I want to convert Maxwell-Boltzmann speed distribution to that of frequency distribution. How should I do it?
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2$\begingroup$ What kind of system and frequencies are you talking about? Are you talking about frequencies of energy eigenstates, in which case the distribution would simply be $Z^{-1}\,g_i\,\exp(-\hbar\,\omega_i/(k\,T)$ where $g_i$ is the number of degenerate states of energy $E_i=\hbar\,\omega_i$ and $Z$ the normalizing partition function. $\endgroup$– Selene RoutleyCommented Jul 4, 2016 at 6:00
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$\begingroup$ @WetSavannaAnimalakaRodVance aka Rod Vance I want to obtain the frequency distribution of a bath of gaseous particles. Maxwell-Boltzmann distribution is a classical distribution. Is it right to replace energy with hw? $\endgroup$– New DeveloperCommented Jul 4, 2016 at 6:10
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1$\begingroup$ the Boltzmann distribution is certainly valid here. However, I find "frequency" confusing - the only thing I can think of is frequency of energy eigenstates. Or am I wrong? What other frequencies are there? $\endgroup$– Selene RoutleyCommented Jul 4, 2016 at 6:34
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$\begingroup$ Yes. It is the frequency of energy eigenstates. We have speed distribution (see equation 6 link) or energy distribution (see equation 9 link). But I need the frequency distribution. $\endgroup$– New DeveloperCommented Jul 4, 2016 at 6:42
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$\begingroup$ Do you by any chance mean the frequency at which particles hit an area element? $\endgroup$– CuriousOneCommented Jul 4, 2016 at 7:11
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The Boltzmann distribution is the main and fundamental result you use here; see the Wikipedia article on the Microcanonical Ensemble. This is:
$$p(E) = \frac{1}{\mathcal{Z}}\,g(E)\,\exp\left(-\frac{E}{k\,T}\right)\tag{1}$$
where $g(E)$ is the degeneracy factor, or the number of distinct quantum states with energy $E=\hbar\,\omega$ and $\mathcal{Z} = \int_E g(E)\exp\left(-\frac{E}{k\,T}\right)\,\mathrm{d} E$ is the normalizing factor ("partition function") needed to make the total probability sum to unity. The frequency distribution will simply depend on the allowed energies and the number of allowed states at each allowed energy $E_i$.
The Maxwell-Boltzmann speed distribution simply follows from this by setting $E_i = \frac{1}{2} \,m\,v_i^2$ and then using $g_i=4\,\pi\,v^2\,\mathrm{d}\,v$ as being proportional to the number of particles in velocity band $[v,\,v+\mathrm{d} v)$. Then you add the normalization factors.
If we have free particles (no position dependence of potential energy) then $E_i = \hbar\,\omega_i$ can replace the $\frac{1}{2} \,m\,v_i^2$. The velocity interval $[v,\,v+\mathrm{d} v)$ corresponds to an energy interval $[E,\,E+\mathrm{d}E)$ where $\mathrm{d}E/\sqrt{E}\propto\mathrm{d}v$. So the number of particles in the energy band is proportional to $g=\sqrt{E}\,\mathrm{d}E$. Energy being proportional to frequency, we have the following expression for the frequency probability distribution:
$$p(\omega) = \frac{1}{\mathcal{Z}}\,\sqrt{\omega}\,\exp\left(-\frac{\hbar\,\omega}{k\,T}\right)\tag{2}$$
where:
$$\mathcal{Z} = \int_0^\infty \,\sqrt{u}\,\exp\left(-\frac{\hbar\,u}{k\,T}\right)\,\mathrm{d}u\tag{3}$$
answered Jul 4, 2016 at 7:34
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1$\begingroup$ The $\mathcal{Z} $ is normally called the partition function or 'sum over states' . $\endgroup$ Commented Jul 4, 2016 at 8:31
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$\begingroup$ @porphyrin Thanks. Could have sworn I put its name in somewhere, but apparently left it out. $\endgroup$ Commented Jul 4, 2016 at 8:33