# PlanetPhysics/Path Independence of Work

Suppose an object of mass $m$ is free to move in some domain, $D$ (it is assumed that $D\subseteq \mathbb {R} ^{3}$), and let $\mathbf {r} _{1}$ and $\mathbf {r} _{2}$ denote the position vectors of points in $D$. The work required to move the object from $\mathbf {r} _{1}$ to $\mathbf {r} _{2}$ is given by
   $W_{12}=\int _{\mathbf {r} _{1}}^{\mathbf {r} _{2}}\mathbf {F} \cdot d\mathbf {r} ,$
where $\mathbf {F}$ is the total force acting on the object, as a function of position in $D$. If $\mathbf {F}$ is a conservative force, then it can be expressed in terms of a potential function; in particular, if $U$ is taken to denote the potential energy, then
   $\mathbf {F} =-\nabla U,$
where $\nabla$ denotes the gradient operator. Under such conditions, the work required to move the object of mass $m$ from position $\mathbf {r} _{1}$ to $\mathbf {r} _{2}$ in $D$ is path independent. This means that if the object were to move along a straight line connecting $\mathbf {r} _{1}$ and $\mathbf {r} _{2}$, the amount of work done would be in exact equality with any other path.

### Proof of Path Independence

Given the expression for work,
   $W_{12}=\int _{\mathbf {r} _{1}}^{\mathbf {r} _{2}}\mathbf {F} \cdot d\mathbf {r} ,$
and the relation between the conservative force, $\mathbf {F}$ and the potential energy, $U$,
   $\mathbf {F} =-\nabla U,$
it follows that, upon substitution of the later into the former,
   ${\begin{matrix}W_{12}&=&\int _{\mathbf {r} _{1}}^{\mathbf {r} _{2}}\mathbf {F} \cdot d\mathbf {r} \\&=&-\int _{\mathbf {r} _{1}}^{\mathbf {r} _{2}}\nabla U\cdot d\mathbf {r} .\end{matrix}}$
Focus on the integrand, $\nabla U\cdot d\mathbf {r}$, and write it in terms of its components as,
   ${\begin{matrix}\nabla U\cdot d\mathbf {r} &=&\left({\frac {\partial U}{\partial x_{1}}},{\frac {\partial U}{\partial x_{2}}},{\frac {\partial U}{\partial x_{3}}}\right)\cdot \left(dx_{1},dx_{2},dx_{3}\right)\\&=&{\frac {\partial U}{\partial x_{1}}}dx_{1}+{\frac {\partial U}{\partial x_{2}}}dx_{2}+{\frac {\partial U}{\partial x_{3}}}dx_{3}.\end{matrix}}$
Now, recall that for some arbitrary function, $f=f(x,y,z)$, the differential of that function is $df={\frac {\partial f}{\partial x}}dx+{\frac {\partial f}{\partial y}}dy+{\frac {\partial f}{\partial z}}dz.$ Based on this, it immediately follows that
   $\nabla U\cdot d\mathbf {r} =dU.$
Substituting this result back into the work equation,
   ${\begin{matrix}W_{12}&=&\int _{\mathbf {r} _{1}}^{\mathbf {r} _{2}}\mathbf {F} \cdot d\mathbf {r} \\&=&\int _{\mathbf {r} _{1}}^{\mathbf {r} _{2}}dU\\&=&U(\mathbf {r} _{2})-U(\mathbf {r} _{1}).\end{matrix}}$
Therefore, from the final equation, it is clearly seen that the work to move the object from position $\mathbf {r} _{1}$ to $\mathbf {r} _{2}$ is only dependent upon the potential energy at those positions, and not the path taken. Note that in the above, the minus sign in front of the integral has been dropped; this was done to show, in the final result, the amount of work done by the system. That is, if the potential energy at the final position is greater than that at the initial, then $W_{12}$ is positive, and has done work.