# I Why does Hubble sphere expand with time?

1. Oct 16, 2017

### Happiness

Since the universe's expansion is accelerating, let's suppose it will expands at a higher rate of 700 km per s per megaparsec (10 times the present value) in future. Then the Hubble radius would be 430 megaparsecs (0.1 times the present value). So shouldn't the Hubble sphere contract instead?

"the universe expands at about 70 kilometers per second per megaparsec. This means that a galaxy 1 megaparsec away from us is receding at about 70 km/s, another galaxy 2 megaparsecs away from us is receding at 140 km/s, and so on. This is Hubble's law. Following the same logic, one could do the math to compute how far a galaxy has to be in order to move away at the speed of light. It turns out, galaxies 4300 megaparsecs away from us recede faster than light. This distance defines the "Hubble sphere", an imaginary sphere centered at us, outside which everything recedes faster than the speed of light. Note that, since the universe expands at an accelerated rate, the Hubble sphere increases its radius as time goes by."

Source: http://curious.astro.cornell.edu/ab...peed-of-light-disappear-from-our-observations

2. Oct 16, 2017

### Orodruin

Staff Emeritus
This is not what acceleration means. Acceleration means that $\ddot a$ is positive, not that $\dot H$ is. In fact, in a universe dominated by a cosmological constant, $H$ decreases towards a constant value.

Note that since $H = \dot a/a$, $\dot H = \ddot a/a - \dot a^2/a^2$. Thus, it is not sufficient that $\ddot a > 0$ to conclude that $\dot H > 0$.

3. Oct 16, 2017

### Staff: Mentor

Thread locked for moderation.

4. Oct 16, 2017

### Staff: Mentor

An off topic subthread has been deleted. Thread reopened.

5. Oct 17, 2017

### Happiness

Do H(t) and a(t) take into account farther (in distance) galaxies are more back into the past?

Since H(t) decreases over time, it was bigger in the past. Then should a galaxy 2 megaparsecs away be receding at >140 km/s (since a galaxy 2 megaparsecs away is 6.6 million years ago while a galaxy 1 megaparsec away is 3.3 million years ago)?

6. Oct 17, 2017

### Staff: Mentor

Yes.

Strictly speaking, yes, but 2 megaparsecs isn't far enough away to see a measurable change in H(t). You need to look much further away (and much further back in time) to start seeing a non-linear relationship between distance and recession velocity, which is what a change in H(t) over time amounts to.

Note also that none of the quantities you are quoting (distance to the object, time since the object emitted the light we are seeing now, and recession velocity of the object) are directly measurable. What we directly measure are the redshift of the light from the object, the brightness of the light, and the angular size of the object on the sky. The quantities you are quoting have to be calculated from those observations, and the calculations are model-dependent--that is, they depend on the precise way in which H(t) varies with t, which depends on the model being considered. So the whole process of coming up with the numbers that cosmologists quote in everyday communications is a lot more involved than is usually realized.

7. Oct 17, 2017

### Staff: Mentor

Actually, even this has to be qualified. H(t) is not a velocity, it's a fractional rate of expansion; so you have to multiply it by a distance to get a velocity. But the distance changes over time as well as H(t), because of the universe's expansion.

The language cosmologists usually use, unfortunately, obfuscates this. Distances are usually given as distance "now", so when we say we are comparing two galaxies, one 1 megaparsec away and one 2 megaparsecs away, those distances are distances "now", not distances when the light was emitted, if we use the usual conventions of language that cosmologists use. Similarly, the recession velocities of 70 km/s and 140 km/s are recession velocities "now", which means they are the distances "now" multiplied by H(t) "now". That means the same value of H(t) is used for both, which is why the distance-recession velocity relationship is exactly linear.

In the terms @Orodruin explained in post #2, we have $H = \dot{a} / a$, which means that $\dot{a} = H a$, where $\dot{a}$ is recession velocity and $a$ is distance. (There some further qualifications even here, but I won't go into them in this post.) But all three of $a$, $\dot{a}$, and $H$ are evaluated at the same time $t$ in these equations. So if $a$ is twice as large for one galaxy as for another, then $\dot{a}$ will be as well, since $H$ is the same for both at a given $t$.

What you are trying to do is figure out what the values of those quantities were at the time the light was emitted, which, as you say, will be different for objects at different distances from us "now". So, for example, if galaxy #1 is 1 megaparsec from us "now", and galaxy #2 is 2 megaparsecs from us "now", then those are the values of $a$ at time $t_{\text{now}}$. But to figure out $\dot{a}$ at the time the galaxies emitted the light we are seeing now, you can't use that value of $a$. You have to use the value of $a$ at the time of emission, just as you are wanting to use the value of $H$ at the time of emission. So if the times of emission are $t_1$ and $t_2$, then you will have

$$\dot{a} (t_1) = H(t_1) a(t_1)$$

$$\dot{a} (t_2) = H(t_2) a(t_2)$$

It is true that $H(t_2) > H(t_1)$, but it is also true that $a(t_2) < 2 a(t_1)$. So you can't conclude, without more information, that $\dot{a}(t_2) > 2 \dot{a}(t_1)$, which is what you were trying to conclude. What you can conclude, on the information given, is simply that $H(t_2) > H(t_1)$; that's really what my "strictly speaking, yes" answer referred to (but, as I said, in practical terms a few megaparsecs is not a large enough distance "now" to see any change in $H$).

8. Oct 17, 2017

### Happiness

Am I right to say that we currently have no direct empirical evidence that a galaxy "now" 1 megaparsec away is receding at 70 km/s because its light has not yet reached us? Since this galaxy is 1 megaparsec (3.3 million lightyears) away, the empirical evidence will only reach us 3.3 million years later (in a static universe). And taking into account that the universe is expanding (assumed so for the next 3.3 million years), light leaving this galaxy "now" will have to travel a distance of >3.3 million lightyears to reach us, and so will only reach us >3.3 million years later.

Next, with the interpretation that the distance in Hubble's law is the distance "now", then to verify Hubble's law don't we need to collect empirical evidence over a long period (>>1 million years)? Suppose we have the redshifts of 3 galaxies that are all 1000 lightyears (ly) away from us but measured in different years: galaxy A in 1608, B in 1708 and C in 1808. By assuming an appropriate expansion model of the universe, we could then work out the distances $d_A$, $d_B$ and $d_C$ of these galaxies in the year 608$^i$. Next, using the redshift data, we could verify Hubble's law. (In other words, we change the "now" moment in Hubble's law from year 2017 to year 608.) Suppose $d_A=920$ ly, then $d_B\approx912$ ly$^{ii}$ and $d_C\approx904$ ly$^{ii}$. We have managed to find the distances of these galaxies in the same instant of time (year 608). But from a measurement data set of 200 years, we could only verify Hubble's law over a short distance range of only $\approx16$ lightyears. So it seems that to verify Hubble's law over a distance range of billions of lightyears, we need a data set of >billions of years, which we don't currently have.

Footnotes
$^i$Here, we used the simplifying assumption that a galaxy 1000 lightyears away is exactly 1000 years ago, when in fact, it is <1000 years ago, since due to the expansion of the universe, light leaving the galaxy must travel a longer distance than the required distance in a static universe, which is $c$ times the duration between the emission of light from the galaxy and its detection on Earth.

$^{ii}$In the 1000 years from 608 to 1608, Galaxy A is further from us (due to the expansion of space) by 1000-920=80 ly. Assuming a constant expansion rate (over time), then in the 1100 years from 608 to 1708, B must have "moved" by $\approx88$ ly$^{iii}$, and in the 1200 years from 608 to 1808, C must have "moved" by $\approx96$ ly$^{iii}$.

$^{iii}$These are approximate values, because (in year 608) B and C were closer to us than A was and hence should "move" by <88 ly and <96 ly respectively, which are (approximately) the values A would "move" in 1100 and 1200 years respectively.

Last edited: Oct 17, 2017
9. Oct 17, 2017

### Staff: Mentor

No. The distance "now" is just a convenience--cosmologists use it because it's a number easily calculated from the models, not because it's something we actually observe. So when they say some galaxy is 1 megaparsec away, what they mean is that, based on what we actually observe of that galaxy (the redshift and brightness of its light and its angular size), and plugging all that into their current best model of the universe, they come out with a prediction that that galaxy we are seeing is "now" 1 megaparsec away from us (where "now" means "at our current value of coordinate time in the standard coordinates used in cosmology") and is "now" receding from us at 70 km/s. In other words, those are model-calculated values for $a$ and $\dot{a}$ at coordinate time $t_{\text{now}}$ for that galaxy.

One reason why I say that the language cosmologists use obfuscates things is that it is extremely rare to see one actually tell you what I've just told you; they all quote "distance now" numbers as though they were read directly off measuring devices, instead of being the outcome of a fairly complicated, model-dependent calculation.

You should rethink the rest of your questions in the light of what I've just said. Another way of putting the "obfuscation" point is to draw your attention to the fact that, when you start rethinking your questions in the light of what I've just said, you will find that you have gotten yourself into a much more complicated process of investigation and understanding than you thought you were signing up for.

10. Oct 17, 2017

### Happiness

I don't know what coordinate time is. Does "now" mean the present moment? E.g. if the Sun were to stop shining now$^1$, we would only notice it 499 s later because it takes 499 s for light to travel from the Sun to the Earth. Does the cosmologist's "now" has the same meaning as now$^1$? In more technical terms, does the cosmologist's "now" = special relativity's now?

Special relativity's now: the time instant—as measured by an array of synchronised clocks placed at different locations in an inertial reference frame—that is the same as the present/current time instant shown by the clock of the observer under consideration (placed at the observer's location). E.g. suppose the observer's clock read 12:01 PM currently, then the now instant is all the instants (or that one single instant) at which the clocks at the various different locations also read 12:01 PM.

Am I right to say that
1. the empirical data we obtain today from galaxies can only give us information about the past, say year AD 100
2. we then model/guess how the universe would behave from year AD 100 to today
3. from the model, we predict/calculate where those galaxies would be today and how they would behave (e.g. their velocities) today
4. but it's possible that those galaxies that are currently 1 megaparsec away from us could have a different location or velocity from what the model predicts and
5. we could only verify with 100% certainty whether our predictions about those galaxies are correct many years later (approximately >3.3 million years later) when light from those galaxies reaches us?

Last edited: Oct 17, 2017
11. Oct 17, 2017

### Staff: Mentor

Then you probably should not be starting an "I" level thread on this topic. Please take some time to review a basic cosmology textbook, which will explain the standard FRW coordinates used in cosmology. You can get a start on Wikipedia here:

https://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric

But I would not stop there if you really want a good understanding.

Not quite, because special relativity assumes spacetime is flat, and the spacetime of our universe is curved. But the cosmologist's "now" does refer to the closest thing you can get to the "now" of SR in the curved spacetime of our universe--a spacelike "surface of simultaneity" that matches up with the symmetries of the spacetime.

The cosmologist's "now" basically means the clocks are synchronized using proper time since the Big Bang according to "comoving" observers--observers who always see the universe as homogeneous and isotropic.

All of these statements are correct, yes, assuming the meaning of "now" that has been described above.

12. Oct 17, 2017

### Orodruin

Staff Emeritus
While I of course agree that cosmological time defines a foliation of space-time, I do not think it is the closest you can get to the "now" of SR (as in a fixed coordinate time in regular Minkowski coordinates), at least not locally. The closest thing would be a space-like surface that locally looks like the the SR "now" (i.e., the coordinate surface of the time coordinate in local normal coordinates), which the surfaces of constant cosmological time are not. To be honest, I don't think I fully understood metric expansion until I computed what the local normal coordinates were. Hardly surprising, it turns out that nearby comoving observers move apart at a relative speed $v = Hd$ in those coordinates. I found it a very useful exercise.

13. Oct 17, 2017

### Staff: Mentor

Wouldn't it be correct, though, to say that, assuming a comoving observer, the spacelike surface that locally looks like the SR "now" is also tangent, at the event on the comoving observer's worldline that crosses the surface, to a spacelike surface of constant FRW coordinate time? So at least locally, the two surfaces "look the same"--you have to get further away to see the difference between them.

14. Oct 17, 2017

### Orodruin

Staff Emeritus
Yes. They would look the same locally, it is just that the time slice of the local normal coordinates would look the same as SR time to the next order as well. My favourite example of this is to take a 1+1-dimensional space-time with metric $ds^2 = dt^2 - (kt)^2 dx^2$, which is essentially a FRLW universe. It just so happens that it is also Minkowski space since one can make a coordinate change to $\tau = \sqrt{t^2 - x^2}$ and $\xi = \tau \tanh(kx)$ leading to $ds^2 = d\tau^2 - d\xi^2$.

The "now" usually encountered in SR are the surfaces of constant $\tau$ while the surfaces of constant cosmological time (hyperbolae in Minkowski space) would correspond to surfaces of constant $t$. I believe you can also extend this to a 3+1-dimensional space-time.

To cut the long story short, the surfaces would look the same in the same sense as a hyperbola locally looks like a line.