AstronomyGalaxies and the Universe

Standard Candles

Cepheid variables and type Ia supernovae have known luminosities, so how bright they look tells how far away they are.

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Parallax works only for nearby stars. To reach other galaxies, astronomers use standard candles: objects whose luminosity $L$ is known. Measure how bright one looks, $b$, and the inverse-square law gives its distance: $d = \sqrt{\dfrac{L}{4\pi b}}$.

Why it works

The hard part is knowing $L$. Cepheid variables are pulsating yellow supergiant stars that brighten and dim every few days to a few months. In 1912 Henrietta Leavitt, studying Cepheids in the Small Magellanic Cloud (all about the same distance from us), found that the longer a Cepheid's period, the more luminous it is. So timing a Cepheid tells its luminosity, and its brightness then tells its distance. Cepheids are bright enough to see in galaxies tens of millions of light-years away. For greater distances, astronomers use type Ia supernovae, exploding white dwarfs that blow up at nearly the same mass and so peak at nearly the same luminosity, about 5 billion Suns, visible billions of light-years away. Each rung is calibrated by the one below: parallax fixes the nearest Cepheids, Cepheids fix the supernovae.

Key idea

A standard candle has a known luminosity, so its apparent brightness gives its distance: for two of the same kind, $\dfrac{d_2}{d_1} = \sqrt{\dfrac{b_1}{b_2}}$. Cepheids (period tells luminosity) and type Ia supernovae (nearly equal peaks) are the main ones.
A Cepheid's brightness rising and falling, and the period–luminosity relation: from the period, read the luminosity (absolute magnitude).

Worked examples

Worked example: Two Cepheids

A 10-day Cepheid in our galaxy is $1000$ pc away. A 10-day Cepheid in another galaxy looks $10^{8}$ times fainter. How far is that galaxy?
  1. Same period, same luminosity, so $\frac{d_2}{d_1} = \sqrt{\frac{b_1}{b_2}} = \sqrt{10^{8}} = 10^{4}$.
  2. $d_2 = 10^{4} \times 1000 = 10^{7}$ pc $= 10$ Mpc, about 33 million light-years.

Answer: $10$ Mpc.

Worked example: Supernovae

A type Ia supernova in a galaxy $20$ Mpc away peaked $400$ times brighter than one in a distant galaxy. How far is the distant one?
  1. $\frac{d_2}{d_1} = \sqrt{400} = 20$.
  2. $d_2 = 20 \times 20 = 400$ Mpc, about 1.3 billion light-years.

Answer: $400$ Mpc.

Common mistake

Two stars that look equally bright are not necessarily equally far away: one might be a dim star nearby, the other a brilliant star far off. Brightness alone tells nothing about distance unless you know the luminosity. Check: Procyon and Rigel look about equally bright, yet Rigel is about 75 times farther away, so by the inverse-square law it must be thousands of times more luminous.
The cosmic distance ladder: each method covers a range of distances and is calibrated by the one below it.

Practice

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Sources and further reading

  • OpenStax, Astronomy 2e, section 19.3, Variable Stars: One Key to Cosmic Distances
  • OpenStax, Astronomy 2e, section 26.4, The Extragalactic Distance Scale
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