3 Fundamentals of Planetary Materials - …web.gps.caltech.edu/classes/ge131/notes2016/Ch3.pdf · around 3g/cc; iron is near 8 Typically of order one megabar Terrestrial planets,

3. Fundamentals of Planetary Materials 21

3 Fundamentals of Planetary Materials In this chapter, we define the classes of planet forming materials and discuss the conditions under which they exist in planets. In order to do this, we must introduce some basic ideas of high-pressure physics.

3.1 What kinds of Materials Exist in Planets? As we discussed in the last chapter, the abundances of elements are determined by nuclear physics. However, most elements are not stable in elemental form but are found as molecules or in compounds. Additionally, it is important to understand the difference between the chemistry of matter and its phase. Chemistry speaks only about the elements that make up the matter (think of the chemical formula). Phase, on the other hand, refers to the physical properties of the matter. For example, water has the chemistry of H2O, but has three different accessible phases on the Earth’s surface: vapor, liquid, and ice. In fact, water has many high-pressure ice phases, which are all unique in their crystal structure. This means that the way in which the atoms are bonded together into a solid network differ between different solid phases. The phase of matter depends on the conditions of the environment, most importantly the temperature and pressure. In this chapter, we are concerned primarily with material or phase properties. The properties of planet-forming materials are the subject of physical chemistry and condensed matter physics. The cosmically most abundant materials can be divided into three general groups:

(i) “Gases”: those that do not condense (i.e., form solids or liquids) under conditions plausibly reached in the medium from which planets form;

(ii) “Ices”: those that form volatile compounds and condense but only at low temperatures (usually beyond the asteroid belt);

(iii) “Rocks”: those that condense at high temperatures and provide the building blocks for the terrestrial planets.

It is important to understand that these are labels of convenience; nothing is ever so simple that you could so easily subdivide materials. The quotation marks are there to remind you that what we call an ice is sometimes in the gas or liquid phase, etc. But the subdivision proves useful nonetheless because of the large differences in behavior and abundances among these groups. Remarkably, there is considerable correspondence between abundances and classes of materials even though they involve completely unrelated physics: Generally speaking, the gases are most abundant, the ices are next most abundant and the “rock” is least abundant (though not by much). The “gases” are hydrogen, helium, and (to a much lesser extent) the heavier noble gases. (Noble gases on Earth are found in tiny quantities, mostly carried to Earth adsorbed on solid particles; thus neon is of low abundance on Earth despite having much higher


cosmic abundance than silicon.) The gases overwhelmingly dominate the “observable” universe1, the Sun, and (as we shall see) giant planets such as Jupiter.

The hydrogen molecule H2 is the low-pressure thermodynamic ground state of H and it has along-range interaction with other hydrogen molecules and with helium by a very weak van der Waals force—this is why condensation of hydrogen requires very low temperatures. It’s also why it’s easy to squeeze hydrogen (as a liquid, solid, or, of course, as a gas) until you approach densities where the distance between molecules is about the size of a molecule, where the interaction becomes strongly repulsive.

The “ices” are mostly hydrides of the next set of light elements: O, C and N (e.g., H2O water, CH4 methane, and NH3 ammonia). But they also include other combinations among themselves (e.g., N2, CO, CO2, HCN....) As we discussed in chapter 2, hydrides do not necessarily dominate—they don’t seem to in the interstellar medium—but they are thermodynamically favored when the partial pressure of hydrogen is high, and will thus form if temperature or pressure permits reactions to occur. Water is the least volatile of this set because of hydrogen bonding between water molecules, which you can think of loosely as a weak form of ionic bonding arising from the very non-uniform charge distribution around the water molecule. Ammonia also has some hydrogen bonding. Methane and molecular nitrogen rely on van der Waals bonding in the liquid and solid state. CO has a small dipole moment, but also interacts mostly by van der Waals forces. “Rocks” include both metallic materials (iron and iron-nickel alloys), as well as what we might usually call rock (oxides and silicates). For those less familiar, oxides are ionically bonded materials made by mixing oxygen with other cations like Mg, Fe, Al, Ca, etc. Silicates, on the other hand, have the silica tetrahedra—one silicon bonded to four oxygens on the vertices of a tetrahedron—as its primary structural unit at low pressure. (Octahedral structures exist at higher pressure.) In the case of oxides and silicates, we have strong ionic and covalent bonding. Metallic bonding can be thought of as a special case of ionic bonding with electrons providing a spatially distributed charge, rather than the discrete charges of an ionic material such as NaCl. These materials are much more tightly bound than “gases” or “ices”, and are therefore stiffer and less volatile. The main constituents of “rocks” are metallic Fe, MgSiO3 pyroxene (or perovskite at high pressure), Mg2SiO4 olivine, and “FeO” (where the quotation marks refer to several different oxidation states representing FeO wustite, Fe3O4 magnetite, or Fe2O3 hematite). It is particularly important that due to their similar ionic size and valence charge, Fe can substitute for Mg in many structures, forming what is called a solid solution. A common and important example is (Mg,Fe)2SiO4 olivine, which can range smoothly between the magnesium and iron end-members. As you consider the material classifications in order of decreasing volatility,

1 Most of the mass in the universe is not directly observable but is detected through its gravitational influence and cosmological consequences. Hydrogen dominates the part of the universe we see through light emitted from stars and gas clouds. 2 This argument is not as straightforward as it seems and might even be wrong in some cases. The problem is that the approach to uniformity in the electron gas does not automatically imply that the material is a metal. The latter requires availability of nearby


“GAS” → “ICE”→ “ROCK” low pressure density increases, and material stiffness increases. As we shall see later, the densities remain in this order even at very high pressure (e.g., ice is always less dense than rock even though it is more compressible). The parameter that describes stiffness is the bulk modulus. It is defined as follows:

Adiabatic bulk modulus ⇒ KS = −V (∂P ∂V )S = ρ(∂P ∂ρ)S Isothermal bulk modulus ⇒ KT = −V (∂P ∂V )T = ρ(∂P ∂ρ)T

where ρ is mass density, P is pressure, S is entropy and T is temperature. As you can see, bulk modulus has units of pressure. [Reminder: One bar is 106 dynes/cm2 or 105 Pascals, and it is roughly the pressure at Earth’s surface.] The bulk modulus is a measure of the strength of interaction among the molecules in the material—soft (weakly-bound/volatile) materials compress easily, while tightly bound materials are stiff. It is also a guide as to how much density change might arise in a planet due to internal pressure. By changing the differential equation defining K into an approximate difference equation, ∂→ Δ , and rearranging, we can see that Δρ ρ ~ ΔP K . Thus the fractional change in density between surface and deep interior is roughly the pressure in the deep interior divided by the bulk modulus. This assumes negligible surface pressure and that P K . Therefore, if the interior pressure is comparable to the bulk modulus, then you expect the density inside the planet to be considerably larger than at the surface. Finally, we should discuss the difference between the two bulk moduli, which depends on what is being held constant while taking the derivative (i.e., during the compression). For interior processes, we should generally use the adiabatic bulk modulus (as we will see later). In any case, it doesn’t really matter since the difference between the two depends linearly on the temperature and some very small-valued material properties. Thus, at ambient temperature (300K) they differ by less than 1%, and at typical interior terrestrial temperatures (few 1000K) they still differ by only ~10%, so we will not often worry too much about the difference between them. To summarize then, we have:

Table 3.1

Type of bonding Examples Solid Densities

at low P Bulk modulus

of solid Locations

found

Van der Waals

“Gases”, Hydrogen,

helium, methane, N2

e.g., hydrogen is ~0.07 g/cc

e.g., hydrogen is a few kilobars

Giant planets (also CH4 & N2

on icy satellites)

Hydrogen bonding

“Ices”, Water, ammonia Around unity Ten kilobar

(roughly) Giant planets, icy satellites

Ionic and covalent

(including

“Rocks”, metallic iron

Rocks are around 3g/cc; iron is near 8

Typically of order one megabar

Terrestrial planets, cores

of giant


metallic) g/cc planets(?), icy satellites.

The type of bonding refers to low pressure behavior. As we will see later in this chapter, everything becomes a metal at high enough pressure. (In Jupiter, the only material that may fail to metallize is helium, though even helium will metallize if sufficiently hot, Stixrude and Jeanloz, Proc. Natl. Acad. Sci. 105, 11071-11075, 2008).

3.2 What are the Pressures inside Planets? We made a rough guess of this already in Chapter 1. It is now time to do it with somewhat more care (though still approximately!).

Pressures are estimated from the equation of hydrostatic equilibrium. The physical assumption is that the material is in static equilibrium—the weight of each element is fully supported by the pressure difference across the element.

Figure 3.1

For a box of material with density ρ, horizontal area A, and lying between r and r+dr, the mass is M = ρ(r)Adr . The net downward force, equal to zero in equilibrium, is just the sum of the weight and pressure force [P(r + dr) − P(r)]A + ρ(r)g(r)Adr = 0 . By rearranging, we obtain the result dP dr = −ρ(r)g(r) . Later, we shall study the consequences of this equation more fully and also consider the corrections arising from rotation. For now, it suffices to comment that the central assumption is that the material cannot support significant stress, e.g., by being rigid or extremely viscous. Clearly this must fail when the body is sufficiently small and at least partly solid. In practice, materials are surprisingly weak, in the sense that they fail and flow at stresses far below the bulk modulus of the material, thus failure occurs well before significant volume change. This is the reason that even very small objects, like asteroids, are forced to adopt a nearly spherical shape, even though their internal densities are not very different from surface densities. As an important first step to understanding the conditions inside planets, we can estimate the pressure by assuming that the density is roughly constant; obviously this is only a very rough guide but it gives a “warning” (i.e., it tells you whether your assumption was a good one) and is actually not too bad an approximation. From hydrostatic equilibrium, we have

dpdr

= −ρ(r)g(r) (3.1)

but we also have


(3.2) (

where is the mean density. By substituting in for g and integrating (3.1), we get

where we’ve used the fact that surface pressure is negligible . Evaluating at r=0,

For the Moon this predicts about the true central pressure (not surprisingly because it is a small body). For Earth, it predicts around 2 Megabars (true value is about 3.6). For Jupiter, it predicts around 10 Mbar (actual is 40 or more). This crude formula under-predicts the central pressure of differentiated bodies (which is to say, all planets), especially when the core has a density much larger than the mean density. This is, of course, because these objects strongly violate the assumption of uniform density.

3.3 How can we Figure out the Behavior of Materials at High Pressure? If we want to figure out what goes on in a planet, then we need to know how the materials listed above behave at planetary pressures. One approach is experiment. This is extremely important, and we will talk about some experimental constraints in due course. The experimental techniques are of two kinds: shock waves and static compression. Shock waves generate very high pressures for tiny fractions of a second by accelerating samples into a stationary target. Static compression is more ideal for studies of what happens in equilibrium, but attaining extremely high pressures is more difficult. The highest static pressures are obtained by squeezing tiny (~ tens of microns) samples between the tips of two diamonds in what are called diamond anvil cells (remember that pressure is force over area, so if we make the area small enough...). But experimental data alone are not enough for two reasons: (1) Experiments do not usually get to the full range of conditions encountered in planets. For example, pressures deep within Jupiter are unattainable by conventional techniques. (2) Even when experiments reach relevant conditions, they seldom map out enough of the thermodynamic phase space—T, P, and composition—to be sufficient for planetary modeling, where one needs a fine grid of parameter values. So it helps greatly to have a theoretical framework to incorporate experimental results and to extrapolate and interpolate. Many of these frameworks seek to find one of the most basic and important properties of a material, which is its equation of state. This relates how the density of a material depends on the environmental conditions; strictly this includes both pressure and temperature, but in many cases the thermal effects can be safely neglected. Thus an equation of state often takes the

g(r) = GM (r)r2 ≈ 4

3πGρr (3.2)

ρ

p(r) ≈ 43r

R

∫ πGρ2xdx = 2

3πGρ

2(R2 − r2 ) (3.3)

p(R) ≈ 0


formP = f (ρ) , with an added thermal correction (sometimes omitted). The thermal correction is discussed in Chapter 7. Here are the theoretical frameworks one can use:

(i) Parameterizations of convenience: These are non-physics based, simple recipes, fitted to data. Because they are not based on a real theory, they are dangerous when extrapolating. They are extensively used, however, because they are convenient. An example is the Birch-Murnaghan equation of state, heavily used in geophysics:

P = (3 2)K0 (ρ ρ0 )5 /3 (ρ ρ0 )

2 /3 −1⎡⎣ ⎤⎦ (3.2)

where K0 is the zero pressure bulk modulus and ρ0 is the zero pressure density. Another simple form (useful because it expresses density directly in terms of pressure) is the Murnaghan equation of state

ρ = ρ0 1+ nP K0[ ]1n (3.3)

where n is often taken to be four.

(ii) Asymptotic theories: Here one appeals to the simplifications that arise at very high pressure. The free electron gas approach (below) and Thomas-Fermi-Dirac theory are examples. This is an excellent approach for metallic hydrogen, for example (and for brown and white dwarf stars in general). They are discussed more fully below.

(iii) Brute force solution of Schrödinger’s equation: Often you will find this referred to as ab initio calculations. Immense advances in computer power have made this possible for quite complex systems. It is still only really practical for systems that have periodicity (crystals) or for small numbers of atoms (much less than Avogadro’s number!). This is now being extensively done for complex earth materials as well as for more complicated systems. Examples of this approach can be found in Gillan MJ, Alfe D, Brodholt J, Vocadlo, L., Price, G. D. (2006) First-principles modeling of Earth and planetary materials at high pressures and temperatures. REPORTS ON PROGRESS IN PHYSICS 69 (8): 2365-2441.

(iv) Pair potentials: This is the time-honored physical chemists approach of identifying species (atoms, molecules, clusters) and representing the energy as the sum of pairwise interactions among them:

(3.7)

where E is the energy and φ is the pair potential. This is often carried out in a Monte Carlo or molecular dynamics simulation on a computer.

Monte Carlo is a method in which you seek the ground state energy by taking an ensemble of molecules or atoms and then moving one of them at random and deciding

E = 12

ϕ( ri − rji≠ j∑ )


probabilistically (like a gambler!) whether to accept the move. A good move is one that lowers the total energy, but even a “bad” move should be accepted with some finite probability, in accordance with the fundamental rules of statistical mechanics. Molecular dynamics is very simple: You solve F=ma for an ensemble of molecules bouncing off each other, and calculate the total energy. As a compromise between convenient parameterizations, and ones that have some physical basis, we also have the Vinet equation of state, which attempts to model the inter-atomic potential with an ad-hoc but reasonable energy function (it is thus in the same spirit as the pair potentials, but applied to solids). The result is

P = 3K0 ρ ρ0( )2 3 1− ρ ρ0( )−1 3⎡⎣

⎤⎦exp (3 2)(K0

′ −1) 1− ρ ρ0( )−1 3⎡⎣

⎤⎦{ } (3.4)

where K0′ is first pressure derivative of the bulk modulus. This equation is found to

perform very well for materials undergoing large compressions. It is found generally to perform much better than the more common Birch-Murnaghan equation of state for many important terrestrial minerals. However, it is still asymptotically incorrect (i.e., begins to diverge from rteality outside the range of pressures for which it is fitted to data). It is therefore subject to the same caveats as the parameterizations of convenience, where extrapolation is risky business.

An excellent example of the state of the art, incorporating aspects of both (iii) and (iv) is the work of Chiarotti and colleagues, e.g. “Superionic and metallic states of water and ammonia at giant planet conditions,”Cavazzoni C, Chiarotti GL, Scandolo S, Tosatti E, Bernasconi M, Parrinello M; Science 283 44-46 (Jan 1, 1999). In this work, ab initio calculations are used to find a suitable empirical potential and then molecular dynamics is used to determine the behavior a system interacting classically in accordance with that potential.

3.4 What Happens in the Limit of High Pressure? It is a good idea to get some physical understanding of why materials resist compression, and one way to do this is to construct a theory that works at very high pressures. This turns out to be an excellent approximation for the deep interior of Jupiter, as well as being pedagogically valuable.

3.4.1 Why does matter resist compression? Materials resist compression because of quantum mechanical effects. This is very important! One tends to think of materials as consisting of positive and negative

charges, and imagines Coulomb forces as being the reason materials don’t like being squeezed. But a moment’s thought tells you this must be nonsense: since unlike charges attract, having negative and positive charges overlap is energetically favorable. If the Coulomb energy were to dominate, all matter would spontaneously collapse. In order to understand what determines the size of atoms and why they resist compression, we must examine the dominant terms in the energy equation. Recall from basic physics that the size of an atom is determined predominantly by the extent of its electron orbital cloud—the nucleus takes up an insignificant fraction of the total atomic volume. Thus we must

EelEel


concern ourselves with the total energy of the electrons. To simplify things, we will focus on the simple hydrogen atom, with only one proton and one electron. In addition to the Coulomb potential energy discussed above, the electron also has kinetic energy. Thus the total energy Eel is given by

Eel =p2

2me

−e2

r (3.5)

where p is the electron momentum, me is the electron mass, e is the charge on the electron (in cgs charge units of esu), and r is the electron orbital distance. The important idea we must now invoke is to recall from basic quantum mechanics that the Heisenberg uncertainty principle states that there is a limit to how well we can simultaneously determine both the position and momentum of a particle:

ΔpΔx (3.6) where Δp and Δx are the uncertainty in momentum and position, and is Planck’s constant divided by 2π. In the case of the atom, we can think of confining the electron in a box the size of the orbital cloud, and thus its minimum momentum is determined by the uncertainty principle. We can therefore approximately substitute in

Eel ≈

( r)2

2me

−e2

r (3.7)

Since the force on any particle is given simply by F = −∇E = −dE dr , we can obtain the equilibrium orbital distance, where the net force on the electron is zero, by setting the derivative of equation 3.7 to zero and solving for the radius. This yields an atomic radius of

a0 ≈2

mee2 = 0.53 Å (3.8)

Using this simple order of magnitude approach, we have derived that the size of the hydrogen atom, also called the Bohr radius a0 , is of the order of angstroms—1Å = 10−8 cm . (Note that due to a lucky cancellation of terms, we have actually obtained the Bohr radius exactly!) In any case, we can now plug this value into the energy equation to obtain the total energy of a ground state electron in a hydrogen atom

E0 = −me2 22 = −13.6 eV ≡ −1 Rydberg . This is also referred to as the ionization energy, since it is how much energy must be provided in order to ionize the electron and free it. So from this exercise, we see that the reason matter resists compression is not due to the Coulomb force, but because the uncertainty principle forces the electron kinetic energy to climb rapidly as the atom is compressed. Most importantly, the kinetic energy term goes as the inverse-square of distance (and is positive), whereas the Coulomb term is only one inverse power of r; accordingly, the kinetic energy term must dominate for small r (large compression).


3.4.2 The high pressure limit—“The Fermi Gas” Now that we have a basic overview of how quantum mechanics is applied to atomic hydrogen, we can begin to consider the more challenging but useful subject of how materials behave under extreme pressures. Once any material is squeezed “enough”, the spacing between atoms becomes so small that the electron clouds no longer remain associated with a single nucleus. Instead, the probability clouds of all the electrons smear out into a giant “electron-sea”. This partially describes how materials eventually become “metals” with free electrons capable of conducting charge. In the next chapter, we will discuss in detail how hydrogen undergoes this transition, as it does within Jupiter. We can gain an intuition about this system by first focusing on the simpler model in which we assume that the electrons are free (non-interacting) but are inserted into a uniform background sea of positive charge (think of uniformly smearing out all the nuclei). The important consequence of this picture is that it fully decouples the electrons from the smeared-out nuclei. Later we will consider the effect that the nuclei actually have. This model is called the non-interacting Fermi gas, since from quantum mechanics we know that electrons are fermions, which has great implications for their behavior.

In order to understand the Fermi gas, we would like to calculate the behavior of a “large” parcel of it, where surface energies are unimportant (anything large compared to an Angstrom will suffice). We consider a box of volume V=L3 and require that the wave function solutions have the same value on opposing faces of the box. These are called periodic boundaries, since the functions are constructed to be periodic over the length dimension L, and they exactly mimic an infinite system that is filled with side-by-side repeats of the calculated volume. As long as the calculated volume is large enough, this infinite repetition makes little difference to the final answer, and thus the actual box size does not matter. We already know from the exercise above that in quantum mechanics particles do not like to be confined to a small volume, and thus there is a kinetic energy associated with doing so. When applying periodic boundary conditions, we are actually allowing particles to range freely over a large (effectively infinite) volume. There is still, however, a kinetic energy “penalty” associated with densely packing the particles. This arises because electrons—like all quantum particles that make up normal matter—are fermions and thus they obey the Pauli exclusion principle. This principle states that fermions cannot exist in the same location while also occupying the same energy eigenstate and spin (i.e., they cannot have the same quantum numbers). This means that the average kinetic energy of a group of coexisting fermions, including electrons, must increase with particle density—as the particles are packed tighter, more of them are forced to higher kinetic energies.

In order to solve for the behavior of these electrons, we will need to take a look at Schrödinger’s equation, which dictates how the energy of particles depends on their wave function:

−2

2m∇2 +V

⎛⎝⎜

⎞⎠⎟ψ = Eψ (3.9)

where V is the potential, ψ is the wavefunction and E is the energy eigenvalue. (If you recall that in quantum mechanics the momentum operator is p=-i∇, then you realize that


this equation is quite similar to the simple energy equation for atomic hydrogen, where the total energy is related to the sum of potential and kinetic energies.) It is important to remember that the wavefunction gives the probability of finding a particle at a given point in space, where the probability density functions is just P(x) = ψ (x) 2 . Since we are allowing particles to move freely, equation 3.9 simplifies with V=0 inside the box. The simplest solution to the second order differential equation

−(2 2m)∇2ψ (x) = Eψ (x) has the form:

ψ ~ exp(ik ·r ) (3.10)

These solutions represent plane waves with wavelengths of λ = 2π k (where k is the magnitude of the wavevector), which travel through the Fermi gas parcel in the k̂ direction. When we apply the periodic boundary conditions to this general solution, we find that only plane waves where L is an integer number of wavelengths are allowed, with

kx = 2πnx L ky = 2πny L kz = 2πnz Lwhere nx , ny , nz = 0, ±1, ± 2, ± 3, ...

(3.11)

Having solved Schrödinger’s equation with the proper boundary conditions, we can now see that the energy states are quantized and described by the three quantum numbers nx , ny , and nz . Additionally, we see that the energy of a given state is just Ek =

2k2 2m , and thus larger wavenumbers (shorter wavelengths) have higher energy. At zero Kelvin, the system of particles occupies the lowest energy state possible, also called the ground state or zero-point energy. We can easily calculate this energy by imagining all of the available states in “phase space”. As you can see in Figure 3.2, states are quantized and separated by 2π L , and thus each state occupies a volume of Vstate = (2π L)3 in phase space.

Figure 3.2


In the ground state, all particles occupy the lowest energy available to them, and thus they fill the states closest to the origin first, filling a sphere outward in phase space up to the maximum energy, called the Fermi energy EF. We can calculate the Fermi wavenumber, kF, by finding how many states fit within the volume of the Fermi sphere, VFermi = 4πkF

3 /3 , as shown below

Figure 3.3 Fermi Sphere

N = nL3 =2VFermiVstate

=2(4πkF

3 / 3)(2π / L)3 ⇒ n = kF

3 / 3π 2 (3.12)

where N is the number of electrons and n is the electron number density. The factor of 2 in the numerator comes from the two available spins of the electron, which are degenerate in energy. It is convenient to write the number density as n = 1 /Vel = 3/(4πrs

3a03) , so that rs is the radius (in atomic units) of the sphere that

contains one electron on average. Then kF =1.92/rsa0 and the Fermi energy is given by EF≡2kF

2/2m = (50.1eV)/rs2.

We have thus calculated the maximum energy any electron has in a Fermi gas. In order to calculate the average energy, one must integrate over the Fermi sphere to find the total energy, and divide by the total number of states. It can be easily seen that the mean energy of the Fermi gas is

EFermi =35EF =

30.1rs2 eV =

2.21rs2 Ryd (3.13)

and the corresponding pressure is

(3.14) (3.18)

[Note: One Rydberg is the binding energy of the hydrogen atom, namely 13.6 eV. The atomic unit of energy, e2/a0, is 2 Rydbergs; the atomic unit of pressure, e2/a0

4, is about 294 Megabars; and the charge on the electron is just 1 in atomic units.] The parameter rs is known as the electron spacing parameter for obvious reasons. It is often useful to express the Fermi pressure (and energy) in terms of the density of the material, rather than the electron spacing parameter. We can relate these two by assuming that all of the electrons in the material are free and thus

PFermi = − dEFermi

dV= 51.6

rs5 Megabars


ρ ≈Matom

Vatom=Amp

VelZ=

3mp

4πa03

⎛⎝⎜

⎞⎠⎟Ars3Z

=2.69Ars3 Z

g/cm3 (3.15)

where A is the atomic mass and Z is the nuclear charge. Among light elements, A/Z is 1 for hydrogen and close to 2 for others. Using this relation, it is easy to see that Fermi pressure goes as PFermi ∝ ρ5 3 . Since the Fermi energy is strictly repulsive, the resulting pressure is always positive, and thus the Fermi gas cannot exist as a bound state at zero pressure. [Note: We call it a “gas” because it involves non-interacting particles. In fact, it is the very high-density limit of all matter! Do not make the mistake of thinking that “gas” means low density!]

3.5 Does Temperature Matter? At T=0, every level up to the Fermi level is occupied, and the occupancy of higher levels is exactly zero. In accordance with the rules of Fermi-Dirac statistics, there is finite occupancy of higher levels at T≠0, at the expense of occupancy close to but below the Fermi level:

Figure 3.4

From this figure, you can see that only the states within the thermal energy of the Fermi energy are affected. But here’s the point: For situations of interest to us, the Fermi energy is enormous relative to thermal energies, so the electrons can be well approximated as lying in their zero temperature state. Here’s the quantitative reasoning:

1 electron volt ≈ 1.6 x 10-12 ergs Boltzmann’s constant ≈ 1.38 x 10-16 ergs/degree Kelvin kT = 1 eV ⇒ T ≈ 12000 K So Fermi temperatures TF, defined to be Fermi energy divided by Boltzmann’s constant, are many tens of thousands of degrees, much higher than actual temperatures in planets. The Fermi temperature in Jupiter is typically around 200,000 to 300,000K. It is this


inequality, T << TF, that leads us to say that planets are degenerate. Do not confuse this with the issue of “hot” or “cold” in the context of the motions of the atoms (e.g., is it melted?). This is a different issue involving the much lower energies of atomic vibrations and translations, and we will deal with this in due course.

Of course, this is just part of the energy in a system. But it is an important part. Some examples:

Table 3.2

Metal Value of rs Fermi Energy (eV)

Free electron Bulk modulus

(in Mbar)

Actual Bulk mod. (in Mbar)

Li 3.25 4.74 0.24 0.12 Al 2.07 11.7 2.3 0.8 H* ~1 ~30 ~80 ~30

(*typical to the interior of Jupiter). The free electron bulk modulus listed above is obtained by taking the volume derivative of the Fermi pressure. Since Fermi pressure scales as (density)5/3, it is obviously 5/3 of the pressure. Clearly there is something that makes the material more compressible than if it were just a free electron gas. That something also provides stability (a Fermi gas with no attractive forces always has positive pressure and so wants to expand no matter what the density). In the case of the atomic Hydrogen, the quantum mechanical repulsive pressure is opposed by the attractive Coulomb forces—similarly, we must add in corrections to the Fermi gas to obtain a material description that is stable, but these attractive forces (“exchange” and “correlation”, as shown below) will also be governed by the dictates of quantum mechanics.

3.6 What about Coulomb Effects? In reality, there is no uniform background positive charge. Rather this charge is in the form of roughly uniformly spaced nuclei of charge Z. To estimate the total electrostatic energy of a distribution of charges, we can employ a trick that works very well: Place about each nucleus a sphere of electronic charge, uniformly distributed, such that the entity as a whole is neutral. The radius of this sphere is therefore Z1/3rs in atomic units. Then think of the entire system as an ensemble of these “atoms”. The error incurred has to do with the overlapping regions of these spheres, and is small in a close packed system. The electrostatic energy of one of these “atoms” is the sum of the term arising from the attraction between the electron cloud and the nucleus and the term for the repulsive interaction of the electrons with themselves—this is all purely classical electrostatics. The electron-nucleus term is:

Eel−nucl = −ZdNel

r= −

Zr0

Z1 3rs

∫0

Z1 3rs

∫4πr2dr4πrs

3 3= −

3Z 5 3

2rs (atomic units)


where we have calculated the electron nucleus energy from each spherical shell out to the effective radius. [Recall: The atomic unit of energy is 27.2 eV = 2 Rydbergs]. The repulsive interaction between the electrons is the classical result

Eel−el =Nel (r)dNel

r=

3r4drrs6

0

Z1 3rs

∫0

Z1 3rs

∫ =3Z 5 3

5rs(atomic units)

where Nel(r) is the number of electrons in the sphere of radius r. (Note: The factor of 3/5 is exactly the same as you get for the gravitational energy of a uniform density sphere, where the result is –3GM2/5R). To get the energy per electron, we must divide by Z, and to get the energy in Rydbergs, we must multiply by 2. So we find that

ECoulomb = −1.8Z 2 /3

rs Ryd/eln (3.16)

If this calculation is done for a specific lattice then instead of 1.8 you get something else. It is about 1.76 for simple cubic lattice, and 1.79… for fcc (face-centered cubic) and bcc (body centered cubic) and hcp (hexagonal close packed) lattices. This dimensionless constant is often referred to as the Madelung constant.

3.7 What about Exchange? Because electrons are Fermions, the total wave function of the system must be antisymmetric (i.e. the wave function changes sign when two electrons are interchanged). What this does is prevent two electrons of the same spin from being close to each other. This has nothing to do with Coulomb repulsion, it is a purely quantum mechanical effect. Consider just a two-electron wave function (where the electrons have the same spin). Then by definition, antisymmetry of the wave function says that

ψ (r1,r2 ) = −ψ (r2 ,r1) where r1 and r2 are the positions of electrons 1 and 2, respectively. But that means the wavefunction is identically zero when r1 = r2 (i.e., the electrons are on top of each other). Since the probability is proportional to ψ 2 , and since the wavefunction is a smooth function, meaning it will be small even when r1 and r2 are merely similar, this is equivalent to “keeping the electrons apart”. But because this quantum effect automatically keeps the electrons apart, they don’t experience as much repulsive Coulomb interaction as we calculated above. Thus we must reduce the repulsive part of the Coulomb energy with the exchange energy correction, which has the value

Eexch = −0.916rs

Ryd/eln (3.17)

3.8 What is the Total Energy? There are two other important contributions. The first arises from the fact that even electrons of opposite spin avoid each other, because of the Coulomb force. This is called a correlation energy, since the electron states are correlated to avoid each other. This is


also a negative energy, because we can lower the energy by arranging to have the electrons avoid getting near each other. The second correction is an energy that arises from the non-uniformity of the electron gas. This is called band-structure energy, and it is small (and bounded like the correlation energy), provided the density is sufficiently high. But for many purposes, it suffices to retain just the Coulomb and exchange energies for determining pressure:

Etot ≈2.21rs

2 −(1.8Z 2 /3 + 0.916)

rs Ryd/eln (3.18)

Ptot = −dEdV

≈51.6rs

5 [1− (0.407Z 2 /3 + 0.207)rs ] Mbar (3.19)

Notice now that there is a stable state at zero pressure. It is given by:

rs =1

0.407Z 2 3 + 0.207⇒ rs = 1.63 (ρ = 0.62 g/cc) for hydrogen

3.9 What have We Learned? In this chapter, we have seen how to make better estimates of pressures inside planets. More significantly, we have seen that some fundamental ideas of quantum mechanics have led to a picture of the compression behavior of matter giving us an equation of state: the relationship between pressure and density. It even gives us an estimate for the zero-pressure density of the material. However, it is built on the approximation that the electrons can be treated as a “gas”: an ensemble of free, weakly interacting particles. The model was made more accurate by accounting for Coulomb effects and exchange. The concept of “independent electrons” is surprisingly accurate and one of the triumphs of twentieth century physics. It has greatest applicability at high pressures. In the next chapter, we will pursue this further and consider it more fully for the most abundant planetary material in the Universe: high-pressure hydrogen.

Ch. 3 Problems 3.1) Consider a planet that has differentiated into a core of density Aρ0 overlain by a

mantle of density ρ0. Assume A>1. (It would be unstable otherwise!) The planet has radius R and the core has radius xR (so 0<x<1). Calculate the pressure at the planet center and compare it with the pressure for a planet that everywhere has a density equal to the mean density (defined as total mass divided by total volume) of this differentiated body. (If we ignore the effects of compression, this is like comparing the state of the planet before and after core formation). Show that the differentiated body always has higher central pressure. [This is really true... Earth’s central pressure has risen since the inner core began to form a billion or more years ago].

Solution: Integrating from the center outwards, P(xR) = Pc − 2

3πGA2ρ02x2R2 where Pc is the central pressure. Outside the


core, M (r) = 43πρ0[(r3 − x3R3) + Ax3R3] . Using g(r)=GM(r)/r2, the

pressure in the outer layer is therefore

P(r) = P(xR) − 23πGρ0

2 (r2 − x2R2 ) − 43πGρ0

2x3R3(1 xR −1 r)

Now P(R)=0, therefore we get

Pc = 23πGρ0

2R2 (x2A2 + 2(A −1)x2 (1− x) + (1− x2 ))

In the undifferentiated case, we get Pc,undiff=2/3πGρo2R2[x3A + (1-x3)]2. The

difference can be written

Pc –Pc,undiff = 2/3πGρo2R2(A-1)x2[(1-x4)(A-1) + 2(2+x)(1-x)]

which is always positive (because each individual term is positive) hence proving the result.

3.2) Our Moon has an observed mean moment of inertia about an axis of I=0.391MR2, where M is the mass and R is the mean radius. (Assume the Moon is a sphere, so the word “mean” then becomes superfluous). Remember from elementary physics that the prefactor would be 0.4 if the moon had uniform density, so it is evident that the density inside is higher than the density near the surface (though not by a lot).

(a) Suppose the Moon has constant composition. Roughly what bulk modulus must the Moon have to explain the observed I?

(b) The real bulk modulus of relevant rock is at least 1 Megabar (1012dynes/cm2 or 1011 Pascals), considerably larger than what you got in (a) if you did it right. Suppose instead that the observed I arises from a small iron core of density 7g/cm3 (over twice the density of silicate rocks, of about 3.3 g/cm3). What is the radius of that core? What is the mass fraction of the Moon in that core? How does this compare with the cosmic abundance of Fe?

Here are suggestions about how to proceed: Since you’re only looking to explain a small difference from uniform density, an approximation scheme will suffice. You can take the pressure distribution predicted for constant density, then use that and the definition of the bulk modulus (assumed constant) to get an improved estimate of the density distribution with radius. You can then use that to calculate the moment of inertia. The radius of the moon is 1760 km and the mean density is 3.34 g/cm3. Don’t try to use the mean density of the Moon to determine the size of the iron core directly. The reason why that is a bad idea is that it requires very highly precise information about the extent to which the silicate density differs from the mean Moon density. The moment of inertia approach is less susceptible to that kind of uncertainty.

3.3) A cold body of the mass and composition of our Sun has a radius similar to Earth’s radius. Confirm this. I’m only seeking slightly better than order of magnitude here which means you equate the typical internal pressure implied by hydrostatic equilibrium (or dimensional analysis, as in the last problem set) with the Fermi pressure for a cold sphere of hydrogen, mass 2 x 1033 g. Getting


within a factor of two is good enough and if this takes you more than six lines then you are doing something unnecessary.

3.4) (a) In the limit of very high pressure, where the pressure is dominated by the Fermi pressure (the ideal degenerate electron gas), why is the mass density of carbon almost exactly equal to the mass density of oxygen or helium at the same pressure? Why are they not exactly equal? (There’s more than one reason; try to think of two very different reasons). Why is hydrogen’s mass density lower by a factor of two? Note: This very high pressure limit really only applies to white dwarf stars (which are typically not made of hydrogen). (b) By equating this Fermi pressure to the estimate of internal pressure due to hydrostatic equilibrium, show that the radius of a body scales as the inverse cube root of its mass. [This is really true for sufficiently massive degenerate bodies; it explains why brown dwarfs can be smaller than Jupiter despite being more massive]. If this derivation takes you more than two or three lines then you’re trying to do something unnecessarily difficult.

Solution: (a) The Fermi pressure depends only on rs and this is the pressure that dominates at high density (because it has a stronger inverse dependence on rs than the other terms). But the mass density scales as A/Z times the inverse third power of rs. Accordingly, materials of the same A/Z will have the same density at very high P. Carbon-12, oxygen-16 and helium-4 all have A/Z=2. Hydrogen has A/Z=1.The equivalence of densities for C, O and He is not exact because of four effects: (i) There are isotopes with a different A/Z (carbon is not entirely mass 12, helium is not entirely mass 4, oxygen is not entirely mass 16.) (ii) The Coulomb correction depends explicitly on the charge on the nucleus and this is different among these elements. (iii) Nuclear binding energy effects cause “mass 16” to have a slightly different mass than 4/3 of “mass 12”. This is a consequence of E=mc2. (iv) There is also a different extent of non-uniformity of the electron gas (again determined by charge on the nucleus). A highly charged nucleus will cause some clumping of electrons around it. In the high pressure limit this is actually the smallest effect. (The isotope effect actually matters in white dwarfs!)

(b) Fermi pressure ∝ ρ5/3∝ (M/R3) 5/3; this must be balanced by gravity which gives a pressure ~GM2/R4. So M5/3R-5 ∝ M2R-4, whence R ∝ M-1/3.

3.5) Consistent with the simple theory developed in the text, we write the energy per proton for metallic hydrogen in the form

Emet =2.21rs

2 −2.72rs

− 0.20 Ryd/eln

Of course, the volume per electron (or per proton, which is the same thing) is 4/3πrs

3 in atomic units (ao3). Suppose that molecular hydrogen is distributed in

the form of a face-centered cubic lattice structure which means that if there is a lattice site at (0,0,0) in x,y,z Cartesian coordinates, then the 12 nearest neighbors are at (0,±½ , ±½)a , (±½ , 0,±½)a, (±½ , ±½,0)a and the next nearest neighbors


are at the vertices of the unit cells: (0,0,±1)a, (0,±1,0)a, (±1,0,0)a. Assume that the energy of interaction between two molecules spaced distance r apart is φ(r) and that nearest neighbor interactions only are considered. Confirm that the energy per electron (not per molecule because we want it to have the same units as the metallic energy) can be estimated as:

Emol = 3ϕa2

⎛⎝⎜

⎞⎠⎟−1.15 Ryd/eln

and that the average volume per proton is a3/8. (By “confirm” I mean that you have to show where the “3” and the square-root of two and the “8” come from; the -1.15 is trivial, it’s just the energy of a hydrogen molecule per proton relative to infinitely dispersed electrons and protons). For the choice φ(r) = 400r-9 (with r in units of a0 and energy in Rydbergs as always), estimate graphically or otherwise the pressure of the molecular to metallic phase transition of hydrogen. (This is very crude, although the exponent in the potential is locally about right; don’t expect it to work out to better than order of magnitude).

3.6) What would planets be like if the mass of the electron were reduced by a factor of two? Proceed by first answering this question: What would mean the density of ordinary matter (e.g., your body) be like in this case? (If you think that the mean density of matter or planets is only determined by the much more massive nuclei then you have completely misunderstood this chapter!)

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3 Fundamentals of Planetary Materials - …web.gps.caltech.edu/classes/ge131/notes2016/Ch3.pdf · around 3g/cc; iron is near 8 Typically of order one megabar Terrestrial planets,