BELIEVE   ME   NOT!    - -     A   SKEPTIC's   GUIDE  

Conservative Forces

Physicists so love their ENERGY paradigm that it has been elevated to a higher status than the original SECOND LAW from which it was derived! In orer to make this switch, of course, we had to invent a way of making the reverse derivation -- i.e. obtaining the vector force   $\mbox{\boldmath$\vec{F}$\unboldmath }$  exerted "spontaneously" by the system in question from the scalar potential energy  V  of the system. Here's how: in one dimension we can forget the vector stuff and just juggle the differentials in   $dW_{\rm me} = F_{\rm me} \, dx$,  where the   $W_{\rm me}$  is the work I do in exerting a force   $F_{\rm me}$  "against the system" through a distance  dx. Assuming that all the work I do against the system is conserved by the system in the form of its potential energy  V, then   $dV = dW_{\rm me}$. On the other hand, the force  F  exerted by the system [e.g. the force exerted by the spring] is the equal and opposite reaction force to the force I exert:   $F = -F_{\rm me}$. The law for conservative forces in one dimension is then

$$F = - {dV \over dx}$$ (11.15)

That is, the force of (e.g.) the spring is minus the rate of change of the potential energy with distance.

In three dimensions this has a little more complicated form, since   $V(\mbox{\boldmath$\vec{x}$\unboldmath })$  could in principle vary with all three components of   $\mbox{\boldmath$\vec{x}$\unboldmath }$:  x, y and z. We can talk about the three components independently,

$$F_x = - {\partial V \over \partial x} \, , \quad F_y = - {\p . . . . . . d \hbox{\rm and} \quad F_z = - {\partial V \over \partial z}$$

where the notation  $\partial$  is used to indicate derivatives with respect to one variable of a function of several variables [here V(x,y,z)] with the other variables held fixed. We call   $\partial V / \partial x$  the partial derivative of  V  with respect to  x. In the same spirit that moved us to invent vector notation in the first place [i.e. making the notation more compact], we use the gradient operator

$$\mbox{\boldmath$\vec{\nabla}$\unboldmath } \; \equiv \; \ha . . . . . . r \partial y} \; + \; \hat{z} \, {\partial \over \partial z}$$ (11.16)

to express the three equations above in one compact form:

$$\mbox{\boldmath$\vec{F}$\unboldmath } \; = \; - \, \mbox{\boldmath$\vec{\nabla}$\unboldmath } V$$ (11.17)

The gradient is easy to visualize in two dimensions: suppose you are standing on a real hill. Since your height   $h \equiv z$  is actually proportional to your gravitational potential energy  V_g, it is perfectly consistent to view the actual hill as a graph of the function  V_g(x,y)  of East-West coordinate  x  and North-South coordinate  y. In this picture, looking down on the hill from above, the direction of the gradient   $\mbox{\boldmath$\vec{\nabla}$\unboldmath }V_g$  is uphill, and the magnitude of the gradient is the slope of the hill at the position where the gradient is evaluated. The nice feature is that   $\mbox{\boldmath$\vec{\nabla}$\unboldmath }V_g$  will automatically point "straight up the hill" -- i.e. in the steepest direction. Thus   $- \mbox{\boldmath$\vec{\nabla}$\unboldmath }V_g$  points "straight downhill" - i.e. in the direction a marble will roll if it is released at that spot! There are lots of neat tricks we can play with the gradient operator, but for now I'll leave it to digest.