My book Tensors made easy by Giancarlo Bernacci makes the following claim:
$\frac {D \vec V} {d \tau}$ is referred to as the covariant derivative along a line ... It can be expressed as:
$\frac {D V^b} {d \tau} := U^a \nabla_a V^b = U^a (\frac {\partial V^b} {\partial x^a} + \Gamma _{a \lambda} ^b V^\lambda) = \frac {d V^b} {d x^a} \frac {d x^a} {d \tau} + \Gamma _{a \lambda} ^b V^\lambda U^a = \frac {d V^b} {d \tau} + \Gamma _{a \lambda} ^b V^\lambda U^a$
End quote.
Here the covariant derivative of $\vec V$ is being taken along a line parameterized as $x^a = x^a(\tau)$ with tangent $\vec U = \frac {d x^a} {d \tau} \vec e_a$.
What I don't understand is how it is that the chain rule gives $\frac {d V^b} {d x^a} \frac {d x^a} {d \tau} = \frac {d V^b} {d \tau}$ since the LHS is a sum over the abstract index $a$ and the RHS appears to be obtained by canceling abstract indices.
(My naive expectation would be that the sum expands as follows: $\frac {d V^b} {d x^a} \frac {d x^a} {d \tau} = \frac {d V^b} {d x^1} \frac {d x^1} {d \tau} + ... + \frac {d V^b} {d x^n} \frac {d x^n} {d \tau} = n \frac {d V^b} {d \tau} $ which differs by a factor of $n$ from what the book claims)
That should be all that is needed to explain to me how the book's definition actually works and how I am misunderstanding it. However, here is some more information that may come in useful. I'm doing problem 38:
A particle follows a helical path on the surface of a cylinder parameterized as:
$l = \begin{cases} x = \rho \space cos \space \omega t \\ y = \rho \space sin \space \omega t \\ z = h t \end{cases}$
Compute:
a) the vector tangent to the path
b) the derivative along the line of the tangent vector itself
My answer to a) is
$\vec T = (\frac {d x} {d t}, \frac {d y} {d t}, \frac {d z} {d t}) = (-\rho \omega \space sin \space \omega t, \rho \omega \space cos \space \omega t, h)$
My answer to b) is
$\frac {d \vec T} {d t} = <\tilde\nabla \vec T, \vec T> = \nabla _\vec T \vec T \rightarrow ^{comp} T^a \nabla_a T^b = T^a \partial_a T^b$
(last equality holds because $\Gamma$ are all zero in the standard Euclidean basis of $\mathbb{R}^3$)
With $b=x$ we have $T^a \nabla_a T^x = T^x \partial_x T^x + T^y \partial_y T^x + T^z \partial_z T^x$
Now $\partial_x T^x = - \partial_x \rho \omega \space sin \space \omega t$
$= - \partial_x (\sqrt{x^2 + y^2} \rho \omega \space sin \space \omega t$ (using the identity that $\rho = \sqrt{x^2 + y^2}$)
$= - \omega \space cos \space \omega t \space sin \space \omega t$ (differentiating with respect to $x$ and substituting again)
Likewise $\partial_y T^x = -\omega \space sin^2 \space \omega t$
And $\partial_z T^x = 0$
With that we can find the $x$ coordinate of the derivative along the line of $\vec T$
$T^x \partial_x T^x + T^y \partial_y T^x + T^z \partial_z T^x = \rho \omega^2 \space cos \space \omega t \space sin^2 \space \omega t - \rho \omega^2 \space cos \space \omega t \space sin^2 \space \omega t = 0$
The correct $x$ coordinate is $-\rho \omega^2 \space cos \space \omega t$. The two ways of solving the problem (expanding and summing over abstract indices v.s. canceling them via the chain rule) should yield the same result, but they do not. What gives?