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Why can't SR explain why electrons do not crash into the nucleus?

  1. Mar 17, 2005 #1

    Andrew Mason

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    In a recent discussion in this thread I wrote:
    Using classical mechanics and electomagnetism, work out the speed that an electron would have to have in order to orbit a hydrogen nucleus at a distance of 10^-12 m:

    [tex]F_c = \frac{m_ev^2}{r} = \frac{kq_e^2}{r^2}[/tex]

    [tex]v = \sqrt{\frac{kq_e^2}{mr}}[/tex]

    where:
    [itex]r = radius of orbit = 1e-12 m[/itex]
    [itex]k = 9e9 Nm^2/C^2[/itex]
    [itex]q_e = 1.602e-19 C.[/itex]
    [itex]m_e = 9.1e-31 kg[/itex]

    v works out to 1.6e7 m/sec or about 5% of the speed of light.

    Now work out what the radius of orbit could be if the electron travelled at the speed of light. This would obviously be the minimum orbital radius permitted by relativity. It would take an infinite amount of energy for an electron to get arbitrarily close to the speed of light.

    I get r = 2.5e-15 m. or 2.5 Fermi units

    The radius of a proton is about .5 Fermi. To reach a 2.5 Fermi radius of orbit, the electron would need an infinite amount of energy. So an orbiting electron simply can't get enough energy to crash into the nucleus!

    I assumed that this non-QM explanation was wrong and that I was missing something obvious somewhere. I have tried to figure out why this is not at least a plausible explanation. I can't.

    We know from celestial mechanics that when orbiting masses trade potential for kinetic energy, they adopt a smaller radius of orbit and speed up. Since the object that they are orbiting is large compared to the radius of orbit, a sufficiently reduced orbit radius means that they crash.

    But in the case of gravity (black holes excepted) the speed of the orbiting object is not enough to change its mass appreciably (ie v<<c). So there is no minimum limit to the radius of orbit.

    It is a very different matter with an electron orbiting a proton. The orbital speed at an atomic radius [itex] \approx 10^{-12} m.[/itex] is relativistic, as I have shown. If it trades potential energy for kinetic, it speeds up but its mass increases as [itex]\gamma = (1 - v^2/c^2)^{-1}[/itex] increases so the radius of orbit approaches a limit that is greater than the radius of the proton.

    What is wrong with this argument?

    AM
     
  2. jcsd
  3. Mar 17, 2005 #2

    Tom Mattson

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    The problem with your argument is that at no point to you address the real problem, which is that the electron loses KE due to electromagnetic radiation (accelerating charges lead to EM waves, which carry away energy). So at some point, according to the classical model, the electron will simply stop moving around the nucleus and fall in.

    According to GR, a similar thing happens with orbiting masses, except the waves are gravitational. But they carry away so little energy that the existence of long-lived solar systems is compatible with this prediction.
     
  4. Mar 17, 2005 #3

    learningphysics

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    From what I see, I think you're falsing assuming centripetal motion. Meaning, for the electron to crash to the nucleus, it isn't necessary that its velocity satisfy the relationship:
    [tex]F_c = \frac{m_ev^2}{r} = \frac{kq_e^2}{r^2}[/tex]

    This relationship can be assumed when the electron is in a specific orbit, but while crashing down into the nucleus this relationship does not need to be maintained. There will be a spiraling motion, and I'm not sure how exactly the velocity will change as it spirals down.

    I might be wrong. This is what I saw at a first glance.
     
  5. Mar 17, 2005 #4

    Andrew Mason

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    This objection is based on a questionable assumption that an electron orbiting a proton will radiate energy because it is accelerating (See: my post here). I don't believe that that has ever been proven. It seems not to be the case for a charge in gravitational orbit, although as you point out, it is very difficult to measure this because the radiation would be so small. But, given the equivalence of gravitational acceleration and inertia under GR, if an electron in a gravitational field radiated because it was accelerating it would also have to radiate if it was not moving at all, which seems to be contrary to em theory.

    AM
     
  6. Mar 17, 2005 #5

    ZapperZ

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    No, it is WELL-KNOWN that an electron is a CIRCULAR motion will radiate EM radiation - refer to cyclotron and synchrotron radiation. This is NOT of the type "electron in gravitational field does not radiate" problem. There's ZERO ambiguity as far as confirming this experimentally, or else all those synchrotron centers all over the world are imagining the radiation they are using.

    Zz.
     
    Last edited: Mar 17, 2005
  7. Mar 17, 2005 #6

    Tom Mattson

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    It definitely has been proven that a charge in orbit radiates. The rate at which energy is radiated is derived in Appendix B of Eisberg and Resinck's Quantum Physics, in case you'd like to look it up.

    In fact it is a standard exercise in an upper-level undergraduate EM course to derive the lifetime of an atom in classical EM theory. It is about 10-12 seconds, if memory serves.
     
  8. Mar 17, 2005 #7
    Hi,

    I think that it may be possible to assume that the electron in orbit gets its energy replenished by the fields emanating from the nucleus.

    juju
     
  9. Mar 17, 2005 #8

    Tom Mattson

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    I don't see how. In the lab frame, the nuclear EM field is electrostatic and provides a central force. And as is well known, central forces don't change the speed of an orbiting body. So that begs the question, what is it in an atom that you think can replace the KE lost to EM radiation?
     
  10. Mar 17, 2005 #9

    Chronos

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    I might be math challenged here, but I'm coming up short on the orbital velocity. If I take the Bohr radius, 5.292E-11, and electron mass, 9.11E-31 and do a couple quick substitutions:

    [tex]r = \frac{\lambda}{2\pi}[/tex] where

    [tex]\lambda =\frac{h}{mv}[/tex] converts to

    [tex]v = \frac{h}{2\pi rm}[/tex]

    which yields 2.188E6 m/s.
     
  11. Mar 17, 2005 #10
    Hi,

    The central electrostatic field may not change the speed of the orbiting electron, however it may be that it is this field that keeps the electron orbiting at a constant velocity by replenishing any energy lost to radiation.

    Alos, it would seem that the dynamics of the central force interacting with the electron charge would have to keep the electron in a constant orbit, in spite of any EM radiation by the orbiting electron.

    juju
     
  12. Mar 17, 2005 #11
    Can't we just say this : the electron closest to the nucleus has the highest kinetic energy and the lowest (most negative) potential energy. It is most tightly bound to the nucleus but it moves around at the highest speeds compared to the other electrons in higher energy levels.

    Now, i don't really know the calculations that prove this, i am sure i once saw them in a QM-course in college but i should look them up. There is no radiation because there is an equilibrium between the two energies. Isn't this correct ? Anyone knows the exact calculations that prove this.

    marlon
     
  13. Mar 17, 2005 #12

    Tom Mattson

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    I seriously doubt that you are talking about the same electrostatic force that is described in the Maxwell theory. If you are, then I would request that you show the mathematical form of the interaction that you think supplies the energy.

    I don't think so. If you treat the EM radiation as damping, then the orbital radius should shrink. And even if the radius doesn't shrink, the KE bleeds away so that the electron eventually stops. Certainly at that point it would just fall straight into the nucleus.
     
  14. Mar 17, 2005 #13

    Andrew Mason

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    There seems to be controversy over this. See:
    http://ernie.ecs.soton.ac.uk/opcit/cgi-bin/pdf?id=oai:arXiv.org:gr-qc/9303025
    http://ernie.ecs.soton.ac.uk/opcit/cgi-bin/pdf?id=oai:arXiv.org:gr-qc/9711027

    The disagreement seems to be over whether a uniformly accelerated charge radiates. It may be noted that in a synchrotron, cyclotron or betatron or linear accelerator, the acceleration is not uniform. Nor is it strictly a central force.

    AM
     
  15. Mar 17, 2005 #14

    ZapperZ

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    So how is this different than an electron "orbiting" a nucleus? It is a "uniform" acceleration? And how is a cyclotron not "strictly a central force"? This "non-strict" property is ALL there is that would cause such a system to radiate?

    We have gone through something like this a long time ago. Again, I would seriously question the validity of the assumption of an electron in an "orbit" around the nucleus. All you need to do is tell me how I can have something like that having ZERO angular momentum per the s-orbital. And this is the SIMPLEST case since I haven't yet brought up how, for example, you would explain not only the geometry of the d-orbitals, but the fact that such orbitals have a PHASE associated with different "lobes" that strongly affects how such an atom forms bonds with other atoms.

    Zz.
     
  16. Mar 17, 2005 #15

    Hans de Vries

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    Your calculation is OK. There's a shorter way to write this:

    v = αc = 2187691.26 m/s

    where α = 1/137.03599911


    Regards, Hans
     
  17. Mar 17, 2005 #16

    learningphysics

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    You used r=5.292E-11. Andrew used r=1E-12. Hence the discrepancy.
     
    Last edited: Mar 17, 2005
  18. Mar 17, 2005 #17

    Hans de Vries

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    You can find a derivation of the E field of a moving, accelerating charge here:

    http://fermi.la.asu.edu/PHY531/larmor/

    See formula 19 which is equal to Jacksons Equation 14.14. which is
    also the one used in your refs.

    If you now simplify (19) to that of a non moving particle then you get this:

    [tex]E = \frac{q\hat{r}}{4 \pi \epsilon_0 r^2} - \frac{q}{4 \pi \epsilon_0 r} \frac{\vec{a}}{c^2} cos (\phi)[/tex]

    Where the first term is the usual Coulomb term and the second term
    is caused by the acceleration a. Now if you go back to (19) then you'll
    see that there's no way to get rid of the second term by choosing an
    arbitrary speed.

    Non radiating charges would need a modified EM theory at small distances.


    Regards, Hans.
     
  19. Mar 17, 2005 #18

    Andrew Mason

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    In a cyclotron the electron has a linear acceleration as it crosses between the Ds (and then a uniform acceleration due to the magnetic field that is perpendicular to the direction of motion).

    When you go beyond one electron orbiting a proton, there is definitely a problem. I am just concerned with the simplest case of the H atom.

    AM
     
  20. Mar 17, 2005 #19

    ZapperZ

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    So you DO have a "uniform acceleraton" from a cyclotron. Yet, you are arguing that such things do not "radiated".

    And you are not bothered by this "problem"? And unless you have forgotten, such issues are ALSO relevant to H atom! You will have a hell of a time trying to explain the H spectral lines (shall we go over the selection rules?). And take 2 H atoms together and BAM! You have bonding-antibonding bonds! I'd like to see you take your orbit and explain that.

    Zz.
     
  21. Mar 17, 2005 #20

    Tom Mattson

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    Not quite. He's saying that there is a nonuniform acceleration between the D's.

    Let's not forget that the electron orbit is not going to be exactly circular. That is only the case if the nucleus is infinitely heavy.

    Those papers you linked me to are a lot to go through, and I'd like to take my time with them. I think I'm overdue for my periodic review of Jackson anyway, so I'll print them out and read them alongside for comparison.
     
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