Compute Commutator of Covariant Derivative & D/ds on Vector Fields

[Pentaquark6]

Hi, let $\gamma (\lambda, s)$ be a family of geodesics, where $s$ is the parameter and $\lambda$ distinguishes between geodesics. Let furthermore $Z^\nu = \partial_\lambda \gamma^\nu$ be a vector field and $\nabla_\alpha Z^\mu := \partial_\alpha Z^\mu + \Gamma^\mu_{\:\: \nu \gamma} Z^\gamma$ be the covariant derivative. Let us lastly define
$$ \frac{\mathrm{D}}{\mathrm{d}s} Z^\mu (s):= \frac{\partial (Z^\mu \circ \gamma)}{\partial s}+\Gamma^\mu_{\:\: \alpha \beta} \dot{\gamma}^\beta Z^\alpha \\
=\dot{\gamma}^\beta \nabla_\beta Z^\mu$$.

Do the covariant derivative and $\frac{\mathrm{D}}{\mathrm{d}s}$ applied to a vector field commute in this case?

I tried verifying that they do, but I don’t know how to compute $\frac{\mathrm{D}}{\mathrm{d}s}$ applied to a tensor.

 

[kent davidge]

Seems like you used wrong indices in defining $Z$. The correct would be $\nabla_\alpha Z^\nu = \partial_\alpha Z^\nu + \Gamma^{\nu}{}_{\alpha \sigma} Z^\sigma$. Considering you last defined quantity we would have a term of the type $$\nabla_\alpha \frac{\partial}{\partial s} \gamma_{\beta} \nabla^{\beta}$$ when calculating the commutator. And you would need to operate with the covariant derivative on the field $\gamma$ on that term. So it seems that they will not commute. PF members please tell me if I'm wrong.

 

[PeterDonis]

Let furthermore $Z^\nu = \partial_\lambda \gamma^\nu$ be a vector field

This can't be correct because the indexes don't match. If $\gamma$ is a scalar field, then $\partial^\nu \gamma = Z^\nu$ is a vector field. (Strictly speaking, the $\partial$ should have a lower index, so you would have a covector field $\partial_\nu \gamma = Z_\nu$, and you could raise an index with the inverse metric to get the corresponding vector field.)

With the correct definition of a vector field, its covariant derivative looks like what @kent davidge posted in post #2.

However, even before we get to this point, I'm confused about what you intend $\gamma$ to be. You define it as:

let $\gamma (\lambda, s)$ be a family of geodesics, where sss is the parameter and λλ\lambda distinguishes between geodesics.

But with this definition, $\lambda$ and $s$ aren't coordinates, they're parameters, so $\gamma$ isn't even a scalar field, and you can't compute its covariant derivative at all. A scalar field would be $\gamma(x^\nu)$, i.e., a function that takes coordinates as input and outputs a number. To obtain that from your definition you would need functions $\lambda(x^\nu)$ and $s(x^\nu)$, i.e., for every point in spacetime, you would need to know which geodesic in the family it was on and what the curve parameter of that geodesic was at that spacetime point. Without this information you can't proceed.

Considering you last defined quantity we would have a term of the type
$$
\nabla_\alpha \frac{\partial}{\partial s} \gamma_{\beta} \nabla^{\beta}
$$

when calculating the commutator.

No, you wouldn't. As I noted above, $\gamma$ shouldn't have an index. Also, $s$ is a curve parameter, not a coordinate, so there is no meaning to a partial derivative with respect to it. The whole scheme in the OP needs to be reworked in the light of the above comments.

 

[Pentaquark6]

Kent davidge: yeah you’re right, I’ve got a typo in the definition of $\nabla_\alpha Z^\mu$.Ok, it seems that there is a conflict of notations here. In the text I am using, the points along a curve $\gamma$ are denoted as $\gamma^\mu \partial_\mu$ with $\partial_\mu$ being the basis vectors. This renders $\gamma^\mu$ (the components of) a vector field.

Furthermore $Z^\mu=\partial_\lambda \gamma^\mu$ shouldn't be a conflict of indices, as $\lambda$ is not an index denoting components, but a parameter by which $\gamma$ is differentiated. If this notation irritates you, we can write $Z^\mu=\dot{\gamma}^\mu$ with $\dot{}=\frac{\partial}{\partial \lambda}$.

I hope this makes my question a little bit clearer.

 

[PeterDonis]

In the text I am using

What text? And what part of it are you getting all this from?

 

[martinbn]

May be this will help. The derivative $\frac D{ds}$ is the induced connection along the curve from $\nabla$. So, at least for tensors that are restrictions along the curve from tensors on the manifold, you have that $\frac D{ds}T=\nabla_{\dot{\gamma}}T$. Your question really is whether $\nabla_X\nabla_{\dot{\gamma}}T=\nabla_{\dot{\gamma}}\nabla_XT$.