Astrophysical Fluids Lecture 5

Peter Ma | Jan 1st 2025

In this lecture we cover instabilities that arise in various astrophysical settings. This includes the collapsing stars (Jean's instabilities), Stellar Jets (Rayleigh Taylor and Kelvin Helmholtz instabilities) as well as stabilities of rotation disks like galaxies. Instabilities are a critical component in understanding dynamical systems and how they behave over long time horizons.

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Instabilties

The key to understanding the physics of instabilities is to look at the dispersion relation of a model system. Dispersion relation relates how length-like scales to time-like scales which reveal places of stabilities. The usual framework is take a complex set of linearly perturbed PDE's and to combine them into one equation and then transforming it into Fourier space to look at how spatial wavelengths relate to time scales. A complex time scale indicates solutions are real and thus have exponential trends which lead to instabilities while real time scales result in ocillator solutions which are stable. Let us put this into practice...

Jeans' Instabilities (Collapsing Gas/Stars)

Jean's instabilities arise from the UNBALANCED force between pressure and self gravtiy. Consider the following perturbed quantities to which we can then plug into our fluid equations: \[\rho = \rho_0 + \rho_1, \quad v = v_1 \quad P = P_0 + P_1 \quad \Phi = \Phi_0 + \Phi_1\] Mass conservation \[\frac{\partial }{\partial t}[\rho_1+\rho_0] + \nabla [(\rho_1+\rho_0)v_1] = 0\] Zeroth order is constant and multiplication of first order term is zero \[\boxed{\frac{\partial }{\partial t}\rho_1 + \nabla \rho_0 v_1 = 0}\] Momentum conservation \[\frac{\partial v_1}{\partial t} + v_1 \cdot \nabla v_1 = - \frac{1}{\rho_0+\rho_1}\nabla (P_1+P_0) - \nabla (\Phi_0+\Phi_1)\] Kill off the constant term and the higher order multiplication \[\boxed{\frac{\partial v_1}{\partial t} = - \frac{1}{\rho_0}\nabla P_1 - \nabla \Phi_1}\] Poission equation \[\boxed{\nabla^2 \Phi_1 = 4\pi G\rho_1}\] We can merge the mass and momentum equations take the gradient of momentum eq and take the time derivative of the mass \[\frac{\partial}{\partial t} \frac{\partial }{\partial t}\rho_1 = - \rho_0\frac{\partial}{\partial t}\nabla v_1 \] \[\frac{\partial }{\partial t} \nabla v_1 = - \frac{1}{\rho_0}\nabla^2 P_1 - \nabla^2 \Phi_1\] The two gives: \[\frac{\partial^2 \rho_1}{\partial t^2} = - \nabla^2 P_1 - \rho_0\nabla^2 \Phi_1\] Recall that the speed of the fluid is \(v_s^2 = \frac{\partial P_1}{\partial \rho_1} \Rightarrow v_s^2 \nabla^2 \rho_1 = \nabla^2 P_1\) \[\boxed{\frac{\partial^2 \rho_1}{\partial t^2} = v_s^2\nabla^2 \rho_1 + 4\pi G \rho_1\rho_0}\] We also added the poission equation in. This gives us a driven wave equation! We can apply the fourier transform and obtain the dispersion relation \[i^2w^2 \hat\rho_1= v_s^2 i^2 k^2 \hat\rho_1 + 4\pi G \rho_0\hat\rho_1 \] \[-w^2 = -v_s^2k^2 + 4\pi G \rho_0 \] \[\boxed{w^2 = v_s^2k^2 - 4\pi G \rho_0 }\] We see there are two settings \[\] Instability: occurs when \(w\) is complex because this would mean the exponential term of \(\rho\) will have a real value power which determines growth over time. Thus a critical value is \[w^2 = 0 \Rightarrow k_{J} = \sqrt{\frac{4\pi G \rho_0}{v_s^2}} \] and so when \(k> k_{J} \Rightarrow w^2 < 0\) thus unstable. \[\] Stability occurs when there is real value of \(w\) \[k< \sqrt{\frac{4\pi G \rho_0}{v_s^2}} \] Here \(k\) is the wavenumber which converted to physical units is \(\lambda_J = \frac{1}{k_J}\) is the Jean's length. This admits an oscillatory behaviour which makes it stable. \[\] Physical Intuition: We can also derive this without solving the full fluid equations: Consider the force density of gravity: \[\frac{F_g}{V} \approx \frac{GM}{R^2} \rho \approx \frac{G}{R} \rho^2\] Now consider the force density due to pressure \[\frac{F_p}{V} \approx \nabla P \approx v_s^2 \nabla \rho \approx v_s^2 \frac{\rho}{R}\] We compare the size of each and reconize that gravity is dominate when \(R > \sqrt{\frac{v_s^2}{G\rho }}\) which is close! \[\] We can also look at the free fall timescales \[t_f \approx \sqrt{2R/a} = \sqrt{R^3/GM} \approx 1/\sqrt{G\rho}\] While the time for the sound is to travel is \[t_s = R/v_s\] And thus gravity wins when the time of free fall is larger than the time of sound travel. This makes sense because

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Rotational Instabilities

Rotating discs are everywhereeeeeeeee in the universe. And thus its important to understand the physics behind rotating discs of gas. Let us consider a thin razor blade of a disc only in 2d plane: We define density as \(\Sigma\) \[\Sigma = \Sigma_0+\Sigma_1 \quad v = v_1 \quad P = P_0 + P_1 \quad \Phi = \Phi_0 + \Phi_1\] We can then plug into our equations yet again! (We follow the same way of cutting off terms as we just did) Mass conservation \[\boxed{\frac{\partial }{\partial t}\Sigma_1 + \nabla \Sigma_0 v_1 = 0}\] Momentum conservation but we include the Corriollis force! \[\frac{\partial v_1}{\partial t} = - \frac{1}{\Sigma_0}\nabla P_1 - \nabla \Phi_1 - 2\Omega \times v_1 +\Omega^2 (x\hat x + y \hat y)\] We can kill of the last term because we are \[\boxed{\frac{\partial v_1}{\partial t} = - \frac{1}{\Sigma_0}\nabla P_1 - \nabla \Phi_1 - 2\Omega \times v_1 }\] Poission equation we make it a razor blade by multiplying it be \(\delta\) function. \[\boxed{\nabla^2 \Phi_1 = 4\pi G\Sigma_1\delta(z)}\] We can then consider taking the fourier transform in one of the cartessian coordinate directions, let us choose \(x\) This means that for each solution either \(\Sigma_1, v_1, \rho_1, P_1 \) will have the following form: \[X_1 = \hat X_1 e^{i(kx-wt)}\] All except for gravity which has the following form \[\Phi_1 = \Phi_1(k) f(z) e^{i(kx-wt)}\] We can hit it with the Laplace equation and get that \[\frac{\partial^2 }{\partial z^2}\Phi_1 = \Phi_1(k) \frac{\partial^2 }{\partial z^2}f e^{i(kx-wt)} + \Phi_1(k)f \frac{\partial^2 }{\partial z^2} e^{i(kx-wt)} = 0\] \[\frac{\partial^2 }{\partial z^2}\Phi_1 = \frac{\partial^2 }{\partial z^2}f -k^2f = 0\] Which has a known solution of the following form: \[f \propto e^{\pm k |z|}\] But since we want gravity to die as we go to far distances we only keep the negative term and keep the constatn inside the \(\Phi(k)\) term. \[\Phi_1 = \Phi_1(k)e^{i(kx-wt)} e^{-k|z|}\] Now we need to satisfy the boundary condition when we get close to the plate the difference in the fields must be \(2E\) like the parrallel plates in Electromagnetism \[4\pi G \Sigma_1(k)e^{i(kx-wt)} = \frac{\partial \Phi_1}{\partial z}|_{0+} - \frac{\partial \Phi_1}{\partial z}|_{0-}\] \[4\pi G \Sigma_1(k)e^{i(kx-wt)} = -2k\Phi_1(k) e^{i(kx-wt)}\] \[\Phi_1(k) = -2\pi G/k \Sigma_1(k)\] Fourier Space \[-iw \Sigma_1+ ik\Sigma_0 v_1x = 0 \tag{mass conser.}\] \[-iw v_{1x}+ \frac{ikv_s^2}{\Sigma_0} \Sigma_1 - 2\pi iG \Sigma_1 - 2\Omega v_{1y} = 0 \tag{Momentum}\] \[-iw v_{1y}+2\Omega v_{1x} = 0\] Do some algebruhhhhhh to get the dispersion relation which is a quadratic form. \[\boxed{w^2 = v_s^2-2\pi G\Sigma_0k + 4\Omega^2}\] Physics: There are a number of interesting settings:

  1. No Gravity: We see that dispersion relation always omits stable solutions at all wavelengths
  2. Gravity + Pressure, we see it goes back to the Jeans instability setting (short wavelengths)
  3. Gravity + Rotation, Disk is unstable for long wavelengths
  4. Only Gravtiy: disc is unstable for long wavelengths
  5. Everything: Pressure stablises short while rotation stablises long wavelengths. The stability criterion is \[\boxed{\frac{v_s\Omega}{G \Sigma_0} > \pi/2}\] This comes from the fact that we can complete the square in our quadratic expression.
  6. Differential rotating disc we can replace \(\Omega(R)\) with epicyclic frequency \(\kappa \) where its defined as \[\kappa^2 = R \frac{\partial \Omega}{\partial R}\Omega^2 + 4 \Omega^2\] We chose this \(\kappa\) to basically subtract off the differential term in the expansion of \(\Omega(R)\) . This gives a new criterion of \[\boxed{Q = \frac{v_s\kappa}{\pi G \Sigma_0}> 1}\]

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Rayleigh-Taylor Instabilities (Stellar Jets, Supernova etc)

Studying how fluids of different densities interact with one another is the focus of Rayleigh-Taylor and Kelvin Helmholtz instabilities. Let us define a top fluid with \(+\) and the bottom fluid as \(-\). We also assume the fluid is incompressible and steady. Together we have our usual suspects (this time we assume its in hydrostatic equilibrium): \[\rho_+ = \rho_{1+}+\rho_{0+} \quad P_+ = P_{1+}+ P_{0+} \quad v_+ = v_{1+}+ v_{0+} \] \[\rho_- = \rho_{1-}+\rho_{0-} \quad P_- = P_{1-}+ P_{0-} \quad v_- = v_{1-}+ v_{0-} \] To balance gravity we have \[P_{0+} = \rho_{0+} g z + C_+ \] \[P_{0-} = \rho_{0-} g z + C_+ \] Consider the equation for the first perturbed quantity \[\] Mass Equation: it is steady and so the time derivative vanishes and we can divide by sides by \(\rho_{0\pm}\) \[\nabla v_1 = 0\] Momentum equation: we have no higher order terms and thus \[\frac{\partial v_1}{\partial t} = \frac{1}{\rho_0}\nabla P_1\] \[\nabla\frac{\partial v_1}{\partial t} = \nabla\frac{1}{\rho_0}\nabla P_1\] use mass equation \[\frac{\partial \nabla v_1}{\partial t} = \frac{1}{\rho_0}\nabla^2 P_1\] \[ \nabla^2 P_1 = 0\] Solving for the Laplace equation for pressure gives possible solutions: \[P_1 = Ae^{i(kx-wt)}(Be^{-k|z|}+ Ce^{k|z|})\] Then simply taking the derivative and then time integrating \[v_1 = \frac{ki}{w\rho_0}Ae^{i(kx-wt)}(Be^{-k|z|}+ Ce^{k|z|})\] We can then assume that the pressure term dies at large distances away from the surface. So for the top and bottom we have \[P_{+1} = Ae^{i(kx-wt)}(Be^{-k|z|})\] \[v_{+1} = \frac{kiP_{+1}}{w\rho_{0+}}\] \[P_{-1} = Ae^{i(kx-wt)}(Be^{k|z|})\] \[v_{-1} = \frac{kiP_{-1}}{w\rho_{0}}\] The vertical displacement is defined by \[\frac{d\xi}{dz} = v_{1z}\] Integrate \[\xi = \frac{i}{w}v_{1z}\] Here we have a surface tension force (which needs to balance the upward push from the bottom) This is the lagrangian form! \[F(\xi) = P_{+1,L} -T \frac{\partial^2\xi}{\partial z^2} = P_{-1,L}\] The left hand lagrangian is : \[P_{+1,L} = P_{+1}+ \xi \cdot \nabla P_{+0}\] \[P_{+1,L} = Ae^{i(kx-wt)}(Be^{-k|z|})+\frac{i}{w}v_{1+z} \cdot \nabla \rho_{0+} g z + C_+\] \[P_{+1,L} = \frac{-iw}{k}\rho_{0+}v_{1+z}+\frac{i}{w}v_{1+z} \cdot \rho_{0+} g \] \[P_{+1,L} = \frac{-iw}{k}\rho_{0+}v_{1+z}(1+\frac{gk}{w^2}) \] Ok now we do the right hand side...(repeat...) \[P_{-1,L} = \frac{iv_{1-z} \rho_{-0}w}{k}(1- \frac{gk}{w^2})\] We can then hit the PDE from before with FT and get our dispersion relation \[-i \rho_{+0}\frac{w}{k}(1 + \frac{gk}{w^2}) + T \frac{ik^2}{w} =i\rho_{-0}\frac{w}{k}(1-\frac{gk}{w^2})\] \[\boxed{(\rho_{+0} +\rho_{-0}) w^2 = (\rho_{-0} -\rho_{+0})gk +Tk^3}\] Some situations:

  1. Fluid on top is less dense, then Fluid always stable (duh)
  2. Fluid on top is very not dense \(\Rightarrow\) deep water waves capillary
  3. Denser fluid ontop- Instability occurs \[k^2 < k_{crit}^2 = \frac{g(\rho_{+0} - \rho_{-0})}{T}\]
  4. No surface tension, \[w^2 = \frac{(\rho_{+0} - \rho_{-0})}{(\rho_{+0} + \rho_{-0})} gk\] Rayleigh Taylor Dispersion and if fluid on top is denser than its always unstable everywhere
  5. The equivalence for accelerating lighter material through denser material also produces an effective \(g\) which can create instabilities! (Crab Nebula)

Kelvin-Helmholtz Instabilities

This is when the fluid has relative velocity with each other at the boundary which addes a phase to the dispersion relation \[\boxed{(\rho_{+0}(w-kv_{+1})^2 +\rho_{-0}(w-kv_{-1})^2) = (\rho_{-0} -\rho_{+0})gk +Tk^3}\] With some algebruhhhhhh we get another relation \[k > \frac{(\rho_{-0}^2 -\rho_{+0}^2 )g}{\rho_{-0}\rho_{+0}(v_{+0} - v_{-0})^2}=k_{crit}\] More physics

  1. With surface tension present shortest wavelength modes are stable
  2. No gravity all modes are UNSTABLE!(like stellar jets!)
  3. when density at the bottom is large the only unstable modes are short modes, and if they are probable on similar orders but the bottom is still larger then Rayleigh taylor is stable but Kelvin Helmholtz unstable example being billow clouds!