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If I integrate $$dW= \vec{ F} \cdot d\vec{\ell}$$ which are the limits?

In $$\int\limits_{W_{inf}}^{W_{sup}}dW= \int\limits_{\vec{\ell}_{1}}^{\vec{\ell}_{2}} \vec{ F} \cdot d\vec{\ell}$$ it is natural to think to the lengths, as the path from $\vec{\ell}_1$ and $\vec{\ell}_2$, but what is the meaning when choosing the limits for the work ($W_{sup}$ and $W_{inf}$)?

I usually set them from $0$ to $W$ without a particular rationality to get $W$, but what is the idea behind? Is it related to the to the boundaries for the RHS integral?

This is probably obvious, but I can not find an insightful or rigorous and enlightening description.

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  • $\begingroup$ If $F=0$ what will the work be? $\endgroup$
    – lemon
    Mar 12, 2015 at 9:41
  • $\begingroup$ 0? What do you mean? $\endgroup$
    – Alex Pacini
    Mar 12, 2015 at 10:12
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    $\begingroup$ Yes. $\int_{W_0}^{W} dW=W-W_0\Rightarrow W=W_0+\int F\cdot dl$ which equals $0$ if $F=0$, therefore $W_0=0$. $\endgroup$
    – lemon
    Mar 12, 2015 at 10:34

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Really, the "lengths" you have chosen should represent positions (i.e. coordinates) in whatever space you're working in. The force integral is an integral over a path, with start and end points which become the limits of the integral. For a non-conservative force field, the integral will depend on the path chosen, so that too goes along with that integral. For this integral, you can call the start and end points whatever you like, including $l_1$ and $l_2$. At each point along this path, conceptually you are calculating the dot product of the force (field) at that point with the (differential) direction vector, and then adding them. $F$ depends in general on position, and so will $dl$.

The work integral is easier to understand if you think of it as $\Delta W$, that is the change in work done, relative to some zero that you're pretty much free to choose. Energies only really care about differences, not absolutes, so some reference energy is usually chosen as 0. With the $\Delta W$ conception you don't then have to worry about the start and end energies in any absolute terms, just the difference or change that it makes. For the left hand integral, you can regard it as a simple integral of 1 with respect to $W$ from $W_i$ to $W_f$, i.e. $\int^{W_f}_{W_i}dW=W_f−W_i=\Delta W$, and this last equality is what we mean when we say $\Delta W$, the change from initial to final value.

Overall then we have the change in work done by the force when it moves from point $l_1$ to point $l_2$.

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  • $\begingroup$ Firstly about the lengths, right, they are coordinates. Thus, can I say "[...] the lengths, as the path from $L_1$ to $L_2$, [...]"? Maybe I should edit the question. Secondly, why it is $\Delta W$ and not $\Delta W /2$ or whatever else? To be clearer: fixing the initial point to $W_{inf}$, the $W_{sup}$ can mathematically change, why should I interpret the result as $\Delta W$ and not as $\Delta W /C$, where $C$ is an arbitrary constant? Is it because it is related to the interval in the LHS integral? I hope to have explained what I mean. $\endgroup$
    – Alex Pacini
    Mar 12, 2015 at 10:10
  • $\begingroup$ For the right hand integral, you can call the start and end points whatever you like, including $L_1$ and $L_2$. At each point along this path, conceptually you are calculating the dot product of the force (field) at that point with the (differential) direction vector, and then adding them. $F$ depends in general on position, and so will $dl$. $\endgroup$
    – danimal
    Mar 12, 2015 at 10:29
  • $\begingroup$ For the left hand integral, you can regard it as a simple integral of 1 with respect to $W$ from $W_i$ to $W_f$, i.e. $\int^{W_f}_{W_i} dW = W_f - W_i = \Delta W$, and this last equality is what we mean when we say $\Delta W$, the change from initial to final value. $\endgroup$
    – danimal
    Mar 12, 2015 at 10:32
  • $\begingroup$ I see. Thus $\Delta W$ means the variation in the LHS related to the boundaries for the RHS integral, right? In this case, related to the path from $l_1$ to $l_2$. Am I missing something? Thanks! $\endgroup$
    – Alex Pacini
    Mar 12, 2015 at 10:41
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    $\begingroup$ Yes, exactly, the change in work done by the force when it moves from point $l_1$ to point $l_2$. $\endgroup$
    – danimal
    Mar 12, 2015 at 10:45

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