Question 6.7: Conservative or not conservative? Show that the field F1 = −...
Conservative or not conservative?
Show that the field F _{1} = −2x i − 2y j − 2zk is conservative but that the field F _{2} = yi − x j is not.
(i) If F _{1} is conservative, then its potential energy V must satisfy
-\frac{\partial V}{\partial x}=-2 x, \quad-\frac{\partial V}{\partial y}=-2 y, \quad-\frac{\partial V}{\partial z}=-2 z,
and these equations integrate to give
V=x^{2}+p(y, z), \quad V=y^{2}+q(x, z), \quad V=z^{2}+r(x, y),
where p, q and r are ‘constants’ of integration, which, in this case, are functions of the other variables. If V really exists, then these three representations of V can be made identical by making special choices of the functions p, q and r. In this example it is clear that this can be achieved by taking p=y^{2}+z^{2}, q=x^{2}+z^{2}, \text { and } r=x^{2}+y^{2}.
Hence F _{1}=-\operatorname{grad}\left(x^{2}+y^{2}+z^{2}\right) \text { and so } F _{1} is conservative.
(ii) If F _{2} is conservative, then its potential energy V must satisfy
-\frac{\partial V}{\partial x}=y, \quad-\frac{\partial V}{\partial y}=-x, \quad-\frac{\partial V}{\partial z}=0.
There is no V that satisfies these equations simultaneously. The easiest way to show this is to observe that, from the first equation, \partial^{2} V / \partial y \partial x=-1 while, from the second equation, \partial^{2} V / \partial x \partial y=+1. Since these mixed partial derivatives of V should be equal, we have a contradiction. The conclusion is that no such V exists and that F _{2} is not conservative.