Solution

Let’s write down the conservation laws of energy and momentum for the case when a photon is emitted in the direction of the atom’s motion:

$$\frac{mv^2}{2} + \varepsilon = \frac{m(v - \Delta v)^2}{2} + h\nu$$

$$mv = m(v - \Delta v) + \frac{h\nu}{c}$$

Similarly, for emission in the opposite direction:

$$\frac{mv^2}{2} + \varepsilon = \frac{m(v + \Delta v')^2}{2} + h\nu'$$

$$mv = m(v + \Delta v') - \frac{h\nu'}{c}$$

From the momentum equations, we find $\Delta v$ and $\Delta v'$:

$$\Delta v = \frac{h\nu}{mc}$$

$$\Delta v' = \frac{h\nu'}{mc}$$

Substitute $\Delta v$ and $\Delta v'$ into the energy equations, remembering that the photon’s momentum is much smaller than the atom’s momentum. We get:

$$\varepsilon = h\nu\left(1 - \frac{v}{c}\right)$$

$$\varepsilon = h\nu'\left(1 + \frac{v}{c}\right)$$

Now, solving for the frequency $\nu'$:

$$\nu' = \frac{1 - \frac{v}{c}}{1 + \frac{v}{c}} \nu$$

Now let's consider the case where the photon is emitted perpendicular to the atom’s motion. Write down the conservation of energy and momentum:

$$\frac{mv^2}{2} + \varepsilon = \frac{mv_1^2}{2} + h\nu''$$

$$(mv)^2 = (mv_1)^2 + \left(\frac{h\nu''}{c}\right)^2$$

Solving for $v_1$ from the momentum equation, we find:

$$\varepsilon = h\nu''\left(1 - \frac{h\nu''}{2mc^2}\right)$$

Since the photon’s momentum is negligibly small:

$$\varepsilon = h\nu''$$

Now compare the change in the atom’s internal energy from the first and third cases. We get:

$$\nu'' = \nu\left(1 - \frac{v}{c}\right)$$

It should be noted that this solution is made using classical mechanics, though strictly speaking, the problem should be solved from the perspective of relativistic mechanics. However, the result obtained here coincides with the relativistic one. :)