# 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 vs and a velocity relative to the place where they are received vr. 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 receive 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 vr = 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 590nm. What is its wavelength, relative to M31?

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

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

Worked Solutions