Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Symetrisation of products of the metric

  1. Nov 19, 2012 #1
    I need to build a tensor from the product of the metric components, like this (using three factors, not less, not more) :

    [itex]H^{\mu \nu \lambda \kappa \rho \sigma} = g^{\mu \nu} \, g^{\lambda \kappa} \, g^{\rho \sigma} + g^{\mu \lambda} \, g^{\nu \kappa} \, g^{\rho \sigma} + ...[/itex],

    however, that [itex]H^{\mu \nu \lambda \kappa \rho \sigma}[/itex] tensor should be fully symmetric under pairs of indices :

    [itex]H^{\mu \nu \lambda \kappa \rho \sigma} \equiv H^{(\mu \nu) \lambda \kappa \rho \sigma} \equiv H^{\mu \nu (\lambda \kappa) \rho \sigma} \equiv H^{\mu \nu \lambda \kappa (\rho \sigma)}[/itex]

    How can I do that ? Someone know what should be that tensor, explicitely ?

    With only two times the metric, it would be easy :

    [itex]H^{\mu \nu \lambda \kappa} = g^{\mu \nu} \, g^{\lambda \kappa} + g^{\mu \lambda} \, g^{\nu \kappa} + g^{\mu \kappa} \, g^{\nu \lambda}[/itex]

    but I don't know how to do it with three times the metric.
     
  2. jcsd
  3. Nov 20, 2012 #2
    You have to go through all possible index pairs. So you'd get something like
    [tex]H^{\mu \nu \lambda \kappa \rho \sigma}= g^{\mu \nu}H^{\lambda \kappa \rho \sigma} + g^{\mu \lambda} H^{\nu \kappa \rho \sigma} + ... [/tex] where the H with 4 indices is as you calculated and you sum over all possible index pairs containing [itex] \mu [/itex].
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...
Similar Threads for Symetrisation products metric Date
I Defining the components of a metric Apr 16, 2018
I Basis vectors and inner product Apr 14, 2018
B Gravity production Jan 10, 2018
B Minkowski metric, scalar product, why the minus sign? Sep 18, 2017
I Wedge Product May 14, 2017