Asymptotic Freedom in QCD and...

Discussion in 'Physics & Math' started by RajeshTrivedi, Sep 14, 2017.

  1. RajeshTrivedi Valued Senior Member

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    The color is mine. This aspect of QCD is opposite and unusual to well understood concept that the bond weakens as the separation increases and vice versa.

    I am attempting to associate this opposite behavior to Gravity, but that will be taken up later; the point for discussion here is, if so how a black hole singularity can form? During the formation of singularity (towards point) the separation between quarks (of a Neutron) will reduce, a kind of continued freeing of quarks, will that not result into release of energy and subsequent loss of mass? I think it will and it will not let the singularity form.
     
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  3. NotEinstein Registered Member

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    I don't understand what you are trying to say by "a kind of continued freeing of quarks"?

    The force between two quarks of different color-charge decreases as they approach each other. The energy release will (classically) be less and less (look at the work-integral: the force is getting smaller, so you need less and less work). So the energy release will at best only be a relatively small "slow down" factor during singularity formation. Note that we probably need a theory of quantum gravity to have a proper quantum description of what happens during the formation of a singularity.

    Putting QCD opposite of gravity is going to be tough. Gravity only has one type of charge (mass), and it can only be positive. QCD has three, and they can be negative, and particles can carry multiple different colors. Gravity doesn't self-interact (to the best of my knowledge), QCD does.
     
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  5. origin Trump is the best argument against a democracy. Valued Senior Member

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    I think the answer is no, because the repulsive force of the quarks is not strong enough to overcome the compressive force of gravity.
     
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  7. RajeshTrivedi Valued Senior Member

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    That is true that repulsive force (Electrostatic and due to uncertainty principle etc) cannot counter compression, so this will lead to gap between quarks reducing to zero. What will be the mass of such structure? For example neutron mass is 940 MeV but rest mass of 3 quarks is only around 12 MeV, so when these quarks are compressed by unstoppable gravity, what will happen to 940-12 = 928 MeV? Can we say that now this 928 MeV is released, because quarks are no longer confined (thanks to Asymptotic Freedom)?
     
  8. origin Trump is the best argument against a democracy. Valued Senior Member

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    No, that energy/mass cannot be 'released' (whatever that means exactly) because it is still within the Event Horizon, so it ain't going anywhere. But keep trying...
    You really hate singularities, don't you?
     
  9. RajeshTrivedi Valued Senior Member

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    Continued freeing means reduction in gap and thus reduction in bond strength as per QCD.
     
  10. origin Trump is the best argument against a democracy. Valued Senior Member

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    Hey Rajesh, you know that The God's ban has been lifted so he can post again, right?
     
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  11. NotEinstein Registered Member

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    Ah, I see. I think "loosening" instead of "freeing" would in that case be a better choice of words.
     
  12. RajeshTrivedi Valued Senior Member

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    That is not the case, the release would happen before the core collapses beneath its Event Horizon.
    Consider a Neutron Star of observed mass of around 1.7 Solar Mass, its radius is around 10-12 Kms.
    As per spherical packing (Kepler Conjecture) if 1.7 Solar Mass equivalent Neutrons are packed rigidly then the sphere radius would be around 7.6 Kms; And the Event Horizon of 1.7 Solar Mass object is around 4.9 Kms.

    If due to gravity this type of structure is compressed below 7.6 kms (tightly packed), then quarks will be squeezed, freeing themselves as per Asymptotic Freedom. This happens outside the Event Horizon [Packing Radius 7.6 Kms being > 4.9 Kms Event Horizon], so the release of energy is not inside event horizon, in fact due to this the core will never fall beneath its event horizon, so no singularity via this route.
     
  13. RajeshTrivedi Valued Senior Member

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    https://www.academia.edu/s/eeb19f43f2/gravity-causes-release-of-energy-in-neutron-star

    On the link above the method is fully discussed.
    My point for discussion is that Asymptotic Freedom mandates weakening of Quark Quark bond, this condition will appear in case of mass accretion by Neutron Star. so surely this weakening of Q-Q strength will release some energy or rather will continue to release till the gravitational pressure on innermost Neutrons becomes within limit again, so that Q-Q bond gap is normal again. This release in energy will reduce the mass of core and if this reduction in mass/rate is more or equal to accretion mass/rate then BH will never form.
     
  14. NotEinstein Registered Member

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    As the Spartans famously replied: if.

    I suggest you calculate it!
     
  15. RajeshTrivedi Valued Senior Member

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    If.

    Point is, the rest mass of quarks is less than 12 MeV while the Neutron mass is 940 MeV, so when the quarks are asymptotically freed (or loosened) due to compression by gravity, then what happens to 940-12=928 MeV.

    My proposal is at least a fraction of 928 MeV (if not 100%) should be released and it should find its way out of the inner core of the Neutron Star (as the neutron Star core state is super conducting type), thus reducing the mass. Some one has to logically account for 928 MeV on compression keeping Asymptotic Freedom Concept in mind. Without accounting for 928 MeV, our claim that all will collapse inside Event Horizon appears fuzzy.
     
  16. NotEinstein Registered Member

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    What arguments do you have for neglecting the energy of the gluon field?
     

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