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Counting Degrees of Freedom in Tensor

  1. May 4, 2015 #1
    I've been thinking about the number of degrees of freedom in a tensor with n indices in 2-dimensions which is traceless and symmetric. Initially, there are [itex] 2^{n} [/itex] degrees of freedom. The hypothesis of symmetry provides n!-1 number of conditions of the form:

    [tex] T_{i_{1}, \ldots i_{n}}- T_{\sigma i_{1}, \ldots \sigma i_{n}} =0 [/tex]

    since there are n! permutations, including identity. So why doesn't this reduce the number of degrees of freedom by n!-1 ? Are these conditions not all independent?

    I think I'm correct that tracelessness reduces the degrees of freedom by [itex] \binom{n}{2} [/itex] since we choose two indices to contract with [itex] g^{ab} [/itex], but maybe not.
     
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  3. May 5, 2015 #2

    Orodruin

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    If a tensor is symmetric in the first and second index as well as in the first and third, is it symmetric in the first and third?
     
  4. May 5, 2015 #3
    Ah right, I did not think about that. So as far as transpositions go, I need only consider the n-1 permutations that interchange 1,2 then 2,3 then 3,4, so on and so forth. I'm trying to think how this generalizes to other permutations. For permutations that are the identity on all but three of the indices, is it enough to consider those that permute 1,2,3 then 2,3,4 then....so on and so forth. Will I simply end up with (n-1)(n-1) independent conditions when all is said and done? I'm not at all confident that I'm thinking about these higher order permutations correctly.
     
  5. May 5, 2015 #4

    Orodruin

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    Think about the following:
    ##
    A_{ijk} = A_{jik} = A_{kij}
    ##
    Which symmetries did I use? What was the final result?

    Once you are done thinking about the permutations, think about how many identities there really are for the trace.
     
  6. May 5, 2015 #5
    It appears you used symmetry under exchange the first two indices as well as cyclic symmetry. Combined, it appears they imply symmetry of the last two indices. Is that right?
     
  7. May 5, 2015 #6

    Orodruin

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    No. For the second equality, the middle index is fixed. I therefore used symmetry between 12 and between 13 to imply cyclic symmetry (the last component is a cyclic permutation of the original one). What does this tell you about the permutation symmetry group?
     
  8. May 5, 2015 #7
    Does it tell you that the full symmetry group of order 3! is generated by the two transpositions you use?
     
  9. May 5, 2015 #8

    Orodruin

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    Yes. So what is the generalisation of this result to a rank n tensor?
     
  10. May 5, 2015 #9
    Well it seems there are n-1 independent transpositions, and since any permutation can be decomposed from a even or odd product of transpositions, is the symmetry group of a rank n tensor generated by n-1 elements?
     
  11. May 5, 2015 #10

    Orodruin

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    Very much so. You can write any permutation using repetitions of n-1 generators. These may be chosen as the swaps of the ith and (i+1)th indices or as the swaps of the 1st and ith indices.

    What does this tell you about the number of independent relations? What does it imply for the trace conditions?
     
  12. May 5, 2015 #11
    Doesn't it just tell us that there are n-1 independent relations?

    Well, assuming we've properly subtracted away all DOF due to symmetry considerations, then we need only take one more away for tracelessness, it seems? We choose two indices to contract with the metric g and once we do this, we can move these indices anywhere we please thanks to the symmetry.
     
  13. May 5, 2015 #12

    Orodruin

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    How many relations is ##A_{ijk} = A_{jik}##?


    How many relations is ##A_{iij} =0##?
     
  14. May 6, 2015 #13
    Right, I forgot index values. So I guess we scrap the cases where i=j since the relations are trivial in this case. So it seems we only have two if we're working in two dimensions.
    [tex] A_{121}=A_{211} [/tex]
    [tex] A_{122}=A_{212} [/tex]

    So maybe in general, for a fixed transposition, we initially have 2^n conditions but we must subtract off those where the two indices are the same. I believe there are [itex] 2\cdot 2^{n-2} [/itex] of these. So I guess there are [itex] 2^{n}-2^{n-1}-2 [/itex] symmetry conditions in total?


    There appears to be 4 of these. Maybe in general there are [itex] 2 \cdot 2^{n-2} [/itex] of these? Assuming that we've already subtracted away the symmetry degrees of freedom.
     
  15. May 6, 2015 #14

    Orodruin

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    Correct, when ##i = j## the relation is just a tautology. This leaves one possible relation for each set of the remaining indices - in this case two.

    Remember that ##A_{12j} = A_{21j}## gives you exactly the same information as ##A_{21j} = A_{12j}##.

    I suggest selecting a set of generators and then, taking one at a time, figure out how many new relations it actually introduces based on what the previous symmetries you have taken into account already tell you.

    No, this is wrong. Remember that the repeated indices ##i## are not free indices, but summation indices.


    On the other hand, I just thought of a simpler way of counting that will not require you to do anything but use the symmetry properties and the trace property. If you first consider the symmetry property, what does this tell you about all of the components with the same number of indices equal to one?
     
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