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Tensor transforming law problem / metric-delta contraction

  1. Aug 20, 2011 #1
    Hey guys it's me again!
    I'm now asking you 2 question, I'm sure you'll be helpful as usual!!

    1) changing basis- changing tensors

    ok, the law of transformation of a (let's make this simple) 02 tensor is [tex]g_{\mu '\nu '}=\frac{\partial x^\alpha}{\partial {x^{\mu}}'}\frac{\partial x^\beta}{\partial {x^{\nu}}'} g_{\alpha\beta}[/tex]where the prime indices indicate the new coordinate system.

    I found an exercise where you're given the tensor [tex]

    in coordinate [itex](x^1=x , x^2=y)[/itex]
    and you're asked to write it down in the new primed coordinate [itex](x^1=x',x^2=y')[/itex] with[tex]x'=\frac{2x} {y}[/tex] and [tex]y'=\frac{y}{2}[/tex]

    So that
    \frac{\partial x^1}{\partial {x^{1}}'}&=&y'\\
    \frac{\partial x^ 2}{\partial {x^{2}}'}&=&x'\\
    \frac{\partial x^1} {\partial {x^{2}}'}&=&0\\
    \frac{\partial x^2}{\partial {x^{2}}'}&=&2\\

    Now using the formula above i can write, for the first element :[tex]S_{1 ' 1 '}= \frac{\partial x^\alpha}{\partial {x^{1}}'}\frac{\partial x^\beta} {\partial {x^{1}}'}S_{\alpha\beta}=(\frac{\partial x^1}{\partial {x^{1}}'})^2S_{11}+2\frac{\partial x^1}{\partial {x^{1}}'}\frac{\partial x^2}{\partial {x^{1}}'}S_{21}+(\frac{\partial x^2}{\partial {x^{1}}'})^2 S_{22}=(y')^2 +(x')^4(y')^2[/tex]

    which is incorrect, becaouse it should come [itex]S_{1'1'}=(y')^2[/itex]

    And all of the rest came up to be wrong, I get[tex]S_{\mu '\nu '=}\left(
    (y')^2 +(x')^4(y')^2&2(y')^2(x')^3\\2(y')^2(x')^3&4(x'y')^2
    \right)[/tex] whil it should be [tex]S_{\mu '\nu '=}\left(
    (y')^2&x'y'\\x'y'&4(x'y')^2 +(x')^2

    what am I missing???

    2) metric-delta contractions

    This question is about the contraction of the metric.
    I know that [tex]g_{\mu\nu}g^{\mu\lambda}=\delta_{\nu}^{\lambda}[/tex] where [itex]\delta_{\nu}^{\lambda}[/itex] is the Kronecker delta, equal to 1 if [itex]\nu=\lambda[/itex], to 0 otherwise.

    But, does it works even for [itex]\delta_{\nu\lambda}[/itex] ?
    I mean, it's still 0 if [itex]\nu=\lambda[/itex] and to 0 otherwise?

    I am pushed to say no, becaouse if I contrac the delta with the metric I get [tex]g^{\kappa\nu}\delta_{\nu\lambda}=\delta^{\kappa}_{\lambda}[/tex] this tensor certainly isn't the kronecker delta (the identity matrix), but will contain some terms of the metric!!


    Thanks for the attention!!
    Last edited: Aug 20, 2011
  2. jcsd
  3. Aug 20, 2011 #2


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    Science Advisor

    To answer your second question,

    \delta_{\mu\nu} = g_{\mu\lambda}\delta^{\lambda}_{\nu} = g_{\mu\nu}

    So a delta with only lower indices IS the metric, you just give it another name.
  4. Aug 20, 2011 #3
    Thanks a lot haushofer for your answes, to both of my posts!!

    now I wait for the first question...:tongue2:
  5. Aug 20, 2011 #4
    But (excuse me for insisting!) Gravitation (the book) and wikipedia says that the kronecker delta IS [itex]\delta_{\mu\nu}[/itex]!!

    It is becaouse actually in this form is NOT a tensor but a normal generalized function and therefore it makes no sense to contrac it with the metric?

    I mean, the kronecker deta, as a tensor, is only [itex]\delta^{\mu}_\nu[/itex] and contracting with the metric gives the metric itself (like you rightfully say)
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