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Directional covariant derivative

  1. Oct 29, 2008 #1
    Is this correct?

    [tex]

    \nabla _{\vec{p}} \vec{p} = (\nabla_a \vec{p} ) p^a

    =< (\nabla_a p^0 ) p^a, (\nabla_a p^1 ) p^a , (\nabla_a p^2 ) p^a, (\nabla_a p^3 ) p^a >

    [/tex]

    (where the a's are summed from 0 to 3)
     
  2. jcsd
  3. Oct 31, 2008 #2

    atyy

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    Science Advisor

  4. Oct 31, 2008 #3
    Hi again, atyy. Perhaps I should re-formulate; I think this is right but I was hoping for confirmation:

    The covariant derivative of a type (a,b) tensor is a type (a,b+1) tensor, but the directional covariant derivative is still type (a,b), since one takes the inner product of the covariant derivative and the directional vector, thereby losing the extra lower rank.

    I was wondering about this because somewhere in Schutz's First Course he writes the geodesic equation simply as [tex]\nabla _{\vec{p}} \vec{p} = 0[/tex], and I wasn't certain what [tex]\nabla _{\vec{p}} \vec{p}[/tex] meant.
     
  5. Oct 31, 2008 #4

    atyy

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    I think it's something like this.

    Expand u in basis vectors (linearity):
    [tex]
    \nabla _{\bold{u}} \bold{v}
    = \nabla _{{u^i}\bold{e_i}} \bold{v}
    ={u^i} \nabla _{i} \bold{v}
    [/tex]

    Expand v in basis vectors (product rule or Leibniz property):
    [tex]
    {u^i} \nabla _{i} \bold{v}
    ={u^i} \nabla _{i} v^j\bold{e_j}
    ={u^i} (v^j\nabla _{i} \bold{e_j}+\bold{e_j}\nabla _{i} v^j})
    [/tex]

    Use definition of Christoffel symbols and covariant derivative of a scalar field:
    [tex]
    \begin{equation*}
    \begin{split}
    \{}{u^i} (v^j\nabla _{i} \bold{e_j}+\bold{e_j}\nabla _{i} v^j}) \\
    &={u^i} (v^j\Gamma^{k}_{ij} \bold{e_k}+\bold{e_j}{\frac {\partial v^j}{\partial x_i}}) \\
    &=({u^i}v^j\Gamma^{k}_{ij} \bold{e_k}+{u^i}\bold{e_j}{\frac {\partial v^j}{\partial x_i}}) \\
    &=({u^i}v^j\Gamma^{k}_{ij} \bold{e_k}+{u^i}\bold{e_k}{\frac {\partial v^k}{\partial x_i}}) \\
    &=({u^i}v^j\Gamma^{k}_{ij} \bold+{u^i}{\frac {\partial v^k}{\partial x_i}}){\bold{e_k}}
    \end{equation*}
    \end{split}
    [/tex]

    For Schutz's equation, let the unknown coordinates of the curve be [tex]x_i=x_i(\tau)[/tex].

    Set [tex]u^i=v^i=({\frac {dx(\tau)}{d\tau}})^i={\frac {dx{_i}(\tau)}{d\tau}}[/tex] as the tangent vector along the curve.

    For a geodesic, set the covariant derivative of the tangent vector along the curve to zero (ie. no acceleration):
    [tex]{\frac {dx{_i}}{d\tau}}{\frac {dx{_j}}{d\tau}}\Gamma^{k}_{ij} \bold+{\frac {dx{_i}}{d\tau}}({\frac {d^2x{_k}}{d\tau^2}}{\frac {d\tau}{dx{_i}}})}={\frac {dx{_i}}{d\tau}}{\frac {dx{_j}}{d\tau}}\Gamma^{k}_{ij} \bold+{\frac {d^2x{_k}}{d\tau^2}}}=0[/tex]

    We can also get the same differential equation by requiring the integrated proper time along the curve to be extremal. A given metric fixes the Christoffel symbols. Solving the differential equation will give you the coordinates of a geodesic [tex]x_i=x_i(\tau)[/tex].
     
  6. Oct 31, 2008 #5
    Need a day to process...
     
  7. Nov 2, 2008 #6
    Thanks, atyy; I follow now. And it turns out that the relationship I was looking for is right here in Schutz's section 6.4, "Parallel-transport, geodesics and curvature":

    [tex]

    \nabla _ {\vec U} \vec U = 0 \Rightarrow [/tex] "in component notation" [tex] U^{\beta} U^{\alpha} _{;\beta}=...=0[/tex]

    I guess in general it pays to read page 166 before page 186.
     
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