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I am unable to obtain the conformal killing equation:

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


Theory:

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:

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

and

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

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


Attempt:

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?

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  • $\begingroup$ You're basically there, right? The term "1" comes from the order zero in $\xi$ in your equation (6). $\endgroup$
    – MannyC
    Mar 28, 2020 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? $\endgroup$ Mar 28, 2020 at 18:49

1 Answer 1

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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 $$ \begin{aligned} (6) &= (\delta^\sigma_\mu +\partial_\mu\xi^\sigma)(\delta^\rho_\nu+\partial_\nu\xi^\rho)\eta_{\sigma\rho} \\&= \eta_{\mu\nu} + \partial_\mu\xi_\rho \delta^\rho_\nu + \delta^\sigma_\mu \partial_\nu\xi_\sigma + O(\xi^2)\,. \end{aligned} $$

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  • $\begingroup$ Thank you very much, it makes perfect sense to me now. I was forgetting $\partial_\mu x^\nu = \delta^\nu _\mu$ . $\endgroup$ Mar 28, 2020 at 19:25

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