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Re: Energy density; the correct one



Why expand d(phi) as (d(phi)/dr)dr if you are not moving the object? d(phi)/dr refers
to changes in phi due to changes in position - arbitrarily change phi for OTHER
reasons negates the whole meaning of the integral.

If I have a function f(x,y) of two variables x and y and integrate it with respect
to only one variable, \int(f(x,y)dx), set the limits of integration to be the same
value of x, the integral has a value of zero. I can't just arbitrarily claim that y
somehow changed during the integration (whatever "during" may mean) resulting in the
integral somehow being some g(y2)-g(y1).

Could you refer me to a calculus book that states otherwise?

Phi itself is actually a function of the position of BOTH the charges we are talking
about. That is, phi(r1,r2), or better phi(r1-r2). The derivative of phi in respect to
r1 gives the force on r1. The derivative of phi with r2 is not a force on charge 1.
The work done on charge one is only the result of the force on charge one and its
displacement. Perhaps the integral you mention can be somehow understood to have a
non-zero value - and perhaps it's related to an overall change in the energy of the
two charge system - but it wouldn't be the work done on charge one.

Traveling at 100 mph for 0 seconds gives zero displacement, even if somehow the speed
was found to actually be 200 mph.

Sorry to be so argumentative, but what you are proposing has very far reaching
consequences. If I'm being dense here and wasting time, please someone else chime in
and let me know. I've been interpreting Cohen's comments as being akin to mine, but
maybe I'm off base there also.

Bob at PC

David Rutherford wrote:



Please check your local calculus book

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

Plug this into the first equation and you'll get the second equation.
The second equation is just as valid as the first and it has no dr's.

Actually performing \int involves a path,

Obviously, not.

not just a change in phi
for reasons independent of dr.

You can change the energy of q by changing potential (by moving other charges),
but it's not work done on q itself, V has changed because of energy input
elsewhere in the q1-field-q2 configuration.

Maybe I'm seeing this totally wrong - I'd appreciate a clarification if I am.

I don't know how I can show you any more clearly. Just look at the
second equation

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

It says that work does not depend on a path (for a field that is the
gradient of the potential). It only depends on the change in potential.
Therefore, work is done on a particle in keeping it in the same position
if the potential at that position changes, even if the particle doesn't
move.

--
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