# Demonstration of relation between geodesics and FLRW metric

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## 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]##

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.

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Orodruin
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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 ?