Download 3 Fundamentals of Planetary Materials

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
3. Fundamentals of Planetary Materials
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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
3. Fundamentals of Planetary Materials
23
“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 ⇒ K S = −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 (weaklybound/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
Van der Waals
Hydrogen
bonding
Ionic and
covalent
(including
Examples
Solid Densities
at low P
Bulk modulus
of solid
Locations
found
Giant planets
(also CH4 & N2
on icy
satellites)
“Gases”,
Hydrogen,
helium,
methane, N2
e.g., hydrogen
is ~0.07 g/cc
e.g., hydrogen
is a few
kilobars
“Ices”, Water,
ammonia
Around unity
Ten kilobar
(roughly)
Giant planets,
icy satellites
“Rocks”,
metallic iron
Rocks are
around 3g/cc;
iron is near 8
Typically of
order one
megabar
Terrestrial
planets, cores
of giant
3. Fundamentals of Planetary Materials
metallic)
24
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
dp
= − ρ(r)g(r)
dr
but we also have
(3.1)
3. Fundamentals of Planetary Materials
g(r) =
where
GM (r) 4
≈ π G ρr
r2
3
25
(3.2)
ρ is the mean density. By substituting in for g and integrating (3.1), we get
R
2
2
4
2
p(r) ≈ ∫ π G ρ xdx = π G ρ (R 2 − r 2 )
3
3
r
(3.3)
where we’ve used the fact that surface pressure is negligible
r=0,
p(R) ≈ 0 . Evaluating at
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 underpredicts 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
3. Fundamentals of Planetary Materials
26
form P = 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)K 0 ( 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
1
ρ = ρ0 [1 + nP K 0 ]n
(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:
E=
1
∑ϕ ( ri − rj ) (3.7)
2 i≠ j
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
3. Fundamentals of Planetary Materials
27
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 = 3K 0 ( ρ ρ0 )
2 3
{
⎡1 − ( ρ ρ0 )−1 3 ⎤ exp (3 2)(K 0′ − 1) ⎡1 − ( ρ ρ0 )−1 3 ⎤
⎣
⎦
⎣
⎦
}
(3.4)
where K 0′ 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?
Eel 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
3. Fundamentals of Planetary Materials
28
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
e2
−
2me 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 e2
−
2me
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
= 0.53 Å
me e 2
(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 2 2 = −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. Fundamentals of Planetary Materials
29
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 2
⎞
⎜⎝ 2m ∇ + 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
3. Fundamentals of Planetary Materials
30
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

 2
point in space, where the probability density functions is just P( x) = ψ ( x) . 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:

(3.10)
ψ ~ exp(ik ·r )
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
k x = 2π nx L
k y = 2π ny L
kz = 2π nz L
where 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 =  2 k 2 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
3. Fundamentals of Planetary Materials
31
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 = 4π kF3 /3 , as
shown below
N = nL3 =
2VFermi 2(4π kF3 / 3)
=
Vstate
(2π / L)3
Figure 3.3 Fermi Sphere
⇒ n = kF3 / 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/(4π rs3a03 ) , 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≡2kF2/2m = (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 =
3
30.1
2.21
EF = 2 eV = 2 Ryd
5
rs
rs
(3.13)
and the corresponding pressure is
PFermi = −
dEFermi 51.6
= 5 Megabars
dV
rs
(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/a04, 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
3. Fundamentals of Planetary Materials
ρ≈
M atom Am p ⎛ 3m p ⎞ A
2.69A
=
=⎜
= 3
g/cm 3
3⎟ 3
Vatom
Vel Z ⎝ 4π a0 ⎠ rs Z
rs Z
32
(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 ∝ ρ 5 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
3. Fundamentals of Planetary Materials
33
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)
Li
Al
H*
3.25
2.07
~1
4.74
11.7
~30
Free electron
Bulk modulus
(in Mbar)
0.24
2.3
~80
Actual
Bulk mod.
(in Mbar)
0.12
0.8
~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:
Z 1 3 rs
Eel − nucl = −
∫
0
ZdN el
=−
r
Z 1 3 rs
∫
0
Z 4π r 2 dr
=
r 4π rs3 3
−
3Z 5 3
2rs
(atomic units)
3. Fundamentals of Planetary Materials
34
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
Z 1 3 rs
Eel − el =
∫
0
N el (r)dN el
=
r
Z 1 3 rs
∫
0
3r 4 dr
3Z 5 3
=
(atomic units)
rs6
5rs
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
Ryd/eln
rs
(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).
2
Since the probability is proportional to ψ , 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 = −
0.916
Ryd/eln
rs
(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
3. Fundamentals of Planetary Materials
35
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:
2.21 (1.8Z 2 / 3 + 0.916)
−
Ryd/eln
rs2
rs
dE 51.6
Ptot = −
≈ 5 [1 − (0.407Z 2 / 3 + 0.207)rs ] Mbar
dV
rs
Etot ≈
(3.18)
(3.19)
Notice now that there is a stable state at zero pressure. It is given by:
rs =
1
0.407Z + 0.207
⇒ rs = 1.63 (ρ = 0.62 g/cc) for hydrogen
2 3
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 zeropressure 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 3 π GA 2 ρ02 x 2 R 2 where Pc is the central pressure. Outside the
3. Fundamentals of Planetary Materials
36
core, M (r) = 4 3 πρ0 [(r 3 − x 3 R 3 ) + Ax 3 R 3 ] . Using g(r)=GM(r)/r2, the
pressure in the outer layer is therefore
P(r) = P(xR) − 2 3 π G ρ02 (r 2 − x 2 R 2 ) − 4 3 π G ρ02 x 3 R 3 (1 xR −1 r)
Now P(R)=0, therefore we get
Pc =
2
3
π G ρ02 R 2 (x 2 A 2 + 2(A − 1)x 2 (1 − x) + (1 − x 2 ))
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
3. Fundamentals of Planetary Materials
37
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 nonuniformity 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 =
2.21 2.72
−
− 0.20 Ryd/eln
rs2
rs
Of course, the volume per electron (or per proton, which is the same thing) is
4
/3πrs3 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
3. Fundamentals of Planetary Materials
38
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:
⎛ a ⎞
Emol = 3ϕ ⎜
− 1.15 Ryd/eln
⎝ 2 ⎟⎠
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!)