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# 10: A Newtonian Homogeneous Expanding Universe

Let's consider a homogeneous expanding universe in a Newtonian manner: space is fixed, and the universe is filled with a fluid that is moving in such a manner as to keep the density spatially uniform. If homogeneity is to be preserved over time, then the motion must be such that the separation between any pair of fluid elements must scale up with time in the same way. That is, all pairs of fluid elements separated at a time we will call $$t_0$$ by a distance $$\ell$$ will, at a later time $$t$$, all have the same increased distance $$a(t)\ell$$.

To visualize what is happening, it will help to choose one point as the center of your coordinate system, which will (in this coordinate system) remain at rest. A fluid element a distance $$\ell$$ away from the center at time $$t_0$$ will, at any other time, be at distance $$a(t)\ell$$ away from the center. It is thus moving away from the center with a speed $$v = \dot a \ell$$.

Box $$\PageIndex{1}$$

Exercise 10.1.1: Derive Hubble's Law in this universe, $$v = H d$$, where $$H = \dot a/a$$.

\begin{equation*} \begin{aligned} d = a\ell \quad \Longrightarrow \quad \ell = \frac{d}{a} \end{aligned} \end{equation*}

Substituting for $$\ell$$ in $$v = \dot a \ell$$, we get

\begin{equation*} \begin{aligned} v = \frac{\dot a}{a}d \end{aligned} \end{equation*}

Now substituting in $$\dot a/a = H$$, we get

\begin{equation*} \begin{aligned} v = H d \end{aligned} \end{equation*}

Now we'll explicitly demonstrate that Hubble's law is valid even if you chose a different point to be the center of the expansion. If the center of the expansion is at point $$\vec x_1$$, then any point $$\vec x$$, has velocity $$\vec v = \left(\vec x - \vec x_1\right) \dot a$$. We'll call this the vector version of Hubble's Law. An observer at rest with respect to the fluid at $$\vec x_1$$ will see a flow as shown in the left panel of the figure below.  Coordinate system with $$\vec x_1$$ at the center of expansion. Image by Bryan Miller. Coordinate system with $$\vec x_2$$ at the center of expansion. Image by Bryan Miller.

What will an observer at rest with respect to the fluid at location $$x_2$$ see? In our original unprimed coordinate system we have

$\vec v_2 = (\vec x_2 - \vec x_1) \dot a \quad ({\rm unprimed} \; {\rm system})$

and we want $$\vec v'_2 = 0$$ via a Galilean transformation to the unprimed system, so we subtract $$\vec v_2$$ from the unprimed velocities everywhere:

\begin{equation} \begin{aligned} \vec v' &= \vec v - \vec v_2 \quad ({\rm Galilean\ velocity\ transformation} \; {\rm rule}) \\ \\ &= (\vec x - \vec x_1) \dot a - (\vec x_2 - \vec x_1) \dot a \\ \\ &= (\vec x - \vec x_2) \dot a \end{aligned} \end{equation}

So we see that we still have Hubble's Law. We see that arranging the fluid so that it all flows away from a given point with speed linearly dependent on distance preserves homogeneity and isotropy.

Although we first derived Hubble's Law in the context of a relativistic description of spacetime, we now see it arising in a Newtonian context. Fundamentally, Hubble's Law follows from the uniformity of the expansion, whether that's a fluid that's expanding or space itself.

# HOMEWORK Problems

Problem $$\PageIndex{1}$$

Demonstrate that a universe obeying a nonlinear version of Hubble's law would violate homogeneity. Take $$v = H_0 d^2$$ for specificity and work in one dimension for simplicity. Show that assuming this velocity pattern about one point leads to a different velocity pattern around other points, thereby demonstrating the violation of homogeneity. You can do this by thinking of 3 points all in one line, with the central point the same distance from the surrounding two.