# Does path independence of line integral imply that the given vector field is conservative (that is, it is negative gradient of some scalar potential)?

For a vector field, is path independence of line integral a necessary and sufficient condition for the field to be conservative or is it just a necessary condition? Please provide proof if possible.

A conservative field is one in which the line integral does not depend on the path, but depends only on the initial and final position. Hence the line integral in a closed path is zero.

Mathematically, the line integral of a vector field $F$ from the point $a$ to $b$ is given by $\int _{ a }^{ b }{ \overrightarrow { F } .\overrightarrow { dl } }$

If $a=b$, the integral $\int _{ a }^{ a }{ \overrightarrow { F } .\overrightarrow { dl } } =0$

This suggests the application of Stokes Theorem which is $\oint { \overrightarrow { F } .\overrightarrow { dl } } =\int { (\triangledown \times \overrightarrow { F) } .d\overrightarrow { a } } \quad$

Hence for a conservative field $\int { (\triangledown \times \overrightarrow { F) } .d\overrightarrow { a } } =0\quad \Rightarrow \quad \triangledown \times \overrightarrow { F } =0$

Hence a sufficient for a field to be conservative is its curl should be zero.

Since curl of a gradient is zero, we can express the field $\overrightarrow { F }$ as the gradient of a scalar potential $-V$ . Note that the minus sign is purely conventional.

Further, using the gradient theorem, we can simplify our calculations. This however requires the curl of the field to be necessarily zero and the only tricky part is finding the potential $V$ such that $\overrightarrow { F } =-\triangledown V=-\left( \frac { \partial V }{ \partial x } \hat { x } +\frac { \partial V }{ \partial y } \hat { y } +\frac { \partial V }{ \partial z } \hat { z } \right)$ . Once the potential is found, the line integral just becomes the difference in potential at the two terminal points i.e. $\int _{ a }^{ b }{ \overrightarrow { F } .\overrightarrow { dl } } =-(V(b)-V(a))$