GRE Physics Companion: Discrete Oscillator Chains and Dispersion
For a monatomic nearest-neighbor chain,
The traveling normal modes satisfy
Figure 1. A compact oscillator-chain strategy: derive the discrete force balance, insert a
lattice-wave ansatz, reduce wave numbers to the first Brillouin zone, then use the dispersion curve
for frequency and propagation questions.
1 High-value GRE facts
- ω = 2
| sin(ka∕2)|.
- Distinct wave numbers may be restricted to −π∕a ≤ k < π∕a.
- k and k + 2π∕a are equivalent on the lattice.
- ωmax = 2
.
- The zone-boundary mode has neighboring masses in opposite phase.
- In the long-wavelength limit, ω ≈ c|k| with c = a
.
- vp = ω∕|k|.
- vg = dω∕dk.
- For 0 < k < π∕a, vg = c cos(ka∕2).
- A fixed-end chain of N masses has exactly N normal modes.
Part I: Original GRE-style problems
Problem 1: lattice equation
For identical masses with nearest-neighbor springs,
- mün = κ(un+1 + un−1)
- mün = κ(un+1 − 2un + un−1)
- mün = −κun
- mün = κ(un+1 − un−1)
- ün = 0
Problem 2: dispersion relation
The monatomic-chain dispersion relation is
- ω2 = (κ∕m)ka
- ω2 = (4κ∕m) sin 2(ka∕2)
- ω2 = (4κ∕m) cos 2(ka∕2)
- ω = κk∕m
- ω2 = κ∕m for all k
Problem 3: maximum frequency
The maximum branch frequency is
- 2
- π
- κ∕m
- unbounded
Problem 4: Brillouin-zone edge
The positive edge of the first Brillouin zone is
- 1∕a
- π∕(2a)
- π∕a
- 2π∕a
- a∕π
Problem 5: zone-boundary pattern
At
the displacement pattern is proportional to
- 1
- n
- (−1)n
- e−n
- n2
Problem 6: continuum speed
In the long-wavelength limit,
equals
- a
- κa∕m
- aκ∕m2
Problem 7: group velocity
For
the group velocity is
- c
- c sin(ka∕2)
- c cos(ka∕2)
- c tan(ka∕2)
- zero for all k
Problem 8: group velocity at zone boundary
At
is
- c
- c∕2
- 2c
- 0
- infinite
Problem 9: periodic chain wave numbers
For a periodic chain of
sites,
is
- sπ∕(Na)
- 2πs∕(Na)
- s∕a
- 2πs∕a
- independent of N
Problem 10: fixed-end frequencies
For
moving masses between fixed ends,
is
- 2
sin[rπ∕2(N + 1)]
r
- 2
cos[rπ∕2(N + 1)]
for every r
- zero
Problem 11: numerical group velocity
If
then at
the group velocity is closest to
- 0.71 m/s
- 1.00 m/s
- 1.41 m/s
- 2.00 m/s
- 2.83 m/s
Problem 12: shortest distinct wavelength
The shortest distinct wavelength of the monatomic lattice is
- a∕2
- a
- 2a
- 2πa
- arbitrarily small
Part II: Complete worked solutions
Solution 1
The two spring forces give the centered second difference:
Answer: (B).
Solution 2
Substituting
gives
Answer: (B).
Solution 3
The sine magnitude reaches one:
Answer: (B).
Solution 4
One complete nonredundant interval is
The positive edge is
Answer: (C).
Solution 5
At the zone boundary,
Answer: (C).
Solution 6
For
Thus
Answer: (B).
Solution 7
Differentiate
for positive
Then
Answer: (C).
Solution 8
At
Therefore
Answer: (D).
Solution 9
Periodic boundary conditions require
Therefore
Answer: (B).
Solution 10
For fixed ends,
Substitution into the infinite-chain dispersion relation gives
Answer: (A).
Solution 11
First,
Then
Answer: (C).
Solution 12
The zone boundary has
Thus
Answer: (C).
2 GRE checklist
- Start with the centered second-difference force law.
- Insert the lattice-wave phase factor carefully.
- Reduce wave numbers modulo 2π∕a.
- Use the long-wavelength approximation only when |ka|≪ 1.
- Distinguish phase velocity from group velocity.
- At the zone boundary, remember the alternating pattern and zero group velocity.
- Match the wave-number quantization to the actual boundary conditions.
- Count exactly one normal mode per independent scalar displacement degree of freedom.
References
References
[1] PhysicsLibrary, M06-05, Small Oscillations and the Normal-Mode Eigenvalue
Problem.
[2] PhysicsLibrary, M06-04, Coupled Oscillators and Normal Modes.
[3] A. P. French, Vibrations and Waves, W. W. Norton, 1971.
[4] C. Kittel, Introduction to Solid State Physics, 8th ed., Wiley, 2005.