Deanna Emery's Astronomy 16 Blog: Worksheet 6, Problem 4: The Stefan-Boltzmann Law

You observe two gravitationally bound stars (a binary pair). One is blue and one is yellow. They yellow star is six times brighter than the blue star. Qualitatively, compare their temperature and radii, i.e. which is hotter, which is smaller? Next, quantitatively compare their radii (to 1 significant figure).

To begin, qualitatively, we know that light from the yellow star has a larger wavelength than light from the blue star. Also, the energy corresponding to photons of larger wavelength is smaller than the energy for photons with smaller wavelength. Since temperature is directly proportional to energy and since energy is inversely proportional to wavelength, we can conclude that the temperature of the yellow star will be smaller than the temperature of the blue star.

We also know that \(L\propto R^2T^4\). So if the temperature of the blue star is greater but the luminosity of the yellow star is greater, then it must be true that the radius of the yellow star is larger.

So we have:

\(T_B>T_Y\) and \(R_Y>R_B\)

Quantitatively, we can use the Stefan-Boltzmann law, which relates luminosity to radius and temperature. Let \(L_Y\) be the luminosity of the yellow star and \(L_B\) be the luminosity of the blue star. We know that \(L_Y=6L_B\), so we can write the Stefan-Boltzmann law for each star:

\(L_B=4\pi R_B^2\sigma T_B^4\)

\(L_Y=6L_B=4\pi R_Y^2\sigma T_Y^4\)

Setting \(L_Y=6L_B\) and canceling like terms, we get:

\(R_Y^2 T_Y^4=6R_B^2 T_B^4\)

Which can be written as a ratio of radii and temperatures:

\(\dfrac{R_Y}{R_B} =\sqrt{6}\left(\dfrac{T_B}{T_Y}\right)^2\)

From Wien's Law, we can solve for an expression for temperature in terms of wavelength:

\(\lambda_{peak}=\dfrac{0.3\text{ cm K}}{T}\Longrightarrow T=\dfrac{0.3\text{ cm K}}{\lambda_{peak}}\)

Plugging this into the expression we found for the ration of radii and temperatures, we get:

\(\dfrac{R_Y}{R_B} =\sqrt{6}\dfrac{\left(\frac{0.3\text{ cm K}}{\lambda_Y}\right)^2}{\left(\frac{0.3\text{ cm K}}{\lambda_B}\right)^2}\)

\(\dfrac{R_Y}{R_B} =\sqrt{6}\left(\dfrac{\lambda_Y}{\lambda_B}\right)^2\)

We can find the wavelengths of yellow light and blue light, which are approximately \(\lambda_Y=570\) nm and \(\lambda_B=450\) nm, as shown in the image below.