Do Black Holes End up as Quark Stars? and quantum gravity

[PeterDonis]

$\sqrt{\Delta}=0$ occurs at the event horizon $(r_+)$

Ah, yes, you're right, I was misremembering which coefficient is which.

the static limit is defined by $g_{tt}=0$

Yes, which means that an integral curve of $\partial / \partial t$ is null at the static limit and spacelike inside it. That is why there can't be any static observers (observers following integral curves of $\partial / \partial t$, i.e., who have constant spatial coordinates) inside the static limit.

$v_s=(\Omega_s-\omega)R/\alpha$ becomes meaningless at the photon sphere

Yes, I see, I was mixing up $v_s$ and what you were calling "frame dragging velocity"; the latter is what becomes meaningless at the static limit (since at and inside the static limit there are no static observers).

Let me rephrase some of this in the terminology I am used to seeing. The angular velocity of a ZAMO (zero angular momentum observer) is given by

$$
\Omega_Z = - \frac{g_{t \phi}}{g_{\phi \phi}} = \frac{2 M a r}{\rho^2 \left( r^2 + a^2 \right) + 2 M r a^2 \sin^2 \theta}
$$

This angular velocity is what you have been calling the "frame dragging rate" $\omega$. In the region outside the static limit, where there are static observers (observers with zero angular velocity, whose spatial coordinates do not change), we can convert $\Omega_Z$ to a velocity, the velocity of a ZAMO relative to a static observer at the same spatial location, by basically the same method you are using to convert $\Omega_s$, the angular velocity required for a stable orbit, to a velocity; that is, we have

$$
v_Z = \Omega_Z \frac{R}{\alpha}
$$

where $R$ and $\alpha$ are the reduced circumference and redshift factor as you have defined them. This could be called the "frame dragging velocity", provided we remember that it is only valid outside the static limit.

This also makes clear that what you are calling $v_s$, the "velocity" required for a stable orbit, is not velocity relative to a static observer; it is velocity relative to a ZAMO (since $\Omega_s - \omega = \Omega_s - \Omega_Z$ is what appears in the formula). Since there are ZAMOs all the way down to the horizon, $v_s$ will be meaningful all the way down to the horizon as you have defined it. But we would only be able to define a velocity for a stable orbit relative to a static observer outside the static limit; that velocity would be

$$
v_S = \Omega_s \frac{R}{\alpha}
$$

(note that $\Omega_s$ appears here instead of $\Omega_s - \omega$).

 

[Bernie G]

The detailed answers here mostly describe what happens in the vacuum outside the singularity or object in a black hole. Does anybody want to add something about the following statements from the original poster's paper: "As a result of the continual gravitational collapse of the black hole, a stage will be reached when ... the entire matter in the black hole will be converted into quark - gluon plasma permeated by leptons. . ... We have an ultrahigh energy particle accelerator in the form of a gravitationally collapsing black hole, which can, in the absence of any physical process inhibiting the collapse of the black hole to a singularity, accelerate particles to an arbitrarily high energy and momentum without any limit."

 

[PeterDonis]

Does anybody want to add something about the following statements from the original poster's paper: "As a result of the continual gravitational collapse of the black hole, a stage will be reached when ... the entire matter in the black hole will be converted into quark - gluon plasma permeated by leptons. . ... We have an ultrahigh energy particle accelerator in the form of a gravitationally collapsing black hole, which can, in the absence of any physical process inhibiting the collapse of the black hole to a singularity, accelerate particles to an arbitrarily high energy and momentum without any limit."

In the classical model of gravitational collapse to a black hole, yes, the collapsing matter will heat up, in principle to unbounded temperatures, and will pass through various states of matter appropriate to those temperatures (quark-gluon plasma permeated by leptons is just the state of matter appropriate to a very high temperature).

Defining the energy of individual particles in the collapsing matter is problematic, however, because the region of spacetime in which the collapse occurs is not stationary; in other words, there is no time translation or symmetry, so there is no natural definition of energy. All we can say is that the stress-energy tensor describing the collapsing matter obeys the local conservation law that its covariant divergence is zero.

 

[Bernie G]

In the classical model of gravitational collapse to a black hole, yes, the collapsing matter will heat up, in principle to unbounded temperatures, and will pass through various states of matter appropriate to those temperatures (quark-gluon plasma permeated by leptons is just the state of matter appropriate to a very high temperature).

Defining the energy of individual particles in the collapsing matter is problematic, however, because the region of spacetime in which the collapse occurs is not stationary; in other words, there is no time translation or symmetry, so there is no natural definition of energy. All we can say is that the stress-energy tensor describing the collapsing matter obeys the local conservation law that its covariant divergence is zero.

Even if your analysis has problems defining energy there should be conservation of energy in a collapsing black hole; energy is not created. If the collapsing contents can not be accelerated to infinite energy the resulting object should have a finite size.

 

[PeterDonis]

Even if your analysis has problems defining energy there should be conservation of energy in a collapsing black hole

If you can't define "energy", how can you check whether it's conserved? In a non-stationary spacetime, there is no global definition of "energy", so there is way to even given meaning to the question of whether energy is conserved.

Locally, energy is conserved--more precisely, as I said before, the stress-energy tensor obeys the local conservation law that its covariant divergence is zero. But that's only local; it doesn't give you any way to define a globally conserved "energy".

If the collapsing contents can not be accelerated to infinite energy

This has no meaning because there is no way to define what "accelerated to infinite energy" means. The only local notion of "energy" is the one I referred to above, that the stress-energy tensor obeys a local conservation law. If you actually look at the underlying math, instead of trying to reason using imprecise ordinary language, you will see why this is the case.

 

[Bernie G]

If you can't define "energy", how can you check whether it's conserved? In a non-stationary spacetime, there is no global definition of "energy", so there is way to even given meaning to the question of whether energy is conserved.

Locally, energy is conserved--more precisely, as I said before, the stress-energy tensor obeys the local conservation law that its covariant divergence is zero. But that's only local; it doesn't give you any way to define a globally conserved "energy".
This has no meaning because there is no way to define what "accelerated to infinite energy" means. The only local notion of "energy" is the one I referred to above, that the stress-energy tensor obeys a local conservation law. If you actually look at the underlying math, instead of trying to reason using imprecise ordinary language, you will see why this is the case.

 

[Bernie G]

No matter what math you use and even if your model can't define energy, there's a good argument that the total kinetic energy is less than or equal to Mc^2

 

[PeterDonis]

No matter what math you use and even if your model can't define energy, there's a good argument that the total kinetic energy is less than or equal to Mc^2

Ok, then please give the argument, in detail, with references. I think you will find that, since "kinetic energy" is frame-dependent, the argument does not prove what you think it does.

 

[nikkkom]

this paper


Do Black Holes End up as Quark Stars ?

R.K.Thakur

(Submitted on 25 Feb 2007)

The possibility of the existence of quark stars has been discussed by several authors since 1970. Recently, it has been pointed out that two putative neutron stars, RXJ 1856.5 - 3754 in Corona Australis and 3C58 in Cassiopeia are too small and too dense to be neutron stars; they show evidence of being quark stars. Apart from these two objects, there are several other compact objects which fit neither in the category of neutron stars nor in that of black holes. It has been suggested that they may be quark stars.In this paper it is shown that a black hole cannot collapse to a singularity, instead it may end up as a quark star. In this context it is shown that a gravitationally collapsing black hole acts as an ultrahigh energy particle accelerator, hitherto inconceivable in any terrestrial laboratory, that continually accelerates particles comprising the matter in the black hole. When the energy \textit{E} of the particles in the black hole is ≥102GeV, or equivalently the temperature \textit{T} of the matter in the black holes is ≥1015K, the entire matter in the black hole will be converted into quark-gluon plasma permeated by leptons. Since quarks and leptons are spin 1/2 particles,they are governed by Pauli's exclusion principle. Consequently, one of the two possibilities will occur; either Pauli's exclusion principle would be violated and the black hole would collapse to a singularity

This looks wrong to me. If we disregard GR for a second, there is no limit how small a ball of degenerate matter supported merely by Pauli exclusion principle can be. It's untrue that it can't be compressed.

For example, if you add more carbon to a carbon white dwarf, it _shrinks_. How is that possible? Easy. If some of degenerate electrons are pushed to states with higher momentum, you now can cram more electrons into same volume, without violating exclusion principle. And since there is no limit how high momentum can be, therefore there is no limit to the possible density of electron degenerate matter.

By the same reasoning, degenerate quark-gluon plasma is compressible too without violating exclusion principle.