0
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ Your title references the broadening of spectral lines and then you proceed to work on the Doppler shift of a wave in a medium (explicitly excluding light from consideration). What gives? $\endgroup$ – dmckee Aug 19 at 16:51
  • $\begingroup$ You state (in point 5) "this is not a Gaussian function of course". I disagree with you - it looks like a Gaussian centered about $E=\overline E$ to me. What am I missing? $\endgroup$ – Floris Aug 19 at 17:00
  • $\begingroup$ The function is not a gaussion function of $\overline{E}$. $\endgroup$ – Tamir Moshe Aug 19 at 17:05
  • $\begingroup$ I start by describing the Doppler shift before discussing the Doppler broadening beacuse I thought that maybe I have a problem with the equation of the Doppler shift I was using for the derivation. This notion was reinforced by the fact that I encountered equation (6) online as I described. $\endgroup$ – Tamir Moshe Aug 19 at 17:06
  • $\begingroup$ Spectral lines generally means light. So why apply the Doppler shift for waves in media? $\endgroup$ – dmckee Aug 19 at 17:39
0
$\begingroup$

Equation (6) makes more sense to me than equation (2). Yet, to first order in $v/c$ the two expressions are equivalent if opposite directions are taken for sign convention for $v$. Whatever the convention you choose, in the end what you need is an expression of $\bar{E}$ as a function of $E$ with first order in $v/c$, which is not what you have in (4).

$\endgroup$
  • $\begingroup$ "Equation (6) makes more sense to me than equation (2)." Here you do not explain why equation (6) may be correct and equation (2) may not so it is not so useful. It would be better if you could refer to an error in my derivation if you find one. In addition, I could not make sense of the rest of your answer, maybe you can try to clarify what did you mean and how it is related to my problem. $\endgroup$ – Tamir Moshe Aug 27 at 7:18
  • $\begingroup$ I meant that (6) looks like the familiar Doppler shift and it leads to the desired gaussian distribution, which makes sense. How could you obtain it? Take your equation (4), pass the term (1-v/c) to the left, then expand 1/(1-v/c) to first order in v/c, which is 1+ v/c. Now apart from the sign, which could correspond to a different convention for counting v positive (from source to observer, or the opposite) this leads to the expected formula. To sum up, I think you might have just missed a 1st order approximation in your derivation. Does it help? $\endgroup$ – user8736288 Aug 27 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.