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Covariant derivative and general relativity

  1. Apr 9, 2007 #1

    cristo

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    I'm not really sure where to put this, so I thought it post it here!

    I'm reading through my GR lecture notes, and have come across a comment that has confused me. I quote
    Now, I don't really see how this is true. For example, consider a scalar field f. The covariant derivative of this is
    [tex]\nabla_af\equiv\partial_af=\frac{\partial f}{\partial x^a}[/tex]

    But, aren't [itex]\partial_af[/itex] the components of a covector?
     
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  3. Apr 9, 2007 #2

    Dick

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    I can only assume that they mean an isolated component of a covariant derivative eg [tex]\nabla_0f[/tex] is a scalar if f is scalar, vector if f is vector etc. Which would not be true for a partial derivative.
     
  4. Apr 9, 2007 #3

    cristo

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    Thanks. So, I would be correct in saying that the covariant derivative of a type (r,s) tensor field is a type (r,s+1) tensor field, right?
     
  5. Apr 9, 2007 #4

    Dick

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    You betcha! And each covariant component of the (r,s+1) field is an (r,s) field - as is always the case.
     
    Last edited: Apr 9, 2007
  6. Apr 9, 2007 #5

    cristo

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    Cheers, Dick!
     
  7. Apr 9, 2007 #6

    robphy

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  8. Apr 9, 2007 #7

    cristo

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    He defines the covariant derivative by setting up the situation of two vector fields u and v ona manifold. Then, picking some point P, u determines a flow line passing through P; parametrised by some path parameter s, such that [tex]\bold{u}_P=\left.\frac{d}{ds}\right|_{s=s_P}[/tex]
    Suppose Q[itex]\neq[/itex]P is another point on the flowline corresponding to [itex]s_Q\neq s_P[/itex]. Then we write
    [tex]\bold{v}(s_Q)\equiv\bold{v}_Q, \hspace{2cm} \tilde{\bold{v}}(s_Q)\equiv \tilde{\bold{v}}_{QP}[/tex] [he actually uses a symbol with two parallel lines over v, like one side of a "norm" sign-- anyway, it means the vector field parallely transported from P, evaluated at Q]
    and define the covariant derivative as
    [tex]\nabla_{\bold{u}}\bold{v}_P=\lim_{s_Q\rightarrow s_P}\frac{\tilde{\bold{v}}(s_Q)-\bold{v}(s_Q)}{s_Q-s_P}[/tex]
     
  9. Apr 9, 2007 #8

    robphy

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    in that case, you are evaluating

    [tex] u^a \nabla_a v^b [/tex], which is a vector.
    Thus, with the above definition,
     
  10. Apr 9, 2007 #9

    cristo

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    Ok, so would I be correct in thinking that whilst [tex]\nabla_{\bold{u}}v^b=u^a\nabla_av^b[/tex] is a vector, [tex]\nabla_bv^a[/tex] is a not a vector, but is a type (1,1) tensor?

    I think this makes sense, since clearly u^a is a vector; if [tex]\nabla_bv^a[/tex] is a type (1,1) tensor, then the product of the two will be a vector.
     
  11. Apr 9, 2007 #10

    robphy

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    Yes.
    Of course, to do the "product", you'd do [tex] (u^b) (\nabla_b v^a)= u^b\nabla_b v^a[/tex].
     
  12. Apr 9, 2007 #11

    cristo

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    Yes. Thanks for your help!
     
  13. Apr 9, 2007 #12

    Dick

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    Yeah, of course they meant [tex]\nabla_{\bold{u}}\bold{T}[/tex] in which case the extra index is contracted. Duh.
     
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