● M
r
m •
|
| rope
|
| ^
| | F force
|
\ o
|
/\ observer
An observer uses a rope to lower a test mass m very slowly close to a mass M. Let us calculate the potential very naively.
The remaining fraction of the mass-energy of a static test mass is
sqrt( 1 - (2 G M / c²) / r ).
The remaining mass-energy is zero at the Schwarzschild radius
rₛ = 2 G M / c².
In the newtonian potential, the remaining fraction of mass-energy is
sqrt( 1 - (G M / c²) / r ).
It is zero at
rₛ / 2 = G M / c².
The horizon radius in the newtonian potential is just 1/2 of the Schwarzschild radius rₛ.
What is the gravitating mass of m?
In the above equations we assumed that the gravitating mass-energy of a test mass m remains m when it is lowered close to M. That is a reasonable assumption if m falls freely toward M and the potential energy which m loses is converted to kinetic energy of m.
But we used a rope to lower the mass m slowly. It did not gain kinetic energy.
The paradox is solved by the fact that the field of M shortens the rope as we lower m downward. The observer can let the force F do work for a surprisingly long additional length of the rope.
The gravitating mass of m and the force F do shrink according to the fraction
sqrt(...),
but that is compensated by the radial metric stretching according to
1 / sqrt(...).
The gravity force F shrinks to zero when m is at the horizon.
A perpetuum mobile? No
Let us lower the test mass m in such a way that first it is allowed to gain kinetic energy. After a while, we let it fall at a constant radial coordinate velocity.
The gravitating mass of m is larger than with the slow lowering procedure. Gravity does more work for the whole trip down. Can we recover more energy this way?
No. If we let m fall freely to some distance r from M, its total mass-energy, measured from far away, is still m c². Lowering that mass-energy from the distance r to the horizon will give the energy m c² to the observer – no more.
Energy is conserved. In this context, the notion of "work done by gravity" is misleading.
If we lower the mass m slowly with a rope, we extract the energy from m at larger radii than in the partial free fall case. That is the difference between the two procedures.
The centripetal force and acceleration
To keep a test mass m on an orbit, gravity must give it enough acceleration. The inertia of the test mass m tries to make the test mass to fly along a straight line.
● M
^
/
• ---------
m
Thus, the centripetal force is gravity. Crucial in this is what is the inertia of m in this configuration.
We have claimed in this blog that the inertia against a tangential acceleration relative to M is
m / sqrt(1 - rₛ / r)
and the inertia against a radial acceleration is
m / (1 - rₛ / r).
Which is the relevant inertia on a circular orbit?
Let us look at circular orbits in the Schwarzschild metric.
There μ = m, because we assume that m is very small. The angular velocity ω in the Schwarzschild metric is the same as in newtonian gravity. Is this a coincidence?
If we believe the Schwarzschild metric, then the gravitating mass of m is its total mass-energy, and its inertia on a circular orbit has to be the same, because m stays on the newtonian orbit.
This is strange. The gravitating mass-energy of a static m is
m sqrt(1 - rₛ / r).
Can its inertia against the acceleration on a circular orbit be that low?
The acceleration on a circular orbit is neither tangential nor radial. The tangential and the radial velocities stay constant. Thus, the inertia might be that low.
The extra inertia in a tangential motion of m relative to M would be a separate "load" that m has to carry, but that load would "naturally" orbit M so that m does not need to exercise a force to keep the load on a circular orbit. The following mechanical model would explain this:
<-- ω
_____ ring of inertia
/ \
| ● M |
\______/
• --> v
m
When m comes close to M, m attaches itself to a "ring of inertia", which slows down the tangential motion of m. Once m is attached to the ring, the extra inertia does not affect a circular orbit of m. The ring rotates around M at an angular velocity ω without any extra effort.
The extra inertia is a property of the common field of m and M. The inertia is not an independent body which could fly around on its own. It is not strange at all that the extra inertia can stay on a circular orbit without any extra effort from m.
Similarly, the extra inertia does not increase the gravitating mass of m.
The circular orbit of a photon around M
Let us calculate the coordinate radius for a circular orbit of a photon around M.
The tangential velocity of a photon is
v = c sqrt(1 - 2 G M / c² * 1 / r).
We have
G M / r² = v² / r
<=>
G M = v² r
= c² r - 2 G M
<=>
r = 3 G M / c²
= 3/2 rₛ.
If we use a newtonian gravity potential, then
v = c (1 - G M / c² * 1 / r),
G M = v² r
= c² r - 2 G M + G² M² / c² * 1 / r
<=>
0 = c² r² - 3 G M r + G² M² / c²
<=>
r = 3/2 G M / c²
+- sqrt(9 G² M² - 4 c² G² M² / c²) / (2 c²)
= 3/2 G M / c²
+- sqrt(5) / 2 * G M / c².
The sensible value is
r = (3/2 + sqrt(5) / 2) G M / c²
= 2.62 G M / c².
Conclusions
The gravitating mass of m close to a mass M is
m sqrt(1 - rₛ / r),
where rₛ = 2 G M / c². The inertia of the gravitating mass alone is the same
m sqrt(1 - rₛ / r).
Also, m feels extra inertia if its tangential velocity or its radial velocity changes. The total inertia of m for a tangential movement is
m / sqrt(1 - rₛ / r)
and for a radial movement
m / (1 - rₛ / r).
It is not a coincidence that circular orbits in the Schwarzschild metric obey the newtonian equation. The gravity force is newtonian, and the gravitating mass of m is the same as its inertial mass, for a circular motion.
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