Directional covariant derivative

[snoopies622]

Is this correct?

$$<br /> \nabla _{\vec{p}} \vec{p} = (\nabla_a \vec{p} ) p^a<br /> <br /> =< (\nabla_a p^0 ) p^a, (\nabla_a p^1 ) p^a , (\nabla_a p^2 ) p^a, (\nabla_a p^3 ) p^a ><br />$$

(where the a's are summed from 0 to 3)

 

[atyy]

I think the "product rule" $$\nabla_a \vec{p} = \nabla_a (p^b \vec{e}_b) = p^b\nabla_a (\vec{e}_b) + \vec{e}_b\nabla_a (p^b)$$ should be used in the intermediate step?
http://en.wikipedia.org/wiki/Covariant_derivative

 

[snoopies622]

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 $$\nabla _{\vec{p}} \vec{p} = 0$$, and I wasn't certain what $$\nabla _{\vec{p}} \vec{p}$$ meant.

 

[atyy]

I think it's something like this.

Expand u in basis vectors (linearity):
$$\nabla _{\bold{u}} \bold{v} <br /> = \nabla _{{u^i}\bold{e_i}} \bold{v} <br /> ={u^i} \nabla _{i} \bold{v}$$

Expand v in basis vectors (product rule or Leibniz property):
$${u^i} \nabla _{i} \bold{v}<br /> ={u^i} \nabla _{i} v^j\bold{e_j}<br /> ={u^i} (v^j\nabla _{i} \bold{e_j}+\bold{e_j}\nabla _{i} v^j})$$

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

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

Set $$u^i=v^i=({\frac {dx(\tau)}{d\tau}})^i={\frac {dx{_i}(\tau)}{d\tau}}$$ 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):
$${\frac {dx{_i}}{d\tau}}{\frac {dx{_j}}{d\tau}}\Gamma^{k}_{ij} \bold+{\frac {dx{_i}}{d\tau}}({\frac {d^2x{_k}}{d\tau^2}}{\frac {d\tau}{dx{_i}}})}={\frac {dx{_i}}{d\tau}}{\frac {dx{_j}}{d\tau}}\Gamma^{k}_{ij} \bold+{\frac {d^2x{_k}}{d\tau^2}}}=0$$

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 $$x_i=x_i(\tau)$$.

 

[snoopies622]

Need a day to process...

 

[snoopies622]

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":

$$<br /> \nabla _ {\vec U} \vec U = 0 \Rightarrow$$ "in component notation" $$U^{\beta} U^{\alpha} _{;\beta}=...=0$$

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