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Conservative fields

Consider a vector field $ {\bf A}({\bf r})$ . In general, the line integral $ \int_P^Q {\bf A}\cdot d{\bf r}$ depends on the path taken between the end points, $ P$ and $ Q$ . However, for some special vector fields the integral is path independent. Such fields are called conservative fields. It can be shown that if $ {\bf A}$ is a conservative field then $ {\bf A} = \nabla V$ for some scalar field $ V({\bf r})$ . The proof of this is straightforward. Keeping $ P$ fixed, we have

$\displaystyle \int_P^Q {\bf A}\cdot d{\bf r} = V(Q),$ (A.82)

where $ V(Q)$ is a well-defined function, due to the path-independent nature of the line integral. Consider moving the position of the end point by an infinitesimal amount $ dx$ in the $ x$ -direction. We have

$\displaystyle V(Q+dx) = V(Q) + \int_Q^{Q+dx} {\bf A}\cdot d{\bf r} = V(Q) + A_x\,dx.$ (A.83)

Hence,

$\displaystyle \frac{\partial V}{\partial x} = A_x,$ (A.84)

with analogous relations for the other components of $ {\bf A}$ . It follows that

$\displaystyle {\bf A} = \left(\frac{\partial V}{\partial x},\,\frac{\partial V}{\partial y},\,\frac{\partial V}{\partial z}\right)\equiv \nabla V.$ (A.85)


next up previous
Next: Rotational coordinate transformations Up: Useful mathematics Previous: Vector identities
Richard Fitzpatrick 2016-03-31