Using that density as the electronic density, we find the following table of information for the requested quantities:

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1 Physics 40 Solutions to Problem Set 11 Problems: Fermi gases! The Pauli Exclusion Principle causes much of the behavior of matter, both of the type you are quite familiar with e.g, the hardness of solids, and things that are more extreme, e.g., the behavior of white dwarf stars. Let s see how all this works: 1. Schroeder Problem 7.19 Think about the electrons around the atoms that form a solid material. In the case of metals, it s pretty clear that some of the electrons are rather free to move around in the material. In fact, you can understand a huge amount about the behavior of metals by treating these electrons as an ideal noninteracting gas of Fermions. To a surprising degree, you can understand the behavior of such a system by assuming that the electrons don t interact! That s partly because a solid is charge neutral, so the electron-electron interactions are screened, and partly good luck! In any case, the first problem has you play with this electron gas model for copper. If you ever take the Solid State course, you ll see that the free electron gas is a hugely important part of our understanding of solids. OK, so everything traces back to the fact that copper contributes a single valence electron to the gas, so if you know the atomic density of copper, then you know the electron density of the gas. The rest follows from the equations of Schroeder s section 7.. So, I looked up the number density of atoms in copper and found it listed in Solid State Physics, by N.David Mermin and N. W. Ashcroft (1976). They have N/V = 8.47x10 atoms/cm Using that density as the electronic density, we find the following table of information for the requested quantities: Dnsty(1/m^) kf (1/m) EF TF (K) PD BF B_measured (Joules) (N/m^) (N/m^) (N/m^) 8.47E+8 1.6E E E E E+11 The bulk modulus is predicted at 6.5 GPa, and is off by about a factor of two from the experimentally determined bulk modulus. It s clear that the Pauli exclusion make a large contribution to the strength of materials! H11sol.0

2 Physics 40 Problem Set 9 Fall 009. Schroeder Problem 7.. elativistic electrons can also be important, say in the interior of stars. Surprisingly, you ll find relativistic electrons in some solids like tungsten!. Here we are asked to consider highly relativistic electrons. In this picture, the wavefunctions are still plane waves, the boundary conditions still lead to a digital k- space, but the relationship between the k-vector and the energy is no longer the classical result. Instead, we have that: k pc kc The argument about the ground state would still have you filling a sphere in k-space, so we still have the requirement that the number of filled states inside the sphere (including the factor of for spin) is given by the number of particles/ and therefore by the k- sphere volume divided by the volume per k-state: 4 N L k F Therefore, the radius of the sphere is still given by: k F N V 1 So far, the argument is completely the same as for the classical case. However, the ENEGY associated with this wavevector would now be different: k k c F F F At zero temperature, the chemical potential is equal to the Fermi energy, so we have: 1 N c V This result is the desired answer. b) Find a formula for the total energy of this system in terms of N and. eally, all that is wanted here is to repeat the calculation of the systems internal energy, U, as Schroeder did in eqs 7.40 through 7.4. The change from a sum on filled states to an integral over k-space is completely the same, so we pick up at Eq. 7.4 with:

3 Physics 40 Problem Set 9 Fall 009 U n _ MAX 0 n n dn However, the energy relationship is now different. We have: hc n kc c kx ky kz n L So, n _ MAX hc 4 MAX 4 L 0 hc U n dn n L One of the factors of the maximum number can be combined to give you the Fermi energy, or equivalently, the chemical potential. The other factors combine to give you the cube of the Fermi wave vector i.e., the density. Here is my approach: 4 nmax c nmax 1 1 L MAX 4 L U n N L. Schroeder Problem 7.. White dwarf stars are charge neutral chucks of matter too, and you won t be surprised to see that a model for the electrons in a white dwarf has a strong overlap with the treatment of electrons in copper. The major difference is that copper is held together mostly by electrostatic attractive forces between the ionic cores of the atoms and the electrons. By comparison, a white dwarf is held together by gravitational forces. The white dwarf. Here, we basically calculate the radius where a charge neutral collection of protons and electrons finds a balance between gravitational pressure and the quantum mechanical Pauli pressure. Since the nuclear energy output has ended in a white dwarf, the radiation pressure (which is what keeps the Sun rather inflated) has dropped out of the problem. Let s start with the gravitational pressure. Schroeder suggests that you use the gravitational energy for a uniform density sphere as a good starting point. Then, the gravitational energy is negative (referenced to infinite separation of all the mass in the white dwarf) due to the fact that gravity is attractive, and is approximately:

4 Physics 40 Problem Set 9 Fall 009 GM U 5 Just for fun, notice that this is the same /5 that appears in the Fermi energy calculation. Schroeder asks you to move on with calculation of the electron energies due to the Pauli principle, but before we do, notice that you can think of the gravitational energy as also creating an inwards pressure on the mass of the white dwarf. The associated gravitational pressure due to gravity is easy to calculate: PG U U V V GM 4 0 b) Next, we assume that the star is composed of equal amounts of protons, electrons, and neutrons. Basically, it s hydrogen, plus neutrons that have been created by the nuclear fusion process. Usually, these neutrons appear in combination with protons as helium nuclei, but Schroeder is not worrying too much about how the whole mess is assembled. In any case, we expect that the star is essentially a box full of electrons and protons and neutrons, all of which are Fermions. Since all of them are held in the same box, they all have the same restrictions on k-vector. Further, if they are nonrelativistic, then they all have the same relationship between k-vector and energy, differing only in the mass: k m However, since the electrons have the smallest mass, they certainly have the highest energy and they will have the most important contribution to the energy of the star (compared to the protons and neutrons). In any case, the average kinetic energy in the ground state is given by: N U KE NF N 5 5 m e V To finish the energy, we need to know the number of electrons. While the electrons certainly dominate the KE, most of the star mass is in the protons and neutrons. Therefore, you can count the electrons by assuming that half the star mass is protons, and that the number of electrons is equal to the number of protons for charge neutrality. Given a stellar mass of M, the number of particles is then,

5 Physics 40 Problem Set 9 Fall 009 N M m p And plugging back into the energy gives (assuming a spherical volume): U KE 5 1 N 5 m V e 5 M m p me This is the result that Schroeder quotes, complete with all the constants. c) OK, so we now have a gravitational energy that is attractive and 1/, along with a Pauli exclusion kinetic energy that is repulsive, and goes as 1/^. The total energy looks like: U TOTAL GM M h m p me 5 For small radii, the repulsive part gets large faster. For large radii, the attractive part dominates, so things are repulsive if the radius is too small, attractive if the radius is too big, and there is an equilibrium point somewhere. The sketch has an attractive potential well. To find the minimum, differentiate: GM M h 1 UTOTAL m p me Or 5 EQ 5 1 h m p me GM 1 This result predicts that as the mass of the star increases, gravitational attraction pulls it in to a smaller radius, the smaller radius causes the Fermi energy to increase until you compensate the larger gravitational attraction with a counteracting kinetic energy. It all makes sense. d) Next we can calculate the properties of a white dwarf with a solar mass. The white dwarf state is widely considered to be the end point of solar evolution for a star

6 Physics 40 Problem Set 9 Fall 009 like the Sun (after the red giant phase). Let s see what things might look like in a few billion years. Let s plug and chug with Mathematica: White Dwarf predictions Let's put down some of the white dwarf equations from HW 11 ugravity = -/5 (gnewton*solarmass )/radius; 5 solarmass protonmass udegeneracy = *hplanck /electronmass*1/radius ; howthingsare = {solarmass*10 0,hPlanck6.6*10-4,protonMass1.67*10-7,electronMass9*10-1,gNewton6.6*10-11 }; Sketch of the energy minimum Here' s the energy function for the solar mass case, so you can see the total energy curve around the minimum utotal = ugravity+udegeneracy/.howthingsare /radius /radius sketch = Plot[uTotal/.howThingsAre, {radius, , }] You can see that there is a minimum around 7 million meters or 7000 km. The equilibrium radius In the solutions, we did the differentiation to find that the equilibrium radius is given by :

7 Physics 40 Problem Set 9 Fall 009 eqadius = 1 protonmass 5 * *hplanck /electronmass*5/(*gnewton* 1 solarmass ); The numerical value for the case of a solar mass is : eqadius/.howthingsare or 7. kilometers The equilibrium density The predicted density is then : density=solarmass/(4/ Pi*eqadius ); density/.howthingsare So, we have something like a billion kg/m^, which is HUGE compared with the density of water (1000 kg/m^) It's a million times as dense as water. Cool. e) The Fermi energy is going to come out larger than it would for, say, a material like copper (see the first problem, above) because the density of the material is larger. The Fermi energy is related to the Fermi k-vector, which is given by the cube root of the density. Therefore, the Fermi energy in the non-relativistic region is going to go as the / power of the density. If the density is a million times larger, then the Fermi energy will be 10^4 times larger. Given that copper has a Fermi energy of something like 10 ev and a Fermi temperature of 100,000K, we expect that the white dwarf energy will be around 100,000 ev, and the Fermi temperature will be around 10^9 K. Given that nuclear fusion temperatures are around 10 million kelvins, we are really quite close to the ground state. f) OK, so suppose that we go relativistic! Then, the energy becomes proportional to the k-vector, rather than the square of the k-vector. That means you go as the density to the 1/ instead of / and THAT is enough to change the dependence on the stellar radius to 1/ instead of 1/^. Now you d have a pair of energy terms, one attractive and one repulsive, but BOTH going as 1/. There is no stable equilibrium point. The star is going to do something interesting g) Mc^ for the electron is around 500,000 ev, but the one solar mass Fermi energy is only 100,000 ev, so you re probably doing OK by predicting a white dwarf end point for the Sun.

8 Physics 40 Problem Set 9 Fall 009 However, the Fermi energy will continue to increase with density. Again, the Fermi energy is: F N M 4 M eq me V Somewhere between and 4 solar masses, you re going to cross the line between the non-relativistic electron and the relativistic electron and things are going to start to lose stability! 4) Schroeder Problem 7.4. Neutron stars! Schroeder s argument about the loss of stability at large enough white dwarf mass assumes that there is SOMETHING that the white dwarf can evolve into. What that something is, is a neutron star. The electrons and protons eventually can react to form neutrons and emit a pile of neutrinos Neutron stars and the Chandrasekhar mass. We expect that the gravitational pressure will eventually squeeze the protons and electrons to a high enough density that the system would be more stable as a collection of neutrons. When the system realizes that it has reached that point, there is a sudden drop in the Pauli pressure as the electrons are gobbled up into neutrons. This sudden pressure drop is a catastrophe for the star as its core suddenly implodes. The resulting shock wave blows off the outer parts of the star and you get a nova or supernova. Left in the middle of the disaster is the neutron stellar core, a neutron star. The mass of the neutron star can vary depending upon the history of the events. Our previous result, just above, can be used to estimate the radius of a solar mass neutron star. The only difference in the derivation is in the use of the neutron mass, rather than the electronic mass. Thus, the radius is just scaled by the neutron to electron mass ratio of roughly 000: M neutron 7000 km / 000.5km Neutron stars are small! Similarly, neutron stars are dense. Since the radius is down by 000X, the density is larger by that radius cubed, or 8 Billion times more dense than the white dwarf. However, because the Fermi energy is going only as the / power of density AND because it also goes as 1/mass of the fermion of interest, the Fermi energy is only scaling as the mass ratio. Therefore the Fermi energy and Fermi temperature are both about 000 times what you get for the white dwarf.

9 Physics 40 Problem Set 9 Fall 009 Finally, the interesting thing about the stability criterion is that the rest mass of a neutron is the rest mass of the neutron, and is therefore 000 times the rest mass of the electron. That means that the neutron star is very close to the same instability as the white dwarf star: Somewhere around to 4 solar masses is where we expect that the neutrons might go relativistic and the star might become unstable. For many years, astrophysicists argued about what a neutron star might become. Yes, the star is predicted to be unstable, but to go into some new form, a new form must be available. For example, classical mechanics tells you that the electron around a hydrogen atom is unstable because it should radiate energy. That does NOT mean that the electrons are unstable. It means that you have a lot to learn about quantum mechanics. Similarly, people were not certain that there is actually anything that a neutron star can decay into. We now expect that neutron stars above a certain size can become black holes.