Stellar radii
A star's colour is a thermometer
- A red star is cool; a blue-white star is scorching.
- A star's colour reveals its temperature — and, with its brightness, its size.
- Two laws turn colour and power into temperature and radius.
Wien's displacement law
- A hot body's spectrum peaks at a wavelength set by its temperature: $\lambda_{\max}T = b$ ($b \approx 2.9 \times 10^{-3}\ \text{m}\cdot\text{K}$).
- Hotter → shorter peak wavelength (blue); cooler → longer (red).
Hotter stars have their spectral peak (Wien's law) at a ____ wavelength.
$\lambda_{\max}T = b$, so a higher $T$ means a smaller $\lambda_{\max}$ — toward the blue.
A star's spectrum peaks at $\lambda_{\max} = 500\ \text{nm}$. What is its surface temperature? ($b = 2.9 \times 10^{-3}\ \text{m}\cdot\text{K}$)
$T = \dfrac{b}{\lambda_{\max}} = \dfrac{2.9 \times 10^{-3}}{500 \times 10^{-9}} \approx 5800\ \text{K}$ — about the Sun.
Stefan–Boltzmann law
- A star's luminosity is $L = 4\pi\sigma r^{2}T^{4}$.
- Strong dependences: $L \propto r^{2}$ and $L \propto T^{4}$ (double $T$ → 16× the luminosity).
The Stefan–Boltzmann law gives a star's luminosity as:
$L = 4\pi\sigma r^{2}T^{4}$ — strongly dependent on both radius and temperature.
If a star's temperature doubles (same radius), its luminosity multiplies by:
$L \propto T^{4}$, so doubling $T$ gives $2^{4} = 16$ times the luminosity.
Estimating a star's radius
- From $\lambda_{\max}$ get $T$ (Wien); from flux and distance get $L$.
- Then solve for the radius: $r = \sqrt{\dfrac{L}{4\pi\sigma T^{4}}}$ — even for a star too far to see as a disc.
Combining Wien's law and the Stefan–Boltzmann law lets us estimate a star's radius.
Wien gives $T$, the flux and distance give $L$, then $r = \sqrt{\dfrac{L}{4\pi\sigma T^{4}}}$.
You've got it
- Wien: $\lambda_{\max}T = b$ — hotter stars peak at shorter wavelengths
- Stefan–Boltzmann: $L = 4\pi\sigma r^{2}T^{4}$ ($L \propto T^{4}$)
- combine the two to estimate a star's radius