Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,862 answers , 20,637 comments
1,470 users with positive rep
502 active unimported users
More ...

Generator of the infinitesimal special conformal transformation

+ 4 like - 0 dislike
2245 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
asked Jun 26, 2014 in Theoretical Physics by CAF (100 points) [ no revision ]
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

2 Answers

+ 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 Trimok (950 points) [ no revision ]
+ 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 Luboš Motl (10,248 points) [ no revision ]
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

Your answer

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):
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$\hbar\varnothing$sicsOverflow
Then 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.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...