Hubble parameter vs Scale factor's derivative

[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

If we set two test particles at a fixed initial distance, and ask how much they have separated a short time thereafter, the answer is given by the Hubble parameter.
If, on the other hand, we pick some fixed source, and ask how it appears to move away from us with time, the answer is given by the change in the scale factor.

How can [itex]\dot{H}[/itex] and [itex]\ddot{a}[/itex] be opposite of each other on the same instance if both are representing the speed of expansion?

 

[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
$$\mathrm{d} s^2=\mathrm{d} t^2 - a^2(t) \left (\frac{\mathrm{d} r^2}{1-k r^2} + r^2 \mathrm{d} \theta^2 + r^2 \sin \theta \mathrm{d} \phi^2 \right).$$

 

[Elnur Hajiyev]

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
$$\mathrm{d} s^2=\mathrm{d} t^2 - a^2(t) \left (\frac{\mathrm{d} r^2}{1-k r^2} + r^2 \mathrm{d} \theta^2 + r^2 \sin \theta \mathrm{d} \phi^2 \right).$$

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

 

[PeterDonis]

How can ##\dot{H}## and ##\ddot{a}## be opposite of each other

Because they're not the same quantity. We have

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

if both are representing the speed of expansion?

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.

 

[Elnur Hajiyev]

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.

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.

 

[phyzguy]

It's fairly easy to see why [itex] H = \frac{\dot a}{a}[/itex]. 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 [itex] \dot x=x_0 \dot a(t)[/itex]. The Hubble constant is defined to be the ratio of the velocity of the object divided by the distance to the object, so [itex] H = \frac{\dot x(t)}{x(t)} = \frac{x_0 \dot a(t)}{x_0 a(t)} = \frac{\dot a}{a}[/itex].

 

[PeterDonis]

I understand your examples related to these concepts, but still struggle to imagine them in the geometrical viewpoint.

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.

 

[Elnur Hajiyev]

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.

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?

 

[PeterDonis]

if both of these quantities(##\dot{H}##, ##\ddot{a}##) represent acceleration of the expansion, how can they differ in ther signs?

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).

 

[Elnur Hajiyev]

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).

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.

 

[PeterDonis]

what does it mean, H is tending towards zero or not?

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.)

 

[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 am have a difficulty in implementing it to real world.

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, any two comoving observers will eventually become causally connected (i.e., within each other's past light cones), and once they are, they will remain so forever. But in the latter scenario (exponential expansion), two comoving observers that are causally connected can become causally disconnected (i.e., outside each other's past light cones), and once they are, they will remain so forever.

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

 

[George Jones]

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

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

 

[Elnur Hajiyev]

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

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?

 

[PeterDonis]

Peter has said, if H is tending toward some positive value

Yes.

(##\dot{H}## is positive)

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.

 

[PeterDonis]

particle horizon is shrinking over time, which means everything near us will be causally disconnected from us one day. Isn't it true?

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.

 

[George Jones]

But in the latter scenario (exponential expansion), two comoving observers that are causally connected can become causally disconnected (i.e., outside each other's past light cones), and once they are, they will remain so forever.

This is impossible.

 

[Elnur Hajiyev]

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.

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

 

[PeterDonis]

This is impossible.

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.