The problem of distance
A star looks faint for two quite different reasons: it may be genuinely dim, or it may be very far away. Separating those possibilities is the central difficulty of astronomy, because everything else depends on knowing how far away things are.
The solution is the standard candle: an object whose true power output — its luminosity — is known from its behaviour rather than assumed. Measure how bright it appears, compare with how bright it should be, and the difference gives the distance through the inverse square law.
Type Ia supernovae are the most useful example. They occur when a white dwarf accumulates matter until it crosses a fixed mass threshold, so they all detonate with very nearly the same energy. Being extraordinarily bright, they can be seen across billions of light years.
- L
- luminositythe true power output, known for a standard candle
- F
- radiant flux intensitythe power per square metre reaching us
- 4πd²
- the sphere's areawhy brightness falls as an inverse square
Reading a star's temperature and size from its light
A star radiates approximately as a black body, and that gives two powerful relationships. Wien's displacement law connects the colour at which it is brightest to its surface temperature — bluer means hotter. Stefan's law then connects that temperature and the star's radius to its total output.
Used together they yield the radius of a star that no telescope can resolve as anything but a point. Measure the peak wavelength to get the temperature, measure the brightness and distance to get the luminosity, and Stefan's law returns the radius.
- λ_max
- peak wavelengthshorter for hotter stars, so blue is hot and red is cool
- σ
- the Stefan constantin the data booklet
- T⁴
- the fourth powerwhy small temperature differences matter enormously
A star has peak wavelength 480 nm and luminosity 4.0 × 10²⁷ W. Find its surface temperature and radius.
- Wien: T = 2.90 × 10⁻³ / λ_max = 2.90 × 10⁻³ / 480 × 10⁻⁹.Converting nanometres to metres before dividing.
- T = 6040 K.Close to the Sun's 5800 K, which makes sense for a star peaking in the visible range.
- Stefan: L = 4πr²σT⁴, so r = √( L / (4πσT⁴) ).Rearranging for the radius, the only unknown left.
- T⁴ = 6040⁴ = 1.331 × 10¹⁵; 4πσT⁴ = 4π(5.67 × 10⁻⁸)(1.331 × 10¹⁵) = 9.48 × 10⁸.Evaluating the denominator in stages reduces calculator errors.
- r = √(4.0 × 10²⁷ / 9.48 × 10⁸) = √(4.22 × 10¹⁸) = 2.05 × 10⁹ m.About three times the Sun's radius — a plausible result for a star of this luminosity.
T = 6040 K, r = 2.05 × 10⁹ m
The fourth power is unforgiving
Forgetting to raise T to the fourth power, or raising it to the second, produces an answer wrong by many orders of magnitude. It is worth writing T⁴ as a separate line in the working. The same applies to r² — the formula contains both, and both must be squared or fourth-powered at the right moment.
Red shift, and a universe that is expanding
Light from distant galaxies arrives with its spectral lines shifted towards longer wavelengths. The lines are recognisable — hydrogen emits at wavelengths we can measure in a laboratory — so the shift can be measured precisely, and it indicates recession.
Hubble's law states that the recession speed is proportional to distance: the further away a galaxy is, the faster it is receding. Crucially, this holds in every direction, which rules out the interpretation that we are at the centre of an explosion. Space itself is expanding, and every observer everywhere sees the same pattern.
Running the expansion backwards implies everything was once together, which is the Big Bang. The reciprocal of the Hubble constant gives an estimate of the age of the universe.
- z
- the red shiftthe fractional change in wavelength
- Δλ
- the wavelength shiftobserved minus laboratory value
- H₀
- the Hubble constantits reciprocal estimates the age of the universe
The chain of reasoning, end to end
- A standard candle has known luminosity, so its brightness gives its distance.
- Wien's law converts peak wavelength into surface temperature.
- Stefan's law then converts temperature and luminosity into radius.
- Red shift of known spectral lines gives the recession speed.
- Hubble's law relates that speed to distance, confirming expansion in all directions.
- The reciprocal of H₀ estimates the age of the universe.
The scale of things, and the units that make it manageable
Astronomical distances are unwieldy in metres, so three units are used according to the scale of the problem, and questions expect you to convert between them.
The astronomical unit is the mean Earth–Sun distance, about 1.50 × 10¹¹ m, and suits distances inside the solar system. The light year is the distance light travels in a year, about 9.46 × 10¹⁵ m, and suits distances between stars. The parsec, about 3.09 × 10¹⁶ m or 3.26 light years, arises from the parallax method of measuring distance and is what professional astronomers use.
A light year is a distance, not a time, despite the name — an error worth avoiding explicitly, since questions sometimes test it. It also means that looking far away is looking back in time: light from a galaxy a billion light years away left it a billion years ago, so every deep image is a picture of the past.
| Unit | Value in metres | Used for |
|---|---|---|
| astronomical unit (AU) | 1.50 × 10¹¹ | distances within the solar system |
| light year (ly) | 9.46 × 10¹⁵ | distances to nearby stars |
| parsec (pc) | 3.09 × 10¹⁶ | professional astronomy; 1 pc = 3.26 ly |
| megaparsec (Mpc) | 3.09 × 10²² | distances between galaxies |
Check which unit H₀ is quoted in
The Hubble constant is often given as about 68 km s⁻¹ Mpc⁻¹, which cannot be used directly with a distance in metres. Converting it to SI gives roughly 2.2 × 10⁻¹⁸ s⁻¹. Mixing the two forms produces an answer wrong by many orders of magnitude, so convert before substituting and check that the units of v = H₀d genuinely cancel to a speed.