# Question On Metric

1. Feb 3, 2007

### astronomia84

Question On Metric!!!

DOES ANYONE KNOW HOW CAN PROVE
THIS METRIC/SPACE IS FLAT?
$$ds^2 = (-dx^2 +dy^2)e^{ax+by}$$

general what is flat space?

how i can find a transformation to make this metric in flat form?
:surprised
:zzz:

2. Feb 3, 2007

### cristo

Staff Emeritus
Calculate the Riemann curvature tensor, and show that it vanishes.

3. Feb 3, 2007

### astronomia84

thanks

thanks!!!!

how i can find a transformation to make this metric in flat form?

4. Feb 3, 2007

### coalquay404

You don't actually need to prove that it's flat - just looking at it will tell you that it's flat. More specifically, the metric you've given above is conformal to two-dimensional Minkowski space, so it's said to be conformally flat.

If you have some $(m\ge3)$-dimensional manifold with metric $g_{ij}$ then you can say that another metric $\overline{g}_{ij}$ is conformally equivalent to $g_{ij}$ if there exists some smooth, strictly positive function $\phi$ such that

$$\overline{g}_{ij} = \phi^{4/(m-2)}g_{ij}$$

The scalar curvature derived from $\overline{g}_{ij}$ is related to that derived from $g_{ij}$ by a very famous relationship called the Lichnerowicz equation:

$$\overline{R} = \phi^{-4/(m-2)}R - \frac{4(m-1)}{m-2}\phi^{-(m+2)/(m-2)}\Delta\phi,$$

where $\Delta=\nabla_m\nabla^m$ is the Laplacian derived from the $g_{ij}$-compatible covariant derivative. Using the Lichnerowicz equation is then a simple way to test for the flatness of a conformally related metric.

In your case, however, I suspect you'd be better off by thinking about exactly what type of metric you have. Can you think of anything special about two-dimensional manifolds? Is there some special (local) relationship between two-dimensional manifolds and flatness?

Hint: yes, there is.

Last edited: Feb 3, 2007
5. Feb 3, 2007

### coalquay404

Well, what would the "flat" form of the metric look like? Show us that you're willing to at least think about the problem and we'll help you out.

6. Feb 4, 2007

### astronomia84

INTEREST coalquay404

I DO NOT HAVE MORE ELEMENTS FOR THIS QUESTION.

NOTE: SORRY FROM MY ENGLISH IS BAD.

7. Feb 4, 2007

### astronomia84

Note: Sorry For My English Is Bad.

8. Feb 4, 2007

### coalquay404

I will help you only if you appear to want to help yourself. I've already given you enough hints to solve this problem. If this is homework, I'm not going to do it for you.

Look, let me give you another hint. You have a metric

$$g_{ij} = \phi^2\eta_{ij}[/itex] where $\eta_{ij}$ is the two-dimensional Minkowski metric and where [tex]\phi^2 = e^{ax+by}[/itex] for some constants $a,b$. We've already told you that this metric is conformally flat. Your questions are i) Is the metric $g_{ij}$ flat? ii) Can you find a clever choice of coordinates so that $g_{ij}$ becomes obviously flat? You've already been told how to answer (i): simply calculate the components of the Riemann curvature tensor and show that they all vanish. This is easy; if you don't know how to do it, or if you don't actually know what the Riemann curvature tensor is, then you shouldn't be attempting to solve the question yet. To answer (ii) I've given you a hint. I asked you the following question: If you had actually found some coordinates in which the metric is manifestly flat, what would the metric look like? As another hint, I'll tell you that if you had some coordinates $U,V$ for which the metric was flat, then you would be able to write it in the form [tex]ds^2 = e^{ax+by}(-dx^2 + dy^2) = -dU^2 + dV^2$$

So, what you have to do to solve this is to find some functions $U(x,y)$ and $V(x,y)$ so that the metric can be written in the above form. Can you do this?

9. Feb 4, 2007

### gvk

It is not correct. The conformal metric can be or can not be flat. The term 'conformally flat' does not have any sense. The flat metric can be only euclidian or pseudo-euclidean.
Sometimes looking on the metric, it is difficult to say that metric belongs to euclidean (pseudo-euclidian) type or not.
Trivial example
$$dl^2 = (dr)^2 + r^2 (d\phi)^2$$;
This metric looks non flat (euclidian), but it is flat euclidian metric of the plane in polar coords. On the contrary, the metric
$$dl^2 = \frac{4}{(1+\frac{r^2}{R^2})^2}((dr)^2 + r^2 (d\phi)^2)$$, where R - number, is not flat (and not euclidian), because it is the metric of sphere with gaussian curvature K=1/R^2.
In such cases, the Riemannian curvature tensor helps. It equals always to zero on the flat metric.
There is the well known theorem that the ANY analytic metric of 2D surface can be expressed in conformal form. In other words, on any smooth surface it is always possible to find the orthogonal system (or even many systems) of coordinates with the conformal metric:
$$dl^2 = f(u,v)((du)^2 +(dv)^2)$$.

Last edited: Feb 4, 2007
10. Feb 4, 2007

### coalquay404

What, precisely, is not correct?

Your point being? I went to great lengths to point out that the curvature of a conformally related metric is governed by the Lichnerowicz equation; nowhere did I say that a conformally equivalent metric needs to be flat. Elements of a conformal equivalence class of metrics can of course generate different scalar curvatures - where did I suggest otherwise?

This is patent nonsense. Conformal flatness is a ubiquitous concept in differential geometry and appears in physics everywhere from electromagnetism to general relativity to Yang-Mills theories to string theory, and many places in between.

''Conformally flat'' - 385,000 hits in Google

Again, your point being? Did I not allude to this in the hints I gave to the OP?

Last edited by a moderator: May 2, 2017
11. Feb 4, 2007

### gvk

Well, it was not clear to me what do you mean saying:
Looking at some metric. Sometimes it looks conformally, but obviously it will not tell you that it's flat.
Example which is similar to posted:
$$dl^2 = \frac{(dx)^2 + (dy)^2}{(1-(x^2 +y^2)^2)^2}$$;
Does it tell that it's flat?
Of course, not.

Let see again. The term "flat metric" means exactly the same as "euclidian" or "psuedo-euclidian" metric.
IT's simply the same thing!
But euclidian or psuedo-euclidian metrics are ALWAYS conformal. What is the sense in using "Conformally flat" term?
May be we know "nonconformally flat" metric? It seems not.
If the people use "Conformally flat" billion times, it does not prove its correctness, from my prospective.

12. Feb 4, 2007

### coalquay404

Of course that metric isn't flat. Generally speaking, anyone worth his or her salt can look at a coordinate representation of a two-dimensional metric and, if given the ranges of the coordinates, tell immediately whether or not it's flat. In fact, looking at the metric you've given here, you can spot pretty much straight away that its scalar curvature is proportional to $-1/r=-(x^2+y^2)^{-1}$. At least, I can...

Quite seriously, you haven't got a clue what you're talking about. You want an example of how a metric can be "conformally non-flat"? Fine. A metric $\overline{g}\in\mathscr{T}^0_2(\mathcal{M})$ over an $m$-dimensional manifold $\mathcal{M}$ is conformally non-flat if there does not exist some $\phi\in\mathcal{F}^+(\mathcal{M})$ such that

$$\overline{g} = \phi^2 \cdot \delta$$

or

$$\overline{g} = \phi^2 \cdot \eta[/itex], where $\delta,\eta$ are the $m$-dimensional Euclidean and Minkowskian metrics, respectively, and where the dot denotes pointwise conformal multiplication. As I've said already, the idea of conformal flatness is fundamentally important in much of differential geometry. Conformal flatness, for example, provides a neat dividing line between the spaces of Riemannian metrics over a given manifold which belong to either the positive or negative Yamabe classes. Just because you are not familiar with the concept does not mean that others are not. And finally, this thread was started by somebody who asked for help with some homework. Don't you think you're doing more harm than good by butting in with comments which are, at best, distracting and at worst completely wrong? Last edited: Feb 4, 2007 13. Feb 4, 2007 ### gvk I ask "nonconformally flat", you did "conformally non-flat". Do you see the difference? 14. Feb 4, 2007 ### coalquay404 Yes. Conformally non-flat is gramatically correct, while "nonconformally flat" is not. But feel free to use "nonconformally flat" if you feel more comfortable with it. Regardless, I've provided ample evidence that your claims are untrue, so the name by which you refer to the problem is unimportant. Last edited: Feb 4, 2007 15. Feb 9, 2007 ### gvk The knowing merely the gramatically correct name of something is the same as not knowing anything at all about it. It seems, you did not get my point about "nonconformally flat". I'll try to explain again and keep in mind that astronomia84 asked about specific 2-D metrics and deviations in the area of high dimentions can only be helpful if it demonstrates some general properties of the subject. Otherwise it's irrelevant, at best, and distracting for learning, at worst. Yes, "conformally flatness" is the wide using term. But, in the same time, its name contains ambiguity and the simple terms "conformal metric" or "conformally equivalent" metric sound more lucid. However, it's not a big deal. The most important that this term is usefull only for the (n>2)-dimensional manifolds, where are not too much conformal metrics and variety of others. It does not make sense to use "conformally flatness" for 1- and 2-dimensional manifolds, because 1- and 2-dimensional manifolds are always "conformally flat". There are no nonconformal manifolds (is that gramatically correct?). This was my point. Now let's come back what Astronomia84 asked There are 3 questions. First was answered by cristo: "Calculate the Riemann curvature tensor, and show that it vanishes." The answer to the second question is: Euclidian or psuedo-Euclidian are the only flat spaces. Third one was unanswered. Your hints for the third question are very vague and did not help. Your deviation in the area of "a very famous relationship called the Lichnerowicz equation:" would be very interested for graduates, but nothing to do with those questions. By the way, you are not quite correct here too. The relation between scalar curvatures of two conformal spaces was received long before Lichnerowicz (1925) by Eisenhart. And the last: No, you can not. The scalar curvature is proportional to $-(x^2+y^2)$ Last edited: Feb 9, 2007 16. Feb 17, 2007 ### coalquay404 Quite honestly, you're determined to ignore my point that conformal flatness is a ubiquitous concept in physics. If you don't believe me, think about what you know about string theory. Consider the Brink-Di Vecchia-Howe action for a bosonic string: [tex]S_{\textrm{BDH}} = -\frac{1}{4\pi\alpha}\int d\tau d\sigma (-\gamma)^{1/2} \gamma^{ab}\partial_a X^\mu\partial_b X^\nu\eta_{\mu\nu}$$

where $X^\mu$ are fields on the world-sheet, $\eta_{\mu\nu}$ is the $D$-dimensional Minkowski metric, and $\gamma_{ab}$ is a metric on the world-sheet. The entire damn point of bosonic string theory is that the world-sheet is conformally flat:

$\gamma_{ab} = e^{2\phi}\eta_{ab}$

Being able to choose this conformal gauge is then crucial in all of the nice results that you're familiar with in elementary bosonic string theory. Seriously, this is basic stuff.

No, no, no, no! There is no ambiguity in the name whatsoever. Conformal flatness is an implicit definition of an equivalence class of metrics, where the equivalence relation is defined by the existence of some smooth positive scalar function. This is not up for debate: the term is accepted by everyone I know and I have never seen it disputed in the literature.

You're correct: ultimately, it's not a big deal. However, for the purposes of the discussion at hand, it is crucial since it demonstrates that what you're claiming is verifiably wrong.

This was not your point at all. Your claim was that conformal flatness is always a meaningless term. I've demonstrated that this is untrue.

It doesn't matter whether it's gramatically correct or not since talking about "nonconformal manifolds" is meaningless. You can talk only about a conformal relationship between metric structures on a given manifold.

On the contrary. My hints were perfectly obvious and, more importantly, in keeping with the guidelines of the forum about homework questions.

Again, no. What I presented was a specific form of the conformal factor, $\phi^{4/(m-2)}$ on an $m$-dimensional manifold. The appearance of the $m$ in the exponent is crucial; Lichnerowicz chose a different, $m$-independent expression for the conformal factor, hence my reference to the equation as the "Lichnerowicz" equation. If you compare it to eq. (28.8) in the 1949 printing of Eisenhart's Riemannian Geometry you'll see that the two expressions are fundamentally different. Please try to keep up.

Regardless, this thread has gone far enough off topic. Should you wish to argue the point more, please open a new thread.

Last edited: Feb 17, 2007
17. Feb 18, 2007

### pmb_phy

Calculating all of the components neccesary to show that the Riemann tensor vanishes is a very cumbersome task. It'd make good practice if you are suicidal. :rofl:

Pete