rboston628
commented
4 months ago After testing some with the DummyStar using Cowling modes, I found it also finds ${\bar{\omega}}^2 = \ell$. Doing the math, with
$$x \frac{d\vec{y}}{dx} = \mathbf{A} \vec{y},$$
and with
$$\vec{y} = \vec{y_0} + \vec{y_2} x^2 + \vec{y_4} x^4 + \cdots$$
and with
$$\mathbf{A} = \mathbf{A_0} + \mathbf{A_2} x^2 + \mathbf{A_4}x^4 + \cdots,$$
then for the DummyStar with $\rho$, $P$ both constant, then $\mathbf{A} = \mathbf{A0}.$
You can then show that solving the system reduces to the eigenvalue equation
$$(2n) \vec{y{2n}} = \mathbf{A0} \vec{y{2n}}.$$
In the Cowling approximation, the coefficient matrix becomes
\mathbf{A_0} = \begin{bmatrix} -\ell-1 & \ell(\ell+1)/\bar{\omega}^2 \\\ \bar{\omega^2} & -\ell\end{bmatrix} has only two eigenvalues, $0$ and $-1-2\ell$. It also has only two eigenvectors,
\vec{y} = \begin{bmatrix} 1 \\\ \bar{\omega}^2/\ell\end{bmatrix}, \qquad
\vec{y} = \begin{bmatrix} -(\ell+1)/\omega^2 \\\ 1 \end{bmatrix}.The first eigenvalue corresponds to $n=0$. The second, $2n = -1-2\ell$, would imply a negative $\ell$. Therefore the only contribution is from $n=0$ with the vector $\vec{y_0} = (1, \bar{\omega}^2/\ell).$ Therefore the solution is a constant vector. At the surface, we impose the condition $y_2 = y_1$. This requires $\omega^2 = \ell$.
Moving to the isopycnic star, the coefficient matrix may be written
\mathbf{A} = \mathbf{A_0} + \frac{2}{\Gamma_1} \begin{bmatrix} 1 & -1 \\\ 1 & -1 \end{bmatrix} \sum_{n=1}^{\infty} x^{2n},where $\mathbf{A_0}$ is the same as for the DummyStar. This suggests the lowest-order solution on the isopycnic star can settle on the solution $\bar{\omega}^2 = \ell$.
According to Cox (1980), in equation 17.80, for the homogeneous model (aka isopycnic, aka n=0 polytrope), the square dimensionless frequencies for $f$-modes follow
$\bar{\omega}^2 = \frac{2\ell(\ell-2)}{2\ell+1}.$
This equation comes from work by Chandrasekhar (1964, eq. 53), which found for a homogeneous model the Kelvin mode frequencies are like
$\sigma^2 = \frac{2\ell(\ell-1)(2\ell-1)}{2\ell+1} \frac{\int P(r) r^{2l-2} dr}{\int \rho(r) r^{2l} dr},$
which, when using $P = Pc(1 - r^2/R^2)$ and $\rho=\rho{avg}$, leads to the equation in Cox.
This equation is apparently validated by numerical data in Cox (1980), Table 17.1, which is also based on a table from Ledoux & Walraven, which only a fool would dare question.
However, I am always finding that $\bar{\omega}^2 = \ell$.
I should try an independent verification that the result of Chandrasekhar is indeed correct for the exact situation I am considering. Or, figure out why I am finding incorrect $f$-mode frequencies.