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I am trying to derive the Fermi normal coordinates formula for a FRW Universe given in Eq. (4) of a paper by Nicolis et al (2008): $$ds^2\approx -[1-(\dot{H}+H^2)|{\bf x}|^2]dt^2+[1-{1\over 2}H^2|{\bf x}|^2]d{\bf x}^2$$

They start with a FRW metric: $$ ds^2=-d\tau^2+a^2(\tau)d{\bf y}^2 $$ and perform the coordinate change $$ \tau=t-{1\over 2}H|{\bf x}|^2, \quad {\bf y}={{\bf x}\over a} [1+{1\over 4}H^2|{\bf x}|^2], $$ To derive their Eq. (4), I use the metric transformation formula given in Eq. (5.69) of Dodelson's Modern Cosmology: $$ \tilde{g}_{\alpha\beta}(\tilde{x}){\partial \tilde{x}^\alpha \over \partial x^\mu}{\partial \tilde{x}^\beta \over \partial x^\nu}=g_{\mu\nu}(x). $$ I take the tilde coordinates as the FRW ones. So for $\mu=\nu=1$, I have: $$ \frac{1}{16} |{\bf x}|^2 \left(9 (x^1)^2+(x^2)^2+(x^3)^2\right) H^4+\frac{1}{2} |{\bf x}|^2 H^2+1=g_{1,1} $$ I then take the limit that $|\vec{x}| H\ll 1$ to get $$ g_{11}\approx1 + {1\over 2} H^2 |{\bf x}|^2 $$ which differs from their result by a minus sign. Any help in explaining this discrepancy would be greatly appreciated.

Update: I was able to get their result using the Mathematica package TensoriaCalc which has a coordinate change command. This confirms Nicolis et al are correct and I am doing something wrong in my calculation.

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  • $\begingroup$ @igael I don't think I ever have $g_{11}\approx 1-{1\over 2}H^2x_1^2$. $\endgroup$
    – Virgo
    Aug 15, 2016 at 17:36
  • $\begingroup$ @Virgo: I was wrong, sorry. With $z =\frac{x_1^2 H^2}{2} , p=|\vec{x}|^2 H^2$ , I get $g_{11} = \frac{p^2}{16}+\frac{p z}{2}+\frac{p}{2}+z^2+1 \approx 1+\frac {p}{2}+(z^2+\frac {p z}{2})$ instead of $\approx 1+\frac{p}{2}$ or $\approx 1-\frac{p}{2}$. If it's not too boring for you, could you please expand your approximation ? $\endgroup$
    – user46925
    Aug 15, 2016 at 20:51
  • $\begingroup$ @Virgo : mathjax doesn't render the same in comments ... without mathjax : g11= p²/16 + pz/2 + p/2 + z² + 1 ( followed by correct renderring ) $\endgroup$
    – user46925
    Aug 15, 2016 at 21:00
  • $\begingroup$ @igael I think you can drop the $z^2$ and $pz$ term as we are assuming $xH\ll1$ $\endgroup$
    – Virgo
    Aug 15, 2016 at 22:20
  • $\begingroup$ formally, it is p which is very small. In p²/16 + pz/2 + p/2 + z², p² is very small, pz too and z² , I don't know ( I don't know the implicit formalism with links $x_1$ to $|\vec{x}$ ). Sorry and thanks $\endgroup$
    – user46925
    Aug 15, 2016 at 22:32

1 Answer 1

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There's a small subtlety. As stated by the authors in the paper, in the given transformation law $a$ and $H$ are evaluated at $t$ rather than at $\tau$. Thus the formula you get is almost right but $a(\tau)$ and $a(t)$ don't cancel, so actually $$g_{11} = \left(\frac{a(\tau)}{a(t)}\right)^2\left(1+\frac{1}{2}H^2|\mathbf x|^2\right)$$ Then using the transformation law above, $$a(\tau)=a\left(t-\frac{1}{2}H|\mathbf x|^2\right)\approx a(t)-\frac{1}{2}H|\mathbf x|^2 \dot a(t)$$ Therefore $$g_{11} = \left(1-\frac{1}{2}H^2|\mathbf x|^2 \right)^2\left(1+\frac{1}{2}H^2|\mathbf x|^2\right)\approx 1-\frac{1}{2}H^2|\mathbf x|^2$$

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