Technical problem of definition of length of a worldline using a metric $g$ In lecture 10 of this series, the professor defines the notions of speed and length of a curve in a smooth manifold equipped with a metric $g$. These definitions are made between 33:10 and 44:10 and are the following:
$$ \text{"speed"} = s(\lambda) := \sqrt{g(v_\gamma, v_\gamma)} \tag{1}$$
$$ \text{"length"} = L[\gamma] = \int_0^1\sqrt{g(v_\gamma, v_\gamma)} \text{d}\lambda \tag{2}$$
where $\gamma: (0,1) \longrightarrow \mathbb{R} $ is a differentiable curve and $v_\gamma :=  \dot{\gamma}^i \partial_i$ is the tangent vector field along $\gamma$. So far so good.
My problem comes in lecture 13 of the same playlist, when at 01:02:50, the professor starts to talk about measurements of time in special relativity.
My problem: is still correct define length of a worldline, for example, as in (2)? If the curve is a particle trajectory and I choose the static gauge $\lambda=t$, t is coordinate time, then talk about an inner product $\sqrt{g(v_\gamma, v_\gamma)}$ makes no longer sense since $v_\gamma = \dfrac{\text{d}\gamma^i}{\text{d}t} \partial_i$ is not a "true vector", anymore since its components no longer satisfies the usual vector components Transformation law.
However, I still know that the more explicit expression
$$\int \sqrt{g_{ij}\dfrac{\text{d}\gamma^i}{\text{d}t} \dfrac{\text{d}\gamma^j}{\text{d}t   }} \text{d}t$$
is correct because most of the Relativity textbook authors use expressions similar to the above to define proper time of a massive particle, even if the argument of the square root is not an inner product  (for coordinate time). How does this proceed?
 A: I believe the problem here is that taking the parameter of the curve to be "the" coordinate time is ambiguous. The coordinates are of course dependent on the choice of chart but the curve parameter is entirely independent of it, so it's under no obligation to follow any coordinate transformation rule.
Let $\gamma : (0,1)\to M$ be a curve with parameter $t$. For some chart $(U,\varphi)$, the coordinate representation of this curve is given by $x=\varphi\circ\gamma : (0,1)\to\mathbb{R}^m$. We define a chart such that the coordinate representation of $\gamma(t)$ is $x(\gamma(t))=(t,x^1,x^2,x^3)$. Clearly, under an arbitrary coordinate transformation $x\to x^\prime$, the variable $t$ parameterizing the curve is unaffected, while the coordinate $t$ transforms like any other coordinate. If you were to then reparametrize the curve in terms of the $0^\text{th}$ (or any) component of $x^\prime$, then that's all it would be: a reparameterization.
A: This is part of the reason why "a vector is something that transforms as a vector" is a bad definition. $v^i = d \gamma^i/dt$ is a perfectly good vector because you've fixed a specific coordinate chart, which $t$ refers to. The problem is that there are two statements that look similar but are crucially different:

*

*"In any coordinate system $(t, x^i)$, define four components by $v^i = d\gamma^i / dt$." This is not a vector: if you use each system's $t$ coordinate to define the components, they won't relate to each other through the correct transformation law.


*"In a specific coordinate system $(t, x^i)$, define four components by $v^i = d\gamma^i / dt$. To find the components in another coordinate system, just use the transformation law." This is a vector, but no one tells you the last part explicitly, and people get confused.
You can always define a vector by giving its components in one coordinate system and using the transformation law to find its components in any other coordinate system. The "a vector transforms like a vector" thing is important when you want a single formula that works in all coordinate systems, which is often the case, but not the case here.
