A Faltering Sun: Stellar Scaling Relationships, Part II

We've talked about the birth of stars in molecular clouds. We also briefly discussed the main sequence, on which stars are in hydrostatic equilibrium owing to energy generated by nuclear fusion in their cores. Now let's investigate what happens when a star like the Sun can no longer support itself via nuclear fusion.

a) At what rate is the Sun generating energy?

This question is really just there to make sure we're awake and paying attention--the rate at which the Sun is generating energy is, of course, the Solar Luminosity, $L_\odot = 4 \times 10^{33}$ ergs s$^{-1}$.

b) If fusion converts matter into energy with a 0.7% efficiency, and if the Sun has 10% of its mass available for fusion (in the core only), how long does it take to use up its fuel supply? What is the general relationship between the mass of a star and its main-sequence lifetime?

Nuclear fusion is a process of converting mass to energy, which makes this kind of a cool question--less so because of what we learn from it than from the fact that we get to use $E=Mc^2$ for it. There's nothing like a physics problem like that--even a pretty easy one like this--to make you feel smart.

Bet it made him feel pretty smart too (source).

$M$ in this case is $0.1 \times M_\odot$, representing the 10% of the Solar Mass available for fusion, while $E$ is the total energy output of the Sun. If we divide both sides by $L_\odot$, we get the total lifetime of the Sun $t_\odot$. This gives us\[t\odot=\frac{0.1M_\odot c^2}{L_\odot}\] Solving for $t_\odot$, we find that the Sun will last around $3\times10^{17}$ seconds, or around nine billion years, on the Main Sequence.

c) The core will collapse until there is a force available to hold it up. One such force is supplied by degeneracy pressure. The pressure inside of a white dwarf star is provided by the motion of electrons. The electrons are in a tough situation: they can't occupy the exact same state, but there's not much room for them to coexist easily inside of a dense white dwarf. As a result, they must always be in motion to avoid other electrons (roughly speaking). This effect becomes important when the inter-particle spacing is of order the de Broglie wavelength $\lambda$, which is related to the momentum $p$ via\[\lambda = \frac{h}{p}\]where $h$ is the Planck constant, $h = 6\times610 ^{27}$ erg s${-1}$. For a stellar core of a given temperature, which particles reach this critical density first: electrons or protons?

This is a slightly unintuitive question, so it's good to define exactly what we're looking for in trying to answer it. The key to it is the de Broglie wavelength, which is a quantum mechanical concept whose details we won't go that far into now (it has to do with the fact that matter can behave as both particles and waves--which is why we're describing a wavelength when talking about electrons). What is important, though, is that a particle becomes degenerate as the inter-particle spacing approaches its de Broglie wavelength--again, degeneracy is a quantum process that we won't examine that closely, but when a particle becomes degenerate it begins to exert degeneracy pressure, which can act in a manner similar to thermal pressure in countering gravitational contraction. What we need to figure out, then, is whether protons or electrons have a longer de Broglie wavelength (the star is contracting, so the inter-particle spacing will reach the longer de Broglie wavelength first). In order to do that, we need to figure out which one has a smaller momentum, since there is an inverse proportionality between the de Broglie wavelength and momentum $p$.

The energy of a gas particle is defined as $k_BT$ (this can be pretty easily confirmed via $P=nK_BT$--$P$ is energy over volume, while $n$ is one over volume--multiply across by $n$ and you get energy equals $k_BT$). It is important to note that this energy will be the same for both protons and electrons, as $k_B$ is a constant, and they are of the same temperature. This means that $k_BT = KE = 1/2MV^2$. This can be rewritten, however, as $\frac{p^2}{2M}$. Since the energies of protons and electrons are equal, we then know that \[\frac{p_{e^-}^2}{2M_{e^-}} = \frac{p_p^2}{2M_p}\] Rearranging this we can then see that \[p_{e^-} = (\frac{M_{e^-}}{M_p})^{1/2}=p_p\] In other words, since $M_{e^-}$ is much, much smaller than $M_p$, $p_{e^-}$ must be much smaller than $p_p$, and the de Broglie wavelength of electrons must be much longer than that of protons. Thus, the collapsing Sun forces electrons into degeneracy long before it does protons.

d) If a typical white dwarf has roughly half the mass of the Sun ($M_\odot = 2\times10^{33}$ g), and the radius of the Earth ($M_\oplus = 6.4\times10^8$ cm), what is the typical density of a white dwarf in grams per cubic centimeter? What is the volume of white dwarf material that weighs as much as a car?

White dwarfs really are that small--incredibly small, compared to the size of the stars they start out as. For scale, here's the Earth (and a WD) compared to a tiny fraction of the Sun: 

(source)

And that's still retaining half of the mass of the original star (sometimes more). So you know right away that this is going to be an absurdly high density. To find it exactly, all we need to do is divide half of $M_\odot$ by $4/3\pi R_\oplus^3$, which gives us about $9\times10^5$ g cm$^{-3}$. If we assume a car weighs around 1500 kg, that means that a chunk of white dwarf a bouillon cube (just over 1cm on a side) would weigh as much as a car.

So yes, pretty dense.

Thanks to Scott Zhuge and Anne Madoff for working through this one with me!

In one of my favorite Hubble photos, the companion white dwarf to Sirius A--the brightest star in our sky--is visible just below and to the left of the brighter star (source).