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A Scalar Function of Vector Coordinates

Let's return to our expression for a total differential of a scalar function, given above:

$\displaystyle df = \grad f \cdot d\vell $

Then

$\displaystyle \int_a^b df = \int_a^b \grad f \cdot d\vell = f(b) - f(b) $

independent of path! The integral depends only on the end points for any total differential that is integrated! Hence we know that:

$\displaystyle \oint_C \grad f \cdot d\vell = 0 $

This should seem very familiar to you. Suppose $ \vF = - \grad U$ for a well-behaved scalar function $ U$ . Then:

$\displaystyle W(a\to b) = \int_a^b \vF \cdot d\vell = -\int_a^b \grad U \cdot
d\vell $

independent of path. In introductory mechanics you probably went from the proposition that the work integral was independent of path for a conservative force to a definition of the potential energy, but as far as vector calculus is concerned, the other direction is a trivial identity. Any vector force that can be written as the (negative) gradient of a smooth, differentiable potential energy function is a conservative force!



Robert G. Brown 2017-07-11