# Fermi coordinates for a Friedman Robertson Walker metric

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.

• @igael I don't think I ever have $g_{11}\approx 1-{1\over 2}H^2x_1^2$. – Virgo Aug 15 '16 at 17:36
• @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 ? – user46925 Aug 15 '16 at 20:51
• @Virgo : mathjax doesn't render the same in comments ... without mathjax : g11= p²/16 + pz/2 + p/2 + z² + 1 ( followed by correct renderring ) – user46925 Aug 15 '16 at 21:00
• @igael I think you can drop the $z^2$ and $pz$ term as we are assuming $xH\ll1$ – Virgo Aug 15 '16 at 22:20
• 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 – user46925 Aug 15 '16 at 22:32

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 $$g_{11} = \left(1-\frac{1}{2}H^2|\mathbf x|^2 \right)^2\left(1+\frac{1}{2}H^2|\mathbf x|^2\right)\approx 1-\frac{1}{2}H^2|\mathbf x|^2$$