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Normal Modes of Vibration One dimensional model # 1 : The Monatomic Chain

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Normal Modes of VibrationOne dimensional model # 1:The Monatomic Chain

- Consider a

Monatomic Chain of

Identical Atoms

with nearest-neighbor,

“Hooke’s Law”

type forces (F = - Kx) between the atoms.

- This is equivalent to a force-spring model, with masses m & spring constants K.

To illustrate the procedure for treating the interatomic potential in the harmonic approximation, consider just two neighboring atoms.

Assume that they interact with a known potential V(r). See Figure. Expand V(r)in a Taylor’s series in displacements about the equilibrium separation, keeping only up through quadratic terms in the displacements:

V(r)

Repulsive

0

a

Attractive

r2

r1

This potential is the same as that associated with a spring with spring constant K:

The Monatomic Chain

a

a

a

a

a

a

Un-2

Un-1

Un

Un+1

Un+2

- This is the simplest possible solid.
- Assume that the chain contains a very large number

(N )of atoms with identical massesm. Let the atomic separation be a distance a.

- Assume that the atoms move only in a direction parallel to the chain.
- Assume that only nearest-neighbors interact with each other (the forces are short-ranged).

Consider the simple case of a monatomic linear chain with only nearest-neighbor interactions.

Expand the energy near the equilibrium point for the nth atom. Then, the

Newton’s 2nd Law

equationof motion becomes:

0

a

a

Un-1

Un

Un+1

l

..

This can be seen as follows. The total force on the nth atom is the sum of 2 forces:

The force to the right is:

a

a

The force to the left is:

Un-1

Un

Un+1

Total Force = Force to the right – Force to the left

..

The Equation of Motion of each atom is of this form.Only the value of ‘n’ changes.

Assume that all atoms oscillate with the same amplitude A & the same frequency ω. Assume harmonic solutions for the displacements unof the form:

Undisplaced

Position:

Displaced

Position:

- Put all of this into the equation of motion:

..

- Now, carry out some simplemath manipulation:

Equation of Motion for the nthAtom

Cancel Common Terms & Get:

After more manipulation, this simplifies to

- Mathematical Manipulationfinally gives:

Solution to the Normal

Mode Eigenvalue Problem

for the monatomic chain.

- The maximum allowed frequency is:

- The physical significance of these results is that, for the monatomic chain, the only allowed vibrational frequencies ω must be related to the wavenumber k = (2π/λ) or the wavelength λ in this way.
- This result is often called the “Phonon Dispersion Relation”for the chain, even though these are classical lattice vibrations & there are no(quantum mechanical)phononsin the classical theory.

“Phonon Dispersion Relations”or Normal Mode Frequenciesorω versus k relation for the monatomic chain.

w

A

C

B

k

k

–л

/

a

0

л

/

a

2

л

/

a

Because of BZ periodicity with a period of 2π/a, only the first BZ is needed. Points A, B & C correspond to the same frequency, so they all have the same instantaneous atomic displacements.

Monatomic Chain Dispersion Relation

Some Physics Discussion

- We started from the Newton’s 2nd Law equations of motion for N coupled harmonic oscillators. If one atom starts vibrating, it does not continue with constant amplitude, but transfers energy to the others in a complicated way. That is, the vibrations of individual atoms are not simple harmonic because of this exchange of energy among them.
- On the other hand, our solutions represent the oscillations of

NUNCOUPLED simple harmonic oscillators.

- As we already said, these are called the Normal Modesof the system. They are a collective property of the systemas a whole & not a property of any of the individual atoms. Each mode represented byω(k)oscillates independently of the other modes.Also, it can be shown that the number of modes is the same as the original number of equations N.Proof of this follows.

To establish which wavenumbers are possible for the one-dimensional chain, reason as follows: Not all values are allowed because of periodicity. In particular, thenth atom is equivalent to the (N+n)thatom. This means that the assumed solution for the displacements:

must satisfy the periodic boundary condition:

This, in turn requires that there are an integer number of wavelengths in the chain. So, in the first BZ, there are only N allowed values of k.

The physical significance of wave numbers koutside of the

- First Brillouin Zone [-(π/a) k (π/a)]?
- At the Brillouin Zone edge:
- This k value corresponds tothe maximum frequency. A detailed analysis of the displacements shows that, in that mode, every atom is oscillating π radians out of phase with it’s 2 nearest neighbors. That is, a wave at this value of k is A STANDING WAVE.

Black: k = π/a

or = 2a

Green:

k = (0.85)π/a

or = 2.35 a

x

x

a

un

Points A and Chave the same frequency& thesame atomic displacements. It can be shown that the group velocityvg = (dω/dk)there is negative, so that a wave at that ω& thatkmoves to the left.

The green curve(below) corresponds to point B in theω(k)diagram. It has the same frequency & displacement as points A and C, but vg = (dω/dk)there is positive, so that a wave at that ω& thatkmoves to the right.

w

C

B

A

0

k

k

π/a

2π/a

-π/a

ω(k)(dispersion relation)

Points A & C are symmetrically equivalent; adding a multiple of 2π/a to k does not change either ωor vg, so point A contains no physical information that is different from point B.T The points k= ± π/a have specialsignificance

x

Bragg reflection occurs atk= ± nπ/a

For visualization purposes, it is sometimes useful to visualize a plane of atoms, made up of a large number of parallel chains like the one we just analyzed.

See the next few slides:

Briefly look in more detail at the group velocity,vg.

- The dispersion relation is:
- So, the group velocity is:

vg(dω/dk) = a(K/m)½cos(½ka)

vg= 0 at the BZ edge [k = (π/a)]

- This tells us that a wave with λ corresponding to a zone edge wavenumber k = (π/a)will not propagate.

That is, it must be a standing wave!

- At the BZ edge, the displacements have the form (for site n): Un= Uoeinka = Uo ei(nπ/a) = Uo(-1)n

vg(dω/dk) = a(K/m)½cos(½ka)

- At the 1st BZ Edge,vg = 0
- This means that a wave with λ corresponding to a zone edge wavenumber
- k = (π/a)Will Not Propagate!
- That is, it must be a
- Standing Wave!

Group Velocity, vg in the 1st BZ

1st BZ Edge

Consider aDiatomic Chain of Two Different Atom Types with nearest-neighbor, Hooke’s Law type forces (F = - kx) between the atoms. This is equivalent to a force-spring model with two different types of atoms of masses, M & m connected by identical springs of spring constant K.

(n-2) (n-1) (n) (n+1) (n+2)

K

K

K

K

M

a)

M

M

m

m

a

b)

Un-1

Un

Un+1

Un+2

Un-2

- This is the simplest possible model of a diatomic crystal.
- a is the repeat distance, so, the nearest-neighbor separation is (½)a

This model is complicated due to the presence of 2 different atom types, which, in general, move in opposite directions.

M

M

M

m

m

Un-1

Un

Un+1

Un+2

Un-2

- The GOALis to find the dispersion relation ω(k) for this model.
- There are 2 atom types, with masses M & m, so there will be 2 equations of motion, one for M & one for m.

Equation of Motion

for M

Equation of Motion

for m

-1

-1

As before, assume harmonic (plane wave) solutions for the atomic displacements Un:

M

M

M

m

m

Un-1

Un

Un+1

Un+2

Un-2

Displacement

for M

Displacement

for m

α = complex number which determinesthe relative amplitude and phase of the vibrational wave. Put all of this into thetwoequations of motion.

Carry out some simplemath

manipulation as follows:

Equation of Motion for the nthAtom (M)

Cancel Common Terms

Equation of Motion for the (n-1)thAtom (m)

Cancel Common Terms

(1)

- The Equation for m becomes:

(2)

- (1) & (2) are two coupled, homogeneous, linear algebraic equations in the 2 unknownsα&ωas functions of k.
- More algebra gives:

- Combining (1) & (2) & manipulating:

Cross multiplying & manipulating with (1) & (2):

The 2 roots are:

So, the resulting quadratic equation for ω2 is:

- The two solutions for ω2 are:

- Since the chain contains N unit cells, there will be 2Nnormal modes of vibration, because there are 2Natoms and 2N equations of motion for massesM& m.

Solutions to the Normal Mode Eigenvalue Problem

ω(k)for the Diatomic Chain

w

A

ω+ = “Optic” Modes

B

C

ω- = “Acoustic” Modes

k

–л

/

a

0

л

/

a

2

л

/

a

- There are two solutions forω2 for each wavenumber k. That is, there are
- 2 branches to the “Phonon Dispersion Relation” for each k.
- From an analysis of the displacements, it can be shown that, at point A,
- the two atoms are oscillating 180º out of phase, with their center of
- mass at rest. Also, at point B, the lighter mass m is oscillating & M is at
- rest, while at point C, M is oscillating & m is at rest.

Diatomic Chain Model: Kittel’s Notation!

The solution is:

K

K

Optic Modes

(Optic Branch)

Near the BZ Center (qa << 1)

TheOptic Modebecomes:

(ω+)2 2K(M1 + M2)/(M1M2)

or ω+ constant

TheAcoustic Modebecomes:

(ω-)2 (½) Kq2/(M1 + M2)

or ω- vsq

vs sound velocityin the crystal.

Just like an acoustic wave in air!

Gap

qa

Acoustic Modes

(Acoustic Branch)

K

K

( = 2a)

Near the BZ edge[q = (π/a)]

(Assuming M1 > M2)

TheOptic Modebecomes:

(ω+)2 2K/M2

TheAcoustic Modebecomes:

(ω-)2 2K/M1

Optic Modes

(Optic Branch)

Gap

So,at the BZ edge, the vibrations of

wavelength = 2afor the 2 modes

behaveas ifthere were 2 uncoupled

massesM1 & M2,vibrating

independently with identical springs

of constantK.

qa

Acoustic Modes

(Acoustic Branch)

Again briefly examine limiting solutions at points 0, A,B & C. In the long wavelength region near k = 0(ka«1), sin(ka/2) ≈ ½ka.

w

A

B

C

k

–л

/

a

0

л

/

a

2

л

/

a

A Taylor’s series expansion, using for, small x:

The root with the minus sign gives the minimum value of the acousticbranch:

w

A

B

C

The root with the positive sign gives the maximum value of the optic branch:

Substituting these values of ω into the expression for the relative amplitude

α and using cos(ka/2) ≈1 for ka«1gives the corresponding value of α:

k

–л

/

a

0

л

/

a

2

л

/

a

OR

Substituting into the expression for the relative amplitude α:

This solution corresponds to long-wavelength vibrations near the center of the BZ at k = 0. In that region, M & m oscillate with same amplitude & phase. Also in that region ω = vsk, where vs is the velocity of sound & has the form:

w

A

Optic

B

C

Acoustic

k

a

–π

/

0

π

/

a

2

π

/

a

Similarly, substituting into the relative amplitude gives:

This solution corresponds to pointAin the figure. This value of α shows that, in that mode, the two atoms are oscillating 180º out of phase with their center of mass at rest.

w

A

Optic

B

C

Acoustic

k

–π

/

a

0

π

/

a

2

π

/

a

The other limiting solutions for ω2 are for ka = π.In this case sin(ka/2) =1, so

w

A

B

C

OR

(B)

(C)

k

–л

/

a

0

л

/

a

2

л

/

a

- At point C in the plot, which is the maximum acoustic branch point, M oscillates & m is at rest.
- By contrast, at point B, which is the minimum optic branch point, m oscillates & M is at rest.

Eigenmodes of chain at q = 0

Optical Mode: These atoms, if oppositely charged, would form an oscillating dipole which would couple to optical fields with

Center of the unit cell is not moving!

< a

Normal modes of chain in 2D space

- Constant force model (analog of TBH):bond stretching and bond bending

- Despite the fact that diatomic chain model is one-dimensional, it’s results for the vibrational normal modes ω

contain considerable qualitative physics that carries over to the observed vibrational frequencies for many real materials.

- So, much of the physics containedin the diatomic chain resultscan teach us something aboutthe physics contained in the normal modes of many real materials.
- In particular, ALL MATERIALS with 2 atoms per unit cell

are observed to have two very different kinds of vibrational normal modes.

These are called The Acoustic Branch & The Optic Branch

- This branch received it’s name because it contains long wavelength vibrations of the form ω = vsk, where vs is the velocity of sound. Thus, at long wavelengths, it’s ω vs. k relationship is identical to that for ordinary acoustic(sound)waves in a medium like air.

The Optic Branch

- This branch is always at much higher vibrational frequencies than the acoustic branch. So, in real materials, a probe at optical frequencies is needed to excite these modes.
- Historically, the term “Optic” came from how these modes were discovered. Consider an ionic crystal in which atom 1 has a positive charge & atom 2 has a negative charge.As we’ve seen, in those modes, these atoms are moving in opposite directions. (So, each unit cell contains an oscillating dipole.) These modes can be excited with optical frequency range electromagnetic radiation.

The vibrational amplitude is highly exaggerated!

A Transverse Optic Mode for the Diatomic Chain

The vibrational amplitude is highly exaggerated!

For the case in which

the lattice has some

ionic character, with

+ & - charges alternating:

A Transverse Acoustic Mode for the Diatomic Chain

The vibrational amplitude is highly exaggerated!

For the case in which

the lattice has some

ionic character, with

+ & - charges alternating:

Which has lower energy? Why?

Acoustic Mode

Optic Mode

Lower Energy

Less Compression of Springs

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