Consider a particular conformal transformation $x^\mu\rightarrow x'^\mu$, and the metric of a flat space transforms in the following way,

$$\eta_{\mu\nu}\rightarrow g'_{\mu\nu}=\Lambda^2(x)\eta_{\mu\nu}.$$

A special conformal transformation (SCF) takes the following form:

$$x'^\mu=\frac{x^\mu-b^\mu x^2}{1-2b\cdot x+b^2 x^2}\tag{4.15d} $$

Here, vector $b^\mu$ parametrizes SCF, and $x^2$ and $b\cdot x$ are defined using $\eta_{\mu\nu}$. It turns out that $\Lambda(x)$ is the denominator.

OK. According to Di Francesco's Conformal Field Theory, a relation is given (Eq.(4.22)):

$$|x'_i-x'_j|=\frac{|x_i-x_j|}{\sqrt{\Lambda(x_i)\Lambda(x_j)}}. \tag{4.22}$$

Here, $x_i,x_j$ collectively represents the coordinates of two arbitrary points in the original coordinate system. That is, $x_i=(x_i^1, x_i^2, ...)$. And $x'_i,x'_j$ are coordinates related to $x_i,x_j$ via the SCF. ${|x_i-x_j|}$ is apparently the distance.

The authors also call the LHS of the above equation a distance, too! But the transformed metric $g'_{\mu\nu}$ is no longer flat. So the distance between any pair of finitely separated points is not well defined, unless a path from one point to the other is specified. So how to understand this distance (the LHS)?

  • 1
    $\begingroup$ The distance between two points is defined in every connected Riemannian manifold, it is nothing but the $\inf$ of the length of the curves joining the two considered points. So it makes sense referring to the metric $g'_{\mu\nu}$, too. However, I do not know if this distance coincides with the right hand side of the identity you quoted. I think so as there exist at most one metric giving rise to a given squared distance and it is evident that the right-hand side of the identity reproduces $g'_{\mu\nu}$ for $x_i$ and $x_j$ sufficiently close to each other. $\endgroup$ – Valter Moretti Jun 24 '16 at 9:06
  • $\begingroup$ @ValterMoretti Hopefully, the $\inf$ coincides with the LHS. Unfortunately, the geodesic equations are really complicated and difficult to solve... $\endgroup$ – Drake Marquis Jun 24 '16 at 16:10

Comments to the post (v2):

  1. Ref. 1 is considering the $d$-dimensional real Euclidean space $(\mathbb{R}^d,|\cdot|^2)$ with the standard norm $$|x|^2~:=~\sum_{\mu=1}^d (x^{\mu})^2~=~\sum_{\mu,\nu=1}^d x^{\mu}\eta_{\mu\nu}x^{\nu}, \qquad \eta_{\mu\nu} ~=~{\rm diag}(1,\ldots, 1),\tag{A}$$ and inner product $$\langle x ,y\rangle~:=~\sum_{\mu,\nu=1}^d x^{\mu}\eta_{\mu\nu}y^{\nu} .\tag{B}$$

  2. We stress that the metric is fixed and the same, induced from the standard norm (A). (However, as always, it takes different explicit forms in various different coordinate systems.)

  3. From the SCT (4.15d), it follows that $$ |x^{\prime}|^2~=~\frac{|x|^2}{\Lambda(x)^2}. \tag{C} $$ Together with a similar calculation of the inner product $\langle x^{\prime} ,y^{\prime}\rangle$, it is possible to derive eq. (4.22): $$ |x^{\prime}-y^{\prime}|^2~=~\frac{|x-y|^2}{\Lambda(x)\Lambda(y)}. \tag{4.22} $$

  4. From eq. (4.22) it follows directly that the SCT (4.15d) is a conformal map

$$ ds^{\prime 2}~=~\frac{ds^2}{\Lambda(x)^2}. \tag{D}$$


  1. P. Di Francesco, P. Mathieu and D. Senechal, CFT, 1997.
  • $\begingroup$ Thanks for editing my post. Concerning your point 3, my question is actually to ask whether this symbol $|x'_i-x'_j|$ represents the distance measured using the new metric $g'_{\mu\nu}$. If so, how to prove the relation (4.22). It seems to me it is difficult. $\endgroup$ – Drake Marquis Jun 24 '16 at 16:14
  • $\begingroup$ Concerning your new point 3, relation (C) is easy to verify, but what is the inner product $\langle x',y'\rangle$? $\endgroup$ – Drake Marquis Jun 24 '16 at 16:54
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Jun 24 '16 at 16:57
  • $\begingroup$ You give me the inner product in the original coordinate system, but not the transformed one. Do you mean $\langle x',y'\rangle=x'^\mu\eta_{\mu\nu}x'^\nu$, instead of $\langle x',y'\rangle=x'^\mu g'_{\mu\nu}x'^\nu$, because of your point 2? (The second expression is of course ill-defined) $\endgroup$ – Drake Marquis Jun 24 '16 at 17:05
  • $\begingroup$ $\uparrow$ Yes. $\endgroup$ – Qmechanic Jun 24 '16 at 17:07

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