Ideal Microscopic Magnetohydrodynamics
$\mu_0 = $ permeability of free space, in $[\frac{kg \cdot m}{C^2}]$
$\epsilon_0 = $ permittivity of free space, in $[\frac{C^2 \cdot s^2}{kg \cdot m^3}]$
$E = $ electrical field, in $[\frac{kg \cdot m}{C \cdot s^2}]$
$B = $ magnetic field, in $[\frac{kg}{C \cdot s}]$
$J = $ electrical current, per volume, in $[\frac{C}{m^2 \cdot s}]$
Microscopic Maxwell Equations (Tells how a charge generates a field around it):
$\vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0}$
$\vec{\nabla} \cdot \vec{B} = 0$
$\vec{\nabla} \times \vec{E} = -\partial_t \vec{B}$
$\vec{\nabla} \times \vec{B} = \mu_0 ( \vec{J} + \epsilon_0 \partial_t \vec{E})$
Ampere's law of the Microscopic Maxwell Equations:
$\vec{\nabla} \times \vec{B} = \mu_0 (\vec{J} + \epsilon_0 \partial_t \vec{E})$, in $[\frac{kg}{C \cdot m \cdot s}]$
${\epsilon^{ij}}_k \nabla_j B^k = \mu_0 (J^i + \epsilon_0 \partial_t E^i)$, in $[\frac{kg}{C \cdot m \cdot s}]$
${\epsilon^{ij}}_k \nabla_j B^k = \mu_0 J^i + \mu_0 \epsilon_0 \partial_t E^i$, in $[\frac{kg}{C \cdot m \cdot s}]$
...replace $\mu_0 \epsilon_0 = \frac{1}{c^2}$
${\epsilon^{ij}}_k \nabla_j B^k = \mu_0 J^i + \frac{1}{c^2} \partial_t E^i$, in $[\frac{kg}{C \cdot m \cdot s}]$
Assume that our electric field is a steady-state, i.e. $\partial_t E^i = 0$
$J^i = \frac{1}{\mu_0} {\epsilon^{ij}}_k \nabla_j B^k$, in $[\frac{C}{m^2 \cdot s}]$
$q = $ electric charge, in $[C]$
$\sigma$ = electrical conductivity, in $[\frac{C^2 \cdot s^2}{kg \cdot m^3 \cdot s}]$
Lorentz force law (Tells how a field influences a charge):
$\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})$, in $[\frac{kg \cdot m}{s^2}]$
$F^i = q (E^i + {\epsilon^i}_{jk} v^j B^k)$, in $[\frac{kg \cdot m}{s^2}]$
$E^i = \frac{1}{q} F^i - {\epsilon^i}_{jk} v^j B^k$, in $[\frac{kg \cdot m}{C \cdot s^2}]$
Let $J^i / \sigma = F^i / q$, in $[\frac{kg \cdot m}{C \cdot s^2}]$ (TODO explain why)
$E^i = J^i / \sigma - {\epsilon^i}_{jk} v^j B^k$, in $[\frac{kg \cdot m}{C \cdot s^2}]$
Maxwell-Faraday equation of the Microscopic Maxwell Equations:
$\partial_t \vec{B} = -\vec{\nabla} \times \vec{E}$
$\partial_t B^i = -{\epsilon^{ij}}_k \nabla_j E^k$
(TODO flip valence of everything below here, to make it match my other EMHD flux eigensystem worksheets ... and double check or just use those).
Magnetic Field
substitute Ampere's law into Lorentz force law:
$E^i = \frac{1}{\sigma\mu_0} {\epsilon^{ij}}_k \nabla_j B^k - {\epsilon^i}_{jk} v^j B^k$
$\eta = \frac{1}{\sigma\mu_0} = $ magnetic diffusivity
$E^i = \eta {\epsilon^{ij}}_k \nabla_j B^k - {\epsilon^i}_{jk} v^j B^k$
substitute this into Maxwell-Faraday equation:
$ \partial_t B_i = -{\epsilon_i}^{jk} \nabla_j (\eta {\epsilon_k}^{lm} \partial_l B_m - {\epsilon_k}^{lm} v_l B_m)$
$ \partial_t B_i = {\epsilon_i}^{jk} {\epsilon_k}^{lm} \nabla_j (v_l B_m) - {\epsilon_i}^{jk} {\epsilon_k}^{lm} \nabla_j (\eta \partial_l B_m)$
Levi-Civita identity: ${\epsilon_i}^{jk} {\epsilon_k}^{lm} = \delta^{ij}_{kl} = \delta^i_k \delta^j_l - \delta^i_l \delta^j_k$
$ \partial_t B_i = (\delta^i_l \delta^j_m - \delta^i_m \delta^j_l) \nabla_j (v_l B_m) - (\delta^i_l \delta^j_m - \delta^i_m \delta^j_l) \nabla_j (\eta \partial_l B_m)$
assume $\eta$ is constant
$ \partial_t B_i = \nabla_j (v_i B_j - v_j B_i) - \eta (\nabla_i \nabla_j B_j - \nabla_j \nabla_j B_i)$
No magnetic monopoles: $\nabla_i B_i = 0$
$ \partial_t B_i = \nabla_j (v_i B_j - v_j B_i) + \eta \nabla_j \nabla_j B_i$
assume $\eta = 0$ (Why, after all this work of assuming it is constant? Why not eliminate it sooner?)
$ \partial_t B_i = \nabla_j (v_i B_j - v_j B_i)$
$ \partial_t B_i + \nabla_j (B_i v_j - B_j v_i) = 0$
Momentum
Lorentz force law:
$F_i = q (E_i + {\epsilon_i}^{jk} v_j B_k)$
assume $E_i = 0$ (why? assuming $\partial_t E = 0$ makes sense for steady-state, but why assume the whole of the field isn't present?)
$\Pi^M_i = F_i =$ Magnetic force on fluid
$\Pi^M_i = {\epsilon_i}^{jk} q v_j B_k$
let $J_i = q v_i$
$\Pi^M_i = {\epsilon_i}^{jk} J_j B_k$
substitute Ampere's law approximation: $J_i = \frac{1}{\mu_0} {\epsilon_i}^{jk} \nabla_j B_k$
$\Pi^M_i = {\epsilon_i}^{jk} (\frac{1}{\mu_0} {\epsilon_j}^{lm} \partial_l B_m) B_k$
$\Pi^M_i = -\frac{1}{\mu_0} {\epsilon_i}^{kj} {\epsilon^{lm}}_j B_k \partial_l B_m$
$\Pi^M_i = \frac{1}{\mu_0} (\delta^i_m \delta^k_l - \delta^i_l \delta^k_m) B_k \partial_l B_m$
$\Pi^M_i = \frac{1}{\mu_0} (B_j \nabla_j B_i - B_j \nabla_i B_j)$
insert additional $\nabla_k B_k = 0$ terms, split $B_j \nabla_i B_j$ into two halves
$\Pi^M_i = \frac{1}{\mu_0} ((\nabla_j B_i) B_j + B_i (\nabla_j B_j) - B_i - \frac{1}{2} (\nabla_i B_j) B_j - \frac{1}{2} B_j (\nabla_i B_j))$
un-distribute gradient:
$\Pi^M_i = \frac{1}{\mu_0} (\nabla_j (B_i B_j) - \frac{1}{2} \nabla_i (B^2))$
$\Pi^M_i = \nabla_j (\frac{1}{\mu_0} (B_i B_j - \frac{1}{2} g^{ij} B_i B_j))$
$\Pi^M_i = \nabla_j \sigma_{ij}$
$\sigma_{ij} = \frac{1}{\mu_0} (B_i B_j - \frac{1}{2} g^{ij} B_i B_j) =$ Maxwell stress tensor for $E_i = 0$
Change in momentum due to magnetic field:
$\Pi^M_i = -\frac{1}{2} \frac{1}{\mu_0} \nabla_j B^2 + \frac{1}{\mu_0} \nabla_j (B_i B_j)$
Total change in momemtum:
$\partial_t \Pi_i = \partial_t \Pi^C_i + \partial_t \Pi^P_i + \partial_t \Pi^M_i$
$\partial_t \Pi_i = -\nabla_j (\rho v_i v_j) - \nabla_i P - \frac{1}{\mu_0} \frac{1}{2} \nabla_i B^2 + \frac{1}{\mu_0} \nabla_j (B_i B_j)$
$\partial_t \Pi_i + \nabla_j (\rho v_i v_j + g^{ij} P + \frac{1}{\mu_0} \frac{1}{2} g^{ij} B^2 - \frac{1}{\mu_0} B_i B_j) = 0$
TODO maybe don't split off pressure so soon, instead maybe start with "equation of state" based on magnetic fluids: $E_{total} = \frac{1}{\gamma-1} P + \frac{1}{2} \rho v^2 + \frac{1}{2} B^2$ and go from there? Since EOS closes the equations.
Let $P_{mag} = \frac{1}{\mu_0} \frac{1}{2} B^2$
$\partial_t \Pi_i + \nabla_j (\rho v_i v_j + g^{ij} (P + P_{mag}) - \frac{1}{\mu_0} B_i B_j) = 0$
Let $P_{total} = P + P_{mag}$
$\partial_t \Pi_i + \nabla_j (\rho v_i v_j + g^{ij} P_{total} - \frac{1}{\mu_0} B_i B_j) = 0$
For $\Pi_i = $ total momentum $ = \rho v_i$ we find
$\partial_t (\rho v_i) + \nabla_j (\rho v_i v_j + g^{ij} P_{total} - \frac{1}{\mu_0} B_i B_j) = 0$
Energy
Change in energy due to magnetic field (could use some better explaining):
$\int_\Omega \partial_t E_{total}^M dV = -\int_{\partial\Omega} B_k (v_k B_j - v_j B_k) dS_j$ TODO where does this come from? I'm guessing $\partial_t B_i$
$\int_\Omega \partial_t E_{total}^M dV = \int_\Omega \nabla_j (B_k (v_k B_j - v_j B_k)) dV$ via Green's theorem
$\partial_t E_{total}^M = \nabla_j (B_k (v_k B_j - v_j B_k))$ at any point
Total change in total energy:
$\partial_t E_{total} = \partial_t E_{total}^C + \partial_t E_{total}^P + \partial_t E_{total}^M$
$\partial_t E_{total} = -\nabla_j (E_{hydro} v_j) - \nabla_j (P v_j) - \nabla_j \frac{1}{\mu_0} (v_j B^2 - B_k v_k B_j)$
$\partial_t E_{total} + \nabla_j ((E_{hydro} + P) v_j + \frac{1}{\mu_0} (v_j B^2 - B_k v_k B_j) = 0$
$\partial_t E_{total} + \nabla_j ((E_{hydro} + P) v_j + \frac{1}{\mu_0} B_k (v_j B_k - v_k B_j)) = 0$
$\partial_t E_{total} + \nabla_j ((E_{hydro} + P) v_j + \frac{1}{\mu_0} B^2 v_j - \frac{1}{\mu_0} B_k v_k B_j) = 0$
$\partial_t E_{total} + \nabla_j ((E_{hydro} + \frac{1}{2} \frac{1}{\mu_0} B^2 + P + \frac{1}{2} \frac{1}{\mu_0} B^2) v_j - \frac{1}{\mu_0} B_k v_k B_j) = 0$
Let $E_{mag} = \frac{1}{2} \frac{1}{\mu_0} B^2$
$\partial_t E_{total} + \nabla_j ((E_{hydro} + E_{mag} + P + P_{mag}) v_j - \frac{1}{\mu_0} B_k v_k B_j) = 0$
Let $E_{total} = E_{hydro} + E_{mag}$
Let $H_{total} = E_{total} + P_{total}$
$\partial_t E_{total} + \nabla_j (H_{total} v_j - \frac{1}{\mu_0} B_k v_k B_j) = 0$