| 1 | == Units in Astrobear == |
| 2 | |
| 3 | Astrobear uses something like rationalized electromagnetic units (extra factor of $\sqrt{4 \pi}$ in the electric and magnetic fields) - or Lorentz-Heaviside but scaling the $E$ field by $c$ and the charge density $\rho$ (and current $J$ ) by $\frac{1}{c}$ |
| 4 | |
| 5 | || $E = c E^{LH} = \frac{c }{\sqrt{4\pi}}E^{G}$ || |
| 6 | || $\rho = \frac{1}{c}\rho^{LH} = \frac{\sqrt{4 \pi} }{c}\rho^{G}$|| |
| 7 | || $J = \frac{1}{c} J^{LH} = \frac{\sqrt{4 \pi}}{c} J^{G}$|| |
| 8 | || $B = B^{LH} = \frac{1}{\sqrt{4\pi}} B^{G}$ || |
| 9 | |
| 10 | Using the approach in the appendix of Jackson, we have |
| 11 | || $k_1 = \frac{c^2}{4 \pi} $ || |
| 12 | || $k_2 = \frac{1}{4\pi}$ || |
| 13 | || $k_3 = 1$ || |
| 14 | || $\alpha = 1$ || |
| 15 | || $\mu_0 = 1$ || |
| 16 | || $\epsilon_0 = \frac{1}{c^2}$ || |
| 17 | This allows us to write Maxwell's equations as |
| 18 | |
| 19 | || $\nabla \cdot \mathbf{E} = c^2 \rho$ || |
| 20 | || $\nabla \times \mathbf{B} = \mathbf{J} + \frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t}$ || |
| 21 | || $\nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}$ || |
| 22 | || $\nabla \cdot \mathbf{B} = 0$ || |
| 23 | |
| 24 | as well as |
| 25 | || Lorentz Force Law || $\mathbf{F} = q \left ( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right)$ || |
| 26 | || Coulomb's Law || $\mathbf{F} = -\frac{c^2}{4 \pi} \frac{q_1 q_2}{r^2}\hat{\mathbf{r}}$ || |
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| 28 | |
| 29 | Most of the time we don't care about $E$, $\rho$, or $J$, however for the Hall MHD terms, we need to determine the electron charge in our system of units. |
| 30 | |
| 31 | For our system of equations, the elementary charge is |
| 32 | |
| 33 | || $e = 4.80320425 \times 10^{-10} \mbox{statC} = 1.70269157 \times 10^{-9} \mbox{hsu} = 5.67956774 \times 10^{-20} \left [\mbox{g}^{1/2} \mbox{cm}^{1/2} \right]$ |
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