The HR Diagram

Section 1 of 6

The Magnitude System

By the end of this reading, you will be able to:

Define
apparent and absolute magnitude and apply the distance modulus to find distances or magnitudes
Convert
between magnitude differences and flux ratios using the Pogson relation
Construct
the observer’s HR diagram ( $M_V$ vs. spectral type) and identify its features
Derive
lines of constant radius on the theorist’s HR diagram from Stefan-Boltzmann
Identify
the main sequence, giant branch, and white dwarf sequence and what places stars in each
Explain
why the main sequence is a mass sequence and connect it to the mass-luminosity relation
Articulate
the questions the HR diagram raises — and why physics (Module 3) is needed to answer them

When Ejnar Hertzsprung and Henry Norris Russell independently plotted stellar luminosity against temperature in the early 1900s, they expected scatter — a random spray of points. Instead, they found structure. Stars cluster into distinct regions: a long diagonal band (the main sequence), a clump of cool luminous stars (giants), and a scattering of hot dim stars (white dwarfs). This pattern was the first evidence that stars are organized by physics. Understanding that physics is the work of Module 3. This reading is about the pattern itself.

Part 1: The Magnitude System — The Astronomer’s Brightness Scale

The Problem: Stars Span a Ridiculous Range

Try the naive thing first: a linear brightness plot with Proxima, Betelgeuse, and the Sun on the same axis. Set Proxima near the bottom and scale upward by equal flux steps. Betelgeuse jumps to tens of millions of Proximas, and the Sun jumps tens of billions beyond that. Proxima and Betelgeuse collapse against zero while the Sun runs off-scale. On a linear axis, the plot is scientifically unusable.

We need a logarithmic ruler: the magnitude system — the last measurement tool before we can build the HR diagram.

Deep Dive: Deep Dive (Optional): Properties of Log Laws

In this module, $\log$ means $\log_{10}$ unless stated otherwise.

Law	Rule	How you will use it
Product law	$\log(AB) = \log A + \log B$	Break numbers into easy factors, like $200 = 2 \times 100$
Quotient law	$\log(A/B) = \log A - \log B$	Convert ratios into differences (magnitude and distance modulus)
Power law	$\log(A^n) = n\log A$	Understand why squared or fourth-power scalings become coefficients
Inverse rule	$\log(10^x) = x$ and $10^{\log x} = x$	Move between log form and exponent form when solving for distance

Useful anchors: $\log_{10}(1)=0$ , $\log_{10}(10)=1$ , $\log_{10}(100)=2$ , $\log_{10}(0.1)=-1$ . Common traps: using $\ln$ rules when the equation uses $\log_{10}$ ; forgetting that numbers between 0 and 1 have negative logs; reversing ratios (which flips the sign).

Why Not Just Use Luminosity?

You already know how to measure a star’s luminosity in physical units: $L$ in $\text{erg}\,\text{s}^{-1}$ , or in $L_\odot$ . So why do astronomers still use magnitudes — logarithmic, inverted (brighter = smaller number), and historically odd? Three reasons:

Dynamic range. Stellar brightnesses span ten orders of magnitude ( $10^{10}$ ). Logarithmic scales compress this into manageable numbers (roughly $-10$ to $+20$ ).
What detectors measure. Eyes, photographic plates, and CCDs respond to flux — energy per unit area per unit time — not luminosity. Magnitudes are tied directly to flux ratios.
Historical inertia. Hipparchus ranked stars from “first magnitude” (brightest) to “sixth magnitude” (faintest visible). Modern astronomy formalized this into a precise logarithmic system but kept brighter = smaller magnitude.

Apparent Magnitude: How Bright It Looks

The apparent magnitude $m$ measures how bright a star appears from Earth. It depends on intrinsic brightness and distance. The formal definition connects magnitude differences to flux ratios (the Pogson relation):

Here $F_1$ and $F_2$ are the fluxes and $m_1$ , $m_2$ the apparent magnitudes. Because the scale is logarithmic, multiplicative flux changes become additive magnitude shifts.

Apparent magnitude

A logarithmic measure of how bright a star appears from Earth ( $m$ ), set by both luminosity and distance. Smaller (more negative) means brighter; 5 magnitudes correspond to a factor of 100 in flux.

Key numbers to remember:

Magnitude difference	Flux ratio
$1~\text{mag}$	$10^{0.4} \approx 2.512$
$2~\text{mag}$	$10^{0.8} \approx 6.31$
$5~\text{mag}$	$10^{2.0} = 100$
$10~\text{mag}$	$10^{4.0} = 10^4$

The anchor: 5 magnitudes = a factor of 100 in flux (by definition — Pogson’s ratio).

Multiple choice

If Star A is 5 magnitudes smaller (i.e. a smaller number) than Star B, is Star A’s flux smaller or larger than Star B’s, and by what factor?

Larger, by a factor of 100Smaller, by a factor of 100Larger, by a factor of 5Smaller, by a factor of 5

Worked Example 1Comparing Two Stars

Problem

Star A has apparent magnitude $m_A = 1.0$ and Star B has $m_B = 6.0$ . How many times brighter is Star A than Star B?

StepMagnitude difference, then flux ratio

$\Delta m = m_B - m_A = 6.0 - 1.0 = 5.0~\text{mag}$ $\frac{F_A}{F_B} = 10^{0.4 \times 5.0} = 10^{2.0} = 100$

Dimensional check

$\Delta m$ is dimensionless (a magnitude difference); $10^{0.4\Delta m}$ is a pure flux ratio ✓.

Result

Star A is $100\times$ brighter than Star B in apparent flux. Sanity check: Star A has the smaller magnitude ( $m_A = 1.0$ , smaller than $m_B = 6.0$ ), so it is indeed brighter ✓.

Absolute Magnitude: How Bright It Actually Is

Apparent magnitude mixes intrinsic luminosity with distance — a nearby dim star can look brighter than a distant luminous one. To compare stars fairly, remove the distance factor. The absolute magnitude $M$ is the apparent magnitude a star would have at a standard distance of $10~\text{pc}$ — a measure of its intrinsic brightness.

Absolute magnitude

The apparent magnitude a star would have if placed at the standard distance of $10~\text{pc}$ ( $M$ ). It expresses intrinsic luminosity in the magnitude system, removing the distance dependence.

Star	$m$ (apparent)	Distance	$M$ (absolute)	Luminosity
Sun	$-26.74$	$1~\text{AU} \approx 5 \times 10^{-6}~\text{pc}$	$+4.83$	$1\,L_\odot$
Sirius	$-1.46$	$2.64~\text{pc}$	$+1.42$	$25\,L_\odot$
Betelgeuse	$+0.42$	$200~\text{pc}$	$-5.85$	${\sim}10^5\,L_\odot$
Proxima Cen	$+11.13$	$1.30~\text{pc}$	$+15.53$	$0.0017\,L_\odot$

The Sun looks blindingly bright only because it’s close; at $10~\text{pc}$ it would be a modest $M = +4.83$ . Betelgeuse looks unremarkable despite being intrinsically $10^5$ times more luminous because it’s far away.

The Distance Modulus: Connecting $m$ , $M$ , and $d$

The relationship between apparent magnitude, absolute magnitude, and distance is the distance modulus:

$m$ is apparent magnitude, $M$ absolute magnitude, $d$ the distance in parsecs; the quantity $(m - M)$ is the distance modulus. Where does it come from? It is the inverse-square law from Lecture 1 ( $F = L/4\pi d^2$ ) rewritten in logarithmic language — the $d^2$ flux dependence becomes a $5\log_{10}(d)$ term. Same physics, different packaging.

Distance modulus

The difference $m - M = 5\log_{10}(d/10\,\text{pc})$ between apparent and absolute magnitude — a logarithmic readout of distance. Zero at $10~\text{pc}$ ; the inverse-square law in magnitude form.

Sanity checks: at $d = 10~\text{pc}$ , $m - M = 5\log_{10}(1) = 0$ , so $m = M$ ✓. At $d = 100~\text{pc}$ , $m - M = 5\log_{10}(10) = 5$ — the star appears $5~\text{mag}$ (i.e. $100\times$ ) fainter ✓. At $d = 1{,}000~\text{pc}$ , $m - M = 5\log_{10}(100) = 10$ — the star is $10^4\times$ fainter ✓.

Worked Example 2Finding Distance from Magnitudes

Problem

A Cepheid has apparent magnitude $m = 14.0$ and absolute magnitude $M = -4.0$ . How far away is it?

StepDistance modulus

$m - M = 14.0 - (-4.0) = 18.0~\text{mag}$

StepSolve for distance

$18.0 = 5\log_{10}\!\left(\frac{d}{10\,\text{pc}}\right) \Rightarrow \log_{10}\!\left(\frac{d}{10\,\text{pc}}\right) = 3.6$ $\frac{d}{10\,\text{pc}} = 10^{3.6} \approx 3{,}981 \Rightarrow d \approx 4.0 \times 10^4~\text{pc} = 40~\text{kpc}$

Dimensional check

The log argument is the dimensionless ratio $d/10\,\text{pc}$ ; solving returns $d$ in parsecs ✓.

Result

A modulus of $18~\text{mag}$ gives ${\sim}40~\text{kpc}$ , a plausible distance for a luminous Cepheid ✓.

Problem

A star has $m = 5.0$ and $M = 5.0$ . How far away is it?
A star at $d = 1{,}000~\text{pc}$ has $m = 10.0$ . What is its absolute magnitude?
The Sun has $M = +4.83$ . What would its apparent magnitude be at $10~\text{pc}$ ? At $100~\text{pc}$ ?
Tricky: Two stars have the same $M$ . Star A is $10\times$ farther than Star B. What is the difference in their apparent magnitudes?

Part 1: The Magnitude System — The Astronomer’s Brightness Scale

The Problem: Stars Span a Ridiculous Range

Why Not Just Use Luminosity?

Apparent Magnitude: How Bright It Looks

Multiple choice

Absolute Magnitude: How Bright It Actually Is

The Distance Modulus: Connecting mmm, MMM, and ddd

Problem

The Distance Modulus: Connecting $m$ , $M$ , and $d$