Appendix

 

 

 

 

 

I. Anisotropy of crystalline tensors.

 

The specific resistance is defined by Ohm’s law:

 

                                  (1.1)

 

in which  is the electric field and  is the current density. The components of  and  in the x- and y-direction are E₁ and E₂, j₁ and j₂. For a isotropic matter we have E= ρ.j (fig. 1.1a.).

If the specific resistance in the x-direction and y-direction are different, resp ρ₁₁ and ρ₂₂ we find  as follows. Decompose j in  and . Then

 

                                   (1.2)

 

If the coordinate system is transformed to x’y’ then the components of  in the new axis system become

 

   (1.3)

 

Further (back transformation of new to old):

 

                          (1.4)

 

By substitution of (1.4) into (1.4) we then find:

 

 

 

or

 

             (1.5)

 

whereby

 

            (1.6)

etc.

 

On a arbitrary coordinate set the relation between E and j is given by (1.6). Our original axis systemwas a special one , where ρ₁₂ = ρ₂₁ = 0. Translated in 3 dimensions (1.5) becomes, whereby the accents are omitted:

 

E₁ = ρ₁₁ j₁ + ρ₁₂ j₂ + ρ₁₃ j₃  = 0.

E₂ = ρ₂₁ j₁ + ρ₂₂ j₂ + ρ₂₃ j₃  = 0.                                         (1.7)

E₃ = ρ₃₁ j₁ + ρ₃₂ j₂ + ρ₃₃ j₃  = 0.

 

This set of 9 quantities is called a tensor of the second order  or shorter:

 

E_i = ρ_ikj_k            or                                       (1.8)

 

These tensors are material quantities. AL these tensor are symmetrical: ρ_ik =ρ_ki ,and the tensor quantities are reduced from 9 to 6. In matrix-form:

 

 

For a antiymmetrical 2^nd order tensor a_ij = - a_ji and this becomes

 

 

1.2. Transformation of 2^nd order tensors.

 

When a new coordinate set is used, the components of a vector relative to this new system obtains new values.

If the tensor a_kl gives the relation between the vectors p and q then:

p_k =a_kl q_l                       (1.9)

 

Transformation of p to the new a new axis-system gives:

 

p'_i =c_ik a_kl q_l                  (1.10)

 

Substitution of (1.9) in (1.10) gives:

 

p'_i =c_jk a_kl q_l                  (1.11)

 

Back-transformation of q:

 

q_l =c_jl q’_j                       (1.12)

 

Substitution of (1.12) into (1.13) gives:

 

p'_i =c_ik c_jl a_kl q’_j             (1.13)

 

Further the relation between p and q in the new coordinate system is given by:

 

p'_i = a’_ij q’_j                    (1.14)

 

so that from (1.13) and (1.14) there follows:

 

a'_ij =c_ik c_jl a_jl a_kl             (1.13)

 

whereby there is summed over k and l. So for example

 

a'₁₁ =c_1k c_1l  a_kl  = c²₁₁a₁₁ + c₁₁ c₁₂a₁₂ + c₁₁c₁₃a₁₃ + c₁₂c₁₁a₂₁ + ….

 

1.3. Transformation on main axis.

 

For symmetrical 2^nd order tensors it is always possible to find a axis system on which the diagonal elements of the tensor are unequal to 0. So:

 

(1.16)

 

Because

 

 

This are 2 equations with 9 unknowns. The c_ij’s are not independent, there are 6 relations between the c_ij ‘s, so that the 3 equations are just enough. These are the main axes of the tensors.

 

1.4. Example: Direction-dependence of the specific resistance.

 

Fig.1.2.

 

The measurement of the specific resistance of a wire-like crystal is as follows. (fig.1.2).  A current is sent through the wire which is measured with an ampere-meter. We choose the x-axis in the direction of the wire. Then j =(j₁,0,0).  E in general makes an angle with j, so E = (E₁,E₂,E₃). The voltmeter gives a voltage V, so E₁=V/l, if l is the length of the wire. Now: E_i =ρ_ik j_k and so: L₁=ρ₁₁ j₁ + ρ₁₂ j₂ + ρ₁₃ j₃  . So ρ₁₁ =E₁/j₁. The measured specific resistance this is a diagonal element of the tensor.

 

Consider a crystal of which we know the axes. (Fig.1.3)

 

Fig.1.3.

 

By cutting wires in the crstal || to x-,y-, and z- directions and measuring the specific resistance we find:

 

 

If we cut a wire in the crystal in a arbitrary direction and let it coincide with the x axis of a new coordinate set, then:

 

 

1.5. Inverse tensors.

 

The relation between the field E and the current density j was given by:

 

E_i = ρ_ik j_k.                     (1.8)

 

We can consider this as 3 equations from which 3 unknown j₁, j₂ and j₃ can be solved. The solution has the form:

 

j_l = σ_lm E_m                     (1.18)

 

The elements of the tensor σ_lm (the inverse tensor) can be expressed in ρ_ik. For isotropic matters ρ=1/σ. In crystalsin which a current is sent:

 

 

and from (1.18):

 

 

1.6. Cystal symmetry and tensor components.

 

For a symmetrical 2^nd order tensor transformed on main axes:

 

 

A tetragonal crystal has 3 mutual perpendicular axes, of which 2 are similar. The 3 axes coincide with the main axes of the tensor. The crystal  has a 4-fold axis (the z-axis).At rotation of 90˚ the crystal goes into itself. SO at tranformation of (xyz) to (x’y’z’) : a₁₁’=a₁₁ etc. The cosinus matrix c_ij between the set of axes then is:

 

 

Then:   and  a₁₁’=a₁₁ so that a₁₁=a₂₂. For cubic crystals, which have three 4-fold axes a₁₁ = a₂₂ = a₃₃. For cubic cystals all symmetrial 2^nd order tensores degenerate into scalars. Below is given a number of cosinus matrices, to be used at  symmetry operations.

 

z-axis 2-fold axis

 

z-axis 3-fold axis

 

z-axis 6-fold axis

 

yz-plane mirror plane

 

inversion symmetry relative to origin

 

1.7. Tensors of higher order.

 

The linear relation between a second order tensor p_ij and a vector q_k is given by:

                   (1.20)

 

a_ijk is a 3^rd order tensor with 3³ =27 components. If  p_ij is symmetrical then a_ijk = a_jik and 6x3=18 components remain.

The linear relation between 2 second order tensors p_ij and q_kl is described by:

 

               (1.21)

a_ijkl is a 4^th oder tensor or bitesor and contains 3⁴=81 components. If p_ij and q_kl are symmetrical then: a_ijkl =a_jikl=a_ijlk=a_jilk., and 6x6=36 elemnts remain.

An example is the relation between mechanical tension σ_ij and the deformation tensor e_kl. The reversed formulas (1.20) and (1.21) are:

 

 

in which the b’s are the inverse tensors of the a’s.

The transformations formulas for 3^rd and 4^th order tensors are:

 

            (1.22)

and

 

 

Suppose a crystal has inversion symmetry. The transformation matrix is:

        

 

Condition:                                 (1.23)

Now:                (1.24)

 

The c’s are onlly unequal to 0 if I=p, j=q, k=r; and then they are each –1.  Thus in the summation of (1.24) only 1 term remains:

 

        (1.25)

 

From (1.23) and (1.25) there follows: a_ijk =0. The simpilfication means that all tensor components are zero.

 

Appendix 1: Transformation of tensors.

 

 

The vector p relative to the xyx has the components (p₁,p₂,p₃).

Fig.1.5.

 

Of the new axis systems the x’-axis is drwan is fig.1.5. The angles of the x’-axis with the old x,y and z axes are resp. α₁₁, α₁₂ ,α_13, the cosines of these are c₁₁, c₁₂ ,c₁₃. The unit vector in the x’-direction is e₁. The components of p along the new axes are (p₁’,p₂’,p₃’), of which p₁’ is given.  Now:

e₁ = c₁₁ e₁ +  c₁₂ e₂ + c₁₃e₃, whereby  e₁ , e₂ ,e₃ are the unit vectorsin the old directions. Further:

 

                     (1.26)

 

These are the transformation formulas for the vector p. Equation (1.26) can be written as:

 

 

whereby the summation rule is applied. The sum-sign is left away, since ther is summed over the repeated index (k).

For back transformation of x’y’x’ to the xyz system:

 

p_i = c_ki p’_k 

 

The 9 direction cosines c_ij are not independent from each other. We can see that

c₁₁² +c₁₂² +c₁₃²=1 and the 2 corresponding equations and

c₁₁c₂₁ + c₁₂c₂₂ + c₁₃c₂₃ and the 2 corresponding equations.

The 2^nd set equations follows from the fact that x’y’ and z’ axis are perpendicular to each other.

 

II. Elasiticity.

 

 

Fig.2.1.

 

When a body is deformed, a arbitrary point of that body moves to a new position. The displacement is given by u. Points O and P are displaced to O’and P’ over OO’= u₀ and PP’= u.  Because P lies close to O for u we develop a series:

 

 

whereby for small displacement the terms of higher order are neglected.The relative displacement of P relative to O are given by u’= u-u₀. The components are:

 

   (2.1)

 

The 9 derivatives in (2.1) form a tensor.They describe the linear relation between the realtive displacement u’ and the position vector relative to O (dx,dy,dz). The tensor contains contributions of the deformation and the rotation. These can de separated. As follows. Let a body which contains P (fig.2.2.) rotate around the z-axis over a smalla angle ф_z. P then dispaces over u, of which the components are:

u₁ = |u | sin α = OP ф_z sin α = y ф_z

u₂ = - |u | cos α = - OP ф_z cos α = -x ф_z

u₃ = 0

 

Fig.2.2.

 

so that the displacement around the z-axis can be described as:

 

 

that is with an anti-symmetrical tensor. Expansion gives that the tensor

 

 

gives the displacements at rotation around x,y and z-axis over ф_x, ф_y, ф_z,

To split the tensor eq.(2.1) in contributions for deformation and rotation we haveto slpit it into a symmetrical and anti-symmetrical tensor. The result is:

 

Symmetrical (deformatuion)                                           anti-symmetrical (rotation-tensor)