Coordinate transformation and conformal transformation?

[ismaili]

I'm confused by the relation of coordinate transformation and conformal transformation. I found a nice note about conformal field theory written by David Tong. It does contain the demonstration related to my question, but I still don't understand. Here it goes,

The definition of conformal transformation is a coordinate transformation \sigma^a \rightarrow \sigma'^a(\sigma), such that the metric changes by
g_{ab}(\sigma) \rightarrow \Omega^2(\sigma)g_{ab}(\sigma)

What I'm confused is, Is the conformal transformation a subset of coordinate transformation? If this is true, then the conformal transformation is just part of the deffeomorphism(two-dimensional coordinate transformation is called deffeomorphism), but it seems that the answer is no.

Tong talked about the dynamical background metric and fixed background metric, he said if the metric is dynamical, the weyl transformation is deffeomorphism; however, if the metric is fixed, the weyl transformation is not deffeomorphism, but a physical symmetry that takes \sigma to \sigma'.

And in a paragraph, he said, "any theory of 2d gravity which enjoys both deffeomorphism and Weyl invariance will reduce to a conformally invariant theory when the background metric is fixed." According to this, if the Weyl transformation is part of the deffeomorphism, then he could just say, ...enjoy deffeomorphism..., unnecessary to mention both deffeomorhpism and Weyl invariance. (I take Weyl transformation same as conformal transformation.)

But, according to the definition, Weyl transformatio is evidently part of coordinate transformation. Anybody solves my confusion please?
Thanks for any instructions!

 

[bapowell]

Hi,

This is a confusing topic because there are lots of definitions circulating about Weyl, conformal, and coordinate transformations. What you have written there is actually a Weyl scaling, not a conformal transformation (although the language is abused often, which adds to the confusion). In a Weyl scaling, you change the metric without changing the coordinates, and so it's distinct from a coordinate transformation. Now, I don't know the full definition of the conformal transformation off hand, but it incorporates both Weyl scalings and diffeomorphisms.

 

[LAHLH]

I don't understand this stuff too well myself, so this may not be the appropriate level, but I do remember seeing a good explanation of conformal trans in one of the appendices of Sean Carrolls book Spacetime and Geometry which is available as a free PDF if you google it.

 

[samalkhaiat]

You can find what you need and more in this thread;

www.physicsforums.com/showthread.php?t=172461

In that thread, I neatly presented a self-containd introduction to this subject so that questions like the above do not arise on these forums. Obviously I have failed!

regards

sam

 

[ismaili]

You can find what you need and more in this thread;

www.physicsforums.com/showthread.php?t=172461

In that thread, I neatly presented a self-containd introduction to this subject so that questions like the above do not arise on these forums. Obviously I have failed!

regards

sam

Thank you so much, Sam! I will study it in detail, I'm eagerly to understand this stuff clearly.

But allow me to ask a question in the beginning of your first post.
According to your definition of conformal transformation, it is a nonlinear coordinate transformation,
f:x\rightarrow \bar{x}=\bar{x}(x)
However, one of the definition of a tensor I learned is that, a tensor is a geometric object which transforms under coordinate transformation as follows, say, for example, the metric tensor should transform like this under coordinate transformation:
<br /> \eta_{ab}(x) \rightarrow \bar{g}_{ab}(\bar{x}) = \frac{\partial x^c}{\partial\bar{x}^a}\frac{\partial x^d}{\partial\bar{x}^b}\eta_{cd}(x)<br />
So that, under general coordinate transformation, the line element is always invariant,
<br /> ds^2 = \bar{g}_{ab}d\bar{x}^ad\bar{x}^b = \eta_{cd}dx^cdx^d<br />

Conversely, in the definition of a conformal transformation, we seem to keep the metric tensor unchanged while we change the coordinate, or seemingly, keeping the coordinate unchanged while changing the metric tensor.
<br /> d\bar{s}^2 = \eta_{ab}d\bar{x}^ad\bar{x}^b = \bar{g}_{cd}(x)dx^cdx^d<br />
This seems to be in conflict with the definition of a metric tensor which I learned before. Could you explain this a little bit?

Thank you so much!

---

By the way, I just finished your first post, now I know why in 2D the conformal group is infinite dimensional! (if the above definition of conformal transformation accepted.)

 

[ismaili]

just to suggest some probably typos so as to make the posts perfect.

(1) RHS of eq(3.2a) is probably \eta^{ac}

(2) Eq(5.8) is exact. RHS of eq(5.8) might be modified to be U(\alpha,b+ae^\beta-be^\alpha)

---

and, wow, you are very good at math, compact and rigorous. Hope I can digest more of latter part of math section tonight.


Ismaili

 

[samalkhaiat]

But allow me to ask a question in the beginning of your first post.
According to your definition of conformal transformation, it is a nonlinear coordinate transformation,
f:x\rightarrow \bar{x}=\bar{x}(x)
However, one of the definition of a tensor I learned is that, a tensor is a geometric object which transforms under coordinate transformation as follows, say, for example, the metric tensor should transform like this under coordinate transformation:
<br /> \eta_{ab}(x) \rightarrow \bar{g}_{ab}(\bar{x}) = \frac{\partial x^c}{\partial\bar{x}^a}\frac{\partial x^d}{\partial\bar{x}^b}\eta_{cd}(x)<br />
So that, under general coordinate transformation, the line element is always invariant,
<br /> ds^2 = \bar{g}_{ab}d\bar{x}^ad\bar{x}^b = \eta_{cd}dx^cdx^d<br />

Conversely, in the definition of a conformal transformation, we seem to keep the metric tensor unchanged while we change the coordinate, or seemingly, keeping the coordinate unchanged while changing the metric tensor.
<br /> d\bar{s}^2 = \eta_{ab}d\bar{x}^ad\bar{x}^b = \bar{g}_{cd}(x)dx^cdx^d<br />
This seems to be in conflict with the definition of a metric tensor which I learned before. Could you explain this a little bit?

You can think of it as transforming the coordinates or transforming the metric! It is the same old "confusing" story;
passive (one point with two coordinate values) versus active ( two points with equal coordinate value) transformation.

In your (passive) transformation law

<br /> \bar{g}_{ab}(\bar{x}) = \frac{\partial x^c}{\partial\bar{x}^a}\frac{\partial x^d}{\partial\bar{x}^b}\eta_{cd}<br />

\bar{x} and x represent two coordinates (lables) of the same geometrical point. There is nothing special about these two lables. I can exchange them and write;

<br /> \bar{g}_{ab}(x) = \partial_{a}\bar{x}^{c} \partial_{b}\bar{x}^{d} \eta_{cd} \ \ \ \ (1)<br />

However, this simple re-naming of the coordinates allows me to give eq(1) a different geometrical meaning! First, I do coordinates transformation, f: \ x \rightarrow \bar{x}, then evaluate (pull) the metric \bar{g} at (to) a point \bar{x} with coordinate value equal to x. So your passive transformation has had the same effect as if you had preformed an active transformation on the metric.

Mathematically, eq(1.1) of the conformal thread, says the followings

1) If f \ : M \rightarrow M is a smooth regular map, we can pull back the metric \bar{g} = \eta \ d\bar{x} \otimes d\bar{x} to a new metric (f^{*}\bar{g}) characterized by the components

<br /> (f^{*}\bar{g})_{ab}= \partial_{a}\bar{x}^{c} \partial_{b}\bar{x}^{d} \eta_{cd} <br />

2) f is called conformal (or angle-preserving) map if it rescales the metric, i.e., there is a positive scalar S(x) on M such that;

(f^{*}\bar{g})_{ab}(x) = S(x)\eta_{ab}

I hope that helps you.

regards

sam

 

[samalkhaiat]

just to suggest some probably typos so as to make the posts perfect.

(1) RHS of eq(3.2a) is probably \eta^{ac}

Thank you for the correction. I think you will find some more typos in there.

(2) Eq(5.8) is exact. RHS of eq(5.8) might be modified to be U(\alpha,b+ae^\beta-be^\alpha)

Yes, but it does not effect the important infinitesimal form of the product. You could even write U(\alpha, a + a \beta - b\alpha)

regards

sam