I am familiar with the formula for a path integral given a parametrisation $\textbf{x}(t)$, $t \in l \subseteq \mathbb{R}$ of a curve $C$, and given some scalar function $f(x,y,z)$, $$\int_C f ds = \int_l f(\textbf{x}(t)) \Big | \Big |\frac{d \textbf{x}}{d t} \Big | \Big| dt $$

Now, my lecture notes say if we are in some curvilinear coordinates $(\xi_1, \xi_2, \xi_3)$, we describe the curve in curvilinear coordinates by $\boldsymbol{\xi}(u), u \in l' \subseteq \mathbb{R}$, we have some transformation $\boldsymbol{\phi}$ such that the curve $C$ is parametrised by $\boldsymbol{\phi}(\boldsymbol{\xi}(u))$, and we have some scalar function $g$ = $g(\boldsymbol{\xi})$, $$\int_C g(\boldsymbol{\xi}) ds = \int_{l'} g(\boldsymbol{\xi}) \Big | \Big |\frac{d \boldsymbol{\phi}(\boldsymbol{\xi})(u)}{d u} \Big | \Big| du $$

My understanding is that usually when we have a scalar function $f(x,y,z)$, for every point in 3D space, there is a (unique) corresponding scalar value. But in the curvilinear case, when we consider scalar functions, do we consider functions that directly create a correspondence between points in parameter space $(\xi_1, \xi_2, \xi_3)$ to scalar values? Or do we first convert these points into cartesian coordinates? Why is the formula not $$\int_C g(\boldsymbol{\phi}(\boldsymbol{\xi})) ds = \int_{l'} g(\boldsymbol{\phi}(\boldsymbol{\xi})) \Big | \Big |\frac{d \boldsymbol{\phi}(\boldsymbol{\xi})(u)}{d u} \Big | \Big| du$$

Which seems to be equivalent to the first formula. When we integrate a scalar field over some curve, don't we want the scalar field to be evaluated over that curve?


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