# Cumulative expansion based on H0 since T = 0

1. Jun 25, 2010

### omnius

The Hubble "constant" is typically given in terms of units of km/s/mpc. If you convert the distances to a common measure like meters then they cancel out and you are left with a "rate", as in 1/s. http://astro.berkeley.edu/~mwhite/darkmatter/hubble.html

This value describes the proportional expansion of the universe, such as .001/s, meaning for every second that passes a distance of 1 increases by .001 (for illustration, my numbers are not even ballpark; they're just intelligible - the real value is insanely small). Equivalently a distance d increases by .001d, meaning the final distance is given by d+.001d or d*1.001

In cumulative terms, therefore, a distance d over a time t expands to d*(1+r)^t, where r is the rate (my .001 above). This is a standard exponential growth function.

My question is, using the actual Hubble constant H0, what is the cumulative expansion of a distance of length 1 (units are irrelevant it turns out) since the beginning of the universe. How much has an individual unit of space expanded since the beginning of expansion?

It seems very simple, simplifying to (1+Hubble constant)^(age of the universe), but I can't work with these enormous and enormously small numbers at the same time on my pocket calculator and get results that I can trust.

I guess some part of me is expecting the answer to be "the diameter of the universe" but that would not sit well, as that would connect the speed of light (which defines, together with the age of the universe, the 'diameter' of the visible universe) to the rate of expansion. A much more reasonable answer would be "42" or perhaps something which can be logically connected back to another fundamental constant.

Thank you.

Last edited: Jun 25, 2010
2. Jun 25, 2010

### Ich

Welcome to PF!

Infinitely. That's called a singularity, where all finite space measures start from zero.
If you take quantum mechanics into account, you'll maybe come up with a finite number. But as of now, there's no theroy to do the calculations.

3. Jun 25, 2010

### omnius

I get that "distance is undefined" in t=0 but the equation doesn't require units of distance. I'd really like to know what this number is but I can't find it online anywhere. Thanks!

4. Jun 25, 2010

### Ich

I didn't say undefined, I said zero. The number you're looking for is infinity, and the formula you're providing is an approximation for bankers, with inconsistent units.
The Hubble constant was infinite at the beginning. If it were constant all the time, there'd be exponential expansion without a beginning.

5. Jun 25, 2010

### Dickfore

Actually, your formula is dimensionally incorrect. The reason is because r has a dimension of T-1 and you add it to a dimensionless number (1), while t has a dimension T and appears as an argument of an exponential function.

The only reason for your sloppy notation I can think of is because you actually plug in the numerical values of these quantities expressed in a common unit of time (the second) in that equation. Namely, your r is actually $r \cdot 1 \, \mathrm{s}$, i.e. it gives the relative dilation of any distance in a 1 second interval. Similarly, your T in the exponent should read $T/(1 \, \mathrm{s})$, i.e. it is the number of seconds in the time interval interval T. There is nothing sacred about the second here. The only important thing is very short compared to cosmological time scales. If we used some other unit of time $\tau$, then your refined formula should read:

$$d(T) = d_{0} \, (1 + r \, \tau)^{\frac{T}{\tau}}$$

Now, use the property that $x \equiv r \, \tau \ll 1$ and your formula has the asymptotic behavior:

$$\frac{d(T)}{d_{0}} = \lim_{x \rightarrow 0} (1 + x)^{\frac{r \, T}{x}} = \left( \lim_{x \rightarrow 0}(1 + x)^{\frac{1}{x}} \right)^{r \, T} = e^{r \, T}$$

which is exactly the exponential function. Notice that the argument in the exponential is dimensionless now and the final result is indeprendent of the intermediate unit of time.

A characteristic feature of the exponential function is that the derivative is:

$$d'(T) = r \, d_{0} \, e^{r T} = r \, d(T)$$

strictly proportional to the value of the function. From here, one can find the rate of expansion as:

$$r = \frac{d'(T)}{d(T)} = \frac{1}{d(t)} \, \frac{d \, d(T)}{dT} = \frac{d}{dT}\left( \ln(\frac{d(T)}{d_{0}}) \right)$$

i.e. either as the ratio of relative expansion ($\Delta d/d$) (what you started with) to the short time interval or as the logarithmic derivative of the relative expansion. But, this is precisely how the Hubble parameter is defined in cosmological models:

$$H(t) = \frac{1}{a(t)} \, \frac{da(t)}{dt}$$

Here t is cosmological time, i.e. the time in the general frame of reference that is co-moving with the local galaxies at any point in space. It is in this frame that the Universe should look isotropic on scales much larger than typical intergalactic distances.

Notice that H(t) is dependent on cosmological time. I can solve the above (differential equation) for a(t) (the scale parameter) as:

$$a(t) = a(t_{0}) \exp\left[\int_{t_{0}}^{t}{H(t') \, dt'}\right]$$

Now, I can get any time dependence as I want. For example, if:

$$H(t) = \frac{n}{t -T}, \ [n] = 1$$

then we actually get a power law:

$$a(t) = a(t_{0}) \, \left(\frac{t - T}{t_{0} - T}\right)^{n}$$

Your conclusion of exponential growth is a consequence of extrapolating the present value of the Hubble constant to epochs in the future or the past of the order of cosmological time scales.

Last edited: Jun 25, 2010
6. Jun 25, 2010

### omnius

Thanks Ich, you're surely right but I want to fit all these pieces together so I understand them, not just "I read that 'Hubble constant was infinite in the beginning' ". Let's take t = .000...0001; a given distance has undergone a finite expansion from then until now, right? I understand that H is not a constant, but this still seems like a good starting point.

I think the footsteps my thought experiments are retracing here is actually the scale factor (http://en.wikipedia.org/wiki/Scale_factor_(universe))

I don't see any charts of it over time though or it's present value. I want to know what this factor is and was and if it is related to the speed of light. I'd like to begin by grounding expansion in a numerical value at t = now. If H is indeed a variable factor, where's the graph? Not being sarcastic I just haven't found these things though there're many abstract articles on the subject.

[EDIT - Posted this before seeing Dickfore's response which does give exactly the formulas I was looking for - Thanks! Now to see if I can use these to derive the numerical values I'm looking for.]

Last edited: Jun 25, 2010
7. Jun 25, 2010

### Calimero

Present value of scale factor is 1. You can go here: http://www.astro.ucla.edu/~wright/cosmo_03.htm" [Broken] to see a graph.

If you are interested to determine value of H for given age, you can do it here: http://www.uni.edu/morgans/ajjar/Cosmology/cosmos.html" [Broken]

Last edited by a moderator: May 4, 2017
8. Jun 25, 2010

### Dickfore

There is something known as a dimensionless deceleration parameter q. It measures the behavior of the Hubble parameter with (cosmological) time. In order to get a dimensionless number, one takes the time derivative of the reciprocal of the Hubble parameter $1/H(t)$. This is a dimensionless number. If the Hubble parameter rises faster than $1/t$ (including being constant), i.e. if $d/dt(1/H) < 1$, then the decelartion parameter is negative. This is the basis for the definition:
$$q = \frac{d}{dt}\left(\frac{1}{H(t)}\right) - 1$$

The reason behind the upper limit choice is, that in that case we have the dependence from my previous post with $n = 1$:

$$a(t) = K \, |t - T|$$

i.e. the scale factor of the Universe is a linear function of cosmological time.

This might seem as random definitions, but, it turns out that cosmological models predict a relation between q and H at any (cosmological) time t:

$$q = \frac{1 + 3 \, w}{2} \left( 1 + \frac{K \, c^{2}}{(a(t) \, H(t))^{2}}\right)$$

where $K = 1, 0, -1$ is a geometric factor depending on whether the Universe is hyperspherical, flat or hyperbolic and $w = p/(\rho c^{2})$ is the so called equation of state of the Universe.

Last edited: Jun 25, 2010
9. Jun 25, 2010

### nicksauce

This is actually really easy to answer. The Friedmann equation gives H as a function of scale factor. It is, roughly

$$H^2 = H_0^2(0.3a^{-3} + 10^{-4}a^{-4} + 0.7)$$

Where H_0 is the Hubble constant today, and a is the scale factor (1 today).

Clearly taking it to be constant is not a good starting point, because it blows up quickly as a->0.

10. Jun 25, 2010

### omnius

Thanks to all for the charts and explanations - what I have been pondering is indeed the scale factor and is pretty aggressively addressed, though I'm sad that it's apparently one of those "hotly contested" areas, given that it is so intertwined with Omega, something we're not "sure" about yet (in the bayesian sense).

http://www.astro.ucla.edu/~wright/cosmo_03.htm
Let me ask a comparative question based on these curves. If accumulated appropriately in any of the three cases (Omega<1, Omega=1, Omega>1), the scale factor contributes some expected size of the universe. Is this greater, less than, or equal to the observed size... or am I still missing the concept here? For example, consider:

Before seeing this, this is actually where I began my thought experiment last night: c/H0. What is the significance of expansion, the actual size of the universe, and is it related to the speed of light? Just asking in an open manner.

11. Jun 25, 2010

### Calimero

Look at the same graph here:"[URL [Broken] (scroll at the middle of page), and look at the curve OmegaM = 0.3, OmegaLambda= 0.7. That it is what we currently beleive is the right one.

Edit: It is same as magenta curve at Ned Wright's pages.

C/Ho just marks the distance from where recesion speed becomes superluminal, and it defines Hubble volume. Note that it is not horizon of any kind because we can see things that are receding faster then light.

Last edited by a moderator: May 4, 2017
12. Jun 25, 2010

### omnius

There are two likely answers to "how much has the universe expanded" (I will assume a Omega=1 universe) - whether or not anyone can/has calculated it to date (though that stands as the figure I am trying to see):

1) Infinity - all of the universe, including that part which is outside of our event cone, began in a "point" in space and is now infinitely large and growing. In the "raisin bread" model, all the raisins occupied a single point.

2) Finite - In the "raisin bread" example, the raisins were all incredibly tightly packed together in an infinitely large dough and then the dough started growing, perhaps giving birth to time and other forces we consider commonplace.

Choosing option #2, my hypothesis is that space farther away from us than the cumulative expansion I keep asking about originated in a different point than our space did.

How could this not be the case? I think this is where Ich's "H(T=0)=Infinity" comes in - it would be necessary to have such a factor in this case to go from a singularity to an infinite universe in 0 time.

If on the other hand the pre-expansion universe was also infinite, and it just started expanding for some reason (decay, etc), then the expansion factor explains the growth of the universe and never need be infinite.

13. Jun 25, 2010

### Calimero

If you believe in initial singularity then Ich gave you answer.

Anyway, nobody can't give you answer to your question for sure because as we get very close to t=0 our equations do not hold anymore. Furthermore, when you put inflation into the picture things are even more complicated.

So you can comfortably answer how much universe expanded since some t>>0, but how much it expanded since t=0 can't be answered.

14. Jun 26, 2010

### omnius

If you integrated these equations, or integrated the a(t) factor from time 0, would that give the cumulative expansion? Wouldn't that be finite?

If infinite, that would require infinity be introduced at some point. Put another way, is this redshift analogous to the Hubble parameter over time?

Which would clearly have infinity at 0. If so, what force is supposed to have caused "rapid deceleration" at t > 0, so much that the expansion scale went from infinity to some finite factor a(t)?

Can the original question at this point be accurately stated as "what is the integral of a(t) from t=0 to t=now"?

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15. Jun 26, 2010

### omnius

In this great graph from atlasoftheuniverse.com, we see that the visible universe is just that only because we can only see light which is presently reaching us since its emission. As the universe ages, our visible universe will increase in size.

Though the light left at t=1, it took 13 periods to reach us due to the expansion of space causing the distance the light had to travel to grow considerably. If space were not expanding, and these galactic lightsources were simply flying apart due to explosive force, then the light would have reached us in 1 period.

This seems to imply a finite expansion from t=1 to t=now of 13x, or something on that order. Is this the answer to the original question?

Doesn't this imply that the diameter of the visible universe (taking into account a dark period of time where there isn't any light emitted from t=0 to t=.0...1) actually is the cumulative expansion?

Imagine for example a distant neighbor particle at t=0 emits a photon and it reaches us in the infinite future. Another photon emitted by a close neighbor after that at t>0 reaches us today. In between these two scales there is a photon which is presently reaching us but was emitted (again ignoring the dark period, maybe I should say "Information" instead of photon) at t~0. It had x>0 distance and x>0 time to go but because of expansion required 14by.

This I hope illustrates what I mean when I'm saying that our visible universe is that portion of the infinite universe which was in our "expansory space" at t=0, and that the superluminal portion was outside our point. Or, in a weaker form, the query is what are these two diameters and how are they related, the "visible universe" and the "expanded space universe".