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What is the most general form of the metric for a homogeneous, isotropic and static space-time?

For the first 2 criteria, the Robertson-Walker metric springs to mind. (I shall adopt the (-+++) signature) $$ds^2=dt^2+a^2(t)g_{ij}(\vec x)dx^idx^j$$

Now the static condition. If I'm not mistaken, it means that the metric must be time-independent and invariant under time reversal $t\to -t$. So does that mean that the most general metric that satisfies all these 3 criteria is $$ds^2=dt^2+g_{ij}(\vec x)dx^idx^j$$ for some spatial metric $g_{ij}(\vec x)$?

Thank you.

(My apologies for the mistaken definition. I may have misunderstood what's been said in class...)

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Staticity isn't defined in terms of $t\rightarrow -t$. GR doesn't have any notion of time-reversal that applies to all spacetimes. A spacetime doesn't even have to be time-orientable. –  Ben Crowell May 4 '13 at 13:23
    
Thanks, @BenCrowell . So is it just equivalent to stationary...? –  Clarice May 4 '13 at 14:16
    
Questions on stackexchange are expected to show research effort. If you need to know the definition of staticity, look it up. –  Ben Crowell May 4 '13 at 16:35
    
The time reversal criterion is given in class. That's why I thought it was true! –  Clarice May 4 '13 at 17:04
    
Time-reversal only makes sense in some spacetimes. Maybe your professor had in mind some more specific context. Wikipedia defines staticity correctly for the general case: en.wikipedia.org/wiki/Static_spacetime –  Ben Crowell May 5 '13 at 14:49
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1 Answer

The FLRW metric can be static, this is the solution that Einstein concocted before Hubble observed the expansion of the universe. The only way that Einstein could make his equations static was by introducing the infamous cosmological constant $\Lambda$. The general FLRW metric has the form $$ \text{d}s^2 = -c^2\text{d}t^2 + a(t)\left[\frac{\text{d}r^2}{1 - kr^2} + r^2\text{d}\theta^2 + r^2\!\sin^2\!\theta\,\text{d}\varphi^2\right], $$ with $k=-1$, $0$, or $1$ the curvature. The Einstein field equations reduce to the Friedmann equations $$ \begin{align} \dot{a}^2 + kc^2 -\frac{\Lambda c^2}{3}\!a^2 &= \frac{8\pi G}{3}\!\rho\, a^2,\\ 2\ddot{a}a + \dot{a}^2 + kc^2 -\Lambda c^2a^2 &= -\frac{8\pi G}{c^2}\!P\,a^2, \end{align} $$ with $\rho$ and $P$ the mass density and pressure density. For ordinary matter, $P=0$, and for a static universe we have $\dot{a}=\ddot{a}=0$, so that $$ \begin{align} \rho &= \frac{kc^2}{4\pi G a^2},\\ \Lambda &= \frac{k}{a^2}. \end{align} $$ moreover, since $\rho > 0$, it follows that $k=1$. In other words, the static universe has the topology of a 3-sphere. It is an unstable solution though: a small value of $\dot{a}$ or $\ddot{a}$ would make it expand or contract. And of course, since Hubble's observations, we know that the universe is in fact expanding.

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