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Meaning of line element vector on work formula

  1. Jan 13, 2014 #1


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    I've been struggling with this for the past hours and I can't find a good answer.
    Using the integral for work [itex] W = \int_a^b \vec{F}\cdot d\vec{s} [/itex], when [itex] a > b [/itex], and the force is directed from a to b, i keep getting a negative result. I am considering [itex] d\vec{s} [/itex] as the infinitesimal difference of the position vector along the integral route, so it has the same direction of the force. Is this wrong? Where am I messing up, and what [itex] d\vec{s} [/itex] should truly mean here?

    E.g The work performed by a gravitational field upon a particle while bringing it from infinity

    [tex] W = \int_\infty^R \vec{F}\cdot d\vec{s} = \int_\infty^R (\frac{-GMm}{r^2})\hat{r}\cdot (-dr)\hat{r} = \int_\infty^R \frac{GMm}{r^2}dr = \left.\frac{-GMm}{r}\right|_\infty^R = \frac{-GMm}{R} [/tex]
    Which is false, since the work is obviously positive.
  2. jcsd
  3. Jan 13, 2014 #2


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    Homework Helper

    Yes. [itex]d\vec{s}[/itex] is shorthand for [itex]\dfrac{d \vec s}{dt}\,dt[/itex] where [itex]t[/itex] parametrizes the path. It can be time, but need not be. Arclength is another possible choice. Expanding the definition of the line integral, one has
    W = \int_C \vec{F} \cdot d\vec{s} =
    \int_{t_0}^{t_1} \vec{F}(\vec{s}(t)) \cdot \frac{d\vec{s}(t)}{dt}\,dt

    You have made a sign error. Here we have purely radial displacement, so [itex]d\vec{s} = r'(t) \hat r\,dt = \hat r \,dr[/itex] (because by definition [itex]r'(t)\,dt = dr[/itex]), not [itex]-\hat r\,dr[/itex] as you have. You will then find that
    \int \vec{F} \cdot d\vec{s} = \int_{\infty}^{R} \frac{-GMm}{r^2}\,dr = \frac{GMm}{R} > 0[/tex]
    as required.

    EDIT: Try to avoid using coordinates as parameters for curves. It leads to exactly this sort of confusion.
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