# How We Know the Distance to a Star: Climbing the Cosmic Distance Ladder One Rung at a Time
> Parallax, Cepheids and Type Ia supernovae form a three-rung ruler. We build each rung from the inverse-square law, find the distance to δ Cephei, and track how errors accumulate.
> https://rikai.mugen-giken.com/en/physics/cosmology/cosmic-distance-ladder
## 0. Key points
- One cannot reach a star with a tape measure. What astronomy does instead is to repeat, many times over, a single manoeuvre: **use a ruler that works nearby to engrave the scale of a ruler that works a little farther out**. The result is called the cosmic distance ladder.
- The first rung is **annual parallax**. Triangulation on a baseline given by the Earth's orbit yields a distance from geometry alone. The formula says nothing more than "the distance in parsecs is the reciprocal of the parallax in arcseconds". Its reach, however, stops at a few thousand parsecs — not even enough to cross our own Galaxy.
- The second rung is the **Cepheid variable**. Measuring the period over which its brightness varies reveals its true brightness (its luminosity), so comparison with the apparent brightness gives the distance. This reaches galaxies 30 million parsecs (about 100 million light-years) away.
- The third rung is the **Type Ia supernova**. The mass that pulls the trigger of the explosion is nearly always the same, so the peak luminosities agree. Each one shines briefly with the light of an entire galaxy, and so can be seen billions of light-years away.
- Distance ratios accumulate multiplicatively, so **in magnitudes (that is, logarithmically) they accumulate additively**. Errors accumulate the same way, and a single rung whose scale is off by 0.2 mag shifts the Hubble constant by a few percent. This is the stage on which cosmology's largest open problem — the Hubble tension — is played out.
## 1. Motivation: on the fact that tape measures do not stretch four light-years
Astronomy textbooks state, without apparent embarrassment, that this star is 430 light-years away and that galaxy 55 million. Nobody has ever gone there. Why is anyone entitled to write a number?
For astronomy this is not a piece of trivia but **the foundation of everything**. A star's true brightness, its size, its mass, the speed at which a galaxy is receding, the age of the universe: not one of them is determined until the distance is. Looking up at the sky, all we ever obtain is "from which direction on the celestial sphere, with what brightness, and in what colour the light arrives". Depth information is simply not contained in that.
The difficulty has been known since antiquity. In the second century BC, Hipparchus estimated the distance to the Moon from the size of the Earth's shadow during a lunar eclipse, and obtained about 60 Earth radii (the modern value is about 60.3, so the accuracy is remarkable). His instrument was triangulation, with the size of the Earth as the baseline. Attempt the same trick on a star, however, and it fails completely.
In the sixteenth century, when Copernicus proposed that the Earth revolves around the Sun, his opponents had a powerful argument. If the Earth moves, nearby stars ought to be seen shifting back and forth slightly, once a year, against the distant ones. No such motion was observed. Tycho Brahe, the finest observer of his age, took this as evidence that the Earth stands still. His reasoning was sound; he was merely missing a premise. The stars were simply unimaginably far away.
That back-and-forth motion — annual parallax — was finally captured in 1838, by Bessel, observing 61 Cygni. The angle he measured was about 0.31 arcseconds, the angle subtended by a one-yen coin placed 13 kilometres away. Some three centuries after Copernicus, astronomy had at last set its first rung firmly in place.
In what follows we assemble the ladder starting from that first rung. The only thing we take for granted is the idea, treated in [Was there really a Big Bang?](/en/physics/cosmology/big-bang-evidence), that light carries information to us.
## 2. Preliminaries: what brightness tells us about distance
Every rung above the first uses "brightness" as its ruler. So let us first sort out the relation between brightness and distance.
The energy that an object radiates in all directions per unit time is its **luminosity**, written $L$ and measured in watts (W). This is a property the object was born with; it does not depend on where the observer stands.
The energy that crosses, per unit time, a unit area placed at the observer's position perpendicular to the direction of propagation is the **flux** (the apparent brightness), written $F$ and measured in $\mathrm{W/m^2}$. This does depend on where the observer stands.
What links the two is the beating heart of the distance ladder.
Suppose an object of luminosity $L$ radiates uniformly in all directions, and that nothing between the object and the observer absorbs or scatters light. Then the flux at distance $d$ from the object is
$$
F = \frac{L}{4\pi d^2}
$$
Consider the sphere of radius $d$ centred on the object. The energy $L$ emitted in one second passes, by the assumption that nothing absorbs it along the way, entirely through this sphere. By the assumption of uniform emission, the energy crossing the sphere is the same at every point of it.
The area of a sphere of radius $d$ is $4\pi d^2$, so the energy crossing unit area of the sphere per second is $L/(4\pi d^2)$. The observer sits on that sphere, so this is precisely $F$.
The formula contains one piece of good news and one piece of bad. The good news is that if both $L$ and $F$ are known, the distance follows as $d = \sqrt{L/(4\pi F)}$. The bad news is that since only $F$ can be observed, no distance emerges until $L$ is procured from somewhere. That procurement is exactly what each rung of the ladder does.
An object whose **luminosity $L$ can be known in advance** from some observable feature — a pulsation period, a spectral type, the shape of a light curve — is called a **standard candle**.
Find one standard candle and gives the distance to it. Cepheid variables and Type Ia supernovae are, in the end, standard candles.
### 2.1. Magnitudes: why astronomers use logarithms
Astronomers express brightness not in watts but in **magnitudes**. The origin is Hipparchus, who classified the brightest stars visible to the naked eye as first magnitude and the barely visible ones as sixth. In the nineteenth century Pogson noticed that this classification in fact corresponds to "a difference of five magnitudes for a factor of about 100 in brightness", and redefined it as follows.
If two objects have fluxes $F_1, F_2$, the difference of their **apparent magnitudes** $m_1, m_2$ is defined by
$$
m_1 - m_2 = -2.5 \log_{10} \frac{F_1}{F_2}
$$
Note that a smaller magnitude means a brighter object.
The apparent magnitude the object would have if it were moved to a distance of $10\ \mathrm{pc}$ (parsecs; see below) is its **absolute magnitude**, written $M$. The absolute magnitude corresponds one-to-one with the luminosity and is intrinsic to the object.
The difference $\mu = m - M$ is called the **distance modulus**.
Let $d$ be the distance to the object, measured in parsecs. If nothing along the way absorbs light, then
$$
\mu = m - M = 5 \log_{10} \frac{d}{10\ \mathrm{pc}}, \qquad
d = 10^{(\mu + 5)/5}\ \mathrm{pc}
$$
Let $F(d)$ be the flux of the object seen at distance $d$, and $F(10)$ the flux it would have at $10\ \mathrm{pc}$. By definition the former corresponds to magnitude $m$ and the latter to $M$. Applying ,
$$
m - M = -2.5 \log_{10} \frac{F(d)}{F(10)}
$$
By we have $F(d) = L/(4\pi d^2)$ and $F(10) = L/(4\pi \cdot (10\ \mathrm{pc})^2)$; since it is the same object, the luminosity $L$ cancels and
$$
\frac{F(d)}{F(10)} = \left( \frac{10\ \mathrm{pc}}{d} \right)^{2}
$$
Substituting this and using $\log_{10} x^{-2} = -2\log_{10} x$,
$$
m - M = -2.5 \times \left( -2 \log_{10} \frac{d}{10\ \mathrm{pc}} \right) = 5 \log_{10} \frac{d}{10\ \mathrm{pc}}
$$
The second formula is just this one solved for $d$: from $\mu/5 = \log_{10} d - 1$ we get $\log_{10} d = \mu/5 + 1 = (\mu+5)/5$.
Why go to the trouble of using logarithms? The reason becomes plain in §6. In one sentence: **distance ratios accumulate multiplicatively, but written in magnitudes they accumulate additively**. For a science that splices rung upon rung, this property is decisively convenient.
The Sun has apparent magnitude $m = -26.74$. Its distance is $1\ \mathrm{au} = 1/206265\ \mathrm{pc} = 4.85 \times 10^{-6}\ \mathrm{pc}$, so by
$$
\mu = 5 \log_{10} \frac{4.85 \times 10^{-6}}{10} = 5 \log_{10} (4.85 \times 10^{-7}) = 5 \times (-6.314) = -31.57
$$
Hence the absolute magnitude is $M = m - \mu = -26.74 + 31.57 = 4.83$.
Sirius, the brightest star in the sky, has $m = -1.46$ at a distance of $2.64\ \mathrm{pc}$:
$$
\mu = 5 \log_{10} \frac{2.64}{10} = 5 \times (-0.579) = -2.89, \qquad M = -1.46 + 2.89 = 1.43
$$
In appearance the Sun outshines Sirius by more than 25 magnitudes, yet placed side by side at the same distance Sirius is brighter by $4.83 - 1.43 = 3.40$ magnitudes. As a luminosity ratio that is $10^{3.40/2.5} = 10^{1.36} \approx 23$.
Apparent brightness, in other words, lies about distance without the slightest hesitation. That is exactly why an independent means of fixing the distance is required.
Finally, let us prepare the "error conversion table" that we shall use repeatedly.
If the estimated distance modulus deviates from the true value by $\delta\mu$, the estimated distance is $10^{\delta\mu/5}$ times the true distance. When $|\delta\mu|$ is well below one magnitude this is approximated by
$$
\frac{\delta d}{d} \approx \frac{\ln 10}{5}\, \delta\mu \approx 0.46 \, \delta\mu
$$
That is, **a shift of 0.1 mag corresponds to about 4.6 % in distance**.
By we have $d = 10^{(\mu+5)/5}$, so replacing $\mu$ by $\mu + \delta\mu$ gives
$$
\frac{d(\mu + \delta\mu)}{d(\mu)} = \frac{10^{(\mu + \delta\mu + 5)/5}}{10^{(\mu+5)/5}} = 10^{\delta\mu/5}
$$
which is the first assertion. For the second, rewrite $10^{x} = e^{x \ln 10}$ and use $e^{y} \approx 1 + y$ for small $|y|$:
$$
10^{\delta\mu/5} = e^{(\ln 10 / 5)\delta\mu} \approx 1 + \frac{\ln 10}{5}\delta\mu = 1 + 0.4605\,\delta\mu
$$
giving $\delta d/d = 10^{\delta\mu/5} - 1 \approx 0.46\,\delta\mu$.
As a check, for $\delta\mu = 0.1$ the exact value is $10^{0.02} - 1 = 0.0471$ against the approximation $0.046$ — the agreement is good.
## 3. Rung one: annual parallax — the only purely geometric ruler in the universe
Hold up a finger and close each eye in turn. The finger appears to jump left and right against the background. What the brain is measuring is the apex angle of a thin triangle whose base is the separation of the eyes and whose height is the distance to the finger. Annual parallax works on exactly the same principle, with the base replaced by the radius of the Earth's orbit.
For a given star, the maximum of the angle between the direction as seen from the Sun and the direction as seen from the Earth is its **annual parallax**, written $p$. Equivalently, it is the angle subtended at the star by the orbital radius $1\ \mathrm{au}$ of the Earth.
The distance at which the annual parallax is exactly $1$ arcsecond ($1'' = 1/3600$ of a degree) is defined to be **1 parsec** (pc). The name is a contraction of *parallax* and *second*.
Let $p['']$ be the annual parallax measured in arcseconds and $d[\mathrm{pc}]$ the distance measured in parsecs. Then
$$
d[\mathrm{pc}] = \frac{1}{p['']}
$$
Here $1\ \mathrm{pc} = 206265\ \mathrm{au} = 3.086 \times 10^{13}\ \mathrm{km} = 3.26\ \text{light-years}$.
Consider the right triangle formed by the Sun, the Earth and the star. The right angle is at the Sun; the side joining the Sun to the Earth has length $a = 1\ \mathrm{au}$, the side joining the Sun to the star has length $d$, and the angle at the star is $p$. Hence
$$
\tan p = \frac{a}{d}
$$
Now $p$ is an exceedingly small angle. For a small angle $x$ in radians we have $\tan x \approx x$ (even at $p = 1''$ the error is below $10^{-11}$), so writing $p[\mathrm{rad}]$ for $p$ measured in radians,
$$
d = \frac{a}{p[\mathrm{rad}]}
$$
Next we convert to arcseconds. Since $180$ degrees $= \pi$ radians and $1$ degree $= 3600$ arcseconds,
$$
1\ \mathrm{rad} = \frac{180 \times 3600}{\pi}\ '' = 206264.8\ ''
$$
Therefore $p[\mathrm{rad}] = p['']/206265$ and
$$
d = \frac{206265 \, a}{p['']} = \frac{206265\ \mathrm{au}}{p['']}
$$
By , the distance at which $p[''] = 1$ is $1\ \mathrm{pc}$, so $1\ \mathrm{pc} = 206265\ \mathrm{au}$. Dividing both sides by this gives $d[\mathrm{pc}] = 1/p['']$.
Let us check the numbers too. Since $1\ \mathrm{au} = 1.496 \times 10^{8}\ \mathrm{km}$, we get $1\ \mathrm{pc} = 206265 \times 1.496 \times 10^{8} = 3.086 \times 10^{13}\ \mathrm{km}$, and since one light-year is $9.461 \times 10^{12}\ \mathrm{km}$, that is $3.086/0.9461 = 3.26$ light-years.
The parsec is a unit invented precisely so that this formula reduces to "one over the parallax". Astronomers prefer parsecs to light-years not out of affectation but because the arithmetic is easier.
Proxima Centauri has an annual parallax measured by the Gaia satellite as $p = 0.768''$. By ,
$$
d = \frac{1}{0.768} = 1.302\ \mathrm{pc}
$$
In light-years that is $1.302 \times 3.26 = 4.25$; in kilometres, $1.302 \times 3.086 \times 10^{13} = 4.0 \times 10^{13}\ \mathrm{km}$, roughly 40 trillion kilometres.
A present-day planetary probe travelling at some $30\ \mathrm{km}$ per second would take about 40 000 years to arrive. And this is the *nearest* star.
Let $\sigma_p$ (in arcseconds) be the measurement error of the parallax. If we wish to determine the distance to within a relative error $\varepsilon$, the largest measurable distance is
$$
d_{\max}[\mathrm{pc}] = \frac{\varepsilon}{\sigma_p['']}
$$
By we have $d = 1/p$, so if $p$ shifts by $\sigma_p$ the relative error in $d$ is
$$
\frac{\sigma_d}{d} = \frac{|1/(p+\sigma_p) - 1/p|}{1/p} = \frac{\sigma_p}{p + \sigma_p} \approx \frac{\sigma_p}{p}
$$
(for $\sigma_p \ll p$). Substituting $p = 1/d$ once more gives $\sigma_d/d \approx \sigma_p \, d$ (with $d$ in parsecs and $\sigma_p$ in arcseconds). The requirement that this be at most $\varepsilon$ reads $d \le \varepsilon/\sigma_p$.
Let us put numbers in. Allowing a relative error of 10 % ($\varepsilon = 0.1$):
| Instrument | Parallax error $\sigma_p$ | Reach $d_{\max}$ |
|---|---|---|
| Ground-based telescopes (20th century) | about $0.01''$ | about 10 pc |
| Hipparcos satellite (1990s) | about $0.001''$ | about 100 pc |
| Gaia satellite (2010s onwards) | about $0.00002''$ | about 5000 pc |
Gaia is astonishing. An accuracy of $0.00002''$ (20 microarcseconds) is comparable to resolving the thickness of a one-yen coin placed on the far side of the Earth. Even so, its reach is 5000 parsecs, about 16 000 light-years.
But the Galaxy is about 30 000 parsecs across, and the Andromeda galaxy lies some 780 000 parsecs away. **The range measurable by geometry alone cannot even cross the galaxy we live in.** The first rung ends here, and we must step onto the second.
## 4. Rung two: Cepheid variables — the rate of blinking reveals the brightness
In 1908, at the Harvard College Observatory, Henrietta Leavitt was employed comparing photographic plates of the Small Magellanic Cloud, on and on, in search of variable stars. At the time women were not permitted at the telescope; their work as "computers" was to measure the plates.
For 25 variable stars she tabulated the period over which the brightness varied together with the mean apparent brightness. A relation emerged with startling cleanliness, as a straight line (when the horizontal axis is the logarithm of the period): **the longer the period, the brighter the star**.
What made this discovery decisive was the stage on which it took place. Every star in the Small Magellanic Cloud is at essentially the same distance from us (the depth of the cloud is far smaller than the distance to it). Hence $d$ in is common to them all, and **differences in apparent brightness are differences in true brightness**. Leavitt grasped the relation between luminosity and period without knowing any distance at all.
For the class of pulsating variables known as Cepheid variables, the period of variation $P$ (in days) and the absolute magnitude $M_V$ satisfy the linear relation
$$
M_V = a \log_{10} \frac{P}{1\ \text{day}} + b
$$
This is the **period–luminosity relation** (Leavitt's law). In visible light the coefficients are roughly $a \approx -2.8$ and $b \approx -1.4$.
Note that $a$ is negative. Since smaller magnitudes mean brighter stars, this says that longer periods go with greater brightness.
Why should such a relation hold? A Cepheid is a "heat engine": the ionisation and recombination of helium make the transparency of its outer layers vary periodically, so the whole star repeatedly swells and contracts. The more a star is swollen, the longer its pulsation period, and at the same time the larger its surface area and hence its luminosity. Period and luminosity are tied together because both are set by the size of the star.
Suppose the period $P$ and the mean apparent magnitude $m_V$ of a Cepheid variable are observed, and that the interstellar extinction $A_V$ (in magnitudes) in that direction is known. Then the distance to the star is
$$
d = 10^{(m_V - A_V - M_V + 5)/5}\ \mathrm{pc}, \qquad M_V = a \log_{10} P + b
$$
First, fixes the absolute magnitude $M_V$ from the observed period $P$. Next, had there been no absorption by interstellar matter, the apparent magnitude would have been $m_V - A_V$ (extinction makes a star fainter, that is, increases its magnitude, so restoring the true value means subtracting).
The extinction-corrected distance modulus is therefore $\mu_0 = (m_V - A_V) - M_V$. Substituting $\mu = \mu_0$ into the second formula $d = 10^{(\mu+5)/5}$ of yields the stated expression.
Let us try the star that gave the class its name, δ Cephei. The observed values are as follows.
- Period $P = 5.366$ days
- Mean apparent magnitude $m_V = 3.95$
- Interstellar extinction in this direction $A_V = 0.28$ mag
First the absolute magnitude. Since $\log_{10} 5.366 = 0.7296$, gives
$$
M_V = -2.8 \times 0.7296 - 1.4 = -2.043 - 1.4 = -3.44
$$
The extinction-corrected distance modulus is then
$$
\mu_0 = (3.95 - 0.28) - (-3.44) = 3.67 + 3.44 = 7.11
$$
and gives
$$
d = 10^{(7.11 + 5)/5} = 10^{2.422} = 264\ \mathrm{pc}
$$
The Hubble Space Telescope, meanwhile, has measured the annual parallax of δ Cephei directly, obtaining $273 \pm 11\ \mathrm{pc}$. **Two values obtained by entirely different principles agree within the errors.** This is the most direct evidence available that the joints of the ladder can be trusted.
Let us see what happens if the extinction correction is forgotten. Taking $\mu = 3.95 + 3.44 = 7.39$,
$$
d = 10^{(7.39+5)/5} = 10^{2.478} = 301\ \mathrm{pc}
$$
about 14 % farther than the true value. Exactly as the conversion in predicts, a shift of $0.28$ mag produced $0.46 \times 0.28 = 0.13$, that is a 13 % shift in distance (the exact figure being $10^{0.28/5} - 1 = 0.137$).
rested on the assumption that nothing absorbs light along the way. Real galaxies contain drifting interstellar dust, which absorbs and scatters light. This is **interstellar extinction**.
Extinction makes stars look fainter, so forgetting the correction always leads to an **overestimate** of the distance. Moreover, dust scatters blue light more strongly (which is why sunsets are red), so distant stars appear redder than they really are. The standard trick is to measure the degree of this "reddening" and thereby infer the extinction $A_V$.
In recent years, observations in the infrared have been preferred, because infrared light passes through dust readily and so the error in the extinction correction itself is smaller.
One of the principal early goals of the Hubble Space Telescope was to find individual Cepheid variables inside M100, a spiral galaxy in the Virgo cluster. In 1994 this succeeded.
Suppose one of the Cepheids found had period $P = 30$ days and, after extinction correction, mean apparent magnitude $m_V = 25.6$. Since $\log_{10} 30 = 1.477$,
$$
M_V = -2.8 \times 1.477 - 1.4 = -4.14 - 1.4 = -5.54
$$
$$
\mu = 25.6 - (-5.54) = 31.14, \qquad d = 10^{(31.14+5)/5} = 10^{7.228} = 1.69 \times 10^{7}\ \mathrm{pc}
$$
That is about $17$ megaparsecs, or in light-years $1.69 \times 10^{7} \times 3.26 = 5.5 \times 10^{7}$, 55 million light-years. The value actually reported was $17.1 \pm 1.8$ megaparsecs.
Pause over that apparent magnitude of $25.6$. The naked-eye limit is 6th magnitude, and $25.6$ is fainter by a factor of $10^{(25.6-6)/2.5} = 10^{7.84} \approx 7 \times 10^{7}$. On the ground such a star drowns in atmospheric turbulence; this is why a space telescope was needed.
In the 1920s Hubble found Cepheids inside the Andromeda galaxy and derived a distance of about 300 000 parsecs. This was the historic calculation that settled the great debate of the day over whether the spiral nebulae were objects within the Galaxy or separate galaxies outside it.
But in 1952 Baade discovered that Cepheids come in two populations obeying different period–luminosity relations. Hubble had picked the wrong population. When the correction was applied, **every extragalactic distance roughly doubled at a stroke**. The universe grew twice as large overnight.
When the scale on a lower rung goes wrong, every rung above it goes wrong together. That lesson applies verbatim to the discussion of the Hubble tension below. Even today, how strongly the Cepheid period–luminosity relation depends on the heavy-element content (metallicity) of the host galaxy is one of the main uncertainties in precision measurements.
## 5. Rung three: Type Ia supernovae — disposable street lamps supplied by the universe
Cepheids reach, at best, some 30 million parsecs (about 100 million light-years). In galaxies farther than that, individual stars cannot be resolved. A brighter standard candle is required.
Enter the **Type Ia supernova**. The mechanism is as follows.
A star of roughly solar mass ends its life as a **white dwarf**, an object with about the mass of the Sun packed into about the size of the Earth. A white dwarf supports itself by the quantum-mechanical pressure of degenerate electrons, but there is a limit to what this support can do. Above about 1.4 solar masses — the **Chandrasekhar limit** — it fails.
If the white dwarf sits in a binary system and keeps drawing gas from its companion, its mass creeps toward that limit. Just short of it, carbon ignites in the core, fusion spreads explosively through the whole star, and the white dwarf is blown apart without remainder. That is a Type Ia supernova.
The decisive point is that **the mass at which the trigger is pulled is nearly always the same**. The same amount of fuel burns in the same way, so the energy released is nearly the same as well. The absolute magnitude at peak is $M_B \approx -19.3$, about five billion times the brightness of the Sun. For a while it rivals the combined light of the hundreds of billions of stars in its host galaxy.
A Type Ia supernova, in short, is a street lamp of standardised specification that the universe has set out on its own — albeit a disposable one that burns out in a few weeks.
First we calibrate the scale. In 2011 a Type Ia supernova, SN 2011fe, exploded in the relatively nearby galaxy M101 (the Pinwheel Galaxy). The distance to M101 has been measured with Cepheids: its distance modulus is $\mu = 29.04$ (about $6.4$ megaparsecs). The supernova reached an apparent magnitude of about $9.9$ at maximum. Hence
$$
M_B = 9.9 - 29.04 = -19.1
$$
Indeed in the $-19$ range. This is the moment at which **the scale is handed over from the Cepheid rung to the supernova rung**.
Now let us use that scale. Suppose a Type Ia supernova is found in a distant galaxy with apparent magnitude $m_B = 20.0$ at maximum.
$$
\mu = 20.0 - (-19.1) = 39.1, \qquad d = 10^{(39.1+5)/5} = 10^{8.82} = 6.6 \times 10^{8}\ \mathrm{pc}
$$
About 660 megaparsecs, or roughly 2.1 billion light-years.
Had there been a Cepheid in the same galaxy, how bright would it appear? A Cepheid of period 30 days has $M_V = -5.54$, so
$$
m_V = 39.1 - 5.54 = 33.6
$$
magnitude $33.6$. The Hubble Space Telescope's limit is around 30th magnitude, so this is hopeless. The numbers make plain why Type Ia supernovae are needed.
### 5.1. Turning "nearly the same" into "very nearly the same"
To be honest, the peak luminosities of Type Ia supernovae are not perfectly uniform. Raw, the scatter is about $0.4$ mag, which by is about 18 % in distance. That is not good enough for precision cosmology.
In 1993 Phillips found a regularity in the scatter: **the brighter the supernova, the more slowly it fades after the explosion**. Writing $\Delta m_{15}(B)$ for the number of magnitudes by which it dims in the B band over the 15 days following maximum, there is a clean correlation between this quantity and the peak absolute magnitude.
Since $\Delta m_{15}(B)$ is measurable by following the light curve (the change of brightness with time), it can be used to correct the scatter. After correction the scatter falls to about $0.15$ mag, some 7 % in distance. Rather than a standard candle, it is more accurate to call this "a candle that can be calibrated if you observe it".
Above we said that the white dwarf draws gas from a binary companion, but this is only one leading scenario (the single-degenerate scenario). A scenario in which two white dwarfs merge and explode (the double-degenerate scenario) is equally viable, and which accounts for what fraction is not settled. In a merger, the total mass at explosion need not coincide with the Chandrasekhar limit.
In other words, why Type Ia supernovae are so homogeneous cannot be derived completely from theory. What we rely on is an **empirical rule** verified on nearby supernovae. The "standard" in standard candle should be understood not as a proved theorem but as a hypothesis under continuing test.
The most dramatic display of this ruler's power came in 1998. Two research teams observed several dozen distant Type Ia supernovae and found them "too faint if the universe expands at a constant rate — that is, farther away than expected". Far from decelerating, the expansion of the universe was **accelerating**. This discovery became the starting point of the puzzle that occupies the second half of [Dark matter and dark energy](/en/physics/cosmology/dark-matter-and-dark-energy) (what accelerated expansion demands is summarised in ).
## 6. Stacking the rungs: distance moduli add, and so do the errors
B --> C --> D --> E
B -.->|"fixes the absolute magnitudes of nearby Cepheids"| C
C -.->|"fixes the absolute magnitudes of supernovae in nearby galaxies"| D`} />
| Rung | Objects / phenomena used | What is taken as "known" | Approximate reach | Main sources of error |
|---|---|---|---|---|
| 0 | Radar ranging, radio observation | The speed of light | Within the solar system (tens of au) | Essentially negligible |
| 1 | Annual parallax | The radius of the Earth's orbit | About 5000 pc | Accuracy of angle measurement |
| 2 | Cepheid variables | The period–luminosity relation | About 30 million pc | Extinction, metallicity, calibration |
| 3 | Type Ia supernovae | Peak luminosity nearly constant | Billions of light-years | Host-galaxy environment, calibration |
Let us write down in formulas what actually happens at a joint of the ladder. This is the heart of the article.
Let objects A and B be standard candles of the same kind (that is, share a common absolute magnitude $M$), with extinction-corrected apparent magnitudes $m_A, m_B$. Then
$$
\mu_B = \mu_A + (m_B - m_A), \qquad \frac{d_B}{d_A} = 10^{(m_B - m_A)/5}
$$
**The quantity $M$ does not appear in these formulas.** That is, without knowing the value of the absolute magnitude, the distance to B is determined as soon as the distance to A is known.
Furthermore, let $\sigma_A$ be the error in $\mu_A$ and $\sigma_{AB}$ the measurement error in $m_B - m_A$. If the two are independent then
$$
\sigma_{\mu_B} = \sqrt{\sigma_A^2 + \sigma_{AB}^2}
$$
If instead the two carry systematic errors $\Delta_A, \Delta_{AB}$ in the same direction, these simply add: $\Delta_{\mu_B} = \Delta_A + \Delta_{AB}$.
By the definition of the distance modulus in , $\mu_A = m_A - M$ and $\mu_B = m_B - M$. Subtracting,
$$
\mu_B - \mu_A = (m_B - M) - (m_A - M) = m_B - m_A
$$
and $M$ vanishes. Rearranging gives the first formula. The second follows because the same computation as in the proof of gives $d_B/d_A = 10^{(\mu_B - \mu_A)/5}$.
As for the errors, $\mu_B$ is written as a sum of two independent quantities, so additivity of variance (the variance of a sum of independent random variables is the sum of the variances) gives $\sigma_{\mu_B}^2 = \sigma_A^2 + \sigma_{AB}^2$. A systematic error is not a random scatter but a fixed offset, so within a sum it simply adds.
The fact that $M$ cancels is what sustains the very idea of a ladder. We need not know how many watts a Cepheid really emits; it suffices to be able to say "the nearby Cepheids and the distant ones belong to the same population". Here lies the reason why astronomy is a science of ratios rather than absolute values.
And the way errors propagate seals the ladder's fate. That distances accumulate multiplicatively means that **relative errors accumulate**. Written in magnitudes (logarithms), that accumulation appears as addition.
Let us do it concretely. Suppose that in a three-rung ladder the statistical errors of the distance moduli at each rung are
- Rung 1 (parallax engraves the Cepheid scale): $\sigma_1 = 0.05$ mag
- Rung 2 (Cepheids in nearby galaxies engrave the supernova scale): $\sigma_2 = 0.10$ mag
- Rung 3 (measuring distant supernovae): $\sigma_3 = 0.15$ mag
Applying twice, the error in the final distance modulus is
$$
\sigma_{\mathrm{total}} = \sqrt{0.05^2 + 0.10^2 + 0.15^2} = \sqrt{0.0025 + 0.01 + 0.0225} = \sqrt{0.035} = 0.187\ \text{mag}
$$
Converting to distance with , $0.46 \times 0.187 = 0.086$, about 8.6 %.
If instead each rung were offset **systematically** by $0.05$ mag in the same direction, the total would be $0.05 \times 3 = 0.15$ mag, about 7 % in distance. And systematic errors do not shrink however many observations one accumulates. That is the frightening thing about ladders.
The ratio of the distance $d$ of a remote galaxy to its recession velocity $v$ (measured via ) gives the **Hubble constant** $H_0 = v/d$ (see ). The age and size of the universe are determined almost entirely by this single number (for details see in [The edge and the age of the universe](/en/physics/cosmology/age-and-size-of-the-universe)).
At present, two independent ways of measuring it disagree.
- The value from the ladder of this article (parallax → Cepheids → Type Ia supernovae): $H_0 = 73.0 \pm 1.0\ \mathrm{km/s/Mpc}$
- The value from the fluctuations of the cosmic microwave background, via a cosmological model: $H_0 = 67.4 \pm 0.5\ \mathrm{km/s/Mpc}$
The difference is $5.6$; dividing by the combined error $\sqrt{1.0^2 + 0.5^2} = 1.1$ gives about $5$. This is a discrepancy that chance cannot explain, and it is called the **Hubble tension** (see also ).
Let us translate the discrepancy into the language of distance. Since $H_0 = v/d$, if the ladder systematically underestimates distances by $8$ %, then $H_0$ comes out $8$ % too large. And $73.0/67.4 = 1.083$, which is exactly of that size.
So how many magnitudes is an $8$ % shift in distance? By ,
$$
\delta\mu = \frac{0.08}{0.46} = 0.17\ \text{mag}
$$
**A mere 0.17 magnitudes.** A 17 % error in brightness hiding in the scale of a single rung would be enough. Which is why many astronomers suspect an undiscovered systematic error somewhere in the ladder and keep re-auditing it. At the same time, since no amount of auditing has turned one up, a growing number think that new physics may be required.
Baade's correction, which we met in , was a mistake that doubled distances. The present discrepancy is on a scale less than a tenth of that. It shows how far the accuracy of the ladder has come.
## 7. Exercises
A star is measured to have annual parallax $p = 0.025''$.
(a) Find the distance to this star in parsecs and in light-years.
(b) If the apparent magnitude of the star is $m = 8.0$, what is its absolute magnitude $M$? Interstellar extinction may be neglected.
(a) By ,
$$
d = \frac{1}{0.025} = 40\ \mathrm{pc}
$$
In light-years, $40 \times 3.26 = 130$.
(b) By ,
$$
\mu = 5 \log_{10} \frac{40}{10} = 5 \log_{10} 4 = 5 \times 0.602 = 3.01
$$
hence
$$
M = m - \mu = 8.0 - 3.01 = 4.99 \approx 5.0
$$
The absolute magnitude of the Sun computed in was $4.83$, so this star is of almost the same brightness as the Sun. Move a star like the Sun 40 parsecs away and it becomes entirely invisible to the naked eye (an 8th-magnitude star).
In an external galaxy a Cepheid variable of period $P = 20$ days is found, with mean apparent magnitude $m_V = 22.9$. The interstellar extinction in that direction is estimated at $A_V = 0.5$ mag. Using the period–luminosity relation $M_V = -2.8 \log_{10}(P/1\ \text{day}) - 1.4$, answer the following.
(a) Find the distance to this galaxy in parsecs and in light-years.
(b) If the extinction correction is forgotten, by what percentage is the distance wrong? Is it an overestimate or an underestimate?
(a) Since $\log_{10} 20 = 1.301$, gives
$$
M_V = -2.8 \times 1.301 - 1.4 = -3.643 - 1.4 = -5.04
$$
Following , we compute the extinction-corrected distance modulus.
$$
\mu_0 = (22.9 - 0.5) - (-5.04) = 22.4 + 5.04 = 27.44
$$
$$
d = 10^{(27.44+5)/5} = 10^{6.488} = 3.08 \times 10^{6}\ \mathrm{pc}
$$
That is about $3.1$ megaparsecs, or in light-years $3.08 \times 10^{6} \times 3.26 = 1.00 \times 10^{7}$, exactly 10 million light-years.
(b) Ignoring the extinction gives $\mu = 22.9 + 5.04 = 27.94$, hence
$$
d' = 10^{(27.94+5)/5} = 10^{6.588} = 3.88 \times 10^{6}\ \mathrm{pc}
$$
The ratio is $d'/d = 3.88/3.08 = 1.26$, an **overestimate** of about 26 %. This agrees with the exact formula of : $10^{\delta\mu/5} = 10^{0.5/5} = 10^{0.1} = 1.259$.
Note that the linear approximation gives $0.46 \times 0.5 = 0.23$, that is 23 %, somewhat off the true 26 %. The reason is that $\delta\mu = 0.5$ can hardly be called "well below one magnitude"; we are at the edge of where the approximation applies.
Suppose the scatter in the peak absolute magnitude of Type Ia supernovae has standard deviation $0.40$ mag before light-curve correction and $0.15$ mag after.
(a) What is the relative error of a distance derived from a single supernova, before and after correction? Use the linear approximation of .
(b) Suppose that within a cluster of galaxies we observe $N$ supernovae of the same kind independently and average the distance moduli obtained. By what factor is the error reduced? For $N = 25$, does the uncorrected case achieve better accuracy than a single corrected supernova?
(c) There is an important caveat attached to the argument in (b). What is it?
(a) By , $\delta d/d \approx 0.46\,\delta\mu$.
- Before correction: $0.46 \times 0.40 = 0.184$, about 18 %
- After correction: $0.46 \times 0.15 = 0.069$, about 7 %
A single correction due to Phillips thus improved the distance accuracy by nearly a factor of 2.7.
(b) The standard deviation of the mean of $N$ independent measurements is $1/\sqrt{N}$ times the original standard deviation. This follows from the composition rule for errors in (variances add): the variance of the mean of $N$ measurements is $\sigma^2/N$.
For $N = 25$ we have $1/\sqrt{25} = 1/5$, so the uncorrected error becomes
$$
\frac{18\ \%}{5} = 3.7\ \%
$$
which beats the 7 % of a single corrected supernova. Looking at the statistics alone, the answer is yes.
(c) The caveat is that **only the statistical error falls as $1/\sqrt{N}$; the systematic error does not fall at all**.
As we saw in the second half of , systematic errors simply add. For example, an error of the form "the absolute magnitudes of the Cepheids used to calibrate the nearby supernovae were off by $0.05$ mag overall" remains a $0.05$ mag offset whether one observes 25 supernovae or 2500.
This is precisely why the Hubble tension of is so awkward. The number of observations is already large and the statistical error is small. What remains is either a systematic error or new physics.
## References
- Leavitt, H. S. and Pickering, E. C., "Periods of 25 Variable Stars in the Small Magellanic Cloud", *Harvard College Observatory Circular* 173 (1912), 1–3. — The original paper on the period–luminosity relation. Three pages that changed how the universe is measured.
- Benedict, G. F. et al., "Astrometry with the Hubble Space Telescope: A Parallax of the Fundamental Distance Calibrator δ Cephei", *The Astronomical Journal* 124 (2002), 1695–1705. — The source of the parallax value used for comparison in .
- Phillips, M. M., "The Absolute Magnitudes of Type IA Supernovae", *The Astrophysical Journal* 413 (1993), L105–L108. — The paper exhibiting the correlation between the decline rate of the light curve and the peak luminosity.
- Freedman, W. L. et al., "Final Results from the Hubble Space Telescope Key Project to Measure the Hubble Constant", *The Astrophysical Journal* 553 (2001), 47–72. arXiv:[astro-ph/0012376](https://arxiv.org/abs/astro-ph/0012376) — A detailed account of the calibration of extragalactic distances with Cepheids and of the error budget of the whole ladder.
- Riess, A. G. et al., "A Comprehensive Measurement of the Local Value of the Hubble Constant with 1 km/s/Mpc Uncertainty from the Hubble Space Telescope and the SH0ES Team", *The Astrophysical Journal Letters* 934 (2022), L7. arXiv:[2112.04510](https://arxiv.org/abs/2112.04510) — The source of $H_0 = 73.0 \pm 1.0$ used in .
- Planck Collaboration, "Planck 2018 results. VI. Cosmological parameters", *Astronomy & Astrophysics* 641 (2020), A6. arXiv:[1807.06209](https://arxiv.org/abs/1807.06209) — The source of the cosmic-microwave-background value $H_0 = 67.4 \pm 0.5$.
## Appendix: How was "one astronomical unit" measured?
**Rung zero of the ladder.** In the main text we used $1\ \mathrm{au}$ as the baseline of as though it were given, but it is itself an object of measurement. Without knowing the length of the baseline, one cannot triangulate.
**The layout of the planets is known up to ratios.** Kepler's third law states that the square of the orbital period is proportional to the cube of the semi-major axis. Periods can be measured by watching the sky, so the **ratios** of the orbital radii of all the planets in the solar system were accurately known by the seventeenth century. All that was missing was a single ruler to attach a unit of length to those ratios.
**The transit of Venus.** In the eighteenth century, following Halley's proposal, an international undertaking was organised to observe the passage of Venus across the face of the Sun from two widely separated points on the Earth (in 1761 and 1769). Observing sites at different latitudes see Venus trace slightly different paths across the solar disc. That displacement is nothing other than triangulation on a baseline given by the distance between the two terrestrial sites. One of the principal aims of Captain Cook's first voyage was to make this observation at Tahiti. The value obtained fell within a few percent of the modern one.
**Radar settled it.** In 1961 radio waves were beamed at Venus and the time for the reflection to return was measured. Since the speed of light is known accurately, the round-trip distance follows directly. The error dropped below $10^{-6}$ at a stroke.
**Today it is a defined value.** The accuracy improved so far that in 2012 the International Astronomical Union fixed $1\ \mathrm{au} = 149\,597\,870\,700\ \mathrm{m}$ as an **exact defined value**. It is no longer a measured quantity.
Because rung zero is thus nailed down, every rung above it has meaning. The cosmic distance ladder joins the size of the solar system, measured with radio waves, to a supernova ten billion light-years away, in one unbroken line.