I am now reading the classic paper by Dicke,

The Effect of Collisions upon the Doppler Width of Spectral Lines

At the very beginning of the paper, Dicke said ''Quantum mechanically, the Doppler effect results from the recoil momentum given to the radiating system by the emitted photon. This recoil momentum implies a change in the kinetic energy of the radiating atom which is in turn mirrored by a corresponding change in the photon's energy. This change in the photon's energy is proportional to the component of the atom's velocity in the direction of emission of the photon and leads to the normal expression for the Doppler effect''

He cited Fermi's RMP

Quantum Theory of Radiation

Indeed, this sounds reasonable. But it would mean the mass of the atom will enter the expression of the energy shift of the photon. But actually, the Doppler shift formula is

$$f = f_0 \left(1+\frac{v}{c} \right ), $$

where the mass of the particle does not appear.

So, what does Fermi or Dicke really mean?


The mass of the isolated emitting enters into the recoil Doppler shift because the atom and the photon must conserve momentum; however, it's a higher-order effect and can be neglected for optical transitions in atoms.

Let's suppose we have an atom with mass $m$ and some excited states with energies $E_2 > E_1$. We'll assume that, like all atomic transitions, this one is nonrelativistic with $E_{2,1} \ll mc^2$. If the atom begins and ends at rest, a transition between these states is associated with a photon with frequency $hf_0 = E_2-E_1$. However if that atom is moving with velocity $\vec v_{2,1}$ before and after the photon interaction, the actual frequency of the emitted photon is \begin{align} hf &= \left(E_2 + \frac12 m v_2^2\right) - \left(E_1 + \frac12 m v_1^2\right) \\ &= hf_0 + \frac12 m \left( v_2^2 - v_1^2 \right). \tag A \end{align} From conservation of momentum, the final momentum of the atom is \begin{align} m\vec v_1 &= m\vec v_2 - h\vec f/c, \end{align} where the vector sign on the frequency reminds us that the photon has some associated direction. Squaring both sides gives \begin{align} (m v_1)^2 &= (m v_2)^2 + \left(\frac{hf}c\right)^2 - 2mv_2 \frac{hf}c \cos\theta \\ 2m\left(hf_0 - hf\right) &= \left(\frac{hf}c\right)^2 - 2mv_2 \frac{hf}c \cos\theta \tag B \\ \frac{f_0}f - 1 &= \frac{hf}{2mc^2} - \frac {v_2}c \cos\theta \end{align} In the line (B) we've used the energy relationship (A). If we compare this to your traditional Doppler transformation $$ \frac f{f_0} = 1 + \frac vc\cos\theta $$ for light emitted from a source with initial velocity $v_2 = v$, we see they are equivalent in the non-relativistic limit $v\ll c$, apart from a term proportional to $hf/mc^2$. However, we explicitly assumed this term was small at the start of the problem, when we used non-relativistic kinetic energies for our atoms! For visible-light transitions in hydrogen its value is roughly $\rm 1\,eV / 1\,GeV = 10^{-9}$, which suggests that that the atomic mass matters in recoil-induced Doppler shifts at the part-per-billion level.

Dicke's paper is about Doppler broadening due to random thermal motion of atoms in a hot, dense gas. His argument seems to be reminiscent of the Mössbauer effect in nuclear physics.

  • $\begingroup$ Rob, on squaring $m\vec v_2=m\vec v_1+h\vec f/c$, why have you written a $\color{red}{-}2mv_1 \frac{hf}{c}\cos\theta$ term instead of a $\color{green}{+}2mv_1 \frac{hf}{c}\cos\theta$? Is this something to do with the vector nature of the frequency of emitted photon? $\endgroup$ – Vishnu Mar 31 '20 at 7:25
  • $\begingroup$ @GuruVishnu It was a labeling error, not a sign error. A pox on four-years-ago-rob for using "$v_2$" to label an initial velocity. I've clarified. $\endgroup$ – rob Mar 31 '20 at 18:20
  • $\begingroup$ @GuruVishnu Your edit changed my meaning, so I reverted it. The two expressions for the Doppler shift are equivalent using the binomial approximation, where $(1+\epsilon)^n \to( 1+n\epsilon)$ in the limit $|n\epsilon| \ll 1$. And by "proportional to" I meant to emphasize the units in the negligible term. $\endgroup$ – rob Apr 1 '20 at 5:18
  • $\begingroup$ Fine. I'm sorry for that. I made that change as it resembled with the expression attained after applying momentum and energy conservation. I felt the problem with this approximation is as $v\to c$, $f\to2f_0$ instead $f\to\infty$ in reality as the wavelength in the direction of motion is heavily compressed $\lambda\to0$. $\endgroup$ – Vishnu Apr 1 '20 at 5:22
  • $\begingroup$ @GuruVishnu Right, but in the limit $v\to c$ the expression (A) is wrong and you have to do the whole argument differently. $\endgroup$ – rob Apr 1 '20 at 5:26

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.