4
$\begingroup$

By definition a conformal transformation of the coordinates is an invertible mapping $x\rightarrow x'$ which leaves the metric invariant upto a scale factor: \begin{equation} g_{\mu\nu}'(x') = \Lambda(x) g_{\mu\nu}(x) \end{equation}

I am facing problems in deriving the scale factor for special conformal transformation (SCT). This transformation has the finite form given as: \begin{equation} {x'}^{\mu} = \frac{x^{\mu}-b^{\mu}x^2}{1-2(b.x)+b^2x^2} \end{equation} I was trying to derive the scale factor for this transformation using the metric condition: \begin{equation} g_{\mu\nu}'(x') =\frac{\partial x^\sigma}{\partial {x'}^\mu}\frac{\partial x^\rho}{\partial {x'}^\nu}g_{\sigma\rho}= \Lambda(x) g_{\mu\nu}(x) \end{equation} I was not able to find out $\Lambda(x)$, however the final form of scale factor $\Lambda(x)$ is given in the book as: \begin{equation} \Lambda(x) = (1-2b.x + b^2x^2)^2 \end{equation}

$\endgroup$

closed as off-topic by ACuriousMind, Kyle Kanos, joshphysics, Brandon Enright, Ali Sep 13 '14 at 18:01

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, joshphysics, Ali
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ And your question is? Plug the transformation in and look at the result! Please, also have a look at our homework policy and our stance on check-my-work questions. $\endgroup$ – ACuriousMind Sep 13 '14 at 14:37
  • $\begingroup$ Hello, welcome to SE. Could you try to clarify your question? What have you tried? Which point is causing trouble? I'm sure if you make it a little clearer, and say what you've tried, somebody might be able to help you. $\endgroup$ – innisfree Sep 13 '14 at 15:40