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I Christoffel symbols transformation law

  1. Jul 5, 2016 #1
    In Carroll's GR book (pg. 96), the transformation law for Christoffel symbols is derived from the requirement that the covariant derivative be tensorial. I think I understand that, and the derivation Carroll carries out, up until this step (I have a very simple question here, I believe- something stupid I'm not seeing):
    [tex] \Gamma^{\nu'}_{\mu'\lambda'} \frac{\partial x^{\lambda'}}{\partial x^{\lambda}}V^{\lambda} + \frac{\partial x^{\mu}}{\partial x^{\mu'}}V^{\lambda}\frac{\partial}{\partial x^{\mu}}\frac{\partial x^{\nu'}}{\partial x^{\lambda}} = \frac{\partial x^{\mu}}{\partial x^{\mu'}}\frac{\partial x^{\nu'}}{\partial x^{\nu}} \Gamma^{\nu}_{\mu\lambda}V^{\lambda} [/tex]

    Since this must be true for any vector [itex] V^{\lambda} [/itex], that can be eliminated. We can then multiply by [itex] \partial x^{\lambda}/\partial x^{\sigma'} [/itex] on both sides, and relabel [itex] \sigma' \to \lambda' [/itex] to get:
    [tex] \Gamma^{\nu'}_{\mu'\lambda'} = \frac{\partial x^{\mu}}{\partial x^{\mu'}}\frac{\partial x^{\lambda}}{\partial x^{\lambda'}} \frac{\partial x^{\nu'}}{\partial x^{\nu}} \Gamma^{\nu}_{\mu\lambda} + \frac{\partial x^{mu}}{\partial x^{\mu'}}\frac{\partial x^{\lambda}}{\partial x^{\lambda'}}\frac{\partial^2 x^{\nu'}}{\partial x^{\mu} \partial x^{\lambda}} [/tex]

    Now, my question is just this: why the hell is there a plus sign in the last equation, instead of a minus sign? If I follow Carroll's steps directly from the first equation, I get a minus sign! :(
    Sorry if it's obvious- any help is appreciated!
     
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  3. Jul 5, 2016 #2
  4. Jul 5, 2016 #3

    George Jones

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    My copy of Carroll's book has a minus sign (in (3.10) on page 96).
     
  5. Jul 5, 2016 #4
  6. Jul 5, 2016 #5

    George Jones

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    As Mr-R noted, be careful with the indices!

    I have to catch my bus now, but I will try to type in the details after I get home tonight.
     
  7. Jul 5, 2016 #6

    George Jones

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    Here are the quantitative details:

    $$\begin{align}
    \frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\frac{\partial x^{\lambda }}{\partial x^{\lambda ^{\prime }}}\frac{\partial ^{2}x^{\nu ^{\prime }}}{\partial x^{\mu }\partial x^{\lambda }} &= \frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\left( \frac{\partial x^{\lambda }}{\partial x^{\lambda ^{\prime }}}\frac{\partial }{\partial x^{\lambda }}\right) \frac{\partial x^{\nu ^{\prime }}}{\partial x^\mu} \\
    &= \frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\left( \frac{\partial }{\partial x^{\lambda ^{\prime }}}\frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu }}\right) \\
    &= \frac{\partial }{\partial x^{\lambda ^{\prime }}}\left( \frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu }}\right) -\frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu }}\left( \frac{\partial }{\partial x^{\lambda ^{\prime }}}\frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\right) \\
    &= \frac{\partial }{\partial x^{\lambda ^{\prime }}}\left( \frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu ^{\prime }}}\right) -\frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu }}\left( \frac{\partial }{\partial x^{\lambda ^{\prime }}}\frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\right) \\
    &= \frac{\partial }{\partial x^{\lambda ^{\prime }}}\left( \delta _{\mu ^{\prime }}^{\nu ^{\prime }}\right) -\frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu }}\left( \frac{\partial }{\partial x^{\lambda ^{\prime }}}\frac{\partial x^{\mu }}{\partial x^{\mu ^{\prime }}}\right) \\
    &= -\frac{\partial x^{\nu ^{\prime }}}{\partial x^{\mu }}\frac{\partial ^{2}x^{\mu }}{\partial x^{\lambda ^{\prime }}\partial x^{\mu ^{\prime }}} .
    \end{align}$$

    Either the expression at the beginning or the term at the end can be used. Not the differing signs (and indices!).
     
  8. Jul 16, 2016 #7
    Sorry for the late response, I had been procrastinating going over the indices :s
    I'm still not really seeing how that affects the original post; the last term you end on is not present, I believe, in Carroll's derivation at all. (It probably is and I'm just not seeing it- in which case I apologize!)
     
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