Blackbody spectrum and HR diagram explorer

What’s happening

A star’s light is close to that of a black body — an ideal object that emits a spectrum set only by its temperature. Planck’s law gives the brightness at each wavelength:

$$B_\lambda(T) = \frac{2hc^2}{\lambda^5}\,\frac{1}{e^{hc/(\lambda k_B T)} - 1}.$$

Two results follow, and both are in every school astrophysics syllabus:

• Wien’s law — the peak wavelength is inversely proportional to temperature: $\lambda_\text{max} = b/T$ with $b = 2.898\times10^{-3}$ m K. Hot stars look blue-white, cool stars red.
• Stefan–Boltzmann law — the total power from a star of radius $R$ is $L = 4\pi R^2 \sigma T^4$. Doubling the temperature multiplies the luminosity by 16.

On the Hertzsprung–Russell diagram, stars are placed by temperature (decreasing to the right) and luminosity. Because $L$ depends on both $T$ and $R$, a cool star can still be very luminous if it is huge (a giant), and a hot star can be faint if it is tiny (a white dwarf).

Assumptions and limits. Real stellar spectra have absorption lines and are not perfect black bodies, so a star’s colour from this model is only approximate.

Try this

Use $L_\odot = 3.828\times10^{26}$ W and $R_\odot = 6.957\times10^{8}$ m.

1. The Sun. With $T = 5772$ K, where does the Sun’s spectrum peak?
   Answer

   $\lambda_\text{max} \approx 502$ nm — in the green part of the visible band.
2. A red dwarf. A star has $T = 3000$ K and $R = 0.3\,R_\odot$. Find its peak wavelength and luminosity.
   Answer

   $\lambda_\text{max} \approx 966$ nm (infrared); $L \approx 0.0066\,L_\odot$.
3. A hot star. A star has $T = 9900$ K and $R = 1.7\,R_\odot$. Find $L$.
   Answer

   $L = 1.7^2 \times (9900/5772)^4 \approx 25\,L_\odot$.
4. Same temperature, different radius. Two stars have the same temperature, but one is 100 times more luminous. How do their radii compare?
   Answer

   $L \propto R^2$ at fixed $T$, so the brighter star is 10 times larger.