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## Measuring the sizes of stars

Last week I blogged about how we measure the brightnesses of stars, and how at optical (and near infrared) wavelengths we use something called the magnitude system. Remember, the magnitude system has its zero point defined by Vega, so there is a conversion between the magnitude of a star and its luminosity. Astronomers tend to observe and measure the brightness of stars through a set of standard filters, for example the Johnson UBV system (U – ultraviolet, B – blue and V – visual). The diagram below shows the transmission curves for these three filters. One of the standard set of filters used in astronomy is the Johnson UBV system.

The conversion between the apparent magnitude of a star when measured in the V-band $m_{V}$ and its actual flux density (energy flowing per second per unit area per unit frequency interval) is $m_{V} = 0 \rightarrow L = 3640 \times 10^{-26} \text{ W/m}^{2} \text{/Hz}$

Referring back to blackbody curves, which I discussed in my blog about the OBAFGKM spectral classification system, we see that to get the total power per unit area from a blackbody we have to first of all sum over all frequencies (or wavelengths). This is just the area under the blackbody curve. The formula for this is the Steffan-Boltzmann law which is $\text{ Power per unit area } = \frac{ P }{ A } = \sigma T^{4}$

where $\sigma$ is the so-called Steffan-Boltzmann constant. So, we can see that the power emitted per unit area by a blackbody, such as a star, is simply related to its temperature. Remember also that the wavelength of the peak of the blackbody curve depends on temperature, as given by Wien’s displacement law $\lambda_{peak} = \frac{ 0.029 }{ T }$

where we need to measure the wavelength $\lambda$ in metres and the temperature $T$ in then given in Kelvin.

The figure below shows three blackbody curves for blackbodies of different temperatures, $T=7000, 5270 \text{ and } 4000 \text{ Kelvin}$. These temperatures correspond to a blue star ( $T=7000 \text{K}$), a yellow star like the Sun ( $T=5270 \text{K}$) and a red star ( $T=4000 \text{K}$. This figure shows the curves for three different blackbodies, at temperatures of T=7000, 5270 and 4000 Kelvin. Notice that not only does the position of the peak change, but also the height of the peak and the total area under the curve.

## Red Giants

Notice that the area under the red star’s curve is much smaller than the area under the other two. Yet, when we calcualted the intrinsic luminosities of stars in my blog about measuring the brightnesses of stars, we saw that Betelgeuse, a red star, is 10,000 times brighter than the Sun. How can it be both cooler and brighter?

The answer is that the area under the curve is the power per unit area, and the area we are talking about is the surface area of the star, given by $A=4 \pi R^{2}$ where $R$ is the radius of the star. Therefore, the only way a cool star can be brighter than a hotter star is if its surface area is bigger, so we know that Betelgeuse must be physically a much larger star than the Sun.

The luminosity of a star, let us call it $L$, is going to be dependent on both its temperature and its surface area (from the Steffan-Boltzmann law), so, we can write $L \propto R^{2} T^{4}$

To compare the luminosity of e.g. Betelgeuse $L_{Bet}$ to that of the Sun $L_{\odot} \text{ (} \odot$ is the standard symbol used in astronomy for the Sun) we can write $\frac{ L_{Bet} }{ L_{\odot} } = \frac{ R_{Bet}^{2} T_{Bet}^{4} }{ R_{\odot}^{2} T_{\odot}^{4} }$

Re-arranging, this gives that the ratio of the radius of Betelgeuse compared to the radius of the Sun $\frac{ R_{Bet} }{ R_{\odot} }$ as $\frac{ R_{Bet} }{ R_{\odot} } = \frac{ T_{\odot}^{2} }{ T_{Bet}^{2} } \sqrt{ \frac{ L_{Bet} }{ L_{\odot} } }$

Betelgeuse has a surface temperature of about $T=3000 \text{K}$ as measured from its blackbody spectrum, and as we saw in this blog, the ratio of its luminosity to the luminosity of the Sun is 10,000 (Betelgeuse is intrinsically 10,000 times brighter than the Sun). The surface temperature of the Sun is $T_{\odot} = 5800 \text{K}$, so plugging these numbers in gives us $\boxed{ \frac{ R_{Bet} }{ R_{\odot} } = \left( \frac{ 5800 }{ 3000} \right)^{2} \sqrt{ \frac{ 1 \times 10^{4} }{ 1 } } = 373.77 \approx 375!!! }$

As we know the distance from the Earth to the Sun, and can measure the angular size of the Sun as seen from the Earth (it is about half a degree), a simple big of trigonometry gives the radius of the Sun as about 700 million metres, so the radius of Betelgeuse is about $2.63 \times 10^{11} \text{m}$. An AU, the average distance from the Earth to the Sun, is about 150 million km, or $150 \times 10^{9} \text{m}$, so the radius of Betelgeuse is approximately $\boxed{ 1.75 \text{AUs!!} }$.

This means that, if we were to replace the Sun with Betelgeuse, even Mars would lie within the outer envelope of the star. This is why we call such intrinsically bright, red stars red giants, they truly are enormous!