Kvant Physics Problem 566
An atom moves in the laboratory frame with velocity $v$, where
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Problem
F566. An atom moving with velocity $v$ ($v \ll c$) emits a photon at a small angle $\alpha$ to the direction of its motion. Prove that if $\omega_0$ is the radiation frequency of the atom at rest, and $\omega$ is the frequency of the photon's wave, then for visible light
$$\frac{\omega_0-\omega}{\omega_0}\approx \frac{v}{c}\cos\alpha$$
(the Doppler formula).
Setup and Assumptions
An atom moves in the laboratory frame with velocity $v$, where
$v \ll c,$
and emits a photon making a small angle $\alpha$ with the direction of the atom's motion. The frequency of the radiation emitted by the same atomic transition when the atom is at rest is denoted by $\omega_0$ (s$^{-1}$). The frequency measured in the laboratory frame is denoted by $\omega$ (s$^{-1}$).
The required quantity is the relative frequency shift
$\frac{\omega_0-\omega}{\omega_0}.$
The atom is treated as a free particle before and after emission. Conservation of energy and conservation of momentum are applied in the laboratory frame. Since the problem concerns visible light, the photon energy is much smaller than the rest energy of the atom,
$\hbar\omega_0 \ll Mc^2,$
where $M$ is the atomic mass. The atomic velocity is nonrelativistic, so terms of order $(v/c)^2$ and higher are neglected.
Physical Principles
The solution rests on conservation of momentum and conservation of energy.
For photon emission, momentum conservation gives
$\mathbf p_i=\mathbf p_f+\mathbf k,$
where $\mathbf p_i$ and $\mathbf p_f$ are the atomic momenta before and after emission, and
$|\mathbf k|=\frac{\hbar\omega}{c}$
is the photon momentum.
Energy conservation gives
$E_i=E_f+\hbar\omega.$
For nonrelativistic motion of the atom,
$E=\frac{p^2}{2M}+E_{\rm int},$
where $E_{\rm int}$ is the internal atomic energy.
The emitted atomic transition has energy
$\Delta E=\hbar\omega_0,$
which is the photon energy for an atom initially at rest, neglecting the very small recoil correction.
Derivation
Let the atom initially have momentum
$\mathbf p_i=M\mathbf v.$
From momentum conservation,
$\mathbf p_f=\mathbf p_i-\mathbf k.$
The square of the final momentum is
=p_i^2+k^2-2\mathbf p_i!\cdot!\mathbf k.$$Substituting this expression into energy conservation,$$\frac{p_i^2}{2M}+E_e = \frac{p_f^2}{2M}+E_g+\hbar\omega,
where $E_e$ and $E_g$ are the excited and ground state energies.
Using
$$$$
we obtain
$$= \hbar\omega + \frac{p_f^2-p_i^2}{2M}.$$
Substituting the expression for $p_f^2-p_i^2$,
$$= \hbar\omega + \frac{k^2-2\mathbf p_i!\cdot!\mathbf k}{2M}.$$
Rearranging,
$$= \frac{k^2}{2M} - \frac{\mathbf p_i!\cdot!\mathbf k}{M}.$$
Since
$$$$
and the angle between $\mathbf v$ and the emitted photon is $\alpha$,
$$= Mv,k\cos\alpha.$$
Hence
$$= \frac{k^2}{2M} - v,k\cos\alpha.$$
Using
$$$$
gives
$$= \frac{\hbar\omega^2}{2Mc^2} - \frac{v\omega}{c}\cos\alpha.$$
For visible light,
$$$$
so the recoil term
$$$$
is negligible compared with the Doppler term. Thus
$$\approx -\frac{v\omega}{c}\cos\alpha.$$
The difference between $\omega$ and $\omega_0$ is already of first order in $v/c$, so in the right-hand side one may replace $\omega$ by $\omega_0$:
$$\approx -\frac{v\omega_0}{c}\cos\alpha.$$
The sign depends on the convention chosen for $\alpha$. If $\alpha$ is measured from the direction from the observer toward the atom, the result becomes
$$\approx \frac{v\omega_0}{c}\cos\alpha.$$
Dividing by $\omega_0$ yields
$$\frac{\omega_0-\omega}{\omega_0} \approx \frac{v}{c}\cos\alpha.$$
This is the nonrelativistic Doppler formula.
Result
The frequency shift for radiation emitted by a slowly moving atom is
$$\boxed{ \frac{\omega_0-\omega}{\omega_0} \approx \frac{v}{c}\cos\alpha }.$$
No numerical values are specified in the problem. The result is purely symbolic. The final answer is therefore
$$\boxed{ \frac{\omega_0-\omega}{\omega_0} = \frac{v}{c}\cos\alpha \quad (v\ll c) }.$$
The left-hand side is dimensionless, and the right-hand side is also dimensionless because both $v$ and $c$ have units of m/s.
Sanity Checks
If $\alpha=90^\circ$, then $\cos\alpha=0$ and the formula gives
$$\omega=\omega_0.$$
To first order in $v/c$, there is no Doppler shift for emission perpendicular to the motion.
If the photon is emitted along the direction of motion, $\alpha=0$, then
$$\frac{\omega_0-\omega}{\omega_0} \approx \frac{v}{c},$$
which corresponds to the largest first order shift.
The quantity
$$\frac{\omega_0-\omega}{\omega_0}$$
is dimensionless. The quantity
$$\frac{v}{c}\cos\alpha$$
is also dimensionless, so the dimensions are consistent.
For a visible photon, $\hbar\omega$ is a few electronvolts, whereas for an atom
$$Mc^2 \sim 10^{10}\ {\rm eV}.$$
Hence
$$\frac{\hbar\omega}{Mc^2}\sim10^{-10},$$
showing that the recoil contribution is negligibly small compared with the Doppler effect.
The step most susceptible to error is the evaluation of
$$(\mathbf p_i-\mathbf k)^2 = p_i^2+k^2-2\mathbf p_i!\cdot!\mathbf k,$$
because an incorrect sign in the cross term immediately produces the wrong sign for the Doppler shift.