Spectral classes of stars and brown dwarfs

Spectral classes of stars and brown dwarfs
Class	Approximate temperature (K)	Color	Typical spectral features
O	> 30 000	Blue	Ionized helium; weak hydrogen lines
B	10 000–30 000	Blue-white	Neutral helium; strong hydrogen lines
A	7 500–10 000	White	Hydrogen lines especially strong
F	6 000–7 500	Yellow-white	Strong ionized metals, including calcium
G	5 200–6 000	Yellow	Metal lines; the Sun is G2
K	3 700–5 200	Orange	Neutral metals; weak hydrogen lines
M	2 400–3 700	Red	Molecular bands, for example titanium oxide
L	1 300–2 400	Red to infrared	Metal hydrides and alkali metals; mostly brown dwarfs
T	700–1 300	Infrared	Methane bands; brown dwarfs
Y	< 700	Infrared	Very cool objects; ammonia features

Class

Approximate temperature (K)

Color

Typical spectral features

> 30 000

Blue

Ionized helium; weak hydrogen lines

10 000–30 000

Blue-white

Neutral helium; strong hydrogen lines

7 500–10 000

White

Hydrogen lines especially strong

6 000–7 500

Yellow-white

Strong ionized metals, including calcium

5 200–6 000

Yellow

Metal lines; the Sun is G2

3 700–5 200

Orange

Neutral metals; weak hydrogen lines

2 400–3 700

Red

Molecular bands, for example titanium oxide

1 300–2 400

Red to infrared

Metal hydrides and alkali metals; mostly brown dwarfs

700–1 300

Infrared

Methane bands; brown dwarfs

< 700

Infrared

Very cool objects; ammonia features

Additional reference tables

Luminosity classes complement spectral type

Normal MK labels; not universal wattage or radius intervals.

Luminosity classes complement spectral type
Class	Name	Interpretation
Ia	Most luminous supergiants	Added to spectral type; assigned from spectral diagnostics
Ib	Less luminous supergiants	Added to spectral type; assigned from spectral diagnostics
II	Bright giants	Added to spectral type; assigned from spectral diagnostics
III	Giants	Added to spectral type; assigned from spectral diagnostics
IV	Subgiants	Added to spectral type; assigned from spectral diagnostics
V	Main-sequence stars	Added to spectral type; assigned from spectral diagnostics

References: OpenStax Astronomy 2e §19.4

Understand and use this reference

O–B–A–F–G–K–M orders stellar spectra approximately from hot to cool; L, T and Y extend the sequence to very cool objects. A spectral type alone is not a complete size or brightness measurement. This guide explains subclasses, luminosity classes and model-based checks of temperature, radius and radiation.

What the classification describes

An observed spectrum contains a broad continuum, spectral lines and molecular bands. Classification compares these features with standards. The table’s temperature column is an approximate guide rather than a defining equation T=f(letter). Near boundaries, different calibrations and luminosity classes can correspond to somewhat different effective temperatures.[1]

Effective temperature T_eff is the temperature of a blackbody with the same total emitted power per area. It does not directly measure core temperature. A star’s visible atmosphere also has no single temperature everywhere; the number summarises a radiative property. Quantitative results should identify the temperature’s model and origin.[6]

Read the sequence and subclasses

In the usual sequence, O and B occupy the hot end, A and F intermediate positions, G and K the cooler side and M the cool end of ordinary stellar spectra. Numbers 0 through 9 subdivide many classes; a higher subclass number typically indicates a cooler type along the normal sequence. G8 is therefore closer to K0 than G2.[1]

Subclass numbers do not form a linear kelvin scale. A subclass step does not represent the same temperature change everywhere. Interpolating between table boundaries therefore does not automatically yield a reliable A4 or M7 temperature. Such precision requires an appropriate empirical calibration, particularly one specifying luminosity class.

Why spectral features change with temperature

Line strength depends on the available atoms, their ionisation state and population of the relevant energy levels. A weak hydrogen line therefore does not demonstrate hydrogen-poor composition. Higher temperatures promote ionisation; lower temperatures allow molecules to persist and form bands. Different diagnostics are consequently useful at different positions along the sequence.[2]

Ionised helium features matter in O spectra, neutral helium in B, and especially strong Balmer lines in A. Cooler types increasingly show metal lines, while M spectra have prominent molecular bands such as TiO. Astronomical “metals” means all elements heavier than helium, not merely electrically conducting materials. Combining diagnostics is more informative than relying on one prominent line.[1]

The second part: luminosity class

A normal complete MK label combines spectral type and Roman-numeral luminosity class, for example F6 V or K3 III. V denotes main-sequence stars, IV subgiants, III giants, II bright giants and I supergiants; Ia and Ib distinguish supergiants. Classification uses spectral diagnostics rather than a universal fixed wattage table.[3]

Stars with similar temperatures can have very different radii and luminosities. Line profiles respond to atmospheric pressure and surface gravity, helping distinguish giants from main-sequence stars. Class V does not mean the fifth-brightest star, and class III does not specify measured distance. Colour alone cannot supply this second part of classification.[2]

Temperature, area and the H–R diagram

L=4\pi R^2\sigma T_{\rm eff}^4,\qquad \frac{L_2}{L_1}=\left(\frac{R_2}{R_1}\right)^2\left(\frac{T_2}{T_1}\right)^4

For spherical isotropic emission, the Stefan–Boltzmann relation connects luminosity, radius and effective temperature. At equal temperature, twice the radius gives four times the luminosity; twice T at equal R gives sixteen times the luminosity.[6]

The H–R diagram plots luminosity against temperature or spectral type. Conventionally temperature increases leftwards and luminosity upwards. Separation between diagram points is not spatial distance. Hot small white dwarfs and cool large giants demonstrate why temperature alone cannot determine total power.[4]

Example: peak wavelength as a model check

Two specified temperatures

Given: Compare spherical thermal models at T₁=4800 K and T₂=9600 K. Use b≈2.897771955×10⁻³ m·K for the wavelength-distribution peak.
Find: Calculate λmax and explain the classification limit.

Apply Wien’s law to the wavelength representation.
$\lambda_{\max}=\frac bT$
For 4800 K the result is 6.037024906×10⁻⁷ m.
$\lambda_{\max,1}\approx603.7\,\mathrm{nm}$
For 9600 K the peak wavelength is half as large.
$\lambda_{\max,2}\approx301.9\,\mathrm{nm}$

Result: The temperatures lie broadly in the table’s K and A ranges. Final spectral typing requires spectral features; a continuum peak is not an MK classification.

Check: λmaxT returns b in both cases. Doubling T halves λmax; the hotter peak is ultraviolet even though visible light is also emitted.

Retained orientation at 5800 K

Given: Thermal model T=5800 K; b≈2.897771955×10⁻³ m·K.
Find: Find the wavelength-representation peak.

Divide b by the temperature in kelvin.
$\lambda_{\max}=b/T$
Convert metres to nanometres.
$\lambda_{\max}\approx499.6159\,\mathrm{nm}$

Result: Approximately 500 nm in the visible range; the model lies broadly in the G range.

Check: λmaxT returns b. Broad continuum emission explains why a peak near green does not imply a green stellar colour.

Example: equal temperature with different luminosity

A radius comparison without assigning classes

Given: Two idealised spherical stars have equal T_eff. Their bolometric luminosity ratio is L₂/L₁=225.
Find: Find their radius ratio and compare observed flux.

With equal temperature and σ, only the area ratio remains.
$\frac{R_2}{R_1}=\sqrt{225}=15$
At equal distance without absorption, F₂/F₁=225.
$F=\frac{L}{4\pi d^2}$
If star 2 is instead fifteen times farther away, d² offsets the luminosity difference.
$\frac{F_2}{F_1}=225\left(\frac1{15}\right)^2=1$

Result: The second radius is fifteen times larger. Apparent brightness may nevertheless match: equal spectral temperature and flux do not imply identical stars.

Check: Substitute R₂=15R₁: 15²=225. This model ratio alone cannot uniquely assign a Roman-numeral luminosity class without spectral diagnostics.

L, T, Y and brown dwarfs

The cool L–T–Y extension uses infrared diagnostics including hydrides, alkali lines and methane bands in T objects. Brown dwarfs cannot sustain long-term hydrogen fusion like ordinary main-sequence stars. Cooling makes temperature and spectral type age-dependent, so class alone cannot uniquely measure mass.[1]

Not every L object is automatically a brown dwarf: the warm end overlaps very low-mass stars. Table ranges are particularly approximate here. A white dwarf and a brown dwarf also denote different physical objects; a compact stellar remnant is not merely another cooler entry after Y.[1][4]

Colour, observation and limitations

Visible colour results from a broad spectrum, the eye or detector and observing conditions. A peak near green does not make a source emit only green light. Interstellar dust changes colour ratios; image processing and white balance can modify them further. Table colour labels are visual guides rather than photometric measurement instructions.[5]

The Wien peaks in wavelength and frequency are different representations of the same broad radiation distribution; converting one peak through ν=c/λ does not locate the other peak. Real atmospheres contain absorption lines and depart from an ideal blackbody. Rotation, binarity, composition and extinction belong to precise observation modelling.[6][2]

Three problems with solutions

Read M4 III: what is specified and what is missing?

Show hint

Read letter, subclass and Roman numeral separately.

Show solution and calculation

M is a cool spectral type, 4 its subclass and III the giant luminosity class.
Exact temperature, mass, distance and current apparent brightness are not supplied.

\text{M4 III: spectral subtype and giant class}

A thermal model peaks at λmax=800 nm. Estimate T_eff using b≈2.897771955×10⁻³ m·K.

Show hint

Convert nanometres to metres first.

Show solution and calculation

T≈3622.21 K, broadly in the M range.
This is a blackbody estimate rather than a standalone spectral classification.

T\approx3.62\times10^3\,\mathrm K

At equal radius, T₂=1.25T₁. Find L₂/L₁.

Show hint

Temperature enters to the fourth power.

Show solution and calculation

L₂/L₁=1.25⁴=2.44140625.
A 25% temperature increase raises power by about 144% in this model rather than by 25%.

\frac{L_2}{L_1}=2.44140625

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 3 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Cite this reference page

Quanta. Spectral classes of stars and brown dwarfs. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/tables/spektralklassen-sterne

When citing a specific scientific statement, also cite the specialist source linked to that statement.

Suggest a correction with a reference

Spectral classes of stars and brown dwarfs
Class	Approximate temperature (K)	Color	Typical spectral features
O	> 30 000	Blue	Ionized helium; weak hydrogen lines
B	10 000–30 000	Blue-white	Neutral helium; strong hydrogen lines
A	7 500–10 000	White	Hydrogen lines especially strong
F	6 000–7 500	Yellow-white	Strong ionized metals, including calcium
G	5 200–6 000	Yellow	Metal lines; the Sun is G2
K	3 700–5 200	Orange	Neutral metals; weak hydrogen lines
M	2 400–3 700	Red	Molecular bands, for example titanium oxide
L	1 300–2 400	Red to infrared	Metal hydrides and alkali metals; mostly brown dwarfs
T	700–1 300	Infrared	Methane bands; brown dwarfs
Y	< 700	Infrared	Very cool objects; ammonia features

Class

Approximate temperature (K)

Color

Typical spectral features

> 30 000

Blue

Ionized helium; weak hydrogen lines

10 000–30 000

Blue-white

Neutral helium; strong hydrogen lines

7 500–10 000

White

Hydrogen lines especially strong

6 000–7 500

Yellow-white

Strong ionized metals, including calcium

5 200–6 000

Yellow

Metal lines; the Sun is G2

3 700–5 200

Orange

Neutral metals; weak hydrogen lines

2 400–3 700

Red

Molecular bands, for example titanium oxide

1 300–2 400

Red to infrared

Metal hydrides and alkali metals; mostly brown dwarfs

700–1 300

Infrared

Methane bands; brown dwarfs

< 700

Infrared

Very cool objects; ammonia features

Additional reference tables

Luminosity classes complement spectral type

Normal MK labels; not universal wattage or radius intervals.

Luminosity classes complement spectral type
Class	Name	Interpretation
Ia	Most luminous supergiants	Added to spectral type; assigned from spectral diagnostics
Ib	Less luminous supergiants	Added to spectral type; assigned from spectral diagnostics
II	Bright giants	Added to spectral type; assigned from spectral diagnostics
III	Giants	Added to spectral type; assigned from spectral diagnostics
IV	Subgiants	Added to spectral type; assigned from spectral diagnostics
V	Main-sequence stars	Added to spectral type; assigned from spectral diagnostics

References: OpenStax Astronomy 2e §19.4

Understand and use this reference

What the classification describes

Read the sequence and subclasses

Why spectral features change with temperature

The second part: luminosity class

Temperature, area and the H–R diagram

L=4\pi R^2\sigma T_{\rm eff}^4,\qquad \frac{L_2}{L_1}=\left(\frac{R_2}{R_1}\right)^2\left(\frac{T_2}{T_1}\right)^4

Example: peak wavelength as a model check

Two specified temperatures

Given: Compare spherical thermal models at T₁=4800 K and T₂=9600 K. Use b≈2.897771955×10⁻³ m·K for the wavelength-distribution peak.
Find: Calculate λmax and explain the classification limit.

Apply Wien’s law to the wavelength representation.
$\lambda_{\max}=\frac bT$
For 4800 K the result is 6.037024906×10⁻⁷ m.
$\lambda_{\max,1}\approx603.7\,\mathrm{nm}$
For 9600 K the peak wavelength is half as large.
$\lambda_{\max,2}\approx301.9\,\mathrm{nm}$

Result: The temperatures lie broadly in the table’s K and A ranges. Final spectral typing requires spectral features; a continuum peak is not an MK classification.

Check: λmaxT returns b in both cases. Doubling T halves λmax; the hotter peak is ultraviolet even though visible light is also emitted.

Retained orientation at 5800 K

Given: Thermal model T=5800 K; b≈2.897771955×10⁻³ m·K.
Find: Find the wavelength-representation peak.

Divide b by the temperature in kelvin.
$\lambda_{\max}=b/T$
Convert metres to nanometres.
$\lambda_{\max}\approx499.6159\,\mathrm{nm}$

Result: Approximately 500 nm in the visible range; the model lies broadly in the G range.

Check: λmaxT returns b. Broad continuum emission explains why a peak near green does not imply a green stellar colour.

Example: equal temperature with different luminosity

A radius comparison without assigning classes

Given: Two idealised spherical stars have equal T_eff. Their bolometric luminosity ratio is L₂/L₁=225.
Find: Find their radius ratio and compare observed flux.

With equal temperature and σ, only the area ratio remains.
$\frac{R_2}{R_1}=\sqrt{225}=15$
At equal distance without absorption, F₂/F₁=225.
$F=\frac{L}{4\pi d^2}$
If star 2 is instead fifteen times farther away, d² offsets the luminosity difference.
$\frac{F_2}{F_1}=225\left(\frac1{15}\right)^2=1$

Result: The second radius is fifteen times larger. Apparent brightness may nevertheless match: equal spectral temperature and flux do not imply identical stars.

Check: Substitute R₂=15R₁: 15²=225. This model ratio alone cannot uniquely assign a Roman-numeral luminosity class without spectral diagnostics.

L, T, Y and brown dwarfs

Colour, observation and limitations

Three problems with solutions

Read M4 III: what is specified and what is missing?

Show hint

Read letter, subclass and Roman numeral separately.

Show solution and calculation

M is a cool spectral type, 4 its subclass and III the giant luminosity class.
Exact temperature, mass, distance and current apparent brightness are not supplied.

\text{M4 III: spectral subtype and giant class}

A thermal model peaks at λmax=800 nm. Estimate T_eff using b≈2.897771955×10⁻³ m·K.

Show hint

Convert nanometres to metres first.

Show solution and calculation

T≈3622.21 K, broadly in the M range.
This is a blackbody estimate rather than a standalone spectral classification.

T\approx3.62\times10^3\,\mathrm K

At equal radius, T₂=1.25T₁. Find L₂/L₁.

Show hint

Temperature enters to the fourth power.

Show solution and calculation

L₂/L₁=1.25⁴=2.44140625.
A 25% temperature increase raises power by about 144% in this model rather than by 25%.

\frac{L_2}{L_1}=2.44140625

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Cite this reference page

Quanta. Spectral classes of stars and brown dwarfs. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/tables/spektralklassen-sterne

When citing a specific scientific statement, also cite the specialist source linked to that statement.

Suggest a correction with a reference

Reference values

Additional reference tables

Luminosity classes complement spectral type

Reference tables as images

What the classification describes

Read the sequence and subclasses

Why spectral features change with temperature

The second part: luminosity class

Temperature, area and the H–R diagram

Example: peak wavelength as a model check

Two specified temperatures

Retained orientation at 5800 K

Example: equal temperature with different luminosity

A radius comparison without assigning classes

L, T, Y and brown dwarfs

Colour, observation and limitations

Three problems with solutions

Read M4 III: what is specified and what is missing?

A thermal model peaks at λmax=800 nm. Estimate Teff using b≈2.897771955×10−3 m·K.

At equal radius, T2=1.25T1. Find L2/L1.

Sources for this explanation

Edition and sources

Reference values

Additional reference tables

Luminosity classes complement spectral type

Reference tables as images

What the classification describes

Read the sequence and subclasses

Why spectral features change with temperature

The second part: luminosity class

Temperature, area and the H–R diagram

Example: peak wavelength as a model check

Two specified temperatures

Retained orientation at 5800 K

Example: equal temperature with different luminosity

A radius comparison without assigning classes

L, T, Y and brown dwarfs

Colour, observation and limitations

Three problems with solutions

Read M4 III: what is specified and what is missing?

A thermal model peaks at λmax=800 nm. Estimate Teff using b≈2.897771955×10−3 m·K.

At equal radius, T2=1.25T1. Find L2/L1.

Sources for this explanation

Edition and sources

A thermal model peaks at λmax=800 nm. Estimate T_eff using b≈2.897771955×10⁻³ m·K.

At equal radius, T₂=1.25T₁. Find L₂/L₁.

A thermal model peaks at λmax=800 nm. Estimate T_eff using b≈2.897771955×10⁻³ m·K.

At equal radius, T₂=1.25T₁. Find L₂/L₁.