# Doppler shift

Jump to navigationJump to search

Course page: Radiative Processes in Astrophysics

## Doppler shift

We will derive both the relativistic and non-relativistic Doppler shift. The easiest way of doing this is to derive the relativistic formula and then take the non-relativistic limit, i.e., ${\displaystyle v\ll c}$.

Consider a light wave with frequency ${\displaystyle f_{0}}$ and wavelength ${\displaystyle \lambda _{0}}$ in the ${\displaystyle S_{0}}$ frame. Then the period of the wave is given by

${\displaystyle \tau _{0}={\frac {1}{f_{0}}}.\,\!}$

Suppose the light wave is traveling in the ${\displaystyle x}$-direction. Now consider a frame, ${\displaystyle S}$, which is moving with velocity ${\displaystyle v_{x}}$ in the ${\displaystyle x}$-direction relative to ${\displaystyle S_{0}}$. There might be ${\displaystyle y}$- and ${\displaystyle z}$-components to the velocity, but these will not concern us for the moment. What does the light wave look like in this frame? In particular, what period is measured in the ${\displaystyle S}$ frame?

Suppose there is an observer in the ${\displaystyle S}$ frame who is sitting at ${\displaystyle x=0}$. In the ${\displaystyle S_{0}}$ frame, this observer will appear to be moving on a trajectory given by ${\displaystyle x_{0}=v_{x}t_{0}}$. First, analyze the situation in the ${\displaystyle S_{0}}$ frame; then transform the result to the ${\displaystyle S}$ frame. Suppose the moving observer is at ${\displaystyle x_{0}=t_{0}=0}$ when the front of the light wave reaches him. At ${\displaystyle t_{0}=\tau _{0}}$, the back of the wave will reach ${\displaystyle x_{0}=0}$. But by this time, the observer is no longer at ${\displaystyle x_{0}=0}$. Instead, he is at ${\displaystyle x_{0}=v_{x}\tau _{0}}$. If ${\displaystyle v_{x}}$ is positive, then the back of the wave still has farther to go to reach the observer. If ${\displaystyle v_{x}}$ is negative, then the back of the wave will already have reached the observer. Let ${\displaystyle {\tilde {t}}_{0}}$ denote the time at which the back of the wave reaches the moving observer. Then

${\displaystyle {\tilde {t}}_{0}=\tau _{0}+{\frac {v_{x}{\tilde {t}}_{0}}{c}}.\,\!}$

The first term is the time it takes for the back of the wave to get to ${\displaystyle x_{0}=0}$ (remember the clock started when the front of the wave got to ${\displaystyle x_{0}=0}$); the second term is the time it takes to get from ${\displaystyle x_{0}=0}$ to the observer’s new position. Note that the second term can be negative. Then we have

${\displaystyle {\tilde {t}}_{0}={\frac {\tau _{0}}{1-v_{x}/c}}.\,\!}$

This is the time between the arrival of the front of the wave and the arrival of the back of the wave at the moving observer (remember that the front of the wave reached the observer at ${\displaystyle t_{0}=0}$).

Now, ${\displaystyle {\tilde {t}}_{0}}$ is the Doppler-shifted period as seen in ${\displaystyle S_{0}}$. What we really want is the period in ${\displaystyle S}$ which we will denote by ${\displaystyle \tau }$. From the time-dilation formula (see Lorentz transformations),

${\displaystyle \tau ={\frac {{\tilde {t}}_{0}}{\gamma }},\,\!}$

where, as usual, ${\displaystyle \gamma =1/{\sqrt {1-v^{2}/c^{2}}}}$ and ${\displaystyle v}$ is now the full 3-dimensional velocity of ${\displaystyle S}$ relative to ${\displaystyle S_{0}}$.

Then

${\displaystyle \tau ={\frac {\tau _{0}}{\gamma (1-v_{x}/c)}}.\,\!}$

More generally, we can think of the ${\displaystyle x}$-direction as the line-of-sight. So ${\displaystyle v_{x}}$ is just the velocity along the line-of-sight away from the observer. Let’s replace ${\displaystyle v_{x}}$ with ${\displaystyle v_{\mathrm {los} }}$. Then, since the wavelength is proportional to the period,

${\displaystyle \lambda ={\frac {\lambda _{0}}{\gamma (1-v_{\mathrm {los} }/c)}}.\,\!}$

This result is valid in the framework of special relativity and depends not only on the motion along the line-of-sight but also on the perpendicular components of the velocity through ${\displaystyle \gamma }$. Since ${\displaystyle \gamma \geq 1}$, the relativistic effects always contribute to a blue-shifting of the spectrum. This boils down to a time-dilation effect. The astronomer is in the same spatial location for both events (the arrival of the front and back of the wave), and so the astronomer’s clock measures the proper time. The proper time is always smaller than the time interval measured in any other reference frame, so the period of the wave is smaller for the astronomer which is a blue-shift. This effect can be overcome by a large velocity in the direction away from the astronomer. This is the contribution of the other term in the denominator which comes from the stretching of the wavelength due to recession.

It is often useful to look at the non-relativistic limit: ${\displaystyle v/c\ll 1}$. Expanding to first order in ${\displaystyle v_{i}/c}$ yields

${\displaystyle \lambda \simeq \lambda _{0}(1+v_{\mathrm {los} }/c).\,\!}$

Here we see that the dependence on the perpendicular components has disappeared. In this limit, a receding object will always appear red-shifted, and an approaching object will always appear blue-shifted.