How are Cepheids used to evaluate their distance?

Question

Cepheids are used to evaluate distances. What is the math and physics behind their use?

Answer 1

The reason Cepheids can be used to determine distance is because the are variable stars that change brightness according to a regular pattern. The stars literally change size, swelling larger and smaller and while doing so change in brightness. It was discovered by Henrietta Leavitt in the early 1900's that the period of the Cepheid variability was directly correlated with it absolute maximum brightness. The longer the period, the larger the absolute luminosity of the star.

So by measuring the period of a Cepheid variable, we know what it absolute magnitude (brightness at a distance of 10 parsecs) should be. Given that we can measure is apparent magnitude and the two are related to each other by distance according the the following formula: $$m - M = 5\log (D/10)$$ where M is the absolute magnitude, m is the apparent magnitude and d is the distance in parsecs.

Answer 2

Cepheids pulse, with periods on the order of days to months. The longer the period, the brighter the star is intrinsically.

This was first known experimentally with parallax studies in the 1910s. "Stellar pulsation theory," which sought to explain the experimental relationship, was a huge field for a long time; the first complete theoretical justifications came decades later.

The first model to get it pretty much right was the celebrated paper in 1953 by Sergei Zhevakin, which explains things in terms of ionization of helium (from He+ to He++). These models are hard, but the short version is this:

He++ is more opaque than He+ to photons. In stellar structure, this effectively means that a shell of He++ surrounding a the interior of a star will trap more photons inside the star than He+ will. This will accelerate the rate of fusion within the core. As the rate of fusion builds, the pressure in the core increases, pushing the He++ outwards. The He++ thus expands. As it expands, it loses energy, like any other gas.

But the expansion of the He++ causes it to fall in temperature. This fall in temperature means there is less energy available, so the ionization fraction drops - eventually, the shell becomes mostly He+. Now, the shell is optically thin! The photons inside can rush out, and the core temperature drops as fusion slows down. The star deflates.

But once it has deflated, the He+ compresses, increases in temperature, and ionizes to form He++. The process then repeats.

We see the pulses in this process because the brightness of the star is proportional to the square of its radius; when the star expands, it gets brighter, when it contracts, it gets dimmer.

The rate of pulsation depends on the total mass of the star, since the total mass of the star governs the rate of fusion in the core. Intuitively, we might expect that a bigger star (which has faster fusion in the core) will be relatively less affected by the presence of He++ or He+ than a dim star which is struggling to stay alive. The real mathematical answer is something like this, but comes in the form of a huge array of many, many, many differential equations and can only be solved on a computer, not analytically.

Answer 3

Cepheid variables have a strong correlation between their pulsation period and luminosity. Thus, as we constrain a given Cepheids pulsation period, we can determine it's luminosity. The distance to any object can be found if we know it's luminosity using the relationship between flux and luminosity.