Newtonian gravitational field energy (long)

[David Bowman <dbowman@TIGER.GEORGETOWNCOLLEGE.EDU>]

I have stayed out of the discussions so far on whether or not energy
is real, localizable, or fluid.  I still do not wish to involve myself
in *that* quagmire, but I thought I would interject (after a long
discussion) a remarkable result which was prompted by Leigh's comment in
the "The reality of charge" thread:

> .... Any other division will result in energy having to be
> instantaneously (in the nonrelativistic sense) transferred from one
> body to the other. This can be got around in the same way as it is
> done in electromagnetism, of course, by putting some or all of the
> potential energy in the field. If one does that one acknowledges
> the nonlocalizability of energy!
>
> Localizing a deficiency of something (gravitational field energy is
> negative) is, I suppose, conceptually possible. I wonder what
> gravitational mass density distribution, rho, might appropriately
> be associated with negative gravitational field energy density u ?
>                                                                g
>                                    u
>                                     g
>                             rho = ----
>                                     2
>                                    c
>
> Proper mass is real in my cosmology, by the way, but I know of no
> instances of negative mass. Eh?

As long as we agree to stick to nonrelativistic quasi-static source
situations such that, in the E&M case (instantaneous) electrostatics
gives an adequate description of the field configurations, and in the
gravitational case, the instantaneous Newtonian gravitational potential
is also an adequate description, then it is straightforward to come up
with an expression for the localized energy density of the gravitational
field analogous to the localized energy density for the electric field.
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.  In the
gravitational case the corresponding energy density is
u_g = (g^2)/(8*[pi]*G) in terms of the gravitational field strength g.
In both cases the energy density of the pure field is quadratic in the
field strength times some constant which depends on both the units used
and on a coupling constant giving the field's coupling strength to the
matter sources.

In *both* cases this pure-field energy density is *positive* (or zero in
a field-free region).  This positivity does *not* depend on whether or
not the interaction (mediated by the field) between static sources is
attractive or repulsive for like-"charge" signed sources.

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.  Since all the q's are positive and V(r) is a
positive function everywhere for only positive charges.  In order to
get finite results for the energy we need to subtract the infinite
constant energies associated with the self-interaction of the charges
with their own field. Once this is done the total positive energy from
these interaction terms is exactly twice the usual electrostatic
potential energy for the system (evaluated without go-between fields).  A
negative contribution of half this amount comes from the spatial integral
of the above energy density integrated over all the space between the
charges.  Now the energy density above is positive definite, but it
includes self-energy contributions (and a surface-at-infinity term) that
are infinite and constant for point charge sources.  Once these offending
constant infinite terms are subtracted, the remainder of the pure field
energy is negative and cancels exactly half of the interaction energy
between the charges and the field.  The resulting total of the pure
field energy plus the interaction energy is just the ordinary positive
electrostatic potential energy of the source charge configuration that
one would evaluate using just Coulomb forces pushing on the charges,
*without* any appeal to an electric field at all.

All of this discussion goes over analogously for Newtonian gravity--
except for some sign changes and different constants multiplying the
relevant terms.  In the case of a quasistatic configuration of
(positive) point masses the solution of the Poisson eqn. for the
gravitational potential is *negative* definite because of a sign change
in its source term.  Thus the interaction terms contribute to the
Hamiltonian negatively because each such term is of the form:
m_i*[phi](r_i) where m_i is a positive mass and [phi](r_i) is the
negative gravitational potential [phi] evaluated at location r_i of the
i-th mass.  Once the (negative) infinite constant self-energy terms are
dropped the rest of the interaction terms give a contribution to the
energy that is negative and, again, *exactly twice* the total Newtonian
potential energy of the system (evaluated without fields just using
the Newtonian inter-mass potential).  To this interaction contribution is
added the pure field energy which comes from integrating the above energy
density over all the space between the point masses.  Again the infinite
constant self-energy (and a surface-at-infinity) contributions are
dropped. But *this* time the finite remainder of the pure field energy is
*still* positive (because of a sign change in some cross terms coming
from a change in the sign of [phi] relative to the electrostatic case).
When this is added to the interaction energies between the gravitational
potential and the masses, (again the pure field part cancels half of the
interaction part) the total sum of both parts is, again, just the usual
negative Newtonian potential energy that one would get if one only used
just the Newtonian gravitational inter-mass potential energy function for
point masses and did not appeal to the any field degrees of freedom in
the Lagrangian Hamiltonian to act as quasi-static intermediaries between
the masses.

With all this background out of the way it is amusing (I think) to
numerically evaluate just how strong the Newtonian field energy density
actually is.  If we just use the value of g = 9.807 m/s^2 for the field
strength at the Earth's surface we get a field energy density of
u_g = (g^2)/(8*[pi]*G) = 5.735 x 10^10 J/m^3 = 57.35 kJ/cm^3.  I think
this value is remarkably high.  *If* there was such a thing as negative
mass that could be used to neutralize a gravitational field and make a
gravity shield, then at the surface of such a gravitational "conductor"
where the field would discontinuously go from 1 g to zero across the
surface, then the gravitational pressure at the "conductor"'s surface
would be a whopping 5.66 x 10^5 atm !

David Bowman
dbowman@georgetowncollege.edu