I am unable to obtain the conformal killing equation:

$$2\kappa(x) \eta_{\mu\nu}= \partial_\mu \xi _\nu + \partial_\nu \xi_\mu\tag{1}$$


I understand that the conformal transformation is:

$$\eta_{\mu\nu} \Omega^2 = \frac{\partial x'^\sigma}{\partial x^\mu} \frac{\partial x'^\rho}{\partial x^\nu} \eta_{\sigma \rho}\tag{2}$$

and that in order to obtain the conformal killing equation we define:

$$x'^\mu = x^\mu +\xi^\mu (x) \tag{3}$$


$$\Omega = 1 + \kappa(x)\tag{4}$$

and that we must ignore $\mathcal{O}(\xi^2)$, $\mathcal{O}(\kappa^2)$, $\mathcal{O}(\xi\kappa)$ as $\xi$ and $\kappa$ are infinitesimal.


I started by expanding the LHS

$$\eta_{\mu\nu} \Omega^2 = (1+2\kappa +\kappa^2)\eta_{\mu\nu}\tag{5}$$

the last term can be ignored.

The RHS would become:

$$\frac{\partial(x^\sigma +\xi^\sigma)}{\partial x^\mu} \frac{\partial(x^\rho + \xi^\rho)}{\partial x^\nu} \eta_{\sigma \rho}\tag{6}$$

But what do I do with the term $1$ in the LHS? I don't understand how to further expand the RHS. I have also seen Conformal transformation equation but I don't understand the answer given.

Where do I go from here? Am I missing a law/rule/equation?

  • $\begingroup$ You're basically there, right? The term "1" comes from the order zero in $\xi$ in your equation (6). $\endgroup$ – MannyC Mar 28 '20 at 18:38
  • $\begingroup$ I am sorry for not understanding your answer, but how will the two $\xi$ in $(6)$ relate to the 1 in $(5)$? Does it involve using the $\eta_{\sigma\rho}$? If so, how? $\endgroup$ – user7077252 Mar 28 '20 at 18:49

From OP one has (the $\kappa^2$ goes away) $$\eta_{\mu\nu} \Omega^2 = (1+2\kappa)\eta_{\mu\nu}\,,\tag{5}$$ and $$\frac{\partial(x^\sigma +\xi^\sigma)}{\partial x^\mu} \frac{\partial(x^\rho + \xi^\rho)}{\partial x^\nu} \eta_{\sigma \rho}\,.\tag{6}$$ Now let's use $\partial_\mu x^\nu = \delta^\nu_\mu$ and keep only the linear order in $\xi$. So $$ \begin{aligned} (6) &= (\delta^\sigma_\mu +\partial_\mu\xi^\sigma)(\delta^\rho_\nu+\partial_\nu\xi^\rho)\eta_{\sigma\rho} \\&= \eta_{\mu\nu} + \partial_\mu\xi_\rho \delta^\rho_\nu + \delta^\sigma_\mu \partial_\nu\xi_\sigma + O(\xi^2)\,. \end{aligned} $$

  • $\begingroup$ Thank you very much, it makes perfect sense to me now. I was forgetting $\partial_\mu x^\nu = \delta^\nu _\mu$ . $\endgroup$ – user7077252 Mar 28 '20 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.