Re: Newtonian gravitational field energy (long)

[Bob Sciamanda <trebor@VELOCITY.NET>]

It seems to me that what is below called the interaction term is already
included in the electrostatic energy whose density is the below given u_e
= epsilon E^2/2 .  This u_e includes the interactions among all the
sources of E.  You are overcounting!

-Bob

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor

----- Original Message -----
From: David Bowman <dbowman@TIGER.GEORGETOWNCOLLEGE.EDU>
To: <PHYS-L@LISTS.NAU.EDU>
Sent: Friday, August 06, 1999 4:06 PM
Subject: Newtonian gravitational field energy (long)


> In the electrostatic case the energy density is (in those sorry SI
units)
> u_e = [epsilon_0]*(E^2)/2 in terms of the electrostatic field E
> . . .
> Consider the electrostatic case of a collection of positively charged
> point sources (for convenience) whose motion is slow enough that the
> electro(quasi)static approximation is valid.  If we look at the
> contributions to the system's Hamiltonian from the interaction between
> the sources and the field, we get a positive contribution  from the
> interaction terms of the form: +q_i*V(r_i) where q_i is the value
> of the i-th charge and V(r_i) is the electrostatic potential at location
> r_i of the i-th charge.