PlanetPhysics/Path Independence of Work

Suppose an [[../TrivialGroupoid/|object]] of [[../Mass/|mass]] $m$ is free to move in some [[../Bijective/|domain]], $D$ (it is assumed that $D\subseteq {ℝ}^3$), and let ${𝐫}_1$ and ${𝐫}_2$ denote the [[../PositionVector/|position vectors]] of points in $D$. The [[../Work/|work]] required to move the object from ${𝐫}_1$ to ${𝐫}_2$ is given by

$W_{12}=\underset{{𝐫}_1}{\overset{{𝐫}_2}{\int }}𝐅\cdot d𝐫,$

where $𝐅$ is the total [[../Thrust/|force]] acting on the object, as a [[../Bijective/|function]] of [[../Position/|position]] in $D$. If $𝐅$ 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 [[../CosmologicalConstant/|energy]], then

$𝐅=-\mathrm{\nabla }U,$

where $\mathrm{\nabla }$ denotes the [[../Gradient/|gradient operator]]. Under such conditions, the work required to move the object of mass $m$ from position ${𝐫}_1$ to ${𝐫}_2$ in $D$ is path independent. This means that if the object were to move along a straight line connecting ${𝐫}_1$ and ${𝐫}_2$, the amount of work done would be in exact equality with any other path.

Proof of Path Independence

Given the expression for work,

$W_{12}=\underset{{𝐫}_1}{\overset{{𝐫}_2}{\int }}𝐅\cdot d𝐫,$

and the [[../Bijective/|relation]] between the conservative force, $𝐅$ and the potential energy, $U$,

$𝐅=-\mathrm{\nabla }U,$

it follows that, upon substitution of the later into the former,

$\begin{array}{ccc}{W}_{12}& =& \underset{{𝐫}_1}{\overset{{𝐫}_2}{\int }}𝐅\cdot d𝐫\\ & =& -\underset{{𝐫}_1}{\overset{{𝐫}_2}{\int }}\mathrm{\nabla }U\cdot d𝐫.\end{array}$

Focus on the integrand, $\mathrm{\nabla }U\cdot d𝐫$, and write it in terms of its components as,

$\begin{array}{ccc}\mathrm{\nabla }U\cdot d𝐫& =& (\frac{\partial U}{\partial x_1},\frac{\partial U}{\partial x_2},\frac{\partial U}{\partial x_3})\cdot (d{x}_1,d{x}_2,d{x}_3)\\ & =& \frac{\partial U}{\partial x_1}d{x}_1+\frac{\partial U}{\partial x_2}d{x}_2+\frac{\partial U}{\partial x_3}d{x}_3.\end{array}$

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

$\mathrm{\nabla }U\cdot d𝐫=dU.$

Substituting this result back into the work equation,

$\begin{array}{ccc}{W}_{12}& =& \underset{{𝐫}_1}{\overset{{𝐫}_2}{\int }}𝐅\cdot d𝐫\\ & =& \underset{{𝐫}_1}{\overset{{𝐫}_2}{\int }}dU\\ & =& U({𝐫}_2)-U({𝐫}_1).\end{array}$

Therefore, from the final equation, it is clearly seen that the work to move the object from position ${𝐫}_1$ to ${𝐫}_2$ 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 [[../SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence/|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.