Solutions to Problems 10: Order of Magnitude Physics 1) Electrical ...

.
Solutions to Problems 10: Order of Magnitude Physics
1) Electrical conductivity σ (in the cgs system) has units of s −1 . Construct an example of
a physical process whose characteristic timescale is σ −1 .
Conductivity describes the rate at which charges spread under the influence of an
electric field. Consider, therefore, what happens when a charge concentration is created in
a conducting medium. The charge flux is better known as the current density ⃗ J = qn〈v〉,
where q is the charge per carrier, n is the volume density of carriers, and 〈v〉 is their mean
drift speed under the influence of the electric field. When the mean free path is short
compared to the length on which the charge density varies, ⃗ J = σE ⃗ = −σ∇V, for electric
potential V.
We also know from Poisson’s Law that ∇· ⃗E = 4πqn, so the conservation equation for
charge is
q ∂n
∂t +∇·⃗J = 0
q ∂n
∂t +4πσnq = 0
if σ is independent of position. As this equation makes clear, the charge density declines
exponentially with timescale σ −1 .
2) There are four quantities with units s −1 relevant to the way electromagnetic waves
interact with metals, of which three are independent: the conductivity, the conduction
electron collision frequency, the conduction electron plasma frequency, and the frequency
of electron transitions from bound states to the Fermi level of the conduction band.
a) Estimate all four in a typical metal.
We begin with the bound state transition frequency. These transitions have an energy
scale corresponding to a slightly inner-shell, i.e., the effective nuclear charge is ≃ 2–3
because a few electrons per atom are out in the conduction band. Atomic transition
energies scale ∝ Z 2 , so instead of being several eV, they are perhaps ≃ 20 eV. The
corresponding frequency (in radian units) is ω tr ≃ 6×10 17 s −1 .
The conduction electron plasma frequency is (4πn e e 2 /m e ) 1/2 , but only the electrons
near enough the Fermi level to be able to move to new states can participate, a fraction
∼ k B T/E F . If the mean volume per atom is ∼ (3Å) 3 and 2 electrons per atom are in the
conduction band, the total number density of conduction electrons is 7×10 22 cm −3 . The
Fermi wavenumber k F ∼ (3n e /8π) 1/3 ≃ 2×10 7 cm −1 (8π instead of 4π because there are
two electron spin states). The Fermi energy E F = (¯hk F ) 2 /2m e = 2 × 10 −13 erg. Then
k B T/E F ≃ 0.2, and
ω p ≃ 7×10 15 s −1 .
Given k F , the typical electron speed at the Fermi surface is v F = ¯hk F /m e ≃ 2 ×
10 7 cm/s (an amusing numerical coincidence). If the scattering length for collisions with

phonons is 30 Å, we find
Lastly, the conductivity is
If ω < ν c ,
On the other hand, if ω > ν c ,
ν c = 7×10 13 s −1 .
σ = ω2 p /4π
ν c −iω .
σ low ≃ 7×10 17 s −1 .
σ high ≃ 5i×10 17 ω −1
14 s−1 ,
where ω has been scaled in units of 10 14 s −1 .
Thus, if the wave frequency is in the near-infrared or lower (ω < ν c ), the transition
frequencyandtheconductivityaresimilar,theplasmafrequencyistwoordersofmagnitude
smaller, and the collision frequency is two orders of magnitude smaller yet. On the other
hand, asthewavefrequencymovesintotheopticalorUVband, theconductivitydiminishes
and becomes mostly imaginary.
b) If their values are ordered according your estimate, discuss the implications for waves
of various frequencies encountering the metal: which pass through, which reflect, which are
absorbed, ...?
At very low frequencies (ω < ν c = 7×10 13 s −1 ), the conductivity is real and greater
than the wave frequency. In this domain, kc ≃ ± √ 2πi(1+i) √ σω. Complex wavenumbers
indicate decay; the wave is absorbed as it enters the conducting medium.
For all ω > ν c , σ = σ high . When the conductivity has this form, the dispersion relation
is ω 2 = k 2 c 2 +ω 2 p . Thus, for ν c < ω < ω p , the wave is also evanescent inside the conductor,
and its energy is absorbed.
On the other hand, for ω p < ω < ω tr , the wave can propagate within the conductor,
but with reduced wave number. Part of its energy will be transmitted and part reflected.
Finally, for ω > ω tr , again there is partial transmission and partial reflection, but
the part that enters the medium will be subject to absorption by boosting electrons from
bound states to the conduction band.
3) Consider the spectroscopy of H 2 O molecules.
a) Estimate their characteristic vibrational frequency.
When we first discussed material properties, we estimated the effective spring constant
ofamolecularbondbyK ∼ 2E b /a 2 . Thecharacteristicscaleofthevibrationalfrequency is
then ω vib ∼ (2E b /a 2 µ) 1/2 , where µ is the reduced mass of the two molecules participating
in the bond. Here, because m H = m O /16, µ ≃ m H . With a binding energy ∼ 3 eV and a
bond length ∼ 1 Å,
ω vib ∼ 2×10 13 s −1 .
2

Also because m H ≪ m O , the oxygen atom hardly participates in the motion, so the two
bonds really can be considered separately. Note here and in the next portion of this
question that the frequency in Hz is ω/2π.
b) Estimate their characteristic rotational frequency. At room temperature, what’s the highest
rotational quantum number you’d expect to be substantially populated? Because water
is an asymmetric molecule, its energy levels also depend on a second quantum number, but,
in the order of magnitude spirit, we’ll ignore it here.
The quantum mechanical expression for the energy of a rotational state with total
quantum number J is ¯h 2 J(J + 1)/2I, where I is the moment of inertia. If it were a
spherical molecule, the energy differences would be (J + 1)¯h 2 /I when J is the quantum
number of the lower state. Water is definitely not spherical, so the axial quantum number
also should be considered, but we’ll ignore that fact in the order of magnitude spirit.
Once again because m H ≪ m O , we’ll take the oxygen atom as coinciding with the
center of mass of the whole molecule. Because the molecule is planar but not linear, two
of the principal moments are the same and the third different, but we’ll ignore that, too.
We’ll simply make I ∼ 2m H (sinφa) 2 ≃ 2×10 −40 gm cm 2 , where sinφ ∼ 1/ √ 2 describes
the degree to which this isn’t a linear molecule. We then find that
ω rot ∼ (J +1)¯h/I = 6×10 12 (J +1) s −1 .
At the highest rotational quantum number substantially populated, ¯h 2 J ∗ (J ∗ +1)/2I ∼
k B T, so
J ∗ ∼ (2Ik B T) 1/2 /¯h ∼ 4.
c) Estimate their cross section for absorbing photons in either rotational or vibrational
transitions when these molecules are in the air.
As shown in class, electric dipole transitions have cross sections in the frame of the
atom or molecule
σ(ω) ∼ α fs a 2 0 |M|2 ω 0 Γ/2
(ω −ω 0 ) 2 +(Γ/2) 2,
where Γ is the deexcitation rate of the upper state and ω 0 is the resonance frequency.
In air, the molecules are all moving with random thermal motions, so it makes sense to
analyze the situation in the lab frame and allow for Doppler shifts due to the thermal
motions. Because of them, even photons that are exactly at resonance in the lab frame
are typically off resonance by ∼ (c s /c)ω 0 ∼ 10 −6 in the molecule frame, or an absolute
frequency offset ∼ 10 7 (ω 0 /10 13 s −1 ) s −1 . On the other hand, the collision rate is
∼ nσ coll c s ∼ 2.5×10 9 s −1 ,
rather larger than the frequency offset due to thermal motions.
Thus, the “pressure broadening” is likely the dominant element, making the cross
section
σ ∼ α fs a 2 0 |M|2 ω 0 /Γ ∼ 1×10 −15 ω 0
10 13 s −1 cm2 .
3

Here we have taken the scaled dipole moment |M| ≃ 1 (it’s actually ≃ 0.8).
4) Microwave ovens generate power at ≃ 6 GHz, and typically produce ∼ 1 kW. It happens
that the lowest frequency rotational transition in water occurs at ≃ 22 GHz. That such
a transition exists should come as a surprise because the characteristic frequency you estimated
in Problem 3 is considerably higher. Its origin turns out to lie in a coincidence:
centrifugal bond-stretching leads to a near match in energy between a particular state within
the J = 5 set and another particular state in the J = 6 set. In any event, the microwave
frequency is well below even this resonance.
a) Estimate how deeply into the food the microwaves penetrate.
As noted in the solution to part c) of the previous problem, the cross section for
absorption of an electromagnetic wave in a rotational transition of water is ∝ [(ω −
ω 0 ) 2 + (Γ/2) 2 ] −1 . In the problem at hand, however, the frequency offset is ≃ 16 GHz∼
1 ×10 11 s −1 . The collision rate is even greater, ∼ 5 ×10 12 s −1 . Consequently, the photon
absorption cross section is still determined primarily by the intrinsic breadth of the
resonance Γ. In this context, it is 5×10 −21 cm 2 because the resonance frequency is only
∼ 10 11 s −1 and the collision rate is rather greater.
The number density of water molecules is 6×10 23 /18 = 3×10 22 cm −3 . However, only
a fraction of them are in the ground state of the relevant transition. If its ground state
is J = 5, its population is reduced by perhaps a factor of 2 because the energy to reach
it is ∼ k B T. More importantly, each angular momentum state has a multiplicity 2J +1,
so that there are 35 states at lower or comparable energy. Thus, the fraction of water
molecules in the J = 5 state might be ∼ 1% of the total. The predicted absorption length
is then ∆r ≃ 0.7 cm.
b) The interaction between microwaves and water molecules can also be viewed in terms
of classical physics because there are very many photons in the microwaves. From that
point of view, the interaction between microwaves and molecules is through the classical
torque ⃗ d · ⃗E that tries to swing the dipole into alignment with the electric field. What is
the instantaneous energy advantage to alignment?
The potential energy of a dipole in an external field is ∆E = − ⃗ E·⃗d. We already know
the magnitude of the permanent dipole in water. The magnitude of the electric field can
be estimated from the power of the oven and its electromagnetic “reverberation time”.
We’ll assume that the food in the oven absorbs any wave energy that strikes it (to be
justified below), but also that it occupies only ∼ 10% of the cross sectional area of the
oven. If so, the waves are absorbed after ∼ 10 crossings of the oven. At c, each crossing
takes 1.6×10 −9 s for a typical oven, 1/2 m across. Consequently, the electromagnetic field
energy in the oven at any given time is ∼ 1×10 −5 J, or ∼ 100 erg. Half of this energy is
in electric field, half in magnetic. Given a volume ∼ (1/8) m 3 ∼ 10 5 cm 3 , that’s an energy
density ∼ 10 −3 erg cm −3 . Setting this energy density equal to E 2 /8π implies an rms field
strength E rms ∼ 0.15 dyne/esu. Applied to a dipole the size of water’s (2×10 −18 esu cm),
the alignment energy is
∆ǫ = −3×10 −19 erg.
Note that this is much smaller than the energy of the rotational transition nearest in
frequency.
4

c) Extra Credit: Those of you with more background in statistical mechanics may want to
pursue the consequences of the fact that ∆ǫ ≪ kT for the rate at which the microwaves
heat the water. Hint: How does the collision frequency compare to the wave frequency?
What follows from that comparison?
Asestimated inthe previousproblem, the collisionfrequency is∼ 5×10 12 s −1 , which is
≫ ω. That means a typical molecule undergoes many many collisions with other molecules
during a time when the electric field hardly changes at all due to its oscillation. If the
field were truly constant, the collisions would lead to a thermodynamic equilibrium in
which the small energy advantage of aligning the dipoles with the field leads to a slightly
greater population pointing in that direction. Approximating the distribution function
of dipole directions as an exactly isotropic component plus an exactly aligned and an
exactly anti-aligned component, the relative probability of the aligned and anti-aligned are
e ±∆ǫ/kT . Because ∆ǫ/kT ≪ 1, the exponential can be expanded to first-order in terms
of its argument, and the population contrast in the aligned and anti-aligned directions
is ∼ ±∆ǫ/kT ≃ 7 × 10 −6 . There are slightly more aligned molecules, and they have
slightly lower potential energy; there are slightly fewer anti-aligned molecules, and they
have slightly higher potential energy. The result is that the system as a whole has a
potential energy ∼ 2(∆ǫ) 2 /kT per molecule lower than it would otherwise.
After what, on the collision timescale, is a very long time, the direction of the electric
field changes sign. At this point, the relative roles of aligned and anti-aligned switch. Now
the effect of the electric field is to swing the newly anti-aligned molecules into alignment.
As they rotate, there are many collisions, and the work done by the field is transformed
into heat. Thus, every sign flip in the field (which happens twice per wave period) injects
∼ ∆ǫ) 2 /kT per molecule of heat into the water. The heating rate per molecule is therefore
H ∼ [(∆ǫ) 2 /kT]4πω,
where the extra factor of two comes from two sign flips per wave period.
Note that because H ∝ (∆ǫ) 2 , and ∆ǫ ∝ | ⃗ E|, the heating rate is proportional to the
microwave power, as it should be.
d) What would you predict for the heating rate if a small sample of ice were placed in a
microwave oven?
Zero. Water molecules in ice can’t rotate, so they cannot be put into an excited state
of rotation.
5