A binary star system Source: http://www.vikdhillon.staff.shef.ac.uk/seminars/lives_of_binary_stars/binary_movie.gif |
A binary star system is one where two stars orbit around a central gravitational mass. In this scenario, we have a bright yellow star and a relatively dim blue star that are orbiting each other. Just by this information, we already can tell some important characteristics about these two stars.
Electromagnetic Spectrum: Blue light has shorter wavelength than yellow light Source: http://www.astronomersgroup.org/images/EMspectrum.jpg |
First of all, the different colors of the stars can tell us that the blue star is hotter than the yellow star. This is because on the electromagnetic spectrum, blue wavelengths of light are shorter than yellow wavelengths of light. Since all electromagnetic waves travel at the speed of light, you can relate wavelength and frequency with $c = \nu \lambda$, where $\nu$ is the frequency and $\lambda$ is the wavelength. As wavelength increases, frequency decreases, and vice versa, when wavelength decreases, frequency increases. But what does wavelength and frequency have to do with temperature of the star? Well, the higher the frequency of the star, the more energy it has, as shown by the Planck-Einstein relation, $E = h\nu$. Since a blue star has a shorter wavelength, and therefore higher frequency than a yellow star, it has more energy than a yellow star. Since temperature is simply a measure of the kinetic (and sometimes potential!) energy, the more energy there is, the hotter something is. Therefore, since the blue star has more energy, it has a higher temperature than the yellow star, and is therefore hotter.
In terms of size, since the luminosities of the stars are different, with the yellow star being much brighter than the blue star, you can look at the equation for luminosity for a blackbody (a star is a good blackbody), $L = 4 \pi R^2 \sigma T^4$, where $R$ is the radius of the star, $T$ is the temperature, and $\sigma$ is the Stefan-Boltzmann constant. Removing the constants, it is apparent that the luminosity of a star is proportional to its radius and temperature, $L \propto R^2T^4$. Since we have established that the blue star is much hotter, the fact that it is much dimmer than the yellow star would mean that the yellow star has to be much larger to compensate for lack of temperature in order to be brighter than the blue star.
In order to show that the yellow star is larger than the blue star, we can quantitatively calculate just how much larger the yellow star is. We again use the luminosity equation, $L = 4 \pi R^2 \sigma T^4$:
- The yellow star's luminosity is $L_{Y}$ with a radius $R_{Y}$ and temperature $T_{Y}$
- The blue star's luminosity is $L_{B}$ with a radius $R_{B}$ and temperature $T_{B}$
Since we know that the yellow star is 6 times as bright as the blue star, we can relate the two as follows:
\begin{align}L_{Y} &= 6L_{B} \\
4\pi R_{Y}^2 \sigma T_{Y}^4 &= 6 \cdot 4\pi R_{B}^2 \sigma T_{B}^4 \\
R_{Y}^2 T_{Y}^4 &= 6 R_{B}^2 T_{B}^4 \\
\frac{R_{Y}^2}{R_{B}^2} &= \frac{6T_{B}^4}{T_{Y}^4}\\
\sqrt{\frac{R_{Y}^2}{R_{B}^2}} &= \sqrt{\frac{6T_{B}^4}{T_{Y}^4}}\\
\frac{R_{Y}}{R_{B}} &= \sqrt{6}\frac{T_{B}^2}{T_{Y}^2}
\end{align}
Hmm... we seem to be stuck here. We don't know the actual temperatures of the two stars. However, we don't need to know their actual temperatures, since we can use the Wien Displacement Law, $\lambda_{max} = \frac{b}{T}$, where $\lambda_{max}$ is the peak of the wavelength, $b$ is the Wien's displacement constant, $0.3$ cm K, and $T$ is the temperature of the blackbody. We know the wavelengths of both blue light (450 nm) and yellow light (580 nm), so we can solve for the temperature using the Wien Displacement Law. By rearranging the Wien's displacement law, we get $T = \frac{b}{\lambda_{max}}$. We can substitute this into our relation of the blue and yellow stars:
\begin{align}
\frac{R_{Y}}{R_{B}} &= \sqrt{6}\left(\frac{T_{B}}{T_{Y}}\right)^2\\
\frac{R_{Y}}{R_{B}} &= \sqrt{6}\left(\frac{\frac{b}{\lambda_{B_{max}}}}{\frac{b}{\lambda_{Y_{max}}}}\right)^2\\
\frac{R_{Y}}{R_{B}} &= \sqrt{6} \left( \frac{\lambda_{Y_{max}}}{\lambda_{B_{max}}}\right)^2\\
\frac{R_{Y}}{R_{B}} &= \sqrt{6} \left( \frac{580 nm}{450 nm}\right)^2\\
\frac{R_{Y}}{R_{B}} &\approx 4
\end{align}
Based on the ratio of the radius of the yellow star to the blue star, the yellow star has a radius that is approximately 4 times as big as that of the blue star!
Another excellent writeup! The explanatory text that helps the reader follow your reasoning. I can imagine a student reading this post and learning a great deal about astronomy. Keep these posts coming!
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