A star's luminosity $L$ is its total power: the energy it sends out each second, in watts. The Sun's is $3.83 \times 10^{26}$ W. From Wien's law you know temperature sets a star's color. Temperature also sets how much power each square meter of its surface gives off, and the size of the surface does the rest.
Why it works
Each square meter of a hot surface radiates $\sigma T^4$ watts (the Stefan–Boltzmann law, with $\sigma = 5.67 \times 10^{-8}\ \text{W/(m}^2\text{K}^4)$). The fourth power is steep: double the temperature and each square meter gives off $2^4 = 16$ times as much. A star's surface is a sphere of area $4\pi R^2$. So $$L = 4\pi R^2 \sigma T^4.$$ Divide by the Sun's values and every constant cancels: in solar units (radius in $R_\odot$, temperature in units of the Sun's $5772$ K), $L = R^2 T^4$. The exponent rules make this quick: $(2R)^2 = 4R^2$ and $(2T)^4 = 16T^4$.
Key idea
$L = 4\pi R^2 \sigma T^4$; in solar units, $L = R^2 T^4$. Luminosity depends on size and temperature together, so a cool star can be very luminous if it is huge, and a hot star can be faint if it is tiny.
Worked examples
Worked example: A giant star
A star is $10$ times the Sun's radius and has half the Sun's surface temperature. Find its luminosity.
- $L = R^2 T^4 = 10^2 \times 0.5^4 = 100 \times 0.0625$.
- $L = 6.25$ solar luminosities: cooler, yet more luminous, because it is so big.
Answer: $6.25\,L_\odot$.
Worked example: A white dwarf
A star is $0.01$ times as luminous as the Sun but twice as hot. What is its radius?
- Solve $L = R^2 T^4$ for $R$: $R = \frac{\sqrt{L}}{T^2}$.
- $R = \frac{\sqrt{0.01}}{2^2} = \frac{0.1}{4} = 0.025$ solar radii.
- That is about $17{,}000$ km, less than three times Earth's radius: a white dwarf.
Answer: $0.025\,R_\odot$.
Worked example: Check with the Sun
Use $R = 6.96 \times 10^{8}$ m and $T = 5772$ K to find the Sun's luminosity.
- Area: $4\pi (6.96 \times 10^{8})^2 = 6.09 \times 10^{18}\ \text{m}^2$.
- Per square meter: $\sigma T^4 = 5.67 \times 10^{-8} \times 5772^4 = 6.29 \times 10^{7}$ W/m².
- $L = 6.09 \times 10^{18} \times 6.29 \times 10^{7} = 3.83 \times 10^{26}$ W.
Answer: $3.83 \times 10^{26}$ W, the measured value.
Common mistake
It seems that a hotter star should always be more luminous, but size matters as much. Sirius B is about $25{,}000$ K, over four times hotter than the Sun, yet it gives off only a few percent of the Sun's power, because it is about the size of Earth. Check with $L = R^2 T^4$: a tiny $R$ can outweigh a big $T$.