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Proving trace of (1,1) Tensor is a scalar

  1. Apr 20, 2009 #1

    trv

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    1. The problem statement, all variables and given/known data
    Given a tensor Mab, Prove that its trace is a scalar.


    2. Relevant equations



    3. The attempt at a solution

    To prove the trace is a scalar, I know I have to prove it doesn't transform under coordinate transformations.

    Now, we can transform M^a_b as follows.
    M'cd=(dx'c/dxa)(dxb/dx'd)Mab

    For the trace Maa, the transformation is as follows. I've used the Kronecker delta for changing Mab to Maa. ( Am I along the right lines in doing so?).
    M'cd=(dx'c/dxa)(dxb/dx'd)Mab(Kronecker deltaab)

    Then I thought using the delta I change dxb to dxa and cancel the two dxa. However that would then the delta away from Mab, leaving it as Mab as before rather than Maa.

    That's the best I can do, please help.
     
    Last edited: Apr 20, 2009
  2. jcsd
  3. Apr 20, 2009 #2

    Dick

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    The way you are doing it leaves your indices way out of balance. You've got three a indices and three b indices on the right side. Multiply both sides by [tex]\delta_c^d[/tex] and use the chain rule for partial derivatives.
     
  4. Apr 21, 2009 #3

    trv

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    Ok, so its,

    To get the transformation equation for the trace, multiply both sides by delta cd[/SUB}

    M'cd(delta cd)=(dxb/dx'd)(dx'c/dxa)Mab(delta cd)

    Thus,
    M'cc=(dxb/dx'c)(dx'c/dxa)Mab

    By chain rule,

    M'cc=(dxb/dxa)Mab

    Now dxb/dxa=delta ab

    Thus,

    M'cc=Maa

    Now, we can change the dummy variable c---> a. (I've seen this being done. Is it allowed only for indices that are being summed over, and thus don't affect other terms?)

    Thus,
    M'aa=Maa
    i.e. Maa is invariant under coordinate transformations. Therefore Maa the trace of Mab is a scalar.

    OK I get it, thanks Dick. Can someone check the dummy variable bit however.
     
    Last edited: Apr 21, 2009
  5. Apr 21, 2009 #4

    Dick

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    Yes, if a variable is summed, it doesn't matter what it's name is. I would put interchange the c and d on your delta though. You've got two c indices up. In the summation convention you usually want one up and one down.
     
  6. Apr 21, 2009 #5

    trv

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    Thanks :)
     
  7. Apr 21, 2009 #6

    trv

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    If I may quickly add another quick question.

    Mab in primed x-coordinates is denoted simply by adding the prime, i.e. M'ab.

    Instead if I was transforming Mab into the y-coordinates. How can I denote that?
     
  8. Apr 21, 2009 #7

    Dick

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    If I understand your question, then you just write the same formula with y instead of x'. But I probably don't. How do you indicate M is a tensor in the y coordinate system?
     
  9. Apr 22, 2009 #8

    trv

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    Hi again Dick. Yeh, you misunderstood my question, but I managed to find out what I wanted.

    Apparently Mab can be written as Mxab for the tensor in the x-coordinate system and Myab for the y-coordinate system.
     
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