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Difficulty in learning tensors

  1. Aug 23, 2011 #1
    Let r[itex]_{\mu}[/itex] be a tensor in coordinates x[itex]^{c}[/itex] and R[itex]_{b}[/itex] be a tensor in coordinates X[itex]^{c}[/itex].
    Then let r[itex]_{\mu}[/itex] = 0.

    Then {[itex]\partial[/itex]X[itex]^{\nu}[/itex]/[itex]\partial[/itex]x[itex]^{\mu}[/itex]}R[itex]_{\nu}[/itex] = 0.
    I read in a book that one can divide both sides of the last equation by the partial derivative to get R[itex]_{\nu}[/itex] = 0.
    I do not understand how this can be done since the partial derivative is summed over together with the R[itex]_{\nu}[/itex].
    Can somebody help me!
  2. jcsd
  3. Aug 23, 2011 #2
    The way I understand to do it is the following:

    {[itex]\partial[/itex]X[itex]^{\nu}[/itex]/[itex]\partial[/itex]x[itex]^{\mu}[/itex]}R[itex]_{\nu}[/itex] = 0

    Multiplying both sides by {[itex]\partial[/itex]x[itex]^{\mu}[/itex]/[itex]\partial[/itex]X[itex]^{\lambda}[/itex]} we get

    {[itex]\partial[/itex]x[itex]^{\mu}[/itex]/[itex]\partial[/itex]X[itex]^{\lambda}[/itex]} {[itex]\partial[/itex]X[itex]^{\nu}[/itex]/[itex]\partial[/itex]x[itex]^{\mu}[/itex]}R[itex]_{\nu}[/itex] = 0

    i.e. {[itex]\partial[/itex]X[itex]^{\nu}[/itex]/[itex]\partial[/itex]X[itex]^{\lambda}[/itex]}R[itex]_{\nu}[/itex] = 0

    [itex]\delta[/itex][itex]^{\nu}_{\lambda}[/itex]R[itex]_{\nu}[/itex] = 0

    R[itex]_{\lambda}[/itex] = 0.

    Is the above method too long?

    Thanks for any help.
  4. Aug 23, 2011 #3


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    It's a theorem in linear algebra. If A is non-singular, and Ab=0, then b=0. If dX/dx is singular, you have other problems.
  5. Aug 23, 2011 #4
    Thanks K^2.

    Is the method I used ok?
  6. Aug 23, 2011 #5


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    Yes it is. Note that what you're doing on the line that starts with "i.e." is to use the chain rule.
  7. Aug 23, 2011 #6
    Is the method above, using Kronecker's delta, considered a long method?
  8. Aug 23, 2011 #7


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    Nothing that only covers a few short lines is ever considered a long method. :smile:

    The only way to do it shorter is to note that you started with a matrix equation in component form, and if you multiply it with the inverse...but that might require some explanation, and it would look a lot like what you just said.
  9. Aug 23, 2011 #8
    Thanks a lot, Fredrik
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