Demonstration of relation between geodesics and FLRW metric

• I

Main Question or Discussion Point

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## :

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 :

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

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.

Related Special and General Relativity News on Phys.org
Orodruin
Staff Emeritus
Homework Helper
Gold Member
Please show us what you did get.

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

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)## :

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

Last edited:
Can't anyone bring a little help ?