Asymptotic Freedom in QCD and...

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

  1. RajeshTrivedi Valued Senior Member

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    Fantastic, thanks for elevating this thread to a nice qualitative discussion.

    Color and bold by me. (pl expand the quote above)
    I agree with your uncolored part in entirety. The Paddoboy's photon hovering link (quite famous once upon a time here on SF) is fine but photon comes inside as soon as even a single additional photon is accreted by BH.

    But I disagree with the color portion. consider the qualitatively most simple density distribution, uniform density spherical symmetry object just inside its Schwarzschild radius. I have given an example say of 1 million solar mass. Except the outer most surface, every other inner part is outside its Schwarzschild radius (Uniform density just at its EH, apply mass / volume formula to get the radius of inner 0.1 - 99.99% mass (any fraction), you will see that fractional sphere is out of its Schwarzschild radius. Now freely apply Newtonian or Birkhoff's (in relativity) and you will see that a photon emitted just inside has no bar in moving away from center. (of course it will remain trapped inside outer EH), and of course I am just going with you, it may be swayed down with the collapsing mass.

    But it does not end here, the object keeps collapsing from 3 million km. And as per my assumptions a stage comes, R(p) = 640 Km, (640 to 3 million is empty and 0 to 640 is densely packed Neutrons, a sphere of uniform density, say), here too as I have shown even when one million solar mass is compressed to 640 Km, the innermost 3.24 Solar Mass is still outside its schwarzschild radius, suggesting that a photon produced anywhere between 0-10 kms (3.24 solar Mass will be within 10 km), has no bar (Birkhoff's or Newtonian) to travel away from r = 0.

    Now You may say that I am questioning the veracity of Kruskal diagram, may be yes, may be no, because this diagram is not talking about any photon produced inside, this diagram is either about the collapse once the object falls below EH or of any incoming particle. I am talking about photons produced at r = 0, while the object is collapsing. Thats why I am also seeking some one else intervention, like you are also seeking.
     
    Last edited: Nov 17, 2017
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  3. RajeshTrivedi Valued Senior Member

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    I am not able to see any argument against my POV, why a photon produced at r = 0, cannot move away from r = 0, under the given circumstances.
    That suggests that either the Kruskal diagram is incomplete or we are not bothered about what happens during the finite proper time of collapse to singualrity once EH is reached.
     
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  5. RajeshTrivedi Valued Senior Member

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    I am not saying quarks are separated. I am saying quarks are brought closer.
     
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  7. NotEinstein Registered Senior Member

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    That's because there are two possible situations:
    1) Spacetime locally is so warped, it's inside an event horizon. In that case, no photon can be produced at r = 0, as per the Kruskal–Szekeres diagram.
    2) No event horizon. In that case, the normal "photon produced inside the core of a celestial body" formulae apply: no need for special treatment.

    I think you can't find anything specific because there's nothing to find: either it's impossible, or it's business as usual.

    Agreed.

    Switching metric is of course a non-trivial thing, so yes, this complicates things a lot.

    Which is (I think) exactly what GR predicts.

    I suppose somebody had to do it...

    Agreed.

    I suppose, but it can't leave the surface of the object. And that's where the Kruskal–Szekeres diagram is valid.

    I think I addressed this in this very post.
     
  8. paddoboy Valued Senior Member

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    I'll E-Mail Professor Bennett Link and ask him for some of his valuable time.
    So you agree that a BH is actually formed, in line with the overwhelming evidence.
    As you have now agreed, obviously a BH is formed, and anything else is rather speculative at best.
    This actually reminds me of an argument I had here a while ago, with expletive deleted, when he railed against my statement and the opinion held by most reputable cosmologists, that the Hulse/Taylor Pulsar binary system was our first real indication of gravitational waves. His skewed view was that it was due to magnetic field interactions. Despite producing at least 6 or 7 arxiv papers showing how and why the conclusion was reached, he claimed that because his skewed invalidated scenario was not mentioned, they [the professional authors of said papers] they therefor did not ever consider it. Obviously it was not considered because either the effect was minimally irrelevant, or did not play any part in the orbital degradation. As we all now know, the H/T orbital degradation and subsequent gravitational wave have of course since been further validated.
    And I showed as per current knowledge that this was similar to what was expected to have occurred, a short instant post BB...that is temperatures were such that quarks existed for a short time as separate entities.

    In summing, it stands that when according to GR, any Schwarzchild radius is reached, then further collapse is compulsory: The Schwarzchild radius/EH is the boundary or perimeter of the massive object whatever that maybe. Every other level of that massive object is within the perimeter or boundary we call the EH and that is the definition of a BH.
     
  9. paddoboy Valued Senior Member

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    In relation to the title of this thread, I'm sure the main conclusion to be reached is the fact that the discovery, known as “asymptotic freedom,” in no way invalidates or even suggests that BHs do not form. The evidence for BHs is strong and decisive and recent discoveries by state of the art equipement that BH deniers obviously have no access to, have shown them in aspects that were previously envisaged...stellar size, Intermediate size, pre BH size [or Neutron/Pulsar] leaving only gravitational waves from SMBHs to be found.
     
  10. RajeshTrivedi Valued Senior Member

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    1,476
    No, Kruskal diagram cannot bar production of photon. That is separate process. What all is happening in the core cannot be determined by this.

    It could be just that it got a miss.
     
  11. RajeshTrivedi Valued Senior Member

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    I proposed as follows..

    1. A BH cannot form via the route of mass accretion by a stable neutron Star. Why? Because the energy release due to Asymptotic Freedom (reduction in gap between quarks) will happen before the core falls beneath its schwarzschild radius.

    And the second point is
    2. If by any means, by extreme dynamic collapse or by any other unknown mechanism, if an object falls below its Schwarzschild radius, then what happens? This scenario is possible only if the fallen core size is > 3.24 solar Mass. In this case the quark gap reduction takes place inside the event horizon [R(s) > R(p)], so huge amount of radiation energy gets created which remains inside the original Event Horizon (kind of fire ball inside).

    I am yet to state how this energy comes out of the original event horizon in the form of huge explosion (astrophysical jets, secondary nova explosion etc), thus clearing the EH in this scenario too.

    Just to give a hint, I will post details soon.
    1. Either the quarks density is less than that of a Neutron.
    2. The gravity vanishes once the gap between quark-quark becomes zero. (Hint: Gravity = Asymptotic Freedom)
    3. Application of simple Newtonian / Birkoff's theorem.
     
  12. paddoboy Valued Senior Member

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    More likely just too unlikely to consider, as per magnetic interactions with the H/T binary Pulsar system, that expletive deleted so raged against, for reasons of a non scientific nature...or an agenda he was harboring.
     
  13. paddoboy Valued Senior Member

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    That has obvious been shown to be false with the observation of gravitational waves from stellar size BHs.
    You do not in actual fact even have a point one. And let me remind you that this is in science and obviously as you are proposing some hypothetical in opposition to mainstream thought, you are in the wrong section.

    And of course whatever it is you are going to post, does not make it fact. People on a public forum, can claim, post or make up scientifically sounding arguments, [some many times here and elsewhere] should firstly be in the correct section, certainly not science, and secondly, be able to answer each and all questions by others re whatever hypotheticals you chose to dream up.
     
    Last edited: Nov 18, 2017
  14. NotEinstein Registered Senior Member

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    572
    Of course it can. In the Kruskal diagram, a photon produced at r = 0 has to instantly be absorbed again, because it literally has no future. In practice this means no photon is ever produced.

    In other words: the singularity doesn't radiate.

    If you disagree with this, please show a light-like path in the Kruskal diagram that starts at r = 0, and goes somewhere else.

    Possibly, but it's also possible the answer to it is so obvious that nobody bothered.
     
  15. NotEinstein Registered Senior Member

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    572
    As you still need to show any of this through GR and QCD calculations, that is pure speculation at this moment. In fact, you haven't even demonstrated it's physically possible! (The squeezing of neutrons.)

    As I've demonstrated (and is very well known), without rotation or charge any object that falls below its Schwarzschild radius will inevitably collapse totally. And it's impossible for anything to come out of the event horizon; this is crystal clear in the Kruskal diagram. The only known way to reduce the event horizon is Hawking radiation, but you haven't even mentioned that.

    What kind of density are you talking about? Mass per volume? Number of particles per volume? Energy per volume?

    Absolute nonsense. QCD and gravity (GR) are violently incompatible: you yourself stated that!

    Please stop using Newtonian physics in a domain where clearly only the application of GR is warranted.
     
  16. James R Just this guy, you know? Staff Member

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    You appear to be posting an Alternative theory in the Science sections. Please choose a more appropriate forum.

    Also, what do black holes and event horizons have to do with QCD?
     
  17. RajeshTrivedi Valued Senior Member

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    1,476
    Yes, GR and QCD are violently incompatible, and thats seeks some changes or further work.

    The Hawking Radiation refers to emission just around EH, I am referring to emission at the center, which is quite likely because during collapse at some stage center will become very dense with extreme interactions happening. You are countering that the photon cannot even produce there, that seems to be untrue.

    I will answer that in a new thread in Alternative.
    But meanwhile under GR what happens to a photon produced at r = 0, when the object is just below its EH and collapsing? I could not get any literature on this.
     
  18. paddoboy Valued Senior Member

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    If a photon could indeed be produced at r=0, it only has one path possible. That is the prime definition of a BH, where obviously you appear doubtful of and subsequently trying to create all manners of weird situations and contradictions to somehow invalidate that accepted scenario.
     
  19. James R Just this guy, you know? Staff Member

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    Nothing happens at $r=0$ under GR, because there is a singularity there. In other words, GR can't describe what is going on at $r=0$ in a black hole.
     
  20. NotEinstein Registered Senior Member

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    572
    So you know that what you are about to post is either groundbreaking, or pseudo-science. And since we both know you can't do GR or QCD calculations, I suggest you take James R's advice about where to post it to heart.

    I know, but Hawking Radiation is the only thing that can "escape" the event horizon. Nothing inside the event horizon matters to the outside (no-hair theorem). In other words, even if you can build stable structures inside the event horizon, they are all insignificant.

    That question has already been answered. If there was a collapse into a singularity (which is inevitable with the Schwarzschild metric), no photons can be produced at r = 0, because there is literally no future for them (Kruskal–Szekeres diagram). If there still is an extended object inside the event horizon, you might produce photons, but they can never breach the event horizon, and, due to mass fluctuations (infalling matter, convection, quantum fluctuations, etc.) they will all fall back to r = 0 eventually.
     
  21. RajeshTrivedi Valued Senior Member

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    1,476
    No, the question is not yet answered. JamesR is also talking about a scenario after the singularity is formed. The question is during the collapse to singualrity. I will retype.

    Consider a core is just at its EH (of the size r = Rs, no more infalling material only the collapsing core).
    A remote observer will see that the core never collapses to singularity (infinite time dilation, eternally collapsing).
    But an observer on the surface will see that the core collapses to singularity in finite time.

    Now the question:
    When the core is at r = Rs, then if a photon is produced near the center (r~0), then what will be the worldline of this photon? Or in simpler terms can this photon at least travel towards the collapsing surface? If yes till what point and if no, why not.


    Note : It is true that once the singularity is formed then no photon can produce at r = 0, but the question refers to a scenario when the singularity is not yet formed, but the core is just fallen inside its EH.
     
  22. paddoboy Valued Senior Member

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    And as we all know according to GR, current thinking, and our models, once any EH is formed and the Schwarzchild radius is reached, whether any photon or anything else, pops into existence, it still only has one path to travel: and that towards the center.
     
  23. NotEinstein Registered Senior Member

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    572
    Ah, okay, then it depends on the density distribution of the energy inside the event horizon. If, for example, you have a collapsing shell, then obviously due to Birkhoff's theorem it will be able to do so. But if the inner regions of said distribution are also inside their own event horizon, then (probably) it won't be possible. At the very least, the photon can never escape that event horizon.

    But let's throw some basic highschool math at the problem, and see what that does.

    The Schwarzschild radius is given by:
    \(R_s=\frac{2MG}{c^2}\)
    where the mass \(M\) is given by the integral over the density up to a specific radius, due to Birkhoff's theorem.

    For a constant density we have \(M=\frac{4}{3}\pi r^3 \rho\), where \(\rho\) now is a constant.

    Combining these two, we get:
    \(R_s=\frac{8G\pi r^3 \rho}{3c^2}\)

    The event horizon condition is thus met when \(R_s\) is larger than \(r\) at that specific \(r\). In other words, when \(R_s\geq r\). Plugging this into the previous equation gives:
    \(r\leq\frac{8G\pi r^3 \rho}{3c^2}\)

    Rewriting a bit:
    \(\rho\geq\frac{3c^2}{8G\pi r^2}\)

    Here we can clearly see one important thing: if density is constant, and there is an event horizon outside the object (or on its outer surface), this condition will not hold for small radii, but it will at some point kick in. A photon produced at r = 0 may be able to travel a little distance, but at some point it will reach a radius where the enclosed mass will be large enough to create an event horizon, and then the photon will not be able to travel to larger radii.

    Note that this is fully compatible with my more hand-wavy derivation that the density needs to drop off if you don't want an event horizon to appear half-way through the object. Also note that this constant density configuration is thus not stable: the entire outer region of the object has met the event horizon condition, and thus must collapse inward.

    And finally note that if I want to be "technically correct", I could point out that since the object is inside its event horizon, a full collapse is inevitable (as demonstrated earlier), and thus every photon eventually has to end up back at r = 0.
     

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