Directional covariant derivative

706
8

Main Question or Discussion Point

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)
 

Answers and Replies

atyy
Science Advisor
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706
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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.
 
atyy
Science Advisor
13,645
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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].
 
706
8
Need a day to process...
 
706
8
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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