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Question

$\partial_\mu \vec{V} = 0$, does this imply that $\nabla_\mu V^\kappa = 0$ ?

My argument

For any vector $\vec{V}$ we can write that $V=V^\kappa \vec{g_\kappa}$ with $\vec{g_\kappa} = \partial \vec{r}/\partial r^\kappa$ such that:

$$0= \partial_\mu \vec{V} = \partial_\mu(V^\kappa \vec{g}_\kappa) = \partial_\mu (V^\kappa) \vec{g_\kappa} + V^\kappa \partial_\mu(\vec{g_\kappa}) = \partial_\mu (V^\kappa) \vec{g_\kappa} + V^\kappa \Gamma_{\mu\kappa}^\rho \vec{g_\rho} = \partial_\mu (V^\kappa) \vec{g_\kappa} + V^\rho \Gamma_{\mu\rho}^\kappa \vec{g_\kappa} = \nabla_\mu (V^\kappa) \vec{g_\kappa}$$

projecting onto the different basis vectors than gives the result that $\nabla_\mu V^\kappa = 0$

Additional question

If this is true, and we consider a translation in space $\vec{x} \rightarrow \vec{x} + \vec{a}$ with $\vec{a}$ constant. Do we than find that $\nabla_\mu a_\nu = 0$?

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  • 3
    $\begingroup$ Are you sure that $\partial_\mu(\vec{e_\kappa}) = \Gamma_{\mu\kappa}^\rho \vec{e_\rho}$. I think $\nabla_\mu(\vec{e_\kappa}) = \Gamma_{\mu\kappa}^\rho \vec{e_\rho}$. $\endgroup$ – Saksith Jaksri May 9 '17 at 14:43
  • $\begingroup$ @SaksithJaksri Yeah I know that equation but it contradicts what is being said in this reference. onlinelibrary.wiley.com/doi/10.1002/9783527618132.app6/pdf I went trough the math in that reference and can't find any holes... $\endgroup$ – gertian May 9 '17 at 15:49
  • $\begingroup$ @SaksithJaksri I have found the full version of that book and it turns out that the $\vec{e_\kappa}$ that he is reffering to are NOT the normalized unit vectors ! $\endgroup$ – gertian May 9 '17 at 16:13
  • $\begingroup$ Isn't the covariant derivative of a function just the directional derivative? Not sure how to interpret the last equal sign. I tried to get an expression for it before which used the koszul formula and it needed two vectors to be computed. $\endgroup$ – Emil May 9 '17 at 18:09
  • $\begingroup$ @Emil, Well in the last equal sign I introduce the usual covariant derivative . $\endgroup$ – gertian May 9 '17 at 18:18
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The initial problem comes when you start with the quantity $\partial_{\mu} \vec{V}$. This is not a valid operation - remember generally vectors are defined in the tangent space - they are linear differential operators. You can only operate on the components of the vector i.e. $\partial_{\mu} V^{\mu}$.

Instead, for some path in a manifold with coordinates $x^i$ and parameterized by $\lambda$, we can ask how does the geometrical object $\vec{V}$ change along the path?

In this case we have,

\begin{align*} \frac{d\vec{V}}{d \lambda} = \frac{d}{d \lambda}(V^i e_i) &= \frac{d V^i}{d \lambda} e_i + V^i \frac{d e_i}{d\lambda} \\ &= \frac{d V^i}{d \lambda} e_i + V^i \frac{d e_i}{dx^b} \frac{dx^b}{d \lambda}\\ &=\frac{d V^i}{d \lambda} e_i + V^i \Gamma_{ib}^c e_c \frac{dx^b}{d \lambda} \end{align*}

If we are just interested in how the components of the vector change, then we can relabel the indices and drop the basis vectors such that

$$ \frac{D V^i}{d \lambda} = \frac{d V^i}{d \lambda} + V^i \Gamma^a_{ic} \frac{d x^c}{d \lambda}$$

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