# What is the general form of the Friedmann-Robertson-Walker metric in 4D?

I have written in some old notes that the FLRW (also known as FRW) metric can be written as:

$$ds^2=dt^2 + a^2 (t) [dr^2 +r^2(d\theta^2 + sin^2\theta d\varphi ^2)] \tag{1}$$

I believe this is its representation in $$4\text{D}$$.

But I have seen in other pages, among them, the Wikipedia page and Robertson-Walker metric and cosmic homogeneity that the equation can be written as:

$$ds^2 = dt^2 - a(t)^2 \Big(\frac{dr^2}{1 - kr^2} + r^2( d\theta^2 + \sin^2\theta \, d\varphi^2 )\Big).\tag{2}$$

Where does the factors under $$dr^2$$ in equation $$(2)$$ come from?

Is the latter the correct form?

• The term k represents the curvature of spacetime. In flat spacetime k=0, so equation 1 holds May 29, 2020 at 12:50
• So the second equation is the correct one? Or can this factor simply be ommited somehow? May 29, 2020 at 12:53
• @Prada $k$ represents spatial curvature. Even when it is zero, the spacetime curvature isn’t zero, because of the $a(t)$. May 29, 2020 at 17:42

The factor $$k$$ is the curvature of space.

Assume that the global geometry of space is that of a sphere of radius $$R$$. Then, $$R^2 = x_1^2 + x_2^2 + x_3^2$$ and the spatial metric can be written $$ds^2 = dx_1^2 + dx_2^2 + \frac{(x_1dx_1 + x_2dx_2)^2}{R^2-x_1^2-x_2^2}.$$

Moving to polar coordinates with $$x_1 = r'\cos \theta$$ and $$x_2 = r'\sin \theta$$, and taking $$r = r'/R$$, one obtains $$ds^2 = R^2 \left(\frac{dr^2}{1-r^2} + r^2 d\theta^2\right)$$ which is the spatial metric of a sphere ($$k=1$$). For a hyperbolic geometry, replace $$R \rightarrow iR$$ in the above giving a metric with $$k=-1$$.

Equation (1) given above the FRW metric assuming that the spatial manifold (i.e. $$t$$=constant hypersurface) is flat meaning that the curvature of the spatial manifold is 0.

While equation (2) is the correct FRW metric. To understand this, the Cosmological principle implies that the spatial manifold is maximally symmetric meaning at all points they have constant curvature (Ricci scalar) everywhere given by $$R=6\kappa$$. Now in order to derive the spatial part of metric $$d\sigma^2=g_{ij}dx^i dx^j$$ (metric $$ds^2$$ when $$dt=0$$), $$(i,j=1,2,3)$$. Since space is maximally symmetric, there will be spherical symmetry: $$$$d\sigma^2= e^{2\beta({r})} dr^2+ r^{2} d\Omega^2$$$$

Using $$R=6\kappa$$ we get $$\beta({r})=-\frac{1}{2}$$ ln$$(1-\kappa{r}^2)$$, thus we get: $$$$d\sigma^2= \frac{d{r}^2}{1-\kappa{r}^2}+{r}^2d\Omega^2$$$$

Thus, at last, we get FRW metric when we consider $$t$$ as cosmic standard time: $$$$d s^{2}=- d t^{2}+R^{2}(t)\left[\frac{d r^{2}}{1-\kappa r^{2}}+r^{2}\left(d \theta^{2}+\sin ^{2} \theta d \phi^{2}\right)\right]$$$$

For $$\kappa > 0$$, the curvature is positive meaning the spatial manifold is 3-sphere which we can check by embedding in 4-dimensions. For $$\kappa < 0$$, the spatial manifold is a hyperboloid by embedding in 4-dimensions.

While $$\kappa=0$$ means that the manifold is flat giving Euclidean space.