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Re: Work; the correct one



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.

David Rutherford wrote:

You don't need to have a displacement in order to do work on a charge q
in an electric field E. If you use

W = \int{F.dr}

as the definition of work W, then in this case, the force F is

F = qE

The electric field can be written E = -d(phi)/dr, where phi is the
electric potential. So

W = \int{F.dr} = \int{qE.dr} = \int{q(-d(phi)/dr)dr} (1)

where "\int" means integral. But

d(phi) = (d(phi)/dr)dr (2)

so if you insert (2) into the last equation in (1), you get

W = \int{q(-d(phi)} (3)

Equation (3) is more fundamental than the last equation in (1) for the
work done on a charged particle in an electric field. Here, you can
see that the work done on the charge q does _not_ depend on it's
displacement, but only on the change in potential at its location. Work
is done on a charge to keep it in place, even if it has not moved, if
the potential at its location has changed.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf

Applications:
"4/3 Problem Resolution"
http://www.softcom.net/users/der555/elecmass.pdf
"Action-reaction Paradox Resolution"
http://www.softcom.net/users/der555/actreact.pdf

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf

Applications:
"4/3 Problem Resolution"
http://www.softcom.net/users/der555/elecmass.pdf
"Action-reaction Paradox Resolution"
http://www.softcom.net/users/der555/actreact.pdf