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A more detailed description of the montecarlo process #310
A more detailed description of the montecarlo process #310
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Hello: Mostly ok - but we don’t (shouldn’t at least!) assume emission is “isotropic in angle for a half-sphere” - we should be using mu = sqrt(z) [as in Abbott & Lucy 1985], which is not isotropic. Typically, packets do work on the ejecta and so we expect on average that they lose energy during interactions. |
@ssim well interestingly enough the escaping packets seem to gain energy. |
No - isotropic would not be sqrt! Isotropic would be mu = z (for positive hemisphere). Overall, work should be done on the ejecta, but it may be that that mainly shows up in the backscattered packets. Also may depend on which frame you calculate the energies in. |
@ssim I just changed the wording. Let me know if we are ready to merge this. |
A more detailed description of the montecarlo process
After talking to Frederik this morning I whipped this up to give a more detailed description of the montecarlo process to the users. @ssim please confirm that this is correct
@fredRos