# A3: Spatially Homogeneous and Isotropic Spacetimes

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[The 'A' series of chapters is an alternative start to the textbook in which chapters 1 through 4 are skipped, so that we dive right into an expanding spacetime. At the end of these A chapters, the student should be ready to move on to chapter 7. A3 is **under construction. **After A3 we might include an A4 which provides optional practice with non-Euclidean spaces -- both the spatial part of FRW and spatial part of Schwarzschild. Oh... somewhere we need to introduce the 'extremizing elapsed time' principle. Maybe in an (also optional) A5.]

[We need to explain homogeneity and istotropy and connect with the 'cosmological principle.']

So far we have worked with spacetimes with just one spatial dimension. But to get to Hubble's law we need to know how to measure distances. And for our particular method of measuring distances, we are going to need to work with more than one spatial dimension, as we will explain later. So the time has come to think about additional spatial dimensions. There appear to be three spatial dimensions, so let's start there.

In Minkowski space, the square of the invariant distance, \(ds^2\), between spacetime point \((t,x,y,z)\) and another one at \((t+dt,x+dx,y+dy,z+dz)\) is given by:

\[ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2.\]

In spherical coordinates the above expression for the invariant distance becomes:

\[ds^2 = -c^2 dt^2 + dr^2 + r^2 \left(d\theta^2 + \sin^2\theta d\phi^2\right). \]

More generally, the invariant distance in a spatially homogeneous and isotropic Universe can be written as:

\[ds^2 = -c^2 dt^2 + a^2(t)\left[\frac{dr^2}{1-kr^2} + r^2 \left(d\theta^2 + \sin^2\theta d\phi^2\right)\right]. \label{eqn:FRWds} \]

Such spacetimes are known as Friedmann-Robertson-Walker (FRW) models, or sometimes just Robertson-Walker, and sometimes Friedmann-Robertson-Walker-Lemaitre models.

### Three-dimensional Homogeneous and Isotropic Spaces

Previously we asserted that one could label a 3-dimensional Euclidean space with coordinates \(r\), \(\theta\), and \(\phi\) such that points separated by \(dr\), \(d\theta\), and \(d\phi\) would be separated by a distance (as one would measure with a ruler) with square given by

\[ d\ell^2 = dr^2 + r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right). \]

A space that can be labeled in this way is homogeneous (invariant under translations) and isotropic (invariant under rotations). The easiest way to see this is to remember that there's a coordinate transformation to Cartesian coordinates for which

\[d\ell^2 = dx^2 + dy^2 + dz^2.\]

Now the homogeneity is more evident, since transforming \(x\) to \(x' = x + L\) would clearly leave the distance rule unchanged. Rotations also leave this distance rule unchanged, although it would take us a little work to show it. So, the space is homogeneous and isotropic. If we choose to label it with spherical coordinates about a particular origin, our labeling obscures the homogeneity and isotropy, but the space itself is still homogeneous and isotropic.

It turns out that whether one can label space in this way or not is a matter to be settled by experiment. It's not necessarily true. Even if we restrict ourselves to completely homogeneous and isotropic geometries, we can mathematically describe spaces that cannot be labeled in this way.

What *is* generally true is that all three-dimensional homogeneous and isotropic spaces can be labeled with coordinates \(r\), \(\theta\), and \(\phi\) such that

\[d\ell^2 = \dfrac{dr^2}{1-kr^2} + r^2\left(d\theta^2 + \sin^2\theta d\phi^2\right)\]

for \(k\) a constant that can be positive, negative or zero. Euclidean space is a special case with \(k=0\).

We can construct the homogeneous and isotropic three-dimensional space and derive its invariant distance rule, at least for the case of \(k>0\), by embedding it in a 4-dimensional Euclidean space. In a 4-dimensional Euclidean space we can have a coordinate system consisting of three dimensions \( x, y, z\), that are all orthogonal to each other, and a fourth we will call \( w \) that is orthogonal to each of the \(x, y,\) and \( z\) directions. Impossible as this is to visualize, we can describe it mathematically. The distance between \(w,x,y,z\) and \(w+dw,x+dx,y+dy,z+dz\) is given by

\[d\ell^2 = dw^2+dx^2+dy^2+dz^2.\]

In this 4-dimensional space, we construct a three-dimensional subspace that is the set of points all the same distance, \(R\), from a common center. Let's center it on the origin so our subspace satisfies this constraint:

\[w^2 + x^2 + y^2 + z^2 = R^2.\]

This subspace is homogeneous (all points are the same) and isotropic (all directions are the same). You can see that this is true by imagining it's two-dimensional analog, a sphere, which is the set of all points satisfying \(x^2+y^2+z^2 = R^2\).

It will be helpful at this point to swap out the Cartesian \(x,y,z\) for the spherical coordinate system \(r,\theta,\phi\) so we have

\[d\ell^2 = dw^2 + dr^2 + r^2(d\theta^2 + \sin^2\theta d\phi^2)\]

and our constraint equation can be written as

\[w^2 + r^2= R^2. \]

From this new version of the constraint equation, we can see that if \(r\) changes by some amount then we will necessarily have to have a change in \(w\) in order to continue to satisfy the constraint. The exact relationship between differential changes you can easily work out to be \(2wdw+2rdr = 0\) (because changing \(r\) by \(dr\) ends up changing \(r^2\) by \(2rdr\) and likewise for \(w\) and \(dw\) and since \(R\) is fixed \(dR^2 = 0\)). Using this relationship to eliminate \(dw^2\) from our invariant distance expression, and using the constraint equation to eliminate \(w^2\) in favor of \(r^2\) and \(R^2\) we get

\[d\ell^2 = \frac{dr^2}{1-r^2/R^2} + r^2(d\theta^2 + \sin^2\theta d\phi^2).\]

We see that our subspace has an invariant distance expression of the form we were intending to derive, and it is exactly the one introduced above if we make the identification \(R^2 = 1/k\).