Spectra of stars

Source: NASA Public Domain

Harvard spectral classification

Spectral classification:
- From hottest to coolest: O B A F G K M
- Mnemonic: Oh, Be A Fine Guy (or Girl) Kiss Me
- Further subdivided: e.g. G0 is hottest in G and G9 is coolest
Luminosity class uses a Roman numeral:
- 0 to IV are giants
- V is main sequence
- VI and VII are dwarfs

Solar values

The symbol ☉ is used to denote solar values.

The Sun is a G2V star.
The surface temperature is 5800 K.
Radius: R_☉ = 696,000 km
Luminosity: L_☉ = 3.85 × 10²⁶ W

[Optional task: If you were to cover the surface of the Sun with enough 100 W light bulbs to match its luminosity, what size would each bulb need to be to all fit on the surface?]

Large numbers

To avoid writing out long numbers, we can use a short hand:

Short	Full	Number of zeroes
10³	1000	3
10⁴	10,000	4
10⁵	100,000	5
10⁶	1,000,000	6

Examples:

2 million = 2,000,000 = 2×10⁶
R_☉ = 696,000 km = 6.96×10⁵ km = 6.96×10⁸ m

HR diagram - classifications

Source: Spacepotato CC-BY SA 3.0

Classification table

Class	Temperature	Colour	Luminosity	% of MS stars
O	≥ 33,000 K	blue	≥ 30,000 L_☉	~0.00003%
B	10,000–33,000 K	blue white	25–30,000 L_☉	0.13%
A	7,500–10,000 K	white	5–25 L_☉	0.6%
F	6,000–7,500 K	yellow white	1.5–5 L_☉	3%
G	5,200–6,000 K	yellow	0.6–1.5 L_☉	7.6%
K	3,700–5,200 K	orange	0.08–0.6 L_☉	12.1%
M	2,000–3,700 K	red	≤ 0.08 L_☉	76.45%

Data source

Luminosity and temperature

Stefan's law: the power radiated by a "black body" at temperature T per unit surface area is proportional to T⁴

In physics, "black body" is an ideal emitter (and absorber) of light
T⁴ = T×T×T×T

Flux and Stefan's law

Power per unit area is often called a flux.
If T is in kelvin, then the flux F is

F = 5.67×10^-8 T⁴

where F will be in W m^-2 (watts per square metre)

Flux at the solar surface

The solar surface is at temperature 5800 K


F	= 5.67×10^-8 × 5800 × 5800 × 5800 × 5800
	= 64,164,532 W m^-2
	= 64,200,000 W m^-2
	= 6.42×10⁷ W m^-2

Since we only started with three significant figures (e.g. 5.67), we round to 3 significant figures in the last step.

Flux and luminosity

The flux is power per unit area.
Luminosity is total power over the entire surface.
So flux times surface area will give luminosity.
L = F × A
where A is the area of the Sun in square metres.

Surface area of the Sun

For a sphere A = 4π R²
where π = 3.141592... but we'll use 3.14
From earlier: R = 6.96×10⁸ m


A	= 4π R²
	= 4 × 3.14 × 6.96×10⁸ × 6.96×10⁸
	= 4 × 3.14 × 4.84 × 10¹⁷
	= 6.08×10¹⁸ m²

Theoretical solar luminosity

Plugging our values for F and A into L = F A, we find


L	= 6.42×10⁷ × 6.08×10¹⁸
	= 3.90×10²⁶ W

This is very close to measured values (see earlier).
This confirms that stars are indeed good approximations to black bodies.

Deducing star radius

L is proportional to R² and T⁴
Double R and L increases by a factor of 2×2=4
Double T and L increases by a factor of 2×2×2×2=16
So is, say, an O type star more luminous than a B type star because R or T is greater?
Answer: both!

Sizes and colours

Source: LucasVB CC-BY SA 3.0

Red giants

Betelegeuse:
- L = 100,000 L_☉
- R = 1000 R_☉
If the Sun was 1000 times larger at the same surface temperature, then its luminosity would be 1000×1000=1,000,000 times larger.
But Betelgeuse is only 100,000 times more luminous than the Sun.
So we can deduce that Betelgeuse must have a lower surface temperature than the Sun.
It is about 3400 K, which corresponds to a red colour.
Hence red and giant.

Stars - Sizes and types