A-level Physics (Advancing Physics)/Doppler Effect

The Doppler effect is a change in the frequency of a wave which occurs if one is in a different frame of reference from the emitter of the wave. Relative to us, we observe such a change if an emitter of a wave is moving relative to us.

All waves travels in a medium. So, they have a velocity relative to this medium v. They also have a velocity relative to their source v_s and a velocity relative to the place where they are received v_r. The frequency at which they are received f is related to the frequency of transmission f₀ by the formula:

$f=\left({\frac {v+v_{r}}{v+v_{s}}}\right)f_{0}$

The Doppler effect can be used to measure the velocity at which a star is moving away from or towards us by comparing the wavelength received, λ, with the wavelength we would expect a star of that type to emit, λ₀. Since the speed of light c is constant regardless of reference medium:

Redshift of spectral lines in the optical spectrum of a supercluster of distant galaxies (right), as compared to that of the Sun (left).

$c=f\lambda =f_{0}\lambda _{0}$

Therefore:

$f={\frac {c}{\lambda }}$ and $f_{0}={\frac {c}{\lambda _{0}}}$

By substitution:

${\frac {c}{\lambda }}=\left({\frac {v+v_{r}}{v+v_{s}}}\right){\frac {c}{\lambda _{0}}}$

${\frac {1}{\lambda }}=\left({\frac {v+v_{r}}{v+v_{s}}}\right){\frac {1}{\lambda _{0}}}$

$\lambda ={\frac {\lambda _{0}(v+v_{s})}{v+v_{r}}}$

In this case, v is the speed of light, so v = c. Relative to us, we are stationary, so v_r = 0. So:

$\lambda ={\frac {\lambda _{0}(c+v_{s})}{c}}$

${\frac {\lambda }{\lambda _{0}}}={\frac {(c+v_{s})}{c}}=1+{\frac {v_{s}}{c}}$

If we call the change in wavelength due to Doppler shift Δλ, we know that λ = λ₀ + Δλ. Therefore:

${\frac {\lambda _{0}+\Delta \lambda }{\lambda _{0}}}=1+{\frac {\Delta \lambda }{\lambda _{0}}}=1+{\frac {v_{s}}{c}}$

So, the important result you need to know is that:

${\frac {\Delta \lambda }{\lambda _{0}}}={\frac {v_{s}}{c}}=z$

This value is known as the red-shift of a star, denoted z. If z is positive, the star is moving away from us - the wavelength is shifted up towards the 'red' end of the electromagnetic spectrum. If z is negative, the star is moving towards us. This is known as blue shift. Note that we have assumed that v is much smaller than c. Otherwise, special relativity makes a significant difference to the formula.

Questions

1. M31 (the Andromeda galaxy) is approaching us at about 120kms⁻¹. What is its red-shift?

2. Some light from M31 reaches us with a wavelength of 590 nm. What is its wavelength, relative to M31?

3. Some light has a wavelength, relative to M31, of 480 nm. What is its wavelength, relative to us?

4. A quasar emits electromagnetic radiation at a wavelength of 121.6 nm. If, relative to us, this wavelength is red-shifted 0.2 nm, what is the velocity of recession of the quasar?

Worked Solutions