A Possible Scalar Term Describing Energy Density in the Gravitational
Field
The
gravitational field of a point mass and the electric field of a point
charge are structurally similar. Each may be represented by a vector
field in which the vectors are directed along radial lines emanating
from the point, and the vector magnitudes decrease as the inverse square
of the [radial] distance from the point. The electric field,
E,
when multiplied by the magnitude of a test charge q in the field, gives
the force f
exerted locally on the test charge
The
gravitational field, g,
when multiplied by the magnitude of a test mass m in the field, gives
the force f
exerted locally on the test mass
In
each case, the force has the same vector sense as the field. The electric
field may point radially toward or away from the source charge, depending
on the sign of the source charge. The gravitational field points toward
the source mass in all known cases. The electric field has a scalar
energy density field (or,following some older texts, a pressure field)
associated with it. When the field vector at a point is E,
then the energy density at the same point is
uE
= ½(0E
. E
) =
½0E2
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where
015
is the permittivity of free space. Energy density is a measure of the
energy stored in the field per unit volume of space. Its unit of measure
is j/m3 (or nt/m2, if it is thought of as a pressure).
While eq. 3 represents the energy density for the
electric field, and a similar expression represents the energy density
for the magnetic field, no such energy density term has ever been defined
for the gravitational field. But one suspects that it could be, and
possibly even should be.
Let
us use the similarity between the gravitational and electric fields
to construct a gravitational energy density term.
We
begin by noting how the permittivity of free space enters into the expression
for E:
where
k = 1/(4p0)
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4a.
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is
a universal constant, Q is the source charge, and r is radial distance
from the source.
The
gravitational field is given by the expression
where
G is a universal constant, M is the source mass, and r is radial distance
from the source.
The
electrical field energy density may be written in terms of k as
uE
= ½ (1/4k)
E2
=
E2/(8k)
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6.
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By
analogy, a candidate gravitational energy density term may now be constructed
and written as
uG
= g2/(8G)
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Near
the surface of the earth
g
= 9.807 m/sec2.
Also
G = 6.672 X 10-11 (nt m2)/kg2
so
that uG = 5.736 X 1010 j/m3.
Using
eq. 7, it might be possible to construct a classical
argument for the rotation of perihelion of a planet around its central
body [sun]. Recall that the rotation of Mercury's perihelion
(43 arc sec per century) was successfully dealt with by general relativity.
Historically, attempts to modify Newton's law of gravitation to account
for the observed motion of Mercury have proved unsatisfactory. So did
the introduction of another [hypothetical] planet called Vulcan,
with an orbit inside that of Mercury. It is reasonable, then, that the
present argument will not be an exact solution, but a rough approximation.
Its usefulness is as a conceptual tool for students, familiar with classical
dynamics, who are just being introduced to Einstein's concepts. It will
serve the function of a bridge between classical orbit theory and general
relativity. Einstein's original paper on the deflection of starlight
passing the sun might be said to serve a similar function. Einstein's
calculation in that paper, is still reproduced as a problem in texts
on special relativity. Throughout this calculation, we will use Newton's
law of gravitation without modification, and incorporate the energy
density term described above to render a qualitative description of
the rotation of perihelion.
First,
we rewrite the term uG by substituting for g in eq.
7.:
g
= GM/r2
so
that uG = GM2/(8r4)
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8.
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Another
scalar field may be obtained from the energy density field by using
the mass-energy equivalence from special relativity; i.e., a scalar
mass density field of the form
uG/c2
= GM2/(8r4c2)
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We
next assume that the mass due to the term uG/c2,
integrated over a suitable volume of space, behaves, gravitationally,
like ordinary matter. For an extended body like the sun (rather than
an ideal point mass), the ramifications of this assumption need to be
explored
i.
In the interior of the extended body, since the interior field contributes
to the overall mass, and, therefore, to its gravitational field, and
ii.
In the "matter free" space surrounding the extended body. Planets
and other objects moving in this region will experience the combined
gravitational effects of the mass called out in i. above, and, also,
the mass contribution of the external field.
The
sun may be considered a sphere of radius r0, and constant
(average) density .
The classical solar mass is then
M0
= 4r03/3
To
find the additional mass contribution due to the interior field, we
must first rewrite eq. 9 for the sun's interior. We
set
M
= 4r3/3
which
is a function of the radial distance r. We next substitute this result
into eq. 9 to obtain
uG/c2
= 2Gr22/(9c2)
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The
corresponding mass contribution Mf is then the volume integral of eq.
10 throughout the sun's interior. The integrand is
dMf
= [2Gr22/(9c2)]
. 4r2
dr
and
the limits of integration are from r = 0 to r = r0. Thus
Mf
= 82G2r05/(45c2)
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11.
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The
total mass of the sun must now be
For
the sun, we know that
M0
= 1.99 X 1030 kg
= 1,410 kg/m3 (avg.)
r0 = 6.912 X 108 m
We
calculate Mf = 4.08 X 1023 kg
As
should be expected, Mf is very small compared to M0.
In fact,
Mf
= (2.05 X 10-7) . M0
and
is, therefore, only slightly greater than one ten-millionth of the classical
solar mass.
Outside
the sun, the local field experienced by an orbiting planet, asteroid,
etc., has contributions from
i.
The mass M = M0 + Mf (a constant in this region),
and
ii.
The mass due to the external field, which will vary if the orbit is
not strictly circular.
Let
a planet be located a distance r from the center of the sun. Associate
with this distance a sphere, concentric with the sun, on whose surface
the planet is always to be found. As r increases, the sphere expands.
As r decreases, the sphere contracts. The motion of the planet is determined
by the total mass inside the sphere. This total mass is made up of the
mass contained in the sun plus the mass due to the external field contained
inside the sphere. For the external field
dMf'
= [GM2/(8r4c2)]
. 4r2
dr
The
volume integral is to be taken throughout the region between the surface
of the sun and the sphere of radius r. Thus
Mf'
= GM2(1/r0 - 1/r)/(2c2)
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It
is the total mass M + Mf' (= M0 + Mf
+ Mf') that attracts the planet, and influences its motion
around the sun. The gravitational field g acting on the planet is
=
G(M + GM2(1/r0 - 1/r)/(2c2))/r2
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This
field represents the acceleration of the planet in its orbit around
the sun. The classical contribution of the sun is represented by
and
results in a stationary, elliptical orbit, as expected. The additional
contribution of the sun, due to its external gravitational field, is
represented by
G2M2(1/r0
- 1/r)/(2c2r2)
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16.
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This
field varies with radial distance r, the more so the smaller the value
of r. Therefore, any planet, near the sun and in a noncircular orbit,
will experience a sideways perturbation from a strict classical orbit.
If, without this perturbation, the planet would follow a classical ellipse,
the perturbation must be such as to divert it sunwards from the ellipse
as it travels from perihelion to aphelion. The reverse must be true
for the other half of the orbit, with the result that the line of apsides
must turn [slowly] with the same directional sense as the orbital
motion of the planet (i.e., clockwise or counterclockwise, viewed from
a suitable vantage point)16.
It
is interesting to calculate the approximate magnitude of the mass contribution
in eq. 13 for the planet Mercury. Accordingly, we
need to know that for Mercury,
r
= 5.75 X 1010 m
Also,
for the sun, M = 1.99 X 1030 kg
r0
= 6.91 X 108 m
And
G = 6.67 X 10-11 nt m2/kg2.
Then,
It
is interesting that the mass calculated in eq. 17
is roughly equivalent to the mass of the earth. It was commented earlier
that when Mercury's rotation of perihelion was first observed, astronomers
attempted to account for it by postulating another planet called Vulcan,
whose orbit was inside that of Mercury and whose gravitational influence
perturbed Mercury's orbit. These observations were made late in the
nineteenth century, before the advent of special relativity. Vulcan
was never found, as we now know, and the perturbation in Mercury's orbit
was finally accounted for satisfactorily by Einstein.
15
0
= 8.8542 x 10-12 F/m.
16
It may be shown that motion, under a law of central attraction of the
general form
results
in the type of motion we have been describing here; i.e., that the perihelion
of a planet, moving under such a law of attraction, will rotate at a
rate proportional to the constant, k1 (Levi-Civita, pg. 396).
The law of attraction stated in eq. 14a, above,
only approximates eq. 18. We may rewrite eq.
14a. as
G(M
+ GM2(1/r0-1/r)/(2c2))/r2
= GM/r2 + [G2M2(r/r0-1)/2c2r3)]
14b.
In
this representation, comparing with eq. 18:
k
= GM
k1 = G2M2(r/r0-1)/(2c2)
The
term k1 is not a constant, as eq. 18 requires,
but is a linear function of the distance, r. Thus, our use of the energy
density term provides, at best, only a crude approximation to the perihelion
problem. None the less, it still seems reasonable that the approach
described above should have instructional value, providing a conceptual
bridge between the worlds of classical physics/special relativity on
the one hand, and general relativity on the other.
Line
of apsides: the line connecting the ascending and descending nodes of
the orbit.
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