Demonstration of relation between geodesics and FLRW metric

[fab13]

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 dependence 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 $\tilde{x}^{\mu}=0$ :

$g_{\mu\nu}=\eta_{\mu\nu}+\dfrac{1}{2}g_{\mu\nu,\alpha\beta}(0)\tilde{x}^{\alpha}\tilde{x}^{\beta}+\,...\quad(eq1)$

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 :

$\dfrac{\text{d}}{\text{d}\tau}\bigg(g_{\mu\nu}(x)\dfrac{\text{d}x^{\nu}}{\text{d}\tau}\bigg)-\dfrac{1}{2}\dfrac{\partial g_{\lambda\nu}}{\partial x^{\mu}}\dfrac{\text{d}x^{\lambda}}{\text{d}\tau}\dfrac{\text{d}x^{\nu}}{\text{d}\tau}=0\quad\quad\mu=0,1,2,3\quad (eq2)$

The author says that, by applying $(eq1)$ into the relation $(eq2)$, one gets, at first order, the following relation :

$\dfrac{\text{d}^2 \tilde{x}^{\alpha}}{\text{d}\tau^2} = -\eta^{\alpha\gamma}\bigg[g_{\mu\gamma,\nu\beta}-\dfrac{1}{2}g_{\mu\nu,\gamma\beta}\bigg]\tilde{x}^{\beta}\dfrac{\text{d}\tilde{x}^{\mu}}{\text{d}\tau}\dfrac{\text{d}\tilde{x}^{\nu}}{\text{d}\tau}\quad\quad(eq3)$

I can't manage to obtain the $eq(3)$ from $eq(1)$ and $eq(2)$, if someone could help me for the details of the demonstration, this would be nice.

Thanks in advance for your help

 

[Orodruin]

Please show us what you did get.

 

[fab13]

For the moment, if I put the definition of $g_{\nu\mu}$ into $eq(2)$, I get :

$\dfrac{\text{d}}{\text{d}\tau}\bigg(g_{\mu\nu}(x)\dfrac{\text{d}x^{\nu}}{\text{d}\tau}\bigg)=\bigg(\dfrac{\text{d}g_{\mu\nu}}{\text{d}\tau}\bigg)\,\dfrac{\text{d}x^{\nu}}{\text{d}\tau}+g_{\mu\nu}\dfrac{\text{d}^2x^{\nu}}{\text{d}\tau^2}\quad\quad eq(4)$

If I separate the two terms on RHS on $eq(4)$ :

$\bigg(\dfrac{\text{d}g_{\mu\nu}}{\text{d}\tau}\bigg)\,\dfrac{\text{d}x^{\nu}}{\text{d}\tau}=\dfrac{\text{d}}{\text{d}\tau}\bigg(\eta_{\mu\nu}+\dfrac{1}{2}g_{\mu\nu,\alpha\beta}(0)\tilde{x}^{\alpha}\tilde{x}^{\beta}\bigg)\dfrac{\text{d}x^{\nu}}{\text{d}\tau}=$

$\dfrac{1}{2}\,g_{\mu\nu,\alpha\beta}(0)\bigg[\dfrac{\text{d}\tilde{x}^{\alpha}}{\text{d}\tau}\,\tilde{x}^{\beta}+\tilde{x}^{\alpha}\,\dfrac{\text{d}\tilde{x}^{\beta}}{\text{d}\tau}\bigg]\dfrac{\text{d}x^{\nu}}{\text{d}\tau}=$

$g_{\mu\nu,\alpha\beta}(0)\bigg[\dfrac{\text{d}\tilde{x}^{\alpha}}{\text{d}\tau}\,\tilde{x}^{\beta}\bigg]\dfrac{\text{d}x^{\nu}}{\text{d}\tau}$

Concerning the second term on RHS of $eq(4)$, 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 $(eq2)$ :

$-\dfrac{1}{2}\dfrac{\partial g_{\lambda\nu}}{\partial x^{\mu}}\dfrac{\text{d}x^{\lambda}}{\text{d}\tau}\dfrac{\text{d}x^{\nu}}{\text{d}\tau}\quad\quad eq(5)$ ??

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 $eq(5)$ , so I don't know how to get it ?

Any help is welcome

 

[fab13]

Can't anyone bring a little help ?