# Meaning of line element vector on work formula

1. Jan 13, 2014

### pfr

I've been struggling with this for the past hours and I can't find a good answer.
Using the integral for work $W = \int_a^b \vec{F}\cdot d\vec{s}$, when $a > b$, and the force is directed from a to b, i keep getting a negative result. I am considering $d\vec{s}$ 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 $d\vec{s}$ should truly mean here?

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

$$W = \int_\infty^R \vec{F}\cdot d\vec{s} = \int_\infty^R (\frac{-GMm}{r^2})\hat{r}\cdot (-dr)\hat{r} = \int_\infty^R \frac{GMm}{r^2}dr = \left.\frac{-GMm}{r}\right|_\infty^R = \frac{-GMm}{R}$$
Which is false, since the work is obviously positive.

2. Jan 13, 2014

### pasmith

Yes. $d\vec{s}$ is shorthand for $\dfrac{d \vec s}{dt}\,dt$ where $t$ 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 $d\vec{s} = r'(t) \hat r\,dt = \hat r \,dr$ (because by definition $r'(t)\,dt = dr$), not $-\hat r\,dr$ as you have. You will then find that
$$\int \vec{F} \cdot d\vec{s} = \int_{\infty}^{R} \frac{-GMm}{r^2}\,dr = \frac{GMm}{R} > 0$$
as required.

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