Velocity Addition and Doppler Effect: Explained

[Kairos]

The Doppler effect of light corresponds to the classical Doppler effect corrected by time dilation, but the first one is obtained with classical velocity additions (c+v at the front of the source and c-v at the back) whereas velocity addition of special relativity gives c at the front and c at the back.. please can you explain my confusion?

 

[Ibix]

Velocity addition does not come into this. Simply observe that the source moves a distance $v/f$ in the time it takes (in your frame) to emit one wave. Thus the wavelength is $c/f\pm v/f$ (i.e., the distance traveled by the wave plus or minus the distance traveled by the source). This is the same as the Newtonian derivation. The only change you need to make is to note that due to time dilation, $f=f_0/\gamma$, where $f_0$ is the frequency measured by the source.

 

[vanhees71]

You don't need the somewhat cumbersome velocity additions (at least not for free em. fields). You just have to consider that the phase of a plane-wave solution, $\omega/c t-\vec{k} \cdot \vec{x}=k_{\mu} x^{\mu}$ is a Lorentz scalar, which implies that $k_{\mu}$ is a Lorentz vector. So you can just using a Lorentz boost to transform the components of the wave-fourvector $k^{\mu}$ wrt. an inertial frame to any other inertial frame.

If you are only interested in the change of the frequency of the light as seen from different observers, you only need to know that it can only depend on the relative velocity between the light source and the observer. So let $u^{\mu}=\gamma(1,\vec{\beta})$ be the four-velocity (normalized to 1) of the light source in the reference frame of the observer. Now it's most convenient to refer to the frequency of the light $\omega_0$ in the restframe of the light source. It is a Lorentz invariant and given by
$$\omega_0/c=u_{\mu} k^{\mu}=\gamma (\omega/c -\vec{\beta} \cdot \vec{k}).$$
Now with $\vec{k}=k \vec{n}$ and the dispersion relation $k=\omega/c$ you get
$$\omega_0=\gamma \omega (1-\vec{\beta} \cdot \vec{n})$$
or
$$\omega=\frac{\omega_0}{\gamma (1-\vec{\beta} \cdot \vec{n})}.$$
Now take the extreme cases: $\vec{\beta}=\beta \vec{n}$, i.e., the light-source travels in the diretion of wave propagation (i.e., towards the observer). This gives $\gamma(1-\beta)=\sqrt{(1-\beta)/(1+\beta)}$ and thus
$$\omega_{\text{max blue}}=\sqrt{\frac{1+\beta}{1-\beta}} \omega_0.$$
This is the maximal blue shift you can get compared to the frequency in the rest frame of the light source. For the maximal redshift you have to set $\vec{\beta}=-\beta \vec{n}$ (source moving away from observer):
$$\omega_{\text{max red}}=\sqrt{\frac{1-\beta}{1+\beta}} \omega_0.$$
Then a specifically relativistic effect is that there's also a shift if the source moves perpendicular to the light-propagation direction, i.e., for $\vec{\beta} \cdot \vec{n}=0$:
$$\omega_{\perp}=\frac{\omega_0}{\gamma}=\sqrt{1-\beta^2} \omega_0,$$
i.e., you have a red shift, which is of course purely due to time dilation between the light-source restframe and the observer's frame. That's the transverse Doppler effect.

 

[Kairos]

that's the transverse Doppler effect.

on this subject, do you know if this transverse Doppler effect has been observed. I have only seen one publication! (Hasselkamp1979). It doesn't seem like much for such an important result.

 

[vanhees71]

Googling empirical confirmation of transverse Doppler effect gives some useful results. The first search result seems to be a nice review article with some references to experiments:

http://elpub.bib.uni-wuppertal.de/servlets/DerivateServlet/Derivate-2663/15tderp5.pdf

 

[Kairos]

thank you I will read this with interest