# Can the age of the universe actually be calculated through Hubble's constant?

I was reading a high-school physics textbook, and it stated that the age of the universe is equal to 1 / Hubble's constant. They even give a derivation: $$v = H_{0} D$$ and $$D = vt$$, so subbing in gives $$t = 1/H_{0}$$. But this can't be correct as this states that a variable equals a constant, aside from that, it would make sense to me to say that the age of the universe was the size divided by Hubbles constant, but why could you assume the size is 1?

I have seen people explain that this is actually just an approximation, but I would appreciate if someone can explain why this approximation has any validity at a level that I (a 12th grade physics student) could understand.

• As a mathematical comment, solving an equation such as $2t=6$ to get $t=3$ does not lead to a contradiction like "a variable equals a constant". What it leads to, instead, is that the only value of that variable which, when substituted, makes the equation true, is the value $t=3$. You can feel free to substitute other values of that variable, like $t=5$, to get false equations like $2 \cdot 5 = 6$. Commented Apr 27 at 14:41

The Hubble constant is called a "constant", because it has the same value everywhere in the Universe. However, it is changing with time, as in the past the Universe was expanding at a different rate than today.

Regarding your second question: if the units of Hubble constant were the units of speed, then indeed to obtain a time (the age of the Universe) we would need to divide a distance by speed. However, the units of the Hubble constant are just reversed units of time, as you can see from it's definition: $$v=HD\quad\Rightarrow\quad H=\frac vD.$$ Basically, the Hubble constant tells you at what speed the objects are moving away from you, if they are located at a unit distance from you. Then, logically, if you assume that the speed of the objects was the same in the past , you can tell when these objects have separated from you: you divide their current distance from you (which is 1), by their current speed (the Hubble constant $$\times$$ 1). So, $$1/H$$ time ago the objects were at the same place as you, that's when the Universe was born. The critical assumption here is that the speed of the expansion of the Universe was the same in the past. Since it is not exactly so, the formula $$1/H$$ gives just an approximation of the age of the Universe.

By the way, there is a significant difference between saying "the speed of the expansion of the Universe was the same in the past" and "the Hubble constant was the same in the past". The former is probably more or less accurate, while the latter is not. We just saw, that to calculate the age of the universe today we can use this formula: $$\mbox{age of the Universe today} = \frac1{H_{\rm today}} \quad\Rightarrow\quad H_{\rm today} = \frac1{\mbox{age of the Universe today}}$$

If we use this formula to calculate the Hubble constant when the Universe was 2 times younger, we'll get 2 times larger value. Why? Consider two objects A, and B, such that today object A is at unit distance from us, while object B is two times further from us (at distance 2 units). Then the object A is running away from us at speed $$H_{\rm today}$$, while object B is running away at speed $$2H_{\rm today}$$. Suppose that the speed of the expansion of the Universe was the same in the past, that is the objects A and B maintained their speeds in the past the same as they are today. Then, when the Universe was 2 times younger, the objects were two times closer to us than they are today. So, back then the object B was at the same place, where the object A is today, at the unit distance from us. If we would like to calculate the Hubble constant back then, we would need to take the speed of the object which was at the unit distance from us at the time, that is the object B. But it is still moving at the speed $$2H_{\rm today}$$, so indeed $$H_{\mbox{when the Universe was 2 times younger}} = 2H_{\mbox{today}}$$

To summarize, "the speed of the expansion of the Universe is constant in time" means that objects maintain their speeds approximately the same since the Big Bang. However, the Hubble constant was larger in the past, because objects, which in the past were at the unit distance from us, are now much further, and are moving (and were moving in the past) at larger speed. Thus, the Hubble constant in the past is approximately inverse proportional to the age of the Universe at that time.

If we plot a graph of the distance between say two galaxies in the universe versus time, then the gradient of that graph is the speed, given by Hubble's law.

If I have a point on that curve now, with distance $$d_0$$, where the gradient is $$H_0 d_0$$, then a straight line with this gradient will intercept the distance axis a time $$H_0^{-1}$$ ago. i.e. The line $$d = (H_0 d_0)t + d_0$$, where $$t=0$$ corresponds to now, will have $$d=0$$ when $$t = -1/H_0$$.

This of course assumes that the expansion of the universe is at a constant rate. Which isn't true and so $$H_0^{-1}$$ is not equal to the age of the universe but gives a rough idea of the timescale on which it has been expanding.

That's not even true for the Hubble parameter $$\rm H(a)$$, let alone the Hubble constant $$\rm H_0$$. The age of the universe is

$$\rm t(a)=\int_0^a \frac{d \bar{a}}{\bar{a} \ H(\bar{a})}$$

where

$$\rm H(a)=H_0 \sqrt{\Omega_R/a^4+\Omega_M/a^3+\Omega_K/a^2+\Omega_{\Lambda}}$$

If the time was $$\rm t=1/H_0$$ the time would be frozen, since the reciprocal of a constant is also constant, so your reference seems to just multiply where you're supposed to integrate, which is a beginner's mistake. If the Hubble parameter was constant as well the universe wouldn't have a beginning by the way, that would equal a De Sitter space which is eternal in both time directions.