C5-2. The Doppler Effect of Light and Galactic Redshift
1. The Doppler Approximation Formula
The Doppler shift for a light-emitting non-relativistic source can be described using the approximation formula:
Notation Key & Units:
- $\Delta f = |f_{\text{observed}} - f_{\text{source}}|$ = change in frequency ($\text{Hz}$)
- $\Delta \lambda = |\lambda_{\text{observed}} - \lambda_{\text{source}}|$ = change in wavelength ($\text{m}$)
- $f, \lambda$ = original emitted (reference) frequency/wavelength
- $v$ = relative speed of the source and observer ($\text{m s}^{-1}$)
- $c$ = speed of light ($3.00 \times 10^8\text{ m s}^{-1}$)
- The Approximation: The $\approx \dfrac{v}{c}$ formula is only valid when the relative velocity is much less than the speed of light ($v \ll c$). For sound, $c$ is the speed of sound, and the general formula must be used for $v \approx c$.
- Redshift: If the source is receding, the wavelength stretches ($\Delta\lambda$ increases) and frequency drops. This shifts visible light toward the red end of the spectrum.
- Blueshift: If the source is approaching, the waves compress ($\Delta\lambda$ decreases) and frequency rises. Light shifts toward the blue end.
2. Galactic Redshift and the Expanding Universe
In space, the Doppler effect of light can be observed when spectra of distant stars and galaxies are observed. Each spectral line represents an element making up the composition of the galaxy. By comparing the light spectrum produced from a close object, such as our Sun, with that of a distant galaxy, we can observe shifts:
- Redshift: Occurs if the object is moving away from the Earth. The wavelength is increasing but the frequency is decreasing. If the speed of the galaxy relative to Earth is negative then the galaxy is moving away from the Earth, which is the case when the observed frequency $f_{\text{obs}}$ is less than the reference frequency $f$. The light from the distant galaxy is shifted towards the red end of the spectrum.
- Blueshift: Occurs if the object is moving towards the Earth. The wavelength is decreasing but the frequency is increasing. If the speed of the galaxy relative to Earth is positive then the galaxy is moving towards the Earth, which is the case when the observed frequency $f_{\text{obs}}$ is greater than the reference frequency $f$.
An Expanding Universe
After the discovery of Doppler redshift, astronomers began to realize that almost all the galaxies in the universe are receding. This led to the idea that the space between the Earth and the galaxies must be expanding. This expansion stretches out the light waves as they travel through space, shifting them towards the red end of the spectrum. The expansion of the universe can be compared to dots on an inflating balloon: as the rubber stretches when the balloon is inflated, space itself is stretching out between galaxies. Just like the dots, the galaxies move away from each other, however, they themselves do not move. The greater the distance to the galaxy, the greater the redshift, meaning the faster the galaxy is moving away from Earth. The furthest galaxies appear to be redshifted the most and are receding the fastest.
3. Examples
Example 1: Basic Spectral Shift
Problem: A stationary source of light is found to have a spectral line of wavelength $438\text{ nm}$. The same line from a distant star that is moving away from us has a wavelength of $608\text{ nm}$. Calculate the speed at which the star is travelling away from Earth.
Solution:
Step 1: List the known quantities
Unshifted wavelength, $\lambda = 438\text{ nm}$
Shifted wavelength, $\lambda_{\text{obs}} = 608\text{ nm}$
Change in wavelength, $\Delta\lambda = (608 - 438)\text{ nm} = 170\text{ nm}$
Speed of light, $c = 3.0 \times 10^8\text{ m s}^{-1}$
Step 2: Write down the Doppler equation and rearrange for velocity $v$ $$ \frac{\Delta\lambda}{\lambda} \approx \frac{v}{c} \implies v = \frac{c\Delta\lambda}{\lambda} $$
Step 3: Substitute values to calculate $v$ $$ v = \frac{(3.0 \times 10^8) \times 170}{438} = \mathbf{1.16 \times 10^8\text{ m s}^{-1}} $$
Example 2: Edge-on Galaxy Rotation
Problem: 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 $656.44\text{ nm}$. The same spectral line from a star on the right-hand side is measured to have a wavelength of $656.12\text{ nm}$. The wavelength of the same spectral line measured on Earth is $656.28\text{ nm}$.
(a) State and explain which side of the galaxy is moving towards the Earth.
(b) Calculate the rotational speed of the galaxy.
Solution:
(a) Direction of Motion
The light from the right-hand side ($656.12\text{ nm}$) is observed to be at a shorter wavelength than the reference line ($656.28\text{ nm}$). Therefore, the right-hand side has been blue-shifted and must be moving towards the Earth.
(b) Rotational Speed
Observed wavelength on LHS, $\lambda_{\text{LHS}} = 656.44\text{ nm}$. Observed wavelength on RHS, $\lambda_{\text{RHS}} = 656.12\text{ nm}$. Reference wavelength, $\lambda = 656.28\text{ nm}$.
Calculate the average change in wavelength
$$
\Delta\lambda = \frac{\lambda_{\text{LHS}} - \lambda_{\text{RHS}}}{2} = \frac{656.44 - 656.12}{2} = 0.32\text{ nm}
$$
Again,
$$
\frac{\Delta\lambda}{\lambda} = \frac{v}{c} \implies v = \frac{c\Delta\lambda}{\lambda}
$$
Therefore,
$$
v = \frac{(3.0 \times 10^8) \times 0.32}{656.28} = 1.46 \times 10^5\text{ m s}^{-1} = \mathbf{146\text{ km s}^{-1}}
$$
(Note: You do not change the wavelengths from nm into m, since it is a ratio and the units will cancel out).
Example 3: Frequency Shift & Direction Analysis
Problem: A particular line in the spectrum of light from a source in the laboratory has a frequency of $4.570 \times 10^{14}\text{ Hz}$. The same line in the spectrum of light from a distant galaxy has a frequency of $4.547 \times 10^{14}\text{ Hz}$. Calculate the speed of the distant galaxy in relation to the Earth. State whether it is moving towards or away from the Earth.
Solution:
Step 1: Write down the known quantities
Observed frequency, $f_{\text{obs}} = 4.547 \times 10^{14}\text{ Hz}$
Reference (source) frequency, $f = 4.570 \times 10^{14}\text{ Hz}$
Shift in frequency, $$\Delta f = |f_{\text{obs}} - f| = |4.547 - 4.570| \times 10^{14} = 2.3 \times 10^{12}\text{ Hz}$$
Step 2: Write down the Doppler redshift equation and Calculate
$$
\frac{\Delta f}{f} \approx \frac{v}{c} \implies v = \frac{c\Delta f}{f}
$$
Hence,
$$
v = \frac{(3.0 \times 10^8) \times (2.3 \times 10^{12})}{4.570 \times 10^{14}} = \mathbf{1.5 \times 10^6\text{ m s}^{-1}}
$$
Step 3: Write a concluding sentence
The observed frequency $4.547 \times 10^{14}\text{ Hz}$ is less than the source frequency $4.570 \times 10^{14}\text{ Hz}$. Because there is a decrease in frequency, the light is red-shifted. Therefore, the source is receding, or moving away from the Earth at $1.5 \times 10^6\text{ m s}^{-1}$.