# A Hubble parameter vs Scale factor's derivative

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1. Oct 2, 2016

### Elnur Hajiyev

How does Hubble parameter and scale factor's derivative differ geometrically? I am reading S. Caroll's GR book. But I cannot get the full representation of these two parameters. On the book, it says
How can $\dot{H}$ and $\ddot{a}$ be opposite of each other on the same instance if both are representing the speed of expansion?

2. Oct 2, 2016

### vanhees71

The Hubble parameter is defined as $H=\dot{a}/a$, where $a$ is the scale parameter in the fundamental observer's reference frame, where the FLRW metric takes the form
$$\mathrm{d} s^2=\mathrm{d} t^2 - a^2(t) \left (\frac{\mathrm{d} r^2}{1-k r^2} + r^2 \mathrm{d} \theta^2 + r^2 \sin \theta \mathrm{d} \phi^2 \right).$$

3. Oct 2, 2016

### Elnur Hajiyev

I know it. I need an explanation from a geometrical perspective.

4. Oct 2, 2016

### Staff: Mentor

Because they're not the same quantity. We have

$$\dot{H} = \frac{d}{dt} \frac{\dot{a}}{a} = \frac{\ddot{a}}{a} - \left( \frac{\dot{a}}{a} \right)^2 = \frac{\ddot{a}}{a} - H^2$$

They aren't; they are representing the acceleration (or deceleration) of expansion, but in different ways. The way I find it easiest to understand the difference is to consider exponential expansion, i.e., $a = e^t$ (in the simplest case). Then we have $\dot{a} = \ddot{a} = e^t$ as well. But we have $H = 1$ and $\dot{H} = 0$. In other words, $H$ is measuring the (inverse) time constant of exponentiation, and $\dot{H}$ is measuring how that (inverse) time constant changes.

Contrast this with, for example, a pure matter-dominated critical density universe where $a = t^{2/3}$. Then $\dot{a} = (2/3) t^{-1/3}$ and $\ddot{a} = - (2/9) t^{-4/3}$. So we have $H = (2/3) t^{-1}$ and $\dot{H} = - (2/3) t^{-2}$. So here $H$ and $\dot{H}$ both tend to zero, $H$ from above and $\dot{H}$ from below, and $\ddot{a}$ is negative the whole time.

In our current universe, which is dark energy dominated but still has significant matter density affecting the expansion rate, $\dot{H}$ is negative even though $\ddot{a}$ is positive, because the expansion is accelerating but the matter density is slowing down the acceleration somewhat; the function $a(t)$ is basically a combination of the two cases I gave above. As the universe continues to expand, the matter density will decrease while the dark energy density remains constant, and $\dot{H}$ will tend to zero from below, while $H$ will tend to a constant value determined by the dark energy density. But $\ddot{a}$ will be positive during all of this.

5. Oct 4, 2016

### Elnur Hajiyev

Thanks for your detailed answer. But I would like to see somewhat generalization in the explanation of this question. I mean, generally, how does Hubble parameter and derivative of the scale factor or derivative of Hubble parameter and second derivative of the scale factor differ from each other. I understand your examples related to these concepts, but still struggle to imagine them in the geometrical viewpoint.

6. Oct 4, 2016

### phyzguy

It's fairly easy to see why $H = \frac{\dot a}{a}$. Suppose some object is at a distance from you of x0 at time t=0 and all distances are growing by a factor of a(t). Then the distance to the object is given by x = x0 * a(t). So the velocity at which you see the object moving away from you is $\dot x=x_0 \dot a(t)$. The Hubble constant is defined to be the ratio of the velocity of the object divided by the distance to the object, so $H = \frac{\dot x(t)}{x(t)} = \frac{x_0 \dot a(t)}{x_0 a(t)} = \frac{\dot a}{a}$.

7. Oct 4, 2016

### Staff: Mentor

I'm not sure I understand what you mean by "in the geometrical viewpoint". $H$ and $\dot{H}$ are dimensionless as far as distance is concerned (their only dimensions are time, more precisely time^-1 for $H$ and time^-2 for $\dot{H}$), so they can be thought of as fractional change. That is the key difference between them and $\dot{a}$ and $\ddot{a}$ conceptually, since the latter two have a dimension of distance in the numerator.

8. Oct 5, 2016

### Elnur Hajiyev

I mean, if both of these quantities($\dot{H}$,$\ddot{a}$) represent acceleration of the expansion, how can they differ in ther signs? Today's universe $\dot{H}$ is negative, while $\ddot{a}$ is positive. What do they mean separately? If the expansion of the universe is accelerating, how can $\dot{H}$ be negative?

9. Oct 5, 2016

### Staff: Mentor

Because they are representing "acceleration of the expansion" in different ways.

In the case of $\ddot{a}$, the representation is simple: positive $\ddot{a}$ means acceleration, negative $\ddot{a}$ means deceleration. So you only need to look at the sign to see which is happening.

In the case of $H$ and $\dot{H}$, the representation is more complex: here acceleration means that $H$ is tending towards some positive value as $t \rightarrow \infty$, whereas "deceleration" means that $H$ is tending towards zero as $t \rightarrow \infty$ (or, if we include the possibility of a closed universe, we will have $H \rightarrow 0$ as $t \rightarrow T$, where $T$ is some finite time).

10. Oct 15, 2016

### Elnur Hajiyev

Yeah, but what does it mean, H is tending towards zero or not? This is the part I have been confusing. Because it is mathematical description, and I have a difficulty in implementing it to real world.

11. Oct 15, 2016

### Staff: Mentor

If $H$ is tending towards zero, there is no limit to how slowly the universe can be expanding; it can even stop expanding and start contracting (if $H$ reaches zero in a finite time).

If $H$ is tending towards some nonzero positive value, then the universe will end up expanding exponentially forever (since a constant $H$ corresponds to exponential expansion).

A key difference between these two scenarios (if we leave out the possibility of the universe starting to contract, and only consider $H \rightarrow 0$ vs. $H \rightarrow H_{\infty} > 0$ as $t \rightarrow \infty$) is that in the former scenario, there is no event horizon in the universe. But in the latter scenario (exponential expansion), there is.

(NOTE: Edited to correct misstatement about event horizon.)

Last edited: Oct 15, 2016
12. Oct 15, 2016

### Elnur Hajiyev

So in our universe, both $\ddot{a}$ and $\dot{H}$ are positive, yes?

13. Oct 15, 2016

### George Jones

Staff Emeritus
No, in our universe, $\ddot{a}$ is positive and $\dot{H}$ is negative, i.e., the Hubble parameter is decreasing,

14. Oct 15, 2016

### Elnur Hajiyev

And this is the confusing part for me. Peter has said, if H is tending towards some positiv value($\dot{H}$ is positive), two comoving observers will eventually be disconnected. I knew, particle horizon is shrinking over time, which means everything near us will be causally disconnected from us one day. Isn't it true?

Last edited: Oct 15, 2016
15. Oct 15, 2016

### Staff: Mentor

Yes.

No. $\dot{H}$ is negative, because the current value of $H$ is larger (more positive) than the positive value towards which $H$ is tending. I explained this in post #4.

16. Oct 15, 2016

### Staff: Mentor

Yes. But that has nothing to do with the sign of $\dot{H}$. It has to do with the fact that $H$ is tending (from above) towards a nonzero positive value instead of towards zero.

17. Oct 15, 2016

### George Jones

Staff Emeritus
This is impossible.

18. Oct 15, 2016

### Elnur Hajiyev

Yeah. Now it has become more clear to me. Thanks for sparing your time to write detailly answer the question.

19. Oct 15, 2016

### Staff: Mentor

Oops, you're right, I misstated it. I should have said there is an event horizon in a dark energy dominated universe and left it at that.