# Generator of the infinitesimal special conformal transformation

+ 4 like - 0 dislike
4926 views

(c.f Di Francesco, Conformal Field Theory chapters 2 and 4).

The expression for the full generator, $G_a$, of a transformation is $$iG_a \Phi = \frac{\delta x^{\mu}}{\delta \omega_{a}} \partial_{\mu} \Phi - \frac{\delta F}{\delta \omega_a}$$ For an infinitesimal special conformal transformation (SCT), the coordinates transform like $$x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2$$

If we now suppose the field transforms trivially under a SCT across the entire space, then $\delta F/\delta \omega_a = 0$.

Geometrically, a SCT comprises of a inversion, translation and then a further inversion. An inversion of a point in space just looks like a translation of the point. So the constant vector $b^{\mu}$ parametrises the SCT. Then $$\frac{\delta x^{\mu}}{\delta b^{\nu}} = \frac{\delta x^{\mu}}{\delta (x^{\rho}b_{\rho})} \frac{\delta (x^{\gamma}b_{\gamma})}{\delta b^{\nu}} = 2 x^{\mu}x_{\nu} - x^2 \delta_{\nu}^{\mu}.$$ Now moving on to my question: Di Francesco makes a point of not showing how the finite transformation of the SCT comes from but just states it. $$x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2}$$ I was wondering if somebody could point me to a link or explain the derivation. Is the reason for its non appearance due to complication or by being tedious?

I am also wondering how, from either of the infinitesimal or finite forms, we may express the SCT as $$\frac{x'^{\mu}}{x'^2} = \frac{x^{\mu}}{x^2} - b^{\mu},$$ which is to say the SCT is an inversion $(1/x^2)$ a translation $-b^{\mu}$ and then a further inversion $(1/x'^2)$ which then gives $x'^{\mu}$, i.e the transformed coordinate.

This post imported from StackExchange Physics at 2014-06-27 11:33 (UCT), posted by SE-user CAF
The expressions $\frac{x'^{\mu}}{x'^2} = \frac{x^{\mu}}{x^2} - b^{\mu}$ and $x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2}$ are equivalent (just express $x'^2$ as a function of $x$), and the infinitesimal form is clearly $x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2$ (because only quadratic terms in $x$ are allowed in infinitesimal conformal transformations)

This post imported from StackExchange Physics at 2014-06-27 11:33 (UCT), posted by SE-user Trimok

+ 1 like - 0 dislike

However, I don't really see how to exponentiate the infinitesimal form to actually get to the finite form

In addition to the detailed answer of @LubošMotl, you may notice that conformal transformations transform light cones in light cones.

This means the following. Starting from a light cone :

$$(x'-a')^2 = x'^2-2x'.a' + a'^2=0\tag{1}$$

you must find that it is also a light cone for $x$, that is, it exist $a$ such as :

$$(x-a)^2 = x^2-2x.a + a^2=0\tag{2}$$

To do that, replace simply $x'$ by its value in function of $x$, by :

$$x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2} \tag{3}$$

and, with some algebra, you will find that it is possible to exhibit $a(a',b)$ ($a$ as a function of $a'$ and $b$) , which satisties ($2$).:

$$a = \dfrac{a' + a'^2b}{1-2a'.b+a'^2b^2}\tag{4}$$

This means that the transformation $(3)$ is a conformal transformation whose infinitesimal expression is clearly: $$x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2 \tag{5}$$ (because only quadratic terms in x are allowed in infinitesimal conformal transformations)

Any other expression of the global special conformal transformation $(3)$, having infinitesimal expression $(5)$, will not transform light cones in light cones.

This post imported from StackExchange Physics at 2014-06-27 11:33 (UCT), posted by SE-user Trimok
answered Jun 27, 2014 by (955 points)
+ 0 like - 0 dislike

The two questions posed above are exactly the same and one may prove both simply by substituting $x'_\mu$ and calculating. If of not showing how the finite transformation of the SCT comes from but just states it. $$x'^{\mu} = \frac{x^{\mu} - b^{\mu}x^2}{1-2x\cdot b + b^2 x^2}$$ then $$(x')^2 = \frac{(x_\mu-x^2 b_\mu)(x^\mu-x^2 b^\mu)}{(1-2xb+b^2x^2)^2}=\frac{x^2(1-2bx+x^2b^2) }{(1-2xb+b^2x^2)^2}=\frac{x^2}{1-2xb+b^2x^2}$$ and therefore $$\frac{x'}{(x')^2} = \frac{x-x^2b}{x^2}$$ (some indices were omitted in a self-explanatory way) which is exactly the second equation. So the two equations you are asking about are exactly equivalent.

The special conformal transformations may be uniquely defined as the conformal transformations that map the infinity to a different point and that map a line of the multiples of $b^\mu$ onto itself. The inversion is really needed to get the point at infinity to a finite place, so that it may be moved elsewhere by the translation in the middle.

If you use a different definition of the special conformal transformation than the definition "inverse times translations time inversion" or the definition from the previous paragraph, you would have to specify what's your definition is and one could easily show the equivalence with the definitions above, too.

This post imported from StackExchange Physics at 2014-06-27 11:33 (UCT), posted by SE-user Luboš Motl
answered Jun 27, 2014 by (10,278 points)
Many thanks for your answer. I can show that the infinitesimal form of the finite SCT is $x'^{\mu} = x^{\mu} + 2(x \cdot b)x^{\mu} - b^{\mu}x^2$ by simply expanding the finite form for $|b^{\mu}| \ll |x^{\mu}|$. However, I don't really see how to exponentiate the infinitesimal form to actually get to the finite form. Thanks.

This post imported from StackExchange Physics at 2014-06-27 11:33 (UCT), posted by SE-user CAF

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\varnothing$ysicsOverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.