The Doppler effect for a moving source and for a moving observer, for sound and for light

Redshift and blueshift: z=Δλ/λ≈Δf/f≈v/c, and when the approximation holds

Hubble's law v=H₀d, Hubble graphs and the units of H₀

The age of the universe from 1/H₀, and why it is uncertain

Critical density and the possible fates of the universe (described qualitatively)

Galaxy rotation curves as evidence for dark matter

1. When a source moves towards you its waves are bunched, so the wavelength is shorter and the frequency higher; moving away, the wavelength is longer and the frequency lower, while the wave speed does not change.

2. For light from a receding galaxy every spectral line moves to a longer wavelength: redshift. z=Δλ/λ≈Δf/f≈v/c, valid when v≪c.

3. Hubble's law: recession speed is proportional to distance, v=H₀d. The gradient of a v–d graph is H₀.

4. If the expansion rate had always been the same, the age of the universe would be 1/H₀, about 14 billion years. Convert H₀ to s⁻¹ first.

5. Whether the expansion continues for ever depends on whether the mean density is above or below the critical density, which is hard to pin down because of dark matter.

6. Stars far from a galaxy's centre orbit much faster than the visible mass can explain: the flat rotation curve is the main evidence for dark matter.

This page follows one chain of reasoning: from a shifted line in a spectrum to the age of the universe. Measure a wavelength, turn it into a speed with z=v/c, combine speeds with distances to get the Hubble constant, and invert the Hubble constant to get an age. It starts with a familiar sound.

A moving source

Each wavefront spreads out from the point where the source was when it was emitted, at the wave speed. The source's motion does not change the wave speed.

Between one wavefront and the next the source moves forward, so it emits each new wavefront closer to the one ahead.

Ahead of the source the wavefronts are bunched: shorter wavelength, so a higher frequency reaches the observer (f=v/λ). Behind the source they are spread out: longer wavelength, lower frequency.

A moving observer

If the source is still and the observer moves towards it, the wavelength in the medium is unchanged, but the observer meets the wavefronts more often, so the observed frequency is higher. Moving away, the observer meets fewer wavefronts per second: lower frequency.

Only the relative motion along the line joining source and observer matters; motion at right angles to that line produces no shift (at everyday speeds).

For light the same idea applies: light from a source moving away is shifted to longer wavelength (redshift); from a source moving towards us, to shorter wavelength (blueshift).

Doppler shifts are used in radar speed measurement and in Doppler ultrasound to measure blood flow, as well as in astronomy.

Atoms of each element absorb and emit at fixed wavelengths, measured in the laboratory: hydrogen, for example, at 410.2, 434.0, 486.1 and 656.3 nm.

In the spectrum of a star or galaxy the pattern of lines is recognisable, but the whole pattern is shifted. Shifted to longer wavelength: the source is moving away (redshift). Shifted to shorter wavelength: moving towards us (blueshift).

z is the redshift (no unit), Δλ = observed wavelength − laboratory (emitted) wavelength, λ the laboratory wavelength, Δf the change in frequency, f the emitted frequency, v the speed of the source along the line of sight and c=3.00×10⁸ m s⁻¹. This is in the formulae list. It is valid only when v is much less than c (in practice, z up to about 0.1). Δλ and λ must be in the same unit; nanometres cancel, so there is no need to convert to metres.

WORKED EXAMPLE · QUESTION

A hydrogen absorption line has a laboratory wavelength of 656.3 nm. In the spectrum of a star the same line is at 655.62 nm. Calculate the speed of the star along the line of sight and state its direction of motion.

SOLUTION

The shift (observed − laboratory): Δλ=655.62-656.3=-0.68 nm

v=c Δλ/λ=3.00×10⁸×0.68÷656.3=3.11×10⁵≈ 3.1 × 10⁵ m s⁻¹

The wavelength is shorter, so this is a blueshift: the star is moving towards us. Δλ is known to only 2 significant figures (0.68 nm), so the speed is given to 2 s.f. even though the wavelengths have 4 and 5.

Rotating objects

When a galaxy or star rotates, one edge moves towards us and the other away, so lines from the two edges are shifted by different amounts. The average of the two shifts gives the speed of the whole object; half the difference gives the rotation speed.

WORKED EXAMPLE · QUESTION

A spiral galaxy is seen edge-on. The hydrogen line of laboratory wavelength 656.3 nm is observed at 659.41 nm from one edge of the disc and at 658.44 nm from the opposite edge. Determine (a) the recession speed of the galaxy as a whole and (b) the orbital speed of the stars at the edge of the disc.

SOLUTION

Speed of each edge along the line of sight: v₁=3.00×10⁸×3.11÷656.3=1.42×10⁶ m s⁻¹; v₂=3.00×10⁸×2.14÷656.3=9.78×10⁵ m s⁻¹.

(a) The galaxy's own motion is the average: (v₁+v₂)÷2= 1.20 × 10⁶ m s⁻¹ away from us

(b) Rotation adds to one edge and subtracts from the other: (v₁-v₂)÷2=2.22×10⁵≈ 2.2 × 10⁵ m s⁻¹

Both edges are redshifted because the recession is larger than the rotation.

Almost every galaxy beyond our local group shows a redshift: galaxies are moving away from us.

Plotting recession speed (from redshift) against distance (from standard candles) gives a straight line through the origin. The further away a galaxy is, the faster it is receding: the pattern expected if the universe is expanding.

Every observer, in any galaxy, would see the same pattern; it does not mean we are at the centre.

v is the recession speed, d the distance and H₀ the Hubble constant. Astronomers quote H₀ in km s⁻¹ Mpc⁻¹; in SI units it is s⁻¹. Hubble's law is in the formulae list. It applies to galaxies at cosmological distances (beyond a few megaparsecs); nearby galaxies have their own motions that can be larger than the Hubble recession, and a few are even blueshifted.

Converting the units

1 pc = 3.09 × 10¹⁶ m, so 1 Mpc = 3.09 × 10²² m. Then H₀ in s⁻¹ = (value in km s⁻¹ Mpc⁻¹) × 10³ ÷ (3.09 × 10²²). For example, 70 km s⁻¹ Mpc⁻¹ = 70 × 10³ ÷ 3.09 × 10²² = 2.27 × 10⁻¹⁸ s⁻¹.

WORKED EXAMPLE · QUESTION

The spectrum of a galaxy has a redshift of 0.0284. Take H₀=70 km s⁻¹ Mpc⁻¹ and 1 Mpc = 3.09 × 10²² m. Calculate (a) the recession speed, (b) the distance to the galaxy in Mpc and in metres and (c) roughly how long ago the light left the galaxy, in years (1 year = 3.15 × 10⁷ s).

SOLUTION

(a) v=zc=0.0284×3.00×10⁸= 8.52 × 10⁶ m s⁻¹ = 8520 km s⁻¹

(b) Keep v in km s⁻¹ to match the units of H₀: d=v/H₀=8520÷70=121.7 Mpc≈ 122 Mpc =121.7×3.09×10²²= 3.76 × 10²⁴ m

(c) time = distance ÷ c =3.76×10²⁴÷3.00×10⁸=1.25×10¹⁶ s≈ 4.0 × 10⁸ years. This is an estimate: the universe has expanded a little while the light was travelling.

Run the expansion backwards. A galaxy now at distance d, receding at speed v, was at our position a time d/v ago. From Hubble's law, d/v=1/H₀: the same for every galaxy, which is why it can be read as the age of the universe. This assumes the galaxies have always moved at their present speeds (a constant rate of expansion).

Not in the formulae list (No — derive it from v=H₀d and time = distance ÷ speed). H₀ must be in s⁻¹ to give the age in seconds.

WORKED EXAMPLE · QUESTION

Use the Hubble graph above to determine the Hubble constant, and estimate the age of the universe in years. 1 Mpc = 3.09 × 10²² m; 1 year = 3.15 × 10⁷ s.

SOLUTION

Gradient from the large triangle: H₀=10 500÷150= 70.0 km s⁻¹ Mpc⁻¹

Convert to SI: H₀=70.0×10³÷3.09×10²²=2.27×10⁻¹⁸ s⁻¹

Age ≈ 1/H₀=1÷2.27×10⁻¹⁸=4.41×10¹⁷ s= 1.40 × 10¹⁰ years (about 14 billion years).

Read the gradient from the best-fit line, not from one data point: individual galaxies scatter about the line because of their own motions and errors in their distances.

Why the age is uncertain

The value of H₀. Distances to galaxies depend on standard candles whose luminosities are themselves uncertain, so H₀ has been revised many times. Different methods today give values from about 67 to 73 km s⁻¹ Mpc⁻¹: ages from 1/H₀ of 13.4 to 14.6 billion years.

The expansion rate has not been constant. Gravity between all the matter in the universe slows the expansion. If the galaxies were moving faster in the past, they reached their present separations in less time than 1/H₀, so 1/H₀ overestimates the age.

How much the expansion has slowed depends on the mean density of the universe, which depends on how much dark matter there is.

Any estimate of the age must be at least as large as the ages of the oldest stars, which provides a cross-check.

The expansion is opposed by the gravitational attraction of all the mass in the universe, like a ball thrown upwards slowing as it rises. Whether the ball comes back depends on how hard it was thrown compared with how strong gravity is; for the universe, the deciding quantity is the density.

For H₀=70 km s⁻¹ Mpc⁻¹ the critical density is about 9.2 × 10⁻²⁷ kg m⁻³, the mass of about 5.5 hydrogen atoms in every cubic metre. (You do not need to calculate it.)

The fate depends on the density relative to the critical density, and the critical density itself depends on H₀, so uncertainty in H₀ feeds into the prediction as well.

The visible matter (stars, gas, dust) adds up to only a few per cent of the critical density.

Whether the universe is open, flat or closed therefore depends on how much dark matter exists, which can be detected only through its gravitational effects.

Beyond this page: observations of distant supernovae suggest that the expansion is in fact speeding up, which needs something beyond matter and gravity ("dark energy"). This is why the fate is still described as uncertain.

A star in a circular orbit about a galaxy's centre has its centripetal force provided by gravity from the mass M inside its orbit: GMm/r²=mv²/r, so v=√(GM/r).

Most of the visible matter is concentrated near the centre. Beyond it, M should stay almost constant, so the orbital speed should fall as 1/√r.

Measured with Doppler shifts, the orbital speeds of stars and gas clouds stay roughly constant out to large distances: the rotation curve is flat.

A flat curve needs M to increase in proportion to r: there must be large amounts of unseen mass far out from the centre.

WORKED EXAMPLE · QUESTION

Gas clouds orbit a spiral galaxy at 25 kpc from its centre with a speed of 220 km s⁻¹, measured from the Doppler shift of their 21 cm radio emission. The visible stars and gas have a total mass of 1.0 × 10¹¹ solar masses, almost all of it well inside this radius. 1 kpc = 3.09 × 10¹⁹ m; solar mass = 1.99 × 10³⁰ kg; G=6.67×10⁻¹¹ N m² kg⁻². (a) Calculate the mass inside the orbit. (b) Calculate the orbital speed expected if only the visible mass were present. (c) Estimate the fraction of the mass that is not visible.

SOLUTION

(a) Gravity provides the centripetal force, so M=v²r/G. r=25×3.09×10¹⁹=7.725×10²⁰ m; M=(2.2×10⁵)²×7.725×10²⁰÷6.67×10⁻¹¹=5.61×10⁴¹ kg= 2.82 × 10¹¹ solar masses

(b) v=√((6.67×10⁻¹¹×1.0×10¹¹×1.99×10³⁰)/(7.725×10²⁰))= 1.31 × 10⁵ m s⁻¹ (about 131 km s⁻¹)

(c) Fraction not visible =1-1.0×10¹¹÷2.82×10¹¹= 0.64, about two-thirds

The measured speed is far above the 131 km s⁻¹ that the visible mass could hold in orbit; without extra unseen mass the gas would escape.

8 Stars, distances and the HR diagram

Hubble's law needs distances, and those come from parallax and standard candles, using I=L/4πd². Errors in those distances are the main reason H₀ is uncertain.

7 Gravitational fields and orbits

The rotation-curve argument is the orbit equation v=√(GM/r) applied to stars around a galaxy instead of satellites around a planet.

Photons, line spectra and progressive waves (Unit 2)

Redshift is measured on the line spectra produced by electron transitions between energy levels; the laboratory wavelengths come from hf=E₁-E₂. The Doppler argument uses only v=fλ: the wave speed is fixed, so a change in wavelength must mean a change in frequency.

1. A car sounding its horn drives past you at constant speed. Describe how the pitch you hear changes, and explain why.

2. A line of laboratory wavelength 500.0 nm is observed at 501.5 nm in light from a galaxy. Calculate the galaxy's speed and state its direction.

3. A galaxy has a redshift of 0.0125. Taking H₀ = 68 km s⁻¹ Mpc⁻¹, calculate its distance in Mpc.

4. Estimates of H₀ once ranged from 60 to 80 km s⁻¹ Mpc⁻¹. What range of ages does 1/H₀ give?

5. Explain how the rotation curves of spiral galaxies suggest that dark matter exists.

1. The pitch is higher while the car approaches and drops to a lower pitch as it passes and moves away. Approaching, the wavefronts are bunched (shorter wavelength, higher frequency); receding, they are spread out (longer wavelength, lower frequency). The change happens as the car passes you, when its motion along your line of sight reverses.

2. v=3.00×10⁸×1.5÷500.0= 9.0 × 10⁵ m s⁻¹, moving away (redshift).

3. v=0.0125×3.00×10⁵ km s⁻¹=3750 km s⁻¹; d=3750÷68= 55 Mpc.

4. 1/H₀= 1.6 × 10¹⁰ years for 60 and 1.2 × 10¹⁰ years for 80 km s⁻¹ Mpc⁻¹. A larger Hubble constant means a faster expansion and a younger universe.

5. Orbital speed depends on the mass inside the orbit, v=√(GM/r). Most visible mass is near the centre, so speeds should fall with distance beyond it. Measured (Doppler) speeds stay roughly constant, so the mass inside the orbit must keep increasing with r: there is mass far out that we cannot see.