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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?

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Is that subscript supposed to be a zero or a letter o? It's kind of odd to have it as a zero. – David Z Apr 23 '13 at 1:12
It was supposed to be o, I mistakenly put 0 in haste :) – Comp_Warrior Apr 23 '13 at 21:48
up vote 7 down vote accepted

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:

$$ \tau' = \gamma t + \gamma v t/c $$

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!

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Very clear and understandable! However one thing that worries me is that you say the observer is at the origin. Suppose you define a frame where the observer is not at the origin (but still at rest) - will the Doppler factor change then? The answer is no of course, but I'm not sure if this is explicit in this derivation. – Comp_Warrior Apr 23 '13 at 21:47
It's true that I've calculated the doppler shift for the observer in the same place as the emitting object at time 0. For an observer in a different place use two transformations, first the Lorentz transformation into the observers frame, then a second linear transformation in the observers frame to calculate what happens at a different position in that frame. This second transformation is trivial as it's just a translation within the frame and it can't change the observed frequency. – John Rennie Apr 24 '13 at 5:54

As for me, there's no assumption that you can do (as in classical doppler shift) like some velocity $c+u$. The speed-of-light should be a constant in every reference frame (i.e) strictly no other information between observers other than EM waves. In this case, if you draw a spacetime diagram, the $c+u$ will not be at the 45-degreee between spatial $x$ and time $ct$ axis. In case of relativistic doppler effect, we use light to synchronize clocks by transmitting it between inertial observers.

But, you say that they're relativistically transforming the time period. They aren't using direct velocities $c+u$, instead they should've assumed that when the crests are shifted over a distance $(c+v)T_0$, the time period is dilated to some value in order for $c$ to remain constant (simultaneously). It's totally alright to use such an assumption...

Basically, relativistic doppler shift can be easily derived using spacetime diagrams (similar to John, but this was taught to me by a guy before I got into Lorentz transformations). It looks cool...

Here's a rough spacetime figure. There are two observers $\mathrm{A}$ (at rest) and $\mathrm{B}$ (at some relative motion $v$). After some time, $\mathrm{A}$ shoots a light pulse to $\mathrm{B}$ and it returns back simultaneously. Thus, the inertial observers synchronize their clocks in this way. If $T$ is the time for $\mathrm{A}$, then $T_B=kT$ ($k$ is some factor). After plugging several things from the diagram, we arrive at a stage where the distance between $\mathrm{A}$ and $\mathrm{B}$ at the time of synchronization would be, $$ct_1=c\Biggl(\frac{T(k^2-1)}{2}\Biggr)=d\tag1$$ $$vt_2=v\Biggl(\frac{T(k^2+1)}{2}\Biggr)=d\tag2$$

Equating both, we simply get $$v(k^2+1)=c(k^2-1)\implies k=\sqrt{\frac{c+v}{c-v}}$$

So, we've found the factor by which both $T$ and $T_B$ are related. Their frequency is simply the inverse of their time period. Hence, the light emitted by $\mathrm{A}$ is red-shifted by $1/k$ and the observed frequency is $$f_B=\sqrt{\frac{c-v}{c+v}}f$$

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Thanks for the response! I am the best at understanding spacetime diagrams, but I will consider this approach as well. – Comp_Warrior Apr 23 '13 at 21:45

Neither origins nor emitters nor receivers are required: consider a photon with momentum $\vec{p}$: it has 4-veclocity $p_{\mu}=(||p||/c, \vec{p})$. To compute what another observer moving at $\vec{v}$ sees, do a Lorentz transform: $p'_{\mu} = \Lambda_{\mu}^{\ \nu}p_{\nu}$. From this you'll get the relativistic (time-dilated) Doppler effect and stellar aberrations.

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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 ($\lambda^{\prime}/\lambda = k/k^{\prime}$).

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