Correct the wavelength distributions
After some discussions with @jingli and @schwab, we realized that the peak of the wavelength distribution \phi(\lambda)
was not at the same position as the energy distribution \phi(E)
. This was because, after the change of variables, the corresponding jacobian was missing. This was properly implemented for the angular distribution.
Consider that |\phi(E) dE| = |\phi(\lambda) d\lambda|
, and \lambda [\textup{~\AA}] = \sqrt{81.82/E [\mathrm{meV}]}
, it is possible to see that dE = |81.82 \cdot (-2) \cdot \lambda^{-3}| d\lambda
.
So, for each bin between \lambda_{\mathrm{min}}
and \lambda_{\mathrm{max}}
, it would be necessary to divide by the integral of the jacobian, i.e. |81.82\cdot(\lambda^{-2}_{\mathrm{min}}-\lambda^{-2}_{\mathrm{max}})|
.
For doing this, a new WavelengthAxis
class was implemented to plot the distributions with the proper normalization (same idea as AngleAxis
). Also, the tracks-vs-tallies.inp
example was changed to see the new shape of the wavelength distributions (see attached).
And, with these implementations, now the peak of both distributions is at the same position