Terms of "second order" and "fourth order"...what does this MEAN?!

**AxiomOfChoice** · Jun15-09, 03:42 PM

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?

**tiny-tim** · Jun15-09, 03:54 PM

Originally Posted by AxiomOfChoice

"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?)

Hi AxiomOfChoice!

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

then the nuclear vibrations are proportional to a₂,

the rotations are proportional to a₄,

and a₁ = a₃ = 0.

**maze** · Jun15-09, 07:05 PM

Physicists take taylor series _all the time_ and don't think twice about it.

**AxiomOfChoice** · Jun17-09, 01:47 PM

Thanks guys. On this same subject, when someone notes that "error estimates are $LaTeX Code: \\mathcal{O}(\\alpha)$ " for some parameter $LaTeX Code: \\alpha$ , does this translate into English as "error estimates are order $LaTeX Code: \\alpha$ ?" And, if it does, what does that mean? Does it mean there is a constant

such that the magnitude of the error is less than $LaTeX Code: c|\\alpha|$ as $LaTeX Code: \\alpha \\to 0$ ? And is it understood that

, such that all it amounts to is that, if $LaTeX Code: \\Delta x$ is the error, we have $LaTeX Code: |\\Delta x| \\leq |\\alpha|$ ?