# Why is the reciprocal of the Hubble constant equal to the age of the universe?

I understand that the Hubble constant is the gradient of the line of best fit when we plot Redshift against distance. I understand why the reciprocal of the gradient would give a value for time. But why do we know (or assume) that this value of time is equal to the age of the universe? How do we know it isn't equal to something else, or it isn't just an arbitrary value?

• – J.G.
Jan 3, 2018 at 16:21
• If $H_0$ is the Hubble constant of today, I do not believe cosmology claims $\frac{1}{H_0}$ is the exact age of the universe. But according to models, it gives a good rough estimate of the age. But it is not obvious how good it will be. Jan 3, 2018 at 16:28
• en.wikipedia.org/wiki/Hubble%27s_law#Hubble_time
– user4552
Jan 3, 2018 at 16:54

Hubble's law shows that the redshift velocity of an object is proportional to the distance to the object:

$$v=H \cdot D$$

The redshift velocity being the velocity that would give the observed redshift. At low velocities the amount of redshift is proportional to the redshift velocity.

If you assume the universe to expand linearly, i.e., that the distance between comoving objects grows linearly in time, then the apparent velocity at which a given object seems to move away due to the expansion of space remains constant over time. Assuming the object moved at this same velocity since the big bang, you can calculate how long ago the distance to the object was zero. This time is given by:

$$t=\frac{D}{v}=\frac{1}{H}$$

Note that a consequence of this assumption is that Hubble's constant is actually not a constant, but changes over time: $H=\frac{1}{t}$

Note also that it is far from trivial that this assumption would be legitimate. It depends on the amount of matter and energy in the universe, whether the rate of expansion is constant or not.

As far as we know, it's a coincidence.

$$1/H = a(t)/a'(t)$$. In the early radiation-dominated era, $$a(t) \propto t^{1/2}$$, so $$1/H$$ is 2 times the actual age of the universe. In the later (but pre-modern) matter-dominated era, $$a(t) \propto t^{2/3}$$, so $$1/H$$ is 1.5 times the actual age. In the future dark-energy-dominated era, $$a(t) \propto e^{t/t_0}$$ where $$t_0$$ is roughly 17 billion years, so $$1/H$$ is about 17 billion years regardless of the actual age.

The present era happens to be near the crossover point where $$1/H$$ goes from being too large to too small. Unless our cosmological model is completely wrong or there's some anthropic reason why we have to live in this era, it's just a coincidence.

• Interesting. Do you have a source for the equations you mentioned that I can read up on? Aug 5, 2020 at 10:49
• @user43712 Not a great source, but it's covered in Wikipedia: equation of state (cosmology). Aug 5, 2020 at 11:12
• thanks! I'll read up on it Aug 5, 2020 at 11:30

The Hubble constant is the reciprocal of the age of the universe for any constant rate of expansion. But in the standard cosmological model the rate of expansion is not well defined, we only have an increasing scale factor.

Actually the universe is also modelled rather well as a 3-sphere. The diagram, if it shows, is a one-to-one projection of a 3-sphere onto infinite Euclidean space (of course there is also an inverse projection of infinite Euclidean space onto a 3-sphere). A 3-sphere is a 3-dimensional object embedded in four dimensions. If the four dimensions are spacetime then the 3-sphere must expand, which allows the conversion of time to length, and the rate of expansion is overwhelmingly likely to be the speed of light. In this projected universe, the expansion rate is constant and the Hubble constant is the reciprocal of the age. A number of other parameters can be calculated rather easily in this projected universe which could be viewed as an analogue computer for our universe.

• You should probably add some details to improve the clarity of the answer. Apr 22, 2020 at 14:41