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Gravity in the Nucleus

  1. Sep 26, 2004 #1

    Andrew Mason

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    Can anyone explain to me why gravity would not be a significant force on the 'surface' of a proton or neutron? A quick calculation shows that the acceleration of a neutron toward another neutron or a proton separated by less than the radius of a neutron, is very large (compared to the radius of the neutron). The acceleration is several orders of magnitude greater than the radius of the neutron [itex]/ sec^{2}[/itex]:

    [tex]G = 6.67 \times 10^{-11} Nm^2 /kg^{2}[/tex]

    1) diameter of nucleus of H is ~ [itex]10^{-15} m [/itex]
    radius of nucleus is: [itex]5 \times 10^{-16} m[/itex]

    2) mass nucleus of H is [itex]1.66 \times 10^{-27} kg.[/itex]

    3) gravitational force and acceleration between two protons in He nucleus is:

    [tex]F = GmM/r^2[/tex]

    [tex]F = 6.67 \times 10^{-11} \times (1.66 \times 10^{-27})^2 \div (5 \times 10^{-16})^2[/tex]

    [tex]F = .735 \times 10^{(-65+32)}[/tex]

    [tex]F = 7.35 \times 10^{-34} N [/tex]

    [tex]F = m a[/tex]

    [tex]a = F / m[/tex]

    [tex]a = 7.35 \times 10^{-34} \div 1.66 \times 10^{-27}[/tex]

    [tex]a = 4.43 \times 10^{-7} m/sec^2[/tex]

    Since the radius of the proton is ~ [itex]10^{-15} m,[/itex] this seems like a significant acceleration, or am I missing something?

    Andrew Mason
     
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  3. Sep 26, 2004 #2

    mathman

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    I won't claim any expertise here, but I understand at the level (subatomic particles) being considered, things don't act like solid balls. Quantum theory rules here.
     
  4. Sep 26, 2004 #3

    Nereid

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    Try doing a similar calculation Andrew, for the acceleration between two protons in a nucleus (make similar, classical, assumptions) ... let us know what you find! :smile:
     
  5. Sep 27, 2004 #4

    Andrew Mason

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    Since the mass of the proton and neutron are the same, the gravitational force is the same. Of course we ignore how we get two protons together and overcome the repulsive electrical forces.

    Andrew Mason
     
  6. Sep 27, 2004 #5
    Of course... That might be what Nereid meant !

    Beside, you cannot only compare an acceleration to a distance, at first it does not make sens. You need either to know either a speed, or a time scale. Gravity must come out negligible in any case. Besides, the strong interaction is named because it is even larger than the EM interaction.
     
  7. Sep 27, 2004 #6

    Nereid

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    Indeed; my point was to compare the numbers for gravitation and EM, using the same calculations: "A quick calculation shows that the acceleration of a [proton] [away from] another [proton] separated by less than the radius of a [proton], is very large (compared to the radius of the [proton])." Within this very restrictive (and unrealistic) set of assumptions, by how many OOM (orders of magnitude) is the EM acceleration greater than the gravitational one?

    (Once you've given us the calculations Andrew, you might like to provide an operational definition of 'negligible' :wink: )
     
  8. Sep 27, 2004 #7

    arivero

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    Hmm one should compare forces to forces, potatoes to potatoes. So nuclear (pions) fermi force is the one to check here.
     
  9. Sep 27, 2004 #8
    I think the easiest way to do so, is to compare potential energies.
     
  10. Sep 27, 2004 #9

    Andrew Mason

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    What is important, it seems to me, is the time required for significant changes in separation to occur. When you are contemplating protons and neutrons in the nucleus, the separation distances are very small. While the earth produces much greater gravitational acceleration at its surface than a proton does at its surface, the time required for signficant changes in separation to be reduced by gravity is much greater:

    Example:
    For an object that is .01 earth radius above the earth (about 60 km), the time required to return to the surface (ignoring friction) is:

    [tex]t = \sqrt{2s/g}[/tex]

    [tex] g \approx 10 m/sec^2[/tex]

    [tex] t = \sqrt{2 \times 120 \times 10^3/10} = 154 seconds[/tex]

    For a proton separated from a neutron by .01 radius of a neutron, the time required to return to the surface of the neutron is:


    [tex] \therefore t = \sqrt{5 \times 10^{-18}/4 \times 10^{-7}} = 3.5 \times 10^{-5} sec.[/tex]

    ([itex]s = .01 \times [/itex] radius of neutron [itex] = 5 \times 10^{-18}m. [/itex])

    ([tex] a \approx 4 \times 10^{-7} m / sec^2[/tex])

    If we are interested in identifying what keeps the nucleus together (as opposed to what keeps a proton together) the essential question is: what are the forces that work against it?

    The magnitude of gravity within the nucleus may be small by comparison to the EM interaction, but how do we know that the EM interaction applies within the nucleus? If it does, then obviously gravity would not be sufficient to keep the nucleus together. But I am not sure that it does. I was hoping someone out there might be able to explain why gravity is not sufficient.

    Andrew Mason
     
  11. Sep 27, 2004 #10
    No, gravity by itself is far not sufficient to overcome EM repulsion. It is only the residual strong interaction potential which keeps the nucleons together. Have you calculated EM repulsion ?
     
  12. Sep 27, 2004 #11

    Andrew Mason

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    I never said gravity by itself was sufficient to overcome EM repulsion. It is dozens of orders of magnitude smaller. But I am not assuming that it has to overcome EM repulsion in order for the nucleus to stay together (ie. once it is together).

    If you put H nuclei together create He, one has to use alot of energy to overcome the EM repulsion (ie. it requires the energy inside a star). But once fusion occurs, do we know that the EM repulsion continues to operate between protons in the nucleus? That would be my question.

    EM repulsion certainly doesn't continue when two protons fuse to produce deuterium (and emit a positron). Does EM repulsion continue when an extra proton is added to the nucleus?

    Andrew Mason
     
    Last edited: Sep 27, 2004
  13. Sep 27, 2004 #12
    Of course EM repulsion continues within the nucleus. That is why heavier nuclei with a lot of protons are unstable - the repulsion is stronger than the nuclear attractive forces. That's also why no elements with more than 100 protons exist in nature - they're very unstable.
     
  14. Sep 27, 2004 #13

    Andrew Mason

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    That doesn't necessarily mean that EM repulsion continues within the nucleus of He. It may be that EM forces have a minimum range: they do not apply within a region of space that is larger than a He nucleus but smaller than a nucleus of Einsteinium.

    I am not saying this is actually the case. I am wondering if anyone can explain why it is not the case.

    Andrew Mason
     
  15. Sep 27, 2004 #14

    Nereid

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    I think you've heard of scattering experiments Andrew, in which a beam of electrons (or protons) hits a target of protons (H nuclei); if you read up on those, I think you'll find that there's very clear experimental data to show that the EM force doesn't weaken at short distances; at least to the experimental limit.
     
  16. Sep 28, 2004 #15

    Andrew Mason

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    Scattering experiments are, for the most part, done with electrons, not protons. The result is that we do not measure electrical repulsion between protons. Perhaps you can explain to me how electron scattering shows that EM force exists within the nucleus and, in particular, that proton-proton repulsion exists within the nucleus.

    If electron-proton EM attraction exists to within a very small distance from the nucleus, why is there not some electron energy at which the electron reaches the nucleus but cannot escape it - ie it joins the nucleus? The quantum mechanical explanation of the electrons in the atom based on the uncertainty principle may be quite correct but begs the question: does EM force really have any meaning when elementary charged particles get very close?

    Andrew Mason
     
  17. Sep 28, 2004 #16

    ZapperZ

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    Andrew,

    When you look at the Schrodinger equation for the hydrogen atom, for example, what do you think that "V" term is?

    Furthermore, you don't use the "uncertainty principle" to accurately solve for the energy states, etc. of an atom.

    As for electron not reaching the nucleus, who said that? There is a difference between the BOUND STATE solution of an atom, where there is a minium, ground state for an electron-atom system, and free electron colliding with a bare proton/nucleus, which can induce an inverse beta decay! In the latter case, you CAN have an electron capture with the appropriate momentum conservation condition.

    Zz.
     
  18. Sep 28, 2004 #17

    Nereid

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    You might like to google on 'proton-proton scattering'; it would seem that there have been quite a few such experiments, over a wide range of energies.

    I'll leave it to a PF member more familiar with this work than I am to say something about the results of scattering experiments wrt your idea that the EM force may have a different behaviour either in nuclei or over short ranges (or both).
     
  19. Sep 28, 2004 #18
    We use quantum electrodynamics to probe the proton structure at very small distance. For instance at 6 GeV, we are under [tex]2\times10^{-17}[/tex]m.
    If course if you insist in saying "what about thousand times smaller than any actual accessible distance" we would have to give up.
     
  20. Sep 28, 2004 #19

    Andrew Mason

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    The Schrodinger equation provides an accurate mathematical model for the quantum mechanical behaviour of the atom, in which EM potential is obviously important. Inside the nucleus may be another matter.

    I agree. But one does use it to explain why the electron doesn't simply 'fall' into the nucleus due to EM attraction.

    But it is quite rare and it is not stable. One might think (classically) that the EM attraction would bring it into the nucleus and keep it there, if EM attraction was that strong inside the nucleus.

    It is assumed that strong nuclear attractive force works against the coulomb repulsion force that exists between protons. This means that the nuclear force is strong only in the region very close to the nucleus. I am looking for the evidence that this is in fact the case. I am suggesting that the same result would occur if the coulomb force had a minimum range (ie did not operate inside a certain the region close to the proton) so that the force which keeps protons together is something much weaker. I am suggesting, if that is the case, that gravity might actually be the dominant force inside the nucleus.

    Andrew Mason
     
  21. Sep 28, 2004 #20

    ZapperZ

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    I'm sorry, but is this still in doubt? A nucleus consists of a bunch of positively charged protons, and neutral neutrons. If there's nothing else that not only counter the coulombic repulsion, but also is way stronger than the coulombic repulsion, don't you think the nucleus would fly apart?

    You can suggest anything you like, but without (i) a self-consistent theory and/or (ii) experimental impetus to suggest that, then you might as well propose that bored angels in their spare time pushes the nucleons together. To allow for what you are proposing, you have to rewrite the whole of Maxwell Equations, since the 1/r potential obviously have to be corrected.

    What I'm puzzled with is that there ALREADY is a verified, consistent explanation/description for the strong force. What is WRONG with it that is causing you to come up with a whole new speculation on why nucleons can stick together in spite of the coulombic force? Did you find a flaw in the Glashow/Salam/Weinberg model that is causing you to refute their theory? Are you proposing that QCD be dumped in favor of your "gravity"?

    Zz.
     
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