Doppler Shift for Light and Radiation

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Following are formulas for the Doppler shift for light and radiation.

\frac{E}{E_{0}} = \frac{\sqrt{1-\beta^{2}}}{1-\beta \cos{\theta}}
$\beta \equiv (v/c)$,
$E_{0}$ is the photon energy in the emitted frame,
$E$ is the the photon energy in the observed frame,
$\theta$ is the angle between the velocity vector and the photon
direction ($180^{\circ}$ corresponds to recession).

The energy shift can be found from $\Delta E = E – E_{0}$.

Note that for $\theta$ equal to $0^{\circ}$ or $180^{\circ}$, the expression reduces to

\left(\frac{E}{E_{0}}\right)_{\theta=0} = \left(\frac{1+\beta}{1-\beta}\right)^{\frac{1}{2}}


\left(\frac{E}{E_{0}}\right)_{\theta=\pi} = \left(\frac{1-\beta}{1+\beta}\right)^{\frac{1}{2}}

since $(1-\beta^{2}) = (1-\beta)(1+\beta)$.

For frequency, simply replace $E_{0}$ and $E$ by $\nu_{0}$ and $\nu$ respectively.
For wavelength, use $E=(hc/\lambda)$, but note that $(\Delta \lambda/\lambda)=
(\Delta E/E)$ because $\Delta E = -hc\Delta \lambda/ \lambda^{2}$.

In the nonrelativistic limit ($\beta \ll 1$),

\frac{\Delta E}{E_{0}} \sim \beta = \left(\frac{v}{c}\right)

to first order in $\beta$.

Tip: If you are ever confused about the correct sign in the denominator in any incarnation of a Doppler shift formula, remember that the equation has to blow up if the light source is coming towards the receiver at the speed of light. In other words, blueshift blows up at $v=c$ but redshift doesn’t (it goes to zero).

Tip: The factor $\sqrt(1-\beta^{2})$ in the very first equation is basically the relativistic time dilation effect and is the same no matter what the relative direction of the emitter and reveiver is. In particular, the time dilation factor gives a Doppler redshift even for transverse relative motion, and it is second order in $(v/c)$. The time dilation factor has no counterpart in the Doppler effect for sound. Only the rest of the formula is due to actual relative motion and apparent change of wavelength because of the motion.