I have a problem with the derivation of the gaussian behavior of the doppler broadening of gas atoms specrtal lines. I will describe what I have done so the problem along the derivation could be referred.
Derivation of the Doppler shift of frequency for moving emitter and static observer:
Let us consider an emitter of a wave which has a velocety component of $v$, in the wave's propagation axis, relative to an observer. The relation between the period between the emission of two successive peaks of the wave in the emitter system, $t$, and the period between the arrival of those two peaks in the observer system, $\overline{t}$, is given by,
$$\overline{t}=t(1-\frac{v}{c}), \tag{1}$$
where $c$ is the wave velocity relative to the wave's medium if we do not consider light. In the presence of medium our derivation is for an observer fixed relative to the medium. From equation $(1)$ we get that the relation between the frequency of the wave in the emitter system, $f$, and the frequency of the wave in the observer system, $\overline{f}$, is,
$$f=\overline{f}(1-\frac{v}{c}).\tag{2}$$
Derivation of the Doppler broadening for emitting free atoms in gas:
For gas of free atoms of mass $m$ with temperature $T$, the probability density of a component of the atom's velocity in some axis, $v_i$, is given by,
$$\rho_{v_i}(v_i)=\sqrt{\frac{m}{2\pi k_bT}}e^{-\frac{m}{2k_bT}v_i^2}.\tag{3}$$
If the gas atoms are excited to the same state they will emit photons of the same energy relative to their own systems. From equation $(2)$ we get that the relation between the energy of a photon in the system of an emitter atom, $E$, and the energy of the photon measured by a fixed observer (relative to the observer the atoms velocities are measured), $\overline{E}$, is,
$$E=\overline{E}(1-\frac{v}{c}).\tag{4}$$
From equations $(3)$ and $(4)$ I got the following probability density for the emitted photons energy in the observer system:
$$\rho_\overline{E}(\overline{E})=\sqrt{\frac{m}{2\pi k_bT}}e^{-\frac{mc^2}{2k_bT}(1-\frac{E}{\overline{E}})^2}.\tag{5}$$
This is not a gaussian function of course, but again I am not sure where is my mistake. I saw in some websites a different version of equation $(2)$ of my derivation for the same described system of the form,
$$\overline{f}=f(1-\frac{v}{c}),\tag{6}$$
which will result with a gaussian probability distribution of the emitted photons energy in the observer system, but it contradicts the derivation of equation (2).
