Dilemma of Classical Physics
1
1.1
Energy quantization
Case 1: Classical mechanics:
collection of particles with mass mi, momentum p~i
H=
X
i
p2i
X
X
+ V (~ri) + V (~ri − ~rj ) + ...
2mi
i
i6=j
(1)
allowed energies E continuous, e.g. pi can have any value
Case 2: Quantum mechanics:
object has possible discrete set of energy values: The Ei depend on
nature of object, determined by Schrödinger’s equation.
Early indications of quantized energy spectra:
• Blackbody spectrum (Planck 1900)
• Photoelectric effect (Einstein 1905)
1
• Heat capacities of solids (Einstein 1907)
1.2
Boltzmann distribution
Object loosely coupled to heat reservoir (massive, spectrum of energy
levels dense). Thermal flucts.=⇒ E of object takes on different (allowed)
values, close to mean Ē
P (Ei) ≡ probability object is in ith level
Assume P (E) function only of E and T
Now couple two objects weakly:
Energies of object are
Eij = Ei(1) + Ej (2) ∀ {i, j}
Probability of finding composite object with energy Eij is
P (E) = P (E1)P (E2), E = E1 + E2
2
(2)
(probabilities multiply!) Now differentiate wrt E1, divide by P (E1),
same for E2. Find
1 dP (E1)
1 dP (E2)
1 dP
=
=
≡ −β
P dE P (E1) dE1
P (E2) dE2
(3)
So β indep. of E =⇒
P ∝ exp(−βE)
(4)
Now define β ≡ 1/kT , with k =Boltzman’s const.= 1.38 × 10−16 erg/K
Const. of prop. fixed by conservation of probability:
X
allj
P (Ej ) = 1
(5)
So normalized distr. is
Pi = exp(−Ei/kT )/
1.3
X
j
exp(−Ej /kT )
(6)
Harmonic oscillator at temperature T
Classical oscillation frequency is ν. Einstein assumed à la Planck energy
spectrum
3
E = nhν
n = 0, 1, 2, 3...
h = Planck0s const. = 6.62 × 10−27
(7)
(8)
Assume the SHO weakly coupled to heat bath at temperature T . Mean
energy is
Ē =
X
n
P∞
PnEn =
−nhν/kT
n=0 (nhν)e
P∞
−nhν/kT
n=0 e
Now evaluate sum. Let α = hν/kT ; “partition function” Z =
Use
−
∞
dZ
X
=
ne−αn,
dα n=0
Ē = −hν
so
d
1 dZ
= −hν
log Z
Z dα
dα
(9)
P∞
n=0 e
−αn
.
(10)
(11)
Z may be written
Z = 1 + e−α + e−2α + ...
= 1 + e−α (1 + e−α + e−2α + ...)
= 1 + e−α Z = 1/(1 − e−α )
(12)
So
d
e−α
−α
Ē = hν
log(1 − e ) = hν
dα
(1 − e−α )
hν
= hν/kT
e
−1
4
(13)
the mean energy of an ensemble of SHO’s with oscillation freq. quantized,
units of ν (assumption!).
1.4
Classical limit & equipartition
When energy spacing ν small, must find classical result!
hν ¿ kT =⇒ α ¿ 1, e−α − 1 ' α
(14)
Ē ' kT
(15)
so
Special case of equipartition theorem of classical stat. phys.: for every
quadratic term in the Hamiltonian get mean energy 12 kT . Here 2 such
√
p2
terms: H = 2m
+ 12 kx2, Ē → kT .
1.5
Einstein & Debye Models of Specific Heat of
Solids
Dilemma faced by late 19th-cent. physicists: Classical theory of specific
heat of gases applied to solids says
1
CV = (dE/dT ) = mN k (const!),
2
(16)
with
N = no. atoms, m=no. degs. freedom/atom
(Dulong-Petit “law”)
Experiments (Nernst) found:
i.e, C → 0 roughly as T 3 as T → 0, approaches 3N kB at higher T
(although not nec. asymptotically high T , see below). Note we get the
3N k result from equipartition with m = 6 for 3 Cartesian directions times
5
2 for both potential and kinetic energy. (Ignored rotational & vibrational
degrees of freedom so far). But what about low-T behavior? It is definitely
not constant in T , so no way Dulong-Petit law holds.
Einstein (1907): If “matter oscillator” had Planck spectrum with
allowed energies nν, low-T sp. ht. went to zero as T → 0
kB T ¿ hν =⇒ E ∼ hνe−hν/kB T , C ∼ (hν/kB T )2e−hν/kB T
(17)
Now Einstein sp. ht. result C ∼ e−hν/kB T , kB T ¿ hν, which qualitatively explains the deviation from the classical result, but in fact the
expt. shows C ∼ T 3. Debye: SHO’s not atoms but groups of atoms
vibrating collectively: “phonons”. A complicated solid can have many
“normal” modes of vibration and, if the solid is infinite, long-wavelength
normal mode frequencies can go to zero (sound!). Debye said just think
of standing sound waves: a sound wave is a time-dependent displacement
of ions, e.g. a solution to the wave equation for ionic displacement u such
that
u(x, y, z, t) = A cos ωt sin kxx sin ky y sin kz z
(18)
where ω is 2π times the frequency of the sound wave, and k = 2π/λ is
the wave vector. We would like the displacement to vanish on the surface
of the solid of size L so that we are describing a standing wave. This can
be done if we allow only discrete values of the components of ~k, e.g.
πnx
kx =
, nx = 1, 2, 3...,
(19)
L
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and the same for ky and kz . Check that this means that the wave vanishes
on the boundary always. This means that the standing wave frequencies
ω = 2πν = csk,
(20)
where cs is the speed of sound, are quantized if we only allow ki to take
on discrete values; in fact
πcs r 2
ωn = cs|(kx, ky , kz )| =
nx + n2y + n2z .
(21)
L
Now we go back to our expression Eq. 13 which is the mean energy for
an oscillator at one frequency, and generalize it to a system with a lot of
different oscillators with different frequencies:
h̄ωn
X
,
(22)
E=
h̄ω /k T
nα ≥0 e n B − 1
where α = x, y, z, h̄ = h/2π, & I’ve renamed Boltzmann’s const. kB to
avoid confusion with the wavevector k.
This is a discrete sum, but as L → ∞, the ωn get so close together
that we can approximate it as an integral. The difference between neighboring quantized kx’s is for example ∆kx = Lπ , so that the sum can be
approximated as
LX
LZ∞
∆kα ∼
dkα .
(23)
nα
π
π 0
We need to sum over all three Cartesian coordinates, so the mean energy
is1
V Z 3
h̄ωn
E =
d
k
(24)
(2π)3
eh̄ωn/kB T − 1
h̄csk
V Z∞
2
(4π)k
dk
(25)
=
(2π)3 0
eh̄csk/kB T − 1
X
=
Debye used a typical theoretical physics trick to solve this integral, introducing a dimensionless variable y = h̄ω/kB T , and scaling out the
1
See if you can figure out where the (2π)3 came from!
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physical quantities with dimensions. The result is
V (kB T )4 Z ∞
y3
.
E =
dy y
(2π)3(h̄cs)3 0
e −1
(26)
Now, even if we couldn’t solve the integral, we would have all the important
information, since we can tell it is a dimensionless number of order unity2;
the important physical quantites are all contained in the prefactor, and
we see, e.g. that E ∼ T 4. As it turns out, Maple can do the integral,
R∞ 3
y
4
0 y /(e − 1) = π /15, so the final answer is
U
π 2 (kB T )4
=
,
V
10 (h̄cs)3
(27)
dE π 2 kB4
3
CV =
=
T
dT
5 (h̄cs)3
(28)
so that the specific heat is
which works very well to explain the experimentally observed specific heat
of normal metals.
1.6
Specific heat of molecular hydrogen
Here’s another example which should have immediately provided a clue
that classical physics wasn’t working. An H2 molecule is basically a pair
of balls connected by a spring which can rotate and translate freely:
2
do a quick plot of the integrand to convince yourself of this
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Equipartition:
1
Ē = N kB T {
2
+
+
+
3 (kinetic energy of translation 3D)
2 (rotation of “dumbbell” around 2 coordinate axes)
2 (vibration along x axis (K.E.+P.E.! ))
1 (rotation about x axis (extended atoms!))...} (29)
The measured specific heat of molecular H gas looks like
There is a hierarchy of regimes where thermal energy kB T becomes
large enough to excite discrete energy levels corresponding to diff. types
of motions. Energies of rot., vibr., etc. are above the ground state of
the molecule because they correspond to “excitations”. As you raise the
temperature, you first excite the low-lying translational states of the gas,
giving you roughly the classical result from equipartition for a gas of particles with no internal degrees of freedom. As you raise the temperature,
eventually you may populate the higher states corresponding to rotation
and vibration. For example, the energy of 1st rot. excited level exceeds
lowest (“ground”) state by
δErot ' kB Trot/h,
Trot ∼ 200K
(30)
Later on we’ll find that angular momentum is quantized in units of h̄,
and that this energy corresponds roughly to h̄2/2I, where I is the moment
of inertia of the Hydrogen “dumbbell”.
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1.7
Blackbody radiation–historical
Blackbody = by def. object in equil. at temp. T emits radiation from
surface only (no reflectance). At T = 0 looks black. Model: cavity with
small hole, rad. density uν .
energy in vol. elt. δV
} δE = uν δνδV
at freqs. betw. ν and ν + δν
Known by 1900:
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(31)
1. uν is universal function of ν and T . If it depended,e.g. on materials
properties, could use bandbass filter to get heat to flow spontaneously
from one cavity to another at temp T , violate 2nd law.
2. Stephan-Boltzman law:
net energy density u =
Z ∞
a = 7.56 × 10
uν dν = aT 4,
0
−15
−3 −4
(32)
erg cm K
3. Rayleigh-Jeans law
uν = 8πkB T ν 2/c3 valid for small ν
(33)
Rayleigh & Jeans derived from classical stat mech, which turned out to
be wrong for ν → ∞. Blackbody spectrum was found experimentally
pre-1900 to be quite different from the classical prediction.
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