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Simplify docstrings

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Nuwan Yapa 2024-06-18 17:42:16 -04:00
parent 89ef87d4fe
commit ea69624e31
1 changed files with 11 additions and 18 deletions
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17
dirac.jl
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@ -9,21 +9,18 @@ M_p = 938.2720813 # Proton mass in MeV/c2
κ is the generalized angular momentum,
M is the mass in MeV/c2,
E in the energy in MeV,
Φ0(r) & W0(r) are functions corresponding to scalar and vector potentials in MeV,
Φ0, W0 are the mean-field potentials (couplings included) in MeV as functions of r in fm,
r is the radius in fm.
Reference: P. Giuliani, K. Godbey, E. Bonilla, F. Viens, and J. Piekarewicz, Frontiers in Physics 10, (2023)."
Reference: P. Giuliani, K. Godbey, E. Bonilla, F. Viens, and J. Piekarewicz, Frontiers in Physics 10, (2023)"
function dirac!(du, (g, f), (κ, M, E, Φ0, W0), r)
du[1] = -(κ/r) * g + (E + M - Φ0(r) - W0(r)) * f / ħc
du[2] = (κ/r) * f - (E - M + Φ0(r) - W0(r)) * g / ħc
end
"Solve the Dirac equation and return g(r=r_max) for given scalar and vector potentials where
κ is the generalized angular momentum,
M is the mass in MeV/c2,
E in the energy in MeV,
Φ0(r) & W0(r) are functions corresponding to scalar and vector potentials in MeV,
r_max is the outer boundary in fm,
r_min (=r_max/1000) is inside boundary in fm which cannot be 0 due to the centrifugal term."
r_min (=r_max/1000) is inside boundary in fm which cannot be 0 due to the centrifugal term,
the other parameters are the same from dirac!(...)."
function boundaryValue(κ, M, E, Φ0, W0, r_max, r_min=r_max/1000)
prob = ODEProblem(dirac!, [0, 1], (r_min, r_max))
sol = solve(prob, RK4(), p=(κ, M, E, Φ0, W0))
@ -31,11 +28,7 @@ function boundaryValue(κ, M, E, Φ0, W0, r_max, r_min=r_max/1000)
end
"Find all bound energies between E_min (=0) and E_max (=M) where
κ is the generalized angular momentum,
M is the mass in MeV/c2,
Φ0(r) & W0(r) are functions corresponding to scalar and vector potentials in MeV,
r_max is the outer boundary in fm,
r_min (=r_max/1000) is inside boundary in fm which cannot be 0 due to the centrifugal term."
the other parameters are the same from dirac!(...)."
function findEs(κ, M, Φ0, W0, r_max, r_min=r_max/1000, E_min=0, E_max=M)
f(E) = boundaryValue(κ, M, E, Φ0, W0, r_max, r_min)
return find_zeros(f, (E_min, E_max))