Arbitrary Distribution functions

[konghz]

Hi,
I use the Arbitrary Distribution functions method to initialize the drift-ring-beam distribution of particles. Are there any other methods for initializing the drift-ring-beam distribution of particles?

drift-ring-beam distribution:

input.deck
begin:constant
malpha = 1836.04me
vp =malpha* 0.0431618 * c #the unique perpendicular speed
vpt=0.001vp #the perpendicular velocity spread of the fast ions
vparat=malpha1000 #the parallel velocity spread of the fast ions

vrange =3.0* vp
end:constant

begin:species
name =alpha
charge = 2.0
mass = 1836.0*4
npart = 100 * nx
number_density = 1.e16

#ring beam distribution in velocity space
dist_fn =exp(-(sqrt(px^(2)+py^(2))-vp)^(2)/vpt^2)*exp(-pz^2/vparat^2)

dist_fn_px_range = (-vrange, vrange)
dist_fn_py_range = (-vrange, vrange)
dist_fn_pz_range = (-3vparat, 3vparat)
end:species

[Status-Mirror]

I think this would be the easiest method for your distribution.

If you wanted to specify the start position, momentum and weight of each macro-particle, you could use the particles_from_file block, but this would be more involved than what you're currently doing.

Hope this helps,
Stuart

[konghz]

Thank you very much for your response! It has been very helpful to me. One more thing I need to confirm with you is whether the initialization distribution function does not include the normalization factor, like in the following form:

[Status-Mirror]

When sampling a generic distribution, EPOCH will randomly choose a $p_x$, $p_y$ and $p_z$ within the limits you give it. EPOCH then calculates your $f$ using these momenta, which it expects to be a number between 0 and 1. It then draws a uniformly distributed random number between 0 and 1, and if the number is lower than your $f$, it saves these momenta, otherwise the momenta are dismissed. The code then samples a new set of momenta and repeats the process.

Hence, EPOCH will generate momenta most efficiently when your maximum $f$ is 1. If your maximum $f$ was 0.5, then EPOCH would produce the same distribution, but it would take twice as long because half the samples would always be dismissed.

Hope this helps,
Stuart