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 received, &lambda;, with the wavelength we would expect a star of that type to emit, &lambda;0. Since the speed of light c is constant regardless of reference medium:



$$c = f\lambda = f_0\lambda_0$$

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 &Delta;&lambda;, we know that &lambda; = &lambda;0 + &Delta;&lambda;. 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/