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

1. Jun 15, 2009

### AxiomOfChoice

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. Jun 15, 2009

### tiny-tim

Hi AxiomOfChoice!

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.

3. Jun 15, 2009

### maze

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.

4. Jun 17, 2009

### AxiomOfChoice

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 $\mathcal{O}(\alpha)$" for some parameter $\alpha$, does this translate into English as "error estimates are order $\alpha$?" And, if it does, what does that mean? Does it mean there is a constant $c$ such that the magnitude of the error is less than $c|\alpha|$ as $\alpha \to 0$? And is it understood that $c = 1$, such that all it amounts to is that, if $\Delta x$ is the error, we have $|\Delta x| \leq |\alpha|$?

Last edited: Jun 17, 2009