Re: Newtonian gravitational field energy (very long)

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

I'm sorry for taking so long to respond on this thread.  I have been
unavailable.

Since my last long post I went back and checked out the literature (which
I should have done *before* submitting the post rather than 'shoot from
the hip' relying on a memory that had faded over the many years since I
had last worked out this stuff).  It ends up that I was wrong about a
couple of my claims for the E&M case for which the full potential field
is a spin-1, 4-vector (or at least a 1-form) in 4-d spacetime. (This has
no effect on what I said about Newtonian gravity, however.)  In
particular, the main incorrect thing I said was:

> .... The reason for the sign change between the gravitational
> and E&M cases is that in the electrostatic case the pure [phi] field
> term in the energy enters the Lagrangian (but not the Hamiltonian) with
> the opposite sign as in the gravitational case.  This is because in the
> E&M case the [phi] field in not a scalar nor a time-time component of a
> 2-tensor, but the time-component of a *vector* field, and as such, the
> |grad([phi])|^2 term in the energy is not really part of a potential
> energy-like term that switches signs between the Hamiltonian and the
> Lagrangian, but is part of a kinetic energy-like term that has the
> *same* sign for the Lagrangian and the Hamiltonian.  If the
> quadratic-in-field pure-field term enters the Lagrangian with the
> opposite sign as before, but the linear-in-field interaction term enters
> the Lagrangian with the same sign as before, then the E-L equation that
> results from Hamilton's principle causes a sign change between the
> potential field and its sources relative to the other case.

In the E&M case things are significantly trickier than I had
(mis)remembered when I made this claim.  Below I outline the correct
derivation for the E&M case including its static limit.  Before doing so
let me clear up a misleading generalization I also made which was based,
in part, on the errant claim above.  I also said:

> In general, if an interaction between sources is mediated by an
> even-integer-spin field (e.g. scalar or 2-tensor potential field--like
> gravity which has the |grad([phi])|^2 term keep the same sign between
> the Hamiltonian and the Lagrangian) then it causes like-signed sources
> to *attract* each other, and if it is an odd-integer-spin field (e.g.
> vector potential field like E&M which has the |grad([phi])|^2 term
> switch signs between the Lagrangian and the Hamiltonian) then it causes
> like-signed sources to *repel* each other.

Although the conclusion here (i.e. that even-integer spin fields cause
attraction between like-signed sources and odd-integer spin fields cause
repulsion between like-signed sources), I believe (still from memory
though, so you may want to discount this claim), is correct, the *reason*
for this is more complicated than I said above.  It is not merely a
matter of different even/odd cases of signs switching between
corresponding Lagrangian and Hamiltonian terms.  It's more a case of
switching signs (between even and odd spin cases) in *both* the
Lagrangian *and* the corresponding Hamiltonian terms for the pure-field
gradient, |grad(phi)|^2, static limit-surviving field component (i.e.
the time-time component for a 2nd rank 4-tensor field, the time-component
for a 4-vector field, or just the whole potential field itself for a
scalar field).  In particular, for the E&M vector potential field case,
even though the |grad(phi)|^2 appears as part of a pure field
generalization of a kinetic energy-like term (actually it is in the
|E|^2 term of the Lagrangian density), it *still* switches its sign
between the Lagrangian and the Hamiltonian because it does not have any
explicit time dependence in the Lagrangian density--even in the fully
relativistic formulation, and effectively acts like a potential energy
term which switches sign between H and L --no matter whether the field's
spin is even or odd integer-valued.

For the E&M case (massless 4-vector potential field) the Hamiltonian
actually has a *negative* square gradient term for the pure time-
component for the potential.  When this negative term is added to the
corresponding terms from the interaction between the field and the
sources the square gradient term, again, cancels against 1/2 of the
interaction terms yielding a net postive contribution for this sum.
Apparently, this sum is often lumped together and called the pure field
contribution in the Hamiltonian because when the Legendre transformation
is done which transforms the Lagrangian to the Hamiltonian, the purely
noninteracting kinetic part of the free *matter* part of the Lagrangian
is transformed to an expression that includes *new different*
interactions expressions between the sources and the space-components of
the A-potential field via the so-called 'minimal' substitution involving
the expression: p_i - q_i*A(r_i) where q_i and p_i are the canonical
charge and momentum of the i-th particle respectively, and A(r_i) is the
value of the A field at the location of the i-th particle.  The original
interaction terms (for the spatial-components, i.e. those involving the a
3-vector A) in the Lagrangian *cancel out and vanish* when the
transformation to the Hamiltonian is performed.

To see how this works for E&M let's start with an expression for the
system's (Lorentz scalar) Lagrangian *density* (imagine a script font) L:

L = L_(free matter) + L_(interact) + L_(pure field)

Here L_(free matter) is the purely kinetic contribution to the Lagrangian
density coming from the matter's mass density [rho_mass] involving the
usual free matter relativistic sqrt(1 - v^2/c^2) (which makes it an
invariant scalar under Lorentz transformations).  In particular:

L_(free matter) = - (c^2)*[rho_mass]*sqrt(1 - v^2/c^2) where v is the
local speed of the matter and c is the speed limit of causation.

The L_(interact) term gives the interaction between the matter sources
and the potential field components as a 4-scalar product of the 4-vector
([rho_charge]*c, j) and the 4-vector (V/c, A).

L_(interact) = j(dot)A - [rho_charge]*V

where j is the charge current density vector, A is the 'magnetic vector'
potential (i.e. the spatial components of the EM 4-vector potential),
(dot) is the spatial 'dot' product, [rho_charge] is the density of
electric charge, and V is the electric 'scalar' potential (i.e. V/c is
the time-component of the 4-vector potential).

The pure field contribution to the Lagrangian density is the Lorentz
4-scalar expression:

L_(pure field) = (1/2)*([eps_0]*|E|^2 - |B|^2/[mu_0])

Here [eps_0} and [mu_0] are the permittivity and permeability of free
space since we are, alas, reluctantly, using SI units.  E and B are the
usual electric field and magnetic induction *defined* as:

E == -dA/dt - grad(V) with the d()/dt notation being a partial derivative

and

B = curl(A)

Note that L depends explicitly on dA/dt, and as such, A is a vector
involving *dynamical* field degrees of freedom. In particular, A has
associated with it dynamical canonical momentum degrees of freedom.  But
L does *not* depend explicitly on dV/dt, and so V is *not* dynamical.  It
merely acts instantaneously to always enforce a constraint.  Actually,
what I just said here is only really true *if* an appropriate gauge is
chosen that does not act to give V a dynamical time dependence.  The
complication of gauge freedom in the theory makes formulating the theory
trickier than it would be without that freedom, and the explicit
formulation of the theory and the Hamiltonian *does* depend, to an
extent, on just which gauge is chosen.  If we choose a gauge which
enforces a gauge constraint involving only the components of A (such as
the Coulomb gauge, div(A)==0, or the axial gauge, A_z ==0) then V will
remain nondynamical.  This is not the case, however, if a gauge is chosen
(such as the Lorentz gauge, div(A) + (1/c^2)*dV/dt = 0) that connects
components of A to the time derivative of V.

If we take the sum of all of these contributions to the Lagrangian
density and integrate them over the 4-dimensions (4-d 'volume' element)
of space-time we get the total action for the system of changed matter
interacting with the EM field.  If we extremize this action w.r.t. the
matter degrees of freedom we get the relativistic equations of motion
for the charged matter.  In particular, we get (for each charged particle
i) the usual Lorentz force law for the E-L equation of motion for the
matter:

d(m_i*v_i/sqrt(1 - |v_i|^2))/dt = q_i*(E(r_i) + v_i(cross)B(r_i)) .

If we extremize the action w.r.t. the potential V we get the Coulomb law
as the E-L equation:

0 = - [rho_charge] - [eps_0]*div(dA/dt + grad(V))

or equivalently,

[eps_0]*div(E) = [rho_charge]

(giving the first of Maxwell's eqns.).

If we extremize the action w.r.t. the potential A we get Ampere's law:

0 = j - [eps_0]*d(dA/dt + grad(V))/dt - curl(curl(A))/[mu_0]

or equivalently,

curl(B) = [mu_0]*(j + [eps_0]*dE/dt)

(giving Maxwell's other inhomogeneous equation).  The remaining 2
homogeneous Maxwell equations follow from the definitions of E & B in
terms of the potentials A & V.  The definition of E == -dA/dt - grad(V)
implies Faraday's law: curl(E) = -dB/dt and the definition of
B == curl(A) implies the tranversality of B, i.e.: div(B) = 0.

-------interlude on nonrelativistic (quasi-static) limit---------

It can be shown (by solving Maxwell's equations in that limit using an
appropriate gauge) that in the nonrelativistic limit that (relative to V)
A is reduced by a factor of 1/c^2, and, thus, A can be legitimately
dropped all together when the electro(quasi)static limit is taken.  So
taking the nonrelativistic limit of the Lagrangian density makes
L_(free matter) become L_(rest energy) + L_(NR kinetic) where

L_(rest energy) = -[rho_mass]*c^2     and
L_(NR kinetic) = (1/2)*[rho_mass]*v^2 .

The first term just contributes a (negative) infinite constant to the
overall Lagrangian and is an infinite total derivative term which can
without harm be neglected in the nonrelativisitic limit.  The second term
is just the usual kinetic energy term for the massive matter.  When the
nonrelativistic limit of the other terms in L is take we get:

L_(interact) = - [rho_charge]*V and

L_(pure field) = (1/2)*[eps_0]*|grad(V)|^2

When the Legendre transformation is made and conversion of these terms
into their corresponding terms in the quasi-static Hamiltonian density
is done we get:

H = H_(rest energy) + H_(NR kinetic) + H_(interact) + H_(pure field)

where

H_(rest energy) = + [rho_mass]*c^2   and
H_(NR kinetic) = (1/2)*[rho_mass]*v^2

The spatial integral of the first term is just the total infinite
rest energy of all the massive particles and the spatial integral of the
2nd term is the sum of the kinetic energies of the massive particles--
each of which has the form: |p_i|^2/(2*m_i) where p_i is the canonical
momentum for the i-th particle and m_i is the particle's mass.  The other
terms in the  Hamiltonian density are:

H_(interact) = + [rho_charge]*V
H_(pure field) = - (1/2)*[eps_0]*|grad(V)|^2 .

-------end interlude on nonrelativistic (quasi-static) limit---------

Now if we *keep* the original Lagrangian density fully relativistic
we can make a transformation to the relavitistic Hamiltonian density so
we can see what the various contributions to the energy are in the
relativistic case.  The precise expression obtained for the Hamiltonian
(density) depends, to a certain extent, on just which choice is made for
restricting the gauge freedom of the theory.  Now if we make a gauge
choice whose enforces gauge constraint does not involve V, then V keeps
its manifest non-dynamical nature.  But because of the gauge constraint
this means that at each point in space the A field does *not* have its
ostensible 3 independent degrees of freedom; it only has 2.  This is
because the gauge condition tends to determine the 3rd component of A
once 2 independent components of A are given.  Thus, even though the EM
4-potential ostensibly has 4 component degrees of freedom at each point
in space, it really only has 2 independent dynamical degrees of freedom
that have canonical momenta associated with them.  These 2 degrees of
freedom can be taken to be the 2 independent polarization states of the
EM field (or the +/- 1 helicity states along the Poynting vector if the
Coulomb gauge is chosen).  Suppose we choose such a gauge (i.e. one whose
gauge constraint doesn't involve V), then we can find the canonical
momenta of the theory and use them in making the canonical construction
of the Hamiltonian, and we can find each of the terms in the Hamiltonian
density derived from each of the corrresponding terms in the Lagrangian
density.  The resulting Hamiltonian density is then:

H = H_(free matter) + H_(interact) + H_(pure field)  .

In terms of the particle velocities the spatial integral of
H_(free matter) is just the sum over each of the massive particles i of
their relativistic free-particle energies of the form:

(m_i)*(c^2)/sqrt(1 - |v_i|^2/c^2)

or, if these same terms are written in terms of their canonical momenta
p_i, then the spatial integral of the H_(free matter) term is the sum
over the particles i of terms of the form:

sqrt((m_i*c^2)^2 +(c^2)*|p_i - q_i*A(r_i)|^2).

Both of these expressions for the free-matter part of the Hamiltonian are
numerically equal even though they look different.  In particular, the
2nd expression has some interaction contributions involving the dot
product of a particle's canonical momentum and the A potential at the
particle's location.  The reason for this is that the canonical momentum
p_i involves A(r_i) and is given by:

p_i = m_i*v_i/sqrt(1 - |v_i|^2/c^2) + q_i*A(r_i)  .

The expression for the interacting Hamiltonian density H_(interact)
(generated by L_(interact)) is:

H_(interact) = + [rho_charge]*V .

Notice that the current term j(dot)A in

L_(interact) = j(dot)A - [rho_charge]*V

cancels out and vanishes (against part of the p*v terms of the
Legendre transformation).  The only thing left of the velocity-dependent
(i.e. current dependent) interaction terms in the Lagrangian is just the
time-component of the 4-divergence and the remaining [rho_charge]*V term
is just the electric potential energy of the charged matter given by the
potential V at the location of the charged sources.

Now given our (V-independent) choice of gauge the expression for the
part of H coming from L_(pure field), i.e. H_(pure field) is:

H_(pure field)=(1/2)*([eps_0]*(|dA/dt|^2 - |grad(V)|^2) + |curl(A)|^2/[mu_0])
              =(1/2)*([eps_0]*|E|^2 + |B|^2/[mu_0])) + [eps_0]*E(dot)grad(V)

Notice that if we set A==0 (as we do in the non-relativistic limit) in
this expression we are left with just the negative pure-field energy term:

-(1/2)*[eps_0]*|grad(V)|^2

which agrees with what we said in the interlude.  Also notice from the
2nd line above that H_(pure field) includes not only the familiar
"|E|^2 + |B|^2" terms, but *also* the [eps_0]*E(dot)grad(V) term as well.
It is this extra term that combines with the |E|^2 term to make the
negative contribution for the |grad(V)|^2 term.

Let's look more closely at the 3rd term: [eps_0]*E(dot)grad(V) above in
H_(pure field).  By using a vector identity we can write:

E(dot)grad(V) = div(E*V) - V*div(E) .

Now if we use the E-L equation obtained by extremizing the action w.r.t
V, (a.k.a Coulomb's law, i.e. 0 = - [rho_charge] + [eps_0]*div(E) )
we can change the last div(E) term into a [rho_charge] term.  Thus,

[eps_0]*E(dot)grad(V) = div([eps_0]*E*V) - [rho_charge]*V .

This means we can write H_(pure field) as these 4 terms:

H_(pure field)=(1/2)*([eps_0]*|E|^2 + |B|^2/[mu_0])) - [rho_charge]*V +
                + div([eps_0]*E*V) .

Notice that the third term here when added to the H_(interact)
contribution above *cancels* it out completely (i.e. both [rho_charge]*V
terms cancel each other out.  Also if we integrate the 4th total
divergence term over all space we get a surface-at-infinity term that
gives no contribution for a localized system of sources.  Thus combining
these two parts of the Hamiltonian density together and dropping the
surface/divergence term gives:

H_(interact) + H_(pure field) = (1/2)*([eps_0]*|E|^2 + |B|^2/[mu_0]))

which is the usual form for the energy in the EM field.  But it should
be emphasized that that this |E|^2 + |B|^2 type expression represents the
*combined sum* of the original contributions to the energy that came from
the pure field and from the interaction between the sources and the
field.  IOW, the [rho_charge]*V interaction contribution is *included* in
the |E|^2 + |B|^2 expression, and this inclusion is what makes the
|grad(v)|^2 term in the electrostatic formulation positive, rather than
the negative value it would have if the [rho_charge]*V interaction term
was not included.

*But* because we *already* used the E-L Coulomb law in combining these
terms in simplifying the expression for the Hamiltonian it means that the
Hamiltonian thus obtained can *not* be used to generate Coulomb's law as
one of Hamiltons' eqn.s derived from this combined Hamiltonian.  In this
situation V is not any kind of degree of freedom at all--dynamical or
otherwise.  In this case it is just a mathematically useful supplementary
expression.  With such a Hamiltonian for the system the Coulomb law is
just taken as a supplementary given condition, much like the
supplementary gauge condition is also a given.  If we want our
Hamiltonian formulation to actually generate *all* of Maxwell's equations
besides the dynamical Lorentz force law for the charged matter we can
*not* a priori combine the terms of H_(interact) & H_(pure field) in such
a way as to eliminate the [rho_charge]*V terms by using Coulomb's law in
the combination/simplification process.  And if we do not make such an a
priori combination, then the electrostatic limit of H_(pure field) is the
*negative* expression: - (1/2)*[eps_0]*|grad(V)|^2  and the H_(interact)
contribution is the expression [rho_charge]*V (which numerically is
positive for a charge distribution of all one sign and happens to be
twice as large as the negative pure-field value).  With such an explicit
dependence the Hamilton's eqn. for V minimizes the energy, and the
process then yields the eqn.:

0 = [rho_charge] + [eps_0]*div(grad(V))

which we recognize as the electrostatic version of Coulomb's law because
in this case V is a (static, instantaneously responding) degree of
freedom.

So if we wish to have the interaction between the charges mediated
through the electrostatic potential V in the electrostatic limit (even if
that interaction *is* instantaneous), then we need to keep the

H_(interact) = + [rho_charge]*V

term distinct from the

H_(pure field) = - (1/2)*[eps_0]*|grad(V)|^2

term in the total Hamiltonian density rather than lump them together
and have one of them cancel against half of the other one.

In the gravitational field case everything I said before is, I believe,
still valid.  If we take the weak field, linear, and nearly static
Newtonian limit of the Hilbert action of general relativity (and drop all
the corrections of non-leading order in 1/c^2), I'm pretty sure we get
what I said before.  If, instead of using the tensor theory of general
relativity, one uses the much simpler scalar theory of gravity of
Nordstrom (that has gravity as a simple special relativistic scalar field
in a flat Minkowski spacetime), then that theory, too, boils down to the
same Newtonian scalar limit I previously discussed.  A nice thing about
the Nordstrom theory of gravitation is that it is *much* easier to deal
with than general relativity and is even much simpler than E&M.  It's
really too bad its predictions conflict with the experimental evidence.
In particular, the Nordstrom theory of gravity only obeys the weak
equivalence principle where all kinds massive matter accelerate at the
same rate in a gravitational field, but there are still physical
differences between the local effects of an acceleration of a frame of
reference and a gravitational field, e.g. the Nordstrom theory predicts
that massless fields like light/E&M are oblivious to gravitational fields
and light rays are not deflected at all by strongly gravitating bodies
(in direct contradiction to careful observation).

I thank those others who posted their objections to what I previously
wrote, since it caused me to go back and look at the situation more
carefully.

David Bowman
dbowman@georgetowncollege.edu