If we convert a central mass M into an expanding dust shell, then the gravity attraction at a distance R seems to stay constant.
What if the dust particles do not all fly outward, but bump into each other?
That corresponds to a "back and forth" movement. Earlier we have argued that the gravity in that case is equal to γ M, that is, the mass-energy.
A mass M bouncing back and forth radially relative to the test mass m
wall s wall
• | ● --> v |
m R M R'
r = distance (m, M)
Let us have a mass M which bounces back and forth between walls at distances R and R'.
After we corrected the September 9, 2023 calculation, the coordinate acceleration of the gravity of M is
1 / γ⁵ * G M / r²
= (1 - 5/2 v² / c²) * G M / r²,
if v << c. That is, the gravity of a moving M is substantially weaker than a for a static M.
Let us calculate the average force from the bouncing of the extra inertia of m. The extra inertia of m "held by" M is
2 m G M / (c² r).
The impulse from a bounce is
4 v m G M / (c² r).
Let us assume that s = R' - R is small relative to r.
A cycle back and forth for M takes the time
t = 2 s / v.
The net impulse for a cycle is
p = 4 v * s m G M / (c² r²)
and it points to the right in the diagram. The force
p / t = 2 v² m G M / (c² r²)
= 2 v² / c² * m G M / r².
The acceleration is
2 v² / c² * G M / r².
That does not seem right. We would like the total gravity to be
(1 + 1/2 v² / c²) G M / r².
When we evaluated a sideways bouncing of M relative to m, the key was to remove the acceleration caused by the stretching of the radial spatial metric. That acceleration is "reversible" as M moves back and forth.
Let us calculate the effect of the stretching of the spatial metric. Let m move at a local velocity V a distance Δr toward M. The spatial metric slows down the coordinate velocity V' through the formula:
V' ~ V * (1 - 1/2 rₛ / r).
The slowdown from the spatial metric is
Δr * V * G M / (r² c²)
= Δt V² / c² * G M / r²,
which corresponds to an acceleration of
V² / c² * G M / r².
Let us sum all these:
(1 - 5/2 v² / c²) * G M / r²
+ 2 v² / c² * G M / r²
+ v² / c² * G M / r²
= (1 + 1/2 v² / c²) G M / r².
The sum is reasonable.
On October 18, 2023 we realized that in certain cases, a linear sum of the metric perturbations is not the correct way to obtain the solution for the combined system. We have to keep this in mind.
Pressure does not cause gravity?
Our calculations above suggest that pressure does not produce a gravity attraction, after all. How is this possible? If we move the test mass m closer to M, the test mass makes the radial metric to stretch, releasing energy from pressure. There should be some attraction.
A possible explanation: manipulating the pressure inside M usually causes an acceleration of mass in M. And we saw on October 17, 2023 that general relativity cannot handle accelerating masses.
_____
/ P \
• |<---------->| M
\______/
m
Let us consider a shell of mass M, held static by an internal pressure P. If we suddenly remove the pressure, then the shell starts to contract.
M attracts m less since the pressure was released?
However, we also have to consider the fact that the extra inertia "held" by M starts to accelerate toward the center of M. Since the near side of M holds more inertia than the far side, m accelerates toward M.
Let us calculate the effect. Let M be very small, so that its internal gravity is negligible. We suddenly increase the pressure P from zero to a large value.
r s
• | P | A = area of
m dm / 2 dm / 2 wall
We assume s << r.
If we move m closer a distance dr, we release pressure energy worth
dW / dr = 1/2 * 1 / (1 - rₛ / r)³/² * rₛ / r²
* s A P
= G m / (r² c²) * s A P,
where we assumed rₛ / r is small. The acceleration of m is
a = G / (r² c²) * s A P,
and its is attractive.
In the configuration, the near wall "holds"
2 s * 1/2 m G dm / (r² c²)
more inertia of m than the far wall. The acceleration of the near wall is
2 P A / dm,
which translates to a repulsive acceleration of
2 G / (r² c²) * s A P.
The net effect is
G / (r² c²) * s A P.
If the pressure is uniform, it makes the shell to expand up and down in the diagram, too. And that might cause an acceleration of m. Let us analyze.
^ a
|
• dm
^
| F | 1/2 s
v v
• r
m
^ y
|
-----> x
There dm is an element of M. How does m react to dm being accelerated upward in the diagram? If we imagine that dm would be static, and m would be pushed down by a force F, what would the path of m be?
The test mass m has some extra inertia which is double in the radial direction. The radial component of the acceleration caused by F is approximately
a * 1/2 s / r,
and it is reduced by a factor 1 - 1/2 rₛ / r, because the inertia is larger to the radial direction. We get a correction to the x acceleration
1/4 rₛ / r * s / r * 2 P A / dm
= 1/4 * 2 G dm / (r² c²) * s / r * 2 P A / dm
= G / (r² c²) * s A P.
The acceleration is attractive.
What is the average correction? The pressure component Pₓ produces a repulsion
1/2 G / (r² c²) * s A P
for an element of the sphere whose projected area is A in the y, z plane.
The repulsion is dominated by the corresponding attractive effects of Py and Pz:
G / (r² c²) * s A P.
But here we forgot the fact that we cannot sum individual perturbations like this. How to correct?
Let us consider the electromagnetic analogue.
____
/ \
• | |
\______/
-q +Q
If the charged shell Q expands, then some excess energy density between q and Q is erased and some reduced energy density behind Q is erased, too. But is there a reason why this energy shipment should produce a force on q?
We have assumed vaguely that in gravity the field energy is shipped "on the average" from the distance (m, M).
In the case of a static electric field it is more complicated. If Q expands, there is an energy shipment.
Expanding dust shells
In the (corrected) blog post on October 11, 2023 we studied a dust shell which expands at a constant velocity. It looked reasonable that the gravity field of such a shell is equivalent to a static mass M which has the same mass-energy. Birkhoff's theorem would hold in that case.
What about a shell whose expansion accelerates?
That boils down to the question, "where" the extra inertia of the test mass m is "stored". Birkhoff's theorem suggests that it is, on the average, stored at the center of the dust shell. If not, then dust shells of the same mass-energy but different radii would have differing gravity on a test mass m approaching them.
Assumption. The extra inertia of m, "on the average" is stored at the center of the dust shell.
If we accept the assumption above, then the pressure P does generate an extra gravity attraction. Birkhoff's theorem fails, after all.
Conclusion
Our best guess is that pressure does gravitate. That, in turn, means that Birkhoff's theorem fails for gravity.
It may be that general relativity does not understand a pressurized configuration and the metric calculated from it satisfies Birkhoff's theorem. Then we have a failure of the geodesic equation.
The only accelerating solution of general relativity which we know well is the Oppenheimer-Snyder dust collapse (1939). The authors were able to match the solution to a static Schwarzschild vacuum solution outside the collapsing dust ball. In the comoving coordinates the dust does not move, and the pressure component in the stress-energy tensor is zero. However, in semi-static coordinates the dust moves and the component is non-zero.
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