Collapse of a spherical shell M of dust
M
--> v <-- v
• • • • • × • • • • •
| |
rs rs
Let us assume that the initial mass-energy of the shell is M, measured from far away. The shell is thin.
The outer edge of the shell M approaches the radius rs(M) where the redshift of a photon sent upward into space becomes infinite. The speed of light, as observed by a faraway observer, is zero at rs(M).
The radius rs(M) is the Schwarzschild radius of M in general relativity. However, in this blog post we do not assume that general relativity is true. The gravity model can be different.
Definition of the gravity "potential". The gravity "potential" of a test mass m at a radius r is
V(r) = E / m,
where E is the energy which can be recovered if we convert m to photons and send them to faraway space. The potential is not defined if photons cannot be sent to faraway space.
The shell is thin. It is reasonable to assume that the outer edge approaches rs(M) earlier than some inner subset M' ⊆ M approaches its respective radius rs(M').
Approach of the outer edge to rs(M) takes an infinite time
Let
V(r)
be the potential at r once the entire shell M has falled inside the radius r.
Assumption 1. The potential V(r) falls essentially linearly at rs(M).
Then the local speed of light c', as measured by a faraway observer, is:
c' ~ r - rs(M).
A faraway observer measures that the time to reach rs(M) is
rs(M)
~ ∫ 1 / (r - rs(M)) dr
r₀
= ∞.
Local versus global time inside M
The coordinates (t, r, φ, θ) that we use can be imagined as the view of a faraway observer.
Assumption 2. The local proper time inside the outer edge of the shell M runs at most at the speed of the empty space just out of the outer edge. The "speed" of the local time is defined relative to the global time coordinate t.
As long as one can send photons to faraway space, Assumption 2 is true. A photon sent from inside the shell M necessarily has a larger redshift than a photon sent from the outer surface of the shell M.
If photons cannot be sent, then we take the assumption as an axiom. One can imagine that something "happens" inside the shell M, even if a photon never can reach the space outside M. An analogous case is a box made of metal. One cannot send a photon out of the box, but there certainly happens something inside the box.
The local speed of light c' can approach zero so fast that a photon from an event X cannot ever reach faraway space. But we can still develop the system forward in time.
Also, we may be able to define the global time coordinate from local phenomena. If a photon can be sent to a close location y from a location x, and back from y to x, then we can synchronize the global time at x and y. If local time runs slower and slower inside M, we may still be able to synchronize global time at locations
x₀, x₁, ... xn,
where x₀ is inside M, and xn in faraway space.
The fate of the collapsing shell: a frozen star
It takes the outer surface of the shell M an infinite time to reach rs(M). The speed of light c' inside M approaches zero extremely fast.
The interior of M is essentially frozen. No singularity can form.
A collision of two frozen stars: a collision of two chunks of "syrup"
In this blog we have earlier investigated the collision of two frozen stars. We had the "syrup" model of gravity, in which two large chunks of syrup can move fast, even though a small test mass m is frozen at its position inside the syrup. We wrote about the merger of two black holes on June 26, 2023.
The frozen star of the section above can be imagined as syrup whose viscosity is increasing extremely quickly.
But we can still accelerate the star, and make the entire star to move very fast. What happens if two such chunks M of syrup collide?
__ __
/ \ / \
| | | |
\__/ \__/
--> v v <--
M M
The speed of both chunks is relativistic at the collision. They have length contracted.
Let us call an event horizon a surface around both M, such that the redshift from the surface is extremely large.
Is it so that both chunks of syrup will sink quickly below, or extremely close their common event horizon?
|
• • • • • --> particles falling in
• • • • •
|
horizon
The event horizon cannot "support" anything a large coordinate distance above the horizon?
The radial spatial metric is very much stretched close to the horizon in general relativity.
In the diagram above, each particle • is falling at a relativistic local proper speed. The local speed of light in global coordinates t and r becomes extremely slow close to the horizon. The distance of each particle from the horizon probably falls exponentially in t.
Thus, all the particles will very quickly come extremely close to the horizon, which also moves outward as more particles fall in. Many particles end up below the horizon.
We conclude that both masses M very quickly end up either inside the horizon, or extremely close to the horizon.
What is the form of the horizon, after some global time t has passed?
An extremely slow speed of light c' close to the horizon distort greatly the lines of force of the (newtonian) gravity field. In many gravity models, the horizon will be perfectly spherical. This is because the distortion tends to pull oblique lines close to the horizon. Outgoing lines of force must go radially straight from the center of the system. The energy of the field in many models is then minimized by uniformly spaced lines of force leaving a perfectly spherical event horizon.
In particular, the event horizon is spherical in general relativity.
There may be event horizons embedded inside each other. The horizons of the individual masses M are, in some sense, preserved under the larger, common event horizon.
The field of a rotating merger of frozen stars
^
| M
● ●
M |
v
Let us then analyze what happens in a collision where two frozen stars M initially orbit each other.
This is a much more complicated setup than a head-on collision. Some aspects:
1. How does gravitomagnetism or frame dragging affect the process?
2. What is the formula for gravitomagnetism?
3. What formula should be used for the centrifugal "force"?
4. Is the gravity potential newtonian, or is it steeper as in general relativity?
5. Once the stars have merged, is the gravity field rotationally symmetric, and does not radiate quadrupole waves any more?
The LIGO group used computer models to calculate approximate solutions. Our May 21, 2024 results show that the Einstein field equations do not have a solution for any "dynamic" system. Thus, the LIGO group is not solving the Einstein equations, but something else. They are, at least, calculating corrections to the newtonian orbit from the Schwarzschild metric around each orbiting mass.
Each frozen star M is very viscous "syrup", but as a whole, each star can move (and probably spin) like any other body floating in space. Therefore, the initial approach of the stars can be handled with post-newtonian approximations and the slowdown which comes from the loss of energy and momentum to gravitational waves.
As the stars come closer to each other, so that they would be partially inside their common event horizon, what happens then?
In a preceding section we argued that any matter outside an event horizon can approach the horizon extremely quickly, so that the distance shrinks exponentially, as measured by a faraway observer.
If that is the case, we still have to prove that the gravity field approaches very quickly a rotationally invariant form. Otherwise,we would not have the quick "ringdown" in the LIGO observations.
The gravity field of a nonrotating frozen star is spherically symmetric
In our blog we have earlier argued in the following way for a non-rotating black hole: Since the potential is very low close to the event horizon, or inside it, we can essentially draw the lines of force of the newtonian gravity field as we like inside that volume. Twisted lines inside add very little to the energy of the field, since they are at an almost zero potential (the redshift is extremely large).
Then, the way to minimize the energy of the newtonian field is to draw the lines of force at a uniform density from a spherical surface enclosing the mass. The energy of the field is dominated by the lines of force which are outside the event horizon.
The event horizon cannot be inflated arbitrarily far from the black hole because the gravity potential has to be zero at its surface.
This is similar to adding an electric charge to a flexible metal mesh. The smallest energy is attained when the mesh is inflated to a spherical shape, and the electric field lines emanate at a uniform density from the surface.
This analysis holds for many gravity models. The essential thing is that the field wants to find its minimum energy just like a Coulomb field, and the gravity potential is very low at the surface of the mass configuration. In general relativity, a very heavy, and an extremely rigid neutron star already has a gravity field which is spherically more symmetric than the star itself.
The rotating gravity field of two merged frozen stars
Let us assume that a common event horizon already encloses the two merged stars.
If the rotation is slow, then it seems plausible that the event horizon is almost spherical and the gravity field is almost spherically symmetric.
For a rapidly rotating merged frozen star, or a merged black hole, the energy of the gravitomagnetic field has to be added to the energy of the Coulomb-like newtonian gravity field. The newtonian gravity field lines should be less dense at the equator.
What about the rotational symmetry?
The field energy outside the horizon is probably minimized by making the field symmetric around the rotation axis. The field lines "repulse" each other, and the horizon forms where the redshift approaches infinity.
ω
-->
●● merged frozen stars
The merged frozen stars may form a rotating dipole inside the common event horizon, like in the diagram above. But their gravity field will be rotationally symmetric outside the event horizon. Therefore, very soon after the merger, gravitational waves are no longer emitted.
Since the newtonian gravity field lines are sparser along the equator of the rotating system, the event horizon has a smaller radius at the equator than at the poles of the rotating frozen star.
The effect of the centrifugal "force"
The analysis is complicated by the centrifugal "force" at the equator of a rotating frozen star. How does it affect the event horizon?
If we shoot a rocket of a mass m out vertically from a launchpad on the equator of Earth, the rocket gets
1/2 m * (460 m/s)²
of kinetic energy from the rotation of Earth. The escape velocity is reduced, thanks to the extra kinetic energy.
If we shoot a photon vertically from the launchpad, the Doppler effect blueshifts it.
The event horizon has an infinite redshift. Can the Doppler effect affect the radius of the horizon at all?
Yes, it can. The horizon is at the radius where the potential of a test mass is zero. The centrifugal "force" can cancel a significant part of the gravity attraction.
Another aspect is the time dilation. If the source of a signal moves at a large velocity on the event horizon of a frozen star, then there is a significant "redshift" from the time dilation.
What is the local rate of time in an accelerating system? Where is the event horizon located for a rotating mass M?
Let us make the following thought experiment:
v <-- • ● ----> rapid acceleration
m M
We have a test mass m close to a nonrotating frozen star M. Then we pull M extremely fast to the right. The escape velocity of m to the left is then surprisingly slow. Does this mean that the "potential" at m was surprisingly high, and the local time at m ran surprisingly fast.
Obviously, not.
In a dynamic system like above, it is hard to define the gravity "potential". The local rate of time cannot then be bound to the potential.
In the static Schwarzschild metric, the rate of local time is determined by the potential. Why? Is it a coincidence?
According to our Minkowski & newtonian model, the slow rate of a mechanical clock close to M is due to two factors:
1. the inertia of the clock parts is larger, and
2. the energy stored into a spring in the clock is less because the spring is at a low gravity potential.
Now we face a dilemma: how do we define item 2 in a dynamic system?
Maybe we should calculate the "current" potential of m in the field of M, and ignore any acceleration.
What if M is rotating? Frame dragging may force m to move along with the surface of M. Can we still ignore the acceleration of the mass elements in M?
For a rotating electric charge Q, the rotation creates a magnetic field B, but it does not affect the electric potential.
• --------> • rotating heavy mass M
^ |
| v
• <--------- • dm
<-- • m test mass
Let us analyze the diagram above. The rotating mass M is very heavy and the local coordinate speed of light c' is very slow. The test mass m must move along the surface of M.
What is the gravity "potential" at m? Let m carry a mechanical clock. Its parts have a lot of extra inertia. What is the energy available for the springs of the clock to move the clock parts?
The simple guess is that we have to add the negative potentials of each mass element dm in the large mass M, to obtain the potential at the test mass m.
In our Minkowski & newtonian gravity model we are working in an inertial coordinate system (= canonical coordinates) of the underlying Minkowski space. The kinetic energy of each dm probably has to be added to the gravitating mass of each dm.
In this simple model, the gravity potential is indifferent to the acceleration of the elements dm.
M
<----- • dm
<----- • m
The test mass m flying along with dm at the surface of M does not "see" the kinetic energy of dm.
An option is to calculate the mass-energy in M in inertial, for the moment, comoving coordinates of m. That might give at least an approximately correct gravity potential.
Where is the event horizon of a rotating mass M located:
1. The event horizon can be defined as the largest radius where the coordinate speed of light, c', is extremely slow in the radial direction toward the center of M.
2. Another definition is where the local time runs extremely slowly relative to the coordinate time.
3. Yet another definition might be that a photon sent from the horizon obtains an extremely large redshift in faraway space.
The definitions 2 and 3 are equivalent. Though, one photon at the sender may be many photons at the receiver.
Conclusions
We outlined what happens in the collapse of a nonrotating star. The end state is a "frozen star", where the coordinate speed of light is extremely slow, and becomes ever slower as time progresses. The end state is not ordered. It is not a "crystal", but an amorphous solid. Like glass at an extremely low temperature.
We also described why the event horizon of a nonrotating frozen star is perfectly spherical – not just in general relativity, but in a wide range of models of gravity.
We sketched how two orbiting frozen stars merge. The event horizon will be rotationally symmetric, for the same reason why the horizon is spherical for a nonrotating frozen star.
We tried to determine the shape of the event horizon of a rotating frozen star, but were not able to find how it should be calculated exactly.
No comments:
Post a Comment