4 | | psi − ln(psi) = 4 ln(xi) + 4/xi − 3, where psi = (v/v,,s,,)^2^ and xi = r/r,,s,,. |
| 3 | == Differential Equation == |
| 4 | |
| 5 | The Parker solution to the solar wind assumes a spherically symmetric radial outflow (with velocity u(r)). Then by conservation of momentum, |
| 6 | |
| 7 | {{{ |
| 8 | #!latex |
| 9 | $\rho u \frac{d u}{d r} = -\frac{d p}{d r} -\rho \frac{G M}{r^2}$ |
| 10 | }}} |
| 11 | |
| 12 | Continuity (conservation of particle #) gives |
| 13 | |
| 14 | {{{ |
| 15 | #!latex |
| 16 | $\frac{1}{r^2} \frac{d (r^2 \rho u)}{d r} = 0$ |
| 17 | }}} |
| 18 | |
| 19 | If we assume an isothermal corona, the ideal gas law is |
| 20 | |
| 21 | {{{ |
| 22 | #!latex |
| 23 | $p = \frac{2 \rho T}{m}$ |
| 24 | }}} |
| 25 | |
| 26 | Continuity shows that $r^2 \rho u = C$, where C is a constant. If we substitute for p and rho into the momentum equation, we get a differential equation in u and r: |
| 27 | |
| 28 | {{{ |
| 29 | #!latex |
| 30 | $\frac{1}{u} \frac{d u}{d r}(u^2 - \frac{2 T}{m_p}) = \frac{4 T}{m_p r} - \frac{G M}{r^2}$ |
| 31 | }}} |
| 32 | |
| 33 | Define $r_c = \frac{G M m_p}{4 T}$ so that $u_c^2 = \frac{2 T}{m_p}$ (i.e., sonic radius); substitute and rearrange to find |
| 34 | |
| 35 | {{{ |
| 36 | #!latex |
| 37 | $\frac{d u^2}{u_c^2} - \frac{d u^2}{u^2} = \frac{4 d r}{r} - \frac{4 r_c d r}{r^2}$ |
| 38 | }}} |
| 39 | |
| 40 | Integrate both sides to find the Parker wind equation, |
| 41 | |
| 42 | {{{ |
| 43 | #!latex |
| 44 | $\psi - \log(\psi) = 4 \log(\xi) + \frac{4}{\xi} + C$, where $\psi = (\frac{u}{u_c})^2$ and $\xi = \frac{r}{r_c}$ |
| 45 | }}} |
| 46 | |
| 47 | == Plotting == |
| 48 | Solving the Parker wind equation: |
| 49 | |
| 50 | {{{ |
| 51 | #!latex |
| 52 | $\psi - \log(\psi) = 4 \log(\xi) + \frac{4}{\xi} - 3$ |
| 53 | }}} |