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

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

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

Hence,

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

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

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