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Partial derivative of ##x^2##

  1. Aug 31, 2015 #1
    How can I figure out ##\partial_\mu x^2## on the manifold ##(M,g)##? I thought that it should be ##2x_\mu##, but I think I'm wrong and the answer is ##2x_\mu+x^\nu x^\lambda \partial_\mu g_{\nu\lambda}##, right?! In particular, it seems to me, we can't write ##\partial_\mu=g_{\mu\nu}\partial^{\nu}##. However, we can raise or lower the indices of the covariant derivative with metric, I mean ##\nabla^\mu=g^{\mu\nu}\nabla_\nu##. Is this true because partial derivative is not a tensor but covariant derivative is? Could you please explain it to me?
     
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  3. Aug 31, 2015 #2

    Orodruin

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    What do you mean by ##x^2##? If ##x^\mu## are coordinates, then ##x^2## is a coordinate dependent statement and you must specify the coordinate system in which this is true in order to make any sense of it. Generally, there is no such thing as a position vector on a manifold.
     
  4. Aug 31, 2015 #3
    Can one consider ##x^2## as the distance between two points ##x^\mu## and ##x^\mu_0=(0,0,\dots)##?
     
  5. Aug 31, 2015 #4

    Orodruin

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    No.
     
  6. Aug 31, 2015 #5
    Sorry, I meant square of distance! ##(x^\mu x_\mu)^{1/2}## is just the Euclidean distance?!
     
  7. Aug 31, 2015 #6

    Orodruin

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    There is no such concept in a general manifold.
     
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