How Do Radial Oscillations Affect Star Stability?

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Radial oscillations play a critical role in the stability of stars, as indicated by the equations derived in the discussion. The equations focus on the relationship between pressure, density, and radial displacement, suggesting that oscillations can be modeled using linearized forms around an equilibrium state. The variable ##\xi_r## represents small radial displacements, and the presence of an oscillation frequency term indicates a dependence on time and radial position. The discussion highlights the complexity of deriving the governing equations, particularly in maintaining linearity while eliminating mixed derivatives. Overall, understanding these dynamics is essential for analyzing star stability under oscillatory conditions.
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Homework Statement
The spherically symmetric oscillations of a star satisfy the Sturm-Liouville equation\begin{align*}
\frac{d}{dr} \left[ \frac{\gamma p}{r^2} \frac{d}{dr}(r^2 \xi_r) \right] - \frac{4}{r} \frac{dp}{dr} \xi_r + \rho \omega^2 \xi_r = 0
\end{align*}
Relevant Equations
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So far I have not made much meaningful progress beyond two equations; \begin{align*}
\rho \frac{D\mathbf{u}}{Dt} = - \nabla p \implies \rho \left( \frac{\partial}{\partial t} + u \frac{\partial}{\partial r} \right)u = - \frac{\partial p}{\partial r}
\end{align*}and thermal energy:\begin{align*}
\frac{Dp}{Dt} = -\gamma p \nabla \cdot \mathbf{u} \implies \left( \frac{\partial}{\partial t} + u \frac{\partial}{\partial r} \right)p = -\gamma p \frac{\partial}{\partial r}(r^2 u)
\end{align*}I'm somewhat confused why there's no time dependence in the form given. What exactly is ##\xi_r##?
 
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You are dealing with oscillations, and there is an \omega^2 in the ODE. That suggests variables are assumed to be functions of r alone times e^{i\omega t}. I suspect \xi_r denotes (small) radial displacement from an equilibrium position, in which case <br /> u_r = \frac{\partial}{\partial t}(\xi_r(r)e^{i\omega t}) = i\omega \xi_r(r) e^{i\omega t}. The ODE you are trying to derive is linear and the governing PDEs are not, which suggests that the system has been linearised about the static equiblibrium state.
 
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Thank you. I applied ##\partial/\partial t## to my first equation and ##\partial/\partial r## to my second equation, and subsequently eliminated ##\partial^2 p / \partial r \partial t##. Then I substituted for ##u = i\omega \xi_r e^{i \omega t}## and kept only terms to linear order in ##\xi_r##. After some manipulation,
\begin{align*}
(\xi_r p')' + (\gamma p(r^2 \xi_r)')' + \rho \omega^2 \xi_r = 0
\end{align*}Seems likely that I made at least one slip. I will check again tomorrow with fresher eyes!
 
ergospherical said:
I will check again tomorrow with fresher eyes!
Have you cleared that with the fresher? :eek:
 
pbuk said:
Have you cleared that with the fresher? :eek:
That would imply that they have a choice!
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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