In this lecture we turn on Electromagnetism and combine it with the full Navier Stokes equations. Nearly all fluids in astrophysics have magnetic properties and the role of EM is crucial in understanding the physics of many of our problem domains. To add in EM we incorporate Maxwell's equations into the fluid equations.
MHD is the study of the dynamics of electrically conducting fluids, this includes things like plasmas, liquid metals, and electrolytes, in the presence of magnetic and electric fields. To begin, we start off with Maxwell's equations.
For this section we will be using CGS units. This means Maxwells equation will look a little different than the SI unit counter part (dropping the various \(4\pi \epsilon_0\) and what not). As such the electric and magnetic fields all have the same units. \[\nabla \cdot E = 4\pi \rho_e \tag{Gauss}\] \[\nabla \times E = -\frac{1}{c}\frac{\partial B}{\partial t} \tag{Faraday}\] \[\nabla \cdot B = 0 \tag{no mag monopoles!}\] \[\nabla \times B = 4\pi J + \overbrace{\frac{1}{c}\frac{\partial E}{\partial t}}^{\text{displacement current added by Maxwell}} \tag{Ampere}\] We can also obtain the conservation laws. Consider hitting Gauss's law with the time derivative: \[\frac{\partial }{\partial t}\nabla \cdot E = 4\pi \frac{\partial }{\partial t}\rho_e \] \[\nabla \cdot \frac{\partial E}{\partial t} = 4\pi \frac{\partial }{\partial t}\rho_e \] Then we rearrange Ampere's law \[c\nabla \times B- c4\pi J = \frac{\partial E}{\partial t}\] Replace \[\nabla \cdot [c\nabla \times B- c4\pi J ] = 4\pi \frac{\partial }{\partial t}\rho_e \] \[\nabla \cdot \nabla \times B- c4\pi\nabla \cdot J ] = 4\pi \frac{\partial }{\partial t}\rho_e \] The left most term is zero divergence of a curl is zero \[\boxed{ \frac{\partial }{\partial t}\rho_e- \nabla \cdot J =0} \] We further see that if there is no current then we note the \[\nabla \times \nabla \times B = -\nabla^2 B +\nabla(\nabla\cdot B)\] Since there is no monopoles last term drops \[\nabla \times \nabla \times B = -\nabla^2 B \] But we also reconize that \[\nabla \times \nabla \times B = \frac{1}{c}\frac{\partial}{\partial t}\nabla \times E = -frac{1}{c^2}\frac{\partial^2 B}{\partial t^2} \] Combine the two we get \[\boxed{\nabla^2 B = \frac{1}{c^2}\frac{\partial^2 B}{\partial t^2}} \tag{wave eq.}\]
In high school we learned about Ohms law, Voltage and Current. Current density is \(J = \frac{I}{A}\). We also see that \(E \approx \frac{V}{L}\) since \(E = \nabla V\) and \(V = \frac{L}{A\sigma}\) where \(\sigma\) is the conductivity. Reistance is thus inverse of conductivity. We can then relate the two (current density and electric field) in the FLUID frame \[J' = \sigma E'\] We can transform into the LAB frame \[J = \sigma (E + \frac{v}{c}\times B)\] We can then see for a perfectly conducting fluid then \(\sigma \to \infty\) but \(E + \frac{v}{c}\times B \to 0\) to keep the current density finite. Which means that \(E \perp B\)! Okay but let us rearrange the equation \[E =\frac{J}{\sigma} -\frac{v}{\sigma c}\times B)\] We can replace \(J \) with \(\nabla \times B\) term. \[E = \frac{c}{4\pi \sigma}\nabla \times B-\frac{v}{\sigma c}\times B\] We can then plug this back into Faraday's equation \[\nabla \times E = -\frac{1}{c}\frac{\partial B}{\partial t}\] \[\nabla \times (\frac{c}{4\pi \sigma}\nabla \times B-\frac{v}{\sigma c}\times B) = -\frac{1}{c}\frac{\partial B}{\partial t}\] \[(\frac{c}{4\pi \sigma}\nabla^2 B-\frac{1}{\sigma c}\nabla \times v \times B) = -\frac{1}{c}\frac{\partial B}{\partial t}\] If we define \(\eta = \frac{c^2}{4\pi \sigma}\) \[\frac{\partial B}{\partial t} = \eta \nabla^2 B - \nabla \times (v\times B)\] We notice these two terms and we can define a magnetic Reynolds number! \[\mathcal{R_{mag}} = \frac{\nabla \times (v\times B)}{\eta \nabla^2 B } \approx \frac{vL}{\eta}\] When it is small then magnetic diffusion dominates. When the number is large then we are in the perfect conductor limit! Since \(\sigma \to \infty\) as \(\eta \sim \sigma^{-1}\).
Let us finally arrive at the full incorporated MHD equation into Navier Stokes. We need to get is the magnetic force density. Lorentz force gives us the following: \[F = q(E + \frac{v}{c}\times B)\] Density of the force is thus \[F = \rho_eE + \frac{\rho_ev}{c}\times B)\] \[F = \rho_eE + \frac{J}{c}\times B)\] We assume that the first time is super small and that the dominate force is the magnetic. \[F =\frac{J}{c}\times B)\] We know that through Amperes law we get \[J = \frac{c}{4\pi }\nabla \times B \] We plug this back into the force equation \[F =\frac{1}{4\pi }\nabla \times B \times B\] Hit this with vector calc identity \[F =\frac{1}{4\pi } (-\nabla^2 B + (B\cdot \nabla)B)\] Okie we can now add this to the Navier Stokes \[\boxed{\frac{\partial v}{\partial t} + v \cdot \nabla v = -\frac{1}{\rho}\nabla(P+ \frac{B^2}{8\pi})-\nabla \Phi + \frac{1}{4\pi \rho}(B \cdot \nabla)B + \nu (\nabla^2 v + \frac{1}{3}\nabla(\nabla \cdot v))}\] Note we split up the terms! We collected the gradient with the pressure term. And so we can reconize the \(B^2/8\pi\) as a magnetic pressure term. The other term is the magnetic tension, which is a restoring force of a bent field into straight lines. This gives us the chance to define a new \(\beta\)-Plasma parameter \[\boxed{\beta = \frac{P}{B^2/8\pi}}\] Which determines regimes of strong or weak B field dominace.
Let us solve some wavessss. Let us consider perturbations without thermal pressure and gravity \[\rho = \rho_0 + \rho_1 \quad B = B_0 +B_1 \quad v = v_1\] We can then compute the first order terms \[\frac{\partial \rho_1}{\partial t} + \rho_0 \nabla v_1 = 0\] \[\frac{\partial v_1}{\partial t} = - \frac{v_s^2}{\rho_0}\nabla \rho_1 + \frac{1}{4\pi \rho_0}(\nabla \times B_1) \times B_0\] We also have \[\frac{\partial B}{\partial t} = \nabla \times (v_1 \times B_0)\] We now apply the usual trick, we hit the conitunity equation with a gradient and then hit the momentum equation with the time derivative \[\frac{\partial \nabla\rho_1}{\partial t^2} + \rho_0 \nabla^2 v_1 = 0\] \[ \frac{\partial^2 v_1}{\partial t^2} = - \frac{v_s^2}{\rho_0} \frac{\partial}{\partial t}\nabla \rho_1 + \frac{1}{4\pi \rho_0} \frac{\partial}{\partial t}(\nabla \times B_1) \times B_0\] We sub in the continuity equation into this guy \[ \frac{\partial^2 v_1}{\partial t^2} = v_s^2 \nabla^2 v_1 + \frac{1}{4\pi \rho_0} \frac{\partial}{\partial t}(\nabla \times B_1) \times B_0\] We shove in Faraday. \[ \frac{\partial^2 v_1}{\partial t^2} = v_s^2 \nabla^2 v_1 + \frac{1}{4\pi \rho_0} (\nabla \times \frac{\partial}{\partial t}B_1) \times B_0\] \[ \frac{\partial^2 v_1}{\partial t^2} = v_s^2 \nabla^2 v_1 + \frac{1}{4\pi \rho_0} (\nabla \times (\nabla \times (v_1 \times B_0))) \times B_0\] This looks atrocious so we define Alfven velocity to make it look less icky \[\boxed{v_{A} = \frac{1}{\sqrt{4\pi \rho_e}}B_0}\] We can rewrite as the following: \[ \boxed{\frac{\partial^2 v_1}{\partial t^2} - v_s^2 \nabla^2 v_1 + v_A\times (\nabla \times (\nabla \times (v_1 \times v_A))) =0}\] Ok now that it is interms of just \(v_1\) we can hit this with a FT to get the dispersion relation \[\boxed{-w^2v_1 +(v_s^2+v_A^2)k(k\cdot v_1)+[(v_a\cdot k)v_1-(v_1\cdot k)v_A-(v_A\cdot v_1)k](k\cdot v_A)=0}\] This loooks dis-cos-tan so we look at special cases when things are parallel or perpendicular.
Consider we rotate all of this. In a cylinder because why not? In the corotation frame our velocities look like this \[v_R = v_R, \quad v_\phi = v_\phi -R \Omega \quad v_z = v_z\] Consider the field is vertical \(B_0 = B_0\hat z\). We can get the perturbations \[B = B_0\hat z + B_1\] We can get the dispersion relation when \(v_1 \perp k\) and \(k || B_0\) and \[\boxed{w^4 - w^2[\kappa +2 (kv_A)^2] + (kv_A)^2 [(kv_A^2) + \frac{\partial \Omega^2}{\partial \ln R}] = 0}\] Physics