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[General Relativity] Prove that a tensor is a co-tensor

  1. Oct 6, 2014 #1
    Hello! I'd appreciate any help or pokes in the right direction.

    1. The problem statement, all variables and given/known data
    Show that a co-tensor of rank 2, ##T_{\mu\nu}##, is obtained from the tensor of rank 2 ##T^{\mu\nu}## by using a metric to lower the indices:
    $$T_{\mu\nu} = g_{\mu\alpha}g_{\nu\beta}T^{\alpha\beta}$$

    2. Relevant equations
    I know that a vector is a covector if its components transform from one frame to another as:
    $$B_\alpha '= \frac {\partial x^\beta}{\partial x'^\alpha } B_\beta$$

    3. The attempt at a solution
    Analogous to a co-vector, I figure that a tensor of rank 2 is a co-tensor if
    $$T_{\alpha\beta} '= \frac {\partial x^\mu}{\partial x'^\alpha } \frac {\partial x^\nu}{\partial x'^\beta } T_{\mu\nu}$$

    So I'll start on the right-side and try to simplify it to the left side...
    $$
    RS = \frac {\partial x^\mu}{\partial x'^\alpha } \frac {\partial x^\nu}{\partial x'^\beta } T_{\mu\nu}
    = \frac {\partial x^\mu}{\partial x'^\alpha } \frac {\partial x^\nu}{\partial x'^\beta } g_{\mu\delta} g_{\nu\epsilon} T^{\delta\epsilon}
    $$
    Now I'll substitute in the Jacobians...
    $$
    = ({\Lambda^{-1}})^\mu_\alpha ({\Lambda^{-1}})^\nu_\beta g_{\mu\delta} g_{\nu\epsilon} T^{\delta\epsilon}
    $$

    And I'm not sure where to go from here. I suspect I can contract the jacobians with the metric tensors but I'm not sure how to handle that.

    Thanks,
    mef
     
  2. jcsd
  3. Oct 6, 2014 #2

    Matterwave

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    Gold Member

    So I think what the problem is asking you is to show that ##T_{\mu\nu}\equiv g_{\mu\alpha}g_{\nu\beta}T^{\alpha\beta}## defined in this way is in fact a co-vector. Is that your interpretation of the problem as well?

    In that case, by changing a coordinate system, we should have still ##T_{\mu'\nu'} = g_{\mu'\alpha'}g_{\nu'\beta'}T^{\alpha'\beta'}##. What does the right hand side of this equation look like in terms of un-primed indices?
     
  4. Oct 6, 2014 #3
    Ok so then we can express the right-side in terms of the unprimed terms using
    $$g_{\mu'\alpha'}=(\Lambda^{-1})_{\mu}^{\delta}(\Lambda^{-1})_{\alpha}^{\epsilon}g_{\delta\epsilon}$$
    $$g_{\nu'\beta'}=(\Lambda^{-1})_{\nu}^{\eta}(\Lambda^{-1})_{\beta}^{\kappa}g_{\eta\kappa}$$
    And we'll get
    $$
    T_{\mu'\nu'}=g_{\mu'\alpha'}g_{\nu'\beta'}T^{\alpha'\beta'}=(\Lambda^{-1})_{\mu}^{\delta}(\Lambda^{-1})_{\alpha}^{\epsilon}(\Lambda^{-1})_{\nu}^{\eta}(\Lambda^{-1})_{\beta}^{\kappa}g_{\delta\epsilon}g_{\eta\kappa}T^{\alpha'\beta'}=(\Lambda^{-1})_{\mu}^{\delta}(\Lambda^{-1})_{\nu}^{\eta}g_{\delta\epsilon}g_{\eta\kappa}(\Lambda^{-1})_{\alpha}^{\epsilon}(\Lambda^{-1})_{\beta}^{\kappa}T^{\alpha'\beta'}
    $$ $$
    =(\Lambda^{-1})_{\mu}^{\delta}(\Lambda^{-1})_{\nu}^{\eta}g_{\delta\epsilon}g_{\eta\kappa}T^{\epsilon\kappa}=(\Lambda^{-1})_{\mu}^{\delta}(\Lambda^{-1})_{\nu}^{\eta}T_{\delta\eta}
    $$

    So we see that ##T_{\mu'\nu'}## transforms as ##T_{\mu'\nu'}=(\Lambda^{-1})_{\mu}^{\delta}(\Lambda^{-1})_{\nu}^{\eta}T_{\delta\eta}## which means that it's a co tensor!
     
  5. Oct 6, 2014 #4

    Matterwave

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    Some of your indices should still be primed. Each ##(\Lambda^{-1})^\mu_{~~\nu'}## should have a prime on the bottom index (or else they wouldn't sum over with the T indices in the 3rd and 4th parts of the expression, and they wouldn't match the primped indices on the left hand side of the equation),. Also, usually one would be a bit more general and just leave it in terms of ##\frac{\partial x^\mu}{\partial x^{\nu'}}## so that we can consider more than just Lorentz transformations. But other than that, it looks fine to me.
     
  6. Oct 9, 2014 #5
    You're right.
    My prof has a convention where you only write the primes on the object instead of the indices (so ##T_{\mu\nu}'## instead of ##T_{\mu'\nu'}## and he also defined the transformation ##\Lambda## so that ##\Lambda## has a prime on the top and ##\Lambda^{-1}## has a prime on the bottom:
    ##\Lambda_{\nu}^{\mu}=\frac{\partial x'^{\mu}}{\partial x^{\nu}}## and ##(\Lambda^{-1})_{\nu}^{\mu}=\frac{\partial x^{\mu}}{\partial x'^{\nu}}##

    I prefer priming the indices instead because then you don't have to remember that ##\Lambda^{-1}## has primes on the bottom and vice versa
     
    Last edited: Oct 9, 2014
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