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

Can I simply reverse the indices in a contraction?

  1. Jun 4, 2015 #1
    Suppose I have something like
    [tex] \left( \nabla_\mu \nabla_\beta - \nabla_\beta \nabla_\mu \right) V^\mu = R_{\nu \beta} V^\nu [/tex]
    Can since all the terms involving ##\mu## on the left and ##\nu## on the right are contractions, can I simply do:
    [tex]\left( \nabla^\mu \nabla_\beta - \nabla_\beta \nabla^\mu \right) V_{\mu} = R^\nu_{\beta} V_{\nu} [/tex]?
     
  2. jcsd
  3. Jun 4, 2015 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    As long as the connection is metric compatible.
     
  4. Jun 4, 2015 #3
    But my concern is that in the first term on the LHS, the gradient operator is acting on the gradient operator via ##\nabla_\mu (\nabla_\beta) V^\mu##, and not on the vector ##V^\mu##. I thought for contractions, these have to be acting on one another?
     
  5. Jun 4, 2015 #4

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No, this is not how to interpret the LHS. The operators are each acting on ##V^\mu##, one after the other.
     
  6. Jun 4, 2015 #5
    Ok, now I'm confused. I thought the chain rule applies:
    [tex]\nabla^\mu \left( \nabla_\mu A_\nu \right) = \nabla^\mu \left( \nabla_\mu\right) A_\nu + \nabla_\mu \nabla^\mu A_\nu[/tex]
     
  7. Jun 4, 2015 #6
    If I have something like ##F_{\mu \nu} = \nabla_\mu A_\nu - \nabla_\nu A_\mu##, what will ##\nabla^\mu F_{\mu \nu}## look like?
     
  8. Jun 5, 2015 #7

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    What do you mean by ##\nabla_\mu(\nabla^\mu)##? This is an operator applied to an operator. The only way to make sense of this is to apply the right operator to something first and then apply the left to the result.
     
  9. Jun 5, 2015 #8

    haushofer

    User Avatar
    Science Advisor

    You should write out this covariant derivative first in terms of F, and then in that expression write out F in terms of A. If your connection is symmetric, F can be expressed in terms of partial derivatives only.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Can I simply reverse the indices in a contraction?
  1. Can time contract? (Replies: 29)

  2. Contracting Indices (Replies: 5)

Loading...