homework and exercises - Relativistic Doppler effect derivation

This is about a step in a derivation of the expression for the relativistic Doppler effect.

Consider a source receding from an observer at a velocity $v$ along the line joining the two. Light is emitted at frequency $f_s$ and wavelength $\lambda_s$. The frequency $f_0$ and wavelength $\lambda_0$ received by the observer will be different.

Some textbooks now argue that the wavelength received by the observer is given by $\lambda_0=(c+u)T_0$, where $T_0$ is the time period of the wave in the observers frame. The argument given is that the successive "crests" will be an extra distance $uT_0$ apart due to movement of the source. Upon relativistically transforming the time period in the source frame to the observer frame, the correct result $f_0=f_s\sqrt{\frac{c-u}{c+u}}$ is obtained.

But the argument for the wavelength seems reminiscent of the classical Doppler effect for sound. Is it really applicable here through this argument? And is there a way to show the same result for $\lambda_0$ mathematically?

Answer

You can derive the relativistic Doppler shift from the Lorentz transformations. Let's start in the frame of the moving rocket, and let's take two events corresponding to nodes in the emitted wave (i.e. 1/$f$). Then in the rocket's frame the two events are (0, 0) and ($\tau$, 0), where $\tau$ is the period of the radiated wave. To see what the period of the radiation is in our frame we just have to use the Lorentz transformations to transform these two spacetime points into our frame.

For simplicity we'll take our rest frame and the frame of the rocket to coincide at $t = 0$. This is convenient because then the first event is just (0, 0) in both frames. Now the Lorentz transformations tell us:

$$ t' = \gamma \left( t - \frac{vx}{c^2} \right ) $$

$$ x' = \gamma \left( x - vt \right) $$

If we're tranforming from the rocket's frame to ours, and the rocket is moving at velocity $v$ wrt us, then we have to put the velocity in as $-v$, and we're transforming the point ($\tau$, 0). Putting these in the Lorentz transformations we find that the point ($\tau$, 0) in the rocket's frame transforms to the point ($\gamma \tau$, $\gamma v \tau$) in our frame.

The last step is to note that if we're sitting at the origin in our frame the light from the event at ($\gamma \tau$, $\gamma v \tau$) takes a time $\gamma v \tau/c$ to reach us. So the time we see the second event is $\gamma \tau + \gamma v \tau/c$ and this is equal to the period of the radiation, $\tau'$ in our frame:

$$ \tau' = \gamma \tau + \gamma v \tau/c $$

We just need to rearrange this to get the usual formula. Noting that $f'$ = 1/$\tau'$ and $f$ = 1/$\tau$ we take the reciprocal of both sides to get:

$$ f' = f \frac{1}{\gamma(1 + v/c)} $$

To simplify this note that:

$$\begin{align} \frac{1}{\gamma} &= \sqrt{1 - \frac{v^2}{c^2}} \\ &= \sqrt{(1 - \frac{v}{c})(1 + \frac{v}{c})} \end{align}$$

and substituting this back in our expression for $f'$ we get:

$$\begin{align} f' &= f \frac{\sqrt{(1 - v/c)(1 + v/c)}}{1 + v/c} \\ &= f \frac{\sqrt{(1 - v/c)}}{\sqrt{1 + v/c}} \\ &= f \sqrt{\frac{c - v}{c + v}} \end{align}$$

and presto it's proved!