Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a generic 2-dimensional metric $$ ds^2=E(x,y)dx^2+2F(x,y)dxdy+G(x,y)dy^2 $$ what is the change of coordinates that move it into the conformal form $$ ds^2=e^{\phi(\xi,\zeta)}(d\xi^2+d\zeta^2) $$ being $\xi=\xi(x,y)$ and $\zeta=\zeta(x,y)$? Is it generally known? Also a good reference will fit the bill.

Thanks beforehand.

share|improve this question
    
Geodesic polar coordinates give a diagonal metric, but it's not clear to me that we can always achieve a "conformal" metric... –  Zhen Lin Jul 19 '12 at 11:04
    
Existence of (local) isothermal coordinates on surfaces is a theorem. –  Willie Wong Jul 19 '12 at 11:29
    
@WillieWong: I am aware of this. But can they always be given explicitly in this case? –  Jon Jul 19 '12 at 11:56
    
Jon: that link was more for @ZhenLin than for you. But since you asked: define "explicit". Noting that $\xi$ and $\zeta$ are harmonic functions w.r.t. the Laplace-Beltrami operator, I think for reasonable definitions of "explicit" the answer would in general be "no". –  Willie Wong Jul 19 '12 at 12:04
    
@WillieWong: My problem is that I have the metric given and I would like to turn it into a conformal shape in order to apply a theorem on Cramer-Rao optimal estimators. So, if I would have a quite general result, I should be able to accomplish the task. –  Jon Jul 19 '12 at 12:16
show 1 more comment

2 Answers

up vote 2 down vote accepted

This problem is equivalent to solving the Beltrami equation $f_{\bar z}=\mu f_{z}$ where the coefficient $\mu$ comes from the given metric, as explained on the Wikipedia page linked by @WillieWong. A solution can be sometimes semi-guessed when the coefficient is really simple. You should at least try it. But in general the solution comes as an infinite series involving singular integral operators. This is carefully written out in the book by Astala, Iwaniec, Martin.

share|improve this answer
    
Thanks Leonid. I will check the approach devised in Wikipedia. The reference you gave seems really interesting. –  Jon Jul 20 '12 at 8:35
add comment

Please check Chandrashekhar's Mathematical Theory of Black holes, section 11 of chapter 2.

share|improve this answer
    
Whoever flagged this as spam, it hardly seems that way to me. I don't know anything about this topic, but the cited section of the book appears to be relevant. –  Zev Chonoles Jul 22 '12 at 15:55
1  
Now, ricci1729, since you're new I'd like to give you some advice: make your answer self-contained. Perhaps scan the relevant pages from the book, or link to a copy of the book (as I've done in my comment above), or even better, explain the contents of that section yourself in your answer. –  Zev Chonoles Jul 22 '12 at 15:57
    
@ZevChonoles Sorry, Zev, I was too quick to flag it as spam. However, I did not downvote. –  Matt N. Jul 22 '12 at 16:18
1  
@Zev: relevant, but only a special case. In the case of stationary axisymmetric blackholes, the reduced two-dimensional Riemannian manifold comes equipped with a canonical harmonic function (the so-called Ernst potential); in physics literature this is related to the "hidden symmetry" of the space-time. Once you have a harmonic function, finding its harmonic conjugate (and hence picking out an isothermal coordinate system) is relatively simple. The hard part is in general how to find the first harmonic function (which is related to Leonid's answer). –  Willie Wong Jul 22 '12 at 16:44
    
Yes, the content of this answer is not worth downvote being very near to my aims and of course, flagging it as spam is blatantly wrong. I understand that Chandra was a physicist and the cited book is about physics, but this guy should be treated somewhat better. –  Jon Jul 22 '12 at 17:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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