f-f scaling with temperature? about fiasco HOT 4 CLOSED

I'm not sure exactly why these expressions are different, but I am extremely confident that the output of fiasco matches that of the IDL code: https://fiasco.readthedocs.io/en/latest/generated/gallery/idl_comparisons/continuum_comparison.html#sphx-glr-generated-gallery-idl-comparisons-continuum-comparison-py.

from fiasco.

Oh I know. It's because of the factor of $du$. You can rewrite it as $du=\hbar \mathrm{d}\omega / k_B T$. I've not checked but I assume with that extra factor of $\hbar/k_B$, the constant should work out to be the same as well.

from fiasco.

Aha! That must be it. Thank you!

I guess then I have to wonder about the physical interpretation. It should increase, no? I was under the impression that the collision rate should increase, so there should be more emission. Perhaps the actual cross-section decreases though? Curious.

from fiasco.

For posterity's sake, the $T^{-1/2}$ scaling is correct for the intensity differential in wavelength, but the total radiative loss (integrated over wavelength) does in fact scale as $T^{1/2}$.

The intensity equation
$I_{ff}(T, \lambda) \propto \frac{1}{\lambda^{2} T^{1/2}} \exp{(\frac{-hc}{\lambda k T})}$
which integrated in wavelength space is
$R_{ff}(T) \propto T^{1/2} \exp{(\frac{-hc}{\lambda k T})}$.

Much of my confusion stemmed from mixing up the two forms.

from fiasco.