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The Friedmann–Lemaître–Robertson–Walker (FLRW) metric is as follows, in natural units: $$\mathrm{d}s^2=-\mathrm{d}t^2+a(t)^2\left(\frac{\mathrm{d}r^2}{1-\kappa r^2}+r^2\mathrm{d}\theta^2+r^2\sin^2\theta\,\mathrm{d}\phi^2\right)$$ For the sake of my own visualization, I am attempting to embed a slice of this manifold in $\mathbb{R}^3$, where $\theta=\frac{\pi}{2}$ and $t=$ constant. This transforms the metric into $$\mathrm{d}s^2=a(t_0)^2\left(\frac{\mathrm{d}r^2}{1-\kappa r^2}+r^2\mathrm{d}\phi^2\right)$$ Using the pullback, we know that ($\hat{g}_{\mu\nu}$ is the Euclidean metric) $$g_{\mu\nu}=(\phi^*\hat{g})_{\mu\nu}=\frac{\partial y^\alpha}{\partial x^\mu}\frac{\partial y^\beta}{\partial x^\nu}\hat{g}_{\alpha\beta}$$ From this, the following system follows, solving for $y_1$, $y_2$, and $y_3$: $$ \begin{align} \frac{a(t_0)^2}{1-\kappa r^2}&=(\partial_r y^1)^2+(\partial_r y^2)^2+(\partial_r y^3)^2 \\ r^2a(t_0)^2&=(\partial_\phi y^1)^2+(\partial_\phi y^2)^2+(\partial_\phi y^3)^2 \end{align} $$ These equations elude me. Is there any way to solve this analytically, or will I have to solve numerically?

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  • $\begingroup$ Your metric is missing some squares: $$\frac{dr^2}{1 - k r^2}$$. $\endgroup$ – Cham Dec 2 '19 at 19:34
  • $\begingroup$ Ah, bless you. It has been edited. $\endgroup$ – Tesseract Dec 2 '19 at 19:43
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You're considering some 3D space, of euclidian metric in cylindrical coordinates: $$\tag{1} ds^2 = dx^2 + dy^2 + dz^2 = dr^2 + r^2 \, d\varphi^2 + dz^2. $$ Now introduce a surface in that space, of height $z = f(r)$ (assuming isotropy in the $x \, y$ plane). Then $dz = f^{\prime} \, dr$ and (1) becomes $$\tag{2} ds^2 = (1 + f^{\prime \, 2}) \, dr^2 + r^2 \, d\varphi^2. $$ You want this metric to coincide with the FLRW metric, in case of $\theta = \frac{\pi}{2}$: $$\tag{3} ds^2 = \frac{1}{1 - k r^2} \, dr^2 + r^2 \, d\varphi^2. $$ Thus, you need to impose the following differential equation (assuming $k = 1$. It is trivial for $k = 0$ and there's no solution for $k = -1$ with metric (1)): $$\tag{4} \frac{df}{dr} = \pm \, \frac{r}{\sqrt{1 - r^2}}. $$ This imposes $f(r) = C \mp \sqrt{1 - r^2}$. You may choose the negative sign and $C = 1$, so $z(r) = 1 - \sqrt{1 - r^2}$ for $0 \le r < 1$ (so $z(0) = 0$ and $z(1) = 1$).

Notice that this surface is half a sphere of radius 1 and centered on $z_c = 1$, in 3D Euclidian space: $$\tag{5} x^2 + y^2 + (z - 1)^2 = r^2 + (z - 1)^2 = 1. $$ This embedded sphere corresponds to the geometry associated to the space curvature parameter $k = 1$.

For $k = -1$, you need a pseudo-euclidian metric instead of (1) above.

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