Proof Thata Vector Field is Conservative iff Curl of the Field is Zero
TheoremA vector field
\[\mathbf{F}\]
is conservative if and only if \[\mathbf{\nabla} \times \mathbf{F} =0\]
on its domain.Proof
If
\[\mathbf{F}\]
is conservative then \[\mathbf{F} = \mathbf{\nabla} f \]
then \[\mathbf{\nabla} \times \mathbf{F} = \mathbf{\nabla} \times (\mathbf{\nabla} f) =0 \]
automatically.Conversely suppose
\[\mathbf{\nabla} \times \mathbf{F} =( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} )\mathbf{i}- ( \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} )\mathbf{j} + ( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} )\mathbf{k} =0\]
then\[\frac{\partial F_3}{\partial y} = \frac{\partial F_2}{\partial z} \rightarrow F_3 +a(x,z) =F_2 + b(x,y) \rightarrow F_3 a(x) =F_2 + b(x)\]
(1)\[\frac{\partial F_1}{\partial z} = \frac{\partial F_3}{\partial x} \rightarrow F_1 +c(x,y) =F_3 + d(y,z) \rightarrow F_1 c(y) =F_3 + d(y)\]
(2)\[ \frac{\partial F_2}{\partial x} = \frac{\partial F_1}{\partial y} \rightarrow F_2 e(y,z) =F_1 + h(x,z) \rightarrow F_2 +e(z) =F_1 + h(z)\]
} (3)(1)-(2) gives
\[F_2 + b(x)=F_1 +c(y)\]
(4),(4)-(3) gives
\[b(x)-e(z)=c(y)-h(z) \rightarrow b(x)=c(y)=constant\]
.Then
\[a(x)=d(y)=constant\]
Similarly
\[e(z)=h(z)=Constant\]
Hence
\[f(x,y,z)\]
is well defined and \[f\]
exists such that \[\mathbf{\nabla} \times \mathbf{F} = 0 \]
automatically.