\\
    &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m',\mu,\mu'} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,\mu'}
                \exp(i(\mu'-m')\Omega t + i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^3 \sin\theta
                    \hat{\phi}\cdot\nabla Y_{2,\mu}^*(\theta, \phi)\\
    &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m',\mu,\mu'} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,\mu'}
                \exp(i(\mu'-m')\Omega t + i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2
                    \frac{\partial Y_{2,\mu}^*(\theta, \phi)}{\partial\phi}\\
    &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m',\mu,\mu'} i\mu\mathcal{U}_{m,m'} \mathcal{U}_{\mu,\mu'}
                \exp(i(\mu'-m')\Omega t + i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2 Y_{2,\mu}^*(\theta, \phi)
\f}
Using the fact that
\f$\delta\bar{\rho}_{m,m'}(\mathbf{r})\propto\exp(im\phi)\f$, we must have
\f$m=\mu\f$, further if we average over over an orbit we must have
\f$m'=\mu'\f$:
\f[
    T_z= -\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m'} i m \mathcal{U}_{m,m'}^2
                \exp(i\Delta_{m,m'})
                \int d^3 x \delta\bar{\rho}_{m,m'}(\mathbf{r}) r^2
                        Y_{2,m}^*(\theta, \phi)
\f]
The real part of which is:
\f{eqnarray*}{
    T_z&=&\frac{G R^3}{M}\left(\frac{M'}{a^3}\right)^2
            \sum_{m,m'} m \mathcal{U}_{m,m'}^2
                \sin(\Delta_{m,m'})
                \int d^3 x \delta\bar{\rho}_{m,m'}(\mathbf{r}) r^2
                        Y_{2,m}^*(\theta, \phi)\\
    T_z&=&T_0 \sum_{m,m'} \mathcal{U}_{m,m'}^2 m \kappa_{m,m'}
            \sin(\Delta_{m,m'})\\
    T_0&\equiv& G R^5\left(\frac{M'}{a^3}\right)^2\\
    \kappa&\equiv&\frac{1}{MR^2}
            \int d^3 x \delta\bar{\rho}_{m,m'}(\mathbf{r}) r^2
                        Y_{2,m}^*(\theta, \phi)
\f}

The torque in the x direction
-----------------------------

Now we dot with \f$\hat{x}=\sin\theta\cos\phi\hat{r} +
\cos\theta\cos\phi\hat{\theta} - \sin\phi \hat{\phi}\f$. Again, the
\f$\hat{r}\f$ term does not contribute:
\f{eqnarray*}{
    T_x &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m',\mu,\mu'} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,\mu'}
                \exp(i(\mu'-m')\Omega t + i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2 (\cos\theta\cos\phi\hat{\theta} -
                         \sin\phi\hat{\phi}) \left[
                        \mathbf{r}\times
                        \nabla Y_{2,\mu}^*(\theta, \phi)
                    \right]\\
    &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m',\mu,\mu'} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,\mu'}
                \exp(i(\mu'-m')\Omega t + i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^3 (\cos\theta\cos\phi\hat{\phi} +
                         \sin\phi\hat{\theta})\cdot
                        \nabla Y_{2,\mu}^*(\theta, \phi)\\
    &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m',\mu,\mu'} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,\mu'}
                \exp(i(\mu'-m')\Omega t + i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2 \left(
                        \cot\theta\cos\phi\frac{\partial}{\partial \phi} +
                         \sin\phi\frac{\partial}{\partial \theta}
                    \right) Y_{2,\mu}^*(\theta, \phi)
\f}
Averaging over an orbit:
\f{eqnarray*}{
    T_x &=& \left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m',\mu} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,m'}
                \exp(i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2 \left(
                        i\mu\cot\theta\cos\phi Y_{2,\mu}^*(\theta, \phi) -
                         \sin\phi\frac{\partial Y_{2,\mu}^*(\theta, \phi)}
                                      {\partial \theta}
                    \right)\\
\f}
Now we use:
\f[
    \frac{\partial Y_{2,\mu}^*}{\partial \theta}=\mu\cot\theta Y_{2,\mu}^* +
    \sqrt{(2-\mu)(3+\mu)}\exp(i\phi) Y_{2,\mu+1}^*
\f]
To get:
\f{eqnarray*}{
    T_x &=& \frac{T_0}{MR^2}
            \sum_{m,m',\mu} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,m'}
                i\exp(i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2 \left(
                        \mu\cot\theta\exp(i\phi) Y_{2,\mu}^*(\theta, \phi) +
                         i\sin\phi\sqrt{(2-\mu)(3+\mu)}\exp(i\phi) Y_{2,\mu+1}^*
                    \right)\\
        &=& \frac{T_0}{MR^2}
            \sum_{m,m',\mu} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,m'}
                i\exp(i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2 \left(
                        \mu\cot\theta\exp(i\phi) Y_{2,\mu}^*(\theta, \phi) +
                         \sqrt{(2-\mu)(3+\mu)}[\exp(2i\phi)-1] Y_{2,\mu+1}^*
                    \right)\\
\f}
Since \f$\delta\bar{\rho}_{m,m'}(\mathbf{r})\propto\exp(im\phi)\f$ the real
part of the above expression is:
\f{eqnarray*}{
    T_x &=& -\frac{T_0}{MR^2}
            \sum_{m,m',\mu} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,m'}
                \sin(\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2 \left(
                        \mu\cot\theta\exp(i\phi) Y_{2,\mu}^*(\theta, \phi) +
                         \frac{\sqrt{(2-\mu)(3+\mu)}}{2}[\exp(2i\phi)-1] Y_{2,\mu+1}^*
                    \right)\\
        &=& T_0
            \sum_{m,m'} \mathcal{U}_{m,m'} \sin(\Delta_{m,m'})(
                \kappa_{m,m'}^-\mathcal{U}_{m-1,m'}+
                \kappa_{m,m'}^+\mathcal{U}_{m+1,m'})\\
    \kappa_{m,m'}^-&\equiv& \frac{\sqrt{(3-m)(2+m)}}{2MR^2}
        \int d^3 x \delta\bar{\rho}_{m,m'}(\mathbf{r})
        r^2 Y_{2,m}^*=\frac{\sqrt{(3-m)(2+m)}}{2}\kappa_{m,m'}\\
    \kappa_{m,m'}^+&\equiv&-\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{m,m'}(\mathbf{r}) r^2\left[
            (m+1)\cot\theta\exp(i\phi) Y_{2,m+1}^*(\theta,\phi)+
            \frac{\sqrt{(1-m)(4+m)}}{2}\exp(2i\phi) Y_{2,m+2}^*
        \right]
\f}

We have already expressed \f$\kappa_{m,m'}^-\f$ in terms of
\f$\kappa_{m,m'}\f$, and just as in Lai (2012), we only need \f$\kappa_{m,m'}^+\f$ for
\f$m=0, \pm 1, 2\f$.
\f{eqnarray*}{
    \kappa_{0,m'}^+&=&-\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{0,m'}(\mathbf{r}) r^2\left[
            \cot\theta\exp(i\phi) Y_{2,1}^*(\theta,\phi)+
            \exp(2i\phi) Y_{2,2}^*(\theta,\phi)
        \right]\\
    &=&-\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{0,m'}(\mathbf{r}) r^2\left[
            -\frac{1}{2}\sqrt{\frac{15}{2\pi}}\cos^2\theta+
            \frac{1}{4}\sqrt{\frac{15}{2\pi}}\sin^2\theta
        \right]\\
    &=&\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{0,m'}(\mathbf{r}) r^2
        \frac{1}{4}\sqrt{\frac{15}{2\pi}}(3\cos^2\theta-1)\\
    &=&\frac{\sqrt{3/2}}{MR^2}\int d^3 x
        \delta\bar{\rho}_{0,m'}(\mathbf{r}) r^2 Y_{2,0}^*(\theta, \phi)\\
    \Rightarrow \kappa_{0,m'}^+&=&\sqrt{3/2}\kappa_{0,m'}\\

    \kappa_{-1,m'}^+&=&-\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{-1,m'}(\mathbf{r}) r^2
        \frac{\sqrt{6}}{2}\exp(2i\phi) Y_{2,1}^*\\
    &=&\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{-1,m'}(\mathbf{r}) r^2
        \frac{\sqrt{6}}{2}\exp(2i\phi) Y_{2,-1}^*\\
    \Rightarrow\kappa_{-1,m'}^+&=&\sqrt{3/2}\kappa_{-1,m'}\\
    \kappa_{1,m'}^+&=&-\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{1,m'}(\mathbf{r}) r^2
        2\cot\theta\exp(i\phi) Y_{2,2}^*(\theta,\phi)\\
    &=&-\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{1,m'}(\mathbf{r}) r^2
        \frac{1}{2}\sqrt{\frac{15}{2\pi}}\sin\theta\cos\theta\exp(-i\phi)\\
    &=&\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{1,m'}(\mathbf{r}) r^2 Y_{2,1}^*(\theta,\phi)\\
    \Rightarrow\kappa_{1,m'}^+&=&\kappa_{1,m'}\\
    \kappa_{-2,m'}^+&=&-\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{-2,m'}(\mathbf{r}) r^2\left[
            -\cot\theta\exp(i\phi) Y_{2,-1}^*(\theta,\phi)+
            \sqrt{\frac{3}{2}}\exp(2i\phi) Y_{2,0}^*
        \right]\\
    &=&-\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{-2,m'}(\mathbf{r}) r^2\left[
            -\frac{1}{2}\sqrt{\frac{15}{2\pi}}\cos^2\theta\exp(2i\phi)+
            \frac{1}{4}\sqrt{\frac{15}{2\pi}}\exp(2i\phi)(3\cos^2\theta-1)
        \right]\\
    &=&-\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{-2,m'}(\mathbf{r}) r^2
        \frac{1}{4}\sqrt{\frac{15}{2\pi}}\exp(2i\phi)\left[
            -\cos^2\theta+3\cos^2\theta-1
        \right]\\
    &=&\frac{1}{MR^2} \int d^3 x
        \delta\bar{\rho}_{-2,m'}(\mathbf{r}) r^2 Y_{-2,m}^*(\theta, phi)\\
    \Rightarrow \kappa_{-2,m'}^+&=&\kappa_{-2,m'}\\
\f}

The torque in the y direction
-----------------------------
Now we dot with \f$\hat{y}=\sin\theta\sin\phi\hat{r} +
\cos\theta\sin\phi\hat{\theta} - \cos\phi \hat{\phi}\f$. Again, the
\f$\hat{r}\f$ term does not contribute:
\f{eqnarray*}{
    T_y &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m',\mu,\mu'} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,\mu'}
                \exp(i(\mu'-m')\Omega t + i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2 (\cos\theta\sin\phi\hat{\theta} -
                         \cos\phi\hat{\phi}) \left[
                        \mathbf{r}\times
                        \nabla Y_{2,\mu}^*(\theta, \phi)
                    \right]\\
    &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
            \sum_{m,m',\mu,\mu'} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,\mu'}
                \exp(i(\mu'-m')\Omega t + i\Delta_{m,m'})
                \int d^3 x
                    \delta\bar{\rho}_{m,m'}(\mathbf{r})
                    r^2 \left(
                        \cot\theta\sin\phi\frac{\partial}{\partial \phi} +
                         \cos\phi\frac{\partial}{\partial \theta}
                    \right) Y_{2,\mu}^*(\theta, \phi)
\f}
Averaging over an orbit:
\f{eqnarray*}{
    T_y &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
        \sum_{m,m',\mu} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,m'}
            \exp(i\Delta_{m,m'})
            \int d^3 x
                \delta\bar{\rho}_{m,m'}(\mathbf{r})
                r^2 \left(
                    i\mu\cot\theta\sin\phi Y_{2,\mu}^*(\theta, \phi) +
                     \cos\phi\frac{\partial Y_{2,\mu}^*(\theta, \phi)}
                                  {\partial \theta}
                \right)\\
    &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
        \sum_{m,m',\mu} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,m'}
            \exp(i\Delta_{m,m'})
            \int d^3 x
                \delta\bar{\rho}_{m,m'}(\mathbf{r})
                r^2 \left\{
                    i\mu\cot\theta\sin\phi Y_{2,\mu}^*(\theta, \phi) +
                    \cos\phi\left[\mu\cot\theta Y_{2,\mu}^* +
                        \sqrt{(2-\mu)(3+\mu)}\exp(i\phi) Y_{2,\mu+1}^*
                    \right]
                \right\}\\
    &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
        \sum_{m,m',\mu} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,m'}
            \exp(i\Delta_{m,m'})
            \int d^3 x
                \delta\bar{\rho}_{m,m'}(\mathbf{r})
                r^2 \left[
                    \mu\cot\theta\exp(i\phi) Y_{2,\mu}^*(\theta, \phi) +
                    \cos\phi\sqrt{(2-\mu)(3+\mu)}\exp(i\phi) Y_{2,\mu+1}^*
                \right]\\
    &=& -\left(\frac{G M'}{a^3 \omega_0}\right)^2
        \sum_{m,m',\mu} \mathcal{U}_{m,m'} \mathcal{U}_{\mu,m'}
            \exp(i\Delta_{m,m'})
            \int d^3 x
                \delta\bar{\rho}_{m,m'}(\mathbf{r})
                r^2 \left[
                    \mu\cot\theta\exp(i\phi) Y_{2,\mu}^*(\theta, \phi) +
                    \frac{\sqrt{(2-\mu)(3+\mu)}}{2}(\exp(2i\phi)+1)
                    Y_{2,\mu+1}^*
                \right]\\
    &=& -\frac{T_0}{MR^2}
        \sum_{m,m'} \mathcal{U}_{m,m'}
                        \left(\kappa^-_{m,m'}\mathcal{U}_{m-1,m'}+
                         \kappa^+_{m,m'}\mathcal{U}_{m+1,m'}\right)
                        \exp(i\Delta_{m,m'})
\f}
With the same \f$\kappa^+_{m,m'}\f$ and \f$\kappa^-_{m,m'}\f$ as for
\f$T_x\f$. So taking the real part gives:
\f[
    T_y = -T_0
        \sum_{m,m'} \mathcal{U}_{m,m'}
                        \left(\kappa^-_{m,m'}\mathcal{U}_{m-1,m'}+
                         \kappa^+_{m,m'}\mathcal{U}_{m+1,m'}\right)
                        \cos(\Delta_{m,m'})
\f]

So the real part is proportional to \f$\cos\Delta_{m,m'}\f$ and so for small
tidal dissipation it is independent of the dissipation, as expected since
this term is responsible for the precession.