Changes between Version 3 and Version 4 of u/adebrech/Matlab


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Timestamp:
09/08/17 10:49:07 (7 years ago)
Author:
adebrech
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  • u/adebrech/Matlab

    v3 v4  
    1 = Parker Wind =
    2 
    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 }}}
    54 
    55 There are 4 types of solutions - combinations of sub/supersonic at the surface and v -> 0/v,,f,, as xi -> infinity. The only physical solution is subsonic at the surface and has a nonzero velocity infinitely far away, and passes through xi = 1, psi = 1.
    56 
    57 First attempted to solve equation in Mathematica - solution appears correct up to the sonic radius, but the incorrect solution (v -> 0) appears for r > r,,s,,. Next attempted to plot approximations for r << r,,s,, and r >> r,,s,, in Matlab, but as would be expected, the approximations fail near r = r,,s,,. Numerically solving the equation in Matlab gives the opposite case of Mathematica - correct solution for r > r,,s,,, but not for r < r,,s,,.
    58 
    59 [[Image(ParkerApprox.jpg)]]
    60 
    61 [[Image(ParkerWrong.jpg)]]
    62 
    63 Looking at Jonathan's code for calculating the Parker wind to find the correct approximation, try to implement it in a different way to test understanding. Works for r > r,,s,,, but not r < r,,s,,.
    64 
    65 Pertinent line is:
    66 yy(i)=vpasolve(y - log(y) == -3 + 4*log(xx(i))+4/xx(i),y,(xx(i) > 1) * xx(i) + (xx(i) <= 1)*exp(3-4*log(xx(i))-4/xx(i)))
    67 
    68 Interpreted as:
    69 
    70     if xx(i) >= 1
    71         yy(i)=vpasolve(y - log(y) == -3 + 4*log(xx(i))+4/xx(i));
    72     else
    73         yy(i)=vpasolve(y - log(y) == exp(3-4*log(xx(i))-4/xx(i)));
    74     end
    75 
    76 Correct & incorrect plots:
    77 
    78 [[Image(parkerwind.jpg)]]
    79 
    80 [[Image(ParkerWrongApprox.jpg)]]
    81 
    82 == Numerical Integration of DE ==
    83 
    84 Write the differential equation for $\frac{d u}{d r}$ as a parameterized system:
    85 
    86 {{{
    87 #!latex
    88 $\frac{d u}{d s} = -2uc^2(r-\frac{GM}{2c^2})$
    89 
    90 $\frac{d r}{d s} = -r^2(u^2-c^2)$
    91 }}}
    92 
    93 Calculate the Jacobian and evaluate at the critical point to linearize the system, then move a small distance along the stable eigenvector (i.e. tangent to the stable manifold, the transonic solution of interest). Integrate from these points out.
    94 
    95 $Jacobian = \begin{bmatrix} 0 & 2c^3 \\ \frac{GM}{2c^3} & 0 \end{bmatrix}$
    96 
    971= Change in Bow Shock with Magnetic Field =
    982If sigma,,*,, and sigma,,p,, are equal, the bow shock radius is unchanged with or without magnetic field - ratio of radius to orbital separation, chi,,bow,, = 0.240468. With sigma,,*,, = 1, sigma,,p,, = 0.1, chi,,bow,, = 0.148204; sigma,,*,, = 0.5, sigma,,p,, = 0.1, chi,,bow,, = 0.187300; sigma,,*,, = 0.1, sigma,,p,, = 0.5, chi,,bow,, = 0.302483; sigma,,*,, = 0.1, sigma,,p,, = 1, chi,,bow,, = 0.363674; and with sigma,,*,, = 0.5 and sigma,,p,, = 1, chi,,bow,, = 0.297793 ≈ chi,,Coriolis,,.