# Why does the Hubble parameter keep getting smaller if the expansion of the universe accelerates?

Suppose a region of space at a distance D from Earth is escaping from us with the velocity v. Since it seems like the expansion of the universe is accelerating, things at D from Earth should be receding faster and faster. It would seem from the v = HD equation, that H is then getting bigger in time. However, in other sources we can read, that the Hubble "constant" it's getting smaller with time. How come?

Let's first derive Hubble's Law. Consider a galaxy at present distance $x$ from us. Then, if the local velocity of the galaxy within its cluster is ignored, its cosmic distance will change over time as $$D(t) = a(t)\,x,$$ where $a(t)$ is the so-called scale factor. If we take the time derivative of this equation, we get $$\dot{D} = \dot{a}\,x = \left(\frac{\dot{a}}{a}\right)\,ax,$$ which we can write as $$v = HD,$$ where $v = \dot{D}$ is the recession velocity and $H = \dot{a}/a$ is the Hubble parameter. This is the famous Hubble Law. In other words, the Hubble parameter is a ratio of two quantities. The expansion of the universe implies that $a(t)$ increases over time, and $\dot{a}>0$. In fact, the expansion of the universe is accelerating, which means that $\dot{a}(t)$ increases over time as well. However, the ratio $H = \dot{a}/a$ decreases, because $a(t)$ increases more rapidly than $\dot{a}(t)$. The expansion rate of the universe is not high enough for $H(t)$ to increase.
Note that if the expansion were exponential, $a(t)\sim\exp(Ht)$, then $H$ is constant. In the standard cosmological model, the universe is evolving towards a state of exponential expansion, dominated by a constant dark energy density. In other words, if the standard model is correct, the Hubble parameter is slowly converging towards a constant value, and the expansion rate of the universe is increasing towards an exponential rate.
• And a following question, supposing that I'm right above. What's with the third derivative of the scale factor? If we differentiate $v = \dot{a}\,x$ we get $\alpha = \ddot{a}\,x$ where by $\alpha$ I mean the derivative of velocity, acceleration. If the second derivative of the scale factor is constant, then it appears to me, that at a distance $x$ from Earth the acceleration will always be the same. Yet I don't have any proof, that $\dddot{a} = 0$. – neverneve Jun 2 '14 at 14:00
• @neverneve You need to be careful about how you think about "distance". You can talk about fixed "comoving distance" or fixed "physical distance". Two points separated by a fixed comoving distance get further apart at exactly the rate of expansion of the Universe (but if you transform away the expansion, their relative motion is in one sense zero). Two points separated by a fixed physical distance is the scenario where you hold a rigid rod between the two points and each stays at one end of the rod. Now have a good look at $D(t) = a(t)x$ with this in mind and go from there. – Kyle Oman Jun 2 '14 at 16:20
• Thanks, I may have messed up the distances in the last comment. So $x$ is the comoving distance, and $D(t)$ the proper distance. We differentiate $D(t) = a(t)x$ twice and get $\ddot{D}(t) = \ddot{a}(t)\,x$, which we can transform into $\alpha = \frac{\ddot{a}}{a} \,D$. Since $\ddot{a}$ is positive, than objects further away accelerate faster. However, I'm still wondering if $\frac{\ddot{a}}{a}$ changes in time, i.e. if after a billion years objects at a proper distance D from Earth will accelerate away from us slower, faster, or as fast as today? – neverneve Jun 2 '14 at 17:19
• @neverneve The acceleration $\ddot{a}$ does change over time. You can find the equation in this post and this post. – Pulsar Jun 2 '14 at 18:32