An electron and a proton can not form a neutron because they are short

  1. An electron and a proton can not form a neutron because they are short about 800KeV of energy. If we place an electron 200 fermi from a proton there is a strong electrostatic attraction. As the electron moves towards the proton electrostatic potential energy is converted to kinetic energy of the electron. About 900KeV of potential energy is converted to kinetic energy more than enough to allow the production of a neutron.

    On the against side:
    the accelerating electron radiates photons/energy
    normal matter is made of electrons and proton and neutron we do not see electrons and protons collapsing

    On the for side:
    in normal matter there are no free electrons, they are bound, only in metals and glow discharge tubes do we have many free electrons

    Is this a possible reaction? It looks like it to me. Anyone have any references? Thanks.

    Ed Pell

    p.s. I know the neutron is not stable and will decay in about 13 minutes.
  2. jcsd
  3. Re: e+p->n+neutrino

    You don't gain any energy from converting potential energy to kinetic. The total energy of the electron remains the same - still insufficient to combine with a proton and make a neutron.
  4. Re: e+p->n+neutrino

    The total energy is conserved. That is potential plus kinetic energy is constant. The total energy of the system is the same but the total energy of the electron and proton is higher and the energy of the electric field is lower.

    As the positive proton and the negative electron get closer the field collapses (goes to zero outside a sphere larger than say 10 proton/electron separations in radius).
  5. PAllen

    PAllen 5,890
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    Re: e+p->n+neutrino

    Electron capture is a known decay mode for radio-isotopes. It is this reaction, with energy supplied (ultimately) by greater binding energy of daughter nucleus.

    If you had low energy protons trapped in some hypothetical magnetic bottle, and shot a beam of low energy electrons at them, I would think there would an extremely small but nonzero chance of this reaction occurring. The sources of extreme low probability are (at least) twofold: the likelihood of forming hydrogen (main factor); the fact that it is not energetically favored (neutron decays).
  6. Re: e+p->n+neutrino

    With the proton at rest, this reaction becomes possible if the electron has (total) energy greater than about 1.3 MeV, which means, in fact, that it needs about .8 MeV of kinetic energy. This is slightly higher than the most naive calculation would imply due to the need for the reaction to conserve momentum.

    As for "normal" matter, we don't see this happen often in Hydrogen because the electrons don't have enough energy. As noted above, it does happen spontaneously in certain heavier elements where the nuclear binding energy makes the reaction energetically favorable.
  7. Re: e+p->n+neutrino

    Parlyne, thank you for the answer.
  8. Re: e+p->n+neutrino

    This is only true if the electrostatic field source is external to the two particles. The original post appeared to be talking about the electron being accelerated by the proton's electrostatic force. If the energy did come from that, it would then come off the total energy of the proton. So the particles would still not have sufficient combined energy to form a neutron.
  9. Re: e+p->n+neutrino

    What is the total energy of the proton to start with? It has mass 938.3MeV and the energy of the electric field surrounding it. What is the energy of the field?
  10. PAllen

    PAllen 5,890
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    Re: e+p->n+neutrino

    This is not correct. Consider the COM frame, which is almost the same as the proton frame. Then, when the electron is close to the proton it has high total energy (KE + rest mass). There is no concept that in some frame of reference "you don't count KE because of where it came from". As for conservation of energy, edpell is correct. The resolution is that the system of separated proton and electron has potential energy that can be converted to KE.

    Put a different way, the total energy (including potential energy) of a slow electron close to a proton is less than a slow electron farther from the proton, due to potential energy. You would have to add energy to the former system (slow, close) to get the latter system (slow, far).

    Note, I have not verified any of edpell's calculations of electron KE, but his conceptualization is correct.
  11. Re: e+p->n+neutrino

    But what about the negative potential energy it then has from having fallen down the proton's electrostatic potential well? I'm not suggesting you don't count KE, but rather that you must count the negative PE too.

    Suppose we started with a proton at rest at (0,0,0) and an electron at rest at infinity. Then, our initial energy tally for this system is 938.3 for the proton, 0.511 for the electron, 0 PE, and 0 KE. We can let the electron accelerate right up to the proton, but however much PE we convert to KE we still only have 938.8MeV - not enough to make a neutron (other than very transiently courtesy of Heisenberg).

    And this is a higher energy state that the one originally suggested, with an electron "placed" (which I take to mean at rest (or as close to as feasible)) at 200 fermi. At that location the electron will have the same negative PE as it would had it got there by 'rolling down' from infinity, but without the KE it would otherwise have acquired.

    When I wrote about an external field, I was really thinking of something like a particle accelerator. Certainly if you accelerated the electron enough in this way and then fired it directly at the proton, you could indeed make a neutron (+ neutrino). But then the additional energy is supplied by the accelerator, not the proton's field.
    This is the bit of conceptualisation I'm unhappy with. My reading of the above statement is that you are suggesting that the field energy of the proton's own electrostatic field is additional to its 938.3MeV rest energy. Surely not? Surely the 938.3Mev includes the 'whole' proton, including the energy of its electrostatic field?
  12. PAllen

    PAllen 5,890
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    Re: e+p->n+neutrino

    Consider two scenarios.

    1) An electron moving slow near a proton is hit with a high energy photon giving it a high KE towards the proton.

    2) An electron starting from far away from the proton, gaining high KE
    toward the proton from Coulomb attraction.

    The possible reactions between the electron and proton do not distinguish an any way between these two cases.

    Potential energy is important for energy conservation and binding energy (= mass deficit of bound system). It has no bearing on interactions. If, in the COM frame electron KE is 900 kev near the proton, e+p-> n + neutrino can proceed. It does not matter how it came to be, at all. Period.
  13. PAllen

    PAllen 5,890
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    Re: e+p->n+neutrino

    Another way to look at this. Noting the relativistic formula:

    total 4-momentum = sum (4 momenta of particles)

    and norm of total 4-momentum is invariant mass = system rest mass of particles.

    Then, in COM frame, total 3-momentum is by definition zero. Total invariant mass is proton (KE + electron KE) / c^2. If this is enough to make a neutron, then the reaction can proceed. Invariant mass (as the name implies) is a scalar invariant. It is the same in all frames of reference.

    The reason you don't see e+p -> n + neutrino to any detectable extent (other than isotopes that decay by electron capture or accelerator experiments) is:

    - classically, if the electron has any momentum, even a tiny amount, orthogonal to the line between electron an proton, you get hydrogen (orbit) instead of this reaction.

    - QM: the cross section for this reaction is vanishingly small unless the electron has substantial excess energy.
  14. Re: e+p->n+neutrino

    I used 200 fermi as a starting point because almost all the PE gain is within 200 fermi. The potential energy at 200 fermi is only 7000eV. If it feels more straight forward please feel free to start the electron at infinite, well, 1 meter, so we do not need to wait an infinite amount of time.
  15. Re: e+p->n+neutrino

    In my electron-from-infinity scenario, then, how long would the neutron exist given that, when the total available energy including PE is counted, this is less than it's rest mass - ie the n is off mass shell?

    I did acknowledge that a neutron could be transiently created in this scenario, but wouldn't it then β-decay back to p+e+antineutrino within the timescales enforced by the uncertainty principle?

    Your scenario (1) is different because there there is an external source of energy, the photon. That's the analogue of my accelerator scenario.
  16. PAllen

    PAllen 5,890
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    Re: e+p->n+neutrino

    My scenario (1) is indistinguishable in every way from (2). This is a fact. Think again about my example that a slow electron near the proton represents a system with less total energy than a slow electron far from the proton. Scenario (1) gets to the state of high KE electron near proton via the lower energy state plus a photon; Scenario (2) gets there via converting the PE of distant slow electron to high KE of near electron. Once you have high KE electron near proton, the history of how it came to be is irrelevant. This is a simple fact, not subject to 'opinion'.
  17. Re: e+p->n+neutrino

    You are wrong.

    Any particle creation process is subject to considerations of total energy. In collider physics, we tend to ignore electric potential energy between colliding particles because it is small compared to the kinetic energy at large distance. However, in your proposed scenario 2, the potential energy is of exactly the same magnitude as the kinetic and cannot be ignored. Simply put, the system as a whole does not have enough energy to support the creation of a neutron.

    Think of it this way. Once the proton and electron fuse, what is left to carry the potential energy? That energy was due to the EM interaction between them. Once they're gone, so is that interaction.

    Perhaps it would be clearer to point to the need for minimal substitution in dealing with EM fields in the context of relativity, which automatically includes the potential energy in the energy component of the 4-momentum sum.
  18. Re: e+p->n+neutrino

    I think you are making an unstated assumption that the energy imparted by the photon in (1) is the same amount as that gained by the electron in falling through the electrostatic potential in (2). In that instance, I can happily agree that it doesn't matter where the KE has come from. In the more general case, however, the photon-imparted KE in (1) could be greater (or smaller) than that acquired in (2). Sorry if this sounds a bit pedantic.
  19. PAllen

    PAllen 5,890
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    Re: e+p->n+neutrino

    I thought it was a stated assumption. Looking back I see that it was implied, but not stated.
  20. PAllen

    PAllen 5,890
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    Re: e+p->n+neutrino

    There two aspects to my statement:

    1) If an electron near a proton has a certain KE by either:

    a) an electron near proton with low KE is hit by a photon
    b) reaching the same KE by coulomb attraction from further away

    then these systems are in every way equivalent.

    This is absolutely true. Further, if a certain 'near proton KE' is adequate for a reaction, that is all you need to know.

    As to the total energy balance, agree I made some erroneous statements. The simplest way to state why the energy balance cannot allow electron capture in isolation starting with a distant slow electron is that neutron decay produces distant electron plus anti-neutrino, together carrying high KE. From this alone, you can deduce that 'near proton' electron KE required is much higher than the distant KE because the electron must escape coulomb attraction with 'high' KE remaining.
    Last edited: Jan 7, 2012
  21. Re: e+p->n+neutrino

    b) wouldn't be equivalent to a) as stated, that is considering the electron KE is attained just by the coulomb attraction as part of the system proton-electron, as Parlyne explained there wouldn't be any energy left once the EM interaction is over. I hope what you mean is that a free charge that may have gotten a certain KE by whatever means like magnetic acceleration from some other interaction or similar that include other parties in the total system is of course capable to produce neutrons when interacting with an opposite charge, for instance like would be the case with cosmic rays colliding on earth and producing neutrons.
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