Why when taking the velocity vector we make
$$u^{\nu}=\frac{d}{d\tau}x^{\nu}$$
and not
$$u^{\nu}=\frac{\nabla}{d\tau}x^{\nu}$$
where in the last equation I meant the covariant derivative. Why?
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$\begingroup$ The covariant derivative acts upon n-forms. $x^\nu(\tau)$ is a curve in spacetime. How would you even define its covariant derivative? $\endgroup$– ACuriousMind ♦Commented Aug 2, 2014 at 21:07
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1$\begingroup$ @ACuriousMind: Yes, except: The covariant derivative acts upon n-forms. This doesn't quite make sense to me. Do you really mean n-forms here? I would have just said tensor fields. $\endgroup$– user4552Commented Aug 2, 2014 at 22:14
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4$\begingroup$ Here's another way to see that defining four-velocity in terms of a covariant derivative wouldn't make sense. Write out the definition of the covariant derivative. You're going to have a Christoffel symbol with an index that stands for $\tau$, but $\tau$ isn't a coordinate. Also, you'll have a vector $x^\nu$, but a 4-tuple of coordinates isn't a vector. $\endgroup$– user4552Commented Aug 2, 2014 at 22:17
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$\begingroup$ With this question I meant this. The way in which the covariant derivative was introduced to me was using this formula $\frac{\nabla}{d\tau} A^{\lambda} = \frac{d}{d\tau} A^{\lambda} + \Gamma_{\mu\nu}^{\lambda} A^{\mu} \frac{dx^{\nu}}{d\tau}$. So, if we (maybe naively) set $A^{\lambda}=x^{\lambda}$ we might define a covariant derivative of $x^{\lambda}$. why is this not legit? $\endgroup$– YossarianCommented Aug 2, 2014 at 22:21
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5$\begingroup$ It is not legit to set $A^\mu = x^\mu$ because the $A^\mu$ is meant to be a vector field along the curve $x^\mu(\tau)$ (a section of the tangent bundle along the curve, in some dictions), but $x^\mu$ is not a vector field. (@Ben Crowell: You're right, in GR, the thing the covariant derivative acts upon are indeed vector fields (and, by extension, arbitary tensor fields). I tend to mix this up with the gauge covariant derivative, which acts more naturally upon n-forms.) $\endgroup$– ACuriousMind ♦Commented Aug 2, 2014 at 22:27
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1 Answer
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Consider an $n$-manifold $M$. A curve is simply a continuous map $\gamma : \mathbb{R} \to M$. For simplicity, suppose $M$ is covered by a single coordinate chart (diffeomorphism) $\varphi : M \to \mathbb{R}^n$. Putting these together, we have \begin{eqnarray} \mathbb{R} & \stackrel{\gamma}{\longrightarrow} & M & \stackrel{\varphi}{\longrightarrow} & \mathbb{R}^n \\ \tau & \longmapsto & p & \longmapsto & (x^0, x^1, \ldots, x^{n-1}) \end{eqnarray} Thus we can view $x^0$ as a simple map $\mathbb{R} \to \mathbb{R}$, and likewise for $x^1$, etc. Each such function is differentiated in the normal way (partial or total differentiation, the two being the same for functions from $\mathbb{R}$ to $\mathbb{R}$), and really there is no other derivative to define.
The concatenation $\varphi \circ \gamma$ circumvents the manifold entirely, relying only on it having some minimal topological structure, not differential (metric) structure. Another way of looking at the situation is to think of $\gamma(\mathbb{R})$ as a one-dimensional submanifold of $M$. As a 1D manifold, it doesn't have intrinsic to itself all the curvature it gets as being embedded in $M$. In fact, as a 1D manifold, it has no intrinsic curvature.
The covariant derivative is only defined for tensors. Let's focus on a fixed $p$ in $M$, for which there is a tangent space $T_p(M)$. Assuming we've defined coordinates, the natural choice of basis for this $n$-dimensional vector space is the set of partial derivatives with respect to those coordinates: $\{\partial/\partial x^0, \partial/\partial x^1, \ldots, \partial/\partial x^{n-1}\}$. And of course tensors are just linear maps from products of $T_p(M)$ and its dual into $\mathbb{R}$.
If you want to covariantly differentiate the object with components $x^\mu$, then you should be able to write it as a linear combination of $\{\partial/\partial x^0, \partial/\partial x^1, \ldots, \partial/\partial x^{n-1}\}$. But this can't be done sensibly -- coordinates and the directional derivatives they induce are entirely different beasts.
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1$\begingroup$ I could have defined a vector $\tilde x_0(x_0)= \sum\limits_i x_0^i (\frac {\partial} {\partial x^i})_{x=x_0}$ $\endgroup$– TrimokCommented Aug 4, 2014 at 9:34
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$\begingroup$ @Trimok yeah but those $x_0^i$ of your vector candidate wouldn't transform as vector components should when changing coordinates because you have absolute freedom to change coordinates the way you want. Thus your vector candidate is no vector $\endgroup$ Commented Aug 4, 2014 at 13:11