Under a local change of coordinates $x\to x'=x+\delta x$, the metric transforms as $$g_{\mu \nu}^{\prime}\left(x^{\prime}\right)=g_{\lambda \rho}(x) \frac{\partial x^{\lambda}}{\partial x^{\prime \mu}} \frac{\partial x^{\rho}}{\partial x^{\prime \nu}}$$ I am trying to show $$\delta g_{\mu \nu}=-\frac{1}{2}\left(g_{\mu \lambda} \partial^{v} \delta x_{\lambda}+g_{\lambda \nu} \partial^{\mu} \delta x_{\lambda}+\partial^{\lambda} g_{\mu \nu} \delta x_{\lambda}\right)$$

Other references, including a few answers I found in this website, are missing this factor of one half. My own derivation also misses it. I wonder where this come from. Here is my approach:

$$g'_{\mu\nu}(x+\delta x)=g'_{\mu\nu}(x)+\delta x^\lambda\partial_\lambda g'_{\mu\nu}\approx g'_{\mu\nu}(x)+\delta x^\lambda\partial_\lambda g_{\mu\nu}$$ where in the last equation, I neglected $\delta x^\lambda\partial_\lambda\delta g_{\mu\nu}$ as it is a second order change.

Since $x'^\mu=x^\mu+\delta x^\mu$, we have $$\frac{\partial x'^\mu}{\partial x^\nu}=\delta^\mu_\nu+\frac{\partial \delta x^\mu}{\partial x^\nu}\implies \frac{\partial x^\nu}{\partial x'^\mu}=\delta^\nu_\mu-\frac{\partial \delta x^\nu}{\partial x^\mu}$$ Hence $$g'_{\mu\nu}(x)+\delta x^\lambda\partial_\lambda g_{\mu\nu}=g_{\lambda\rho}\left(\delta^\lambda_\mu-\frac{\partial \delta x^\lambda}{\partial x^\mu}\right)\left(\delta^\rho_\nu-\frac{\partial \delta x^\rho}{\partial x^\nu}\right)$$ Or equivalently $$\delta g_{\mu\nu}=g'_{\mu\nu}-g_{\mu\nu}=-\left(g_{\lambda\nu}{\partial_\mu \delta x^\lambda}+g_{\mu\lambda}{\partial_\nu \delta x^\lambda}+\partial_\lambda g_{\mu\nu}\delta x^\lambda\right)$$ which, as one can see, misses the factor of $\frac 12$.

  • $\begingroup$ There isn't supposed to be a factor of 1/2. $\endgroup$
    – Prahar
    May 11, 2022 at 14:25

1 Answer 1


I think that you are trying to compute the Lie derivative of the metric. If so, there should be no factor of 1/2. Under an infinitesimal shift $x^\mu\to x^\mu +\eta^\mu$ we have $g \to g+ {\mathcal L}_\eta g$ $$ ({\mathcal L}_\eta g)_{\mu\nu}= \eta^\lambda\partial_\lambda g_{\mu\nu}+ g_{\mu\lambda}\partial_\nu \eta^\lambda + g_{\lambda\nu }\partial_\mu \eta^\lambda $$ see section 11.2.2 on Lie Derivatives in https://goldbart.gatech.edu/PostScript/MS_PG_book/bookmaster.pdf and in particular eq 11.38 and its its dervation.

The factor of 1/2 does appear in the definition of the strain tensor which, for historical reasons, is defined as 1/2 of the Lie derivatve. See exercise 11.3 on the next page of the book.

  • $\begingroup$ Exactly so! It looks like you are one of the authors of the book, I do own a physical copy of this book and I am current reading it. So far it has been great! There is a typo on page 745 (Appendix A), part (iv) of Exercise A.1, which reads "Given $x, y$ in $V$ , there is a unique $z$ such that $x+z = y$, to whit $z = x-y.$", I think "whit" is supposed to mean "wit", and $z=y-x$. I didn't find this in the errata, so it might be good to know. (I think I am abusing comments here, feel free to delete it if it's irrelevant). $\endgroup$
    – Sofvar
    May 11, 2022 at 14:36
  • $\begingroup$ Yes that's me --- and I am still embarrassed by that whit-wit error. It's not a typo, but just grammatical ignorance. $\endgroup$
    – mike stone
    May 11, 2022 at 18:45
  • $\begingroup$ +1 for the fantastic book suggestion $\endgroup$
    – Rob Tan
    Aug 16, 2022 at 14:08

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