Components of contravariant velocity

Question

The question is from Grinfeld's "Tensor Analysis and Calculus of Moving Surfaces" - exercise 81.

QUESTION:

Consider a particle moving along a curve with parameterization with respect to time given by $$ x^i \equiv x^i(t). $$ The velocity is defined as $\vec{v}(t) = \dot{\vec{r}}(t)$. Show that the component $v^i$ of $\vec{v}$ is given by $$ v^i(t) = \frac{dx^i}{dt}. $$

ATTEMPT AT SOLUTION:

Defining a covariant basis $\{ \vec{e}_i \} $, we have $$ \vec{v} = v^i \vec{e}_i = \dot{\vec{r}} = \frac{d}{dt} \left(x^i \vec{e}_i \right) = \dot{x}^i \vec{e}_i + x^i \frac{d\vec{e}_i}{dt}. $$ At this point, judging from the answer the second term somehow needs to be zero, but I couldn't figure out how. Since the covariant basis is defined by $$ \vec{e}_i = \frac{\partial\vec{r}}{\partial x^i}, $$ as we move along the curve the basis vectors are subject to change. So the time derivative $d\vec{e}_i/dt$ may be non-zero. Using the chain rule didn't lead me anywhere: $$ x^i \frac{d\vec{e}_i}{dt} = x^i \frac{\partial \vec{e}_i}{\partial x^j} \frac{dx^j}{dt} = x^i \dot{x}^j \vec{e}_k \Gamma^k_{ij} = \ldots? $$ I know I must be missing a very basic point here, would appreciate any help.

Answer 1

My guess is that the issue is that the notation is confusing, and $x$ should be seen as arguments to $R$, but not as component values. That is, $x$ is being used in the $R$ function ($R = R(x)$) and not in the component decomposition $R = R^i e_i \neq x^i e_i$.

This is how I see a solution:

$$ V = \dot{R} = \dot{R}(x) = \frac{\partial R(x)}{\partial x^i} \frac{\partial x^i}{\partial t} = e_i \dot{x}^i $$

Answer 2

You could have coordinates that change with time (for instance, a rotating polar coordinate grid in the plane). However, in that case, it's usually easier to introduce a time dimension to your space and have a static coordinate grid on that space (the rotating polar grid will now look like a winding staircase, but stand perfectly still), just like how special and general relativity does.

Which is to say, the entire theory is built on a fixed coordinate grid and fixed coordinate functions. They do not change. As a consequence, the basis vectors do not change either.