1
$\begingroup$

I am studying the book "Modern canonical quantum general relativity" written by Thomas Thiemann. In this book, it has been written

since $\vec{N}$ is an asymptotic Killing field of $δ_{ab}$ we have $\mathcal{L}_{\vec{N}}q_{ab} = O(r^{−2})$ odd or $O(r^{−1})$ even respectively for asymptotic translation or boost and rotation while $\delta P^{ab}$ is $O(r^{-2})$ odd.

The only thing I want to know is what the meaning of "$O(r^{-n})$ odd/even" is. This is the first time I encounter this thing and I don't know what it is. Any help would be very much appreciated.

Edit:

Actually I know about odd and even functions but lie derivative of the metric is a tensor, what does even/odd tensor mean?

$\endgroup$
2
$\begingroup$

Odd functions have the property that $f(-x) = -f(x$), and even functions have the property that $f(-x)=f(x)$. Geometrically, even functions are a reflection about the $y$ axis, and odd functions are either a half rotation about the origin or two reflections about the $x$ and $y$ axes. This relationship comes from the quality that $x^n$ which have even ($2m=n$) exponents are even, and $x^n$ with odd exponents ($2m+1=n$) are odd. A function may be even, odd or neither.

The Big-$O$ notation indicates the order of the next term in an expansion, i.e. $O(r^{-2})$ which is typically the final term considered and ignored because it is too small or doesn't affect the calculation (due to its evenness, for example). This is the less rigorous use of Big-$O$ notation used by physicists.

$\endgroup$
  • $\begingroup$ Actually I know about odd and even functions but lie derivative of the metric is a tensor, what does even/odd tensor mean? $\endgroup$ – Sepideh Bakhoda Mar 23 '17 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.