Re: Work; the correct one

[Robert Cohen <Robert.Cohen@PO-BOX.ESU.EDU>]

> -----Original Message-----
> From: David Rutherford [mailto:drutherford@SOFTCOM.NET]
> Sent: Thursday, September 26, 2002 3:41 AM
>
> I think the equation for work in (3) below is more
> fundamental, but you
> could probably also write a general equation for work W as
>
> W = \int_s1^s2{F(x,y,z,t).ds}
>
> where F(x,y,z,t) is the force, ds is the differential displacement in
> spacetime, and s1 and s2 are initial and final events. This would
> include, not only spatial displacements, but also time
> displacements for
> work done on 'stationary' particles.

Now we are getting somewhere.

Suppose I have a proton sitting by itself at rest (KE = 0).  I then hold it
in place while I bring an electron to it (the electron was initially at
rest).  Once the electron/proton pair is created, I let go.  The pair
remains at rest (KE = 0).

As I bring the electron into the picture, I exert a force opposite its
motion (to counter the electric force) so I do negative work on the
electron.  There is an electric force on the electron which acts in the
direction of the motion, so this does positive work on the electron.  The
net work is zero and the electron experiences no change in kinetic energy.

So far so good.

What can we say about the proton?

As I bring the electron into the picture, I need to exert a force on the
proton (to keep it from moving toward the electron) so, according to you, I
do work on the electron.  The force I exert increases with time, so I guess
you would say the work I do is positive.  Meanwhile, the electric force on
the proton (due to the electron) increases also so, according to you, this
also does work on the proton.  Since the force is increasing with time, I
would guess that this work also is positive.

It seems to me that you would say that there was net work done on the proton
during this time.  However, the kinetic energy doesn't change.  Does
anything change about the proton?  What purpose does it serve to keep track
of the work done as defined above?

I am confused.  I am willing to replace the current definition of work with
the "correct one" but I need to see the advantage of doing so.
____________________________________________
Robert Cohen; rcohen@po-box.esu.edu; 570-422-3428; http://www.esu.edu/~bbq
Physics, East Stroudsburg Univ., E. Stroudsburg, PA 18301