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?
Answer
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 ˙D=˙ax=(˙aa)ax, which we can write as v=HD, where v=˙D is the recession velocity and H=˙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 ˙a>0. In fact, the expansion of the universe is accelerating, which means that ˙a(t) increases over time as well. However, the ratio H=˙a/a decreases, because a(t) increases more rapidly than ˙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)∼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.
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