Hubble’s constant YouTube

Hubble's law

Consider two galaxies, separated (along the $x$-axis for simplicity) by the distance $\Delta x$, as measured in co-moving coordinates. As defined, $\Delta x$ remains fixed over time.

Then, the physical distance between them is $$D(t) = a(t) \Delta x$$

Now, we can calculate the speed, $V$ with which these galaxies are moving apart, by differentiating both sides with respect to time (time is identical in both physical coordinates and co-moving coordinates), to get $$V = \dot{D} = \dot{a} \Delta x$$

We can re-write this expression using an old mathematical trick, which is to multiply and divide the right-hand side by $a$, to get $$V = \dot{D} = \frac{\dot{a}}{a} \left(a \Delta x\right)$$ so that we can replace $D(t) = a \Delta x$, $$V = \frac{\dot{a}}{a} D$$

Notice that since the two galaxies were chosen arbitrarily, this relationship holds for any two galaxies. That means that the quotient, $$H = \frac{\dot{a}}{a}$$ called Hubble's constant, is a universal parameter. Whilst it's technically not a constant, since it depends on time, it is constant in space.

Hubble's law is the equation, $$V = H D$$ and it simply states that, for any two arbitrarily chosen galaxies, the further they are apart, the faster they move apart. It has been shown to be very accurate on large enough scales.

Of course, since $H = \frac{\dot{a}}{a}$ does vary with time, in measuring the speed and distance of various galaxies, we have to take into account the fact that we might be seeing them as they were at different times in the past.

In the next lecture we shall look at how $a(t)$ and $\dot{a}$ do vary with time, but for now, we note that it turns out that $H$ varies very slowly indeed - on the scale of the age of the universe - so for smaller times (and distances), we can treat $H$ as a constant in both space and time.

Age of the universe

Suppose, for argument's sake, the speed, $V$, of recession (for any pair of galaxies) was constant over time. We actually think it decreases over time, but just for now, suppose it's constant over time for all pair of galaxies.

Now, if we also suppose that at the big bang, everything was packed into a small region, then for a given pair of galaxies, we can write $$D = VT$$ where $T$ is the age of the universe

Thus, $$T = \frac{D}{V} = \frac{1}{H}$$ is a rough approximation of the age of the universe. Since $V$ is actually decreasing over time, $\frac{1}{H}$ becomes an upper bound for the age of the universe. A more accurate statement might thus be $$T \lt \frac{1}{H}$$

Light in co-moving coordinates

In special relativity, we characterised light rays by saying that the proper time measured along a light ray is zero, $$\mathrm{d}\tau = 0$$ (assuming, that the speed of light has been set to be $c \equiv 1$). In other coordinates, we then find that a ray of light travelling along the $x$-axis satisfies $$0 = \mathrm{d}\tau^2 = \mathrm{d}t^2 - \mathrm{d}x^2$$ implying that $$\mathrm{d}t = \mathrm{d}x$$ (there are two possibilities actually, light could be travelling in either the positive or negative $x$-direction).

Suppose that $\mathrm{d}\tau = 0$ stills holds for general relativity, and in particular for co-moving coordinates. It would mean that, for light along the $x$-axis, $$0 = \mathrm{d}\tau^2 = \mathrm{d}t^2 - a(t)^2\mathrm{d}x^2$$ and so $$\mathrm{d}t = a(t) \mathrm{d}x$$ or $$\frac{\mathrm{d} x}{\mathrm{d} t} = \frac{1}{a(t)}$$

Since $a(t)$ is an increasing function of time (due to the expansion assumption), this says that, in co-moving coordinates, light slows down over time. This is. of course, just a peculiarity of co-moving coordinates.