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Magnetic Repulsion between Proton and Electron in Hydrogen

  1. Feb 9, 2005 #1
    Is it possible that the magnetic field of the proton is strong enough to force the electron to stay away from the proton?

    Do neutrons exert more magnetic force than protons?
  2. jcsd
  3. Feb 9, 2005 #2

    Hans de Vries

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    The magnetic force is a factor (v/c)2 lower than the electric force.
    For the lowest hydrogen state that's about 1/1372.

    The reason behind the survival of the atoms can indirectly be found in the rest

    Proton = 938.27203 MeV
    Electron = 0.51099892 MeV

    Neutron = 939.56536 MeV

    So the neutron has more rest mass than Proton and Electron together.
    The energy needed to create a neutron is the same as you need to push
    the electron through an electric field of 782,331 Volt

    Regards, Hans.
  4. Feb 12, 2005 #3
    Doesn't the magnetic repulsion increase exponentially as the two particles approach each other?
    And won't their mutual approach cause photon emission from increased dipole oscillation thus renormalizing the original quantum state?
    Won't an approach also cause an increase in the angular momentum of both particles?
    Last edited by a moderator: Feb 12, 2005
  5. Feb 12, 2005 #4


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    1.What is magnetic repulsion...?

  6. Feb 12, 2005 #5
    The electron and proton both have magnetic moments, therefore they have permanent magnetic fields. The repulsion is between like magnetic poles (fields). Does that explain?
  7. Feb 13, 2005 #6
    I understand that the induced "force" of the apparent electric field of the proton on the electron is 1/1372 larger than the induced "force" of the apparent magnetic field of the mutual interaction. And I'm keeping in mind the fact that the electric and magnetic fields are intimately connected as electromagnetic fields for all massive particles. But I do not understand if this difference in "force" levels has any relevance to the question at hand.

    What I'm asking is: Despite the simultaneous attractive and repulsive "forces" that exist between the electron and the proton in a hydrogen atom, could the induced "force" of the magnetic field be the dominant cause of the stationary state (r=0.53 Ang) of the 1s electron in hydrogen. In other words: The attraction dominates at large distances, but the repulsion dominates at short distance.
  8. Feb 13, 2005 #7
    One thing to remember is that the direction of their magnetic moments depend on their spin direction, and both particles can be in either direction. So there is an orientation where they are repulsive, and an orientation where they are attractive (classically speaking). For QM, this means that there is a higher energy state and a lower energy state because of the proton-electron spin-spin coupling. This gives rise to hyperfine structure in the hydrogen spectrum, see for example: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/h21.html

    So the magnetic repulsion is not responsible for the existence of stationary states in hydrogen, and if you look at the amount of energy involved with the hyperfine splitting, the magnetic interaction causes an extremely small change to the energy levels.
  9. Feb 13, 2005 #8
    I'm thinking time average basis at near relativistic speeds, and I'm also keeping in mind that the electron as a point charge moves at 0.2C as it makes its "cloud", and that the fields of the proton does not always stay in sync with the fields of the electron. I'm assuming that the proton with its "quarks" effectively forms a normal sphere.
  10. Feb 13, 2005 #9


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    That still doesn't change much thing,when it comes to enerrgy levels...Hyperfine interaction energy is much smaller than any correction due to relativity or spin-orbit coupling...


    P.S.What's with that 0.2c ??For what atom...?
  11. Feb 13, 2005 #10
    The question would be easy to answer mathematically using Maxwell's equations if there was an easy way to interperet the current density J due to a proton moving at velocity v.

    Let us try this using dimensional analysis:

    J = q*v/V = p*v where V is colume and p is charge density.

    But what is p for a proton? No one gives me a straight answer on this, and it obviously is pretty critical to the question you are asking. If the proton is a point charge, perhaps it is appropriate to use V = L^3 where L is the plank length. This however would increase current density by a 105 orders of magnitude, a little overkill. (We only need about 14 orders of magnitude).

    The problem is that even if the magnetic force was exactly what magnitude you wanted it to be, you still could not explain atoms this way. Maxwells equations demand an accelerating charge to radiate its energy away, and conservation of energy thus demands the orbit must decay.

    So if you are going to overthrow quantum mechanics, you are going to have to be a bit more creative.
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