# Asymptotic Freedom in QCD and...

Discussion in 'Pseudoscience' started by RajeshTrivedi, Sep 14, 2017.

1. ### James RJust this guy, you know?Staff Member

Messages:
31,713
Also, any photon produced inside an event horizon (at whatever distance) can only propagate towards the centre; it can never propagate outwards.

3. ### RajeshTrivediValued Senior Member

Messages:
1,525
You are technically ok.
But when a core of mass M is at at its event horizon (just beneath), then it does not necessarily mean that all the inside shells (r < EH) will also be inside their event horizon (the gravitational influence is only of mass inner to the point), and thus a photon produced at any such point can travel away from center. This photon is technically inside the bigger object's event horizon but still outside the inner mass event horizon, so there is no obligation that it must travel towards center only. (of course till it encounters a point which makes it beneath inner mass event horizon).

5. ### RajeshTrivediValued Senior Member

Messages:
1,525
True, but under uniform density distribution, as I stated earlier, when an object is just at EH, even the inner 99.99% shell will be out of its EH, only the outer most surface shall be inside EH. . Technically as the object keeps collapsing more and more fraction will fall under EH.

This point is explained with some simple figures in #61 and then in #66.

Yes, you have grasped it correctly. The photon produced at r = 0, may travel outward till a point where due to collapsing core it encounters the now inner shell beneath its event horizon. So technically it can never come out of original EH, but it can surely run away from center when first emitted at r = 0.

This photon description is not there in any literature available, probably due to the fact that under prevalent GR the ultimate fate of this photon is singularity only.

Now the moot question is under known Physics, by applying the known concepts, can we make a plausible case that the entire structure comes out of its event horizon? There appears a possibility, as we could show that a photon could travel away from the center, which was hitherto ignored aspect.

(I have attempted by resorting to my proposed PSMR hypothesis, but as per James that I may take up in alternative section, on this thread only under known Physics. PSMR = Progenitor State of Matter & Radiation.)

Messages:
21,739

How can you know it has been ignored? I see it and all possible contingencies being considered and some just automatically discarded as irrelevant or a negligible affect if any, hence rarely mentioned. Expletive deleted tried to pull that trick with me in our long drawn out debate on the already accepted Hulse/Taylor binary Pulsar system as the first evidence of gravitational waves, which he just refused to accept and resulting in him crying conspiracy.
That is precisely what the alternative section is for.

8. ### NotEinsteinValued Senior Member

Messages:
1,600
Incorrect. Under a specific uniform density distribution this is true, but not in general. For example: I can have a small metal ball-bearing (which has a uniform density distribution) sitting on my desk without it forming any event horizon either at its surface or outside of it. Likewise, with a high enough uniform density, the event horizon will be outside the object.

My formula clearly states that only one specific uniform density distribution will have the event horizon on its outer surface.

(I'm interpreting the "will fall under EH" as "satisfies the event horizon condition".) True.

(I'm not seeing any images in those posts?)

No worldline exists that can cross the event horizon. The only option is evaporating the mass away through Hawking radiation. I'm not familiar enough with it to answer this question, but it seems like something that probably has already been handled in some literature.

Another important question is whether the inside object there long enough for this to happen: I wouldn't be surprised if the inside object collapses faster than the event horizon can be "evaporated away".

With all due respect, if that hypothesis is bereft of as many essential GR and/or QCD calculations as your previous one, it's not a hypothesis but merely baseless speculation. Might I suggest you strengthen that one first, before diving even deeper into material you clearly are not familiar with?

9. ### NotEinsteinValued Senior Member

Messages:
1,600
And now that I think about it, I'm not so sure squeezing a neutron can release energy, because it doesn't make any sense from an energetic point of view. If there's a lower energy state available, the neutron would transition to that on its own. In other words, if squeezing neutrons into "quark balls" were to release energy as smoothly as stated, all neutrons would simply do that automatically, and there would be none left.

There is no such thing as pressure at the QCD-level: there's only particles (fields) interacting. Increasing the pressure can thus only be done by bouncing neutrons against each other more and harder. This will increase the so-called probing energy, and thus there will be more and more gluons involved (as I showed earlier with the PDF in post #15). These gluons carry momentum, and thus energy. I strongly suspect this effect is bigger than any energy release. In fact, I think you'll find this picture excludes the possibility for any energy release. The equilibrium point between this increased energy in the gluons and the strong-interaction binding keeping the particles together is (no doubt) the stable neutron we see today.

In other words: I think the entire premise you based your hypothesis on is wrong. This is why I asked you to do the QCD calculations!

10. ### RajeshTrivediValued Senior Member

Messages:
1,525
No, its not incorrect. The point is when an object (of any mass M) is at its Event Horizon, then under uniform density assumption all other inner fractional mass shells will be out of their respective event horizon. Simple algebra will yield that when an object of mass M is of size Rs (Schwarzschild Radius), then an inner fractional mass xM (0<x<1) will be out of xM event horizon as the event horizon of xM is at xRs, but the radius of xM mass will be (x)^(1/3)Rs, and (x)^(1/3) > x.

Your formula is correct but you missed the cross referencing of M with Rs.

11. ### NotEinsteinValued Senior Member

Messages:
1,600
Ah, I see, I misunderstood your "an object is just at EH". We are indeed in full agreement on this.

12. ### RajeshTrivediValued Senior Member

Messages:
1,525
Ok, so we agree that when the object is of its event horizon size, then photon released at any interior point can travel towards outer surface (for whatever duration). As the collapse continues (well within outer EH), more and more fraction of this object gets inside "that fraction's" event horizon, till it becomes a singularity and no more outward direction of motion. So I am now working on this finite time from first fall beneath EH till singularity is formed, to see if singularity can be avoided by invoking the accepted Physics.

but before that, I have one more equally important point, and could not get the clarification.

Even though escape velocity is irrelevant in GR, but still its a physical parameter quite relevant around a celestial object. So a reference can be made. On the surface of the earth, the escape velocity is around 11.2 km/sec, it means that an object fired at 11.2 km/sec on the surface will leave the gravity of earth, but it does not mean that an object fired at lesser velocity (say 1 km/sec) will not attain any height, it will certainly move away from the earth and will fall back. Now let us move over to the collapsing "object just at its EH", the escape velocity is c; let a photon be fired from its surface (r=Rs), why can it not travel away from this surface ? (may be like that bullet fired from the earth surface at 1 km/sec).

Even if we try this question in GR, then there is nothing untoward about just formed EH, moreover the curvature of spacetime can be made small by considering a very large mass object at EH, furthermore the possible singularity (on account of r = 2M) in case of Schwarzschild metric can be eliminated by changing to different coordinates. So a bullet fired at r = 2M) outwardly just when the object is also of this size) should be able to move at least a little distance away from surface.

Messages:
21,739
Any photon emitted outwards from the EH, will arc back and fall into the BH, unless it is directed exactly radially away...In that case ignoring everything else and in the FoR of a photon [which is impossible anyway] the photon will hover there never getting away but able to resist falling in. It of course has no bearing on the formation of a BH.
This can be seen with the Waterfall analogy of a BH.
https://arxiv.org/pdf/gr-qc/0411060.pdf

The river model of black holes:

This paper presents an under-appreciated way to conceptualize stationary black holes, which we call the river model. The river model is mathematically sound, yet simple enough that the basic picture can be understood by non-experts. In the river model, space itself flows like a river through a flat background, while objects move through the river according to the rules of special relativity. In a spherical black hole, the river of space falls into the black hole at the Newtonian escape velocity, hitting the speed of light at the horizon. Inside the horizon, the river flows inward faster than light, carrying everything with it. We show that the river model works also for rotating (Kerr-Newman) black holes, though with a surprising twist. As in the spherical case, the river of space can be regarded as moving through a flat background. However, the river does not spiral inward, as one might have anticipated, but rather falls inward with no azimuthal swirl at all. Instead, the river has at each point not only a velocity but also a rotation, or twist. That is, the river has a Lorentz structure, characterized by six numbers (velocity and rotation), not just three (velocity). As an object moves through the river, it changes its velocity and rotation in response to tidal changes in the velocity and twist of the river along its path. An explicit expression is given for the river field, a six-component bivector field that encodes the velocity and twist of the river at each point, and that encapsulates all the properties of a stationary rotating black hole.

14. ### NotEinsteinValued Senior Member

Messages:
1,600
For an object with a constant density distribution, which a neutron star has not.

Yes.

This is one of the reasons escape velocity is a poor analog. Look at the Kruskal diagram: there is no worldline possible where the photon moves away from the surface. Spacetime is warped to such a degree, that all lightlike worldliness from r = 1 cannot exceed r = 1. The calculations tell us this is the case. If your intuition tells you differently, it's because this is a GR environment, and your intuition doesn't work properly there. (Don't worry; this is normal. There are very few people (if any at all) that have an intuitive understanding of GR (or QCD)).

If a singularity can be eliminated by a coordinate transformation, it was never a (physical) singularity.

No, this is you using your intuition, instead of GR. A bullet fired at r = 2M (the event horizon) must travel inward, towards r = 0, and faster than a photon. Look at the Kruskal diagram: that's what must happen (according to GR).
And no, working from Schwarzschild coordinates is no good, because then a bullet cannot be fired: time is frozen at the event horizon when viewed from an observer infinitely far away. (That actually also means it will never travel outward, because it doesn't travel at all!)

15. ### RajeshTrivediValued Senior Member

Messages:
1,525
Let us leave intuition aside.
It is not my case to argue here (because of shear futility), that you consider Kruskal which leads you to absurd white hole, over the Newtonian which may give you results with somewhat higher error bars. you cannot say that Newtonian just not applicable there, after all the motion of objects around a Black hole is being determined by Newtonian only.

Anyways, you fail to consider the salient point, a celestial object irrespective of its gravity (mind you an object just at its EH, has a finite gravitational potential energy for a test particle at EH) will have an escape velocity. Escape velocity is a universal parameter which conceptually supersedes the underlying theory, you cannot say that since we are discussing GR so escape velocity is not applicable or meaningless. It is there.

16. ### RajeshTrivediValued Senior Member

Messages:
1,525
You have rightly understood it, the photon has no rest frame, and 'hovering' suggest rest frame for photon, so this is just one of those inaccurate analogies.

Messages:
21,739
"Hovering"may be the wrong word.....See the River analogy...
https://arxiv.org/pdf/gr-qc/0411060.pdf
Gee, I must be having an illusion.

And of course the futility of it is that you over many years, have submitted papers, made many claims here and elsewhere, to basically invalidate GR and like the many other anti GR proponents that have graced us with their presence, you like them have failed.
Perhaps the old "tall poppy" syndrome is more prevelent then we give it credit for.
Well it certainly at this time remains unevidenced, and unobserved, just as worm holes do, but hey, since we cannot be sure of what is at the Singularity region, all options remain open I suggest. Funny, I have yet to see you comment on the validation of gravitational waves another prediction of GR, and/or the Lense Thirring effect for that matter.

18. ### NotEinsteinValued Senior Member

Messages:
1,600
I fully agree. From this point onwards, only proper GR calculation.

Not mere error bars, simply the wrong answer. Newtonian mechanics breaks down in strong gravity; see the precession of Mercury. That's not just "higher error bars", because they are actually very small. It's the wrong value.

No, in the presence of strong gravity Newtonian mechanics give wrong answers. GR is the model to use.

Please stop using Newtonian concepts in a GR environment. "Escape velocity" is the wrong way to think about this, as it is not a fundamental concept in GR. Worldlines are.

False, as argued earlier.

False, as argued earlier.

But seeing as you disagree: calculate for me the escape velocity under the Schwarzschild metric, and then show how it is a well-defined concept in GR. No more intuition, remember? Demonstrate that your claim that escape velocity is applicable and meaningful is true. Proof me wrong.

19. ### RajeshTrivediValued Senior Member

Messages:
1,525
You are not appreciating that the concept of escape velocity is primary (about the motion of a test particle around any celestial object). So any theory of of gravity must consider and must explain the escape velocity aspect. Say even on earth surface GR equations, with initial velocity of test particle at 11.2 km/sec, must lead to a motion trajectory (whatever you name it) which will make this particle unbound. You cannot say >11.2 km/sec in GR will not make this particle unbound. Can you?

Now, as i said a stone thrown at 1 km/sec (or whatever speed less than escape velocity) will travel upward but will not become unbound, it will come back and strike the earth. so when the object is just at EH, not yet collapsed to singularity, it is still in definable space (not spacetime), then if a bullet is fired from the surface of this object, it will move upward; why? simply because this bullet has a finite gravitational potential energy, and a KE is added to it. No intuition, it is simple conservation law.

20. ### NotEinsteinValued Senior Member

Messages:
1,600
I'll ask you again: please demonstrate that escape velocity is primary in GR. Merely repeating your assertion is dodging the question.

No, and I never claimed as such. The gravity on earth's surface is weak enough that Newtonian gravity is a good enough approximation. This is not the case near an event horizon.

If the bullet is fired from the event horizon, it cannot travel outwards (as seen from an observer infinitely far away); see Schwarzschild. If the bullet is fired outside of the event horizon (and has a high enough speed) away from the event horizon, it will be able to travel to infinity. If the bullet is fired from within the event horizon, it will travel towards the center.

I strongly suspect you are once again using Newtonian concepts while they are not valid (we are in a GR environment), as KE is not a fundamental concept in GR, and neither is gravitational potential energy. Please start using the proper descriptive model.

And it is wrong, because you are using Newtonian gravity in a domain where it is not valid.

Please stop using Newtonian gravity in GR environments.

21. ### RajeshTrivediValued Senior Member

Messages:
1,525
I am not interested in arguing against GR, I am proposing what I consider appropriate.
Of course in alternative section.

22. ### NotEinsteinValued Senior Member

Messages:
1,600
That's fully up to you.

But you are, because you keep using non-GR concepts, like escape velocity, in an environment where GR needs to be used.

And I am pointing out some of the issues in it.

Of a forum called "sci forums".

23. ### RajeshTrivediValued Senior Member

Messages:
1,525
because I do not think that concept of escape velocity or gravitational potential energy can be negated or made irrelevant by GR.
If you refer to literature the concept of binding energy is indeed there in GR, but may be quite complex as the energy conservation aspect in GR is quite complex.
and so far you could not find any issue with whatever i stated, you agreed on first two points and we are stuck with the third one:

1. that when the object is just at its EH, the inner fractions are out of their respective eh.
2. during the collapse a photon produced inside can at least attempt to move away from center.
3. Why a particle fired from EH cannot move up even though it has a finite potential energy when the object is just at EH?

You are dismissing pt#3 by taking shelter that in GR the concept of escape velocity or gravitational potential energy is not there. You fail to appreciate that even the stability equations of neutron star refers to gravitational potential energy (inward pressure), you fail to respond that a satellite launched for moon or mars mission requires escape velocity even though it may be irrelevant in GR. You have failed to make an argument, except hand waving that escape velocity is irrelevant in GR, on pt#3.