The Vainu Bappu Observatory stands at Kavalur, in the Javadi Hills of Tamil Nadu. There a spectrograph on the telescope spreads the faint light of a distant galaxy into its colours. Hydrogen, glowing in a lamp in the laboratory, gives a bright red line at \(656\,\text{nm}\). The same red line of hydrogen appears in the galaxy's spectrum, but at about \(669\,\text{nm}\). Every other line of the galaxy is moved towards the red by the same fraction.
Wavefronts from a moving source
The chapter on beats and the Doppler effect showed what happens when a source of sound moves. Each wavefront, once emitted, grows as a circle about the point where the source was at that moment. The source moves on before it sends the next front. So the fronts crowd together ahead of the source and spread apart behind it.
Light behaves in the same way. Play Fig 2.1. The source S sends out a wavefront once every period \(T\). In one period the first front travels \(cT = \lambda\), while the source follows it by \(vT\). The next front starts that much closer.
So the wavelength ahead of the source is \(\lambda' = (c - v)T\), and behind it is \((c + v)T\). The observer ahead sees a shorter wavelength, shifted towards the blue end of the spectrum. The observer behind sees a longer one, shifted towards the red. This change of wavelength by the motion of the source or the observer is the Doppler effect.
For light, only the relative speed matters
Sound travels through air, so the air gives a standard of rest. A moving source and a moving observer at the same speed give different pitches, as the chapter on the Doppler effect found. With the speed of sound \(340\,\text{m/s}\) and \(34\,\text{m/s}\) of motion, a source approaching raises the frequency by the factor \(340/306 = 1.111\). An observer approaching raises it by \(374/340 = 1.100\).
Light needs no medium. Nothing marks which of the two is "really" moving, the source or the observer. Einstein's theory of relativity of 1905 shows that the shift of light depends only on the speed at which source and observer approach or separate. For speeds much smaller than \(c\), both cases give the same simple result.
A star moves towards the Earth at \(3 \times 10^4\,\text{m/s}\). In a second case the Earth moves towards a star at rest at the same speed. How do the Doppler shifts of the starlight compare?
Red shift and blue shift
For a source moving away at speed \(v\), the wavelength the observer receives is \(\lambda' = (c + v)T = \lambda(1 + v/c)\). The change is \(\Delta\lambda\) \(= \lambda' - \lambda\) \(= \lambda v/c\). For a source coming closer, the sign of \(v\) is reversed and the wavelength shrinks.
A lengthened wavelength moves every line towards the red end of the spectrum. This is a red shift, and it means the source is moving away. A shortened wavelength moves the lines towards the violet end, a blue shift, and the source is coming closer. Only the part of the velocity along the line of sight counts. A source moving across the line of sight gives no first-order shift.
Fig 2.2 shows the rule an astronomer relies on. Every line of one source moves by the same fraction \(\Delta\lambda/\lambda\), not by the same number of nanometres. A \(5\,\%\) red shift moves the line at \(434\,\text{nm}\) by about \(22\,\text{nm}\) and the line at \(656\,\text{nm}\) by about \(33\,\text{nm}\). That common fraction identifies the lines and gives the speed.
A galaxy moves away from the Earth at \(3 \times 10^6\,\text{m/s}\). At what wavelength do we see a line that is emitted at \(500\,\text{nm}\)?
The galaxy is receding, so the line is red-shifted. \(v = 3 \times 10^6\,\text{m/s}\) is a hundredth of \(c\), much less than \(c\).
\(\Delta\lambda = \lambda\,v/c\).
\(v/c\) \(= 3 \times 10^6/(3 \times 10^8)\) \(= 0.01\), so \(\Delta\lambda\) \(= 500 \times 0.01\) \(= 5\,\text{nm}\).
The line is seen at \(505\,\text{nm}\), a red shift of \(5\,\text{nm}\).
A line that is \(600\,\text{nm}\) in the laboratory is seen at \(603\,\text{nm}\) in the light of a galaxy. How fast is the galaxy moving away, in units of \(10^6\,\text{m/s}\)?
A star moves towards the Earth at \(6 \times 10^5\,\text{m/s}\). What happens to a line it emits at \(500\,\text{nm}\)?
The rule \(\Delta\lambda/\lambda = v/c\) is the first step of an exact result. Relativity gives, for motion along the line of sight, \(\lambda'/\lambda\) \(= \sqrt{(1 + v/c)/(1 - v/c)}\). At \(v = 0.01c\) this is \(1.01005\), and the simple rule gives \(1.01\): no difference worth measuring. At \(v = 0.5c\) the exact ratio is \(\sqrt3 = 1.73\), while the simple rule gives \(1.5\). So the simple rule is kept only for speeds much less than \(c\).
Why is \(\Delta\lambda/\lambda = v/c\) used only for speeds much smaller than \(c\)?
The rule keeps only the first power of \(v/c\). The exact result from the theory of relativity has further terms in \(v^2/c^2\) and higher powers, and it includes the slowing of the source's clock at high speed. When \(v \ll c\) these extra terms are far smaller than the first, so the simple rule is accurate. For speeds comparable with \(c\), such as \(0.5c\), it would be badly wrong: it gives \(1.5\) where the true ratio is \(1.73\).
The expanding universe
Nearby stars in our own galaxy show small shifts of both kinds, some red and some blue. They move about the centre of the galaxy at a few tens of kilometres per second. Distant galaxies are different. Nearly all of them show red shifts, and the fainter and farther a galaxy, the larger its red shift.
From 1912, Vesto Slipher at the Lowell Observatory in Arizona photographed the spectra of faint spiral nebulae. Most of them were red-shifted, some by more than \(1000\,\text{km/s}\). Nobody yet knew what the nebulae were. In the 1920s Edwin Hubble, at Mount Wilson in California, showed that they are galaxies far outside our own, and measured their distances. In 1929 he put the two together. A galaxy's speed of recession is proportional to its distance. Every distant galaxy is moving away from every other: the universe is expanding.
The rule is Hubble's law, \(v = H_0 d\). Its constant \(H_0\) is about \(70\,\text{km/s}\) for every megaparsec of distance, and a megaparsec is about \(3.26\) million light years. A galaxy with \(\Delta\lambda/\lambda = 0.01\) recedes at \(3000\,\text{km/s}\). It is about \(3000/70 \approx 43\) megaparsecs away, some 140 million light years.
A galaxy shows a red shift \(\Delta\lambda/\lambda = 0.02\). Using \(H_0 = 70\,\text{km/s}\) per megaparsec, roughly how far away is it, in megaparsecs?
On a highway, a traffic policeman aims a radar speed gun at an approaching car. The gun sends out radio waves of a fixed frequency, often about \(24.15\,\text{GHz}\). The car meets the waves as a moving observer, and sends them back as a moving source, so the shift is counted twice: \(\Delta f = 2vf/c\). For a car at \(30\,\text{m/s}\), \(108\,\text{km/h}\), \(\Delta f\) \(= 2 \times 30 \times 24.15 \times 10^9/(3 \times 10^8)\) \(= 4830\,\text{Hz}\). The gun mixes the returning wave with its own and counts the beats, a few thousand per second, then displays the speed. Radio waves are light of a long wavelength, so the same rule holds.
Now return to Kavalur. The red line of hydrogen, \(656\,\text{nm}\) in the laboratory, arrives at about \(669\,\text{nm}\). The wavelength is longer, so the galaxy is moving away. The shift is \(\Delta\lambda/\lambda\) \(= 13/656\) \(\approx 0.0198\). Its speed is \(0.0198 \times 3 \times 10^8\) \(\approx 5.9 \times 10^6\,\text{m/s}\), about \(2\,\%\) of the speed of light. By Hubble's law it is roughly \(85\) megaparsecs away.
- Reading a red shift as a change in the colour we see, so that a blue star turns red.Every line moves by the same small fraction. A red shift of 1 % hardly changes the look of a star; it is found by measuring lines.
- Adding the shift as a fixed number of nanometres to every line.The fraction \(\Delta\lambda/\lambda = v/c\) is the same for every line, so longer wavelengths move by more nanometres.
- Using the two different sound formulas for a moving source and a moving observer.Light has no medium; only the relative speed matters, and for \(v \ll c\) both give \(\Delta\lambda/\lambda = v/c\).
- Using \(\Delta\lambda/\lambda = v/c\) for a source moving at half the speed of light.The simple rule holds only for \(v \ll c\). Near \(c\) the exact relativistic formula is needed.