- #1
Pentaquark6
- 6
- 2
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.
$$ \frac{\mathrm{D}}{\mathrm{d}s} Z^\mu (s):= \frac{\partial (Z^\mu \circ \gamma)}{\partial s}+\Gamma^\mu_{\:\: \alpha \beta} \dot{\gamma}^\beta Z^\alpha \\
=\dot{\gamma}^\beta \nabla_\beta Z^\mu$$.
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.
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