# Relativistic Doppler effect derivation

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!

Yet another way. Consider a wave of a form $f(kx-\omega t)$ in the emitting frame, being observed as some other wave $f^{\prime} (k^{\prime} x^{\prime}-\omega^{\prime} t^{\prime})$ in a frame moving at $v$ along the x-axis.

We actually don't care what $f^{\prime}$ is, only the argument of the function matters. So use the Lorentz transforms$^{*}$ to change $x$ and $t$ to $x^{\prime}$ and $t^{\prime}$. $$kx -\omega t = k\gamma(x^{\prime} + vt^{\prime}) - \omega \gamma (t^{\prime} +vx^{\prime}/c^2)$$ Now use the fact that $\omega =kc$ for an EM wave in vacuum and group together terms in $x^{\prime}$ and $t^{\prime}$. $$kx - \omega t = \gamma(1+v/c)kx^{\prime} - \gamma (1+v/c)\omega t^{\prime} $$

Thus $k^{\prime} = \gamma(1+v/c)k$ and $\omega^{\prime} = \gamma(1+v/c)\omega$.

This immediately gives you the doppler shift formula for frequency ($\omega^{\prime}/\omega$) and wavelength ($\lambda^{\prime}/\lambda = k/k^{\prime}$).

For a source receding from the observer with speed $u$ then the velocity to use would just be $v =-u$ in the above formulae.

$^{*}$ Which are of course that $$x = \gamma (x^{\prime} + vt^{\prime})\ ,$$ $$t = \gamma( t^{\prime} + vx^{\prime}/c^2)\ .$$