Tidal Power for Inclined and Eccentric Orbits

Starting from eq. 21 of Lai (2012)

$$\begin{split}\dot{E}&=&-\int d^3 x \rho(\mathbf{r}) \frac{\partial \mathbf{\xi}(\mathbf{r}, t)}{\partial t} \cdot\nabla U*(\mathbf{r}, t)\\ &=& \int d^3 x \nabla\cdot \left(\rho(\mathbf{r}) \frac{\partial \mathbf{\xi}(\mathbf{r}, t)}{\partial t} \right)U*(\mathbf{r},t)\end{split}$$

Where the surface term is zero because the density is zero at the surface.

$$\dot{E}=-\left(\frac{GM'}{\omega_0 a^3}\right)^2\Omega \sum_{m,m',\mu,\mu'} \mathcal{U}_{m,m'}\mathcal{U}_{\mu,\mu'}im' \exp(i(\mu'-m')\Omega t + i\Delta_{m,m'}) \int d^3 x \delta\bar{\rho}_{m,m'}(\mathbf{r}) r^2 Y_{2,\mu}^*(\theta, \phi)$$

Since $\delta\bar{\rho}_{m,m'}(\mathbf{r})\propto\exp{im\phi}$ we must have $m=\mu$, further, averaging over an orbit we must have the time dependence term in the exponent vanish, $\Rightarrow \mu'=m'$:

$$\begin{split}\dot{E}&=&-\left(\frac{GM'}{\omega_0 a^3}\right)^2\Omega \sum_{m,m'} \mathcal{U}_{m,m'}^2 im' \exp(i\Delta_{m,m'}) \int d^3 x \delta\bar{\rho}_{m,m'}(\mathbf{r}) r^2 Y_{2,m}^*(\theta, \phi)\\ &=&-T_0\Omega \sum_{m,m'} \mathcal{U}_{m,m'}^2im'\exp(i\Delta_{m,m'}) \kappa_{m,m'}\\\end{split}$$

So taking the real part:

$$\begin{split}\dot{E}=T_0\Omega \sum_{m,m'} \mathcal{U}_{m,m'}^2m'\sin(\Delta_{m,m'}) \kappa_{m,m'}\\\end{split}$$