Does knowing the $FRW$ metric equation is required first so as to solve Einstein's field equation ?

$\Rightarrow$ First FRW, then the christoffel symbols, then the riemann tensor, then the ricci tensor, then the ricci scalar, then the einstein tensor, in the end the stress–energy tensor will form the Einstein equations ? Are these enough to solve Einstein equations ?

  • $\begingroup$ If the metric FRW is the solution of the Einstein Field equation then the Einstein Tensor is equal zero , this is the vacuum solution , the stress–energy tensor equal zero $\endgroup$
    – Eli
    May 7, 2021 at 12:07
  • $\begingroup$ Ok . got it ... $\endgroup$ May 7, 2021 at 12:09
  • 1
    $\begingroup$ @Eli What? The FRW metric isn't a vacuum solution... $\endgroup$
    – J. Murray
    May 7, 2021 at 12:17
  • $\begingroup$ I wrote if it is the solution ? $\endgroup$
    – Eli
    May 7, 2021 at 12:18
  • 1
    $\begingroup$ @Eli You said that the Einstein tensor corresponding to the FRW metric is zero, which it isn't. $\endgroup$
    – J. Murray
    May 7, 2021 at 12:20

1 Answer 1


You can think of the Einstein equations as a set of 10 coupled, nonlinear differential equations where the unknown functions are the components of the metric tensor. Writing them out in their entirety is an absolute nightmare:

$$\partial_\rho \left(\frac{1}{2} g^{\rho \sigma} \left( \partial_b g_{\sigma a} + \partial_a g_{\sigma b} - \partial_\sigma g_{ab} \right)\right) - \partial_b \left(\frac{1}{2} g^{\rho \sigma} \left( \partial_a g_{\sigma \rho} + \partial_\rho g_{\sigma a} - \partial_\sigma g_{\rho a} \right) \right) + \left( \frac{1}{2} g^{\rho \sigma} \left( \partial_\lambda g_{\sigma \rho} + \partial_\rho g_{\sigma \lambda} - \partial_\sigma g_{\rho \lambda} \right)\right) \left( \frac{1}{2} g^{\lambda \sigma} \left( \partial_a g_{\sigma b} + \partial_b g_{\sigma a} - \partial_\sigma g_{ba} \right) \right) - \left( \frac{1}{2} g^{\rho \sigma} \left( \partial_\lambda g_{\sigma b} + \partial_b g_{\sigma \lambda} - \partial_\sigma g_{b \lambda} \right) \right) \left(\frac{1}{2} g^{\lambda \sigma} \left( \partial_a g_{\sigma \rho} + \partial_\rho g_{\sigma a} - \partial_\sigma g_{\rho a} \right) \right) - \frac{1}{2} g_{a b} \Big[ \partial_\rho \left(\frac{1}{2} g^{\rho \sigma} \left( \partial_c g_{\sigma c} + \partial^c g_{\sigma c} - \partial_\sigma g^c_{c} \right)\right) - \partial_c \left(\frac{1}{2} g^{\rho \sigma} \left( \partial^c g_{\sigma \rho} + \partial_\rho g_{\sigma }^c - \partial_\sigma g_{\rho }^c \right) \right) + \left( \frac{1}{2} g^{\rho \sigma} \left( \partial_\lambda g_{\sigma \rho} + \partial_\rho g_{\sigma \lambda} - \partial_\sigma g_{\rho \lambda} \right)\right) \left( \frac{1}{2} g^{\lambda \sigma} \left( \partial^c g_{\sigma c} + \partial_c g_{\sigma }^c - \partial_\sigma g_{c}^c \right) \right) - \left( \frac{1}{2} g^{\rho \sigma} \left( \partial_\lambda g_{\sigma c} + \partial_c g_{\sigma \lambda} - \partial_\sigma g_{c\lambda} \right) \right) \left(\frac{1}{2} g^{\lambda \sigma} \left( \partial^c g_{\sigma \rho} + \partial_\rho g_{\sigma }^c - \partial_\sigma g_{\rho }^c \right) \right) \Big] + \Lambda g_{a b} = \frac{8 \pi G}{c^4} T_{a b}$$

($\LaTeX$ due to Jack Fraser)

For a fixed choice of $a$ and $b$, this monstrosity constitutes a second order differential equation for $g_{ab}$. Since $g$ is symmetric, in $3+1$-dimensional spacetime that makes for 10 combinations. These equations are hopelessly unsolveable in their present form. In order to make progress we need to simplify them, which we often do by imposing symmetries on the metric (and the stress-energy tensor $T$).

In the case of the FRW metric, we impose the constraints of spatial homogeneity and isotropy. For the moment I will also assume a spatially flat universe, but this can be generalized to a universe of constant positive or negative spatial curvature. It takes some work to show exactly what this implies, but once we turn the crank and simplify everything it means that with an appropriate choice of coordinates (called comoving coordinates) the stress-energy tensor is of the form $$T_{\mu\nu} \sim \pmatrix{\rho &0&0&0\\0&p&0&0\\0&0&p&0\\0&0&0&p}$$ and the metric is of the form $\mathrm ds^2 = -c^2\mathrm dt^2 + a(t)^2(\mathrm dx^2+\mathrm dy^2+\mathrm dz^2)$, which reduces the Einstein equations to the two(!) Friedmann equations

$$\frac{\dot a^2}{a^2} = \frac{8\pi G \rho + \Lambda c^2}{3}$$ $$\frac{\ddot a}{a} = -\frac{4\pi G}{3}\left(\rho + \frac{3p}{c^2}\right)+\frac{\Lambda c^2}{3}$$

Once we specify $p$ and $\rho$ (the pressure and energy density of the matter/radiation/whatever in the universe), these equations can be solved to yield $a(t)$, the only unknown degree of freedom in the FRW metric.


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