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Terms of second order and fourth order what does this MEAN?

  1. Jun 15, 2009 #1
    Terms of "second order" and "fourth order"...what does this MEAN?!

    I am reading the paper written by Born and Oppenheimer that explains the development of the Born-Oppenheimer approximation. The paper contains the following cryptic (to me) statement:

    "The nuclear vibrations correspond to terms of second order and the rotations to fourth order in the energy, while the first and third order terms vanish."

    What, EXACTLY, is a "term of second order...in the energy?" (Or fourth order, for that matter?) I'm sure this is something I should know from freshman calculus, but this vernacular gets used a lot, and my understanding of it is muddled - it just is. Should I feel bad about this?
     
  2. jcsd
  3. Jun 15, 2009 #2

    tiny-tim

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    Hi AxiomOfChoice! :wink:

    It just means that if you expand it as ∑ anEn ,

    then the nuclear vibrations are proportional to a2,

    the rotations are proportional to a4,

    and a1 = a3 = 0. :smile:
     
  4. Jun 15, 2009 #3
    Re: Terms of "second order" and "fourth order"...what does this MEAN?!

    Physicists take taylor series _all the time_ and don't think twice about it.
     
  5. Jun 17, 2009 #4
    Re: Terms of "second order" and "fourth order"...what does this MEAN?!

    Thanks guys. On this same subject, when someone notes that "error estimates are [itex]\mathcal{O}(\alpha)[/itex]" for some parameter [itex]\alpha[/itex], does this translate into English as "error estimates are order [itex]\alpha[/itex]?" And, if it does, what does that mean? Does it mean there is a constant [itex]c[/itex] such that the magnitude of the error is less than [itex]c|\alpha|[/itex] as [itex]\alpha \to 0[/itex]? And is it understood that [itex]c = 1[/itex], such that all it amounts to is that, if [itex]\Delta x[/itex] is the error, we have [itex]|\Delta x| \leq |\alpha|[/itex]?
     
    Last edited: Jun 17, 2009
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