path independence of work
Suppose an object
of mass
is free to move in some domain,
(it is assumed that
), and let
and
denote the position vectors
of points in
. The work
required to move the object from
to
is given by
 |
(1) |
where
is the total force acting on the object, as a function
of position
in
. If
is a conservative force, then it can be expressed in terms of a potential function; in particular, if
is taken to denote the potential energy, then
 |
(2) |
where
denotes the gradient operator. Under such conditions, the work required to move the object of mass
from position
to
in
is path independent. This means that if the object were to move along a straight line connecting
and
, the amount of work done would be in exact equality with any other path.
Given the expression for work,
 |
(3) |
and the relation
between the conservative force,
and the potential energy,
,
 |
(4) |
it follows that, upon substitution of the later into the former,
Focus on the integrand,
, and write it in terms of its components as,
Now, recall that for some arbitrary function,
, the differential of that function is
Based on this, it immediately follows that
 |
(5) |
Substituting this result back into the work equation,
Therefore, from the final equation, it is clearly seen that the work to move the object from position
to
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
is positive, and has done work.
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As of this snapshot date, this entry was owned by mdo.