4
$\begingroup$

Is it possible to derive the Friedmann equations from Raychaudhuri's equation, assuming only the Weak Energy Condition? The standard method for the derivation of Friedmann's equations is, if I am not mistaken, by assuming the FLRW metric and plugging it in Einstein's field equations.

Now, the Raychaudhuri equation for the expansion can be derived in a geometrical manner without assumptions for the symmetry of the metric, as is done in Wald or Carroll. For example, one can replace in $$ \frac{dθ}{dτ}=-\frac{1}{3}θ^{2}-σ_{μν}σ^{μν}+ω_{μν}ω^{μν}-R_{μν}U^{μ}U^{ν} $$ the evolution of a representative length scale as $ \theta=\frac{3\dot{l}}{l} $, getting a 2nd order differential equation for the length scale l. Under what assumptions would this be enough to reproduce the Friedmann equation for $ \ddot{a} $ ?

$\endgroup$
2
  • 1
    $\begingroup$ You say "without assumptions for the symmetry of the metric." This confuses me. The FLRW spacetimes are homogeneous and isotropic. Without assuming any symmetry, couldn't you have any spacetime you want? If you do assume symmetry, doesn't, e.g., the vorticity vanish? $\endgroup$
    – user4552
    Feb 11, 2018 at 20:41
  • $\begingroup$ Yes you are right. I thought of it better, and realized that by assuming that the vorticity vanishes, physically it is the same as imposing a symmetry in our spacetime. I guess the question perhaps would be better that way, is it always equivalent to make assumptions for shear and vorticity, with assuming a certain symmetric form of the metric? For example, from a certain point of view one could argue it is not, since the metric isn't a directly observable quantity, while vorticity and shear are measurable quantities. $\endgroup$ Feb 12, 2018 at 15:55

1 Answer 1

3
$\begingroup$

You can! Here's how I did it. We're going to need the following:

The Einstein field equations:

\begin{equation} R_{ab} - \Lambda g_{ab} = 8 \pi G \Big(T_{ab} - \frac{1}{2} T g_{ab} \Big) \end{equation}

The energy-momentum tensor for a perfect fluid with four-velocity $u^a$:

\begin{equation} T_{ab} = (\rho + P) u_a u_b + P g_{ab} \end{equation}

The geodesic congruence to the FLRW metric which is $\xi^a = \partial_t x^a = \delta^{a}_{\ t}$ (note that $\xi^a \xi_a = -1$). The expansion is $\theta := h^{ab} \nabla_b \xi_a = 3 \frac{\dot{a}}{a}$, where $h_{ab} := g_{ab} + \xi_a \xi_b$, and $\nabla_b \xi_a = \frac{\dot{a}}{a} h_{ab}$. Lastly, both $\sigma_{ab} = \omega_{ab} = 0$. This is because

\begin{align} \sigma_{ab} &:= \nabla_{(b} \xi_{a)} - \frac{1}{3} \theta h_{ab}\\ \omega_{ab} &:= \nabla_{[b} \xi_{a]} \end{align}

Thus, given the former, we can now deduce the second Friedmann equation. First we contract the Einstein field equations with $\xi^a \xi^b$ which gives

\begin{equation} R_{tt} + \Lambda = 8 \pi G \Big(T_{tt} + \frac{1}{2} T \Big) \end{equation}

We compute $\xi^a \nabla_a \theta = \partial_t \theta = 3 \Big(\frac{\ddot{a}}{a} - \Big(\frac{\dot{a}}{a} \Big)^2 \Big)$. Since $\sigma_{ab} = \omega_{ab} = 0$, using Raychaudhuri's equation we obtain

\begin{equation} R_{tt} = -\partial_t \theta - \frac{1}{3} \theta^2 = -3\frac{\ddot{a}}{a} \end{equation}

Plugging into the Einstein field equations, we have:

\begin{equation} -3 \frac{\ddot{a}}{a} + \Lambda = 8 \pi G \Big(\rho + \frac{1}{2} (-\rho + 3P) \Big) \end{equation}

\begin{equation} \therefore \frac{\ddot{a}}{a} = -\frac{4 \pi G}{3} (\rho + 3P) + \frac{\Lambda}{3} \end{equation}

Which is the second Friedmann equation. Note that $T_{ab} \xi^a \xi^b = \rho \geq 0$, thus the WEC is satisfied. In fact

\begin{equation} T^{ab} \xi_b T_{a}^{\ c} \xi_c = T_{ab} T^{b}_{\ c} \xi^a \xi^c = -\rho^2 \leq 0 \end{equation}

Which means that the DEC is also satisfied.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.