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Lack of Relativistic Consideration

  1. Oct 30, 2009 #1
    This may be a basic question; I'm not sure because I'm just getting into quantum theory. As I understand it, the electron orbits somewhere around the order of n x 10^6 (based on calculations I've seen). Moreover, the wave model of the electron suggests that it moves at velocity c. From this, I draw the conclusion that relativistic effects should apply since the electron is moving at a velocity close to c (I consider n x 10^6 to be "close"; perhaps this is part of my flaw), or at c. Obviously there's something wrong with this conclusion, but I'm not sure what it is.

    I guess a simplified version of what I'm asking is: why don't relativistic effects affect the orbit of the electron in quantum theory. I have a suspicion it has to do with the uncertainty principle, but I really have no clue. I apologize if there are recent threads on this, but I couldn't go through all of them to make sure. Does anyone have a relatively simple answer (I know this isn't a simple matter), or does anyone have any links to some basic explanations of this?
     
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  3. Oct 30, 2009 #2

    jtbell

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    If that's in meters/sec, it's only on the order of 1% of c. Try calculating (for example) the relativistic time dilation factor for that speed.

    Relativistic effects do show up in atomic spectra, but only at fairly high precision. They help produce what is called "fine structure" in the energy levels and spectra.
     
  4. Oct 30, 2009 #3

    Vanadium 50

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    Short answer - they do. Relativistic effects do influence atomic energy levels.

    For hydrogen, though, the effect is small: v = c/137, but v2 is the figure of merit, and (v/c)2 is about 5 x 10-5. You need to get to middleweight-to-heavy nuclei befoe the effect is large.
     
  5. Oct 31, 2009 #4

    alxm

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    Gold is yellow because of relativistic effects.
     
  6. Oct 31, 2009 #5
    That's very interesting, could you explain?
     
  7. Oct 31, 2009 #6
    The binding energy of an electron in the hydrogen atom is:

    E = -(1/2) α2 m0c2,

    where α = 1/137 and m0c2 is the electron rest mass.

    Exact treatment of the electron orbit requires the relativistic Dirac equation.

    Bob S
     
  8. Oct 31, 2009 #7

    jtbell

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  9. Oct 31, 2009 #8
    Interesting, thanks for the link. Are the relativistic effects significant more because of the size of the nucleus or because of the electron configuration? Or since more protons usually translates to more electrons (and thus they fill higher energy levels), is it a combination of both?
     
  10. Oct 31, 2009 #9
    In my earlier post, I wrote only the first term in the expansion for the relativistic solution for the hydrogen atom. For complete relativistic solution see Schiff "Quantum Mechanics" second edition page 337 eqn 44.27.
    For higher Z atoms, the 1s energy level is

    E = -(1/2) α2Z2 m0c2,

    where α = 1/137 and m0c2 is the electron rest mass.

    As E/m0c2 gets larger, the relativistic effects increase.

    [edit] When I had to once calculate the reduced mass correction for pionic atoms (using the Klein-Gordon wave functions), the difference between the relativistic and non-relativistic correction was significant.

    Bob S
     
    Last edited: Oct 31, 2009
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