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I'm doing a multivariable calculus course at the moment. I've seen path integrals in the cartesian coordinate system as the following definition:

Definition. The path integral of $f(x,y,z)$ along the curve $C$ is

$$ \int_C f ds = \int_a^bf(\mathbf{x}(u))||\mathbf{x}'(u)||du$$

where $\mathbf{x}:[a,b]\to\mathbb R^3$ is the parametric representation of $C$. The definition didn't make any reference to the coordinate system.

Question. Suppose that $ (x,y,z)=\mathbf \Phi(\xi_1,\xi_2,\xi_3) $ where the transformation $\mathbf \Phi:U\to V$ is sufficiently differentiable and has a inverse $\mathbf{\Phi^{-1}}:V\to U$ where $U,V$ are open subsets of $\mathbb R^3$.

What would be the definition of a path integral over $C$ when $f$ is a function of curvilinear coordinates ($\xi_1,\xi_2, \xi_3$), and the curve was parameterized in curvilinear coordinates as $\mathbf{\xi}=\mathbf{\xi}(u)$ for $a\le > u\le b$?

Is the definition the same as a regular "cartesian" path integral? If so, can you please derive how you would get a path integral for a function in curvilinear coordinates from the definition above for cartesian?

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Note that the curve $f$ in Cartesian coordinates is given by $\tilde f = f \circ \Phi^{-1}$ and similarly the curve parametrization is given by $\tilde x = \Phi \circ \xi$. So in the Cartesian coordinates the path integral would be

$$ \begin{align} \int_a^b \tilde f (\tilde x(u)) \| \tilde x'(u) \| du &= \int_a^b (f \circ \Phi^{-1} \circ \Phi \circ \xi)(u) \| (\Phi \circ \xi)'(u)\|du \\ &=\int_a^b f(\xi(u)) \|(D\Phi)(\xi(u)) \xi' (u)\|du \end{align}$$

So yes there is no new definition, you just have to write it in terms of a cartesian path integral. You see that the difference is in the Jacobian of the coordinate transformation that we have to respect.

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  • $\begingroup$ Hi thanks for the reply! So later in the lecture slides, they claimed: $\int_C f ds = \int_a^b f(\xi (u))||\mathbf{x}'(u)||du $ Could you explain why your integral is the same as what they have please? $\endgroup$ – user523384 Apr 15 at 12:37
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    $\begingroup$ well the $x$ you just used in your formul is the same as my $\tilde x$ and we have $\tilde x = \Phi \circ \xi$ which results exactly in this equation. To get to this for you just have to expand the first term (that includes $\tilde f$ but not the second term (that includes the norm). $\endgroup$ – flawr Apr 15 at 12:39
  • $\begingroup$ Just to clarify, is the line integral of the function on the LHS ($\tilde{f}$) different to the function those line integral is to be determined ($f$)? I see how they are related, but would the outputs of the two functions be the same in the range $ u\in (a,b) $? Is that why the line integral remain unaffected? $\endgroup$ – user523384 Apr 15 at 12:43
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    $\begingroup$ Yes $\tilde f$ is the same function as $f$ but with transformed input coordinates. The function $\tilde f$ "lives" on the rectilinear coordinates while the function $f$ "lives" on the curvilinear coordinates. The outputs are therefore the same, they just have a different parametrization. If we would only compute directly in the curvilinear coordinates, the velocity term $\xi'$ would have no meaning, however in the cartesian coordinates the length of $x'$ weighs the function values appropriately. $\endgroup$ – flawr Apr 15 at 12:48
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    $\begingroup$ No problem:) It might seem that way but it is actually not, within this chain of equations we always have the same integration variables. Here the Jacobian comes just from the fact that we have a derivative in our formula. In the transformation formula/change of variables/substitution you actually change the integration domain. I hope this makes sense:) $\endgroup$ – flawr Apr 15 at 13:20

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