A qualitative question about Blackbody Radiation

[Rishabh Narula]

would this be a correct understanding
of blackckbody radiation phenomena?
in particular the intensity versus
wavelength curve?
"A Blackbody consists of oscillators of
molecular dimensions.

Intensity is proportional to number of oscillators
with sufficient energy hv
emitting radiation,
and that number of oscillators
is proportional to e^-(hv/kT)=1/e^hv/kT.
now if v(frequency) increases,1/e^hv/kT decreases,and
thus number of oscillators decreases.
Hence there are more oscillators emitting
radiation for lesser values of v,i.e larger
values of lambda(wavelength).Hence more
radiations(more intensity) of shorter
wavelengths are observed since more oscillators
emit these radiations.

And thus at a given temperature T,Intensity
increases with Wavelength in the curve.

If we increase temperature T,value of
fraction 1/e^hv/kT increases and thus consequently
number of oscillators increase.thus even if
wavelengths of radiations are low,
at higher temperature their intensities
will still be higher compared to at
lower temperatures.

Hence at higher T even shorter wavelengths correspond
to high Intensities in the curve.

This explains the curve till maxima,
After the maxima I think the Intensity
falls with increasing wavelength since
now the range of radiations are having too
high a wavelength,the energy for these radiations
is too less,there are not many oscillators
vibrating at this less a frequency...hence
the drop in intensity (bit unsure about this
last para do tell me if wrong)."

-Source,mostly me explaining
some university chemistry study material to
myself.

 

[Charles Link]

You sort of have a handle on it, but not completely. Detailed calculations show that the density of oscillator modes per frequency interval is proportional to $\nu^2$. Meanwhile, the probability of a mode being occupied, [edit:correction=the mean occupancy of a given mode] is the Bose factor $\frac{1}{(e^{h \nu /(kT)}-1)}$.
(This is approximately the Boltzmann factor $e^{-h \nu /(kT)}$ for small $T$. For large $T$, the Bose factor is approximately $kT/(h \nu)$).
In addition, the photon energy is $h \nu$. With a couple more details, the complete Planck blackbody function can be derived.
See also https://www.physicsforums.com/threa...-density-of-a-black-body.956343/#post-6063569

 

[vanhees71]

The most simple way to describe black-body radiation is to think about it in terms as a gas of photons, i.e., the electromagnetic radiation in a finite-volume box described with finite-temperature QED. This leads immediately to the Planck Law of the phase-space distribution:
$$f(\vec{q})=\frac{2}{(2 \pi \hbar)^3} \frac{1}{\exp(\beta c q)-1}, \quad q=|\vec{q}|.$$
The relation to photon energy, frequency and wavelength is [EDIT: corrected factors of $c$; thanks to @Charles Link for pointing them out to me]
$$E=c q = \hbar \omega = 2 \pi \hbar c/\lambda.$$
Then you have to remember how distribution functions transform one one variable to another, i.e., you must keep $\mathrm{d}^3 q f(\vec{q})$ the same when changing the independent variable. E.g., you get the distribution with respect to energy or frequency using
$$\mathrm{d}^3 q=4 \pi q^2 \mathrm{d} q=\frac{4 \pi E^2}{c^3} \mathrm{d} E=\frac{4 \pi \hbar^3 \omega^2}{c^3} \mathrm{d} \omega,$$
and from this finally to wavelength
$$\mathrm{d} \omega=\frac{2 \pi c}{\lambda^2} \mathrm{d} \lambda \; \Rightarrow \; \mathrm{d}^3 q = \frac{32 \hbar^3 \pi^4}{\lambda^4} \mathrm{d} \lambda.$$
Multiplying these elements with the phase-space distribution factor $2/(2 \pi \hbar)^3 f_{\text{B}}$ gives the photon-number-phase-space density. Here
$$f_{\text{B}}=\frac{1}{\exp(\beta \hbar \omega)-1}=\frac{1}{\exp(\beta h \nu)-1}=\frac{1}{\exp(\beta c/\lambda)-1}=...$$
with $\beta=1/(k_{\text{B}} T)$.

 

[Rishabh Narula]

You sort of have a handle on it, but not completely. Detailed calculations show that the density of oscillator modes per frequency interval is proportional to $\nu^2$. Meanwhile, the probability of a mode being occupied, [edit:correction=the mean occupancy of a given mode] is the Bose factor $\frac{1}{(e^{h \nu /(kT)}-1)}$.
(This is approximately the Boltzmann factor $e^{-h \nu /(kT)}$ for small $T$. For large $T$, the Bose factor is approximately $kT/(h \nu)$).
In addition, the photon energy is $h \nu$. With a couple more details, the complete Planck blackbody function can be derived.
See also https://www.physicsforums.com/threa...-density-of-a-black-body.956343/#post-6063569

thanks for replying,will reply further when I know more. :3