The Doppler Effect of Light

The Doppler shift for a light-emitting non-relativistic source can be described using the equation:

$\frac{∆ f}{f} = \frac{∆ λ}{λ} \approx \frac{∆ v}{c}$

Where:

Δf = change in frequency (Hz)
f= reference frequency (frequency of the source) (Hz)
Δλ = change in wavelength (m)
λ = reference wavelength (wavelength of the source) (m)
Δv = relative velocity of the source and observer (m s^–1)
c = the speed of light (m s^–1)

The sign ≈ means 'approximately equals to'
This equation only works if v << c
The change in wavelength Δλ is equal to:

$∆ λ = λ' - λ$

Where:
- λ' = observed wavelength of the source (m)

Since the fractions have the same units on the numerator (top number) and denominator (bottom number), the Doppler shift has no units

The relative speed between the source and observer along the line joining them is given by:

$∆ v = v_{s} - v_{o}$

Where:
- v_s = velocity of the source of the light (m s^–1)
- v_o = velocity of the observer (m s^–1)

Usually, we are calculating the speed of the source of electromagnetic waves relative to an observer which we assume to be stationary
- Therefore v_o = 0, hence ∆v = v_s = v
- Where v is the velocity at which the source of the electromagnetic waves is moving from the observer

Sometimes the reference frequency is written as f₀and the frequency wavelength as λ₀
Hence, the Doppler shift equation can therefore also be written as:

$\frac{∆ f}{f} = \frac{λ' - λ}{λ} = \frac{∆ λ}{λ} \approx \frac{∆ v}{c}$

Spectral Lines

Doppler shift can easily be seen in atomic spectral lines from planets and stars

25-2-redshift

Spectral lines showing red shift

Each line represents an element making up the composition of the galaxy
The lines are identical in that predicted in lab and the light measured from the distant galaxy
Since the lines all move to the left (the red end of the spectrum) this means the galaxy is travelling away from Earth

Worked example

A stationary source of light is found to have a spectral line of wavelength 438 nm. The same line from a distant star that is moving away from us has a wavelength of 608 nm.

Calculate the speed at which the star is travelling away from Earth.

Answer:

Step 1: List the known quantities

Unshifted wavelength, λ = 438 nm
Shifted wavelength, λ' = 608 nm
Change in wavelength, Δλ = (608 – 438) nm = 170 nm
Speed of light, c = 3.0 × 10⁸ m s^–1

Step 2: Write down the Doppler equation and rearrange for velocity v

$\frac{∆ λ}{λ} = \frac{v}{c}$

$v = \frac{c ∆ λ}{λ}$

Step 3: Substitute values to calculate v

$v = \frac{(3.0 \times 10^{8}) \times 170}{438}$ = 1.16 × 10⁸ m s^–1

Worked example

The stars in a distant galaxy can be seen to orbit about a galactic centre. The galaxy can be observed 'edge-on' from the Earth.

Light emitted from a star on the left-hand side of the galaxy is measured to have a wavelength of 656.44 nm. The same spectral line from a star on the right-hand side is measured to have a wavelength of 656.12 nm.

The wavelength of the same spectral line measured on Earth is 656.28 nm.

(a)

State and explain which side of the galaxy is moving towards the Earth.

(b)

Calculate the rotational speed of the galaxy.

Answer:

(a)

The light from the right-hand side (656.12 nm) is observed to be at a shorter wavelength than the reference line (656.28 nm)
Therefore, the right-hand side has been blue-shifted and must be moving towards the Earth

(b)

Step 1: List the known quantities

Observed wavelength on LHS, $λ_{L H S}$ = 656.44 nm
Observed wavelength on RHS, $λ_{R H S}$ = 656.12 nm
Reference wavelength, λ = 656.28 nm
Speed of light, c = 3.0 × 10⁸ m s⁻¹

Step 2: Calculate the average change in wavelength

$∆ λ = \frac{λ_{L H S} - λ_{R H S}}{2} = \frac{656.44 - 656.12}{2}$

$∆ λ$ = 0.32 nm

Step 3: Write down the Doppler equation and rearrange for velocity v

$\frac{∆ λ}{λ} = \frac{v}{c}$

$v = \frac{c ∆ λ}{λ}$

Step 4: Substitute values into the velocity equation

$v = \frac{(3 \times 10^{8}) \times 0.32}{656.28}$

Rotational speed: $v = 1.46 \times 10^{5} m s^{- 1} = 146 km s^{- 1}$

Exam Tip

You need to know that in the visible light spectrum red light has the longest wavelength and the smallest frequency compared to blue light which has a shorter wavelength and higher frequency.

The second worked example didn't change the wavelengths from nm into m, since it doesn't matter in the equation as the units will cancel out.

The Doppler Effect of Light (HL IB Physics)

Revision Note