Today we will cap off the course with some modeling. It should feel good. You have climbed a mountain.
It doesn't really matter what we model. I however chose something you probably haven't heard anything about: nuclei.
Nuclei are a bit old school (the 50's were a hot time in nuclear physics). There are not a lot of physicists at Cornell studying them. Its not that all the nuclear physics problems are solved -- the real difficulty is that there is no really good way to quantitatively understand nuclei. Big time lattice QCD calculations can explain the properties of the neutron -- can't do much bigger. This means there is a bunch of wonky semi-empirical and arm waving stuff. [Along with some really complicated perturbative expansions that you sum to all orders, and end up with a complicated wonky arm-waving calculation, which is not particularly accurate.] For our course this is perfect: we are experts at the wonky stuff.
I am being a bit facetious. There are a few modern and exciting things going on right now in the study of nuclei. The first is the Relativistic Heavy Ion Collider in Brookhaven. It takes massive gold nuclei and smushes them against each-other. When the nuclei pass through one another, they produce this really hot region of space, in which you start "pair-producing" a soup of quarks and anti-quarks. The interactions in this soup are sufficiently strong that it equilibrates, and you get a state of matter known as a "quark-gluon plasma." This is supposed to be the state of matter in the early universe. The experiments found that this soup has a number of interesting properties, including a remarkably small visoscity. [There are some interesting string theory calculations suggesting that the ratio of viscoscity to entropy density is bounded, and the bound appears to be pretty well saturated in these experiments.]
The other "modern" application of nuclear physics is Neutron Stars. Actually, Neutron stars are pretty old, but there are always more interesting things happening in that field. You can kind of think of a Neutron star as a really big nucleus. There may actually also be a quark-gluon plasma in the center of neutron stars, but it would be in a very different regime from RHIC -- it is ``cold" instead of ``hot".
Constituents of Nuclei
Nuclei are made of two things: protons and neutrons. These are all spin-1/2 fermions. We label the number of protons, $Z$. Chemical properties only depend on $Z$. We label the number of neutrons $N$. The total number of nucleons will be $A=N+Z$.
The glue holding together the nuclei is the strong force. Actually it is a bit more subtle than that: the nucleons are color neutral, so they do not directly produce gluons. The standard cartoon one uses for their interaction is that the quarks inside the nucleus produce pions. It is the exchange of the pions which leads to the force.
We know that if you exchange massless particles you get a Coulomb potential. When you exchange a massive particle like a pion, you get a finite range potential. We can estimate the range of the potential via dimensional analysis. The only length scale we can calculate from $m_\pi,\hbar,$ and $c$ is $r_0=\hbar/(m_\pi c)$. This is roughly a fermi ($10^{-15} m$).
A slightly more sophisticated argument is that the Uncertainty Principle lets you cheat energy conservation for a time $\delta t =\hbar/E$. Thus you can exchange a "virtual" pion if you do it quickly. You must cheat by an amount $E=m_\pi c^2$. In time $\delta t$ a relativistic pion will move a distance $r_0$.
We could further give some field theory arguments about this -- and if you go through them, you find
\begin{equation}\label{yukawa}
V(r)\propto \frac{e^{-r/r_0}}{r}.
\end{equation}
This is known as a Yukawa potential.
For us, at least, it is not worth going through the argument because it is a bit of a fib. The nuclear force is not really from the exchange of pions. The long-range part of the force is dominated by pion exchange, but more complicated stuff happens at short range. Thus the potential in Eq.~(\ref{yukawa}) is not really the right potential. You actually also get potentials like this in solid state physics, from screaning.
Regardless, the only thing we need to know about the potential is that it is short ranged.
We can estimate its depth by looking at the mass of a deuteron. A deuteron in a bound state of a proton and a neutron. It has a mass $m_d=1.876GeV$. Interestingly, the mass of a proton and the mass of the neutron are $m_p=938MeV$ and $m_n=940$. The binding energy is $E_b=m_p+m_n-m_d=2.2MeV$. If we treated the potential as a square well of depth $V_0$ and width $r_0$, the binding energy would be $E_b=-V_0+\hbar^2/m_n r_0^2\approx -V_0+m_\pi^2/m_n$. The pion mass is about $m_\pi=140MeV$, so the depth must be around $20MeV$, give or take some factors of $\pi$.
Apparently the binding between two protons is weaker than between either two neutrons or two protons -- as I have never heard of a bound state of two neutrons or two protons. In other words, the nuclear forces depend on isospin. Furthermore, people tell me that the nuclear forces are spin dependent. These subtleties aren't important at the level of our discussion.
How big is a nucleus?
Just knowing that the interactions are short ranged, we can estimate the size of a nucleus, and how it depends on atomic mass. Imagine we put $A$ nucleons in a clump of size $L$. The interaction energy per nucleon will be $V_0$ times the number of nucleons within a distance of $r_0$ of it:
\begin{equation}
\frac{E_{\rm int}}{A}\approx - V_0 \left(\frac{A\times r_0^3}{L^3}\right),
\end{equation}
or writing $g=V_0 r_0^3$, we have
\begin{equation}
E_{\rm in}\approx \left(\frac{g A^2}{L^3}\right).
\end{equation}
The kinetic energy should instead scale as
\begin{equation}\label{kfermi}
E_k\sim \frac{\hbar^2}{m}\left(\frac{N^{5/3}+Z^{5/3}}{L^2}\right),
\end{equation}
where $N$ is the number of neutrons, and $Z$ the number of protons.
This estimate comes from:
\begin{equation}
E_{kin}=2\sum_{n_x, n_y, n_z}\left( \frac{\pi^2}{L^2} \right)\left(n_x^2+n_y^2+n_z^2\right),
\end{equation}
where the sum is taken over all $n_x, n_y,$ and $n_z$ such that
\begin{equation}
\frac{\pi^2}{L^2} \left(n_x^2+n_y^2+n_z^2\right)\leq E_f.
\end{equation}
The number of Neutrons is
\begin{equation}
N=2\sum_{n_x, n_y, n_z} 1.
\end{equation}
Converting the sums into integrals, one arrives at the expression in Eq.~(\ref{kfermi}).
If we have a fixed number of nucleons $A$, Eq.~(\ref{kfermi}) is minimized by taking $N=A/2$ and $Z=A/2$.
So our first stab at the energy of a nucleus is
\begin{equation}
E=\alpha \left(\frac{A^{5/3}}{L^2}\right)-\beta \left(\frac{A^2}{L^3}\right)+A m,
\end{equation}
where $m$ is the nucleon mass.
If we minimize this with respect to $L$ we get $L\sim A^{1/3}$, and hence $E\propto A$. This means that the density is essentially independent of atomic number.
Both these findings are born out in experiment.
Charge
One thing which we did not quite get right is that heavier nuclei actually have fewer protons than neutrons. Take for example Uranium, which has isotopes with 125-150 neutrons, but only 92 protons. The standard argument for accounting for this is the Coulomb interaction between protons,
\begin{equation}
E_c \sim \frac{e^2}{4\pi\epsilon_0} \left(\frac{Z^2}{L}\right).
\end{equation}
If you actually do the integrals, I think you get a factor of $3/5$,
\begin{equation}
E_c=\frac{e^2}{4\pi \epsilon_0} \int d^3 r \int d^3 r^\prime \frac{1}{|r-r^\prime|} \left[\frac{Z}{L^3}\right]^2.
\end{equation}
To figure out how many protons are in a nucleus, one writes $N=A/2+x$ and $Z=A/2-x$. One then Taylor expands the energy in powers of $x$. The nuclear interaction term (at least how we treated it) is independent of $x$. The correction to the kinetic term goes as $x^2$, so not worrying too hard about the constants
\begin{equation}
E_{kin}\propto \frac{A^{5/3}}{L^2}\left[1+c \frac{x^2}{A^2}\right]\sim A+ c \frac{x^2}{A}
\end{equation}
In terms of $x$, the coulomb term should be
\begin{equation}
E_c\sim \frac{A^2}{4 L}- \frac{A x}{L}.
\end{equation}
Putting it together, we get
\begin{equation}
E=\alpha A + \beta \frac{x^2}{A} +\gamma [A^{5/3}/4- A^{2/3} x].
\end{equation}
Minimizing with respect to $x$ we find $x\sim A^{5/6}$.
Miraculously if you look at the data, this works-- in fact the proton-neutron mismatch scales as the nucleon number to the $5/6$ power.
Surface Tension
We have a pretty good empirical form for the energy of a nucleus:
\begin{equation}
E=\alpha A + \beta \frac{x^2}{A} +\gamma [A^{5/3}/4- A^{2/3} x].
\end{equation}
The standard empirical model used for fitting the masses of the nuclei includes one more term. This has to do that the nucleons on the surface do not have as many neighbors, and hence have a little less binding energy. This correction should be proportional to the surface area $L^2\sim A^{2/3}$. Thus a typical model is
\begin{equation}
E=\alpha A + \beta \frac{x^2}{A} +\gamma [A^{5/3}/4- A^{2/3} x] + \delta A^{2/3}.
\end{equation}
One then fits the coefficients $\alpha,\beta,\gamma,\delta$ to data. This works remarkably well.
[This surface tension term is crucial for our next argument.]
Energy per nucleon
We are now ready to think about the following problem. Imagine we put a fixed number of nucleons $X$ in a large box. What is the lowest energy configuration. Clearly we can reduce their energy by binding them into nuclei. What nuclei should we choose?
Well the total energy will be $X$ times the energy per nucleon. Thus we should choose the nuclei which minimize $E/A$. At small $A$ the surface tension term dominates, $E/A\sim A^{-1/3}$. At large $A$ the coulomb term dominates $E/A\sim A^{2/3}$. Thus we have a minimum -- roughly around iron.
From this argument, we expect that the entire universe should be made of iron. Its not. The problem is kinetics. It is not easy to turn Hydrogen into iron. You need to be in a very dense environment (such as a supernova). In highschool you probably learned some of the laws of fusion which describe the pathways for these things.
