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Setting the trace of a tensor equal to zero

  1. Jan 7, 2016 #1
    1. The problem statement, all variables and given/known data
    A tensor t has the following components in a given orthonormal basis of R3

    tij(x) = a(x2)xixj + b(x2) \deltaij x2 + c(x2) \epsilonijk xk (1)​
    where the indices i,j,k = 1, 2, 3.
    We use the Einstein summation convention. We will only consider orthogonal transformations on R3; therefore we will not make a distinction between upper and lower indices.
    a,b,c are real functions of x2 := xixi, i.e. a = a(x2) depends on xi through the rotationally-symmetric combination xixi only.
    The trace of the tensor t is defined as:
    Tr(t) = tii (2)​

    1. Calculate the trace (2) and set it to zero, i.e. find a constraint among the functions a,b,c such that Tr(t) = 0 for all x.

    2. Relevant equations
    We only need to use equation (1) and (2)

    3. The attempt at a solution
    Since the indices go from 1 to 3 we have to do t11 + t22 + t33 = 0
    So the index i has to be equal to j. We get:
    a(x2)x1x1 + b(x2)x2 + a(x2)x2x2 + b(x2)x2 + a(x2)x3x3 + b(x2)x2
    The kronecker delta after the b is going to be one in every case so we can leave it out. The epsilon tensor in the part with c will be zero because we won't be able to get 3 different numbers, making the entire part with c 0.
    We can now simplify to :
    x2(a(x1)2 + a(x2)2 + a(x3)2 + 3b(x2)) = 0
    So either x2 = 0
    or (a(x1)2 + a(x2)2 + a(x3)2 + 3b(x2) = 0

    So x = 0 or (a(x1)2 + a(x2)2 + a(x3)2 = -3b(x2)

    Now my question is, is it the case that x1, x2 and x3 will they become just a normal x together?
    such that we will have ax2 = -3bx2
    a = -3b

     
  2. jcsd
  3. Jan 7, 2016 #2

    Orodruin

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    This should already contain the answer to your question.
     
  4. Jan 7, 2016 #3

    samalkhaiat

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    Contract your tensor equation with [itex]\delta_{ij}[/itex] and set it equal to zero.
    [tex]\mbox{Tr}(T) = \delta_{ij}T_{ij} = 0, \ \Rightarrow \ a = - 3b.[/tex]
    So
    [tex]T_{ij} = -3b X_{i}X_{j} + b X^{2} \delta_{ij} + c \ \epsilon_{ijk} X_{k},[/tex]
    with [itex]b[/itex] and [itex]c[/itex] are arbitrary functions of the invariant [itex]X^{2}[/itex].
     
  5. Jan 7, 2016 #4

    Samy_A

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    Never mind, others already reacted while I was composing my post.
     
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