Worksheet 8, Problem 5: The Kelvin-Helmholtz Timescale

The Virial Theorem states that half of a gravitationally-bound system's potential energy goes into the kinetic energy of the system's constituent particles. In the case of a cloud of gas, the cloud can shrink, but only if it loses energy by radiating. This occurs because as the radius shrinks, the potential energy becomes more negative, and therefore the particle motions must increase (higher kinetic energy). Particles moving faster leads to a higher gas temperature, and an increase in thermal emission. 

We know that the Sun started from the gravitational collapse of a giant cloud of gas. Let's hypothesize that the sun is powered solely by this gravitational contraction, as was once posited by astronomers long ago. As it shrinks, its internal thermal energy increases, increasing its temperature and thereby causing it to radiate. How long would the Sun last if it was thermally radiating its current power output, $L_\odot =4\times 10^{33} \text{erg s}^{-1}$? This is known as the Kelvin-Helmholtz timescale. How does this timescale compare to the age of the oldest Moon rocks (about 4.5 billion years, also known as Gyr)?

Diagram of the layers of the Sun
Image Credit:
http://www.amnh.org/education/resources/rfl/web/starsguide/images/diagram_layers.jpg

The Kelvin-Helmholtz timescale tells us the amount of time the Sun will last if it were only powered by gravitational contraction. We are assuming that the Sun has constant density, $\rho$, and a constant luminosity output. To solve for this time, we begin by finding the energy output over time. Since we know its luminosity, which is its power output, and since we know that it is constant over time, we can write the expression:

$K=L_\odot t$

This is the equation the thermal energy, which is kinetic energy. Now, we need to solve for the gravitational potential energy for the Sun. To do this, we can image the Sun being made up of many spherical shells of thickness $dr$. The gravitational potential between a shell and the mass interior to it is given by:

$U=-\dfrac{G\text{ }m_{shell}\text{ }m_{interior}}{r}$

We know that the density, $\rho$, of the Sun is constant, so we can solve for the masses of the shell and the interior:

$m_{shell}=4\pi r^2 \rho dr$

$m_{interior}=\frac{4}{3}\pi r^3\rho$

Plugging these into our equation for potential energy, we get:

$U=-\dfrac{G(4\pi r^2\rho dr)(\frac{4}{3}\pi r^3\rho )}{r}$

Integrating over r, we get:

$U=-\int_0^R \frac{16}{3}G\pi ^2 r^4\rho ^2 dr$

$U=-\frac{16}{15}G\pi ^2 \rho ^2 R^5$

But we also know that the density is equal to mass over volume, so we can solve for $\rho$ to get:

$\rho =\dfrac{3M}{4\pi R^3}$

So plugging this into our equation:

$U=-\frac{16}{15}G\pi ^2 R^5 \left( \dfrac{3M}{4\pi R^3} \right) ^2$

$U=-\dfrac{3GM^2}{5R}$

Now, we can apply Virial Theorem:

$K=-\frac{1}{2}U$

$L_\odot t=\dfrac{3GM_\odot ^2}{10R_\odot}$

Solving for t, we get:

$t=\dfrac{3GM_\odot ^2}{10R_\odot L_\odot}$

We know that $M_\odot=2\times 10^{33}$ g, $R_\odot =7\times 10^{10}$ cm, and $L_\odot =4\times 10^{33}\text{ erg s}^{-1}$. So we can put these values into the equation above to find the Kelvin-Helmholtz timescale:

$t=\dfrac{3(6.674\times 10^{-8}\text{ cm}^3\text{ g}^{-1}\text{ s}^{-2})(2\times 10^{33}\text{ g})^2}{10(7\times 10^{10}\text{ cm})(4\times 10^{33})}$

$t=2.8\times 10^{14}\text{ s}$

     $\approx 9\times 10^6\text{ years}$

So we have found that the Sun has a thermal timescale of 9 million years, which seems small. The oldest moon rocks are around 4.5 billion years old, which is 500 times the thermal timescale of the Sun. But in reality, the Sun would not continue radiating its current power output constantly. This will change over time, as will the radius, mass, and therefore potential energy of the Sun. Also, there are other factors to be considered, such as stellar nuclear fusion, which produces energy that keeps the Sun still going. The age of the Sun is approximated to be around 4.6 billion years and it is predicted that the Sun will have a lifespan of at least 10 billion years. 

I would like to thank Charles Law and Daniel Chen for their help in solving this problem. 

Citations:

http://en.wikipedia.org/wiki/Sun