# Scaling Factor, Hubble's Parameter, and the Age of the Universe

Scaling factor a(t) and Hubble's Parameter H(t)

This video was produced by David Butler. For a copy of the transcript of this video, visit: http://howfarawayisit.com/documents/

Shortly after the precise quantitative predictions of Einstein’s general relativity concerning the precession of Mercury’s perihelion and the deflection angle of rays of light passing the Sun, Einstein moved beyond investigations of the solar system and applied general relativity to the entire universe. He wondered what the effects of gravity would be due to all the masses and energy in the universe. This might seem like an impossible task, but Einstein greatly simplified matters by assuming that the distribution of all the matter in the universe was spatially uniform. He called this assumption the cosmological principle. This means that the distribution of all mass throughout space is homogenous and isotropic. If the mass distribution is homogenous, then if you draw a line in any direction which extends throughout all of space, all of the mass distribution along that line will be equally spaced; isotropy means that the distribution of mass is the same in all directions. If a distribution of mass is both homogenous and isotropic then it is equally spaced and the same in all direction. Later observations (in particular the Cosmic Microwave Background Radiation) proved that on the scale of hundreds of millions of light-years across space, the distribution of galaxies very nearly (up to very miniscule non-uniformities) is completely homogenous and isotropic; thus on this scale the cosmological principle is a reasonable idealization.

Figure 1: The coordinate value $$x^i$$ assigned to each tick mark in the rectangular coordinate system above remains the same as the coordinate system stretches or contracts. Only the scaling factor $$a(t)$$ changes when the coordinate system stretches or contracts.

Imagine that we draw a line through our galaxy that extends across space for hundreds of millions of light-years. Let’s label this line with equally spaced points which have fixed coordinate values $$x^1$$. Imagine that embedded and attached to those points are point-masses (each having a mass $$m$$) which we can think of as galaxies. If we stretch or contract this line, the point-masses (galaxies) will either move away from or towards one another. The coordinate value $$x^1$$ of each mass does not change since, as we stretch the line, the point embedded in the line and the galaxy remain “overlapping each other.” We shall, for simplicity, consider our galaxy to be located at the origin of the coordinate system at $$x^1 = 0$$ although (as we will soon see) the choice of the origin is completely arbitrary. We define the distance between galaxies on this line to be $$D\equiv{a(t)∆x^1}$$ where, based on this definition, the scaling factor $$a(t)$$ is the distance $$D=a(t)\cdot1=a(t)$$ between two galaxies separated by $$∆x^1=1$$. (I repeat, the coordinate value $$x^1$$ of each galaxies doesn’t change and, therefore, the “coordinate separation” $$∆x^1$$ between galaxies doesn’t change.) We will assume that the masses along this line are homogeneously distributed which just means that all of the masses are, at all times $$t$$, equally spaced. In other words, at all times $$t$$, the distance $$D=a(t)(x^1 - x^1_0) = a(t)$$ (where $$∆𝑥¹=(x^1 - x^1_0)=1$$) between any two galaxies on the line separated by $$∆x^1=1$$ with any coordinates $$x^1$$ and $$x^1_0$$; this is just a mathematically precise way of saying that the distance $$D=a(t)$$ between two galaxies separated by “one coordinate unit” doesn’t depend on where we are on the line ($$x^1$$ and $$𝑥^1_0$$ could be anything, the distance will still be the same.)

Let’s draw another line (at a right angle to the first) through our galaxy which, also, extends for hundreds of millions of light-years across space. Let’s also label this line with equally spaced points where galaxies of mass $$m$$ sit on. We will also assume that the distribution of masses along this line is homogenous (meaning they are all equally spaced) and that the spacing between these points is the same as the spacing between the points on the other line (which means that the total mass distribution along both lines is isotropic). Isotropic just means that the distribution of mass is the same in all directions. The equation $$D=a(t)∆x^2$$ is the distance $$D$$ between two galaxies on the vertical line drawn in the picture. We can find the distance $$D$$ between two galaxies with coordinates $$(x^1_0, x^2_0)$$ and $$(x^1, x^2)$$ using the Pythagorean Theorem. Their separation distance $$D_{x^1}$$along the horizontal line is $$D_{x^1}=a(t)∆x^1$$ and their separation distance $$D_{x^2}$$ along the vertical line is $$D_{x^2-axis}= a(t)∆x^2$$. Using the Pythagorean Theorem, we see that $$D=\sqrt{(D_{x^1})+(D_{x^2})}$$. To make this equation more compact, let $$∆r=\sqrt{(∆x^1)^2 +(∆x^2)^2}$$ which we can think of as the “coordinate separation distance” which doesn’t change. Then we can write the distance as $$D=a(t)∆r$$.

If we drew a third line going through our galaxy (at right angles to the two other lines), we could find the distance between two points in space with coordinates $$x^i_0 = (x^1_0, x^2_0, x^3_0)$$ and $$x^i=(x^1, x^2, x^3)$$, using the Pythagorean Theorem in three dimensions, to be

$$D=\sqrt{(∆𝑥^1)^2 + (∆𝑥^2)^2 + (∆𝑥^3)^2}.\tag{1}$$

Equation (1) gives us the distance $$D$$ between any two points with coordinates $$x^i_0$$ and $$x^i$$. Since the galaxies always have fixed coordinate values, we can simply view equation (1) as the distance between any two galaxies in space. (Later on, we will come up with a “particles in the box” model where, in general, the particles will not have fixed coordinate values and it will be more useful to think of Equation (1) as the distance between coordinate points.)

Although the coordinate separation $$∆r$$ between galaxies does not change, because (in general) the space can be expanding or contracting, the scaling factor $$a(t)$$ (the distance $$D$$ between “neighboring galaxies” whose coordinate separation is $$∆r=1$$ ) will vary with time $$t$$ (where $$t$$ is the time measured by an ideal clock which is at rest with respect to our galaxy’s reference frame). (We shall see later on that the FRW equation determines how $$a(t)$$ changes with $$t$$ based on the energy density $$ρ$$ at each point in space and the value of $$κ$$.) Since $$a(t)$$ is changing with time, it follows that the distance $$D=a(t)∆r$$ between any two galaxies is also changing with time. For example, the distance $$D$$ between our galaxy and other, far off galaxies is actually growing with time $$t$$. The fact that the distance $$D$$ between any two galaxies is changing with time according to the scaling factor $$a(t)$$, this means that there must be some relative velocity $$V$$ between those two galaxies as their separation distance increases with time. To find the relative velocity $$V$$ between any two galaxies, we take the time rate-of-change of their separation distance $$D$$ to obtain $$V=dD/dt$$. $$∆r$$ is just a constant and the scaling factor $$a(t)$$ is some function of time; thus the derivative is

$$V=\frac{dD}{dt}=\frac{d}{dt}(a(t)∆r)=∆r\frac{d}{dt}(a(t)).$$

Let’s multiply the right-hand side of the equation by $$a(t)/a(t)$$ to get

$$V=a(t)∆r\frac{d/dt(a(t))}{a(t)}.$$

$$a(t)∆r$$ is just the distance $$D$$ between the two galaxies moving away at a relative velocity $$V$$; thus,

$$V=D\frac{d/dt(a(t))}{a(t)}.$$

The term $$\frac{d/dt(a(t))}{a(t)}$$ is called Hubble’s parameter which is represented by $$H(t)$$:

$$H(t)=\frac{da(t)/dt}{a(t)}.\tag{2}$$

Substituting Hubble's parameter for $$\frac{d/dt(a(t))}{a(t)}$$, we get

$$V=H(t)D.\tag{3}$$

The value of Hubble’s parameter at our present time is called Hubble’s constant and is represent by $$H(today)=H_0$$ . Thus, at our present time, the recessional velocities between any two galaxies is given by

$$V=H_0D.\tag{4}$$

and the value of Hubble’s constant has been measured to be

$$H_0≈500\text{ km/s/Mpc}=160\text{ km/s}.\tag{5}$$

Since $$H_0$$ is a positive constant, this tells us that (at $$t=today$$, not later times, because $$H(t)$$ varies with time) the farther away a galaxy or object is from us (our galaxy), the faster it’s moving away. The bigger $$D$$ is, the bigger $$V$$ is.

By substituting Equation (5) into Equation (4) and by measuring the separation distance $$D$$ between any two galaxies, we can use Equation (4) to calculate the relative, recessional speeds between those galaxies—today. To determine $$V$$ as a function of time, you must compute $$a(t)$$ from the FRW equation, then substitute $$a(t)$$ into Equation (3); but this will be discussed later on. By substituting sufficiently big values of $$D$$ (namely, values which are tens of billions of light-years) into Equation (4), one will discover that it is possible for two galaxies to recede away from one another at speeds exceeding that of light. This, however, does not violate the special theory of relativity which restricts the speeds of massive objects through space to being less than that of light. This is because it is space itself which is expanding faster than the speed of light and general relativity places no limit on how rapidly space or spacetime can expand or contract.

It might seem unintuitive, but the two coordinate points $$x^i_0$$ and $$x^i$$ are not actually moving through space at all. Of course, the galaxies do have some motion and velocity through space; but it is a useful idealization and approximation to assume that they are "attached" to the coordinate points and not moving through space at all. Sir Arthur Edington’s favorite analogy for this was an expanding balloon with two points drawn on its surface. As the balloon expands, the points are indeed moving away from one another; but those points are not actually moving across the space (which in this example, the space is the surface $$S^2$$.)

Age of the universe

We can use Hubble’s Law to come up with a rough estimate of the age of the universe. If all of the galaxies are moving away from one another then that means that yesterday they must have been closer to one another—and a week ago even closer. If you keep running the clock back far enough, then at some time all of the galaxies and matter in the universe must have been on top of each other. Let’s assume that during that entire time interval (which we’ll call $$t_{\text{age of the universe}}$$) the recessional velocity $$V$$ of every galaxy is exactly proportional to $$D$$ (which, empirically, is very close to being true). Then it follows that the ratio $$D/V = 1/H$$ is the same for every galaxy. Since $$1/H$$ stays the same, it follows that $$1/H = 1/H_0$$. Let’s also assume that during the entire history of the universe the velocity $$V$$ of every galaxy remained constant. Then, according to kinematics, the time $$t_{\text{age of the universe}}$$ that it took for every galaxy to go from being on top of one another (when $$D=0$$) to being where they are today is given by the equation $$t_{\text{age of the universe}}=D/V=1/H_0≈\text{14 billion years}$$. (When this calculation was first performed it gave an estimate for the age of the universe of only about 1.8 billion years. Although Hubble correctly measured the recessional velocities of the galaxies, his distance measurements were off by about a factor of ten. Later astronomers corrected his distance measurements.) To come up with a more accurate age of the universe we have to account for the acceleration/deceleration of the galaxies. When we do this we are able to obtain the more accurate estimate which is given by $$t_{\text{age of the universe}}≈\text{13.8 billion years}$$.