Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    T_{ij} = \lambda \delta_{ij}
  2. jcsd
  3. Jul 29, 2014 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
    T_{ij} = T'_{ij} = R_{ip} R_{jq} T_{pq}

    For example, for the Kronecker delta this is pretty straight forward to show:
    \delta_{ij} = R_{ip} R_{jq} \delta_{pq} = R_{ip}R_{jp} = \delta_{iq}\text{,}
    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


    User Avatar
    Science Advisor
    Homework Helper

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Proof Kronecker delta Date
I Proof of Stokes' theorem Apr 8, 2017
I Confusion on Bianchi Identity proof Sep 21, 2016
Proof of dimension of the tangent space May 21, 2015
Cyclic Graph Proof? Jan 3, 2015
Jacobian for kronecker delta Aug 16, 2014