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Consider the FRW metric for the Universe in the form found in many standard cosmology textbooks:

$$ds^2 = -dt^2 + a(t)^2\left(\frac{dr^2}{1-Kr^2}+r^2(d\theta^2 + \sin^2\theta d\phi^2)\right)$$

I am confused as to what $r$, $\theta$ and $\phi$ represent in this formula. For example, some texts introduce this topic by considering a 2-sphere as opposed to the 3-sphere that is described when $K=1$ in the formula above. For the 2-sphere we have the spatial line element:

$$dl^2 = \frac{dr^2}{1-r^2}+r^2d\theta^2,$$

where $r$ is the distance of a point on the surface of the 2-sphere from the $z$ axis, and $\theta$ is the angle that $r$ makes with the positive $x$-axis. It occurs to me that we have essentially used cylindrical coordinates to describe points on the surface of the sphere in this case. We may then associate $r$ with $\sin \chi$, where $\chi$ is the angle that the position vector of a point on the sphere makes with the positive $z$ axis.

My trouble comes when we then extend this argument to the 3-sphere. What exactly do the parameters now represent? To illustrate my problem: suppose we wish to calculate the volume of a sphere of radius $R_0$ that exists on the surface of the 3-sphere ($K = 1$ Universe). How would we do that using this metric? The volume element would be easy enough to write down, but in order to perform the integration we would need to know what limits to place on $r$, $\theta$ and $\phi$. This is an impossible task if one does not understand the physical significance of the parameters in this more general case.

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The coordinates for a general sphere, with its centre at some arbitrary place, would be quite hard to capture in a single expression. But if you consider a sphere (I mean a 2-sphere located in the 3-sphere manifold) which is centred at the origin of coordinates, then the limits are easy: just pick one value for $r$ and let $\theta$ and $\phi$ vary over their full ranges ($\pi$ and $2\pi$ respectively). Thus you can get the surface area very easily, and for the volume you would do an integral over $r$.

To locate the surface of a 2-sphere (in the 3-sphere manifold) more generally, one way would be first to specify the location $(r_0,\theta_0,\phi_0)$ for the centre, and then find the locus of points at a fixed distance from that centre, by integrating the metric along a set of geodesics going outwards from the centre. I feel certain that this is algebraically not the most straightforward way, and really you would use a bunch of tricks from differential geometry, but unfortunately I do not know them.

Finally, now, an answer to the general question as to the meaning of $r,\theta,\phi$. It is a good question. $r$ is a coordinate which increases as one moves along a line outwards from the chosen origin of coordinates. $\theta$ and $\phi$ take you around circles centred on the origin, and together around a spherical surface at fixed $r$. So these coordinates are very much like the familiar spherical polar coordinates which can be used in Euclidean geometry. But be careful, this is a statement about their role in the 3-sphere manifold itself, not their relation to any embedding of that manifold in a higher-dimensional space.

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  • $\begingroup$ It is the 3-sphere manifold that I am mainly interested in. We talk about two events being separated by $(\delta t, \delta r, \delta \theta, \delta \phi)$, for example, but what are these actually measuring? If we inhabit the surface of a 3-sphere, what is $\delta r$, for example? Is it the radial distance from us to another point on the surface, or the distance from the fourth dimension axis to that point? $\endgroup$
    – wrb98
    Apr 25 at 16:03
  • $\begingroup$ @wrb98 $\delta r$ is the radial coordinate distance from one event to another on/in the 3-sphere manifold at the same $t$. It has nothing to do with anything outside the manifold. The corresponding proper distance is $a \delta r / \sqrt{1 - K r^2}$. $\endgroup$ Apr 25 at 19:06
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A common elementary coordinatization of the $2$-sphere of radius $R$ uses two angles $(\theta,\phi)$ as (co)latitude and longitude. If we embed the sphere into $\mathbb R^3$, then the points $(x,y,z)$ on the surface take the form $$\pmatrix{x\\y\\z} = \pmatrix{R \sin(\theta)\cos(\phi)\\R\sin(\theta)\sin(\phi)\\ R\cos(\theta)}$$

If we restrict $\theta \in (0,\pi)$ and $\phi\in (0,2\pi)$, then this constitutes a coordinate chart which covers all of the $2$-sphere except for the poles and the line $\phi=0$ which connects them. In this chart, the metric takes the form

$$\mathrm ds^2 = R^2 (\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2)$$


An alternative approach is the following. Rather than using the polar angle $\theta$ as a coordinate, we can choose to use the distance from the $z$-axis, given by $R\sin(\theta)$, which we will call $r$. Equivalently, $r$ is the circumference of the circle centered at the north pole divided by $2\pi$. Note that we can only cover the northern (or southern) hemisphere of the sphere in this chart, but that's okay. Embedding the $2$-sphere in $\mathbb R^3$, we would have $$\pmatrix{x\\y\\z} = \pmatrix{r \cos(\phi)\\r\sin(\phi) \\ \sqrt{R^2-r^2}}$$ In this chart, the metric takes the form $$\mathrm ds^2 = \frac{1}{1-kr} \mathrm dr^2 + r^2 \mathrm d\phi^2,\qquad k\equiv \frac{1}{R}$$

This should look familiar.


The extensions to the $3$-sphere are straightforward. The "spherical coordinate" embedding uses three angles $\psi,\theta,\phi$ and takes the form

$$\pmatrix{x\\y\\z\\w} = \pmatrix{R\sin(\psi)\sin(\theta)\cos(\phi)\\R\sin(\psi)\sin(\theta)\sin(\phi)\\R\sin(\psi)\cos(\theta)\\R\cos(\psi)}$$

Defining $r\equiv R\sin(\psi)$, this becomes $$\pmatrix{x\\y\\z\\w}=\pmatrix{r\sin(\theta)\cos(\phi)\\ r\sin(\theta)\sin(\phi) \\ r\cos(\theta)\\ \sqrt{R^2-r^2}}$$

and in this chart, the metric takes the form

$$\mathrm ds^2 = \frac{1}{1-kr} \mathrm dr^2 + r^2(\mathrm d\theta^2 + \sin^2(\theta) \mathrm d\phi^2)$$ where once again $k\equiv 1/R$.


My trouble comes when we then extend this argument to the 3-sphere. What exactly do the parameters now represent?

In the 2D case, the set of points equidistant from the origin constitutes a (generalized) circle (a $1$-sphere); the coordinate $r$ is the circumference of that circle divided by $2\pi$. It is not the radius of the circle, despite the deceptive name.

Generalizing to the 3D case, the set of points equidistant from the origin constitutes a $2$-sphere; the coordinate $r$ now represents the surface area of that sphere divided by $4\pi$. This is the same interpretation as the "radial" Swarzschild coordinate, for example.

Once you've chosen some $r$, you've restricted your attention to a $2$-sphere of points which sit at the same distance from the coordinate origin. The angles $\theta$ and $\phi$ specify a point on this $2$-sphere precisely as they usually do in elementary spherical coordinates.

To illustrate my problem: suppose we wish to calculate the volume of a sphere of radius $R_0$ that exists on the surface of the 3-sphere (K=1 Universe). How would we do that using this metric?

The surface area of a sphere of radius $r$ is simply $4\pi r^2$, which follows immediately from the interpretation provided above. The volume of the ball of radius $R_0$ is then straightforwardly

$$V = \int_0^{R_0} 4\pi r^2 \frac{1}{1-kr} \mathrm dr = 4\pi \frac{kR_0(kR_0+2)+2\log(1-kR_0)}{2k^3}$$ which reduces to $\frac{4\pi R_0^3}{3}$ in the limit as $kR_0 \rightarrow 0$, as expected.

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  • $\begingroup$ I don't think this answers my question. Rather it is just rephrasing what I have already talked about, and does not really address any of the questions I raised. Specifically, no physical interpretation for $r$ is given in the second case, nor even for $\psi$. $\endgroup$
    – wrb98
    Apr 25 at 15:59
  • $\begingroup$ @wrb98 I have expanded my answer to more explicitly address your questions. $\endgroup$
    – J. Murray
    Apr 25 at 16:54
  • $\begingroup$ I think this illustrates my problem a little more: Suppose you are observing an object a fixed comoving distance $S_K(\chi)$ away. Then from the metric, why is the area of the sphere centred on you and passing through that object not $S_K(\chi)^2d\Omega^2$, as opposed to $r^2d\Omega^2$? $\endgroup$
    – wrb98
    Apr 25 at 17:25
  • $\begingroup$ @wrb98 Because the space is not flat. If you consider the 2D case, a circle (centered on the north pole) is a line of constant latitude. That circle has a radius $\rho$ (given by the distance along the sphere to the north pole) and a circumference $C$ (the distance around the circle), but they are not related via $C=2\pi\rho$ precisely because the surface of the sphere isn't flat. However, if we label a circle by $r = \frac{C}{2\pi}$ rather than $\rho$, then its circumference is $C=2\pi r$ by definition. $\endgroup$
    – J. Murray
    Apr 25 at 17:31
  • $\begingroup$ My confusion arose when viewing point 1 on page 13 of physics.bu.edu/~schmaltz/PY555/baumann_notes.pdf, which uses $d_m^2 := S_K(\chi)^2$ as the multiplier for the solid angle in FRW spacetime. Is this a mistake? $\endgroup$
    – wrb98
    Apr 25 at 17:36

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