# The Death of a Star

Second Authors: Nathan, Lauren

Introduction
We know how stars are formed to a certain extent and that while they are on the main sequence, they are supported by hydrostatic equilibrium. However, when a star moves off the main sequence and can no longer support itself, what happens? We assume that at this point, the core of the sun has converted all of its mass to energy and is now undergoing gravitational collapse.

Methods
We know that the Sun generates energy throughout its lifetime at a rate of:

$L_{\odot }=4 \times 10^{33} erg/s$

Assuming that the sun uses up the entire mass of its core (10% of a solar mass) as it undergoes fusion, and converts energy with 0.7% efficiency, we can determine the total energy it produces in its lifetime with

$E=0.007\Delta mc^{2}=0.007\left ( 2 \times 10^{32}g \right )\left ( 9 \times 10^{20} cm^{2}/s^{2} \right )=1.26 \times 10^{51} ergs$

Dividing this number by the rate of energy production, we can determine the time it takes for the Sun to use all of its mass available for fusion. This time is

$\frac{1.26 \times 10^{51} ergs}{4 \times 10^{33} ergs/s}=3.15 \times 10^{17}s=9.99 \times 10^{9} years$

Now we know the core will collapse, but it won't collapse indefinitely.  We find that the core collapses to the point that the interparticle spacing is on the order of the De Broglie wavelength.  Since it is easy to see that electrons have greater momentum compared to protons of equal energy, electrons are the first to reach this critical density.  We can calculate this using the equations:

$\newline E=\frac{1}{2}m_{e}v^{2}\newline \newline v=\sqrt{\frac{2E}{m_{e}}}\newline \newline \lambda=\frac{h}{mv}=\frac{h}{\sqrt{2Em_{e}}}, \: E = kT$

It is easy to tell that we have one molecule per cubic lambda.  So we have:

$N=\left ( \frac{\sqrt{2m_{e}kT}}{h} \right )^{3}$

The actual value is 8 times this, for reasons I can't remember, but Nathan tells me Professor Johnson said the factor of 8 was okay to include in our calculations.  So multiplying this by the mass of a hydrogen atom and using T = the temperature of the sun's core, we get density

$\rho = 8\frac{(2m_{e}kT)^{\frac{3}{2}}}{h^{3}}m_{H}\approx 360 \; g/cm^{3}$

Which is more than twice the current maximum density of the sun's core.

Conclusions
We find that by converting 0.7% of the mass of the sun's core into energy, the sun's lifetime is roughly 10 billion years, which agrees with what scientists have predicted.  The density of the core after collapse will also be far greater than the current density of the sun's core, which is reasonable or it would not be able to support the sun post-collapse.

## One Response to The Death of a Star

1. Nice writeup.

The factor of 8 comes from the fact that degeneracy sets in for interparticle separations of lambda/2, not at lambda.

For your LaTeX coding, use {\rm text} if "text" is to appear as regular font, as in expressing units. Instead of "x" use \times for multiplication.