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Deriving component form of Riemann tensor in general frame

  1. Sep 10, 2015 #1
    How does one derive the general form of the Riemann tensor components when it is defined with respect to the Levi-Civita connection?

    I assumed it was just a "plug-in and play" situation, however I end up with extra terms that don't agree with the form I've looked up in a book. In a general coordinate frame it should be of the form [tex] R_{\kappa\lambda\mu\nu}\equiv g_{\kappa\zeta}R^{\zeta}_{\;\lambda\mu\nu}=\frac{1}{2}\left(\partial_{\mu}\partial_{\lambda}g_{\nu\kappa}-\partial_{\mu}\partial_{\kappa}g_{\nu\lambda}-\partial_{\nu}\partial_{\lambda}g_{\mu\kappa}+\partial_{\nu}\partial_{\kappa}g_{\mu\lambda}\right)+g_{\kappa\zeta}\left(\Gamma^{\zeta}_{\;\mu\eta}\Gamma^{\eta}_{\;\nu\lambda}-\Gamma^{\zeta}_{\;\nu\eta}\Gamma^{\eta}_{\;\mu\lambda}\right)[/tex]
    However, when I naively calculate ##R_{\kappa\lambda\mu\nu}=g_{\kappa\zeta}\left[\partial_{\mu}\Gamma^{\zeta}_{\;\nu\lambda}-\partial_{\nu}\Gamma^{\zeta}_{\;\mu\lambda}+\Gamma^{\zeta}_{\;\mu\eta}\Gamma^{\eta}_{\;\nu\lambda}-\Gamma^{\zeta}_{\;\nu\eta}\Gamma^{\eta}_{\;\mu\lambda}\right]## using that [tex]\Gamma^{\lambda}_{\;\mu\nu}=\bigg\lbrace\matrix{\lambda \\ \mu\nu}\bigg\rbrace=\frac{1}{2}g^{\lambda\eta}\left(\partial_{\mu}g_{\nu\eta}+\partial_{\nu}g_{\eta\mu}-\partial_{\eta}g_{\mu\nu}\right)[/tex] I end up with the following extra contributions [tex]\frac{1}{2}g_{\kappa\zeta}\left(\partial_{\mu}g^{\zeta\eta}\right)\left(\partial_{\nu}g_{\lambda\eta}+\partial_{\lambda}g_{\eta\nu}-\partial_{\eta}g_{\nu\lambda}\right)-\frac{1}{2}g_{\kappa\zeta}\left(\partial_{\nu}g^{\zeta\eta}\right)\left(\partial_{\mu}g_{\lambda\eta}+\partial_{\lambda}g_{\eta\mu}-\partial_{\eta}g_{\mu\lambda}\right)[/tex] which I can't seem to make disappear! What am I doing wrong?
     
    Last edited: Sep 10, 2015
  2. jcsd
  3. Sep 10, 2015 #2

    PeterDonis

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    There's no way to tell what you're doing wrong if you don't show your actual calculations.
     
  4. Sep 10, 2015 #3
    Apologies. Here's what I've done so far:

    [tex]R_{\kappa\lambda\mu\nu}\equiv g_{\kappa\zeta}R^{\zeta}_{\;\lambda\mu\nu}=g_{\kappa\zeta}\left[\partial_{\mu}\left(\frac{1}{2}g^{\zeta\eta}\left(\partial_{\nu}g_{\lambda\eta}+\partial_{\lambda}g_{\eta\nu}-\partial_{\eta}g_{\nu\lambda}\right)\right)-\partial_{\nu}\left(\frac{1}{2}g^{\zeta\eta}\left(\partial_{\mu}g_{\lambda\eta}+\partial_{\lambda}g_{\eta\mu}-\partial_{\eta}g_{\mu\lambda}\right)\right)+\Gamma^{\zeta}_{\;\mu\eta}\Gamma^{\eta}_{\;\nu\lambda}-\Gamma^{\zeta}_{\;\nu\eta}\Gamma^{\eta}_{\;\mu\lambda}\right]\\
    =g_{\kappa\zeta}\left[\frac{1}{2}\left(\partial_{\mu}g^{\zeta\eta}\right)\left(\partial_{\nu}g_{\lambda\eta}+\partial_{\lambda}g_{\eta\nu}-\partial_{\eta}g_{\nu\lambda}\right)-\frac{1}{2}\left(\partial_{\nu}g^{\zeta\eta}\right)\left(\partial_{\mu}g_{\lambda\eta}+\partial_{\lambda}g_{\eta\mu}-\partial_{\eta}g_{\mu\lambda}\right)+\frac{1}{2}g^{\zeta\eta}\left(\partial_{\mu}\partial_{\nu}g_{\lambda\eta}+\partial_{\mu}\partial_{\lambda}g_{\eta\nu}-\partial_{\mu}\partial_{\eta}g_{\nu\lambda}\right)-\frac{1}{2}g^{\zeta\eta}\left(\partial_{\nu}\partial_{\mu}g_{\lambda\eta}+\partial_{\nu}\partial_{\lambda}g_{\eta\mu}-\partial_{\nu}\partial_{\eta}g_{\mu\lambda}\right)+\Gamma^{\zeta}_{\;\mu\eta}\Gamma^{\eta}_{\;\nu\lambda}-\Gamma^{\zeta}_{\;\nu\eta}\Gamma^{\eta}_{\;\mu\lambda}\right][/tex]

    So upon a bit of algebra I end up with what I put in my first post, i.e. the correct expression plus the two extra contributions.
     
  5. Sep 10, 2015 #4

    PeterDonis

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    Also, which book?
     
  6. Sep 10, 2015 #5
    Nakahara's book: "Geometry, topology & physics", in the chapter on Riemannian geometry (chapter 7), section on the Levi-Civita connection.
     
  7. Sep 11, 2015 #6
    To elaborate on this (sorry I was writing the last post on my phone and lost the will to try and LaTeX on such a small screen!), I end up with

    [tex]R_{\kappa\lambda\mu\nu}=\frac{1}{2}\left(\partial_{\mu}\partial_{\lambda}g_{\kappa\nu}-\partial_{\mu}\partial_{\kappa}g_{\nu\lambda}-\partial_{\nu}\partial_{\lambda}g_{\kappa\mu}+\partial_{\nu}\partial_{\kappa}g_{\mu\lambda}\right)+g_{\kappa\zeta}\left(\Gamma^{\zeta}_{\;\mu\eta}\Gamma^{\eta}_{\;\nu\lambda}-\Gamma^{\zeta}_{\;\nu\eta}\Gamma^{\eta}_{\;\mu\lambda}\right)+\frac{1}{2}g_{\kappa\zeta}\left[\left(\partial_{\mu}g^{\zeta\eta}\right)\left(\partial_{\nu}g_{\lambda\eta}+\partial_{\lambda}g_{\eta\nu}-\partial_{\eta}g_{\nu\lambda}\right)-\left(\partial_{\nu}g^{\zeta\eta}\right)\left(\partial_{\mu}g_{\lambda\eta}+\partial_{\lambda}g_{\eta\mu}-\partial_{\eta}g_{\mu\lambda}\right)\right][/tex]

    And the exact reference is: "Geometry, topology & physics", Nakahara. Chapter 7, Section 7.2, Sub-section 7.4.5, page 268.
     
    Last edited: Sep 11, 2015
  8. Sep 11, 2015 #7

    samalkhaiat

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    Here is a quick way to do it. Write
    [tex]R_{\rho\sigma\mu\nu} = \left\{ \partial_{\mu}\left(g_{\rho\tau}\Gamma^{\tau}_{\nu\sigma}\right) - \Gamma^{\tau}_{\nu\sigma} \partial_{\mu}g_{\rho\tau} + g_{\rho \alpha} \Gamma^{\alpha}_{\mu\tau} \Gamma^{\tau}_{\nu\sigma} \right\} - \left\{\mu \leftrightarrow \nu \right\} ,[/tex]
    or
    [tex]R_{\rho\sigma\mu\nu} = \left\{ \partial_{\mu}\left(g_{\rho\tau}\Gamma^{\tau}_{\nu\sigma}\right) - \left( \partial_{\mu}g_{\rho\tau} - g_{\rho\alpha} \Gamma^{\alpha}_{\mu\tau} \right) \Gamma^{\tau}_{\nu\sigma} \right\} - \left\{\mu \leftrightarrow \nu \right\} .[/tex] Now, if you use the identity [tex]\partial_{\mu}g_{\rho\tau} - g_{\rho\alpha} \Gamma^{\alpha}_{\mu\tau} = g_{\alpha\tau} \Gamma^{\alpha}_{\rho\mu} ,[/tex] you get what you need to arrive at the final result
    [tex]R_{\rho\sigma\mu\nu} = \left\{ \partial_{\mu}\left( g_{\rho\tau}\Gamma^{\tau}_{\nu\sigma} \right) - g_{\alpha\tau} \Gamma^{\alpha}_{\rho\mu} \Gamma^{\tau}_{\nu\sigma}\right\} - \left\{\mu \leftrightarrow \nu \right\} .[/tex]
     
  9. Sep 12, 2015 #8
    Cool, thanks for your help. I'm guessing I could rectify my attempt using the identity you specified in your post (although it would be a lot more work)?!
     
  10. Sep 15, 2015 #9

    samalkhaiat

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    Yes, differentation alone does not get you any where. You need to use [itex]\nabla_{\rho}g_{\mu\nu}=0[/itex]. However, doing it your way generates 12 terms of the form [itex](g\partial g)(g \Gamma)[/itex]! Why make life difficult for yourself.
     
  11. Sep 15, 2015 #10
    Yes, you're right. I ended up heeding your advice and doing it the way you suggested - much more straight forward!
     
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