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8.2 Examples of more complicated laws Ohm’s law in an anisotropic medium Now we have defined a second rank tensor, the natural question to ask is whether δ is a tensor or not. We have defined δij to be 1 if i = j and 0 otherwise in all bases and thus we have that δij′ = δij . But using δij = lik ljk this gives The simple form of Ohm’s Law stated above, in which an applied electric field E produces a current in the same direction, only holds for conducting media which are isotropic, that is, the same in all directions. This is certainly not the case in crystalline media, where the regular lattice will favour conduction in some directions more than in others. δij′ = liα lβ δαβ This shows that δ transforms as a second rank tensor, we need no longer speak of the δ–symbol but the δ–tensor. Rotating rigid body The most general relation between J and E which is linear and is such that J vanishes when E vanishes is of the form ω _ Ji = σij Ej where σij are the components of the conductivity tensor in the chosen basis, and characterise the conduction properties when J and E are measured in that basis. Thus we need nine v _ numbers, σij , to characterise the conductivity of an anisotropic medium. The conductivity tensor is an example of a second rank tensor. For suppose we consider an orthogonal transformation of basis. Simply changing basis cannot alter the form of the physical law and so we conclude that ′ Ji′ = σij Ej′ where Ji′ = lij Jj and Ej′ = ljk Ek θ _r O Now consider the angular momentum of the particle about the origin O, defined by L = r × p = r × (mv) where m is the mass of the particle. Using the above expression for v we obtain Thus we deduce that L = mr × (ω × r) = m ω(r · r) − r(r · ω) ′ lij Jj = lij σjk Ek = σij ljk Ek which we can rewrite as ′ (σij ljk − lij σjk )Ek = 0 This must be true for arbitrary electric fields and hence ′ σij ljk = lij σjk Multiplying both sides by llk , noting that llk ljk = δlj and using the sifting property we find ′ that σil = lij λlk σjk or re-labelling Consider a particle of mass m at a point r in a rigid body rotating with angular velocity ω. Recall previously that v = ω × r. You were asked to check that this gives the right direction for v; that it is perpendicular to the plane of ω and r; that the magnitude |v| = ωr sin θ = ω× radius of circle in which the point is travelling (24) where we have used the identity for the vector triple product. Note that only if r is perpendicular to ω do we obtain L = mr 2 ω, which means that only then are L and ω in the same direction. Taking components of equation (24) in an orthonormal basis {ei }, we find that Li = m ωi (r · r) − xi (r · ω) h i = m r 2 ωi − xi xj ωj noting that r · ω = xj ωj h i = m r 2 δij − xi xj ωj using ωi = δij ωj Thus σij = liα ljβ σαβ ′ This exemplifies how the components of a second rank tensor change under an orthogonal transformation and indeed will be taken as our definition of a second rank tensor. The above discussion of the covariance of Ohm’s law is then an example of the quotient theorem. Li = Iij (O) ωj h i Iij (O) = m r 2 δij − xi xj where By the quotient theorem Iij (O) are the components of the inertia tensor, relative to O, in the ei basis. The inertia tensor is another example of a second rank tensor. This may also be shown directly as Iij′ (O) = m[r ′2 δij − x′i x′j ] Interlude: δij as a tensor = m[r 2 liα ljβ δαβ − liα xα ljβ xβ ] = liα ljβ Iαβ (O) 33 34 so I transforms as a second rank tensor. Note also that if a is a pseudovector then a × b is a vector as (det L)2 = 1. Summary of why we need tensors We can also define tensors of rank greater than two by introducing more indices, together with more ls. Indeed ǫ is actually a pseudotensor of rank three. (i) Physical laws often relate two vectors. There are some additonal special properties for rank two tensors only, as the set of nine (ii) A second rank tensor provides a linear relation between two vectors which may be in different directions. (iii) Tensors allow the generalisation of isotropic laws (‘physics the same in all directions’) to anisotropic laws (‘physics different in different directions’) In general a second rank tensor maps a given vector onto a vector in a different direction: if a vector n has components ni then numbers, Tij , representing the tensor T , can T11 T = T21 T31 We can rewrite the generic transformation law for a second rank tensor as follows: ′ Tij = liα ljβ Tαβ = (L)iα Tαβ (LT )βj Thus in matrix form the transformation law is However some special vectors called eigenvectors may exist such that mi = λ ni i.e. the new vector is in the same direction as the original vector. Eigenvectors usually have special physical significance (see later). 8.3 (i) It is wrong to say that a second rank tensor is a matrix; rather the tensor is the funda- Scalars and vectors are called tensors of rank zero and one respectively, where rank = no. of indices in a Cartesian basis. Thus we have φ′ = φ scalar a′i = liα aα φ = liα ljβ Tαβ rank two tensor = det L φ pseudoscalar a′i = det L liα aα Tij′ ′ ′ = det L liα ljβ Tαβ ′ (a , b , c ) ≡ a · (b × c) ′ = (ii) It is wrong to say a matrix is a tensor e.g. the transformation matrix L is not a tensor but nine numbers defining the transformation between two different bases. 8.4 Invariants Trace of a tensor: The trace of a matrix A (with elements aij ) is defined as the sum of its diagonal elements, so pseudosvector tr A = aii rank two pseudotensor An example of a pseudovector is a × b, see section 7.3, and an example of a pseudoscalar is the scalar triple product, for ′ mental object and is represented in a given basis by a matrix. vector We can also have pseudoscalars, pseudovectors and pseudotensors, ′ T ′ = LT LT Notes General properties Tij′ T12 T13 T22 T23 T32 T33 This of course is not true for higher rank tensors (which have more than 9 components). Tij nj = mi , where mi are components of m, the vector that n is mapped onto. be written as a 3 × 3 array a′i (b × c)′i This has the useful cyclic property tr AB = aij bji = bji aij = tr BA Consider the trace of the the tensor in the transformed basis = liα aα det L liβ (b × c)β = det L aα (b × c)β δαβ = det L a · (b × c) ≡ det L (a, b, c) 35 ′ tr T ′ = Tii = liα liβ Tαβ = (LT L)αβ Tαβ = Tαα = tr T 36