Easy question on conformal transformations

[physicus]

A conformal transformation is a coordinate transformation that leaves the metric invariant up to a scale change $g_{\mu\nu}(x) \to g'_{\mu\mu}(x)=\Omega(x)g_{\mu\nu}(x)$.
This means that the length of vectors is not preserved: $g_{\mu\nu}x'^{\mu}x'^{\nu}\not=g_{\mu\nu}x^{\mu}x^{\nu}$
But is $g'_{\mu\nu}x'^{\mu}x'^{\nu}=g_{\mu\nu}x^{\mu} x^{\nu}$ correct?

physicus

 

[jfy4]

well, line elements are related by $ds'^2=\Omega^2 ds^2$. Note that for null vectors, the line element is preserved $(0=0)$.

Edit/Note: my definition has a square for aesthetics, namely, $g'=\Omega^2 g$.

 

[geoduck]

well, line elements are related by $ds'^2=\Omega^2 ds^2$. Note that for null vectors, the line element is preserved $(0=0)$.

Edit/Note: my definition has a square for aesthetics, namely, $g'=\Omega^2 g$.

I thought the line element squared is always the same. If it's physical, why would it change with a chance of reference frame?

Therefore I think gxx=g'x'x', where the indices on position x are contracted with the metric g.

I vaguely recall conformal transformations in complex variables, which means transformations that preserve angles but not lengths. So

gxx=g'x'x'=Ωgx'x' ≠ gx'x'

However, suppose you have two vectors x and y that are orthogonal:

gxy=0

Then it is true that:

gx'y'=0 (they are still orthogonal using the old metric)

since

0=gxy=g'x'y'=Ωgx'y'

so if Ω≠0, then: gx'y'=0

 

[vanhees71]

I guess you talk about general relativity (why then in the quantum physics subforum?). Anyway, the conformal mapping is meant as a symmetry transformation not as a coordinate transformation. Of course, by construction, in general relativity the equations are invariant under arbitrary coordinate transformations (as long as they are local diffeomorphisms, but that's understood implicitly anyway).

The conformal transformation is defined by

$${x'}^{\mu}=x^{\mu}, \quad g_{\mu \nu}'(x)=\Omega^2(x) g_{\mu \nu}.$$

This means of course that

$${g'}^{\mu \nu}=\Omega^{-2}(x) g_{\mu \nu}.$$

The free electromagnetic action is

$$A_{\text{em}}=-\frac{1}{4} \int \mathrm{d}^4 x \sqrt{-g} F_{\mu \nu} F^{\mu \nu},$$

where $F_{\mu \nu}=\mathrm{D}_{\mu} A_{\nu}-\mathrm{D}_{\nu} A_{\mu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}$ and $g=\mathrm{det}(g_{\mu \nu})$.

Then it's easy to see that $A_{\text{em}}$ is invariant under conformal transformations, when we assume that

$$A_{\mu}'(x)=A_{\mu}(x)$$

under conformal transformations. Then also

$$F_{\mu \nu}'(x)=F_{\mu \nu}(x) \; \Rightarrow \; {F'}^{\mu \nu}(x) = \Omega^4(x) F^{\mu \nu}.$$

For the determinant of the metric we have

$$g'(x)=\Omega^8(x) g(x)$$

and thus the free Maxwell equations are invariant under conformal transformations (of course only in 4 space-time dimensions!).

To make this slightly on topic in the quantum-physics forum: Conformal invariance is generally broken when going over to quantum field theory with interacting particles, i.e., the conformal symmetry is anomalously broken: If the Lagrangian of the theory doesn't contain any dimensionful parameter (like QED with massless Dirac or Klein-Gordon fields as matter fields), it's invariant under conformal transformations. When quantizing the theory, one needs to introduce a momentum scale at which the divergent quantum corrections of perturbation theory (i.e., Feynman diagrams with loops), because since all particles are massless in such a theory, all vertex functions have branch cuts starting right at the point where all external momenta are 0. Thus, you cannot subtract the divergences at this point, and thus one necessarily has to subtract at a point where all momenta are space like which necessarily introduces a scale into the game.

A non-perturbative proof of anomalies goes over Fujikawa's path-integral method. In this approach the anomalous breaking of conformal symmetry comes from the fact that the path-integral measure of the fields is not invariant under conformal transformations.

 

[geoduck]

The conformal transformation is defined by

$${x'}^{\mu}=x^{\mu}, \quad g_{\mu \nu}'(x)=\Omega^2(x) g_{\mu \nu}.$$

I'm confused about this. If the coordinates don't change, how can the metric change?

Anyhow, the conformal transformation includes things such as time dilation and time translation, so surely x' must be different from x.

When quantizing the theory, one needs to introduce a momentum scale at which the divergent quantum corrections of perturbation theory (i.e., Feynman diagrams with loops), because since all particles are massless in such a theory, all vertex functions have branch cuts starting right at the point where all external momenta are 0. Thus, you cannot subtract the divergences at this point, and thus one necessarily has to subtract at a point where all momenta are space like which necessarily introduces a scale into the game.

Why must you subtract at a space-like point? Doesn't the scale enter in as log[p^2/μ^2], and since your external momentum p is time-like, wouldn't it make sense to have μ time-like too? You can only do experiments with time-like separations anyways, so you have to be able to make a measurement of the physical parameters at p^2=μ^2, so doesn't μ have to be time-like?

 

[samalkhaiat]

The conformal transformation is defined by

$${x'}^{\mu}=x^{\mu}, \quad g_{\mu \nu}'(x)=\Omega^2(x) g_{\mu \nu}.$$

This is called Weyl re-scaling of the metric. The conformal group is a lot bigger than Weyl transformation:

(1) In Minkowski space $\mathbb{R}^{1}\times \mathbb{R}^{n-1}$, Poincare’ transformation,
$$\{L(\Lambda),T(a)\}: \ \bar{x} = \Lambda x + a,$$
is a conformal map with the conformal factor,
$$\Omega^{2}(x) \equiv (\det |\frac{\partial \bar{x}}{\partial x}|)^{\frac{2}{n}}= 1.$$
This means that Poincare’ transformations form the isometry group $ISO(1,n-1)$ of $\mathbb{R}^{1}\times \mathbb{R}^{n-1}$.

(2) An inversion of the coordinates,
$$I: \ \bar{x} = \frac{x}{x^{2}},$$
is a conformal map with the conformal factor,
$$\Omega^{2}(x) = \frac{1}{x^{2}}.$$

On $\mathbb{R}^{1} \times \mathbb{R}^{n-1}$ with $n > 2$, all possible conformal transformations (i.e., the whole conformal group $C(1,n-1) \simeq SO(2,n)/Z_{2}$) can be generated using only $(1)$ and $(2)$. Indeed, the conformal group $C(1,n-1)$ is the smallest group containing the isometry group $ISO(1,n-1)$ and the inversion $I$.
The proof consists of two parts;
(i) It is almost trivial to see that the combination
$$K(c) \equiv I T(c) I: \ \bar{x} = \frac{x + c x^{2}}{1 + 2 c.x + c^{2}x^{2}},$$
generates a conformal map with

$$\Omega^{2}(x) = (1 + 2c.x + c^{2}x^{2})^{-2}.$$

The transformation $K(c)$ is called a “conformal boosts” or “special conformal transformation”. The latter name was first coined by H.A. Kastrup in 1966.

(ii) It is rather tedious but straightforward to show that the combination
$$D(\frac{1}{1+c^{2}})\equiv T(\frac{-c}{1+c^{2}})K(c)T(c)K(\frac{-c}{1+c^{2}}),$$
generates dilatation,
$$D(\lambda): \ \bar{x}= \lambda x.$$
Dilatation is a conformal map with the conformal factor
$$\Omega^{2}(x) = \lambda^{2}.$$

Then it's easy to see that $A_{\text{em}}$ is invariant under conformal transformations, when we assume that

$$A_{\mu}'(x)=A_{\mu}(x)$$

under conformal transformations. Then also

$$F_{\mu \nu}'(x)=F_{\mu \nu}(x) \; \Rightarrow \; {F'}^{\mu \nu}(x) = \Omega^4(x) F^{\mu \nu}.$$

Under the conformal group, a (Lorentz) vector field of weight $d$ transforms according to
$$\bar{A}^{a}(\bar{x}) = |\frac{\partial \bar{x}}{\partial x}|^{\frac{d-1}{n}}\frac{\partial \bar{x}^{a}}{\partial x^{c}}A^{c}(x).$$
In general, spinor tensor field in the $(j_{1},j_{2})$ representation of the Lorentz group $SL(2,\mathbb{C})$, transforms in the $(d, j_{1},j_{2})$ representation of the conformal group $SU(2,2)$;
$$\delta \Phi^{(d,j_{1},j_{2})}(x) = \left(-f^{a}\partial_{a} + \frac{1}{2}\partial^{[a}f^{b]}\Sigma_{ab}^{(j_{1},j_{2})} + \frac{d}{4}\partial_{a}f^{a}\right) \Phi^{(d,j_{1},j_{2})}(x),$$
where,
$$f^{b}(x) = a^{b} + \omega^{b}{}_{c}x^{c} + \alpha x^{b} + 2(c.x)x^{b}+ c^{b}x^{2},$$
is the conformal Killing vector.

Sam

 

[Emily1]

I know this is coming one-year too late, but if anyone else stumbles across this: physicus was right, $ds^2$ is invariant, and $g'(x')=\Omega(x)g(x)$. Here's a link from an MIT lecture (just check the first paragraph):

http://ocw.mit.edu/courses/physics/8-821-string-theory-fall-2008/lecture-notes/lecture10.pdf