I am reading a book of General Relativity and I am stuck on a demonstration. If I consider the FLRW metric as :
$\text{d}\tau^2=\text{d}t^2-a(t)^2\bigg[\dfrac{\text{d}r^2}{1-kr^2}+r^2(\text{d}\theta^2+\text{sin}^2\theta\text{d}\phi^2)\bigg]$
with $g_{tt}=1$, $\quad g_{rr}=\dfrac{a(t)^2}{1-kr^2}$ and $\quad g_{\theta\theta}=\dfrac{g_{\phi\phi}}{\text{sin}^2\theta}=a(t)^2 r^2$
It is said in this book that, despite of the utility of comoving coordinates, the dependance of time in scale factor $a(t)$ can be better understood if we consider a set of coordinates called "Free Fall coordinates" and noted $(\tilde{x}^\mu, \mu=0,1,2,3)$ with a metric locally Lorentzian near to the origin $\tilde{x}^{\mu}=0$ :
$g_{\mu\nu}=\eta_{\mu\nu}+\dfrac{1}{2}g_{\mu\nu,\alpha\beta}(0)\tilde{x}^{\alpha}\tilde{x}^{\beta}+\,.......\quad\text{(eq1)}$
with $\eta_{00}=-\eta_{11}=-\eta_{22}=-\eta_{33}=1\quad\quad$ and $\eta_{\mu\neq\nu}=0$
and $g_{\mu\nu,\alpha\beta}=\dfrac{\partial^2 g_{\mu\nu}}{\partial \tilde{x}^{\alpha}\partial \tilde{x}^{\beta}}$
Moreover, one takes the expression of classic geodesics :
$\dfrac{\text{d}}{\text{d}\tau}\bigg(g_{\mu\nu}(x)\dfrac{\text{d}x^{\nu}}{\text{d}\tau}\bigg)-\dfrac{1}{2}\dfrac{\partial g_{\lambda\nu}}{\partial x^{\mu}}\dfrac{\text{d}x^{\lambda}}{\text{d}\tau}\dfrac{\text{d}x^{\nu}}{\text{d}\tau}=0\quad\quad\mu=0,1,2,3\quad \text{(eq2)}$
The author says that, by applying $\text{(eq1)}$ into the relation $\text{(eq2)}$, one gets, at first order, the following relation :
$\dfrac{\text{d}^2 \tilde{x}^{\alpha}}{\text{d}\tau^2} = -\eta^{\alpha\gamma}\bigg[g_{\mu\gamma,\nu\beta}-\dfrac{1}{2}g_{\mu\nu,\gamma\beta}\bigg]\tilde{x}^{\beta}\dfrac{\text{d}\tilde{x}^{\mu}}{\text{d}\tau}\dfrac{\text{d}\tilde{x}^{\nu}}{\text{d}\tau}\quad\quad\text{(eq3)}$
I can't manage to obtain the $\text{eq(3)}$ from $\text{eq(1)}$ and $\text{eq(2)}$, if someone could help me for the details of the demonstration, this would be nice.
For the moment, if I put the definition of $g_{\nu\mu}$ into $\text{eq(2)}$, I get :
$\dfrac{\text{d}}{\text{d}\tau}\bigg(g_{\mu\nu}(x)\dfrac{\text{d}x^{\nu}}{\text{d}\tau}\bigg)=\bigg(\dfrac{\text{d}g_{\mu\nu}}{\text{d}\tau}\bigg)\,\dfrac{\text{d}x^{\nu}}{\text{d}\tau}+g_{\mu\nu}\dfrac{\text{d}^2x^{\nu}}{\text{d}\tau^2}\quad\quad \text{eq(4)}$
If I separate the two terms on RHS on $\text{eq(4)}$ :
$\bigg(\dfrac{\text{d}g_{\mu\nu}}{\text{d}\tau}\bigg)\,\dfrac{\text{d}x^{\nu}}{\text{d}\tau}=\dfrac{\text{d}}{\text{d}\tau}\bigg(\eta_{\mu\nu}+\dfrac{1}{2}g_{\mu\nu,\alpha\beta}(0)\tilde{x}^{\alpha}\tilde{x}^{\beta}\bigg)\dfrac{\text{d}x^{\nu}}{\text{d}\tau}=$
$\dfrac{1}{2}\,g_{\mu\nu,\alpha\beta}(0)\bigg[\dfrac{\text{d}\tilde{x}^{\alpha}}{\text{d}\tau}\,\tilde{x}^{\beta}+\tilde{x}^{\alpha}\,\dfrac{\text{d}\tilde{x}^{\beta}}{\text{d}\tau}\bigg]\dfrac{\text{d}x^{\nu}}{\text{d}\tau}=$
$g_{\mu\nu,\alpha\beta}(0)\bigg[\dfrac{\text{d}\tilde{x}^{\alpha}}{\text{d}\tau}\,\tilde{x}^{\beta}\bigg]\dfrac{\text{d}x^{\nu}}{\text{d}\tau}$
Concerning the second term on RHS of $\text{eq(4)}$, maybe I could write :
$g_{\mu\nu}\dfrac{\text{d}^2x^{\nu}}{\text{d}\tau^2}=\eta_{\mu\nu}\dfrac{\text{d}^2x^{\nu}}{\text{d}\tau^2}$ by neglecting the term $\dfrac{1}{2}g_{\mu\nu,\alpha\beta}(0)\tilde{x}^{\alpha}\tilde{x}^{\beta}$ in the expression of $g_{\mu\nu}$.
Another problem, How can I transform the second term on LHS of $\text{(eq2)}$ :
$-\dfrac{1}{2}\dfrac{\partial g_{\lambda\nu}}{\partial x^{\mu}}\dfrac{\text{d}x^{\lambda}}{\text{d}\tau}\dfrac{\text{d}x^{\nu}}{\text{d}\tau}\quad\quad \text{eq(5)}$ ??
Indeed, it seems that we can deduce from this term the wanted term :
$\dfrac{1}{2}\,\eta^{\alpha\gamma}\,g_{\mu\nu,\gamma\beta}\tilde{x}^{\beta}\dfrac{\text{d}\tilde{x}^{\mu}}{\text{d}\tau}\dfrac{\text{d}\tilde{x}^{\nu}}{\text{d}\tau}$
But $\tilde{x}$ coordinates ("Free Fall coordinates") appear in this last expression instead of "Comobile coordinates" $x^{\mu}$ into $\text{eq(5)}$ , so I don't know how to get it ?
Finally, to do the link between "Fre Fall coordinates" and "Comoving coordinates", can I write :
$g_{\mu\nu}\text{d}x^{\mu}\text{d}x^{\nu}=\eta_{\alpha\beta}\text{d}\tilde{x}^{\alpha}\text{d}\tilde{x}^{\beta}$ ??
Any help is welcome
UPDATE 1 : You can find below 2 image captures resuming the different equations (mostly important eq(4.4), eq(4.5); eq(4.6) and eq(4.87)) :
UPDATE 2 : Thanks for your demo. Just a last question, how do you do the link between the definition of coordinates called "Free Fall coordinates" and noted $(\tilde{x}^\mu, \mu=0,1,2,3)$ :
$g_{\mu\nu}=\eta_{\mu\nu}+\dfrac{1}{2}g_{\mu\nu,\alpha\beta}(0)\tilde{x}^{\alpha}\tilde{x}^{\beta}+\,.......\quad\text{(eq1)}$
and the comoving coordinates that you use into your following start point :
$\frac{\partial g_{\lambda\nu}}{\partial x^\mu} = g_{\lambda\nu,\mu\beta} x^\beta$
I mean, you don't use $\tilde{x}$ notation in your demontsration, that makes me confused.
Regards
We have $$\frac{\partial g_{\lambda\nu}}{\partial x^\mu} = g_{\lambda\nu,\mu\beta} x^\beta$$ and $$ \frac{d}{d\tau}\left(g_{\mu\nu}\frac{dx^\nu}{d\tau}\right) = \eta_{\mu\nu}\frac{d^2x^\nu}{d\tau^2}+\frac{d}{d\tau}\left(\frac{1}{2}g_{\mu\nu,\alpha\beta} x^\alpha x^\beta \frac{dx^\nu}{d\tau}\right)\\ =\eta_{\mu\nu}\frac{d^2x^\nu}{d\tau^2} + g_{\mu\nu,\alpha\beta} x^{\alpha} \frac{dx^\beta}{d\tau}\frac{dx^{\nu}}{d\tau} + O(x^2)$$ where we used symmetry in the last line. So plugging into the geodesic equation gives $$ \eta_{\mu\nu}\frac{d^2x^\nu}{d\tau^2} = \frac{1}{2}g_{\lambda\nu,\mu\beta}x^\beta\frac{dx^\lambda}{d\tau}\frac{dx^\nu}{d\tau}-g_{\mu\nu,\alpha\beta}x^\alpha\frac{d x^\beta}{d\tau}\frac{dx^\nu}{d\tau}. $$
Contracting both sides, doing some index relabeling to factor out the x's, and introducing the answer's awkward minus sign placement, $$ \frac{d^2 x^\alpha}{d\tau^2} =-\eta^{\alpha \mu}\left(g_{\mu\nu,\alpha\beta}-\frac{1}{2}g_{\beta\nu,\mu\alpha}\right)x^{\alpha}\frac{dx^\beta}{d\tau}\frac{dx^\nu}{d\tau}.$$
Finally, using the symmetry in the first pair and second pair of indices in the first term, $$ \frac{d^2 x^\alpha}{d\tau^2} =-\eta^{\alpha \mu}\left(g_{\nu\mu,\beta\alpha}-\frac{1}{2}g_{\beta\nu,\mu\alpha}\right)x^{\alpha}\frac{dx^\beta}{d\tau}\frac{dx^\nu}{d\tau} $$ which I believe is a relabeling of dummy indices away from the answer.