UPDATE October 22, 2023: The calculation below about m approaching M contains errors.
We also corrected the corresponding calculation from September 9, 2023. The result there for a radial movement is
1 / γ⁵ * G M / R².
If we believe the tangential calculation below, there
γ⁴ G M / R².
The "average" over x, y, z directions is
(-5 + 4 + 4) / 3 = 1.
This would mean that Gauss's law works: the average gravity of a moving mass M is γ M at the distance R.
----
The most common form of Gauss's law states that if we have static charges, then the flux of the electric field through a closed surface is the sum of the charges inside that surface. The electric lines of force should never break in electromagnetism in a volume where there are no charges. This is a form of Gauss's law for moving charges.
In our previous blog post we suggested that the acceleration of a test charge q also depends on how the electric field of Q affects the inertia of q.
Acceleration of a test mass m close to a moving mass M
We want to find out what is the flux of the gravity field through a closed surface in various situations.
M
● ---> v
R = distance (m, M)
^ ay
|
• m
We assume that
rₛ / R << v² / c²,
and that v is slow. We are interested in how v affects the gravity acceleration. Then, if m and M are both static, we can assume that the acceleration of m is
G M / R².
In our first post on September 20, 2023 we calculated that the acceleration ay in the above configuration is
ay = γ⁴ G M / R²,
where
γ = 1 / sqrt(1 - v² / c²).
M
m • --> aₓ ● ---> v
================ ruler
r
R = laboratory distance (m, M)
Let us then check the acceleration "on the side" of M. We assume that M is carrying along with it a long ruler and m is at a position r on the ruler at the moment we are interested in. Because of length contraction,
r = γ R.
We have to determine the acceleration in the comoving frame of M.
The formula above says that the (pseudo) "kinetic energy" on the left side and the negative "potential energy" G M m / r on the right side have a constant sum. It is like newtonian gravity, with the exception that the time which is used, τ, is the proper time of m.
If v is slow and M / r is small, we can write
dτ² = dt² * (1 - v'² / c²) * (1 - rₛ / r),
where v' is the speed of m as measured a local static observer, and
rₛ = 2 G M / c².
The energy-momentum relation says
E² = p² c² + m² c⁴.
The formula in the square brackets [ ] is
p² / (2 m) + 1/2 m c² - 1/2 m c²
= p² / (2 m),
which for a slow v and small M / r is the kinetic energy of the test mass m if it would fly out of the field of M. In newtonian gravity, the above equation says that the kinetic energy of m is the kinetic energy that m had outside the field of M, plus the energy it gained by descending down in the potential of M.
We obtain
V = dr / dt = sqrt(p² / (2 m) + 2 G M / r)
* (1 + 1/2 v'² / c²)
* (1 + 1/2 rₛ / r).
We have to calculate
dV/ dr
to get the coordinate acceleration aₓ' in the comoving frame of M. Note that
v ≅ sqrt(p² / m² + 2 G M / r),
since v is slow and M / r small.
The second and the third terms contribute the corrections from the change of the rate of the proper time τ of m. When m moves farther away from M, the proper time of m runs faster. The velocity of m increases because of this. The second and the third term contribute equally.
The change dV₃ of V from the third term is
dV₃ / dr = v / (1 - 2 G M / (c² r))³/²
* -1/2 * -2 G M / (c² r²)
≅ v * G M / r² * 1 / c²
= v * m G M / r² * 1 / (m c²),
that is, the relative change in the total energy of m. It is as if m would lose dW / c² of inertia where dW is the work done by gravity on when r changes by dr.
Then
dV₃ / dt = dV₃ / dr * dr / dt
≅ dV₃ / dr * v
= v² / c² * G M / r².
The first term in the definition of V is exactly like in newtonian gravity.
We get
aₓ' = dV / dt
= G M / r² * (1 + 1/2 v² / c²)
- 2 v² / c² * G M / r²
= 1 / γ³ * G M / r²
= 1 / γ⁵ * G M / R².
The Lorentz transformation of the acceleration gives
aₓ = γ² aₓ'
= 1 / γ² * G M / R².
Birkhoff's theorem fails?
● M
--------- "gravimetric" sphere
Suppose that we have a static spherical mass shell whose mass, measured from far away, is M. The sphere is very lightweight.
We put a much larger "gravimetric" sphere symmetrically around M and measure the gravity acceleration on that shell.
Then we use some of the mass-energy of M to make its parts to move at a velocity v toward the center of M. Let dm be a moving part. If the gravity of dm would be
~ γ dm,
where
γ = 1 / sqrt(1 - v² / c²),
then the gravitational pull on the gravimetric sphere would stay constant. Let v be to the x direction. Our calculations in the previous section suggest that the gravity acceleration is
ay = az ~ γ⁴ dm,
and
aₓ ~ dm / γ².
The "average of these is
γ² dm,
not γ dm! This suggests that the gravity M becomes larger by a factor
γ M,
when the parts of M are made to move toward the center. If that is the case, then Birkhoff's theorem fails.
Birkhoff's theorem is a direct result of general relativity. We calculated the gravity using the Schwarzschild metric for each part which moves at a constant velocity. Thus, our result should be a fairly accurate solution of the Einstein field equations. How can then Birkhoff's theorem fail?
The reason might be that general relativity cannot handle the situation when the parts dm of M are accelerated from a zero velocity to the velocity v. The field outside M maybe turns from one vacuum solution to another?
The ADM formalism can be used to prove some kind of Gauss's law for gravity. We have to check how they handle accelerating masses.
It looks like that Birkhoff's theorem holds, after all. See the next section.
The radial stretching of the spatial metric does not affect the average acceleration?
Our "average" of γ⁴ dm and dm / γ² in the previous section is probably a wrong way to calculate. Let us analyze the acceleration which is due to the distorted spatial metric.
\|/ v
--- • --- dm
/|\
• m
Let us have a dm which moves to a random direction from the origin, at a velocity v.
We have a test mass m very far away. What is the average effect of the the stretched radial metric of dm on the acceleration of m?
It is probably zero. We may imagine that dm moves along a circle.
___ ^ v
/ \ |
\___ • dm
• m
After dm has made a full round, its effect on the spatial metric around m is restored. When dm traveled around the circle, the stretched spatial metric made m to accelerate to various directions. It is like dm would be carrying a magnifying glass. When we view m through the glass, its position appears to change as the glass travels a full circle – but the apparent position of m after the full circle is back to the original.
The various locations of dm on the circle have v pointing to any direction: the process does describe the average for different directions of the velocity v.
This suggests that in the average acceleration, we can ignore the distorted spatial metric. The acceleration solely comes from the metric of time.
This also (partially) solves the "back and forth" problem of our first blog post on September 20, 2023. If we have a large number of particles bouncing around inside a mass M, the velocities of the particles have a random direction. Thus, we can ignore the effect of the stretched radial metric around each particle, and only concentrate on the metric of time.
However, there is more to bouncing than a simple circular movement. In bouncing, particles are accelerated, and general relativity seems to have problems in handling the gravity of accelerating masses M.
What is the "pressure" in the stress-energy tensor?
Suppose that we have gas inside a container. Its molecules collide frequently. There is a pressure. But what if we have a single molecule or a small rock bouncing around? Is there "pressure" in the energy-momentum tensor in that case?
What about a cloud of dust moving to one direction? If we add another, overlapping cloud which moves to the opposite direction, is there pressure? We might assume that there are no collisions between dust particles.
Wikipedia does contain the answer to this question. If a particle has an x velocity of vₓ, then the xx "pressure" component is γ m vₓ².
Thus, also one-way movement of matter generates "pressure".
Tolman's paradox of pressure as a source of gravity
J. Ehlers, I. Ozsvath, E. L. Schucking, and Y. Shang (2005) write about Tolman's paradox: can we increase the gravitational attraction of a spherically symmetric mass M by converting it to a spherically symmetric configuration where particles move swiftly, i.e., there is a lot of kinetic energy and a lot of pressure?
If this were possible, it would break Birkhoff's theorem.
We may consider a special case where an initially static M is converted into a spherical shell of particles which move at a large velocity v outward from the center of M.
^
|
•
<-- • • -->
• dm
|
v
A reasonable version of Gauss's theorem for gravity would state that the average gravitational attraction of a particle dm over a spherical shell enclosing the entire system is
γ dm,
where γ = 1 / sqrt(1 - v² / c²). That theorem would make Birkhoff's theorem true.
Has anyone calculated the metric? Did it match the static Schwarzschild metric like Birkhoff's theorem claims?
Pressure does not require much energy. In principle, we could change the pressure p inside M vastly, for a short time, with only a minimal change in the composition of M.
It is believed that a positive pressure attracts test mass m. This makes sense: moving m closer to M stretches the radial distances relative to m, and expands the volume of M.
Charles Misner calculated in 1959 that if we have a static configuration where the positive pressure is balanced with a negative pressure in the walls of the container, then the pressure does not have an effect on the attraction. The paper of Jürgen Ehlers et al. (2005) does the same, but also considers the internal effects of gravity inside M.
This is expected. But what happens in a dynamic case?
If gravity is Lorentz covariant, we should be able to derive the gravity of pressure by Lorentz transforming the Schwarzschild solution around each dm, and summing them, assuming that M is very lightweight and not very compact, and assuming that gravity is reasonably linear.
If Birkhoff's theorem is broken by such a setup, then either general relativity is not Lorentz covariant, or there is magical nonlinearity which miraculously restores the behavior to the one described by Birkhoff's theorem.
We already showed that general relativity does not understand accelerating masses M.
Could it be that the Schwarzschild solution around a spherically symmetric M changes if we accelerate the particles dm?
Then Birkhoff's theorem would remain true, but it would not accept any acceleration inside the system M. But gravity itself causes acceleration in the parts m. The theorem would never be applicable?
A dynamic system almost certainly breaks Birkhoff's theorem
•
/\/\/\/\ springs
• /\/\/\/\/\/\ •
\/\/\/\/
• dm
• m
Consider a spherically symmetric system which is kept from collapsing with a positive pressure p in springs. Besides the mass-energy density ρ, it is commonly believed that the pressure p "generates" gravity and attracts a test mass m.
Let us then release the springs. The pressure is suddenly removed. We expect the attractive force on m to decrease.
Birkhoff's theorem seems to claim that the force must stay constant.
Note that we cannot use our true-and-tested trick of summing Lorentz transformations of Schwarzschild solutions, because the pressure field does not consist of small particles dm.
What are the assumptions in the proof of Birkhoff's theorem?
The derivation of the theorem is quite simple. If we assume that the metric is spherically symmetric and only depends on t and r, then by coordinate transformations, and demanding two Ricci tensor components to be zero in the vacuum area, we see that the metric has to be time-independent.
The only possible time-independent metric is the Schwarzschild metric.
In the previous section we have an example where the metric seems to change with time. This would be a contradiction. How to resolve this?
The proof of Birkhoff's theorem seems to rest on at least the following assumptions:
1. The Einstein field equations have a solution for at least some spherically symmetric configurations. If they would have none, the theorem would be empty of content. Then one could claim that "all solutions" satisfy whatever condition we want to define.
2. The metric of a spherically symmetric configuration is spherically symmetric at all times. This condition is broken in the real world. If we start from a configuration which is not symmetric, then it would take an infinite time to even out all the asymmetry in the metric. Let us assume that the asymmetry in the far field is not significant.
3. The stress-energy tensor outside the system is zero at all times. There is no mystical "longitudinal radiation" which could carry away mass-energy from the system. Neither is there some subtle asymmetry which could do the same trick.
Birkhoff's theorem in our own Minkowski & newtonian gravity model
We do not claim that there exists a "metric" which can capture the gravity interaction. We neither claim that there exists a "field" (like in electromagnetism) which can capture the collective interaction of many particles.
We calculate the interaction "privately" between any pair of particles.
Gravity must preserve the usual conservation laws. We are not sure how that is implemented technically, though. It is as if nature would perform "transactions" which always keep the grand total of energy and some other parameters constant.
Under these assumptions, Birkhoff's theorem may be true for some configurations. For example, if we use a spring to increase the radius of the central mass M very slowly, the force on a static test mass m outside M will probably stay almost exactly constant.
However, if we change the pressure inside M, as described in the previous sections, the gravity force on m will change. It is no problem for us if Birkhoff's theorem is broken in most cases.
Our model does allow kind of "longitudinal waves" in gravity, if we call a spherically symmetric change in the gravity attraction a wave.
We may imagine a general interaction between m and M as a kind of clockwork around M which grabs m, adds or reduces the inertia of m, and exerts a force on m. There is nothing which requires the interaction to satisfy Birkhoff's theorem. Rather, it would be a surprise if the complex clockwork would function in a way that Birkhoff wants!
Conclusions
We should calculate the gravity flux for a moving M through a spherical surface. The calculation is somewhat complex and we will defer it for now. We do not know if gravity satisfies Gauss's law. Birkhoff's theorem would imply such a law.
Tolman's paradox clashes with Birkhoff's theorem. We claimed that by manipulating the pressure inside M we can break Birkhoff's theorem. We will write a new blog post about what it means to general relativity if Birkhoff's law is broken.
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