Version 3 (modified by 9 years ago) ( diff ) | ,
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Parker Wind
Differential Equation
The Parker solution to the solar wind assumes a spherically symmetric radial outflow (with velocity u(r)). Then by conservation of momentum,
Continuity (conservation of particle #) gives
If we assume an isothermal corona, the ideal gas law is
Continuity shows that
, 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:Define
so that (i.e., sonic radius); substitute and rearrange to findIntegrate both sides to find the Parker wind equation,
Plotting
Solving the Parker wind equation:
There are 4 types of solutions - combinations of sub/supersonic at the surface and v → 0/vf 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.
First attempted to solve equation in Mathematica - solution appears correct up to the sonic radius, but the incorrect solution (v → 0) appears for r > rs. Next attempted to plot approximations for r << rs and r >> rs in Matlab, but as would be expected, the approximations fail near r = rs. Numerically solving the equation in Matlab gives the opposite case of Mathematica - correct solution for r > rs, but not for r < rs.
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 > rs, but not r < rs.
Pertinent line is: 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)))
Interpreted as:
if xx(i) ≥ 1
yy(i)=vpasolve(y - log(y) == -3 + 4*log(xx(i))+4/xx(i));
else
yy(i)=vpasolve(y - log(y) == exp(3-4*log(xx(i))-4/xx(i)));
end
Correct & incorrect plots:
Numerical Integration of DE
Write the differential equation for
as a parameterized system: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.
Change in Bow Shock with Magnetic Field
If sigma* and sigmap are equal, the bow shock radius is unchanged with or without magnetic field - ratio of radius to orbital separation, chibow = 0.240468. With sigma* = 1, sigmap = 0.1, chibow = 0.148204; sigma* = 0.5, sigmap = 0.1, chibow = 0.187300; sigma* = 0.1, sigmap = 0.5, chibow = 0.302483; sigma* = 0.1, sigmap = 1, chibow = 0.363674; and with sigma* = 0.5 and sigmap = 1, chibow = 0.297793 ≈ chiCoriolis.
Attachments (4)
- ParkerApprox.jpg (11.8 KB ) - added by 9 years ago.
- parkerwind.jpg (12.3 KB ) - added by 9 years ago.
- ParkerWrong.jpg (12.0 KB ) - added by 9 years ago.
- ParkerWrongApprox.jpg (11.6 KB ) - added by 9 years ago.
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