We assume that a uniform ball consisting of dust particles was initially static, and had a very large radius. Then we let it collapse. It is well known that it will stay uniform in newtonian mechanics.
We study the time-reversed process: the dust ball is expanding.
^ expansion
|
large cloud of dust
environment of m
______
/ \
| • m | × • m'
\_______/ center
R = distance (m, ×)
m = test particle
r = distance (m', ×)
| expansion
v
Current laboratory
time t = 0.
The diagram is an extremely crude description of the process.
In the diagram, the dust ball is large and fills the entire diagram. It is expanding uniformly. Gravity is decelerating the expansion. We have marked a close environment of the test particle m. In that environment m knows the current location of dust particles quite well (current in the laboratory time coordinate).
For the bulk of the dust ball, the gravity field is retarded at m: the field is as if the dust particles would have moved at a constant velocity since m last "saw" them. The field is not aware of the fact that currently (in the laboratory time), the dust particles are located just like in the close environment of m, symmetrically around the center ×.
Let us have a dust shell S whose radius
r < R
at the current laboratory time. Let m' be a dust particle in the shell S.
Let
s = distance(m , m').
The speed of m' at a laboratory time
t(s) = s / c
earlier was larger than now. The deceleration d(t) of the expansion of the dust ball is
d(t) = C M / a(t)²,
where C is a constant, M is the mass of the ball, and a is the scale factor of the ball. We may set a(0) = 1.
On the line going through m and ×, we can calculate an approximate distortion the location of m' between
1. the true current location in the laboratory time, and
2. the retarded location of m', which determines the field of m' at m.
The current position of m' differs by
~ -1/2 d(t) r t(s)²
= -1/2 d(t) r s² / c²
from the retarded position. There r is negative if m' is between m and ×.
Let us have r fixed. The difference in the position of m' is
~ s²,
and its effect on the gravity force of m' on m is at most
~ s² * 2 / s
~ s.
S
"cap" "cap"
• Ω | r × r |
m A center B
R = distance (m, ×)
Let us look at a narrow solid angle Ω drawn from m, such that the angle intersects S. If we draw a straight line through S, then by symmetry, the intersection angle between the line and S is the same at both intersections. The intersection area thus is
~ s².
The gravity force of the intersection area is
~ s² / s² = 1.
That is, the intersection "cap" A close to m has the same gravity force as the intersection "cap" B far from m.
Let us then correct the gravities of A and B for retardation. The correction increases the gravity of A, and reduces the gravity of B.
The correction is ~ s, which means that the correction to B wins: the gravity is smaller in the correct, retarded view than in the naive laboratory view.
We can now easily calculate order-of-magnitude estimates for the retardation correction to gravity.
The dependency of retardation correction on R
Does the correction still keep the expansion of the dust ball uniform?
Let us double R, r, and s. Then the acceleration of m doubles. What about the correction?
The correction to the position of m' becomes 8-fold. It must be divided by s to get the relative correction to gravity: the relative correction is 4-fold. We conclude that the relative effect of the correction on gravity is much larger with large R.
That is, m at a large R will feel a surprisingly weak gravity.
Conclusions
We found a very crude formula for calculating the effect of retardation in a collapse, or expansion process where the force field is analogous to the electric field (newtonian gravity with c the maximum speed of a signal). We did not consider "spacetime geometry" effects, or the mass-energy of kinetic energy, at all.
In a collapsing star, or in the universe, spacetime geometry may affect retardation a lot. We need a more detailed analysis.
Suppose that two electric charges are in a free fall, side by side, in a homogeneous gravity field. The electric field certainly is distorted in some way, because of the acceleration, but in which way? An equivalence principle would suggest that in the accelerating frame, the electric field around each charge looks like it would look like in an inertial frame. That is, the equivalence principle holds for the electric field, too.
The gravity field in a spherically symmetric collapse is not homogeneous. The equivalence principle does not hold there.
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