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Given a force field $F:A\subseteq\mathbb{R^3}\to\mathbb{R^3}$ where A is an open connected set and Given a regular curve $\phi:[a,b]\subseteq\mathbb{R}\to\mathbb{R^3}$ such that $\phi\left(\left[a,b\right]\right)\subseteq A$ my book defines the Work $W$ of $F$ to move a particle from $\phi(a)$ to $\phi(b)$ as: $$W=\int_\phi\langle F,T\rangle dS$$ Where $\langle F(x),T(x)\rangle$ is the scalar product between $F(x)$ and the tangent versor $T(x)$ to the curve at the point $x\in\phi([a,b])$. My problem with this definition is that we defined the tangent versor $T(t)$ to the curve at a point $t\in[a,b]$ as $T(t)=\frac{\phi^{'}(t)}{\Vert\phi^{'}(t)\Vert}$ so $T$ is a function that goes from $[a,b]\subseteq\mathbb{R}$ to $\mathbb{R^3}$

while to consider the line integral $$\int_\phi f dS$$ i should have that $f$ is a function defined in $\phi\left(\left[a,b\right]\right)\subseteq\mathbb{R^3}$ to $\mathbb{R}$

but if the curve is not injective i can't consider the function $\langle F,T\rangle:\phi\left(\left[a,b\right]\right)\subseteq\mathbb{R^3}$ to $\mathbb{R}$.

So i don't get if $\int_\phi\langle F,T\rangle dS$ is an abuse of notation to mean $\int_a^b\langle F(\phi(t)),T(t)\rangle\Vert\phi(t)\Vert dt$

Or if i need to suppose that $\phi$ is injective (or something similar like that $\phi$ is simple)

EDIT: the problem is that if $\phi$ is not injective i could have that for two different points $t_0$ and $t_1$ in [a,b] i have that $\phi(t_0)=\phi(t_1)$ and if $T(t_0)\neq T(t_1)$ i can't consider a function that for every $x\in \phi ([a,b])$ gives me the tangent versor to the curve $\phi$ at the point $x$. (because for the point $\phi(t_0)=\phi(t_1)$ i have two possible tangent versors to consider)

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1 Answer 1

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Your assertion

while to consider the line integral $$\int_\phi f dS$$ I should have that $f$ is a function defined in $\phi\left(\left[a,b\right]\right)\subseteq\mathbb{R^3}$ to $\mathbb{R^3}$

Is not correct. A line integral

$$\int_\phi f dS$$ is perfectly defined for $f : \mathbb R^3 \to \mathbb R$ by $$\int_a^b f(\phi(t)) \Vert \phi^\prime(t) \Vert dt.$$

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  • $\begingroup$ So? The problem is that $\langle F,T\rangle$ is not a definable function unless $\phi$ is injective $\endgroup$
    – irbag
    Jun 10, 2021 at 14:53
  • $\begingroup$ Why do you say that? For any $t \in [a,b]$, $F(\phi(t))$ is well defined as well as $T(t)$. If you have $\phi(t_1) = \phi(t_2)$ ($\phi$ not injective), then $F(\phi(t_1)) = F(\phi(t_2))$ while $T(t_1)$ may be different to $T(t_2)$. $\endgroup$ Jun 10, 2021 at 15:15
  • $\begingroup$ Exactly so i can't consider a function $L:x\in\phi\left(\left[a,b\right]\right)\subseteq\mathbb{R^3}\to L(x)\in\mathbb{R^3}$ that gives me the tangent versor $L(x)$ for each $x \in \phi\left(\left[a,b\right]\right)$ and so i can't consider $\int_\phi\langle F,L\rangle dS$ because i can't define $L$ (remember that i need that $\langle F,L\rangle$ is a function from $\mathbb{R^3}$ to $\mathbb{R^3}$ but i can't define $L$ from $\mathbb{R^3}$ to $\mathbb{R^3}$ do you get the problem? $\endgroup$
    – irbag
    Jun 10, 2021 at 15:22
  • $\begingroup$ Sorry, I still don't understand what you mean. You now speak of $L$ which is not defined in your question. $\endgroup$ Jun 10, 2021 at 16:01
  • $\begingroup$ Do you agree that to define $\int_\phi f dS$ i need that f goes from $\mathbb{R^3}$ to $\mathbb{R^3}$ right? So to define $\int_\phi\langle F,T\rangle dS$ i would need that $\langle F,T\rangle$ goes from $\mathbb{R^3}$ to $\mathbb{R^3}$ but it's not the case since $T$ goes from [a,b] to $\mathbb{R^3}$ and $F$ goes from $\mathbb{R^3}$ to $\mathbb{R^3}$ $\endgroup$
    – irbag
    Jun 10, 2021 at 16:05

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