Astrophysical Fluids Lecture 2

Peter Ma | Dec 22nd 2024

This is the second lecture on astrophysical fluids and will be applying the equations derived in the previous lecture here to obtain the few analytical solutions to the fluid equations. This is assuming hydrostatic equilibrium.

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Applications

Isothermal Atmosphere.

Consider the setting of trying to model the atmosphere that is isothermal and is made of ideal gas on Earth. Let us try and derive the properties, namely the densities of the fluid under equilibrium.

Hydrostatic equilibrium means that the force of the pressure balances the force of the gravity pulling it inwards: \[\frac{-1}{\rho}\nabla P = \nabla \Phi\]

We use the fact that this is an ideal gas: \[P = \frac{nk_BT}{V} = \frac{\rho k_BT}{\mu m_p }\]

We can use this in the hydrostatic equilibrium case \[\frac{-1}{\rho}\nabla P = -\frac{1}{\rho}\nabla( \frac{\rho k_BT}{\mu m_p })\] We can first notice that the only coordinate that varies is the height in the \(z\) direction \[\frac{-1}{\rho}\nabla P = -\frac{1}{\rho} \frac{\rho k_BT}{\mu m_p }\frac{d}{dz}(\rho)\] We can then notice that we can undo the chain rule! Since \(\frac{d}{dz}\ln \rho = \frac{1}{\rho} \frac{d}{dz}\rho\) \[\frac{-1}{\rho}\nabla P = - \frac{ k_BT}{\mu m_p }\frac{d}{dz}(\ln \rho)\] This needs to equal the right hand side \[- \frac{ k_BT}{\mu m_p }\frac{d}{dz}(\ln \rho) = \nabla \Phi\] \[- \frac{ k_BT}{\mu m_p }\frac{d}{dz}(\ln \rho) = \frac{d}{dz}\Phi\] \[- \frac{ k_BT}{\mu m_p }\frac{d}{dz}(\ln \rho) = g\] Integrate \[- \frac{ k_BT}{\mu m_p }(\ln \rho) = gz + C\] \[\boxed{\rho = \rho_0 e^{- \frac{g \mu m_p z }{ k_BT}}}\]

This is how the density of air changes in an hydrostatic atmosphere.

Spherical Star

In the setting of a spherical star we have spherically symmetrical. So let us apply our knowledge of continuity, momentum and equations of state. Let us first recall the mass gradient: \[\frac{dM}{dr} = \frac{d}{dr} \int \rho dV\] \[\frac{dM}{dr} = \int \frac{d}{dr} \rho dV\] \[\frac{dM}{dr} = \int 4\pi \frac{d\rho}{dr} r^2 dr\] \[\frac{dM}{dr} = \int 4\pi r^2 d\rho\] \[\frac{dM}{dr} = 4\pi r^2\rho \] Then we can look at hydrostatic equilibrium \[\frac{-1}{\rho}\nabla P = \nabla \Phi\] We can then use the fact that this is a spherical symmetrical problem and knock out the other derivatives \[\frac{-1}{\rho r}\frac{d}{d r}(P) = \frac{1}{r}\frac{d }{d r} \Phi\] \[\frac{-1}{\rho }\frac{d }{d r}(P) = \frac{d }{dr} \Phi\] We can notice something that rearranging the derivatives \[\rho = -\frac{d P}{d \Phi} \] This means that the pressure is only dependent on density (Barytrope) Let us give the equation of state a polytropic index \[P = K \rho^{1+ \frac{1}{n}}\] Let us use this fact to get some understanding of \(\rho\) by subbing in the pressure \[\frac{-1}{\rho r}\frac{d}{d r}( K \rho^{1+ \frac{1}{n}}) = \frac{-1}{\rho r} (1+\frac{1}{n})K \rho^{1/n} \frac{d\rho}{dr}\] \[\Rightarrow -(1+\frac{1}{n})K \rho^{1/n-1} \frac{d\rho}{dr} = \frac{d }{dr} \Phi\] We can integrate this! \[\rho = [\frac{\Phi_0-\Phi}{(n+1)K}]^n\] Note we have boundary case when the density needs to fall off to 0 (edges) and at the center it needs to equal a constant \[\rho = \rho_c[\frac{\Phi_0-\Phi}{\Phi_0 - \Phi_c}]^n\] Let us define a variable \(\theta = \frac{\Phi_0-\Phi}{\Phi_0 - \Phi_c}\) \[\boxed{\rho = \rho_c\theta^n}\] \[\boxed{P = K\rho_c^{1+1/n}\theta^{n+1}}\] Return to the Hydrostatic equilibrium equation... \[\] \[\frac{-1}{\rho }\frac{d }{d r}(P) = \frac{d }{dr} \Phi\] Now we can take the derivative on both sides again! \[ \frac{d}{dr}[\frac{-1}{\rho }\frac{d}{d r}(P)] = \frac{d}{dr}\frac{d }{d r} \Phi\] \[ \frac{d}{dr}[\frac{-1}{\rho }\frac{d}{d r}(P)] = \frac{d}{dr}\frac{Gm}{r^2}\] Then we apply product rule \[ \frac{d}{dr}[\frac{-1}{\rho }\frac{d}{d r}(P)] = \frac{-Gm}{r^3} - \frac{G}{r^2} \frac{dm}{dr}\] We can then apply the gradient of the mass from before \[ \frac{d}{dr}[\frac{-1}{\rho }\frac{d}{d r}(P)] = \frac{-Gm}{r^3} - \frac{G}{r^2} 4\pi \rho r^2\] We reconize from the hydrostatic equilibrium we have the following \[ \frac{d}{dr}[\frac{-1}{\rho }\frac{d}{d r}(P)] = -\frac{2}{r \rho}\frac{dP}{dr} -G 4\pi \rho\] \[ \frac{d}{dr}[\frac{-1}{\rho }\frac{d}{d r}(P)] + \frac{2}{r \rho}\frac{dP}{dr}= -G 4\pi \rho\] We will now reverse the product rule by multiplying both sides by \(r^2\) \[ r^2\frac{d}{dr}[\frac{-1}{\rho }\frac{d}{d r}(P)] + \frac{2r}{ \rho}\frac{dP}{dr}= -G 4\pi r^2 \rho\] Now we can undo the product rule \[ \frac{d}{dr}[\frac{r^2}{\rho} \frac{dP}{dr}]= -G 4\pi r^2 \rho\] We will now try to replace these terms with information we already know namely the pressure and the density \[ \frac{d}{dr}[\frac{r^2}{\rho_c \theta^n} \frac{d (K\rho_c^{1+1/n}\theta^{n+1})}{dr}]= -G 4\pi r^2 \rho_c \theta^n\] \[ \frac{d}{dr}[\frac{r^2 K\rho_c^{1+1/n}}{\rho_c \theta^n} \frac{d (\theta^{n+1})}{dr}]= -G 4\pi r^2 \rho_c \theta^n\] \[ \frac{d}{dr}[\frac{r^2 K\rho_c^{1+1/n}}{\rho_c \theta^n}\theta^{n} \frac{d \theta}{dr}]= -G 4\pi r^2 \rho_c \theta^n\] \[ \frac{d}{dr}[r^2 K\rho_c^{1/n}\theta^{n} \frac{d \theta}{dr}]= -G 4\pi r^2 \rho_c \theta^n\] New constant... \[\] We will now define a new constante \(\alpha^2 = \frac{(n+1)K\rho_c^{1/n-1}}{4\pi G}\) and define \(r = a \cdot \xi\) \[ \frac{d}{dr}[\frac{(n+1)K\rho_c^{1/n-1}}{4\pi G} \xi^2 K\rho_c^{1/n}\theta^{n} \frac{d \theta}{dr}]= -G 4\pi \frac{(n+1)K\rho_c^{1/n-1}}{4\pi G} \cdot \xi^2 \rho_c \theta^n\] \[ \frac{d}{dr}[\frac{(n+1)K\rho_c^{1/n}}{4\pi G} \xi^2 K\theta^{n} \frac{d \theta}{dr}]= -(n+1)K\rho_c^{1/n} \cdot \xi^2 \theta^n\] \[ \frac{d}{dr}[\frac{1}{4\pi G} \xi^2 K\theta^{n} \frac{d \theta}{dr}]= - \xi^2 \theta^n\] Recall we can also replace the \(dr =\alpha d\xi \) \[\frac{1}{\alpha} \frac{d}{d\xi}[\frac{1}{4\pi G} \xi^2 K\theta^{n} \frac{1}{\alpha} \frac{d \theta}{d\xi}]= - \xi^2 \theta^n\] \[ \frac{d}{d\xi}[\frac{1}{4\pi G} \xi^2 K\theta^{n} \frac{1}{\alpha^2} \frac{d \theta}{d\xi}]= - \xi^2 \theta^n\] \[ \frac{d}{d\xi}[\xi^2 K\theta^{n} \frac{1}{(n+1)K\rho_c^{1/n-1}} \frac{d \theta}{d\xi}]= - \xi^2 \theta^n\] \[ \frac{d}{d\xi}[\xi^2 \theta^{n} \frac{1}{(n+1)\rho_c^{1/n-1}} \frac{d \theta}{d\xi}]= - \xi^2 \theta^n\] \[ \frac{d}{d\xi}[\xi^2 \frac{d \theta}{d\xi}]= - \xi^2 \theta^n\] \[ \boxed{\frac{1}{\xi^2}\frac{d}{d\xi}[\xi^2 \frac{d \theta}{d\xi}]= - \theta^n}\] This is the famous Lane-Emden equation! We can see when the index \(n=2\) this can actually be solved. and like wise this admits analytical solutions. Boundary Conditions We notice that this needs to satisfy a few boundary conditions. \[\] At the center we must have \[\theta(\xi =0) = 1\] \[\frac{d\theta}{d\xi}= 0\] On the surface we must have \[\theta(\xi_{max}) = 0\] Mass-Radius relationship We notice that from the definiton of the radius we have a dependence on density! \[r \propto \rho_c^{1/2(1/n-1)}\] We can extend this to the Mass \[M = \int \rho dV\] \[M = \int_0^R 4\pi r^2\rho dr\] We can add in the density \[M = 4\pi \rho_c \left[ \frac{4\pi G}{(n+1)K} \rho_c^{1-1/n}\right]^{-3/2}\int_0^{\xi_{max}} \theta^n \xi^2d\xi\] The integral is some unitless quantity so we dont care about it \[\boxed{M \propto R^{(n-3)/(n-1)}}\]

Bernoulli's Principle.

This last section is an investigation on Bernoulli's principle! Firstly recall the momentum equations: \[\frac{\partial v}{\partial t} + (v\cdot \nabla v) + \frac{1}{\rho}\nabla P + \nabla \Phi = 0\] We reconize that we can write the following vector identity \[(v\cdot \nabla v) = \nabla(\frac{1}{2}v^2) - v\times \nabla \times v\] Let us then denote a "vorticity" \(\omega = \nabla \times v\) We can also assume that we have steady flow \(\Rightarrow \frac{d}{dt} = 0\) \[\nabla(\frac{1}{2}v^2) - v\times \omega + \frac{1}{\rho}\nabla P + \nabla \Phi = 0\] \[\nabla(\frac{1}{2}v^2 + \frac{1}{\rho}P + \Phi)- v\times \omega = 0\] We see that if we flow ALONG THE STREAM we project into the flow direction \[v \cdot \nabla(\frac{1}{2}v^2 + \frac{1}{\rho}P + \Phi)- v\cdot v\times \omega = 0\] \[v \cdot \nabla(\frac{1}{2}v^2 + \frac{1}{\rho}P + \Phi) = 0\] We notice that the center term is conserved \[\mathcal{B} = \frac{1}{2}v^2 + \frac{1}{\rho}P + \Phi\] We notice that the pressure term can be written as interms of internal energy assuming we are along the flow line since entropy does not change here we can use enthalpy \(dh = TdS + \frac{dP}{\rho}\) \[\mathcal{B} = \frac{1}{2}v^2 + h + \Phi\] Does not change. In general then we see that if we are in the direction of the flow then (fluid does not rotate) \[\boxed{\nabla \mathcal{B} = 0}\]