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Proof? Kronecker delta is the only isotropic second rank tensor

  1. Jul 29, 2014 #1
    It is pretty straight forward to prove that the Kronecker delta [itex]\delta_{ij}[/itex] is an isotropic tensor, i.e. rotationally invariant.

    But how can I show that it is indeed the only isotropic second order tensor? I.e., such that for any isotropic second order tensor [itex]T_{ij}[/itex] we can write
    [tex]
    T_{ij} = \lambda \delta_{ij}
    [/tex]
     
  2. jcsd
  3. Jul 29, 2014 #2

    Fredrik

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    Not sure what the answer is, or even what the question means. So this may or may not be useful.

    What does "rotationally invariant" mean here? Is it something like ##T_{ij}=R_{ik}T_{kl}R_{lj}##? This can be written as a matrix equation: ##T=RTR^T##. If this is supposed to hold for all R, you can try many different choices of R. Each choice gives you a little more information about the components of T. Perhaps you can find a bunch of rotations that give you enough conditions on T to determine all its components.
     
  4. Jul 29, 2014 #3
    Rotationally invariant in this context means that
    [tex]
    T_{ij} = T'_{ij} = R_{ip} R_{jq} T_{pq}
    [/tex]

    For example, for the Kronecker delta this is pretty straight forward to show:
    [tex]
    \delta_{ij} = R_{ip} R_{jq} \delta_{pq} = R_{ip}R_{jp} = \delta_{iq}\text{,}
    [/tex]
    where we have used that [itex]R_{ij}[/itex] are orthogonal matrices, i.e. [itex]\mathbf{R}\mathbf{R}^T = \mathbf{1}[/itex].
     
  5. Jul 29, 2014 #4

    dextercioby

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    Do you know Schur's lemma ?
     
  6. Jul 29, 2014 #5

    Fredrik

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    Ah, yes, that's what I was thinking, but I put two indices in the wrong order. What I wrote corresponds to the matrix equation ##T=RTR##. What you wrote corresponds to ##T=RTR^T##. My idea should still work, but I suspected that there would be a fancier way, and it looks like dextercioby has given you one.
     
  7. Jul 30, 2014 #6
    It is probably worth noting that the uniqueness only applies to (cartesian) tensors in Euclidean space.
     
    Last edited: Jul 30, 2014
  8. Aug 1, 2014 #7
    If a tensor T is rotationally invariant, that means that for every rotation R, that T = RT.T.R. Note that RT = R-1.

    Since pure rotations form a Lie group, we can use its Lie algebra: R = 1 + ε*L for small ε. Since LT = -L, that gives us commutator [L,T] = 0.

    A rotation-algebra generator has form (Lab)ij = δaiδbj - δajδbi to within some multiplicative factor. Its commutator with T is

    [Lab,T]ij = δaiTbj - δbiTaj - δbjTia + δajTib

    If a,b,i,j are all different, then this expression is zero. But let's try ab = 12 and ij = 13 for definiteness. Then it equals T23. Thus, for number of dimensions >= 3, all off-diagonal T is zero.

    Let's turn to ab = 12 and ij = 11. We get T21 + T12 = 0, or Tij ~ εij, the antisymmetric symbol.

    Turning to ab = 12 and ij = 12, we get T22 - T11 = 0

    We thus find two possible invariants: T ~ δ for all numbers of dimensions and also T ~ ε for two dimensions.

    -

    Alternately, we can contract the commutator on b and j, giving δaiTr(T) - Tai - (n-1)*Tia for n dimensions.

    Adding (ai) and (ia) gives 2*δaiTr(T) - n*(Tai + Tia)

    Thus, the symmetric part of T is proportional to δ.

    Subtracting instead gives (n-2)*(Tia - Tai)

    Thus, the antisymmetric part of T vanishes, except for 2 dimensions, where it is proportional to ε.
     
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