2Problem set #2

2.1. Chapter 4, page 103, problem 2, Continuum wave equation

Equation (2) in the textbook is the equation of motion for a 1D atomic chain (same as we discussed in class)

2.2. Chapter 4, page 103, problem 3, Basis of two unlike atoms

2.3. Chapter 4, page 103, problem 5, Diatomic chain

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2.4. Poisson’s ratio

For a cube with linear size L, the Poisson’s ratio is defined as:

(2.27)Ν » DL' � DL

If we assume that the deformation is uniform, we know (from geometry) that

(2.28)¶ux

¶ x

= 2DL

L

and

¶uy

¶ y

=¶uz

¶ z

= -2DL'

L

(2.29)¶ux

¶ y

=¶ux

¶ z

=

¶uy

¶ x

=

¶uy

¶ z

=¶uz

¶ x

=¶uz

¶ y

= 0

We know that the elastic energy is [Eq. (4.21) in the lecture note of Chapter 4]

(2.30)U = á â r®B B -

2

3G

2Iexx

2+ eyy

2+ ezz

2+ 2 exx eyy + 2 exx ezz + 2 eyy ezzM + GIexx

2+ eyy

2+ ezz

2+ 2 exy

2+ 2 exz

2+ 2 eyz

2MF

where B and G are the bulk and shear moduli.

(a). Prove that the elastic energy is

(2.31)U = 4 HDLL2L B 2 B +

2

3G

DL'

DL

2

- 2 B -2

3G

DL'

DL+

B

2+

2

3G F

Hint: the volume of the cube (Ù â r®

) is L3.

(b). Prove that with fixed DL and L, to minimize the elastic energy, the Poisson’s ratio must be:

(2.34)Ν = DL' � DL =3 B - 2 G

2 H3 B + GL

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2.5. Stability condition for a cubic lattice

For a spring with elastic energy U =1

2C ∆L2, we know that the elastic energy must be positive HU > 0L for any nonzero ∆L. This condition is

known as the stability condition for a spring. The physical meaning of this condition is that it always costs energy to deform a spring. For a

spring, this condition implies that C > 0. (If C < 0, the spring can save some energy by deform itself without any external force, which is

unphysical).

For a solid, we have the same stability condition which says that it always costs energy to deform a solid. What does this condition tells us

about the elastic constants? Here, we explore this question by considering a cubic lattice.

For a cubic lattice, we know that the elastic energy is [Eq.(4.20) in the lecture note of Chapter 4]

(2.41)U = à â r®B 1

2CxxxxIexx

2+ eyy

2+ ezz

2M + CxxyyIexx eyy + exx ezz + eyy ezzM + 2 CxyxyIexy2

+ exz2

+ eyz2MF

This elastic energy can be written in a matrix form

(2.42)U = á â r®B 1

2H exx eyy ezz exy eyz exz L

Cxxxx Cxxyy Cxxyy 0 0 0

Cxxyy Cxxxx Cxxyy 0 0 0

Cxxyy Cxxyy Cxxxx 0 0 0

0 0 0 4 Cxyxy 0 0

0 0 0 0 4 Cxyxy 0

0 0 0 0 0 4 Cxyxy

exx

eyy

ezz

exy

eyz

exz

F

Here we have a 6×6 matrix formed by the three elastic constants and some zeros. Mathematicians have proved for us that the stability condition

is equivalent to requiring this 6×6 matrix to be positive definite (In other words, all the eigenvalues of this 6×6 matrix must be positive).

(a). Find all the six eigenvalues of this 6×6 matrix (Hint: Cxxxx + 2 Cxxyy is one of the eigenvalues).

(b). Find the restriction on the elastic constants by requiring all these eigenvalues being positive

Comment (not a question): there is another (equivalent) definition for positive definite matrices: the leading

principal minors are all positive. This definition is widely used in literature because principal minors is often

easier to compute then eigenvalues.

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