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Transform rule for (1,2) tensor

  1. Feb 8, 2004 #1
    What is the transform rule for a (1,2) tensor? Is it:


    T^a{}_{bc} = \bar T^d{}_{ef}\frac{\partial x^a}{\partial \bar x^d}\frac{\partial \bar x^e}{\partial x^b}\frac{\partial \bar x^f}{\partial x^c}


    or is there an added term like in the transform rule for Christoffel symbols of the second kind?
  2. jcsd
  3. Feb 8, 2004 #2
    this is correct. you only need the Christoffel symbols if you are taking a derivative.
  4. Feb 8, 2004 #3
    Re: Re: Transform rule for (1,2) tensor

    Thank you.

    i know, but that transofrm is the same as the one for the christoffel symbols except the christoffel symbol's transformation rule has the added term:
    \frac{\partial x^i}{\partial x^k}\frac{\partial^2 \bar x^m}{\partial x^j \partial x^k}

    That is what i was referring to.
  5. Feb 9, 2004 #4
    Re: Re: Re: Transform rule for (1,2) tensor

    yeah, the Christoffel symbols have an additional term, because they involve taking a derivative.

    since you are not, then you do not need the extra term, and the equation you have in the first post is correct.
  6. Feb 9, 2004 #5
    tensors always transform always like you wrote [tex]T^a{}_{bc} = \bar T^d{}_{ef}\frac{\partial x^a}{\partial \bar x^d}\frac{\partial \bar x^e}{\partial x^b}\frac{\partial \bar x^f}{\partial x^c}[/tex]
    (by definition)
    When you take a normal derivative from a tensor, you don't become a tensor. This is a problem for making diff equations with tensors. Therefor we define a new derivative (covariant derivative)
    We becomes this by putting a second term (connection coefficients) by ten partial derivative. In general relativity we take a connection coëfficient we have derived from the metric. This is the Christoffel connection(are symbols)
  7. Feb 10, 2004 #6


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    The Christoffel symbols have that added term specifically because the Christoffel symbols are not tensors.
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