I am currently reading Baumann's Cambridge lecture slides on cosmology and am confused on the notion of the Hubble radius. In particular, on page 12 is the following diagram. I would really appreciate a detailed, mathematical explanation as to why the Hubble Sphere takes this particular shape. I understand why the light cone takes the shape it does, but not why the hubble sphere follows that particular curve. There must be a mathematical justification of this, but I am just not seeing it. Any help would be appreciated. enter image description here


1 Answer 1


The Hubble radius is the distance at which recession velocities surpass the speed of light; via Hubble's law, $v = Hr$, the Hubble radius ($r_H$) is simply $r_H = 1/H$ (with $v = c = 1$). To relate this proper distance to a comoving distance we must divide it by the scale factor, $r_{H,com} = r_H/a$.

Now, we'd like to understand how $r_{H,com}$ evolves in time. We can work in conformal time, $d\tau = dt/a$, which I believe is what Baumann refers to as "comoving time" in his diagram. So, let's take the conformal time derivative of $r_{H,com}$, $$r'_{H,com} = \frac{r'_H}{a} - r_H H = -\frac{H'}{H^2a} - 1.$$ Now, $H'= a^{-2}(a'' - 2a'^2/a)$, which gives $$r'_{H,com} = 1 - \frac{a''}{a^3H^2}.$$ From this expression we can see that $r_{H,com}$ is increasing in time as long as $a'' < a^3H^2$. If $a'' < 0$, then we must have $\ddot{a} < 0$ and we have a decelerating universe. The first conclusion we can draw then is that the comoving Hubble radius grows in a decelerating universe. This means that the point at which recession velocities exceed the speed of light is moving away (from earth, say) at a greater speed than the recession velocity at that point (namely, the speed of light). Again, this is a property of decelerating space.

If, however, the universe is accelerating, and sufficiently so such that $a'' > a^3H^2$, then we find that $r'_{H,com} < 0$. Now, the rate of expansion is speeding up, and the comoving Hubble radius is shrinking. The background is expanding at such a rate that points on the Hubble sphere are receding faster than the Hubble sphere itself is growing.

As you probably know, the universe today is accelerating, but it wasn't always. This should make clear why the Hubble sphere in the diagram starts off getting bigger, then turns over as the universe begins to accelerate, and then starts to get smaller.

To derive the exact form of the curve, you need to know $a$ as an explicit function of $\tau$, which depends on the stress-energy content of the cosmology.

  • $\begingroup$ Thank you for your response. It's the dependence on $\tau$ that puzzles me. I can't find any mathematical formula that links $a$ to $\tau$ $\endgroup$
    – wrb98
    May 4, 2021 at 19:28
  • $\begingroup$ Ah, you need the Friedmann and continuity equations. For example, for matter domination, the continuity equation tells us that $\rho \propto a^{-3}$. Meanwhile the FE gives $H^2 \propto \rho \propto a^{-3}$. With $H = a’/a^2$, you can solve this to obtain $a(\tau)$. $\endgroup$
    – bapowell
    May 4, 2021 at 23:52
  • $\begingroup$ Thank you, that is very helpful. One final point: I often see the Hubble radius described alternatively as 'the comoving distance particles can travel over one expansion time'. What exactly does this mean? I have never come across the notion of 'expansion time' before and I am not seeing the connection. Any mathematical justification of this would be very helpful. $\endgroup$
    – wrb98
    May 5, 2021 at 14:10

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