- #1

Diferansiyel

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I want to derive the Friedmann equations from Einstein Field Equations. However, I have a problem that stems from the energy-momentum tensor. I am also trying to keep track of ## c^2 ## terms.

FRW Metric:

$$ ds^2= -c^2dt^2 + a^2(t) \left( {\frac{dr^2}{1-kr^2} + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2} \right)$$

EFE:

$$ G_{\mu \nu} \equiv R_{\mu \nu }- \frac{1}{2} R g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} $$

I calculated metric dependent components which are Christoffel symbols, Riemann Curvature Tensor, Ricci Tensor & Scalar. What I've found is

$$ R_{tt} = -3 \frac{\ddot{a}}{a} $$

$$ R_{ii} = \dfrac{g_{ii}}{c^2a^2}( a \ddot{a} +2 \dot{a}^2 +2kc^2 ) $$

$$ R=g^{\mu \nu}R_{\mu \nu} = 6\left[\frac{\ddot{a}}{ac^2}+ \left(\frac{\dot{a}}{ac}\right)^2 +\frac{k}{a^2}\right] $$

For time-time components I want to obtain:

$$ \boxed{\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3c^2} \rho - \frac{kc^2}{a^2(t)}} $$

and for the space-space components:

$$ \boxed{\frac{2\ddot{a}}{a} + \left(\frac{\dot{a}}{a}\right)^2 = -\frac{8 \pi G }{c^2}p - \frac{kc^2}{a^2(t)}} $$

here ## \rho ## is the

**energy density**, not the mass density.

So in order to obtain the boxed equations, energy-momentum tensor ## T_{\mu \nu} ## must be ## diag(\rho c^2, p g_{ij}) ## . However from the expression ## T_{\mu \nu}=\left(\rho + \frac{p}{c^2}\right)u_\mu u_\nu +pg_{\mu \nu} ## I can't get what I want for ## u^\alpha=(c,0,0,0) ## . Am I using wrong form of ## T_{\mu \nu} ## or 4-velocity ## u^\alpha ## ?

Thanks for your help,

K.