# Constants linked to the observable universe, not whole

1. Dec 6, 2014

### liometopum

A common equation (attributable to Fred Hoyle) for the mass of the observable universe is: c3/(2GH0).

Despite that the whole universe is vastly larger than the observable universe, we can create an equation to calculate the mass of the observable universe, using the standard constants of nature, namely c, G, and H.

If you think about it, the ability to calculate the mass of the observable universe is interesting. How is it that these constants give us the mass of the observable universe, instead of the whole? You might expect that these constants would be related to the whole universe, not some tiny fraction of it.

To get the mass of the whole universe, we would have to scale up one or more of these three constants, substantially, suggesting that one or more are actually scaled to the size of the observable universe.

What are the thoughts you Physics Forums people on this?

2. Dec 7, 2014

### Simon Bridge

The derivation of the formula follows from a density argument - oversimplifying: the mass of the observable universe is it's density times it's volume.
You get the Hoyle equation by assuming a steady-state.

We don't know the volume of the total Universe, so we cannot find a similar relationship.

3. Dec 7, 2014

### Chronos

For obvious reasons, we don't have observations of the unobservable universe. So we are stuck with what we can observe to draw inferences.

4. Dec 7, 2014

### liometopum

Thanks.
The observation is that the whole universe is vastly larger than the observable, but we can use c, G, and H to calculate its mass. Although Fred Hoyle was a steady state guy, his derivation was not steady state-based; he was a leading relativist of his day. Another similar mass equation is that of Joel Carvalho,
"Derivation of the mass of the observable universe" (See http://link.springer.com/article/10.1007/BF00670782#page-1). His is different by a factor of two from Hoyle's, and he derived his mass estimate from "solely microphysical arguments". Carvalho's work might be more relevant to this thread, as the point is, again, that we can derive the mass of the observable universe from common constants, given to us by nature. Carvalho's work accentuates the idea that these constants are related to the observable, not the whole, because his approach suggests that the microphysical world is related more to the observable than the whole.

To me, this is a paradox, potentially as important as the dark night sky paradox. At first thought, you don't think about it. I didn't. It seems likes it is no big deal, not an issue, but upon serious reflection, there is something important here. I humbly suggest you ponder it a bit before replying. Nature gives us these constants, and at least one of them exclusively, based on the observable universe, not the whole universe,

5. Dec 7, 2014

### Simon Bridge

There are no observation on the "whole Universe", only that bit we can see, That is what the word "observable Universe" means. Therefore there is no way to tell if the whole Universe is bigger than what we see.
How do you know that all we see is not all there is?
How do you know that the Rest of the Universe is not infinite?

Please summarize the main points of the derivation for those who do not have access to the journal.

It looks like Carvalho is still using a density argument.

... and yet for everyone else it is trivial: an artifact of the calculation process.
The reason the mass formulas come out as some nice combination of universal constants is because they all assume some simplicity in the symmetry of the Universe.

6. Dec 7, 2014

### liometopum

It posted before I was done. I'll be back.

7. Dec 7, 2014

### Simon Bridge

That reasoning just shows you that it makes no sense to suggest an infinite Universe now from a finite universe in the past. It could have been infinite all along...
The cosmologcal expansion need not be an increase in the overall volume of the Universe, just a manifestation of decreasing density.

I am not the only one who thinks you cannot tell if the Universe is finite or infinite.
http://www.esa.int/Our_Activities/S...ite_or_infinite_An_interview_with_Joseph_Silk
(interview - also nicely reinforces your point about the universe being much bigger than the observable, well done.)

The possibility of an arbitrarily large volume for the universe should show you why it is not surprising that we cannot use arguments like you've shown to determine the total mass of the whole universe. It is not surprising that the universal constants are only locally applied because locally is all the calculation that could be done. It is very common to use universal constants in a local context. Or perhaps I have misunderstood the nature of the "paradox" - perhaps you could spell out what it is that you find so paradoxical?

... well, yes. Depending on what he means by "deep", you do somehow have to go from microscopic to macroscopic within any unified theory.

8. Dec 7, 2014

### liometopum

Do you really think that there is no way to tell if the universe is bigger than what we can see?
If current theory is correct, we can only see back to the time light could begin to travel freely, leaving us something like 380,000 years of expansion beyond the observable universe. What was the universe doing for that first 380,000 years?
According to this Wikipedia article (http://en.wikipedia.org/wiki/Observable_universe#The_universe_versus_the_observable_universe) Alan Guth has an estimate for the size of the whole at 3x10^23 times the observable. I think it is about 10^243 times larger. The whole is much larger than what we can see, and I am sure you agree, despite this pedantic question.

That comment treats infinity like a number. If the universe had a start, it cannot be infinite now. How big was the universe when it was half the current age, or a quarter, or one hundreth, etc? The universe had a start, and is growing still. It is big, but not infinite, because it cannot be infinite. Infinite implies it cannot grow more. What are you going to do, multiply infinity by two every time it doubles?

Carvalho starts out discussing the standard density solution, as a point of reference, showing the mass of the observable universe to be c^3/(GH). He then considers the case when the radius of the universe was about the size of the wavelength of an elementary particle.

$$R \sim l_p=\frac{\hbar }{m_\pi c}$$

The density should be the Planck density:

$$\rho =\frac{c^{5}}{G^{2}\hbar }$$

He then computes the mass inside the volume of R, which is $$\rho R^{3}$$
using the above two equations:

$$\rho R^{3}=\left( \frac{c^{5}}{G^{2}\hbar }\right) \left( \frac{\hbar }{% m_{\pi }c}\right) ^{3}=\frac{c^{2}\hbar ^{2}}{G^{2}m_{\pi }^{3}}$$

He inserts the numerical values and gets about 1.5 x 10*52 kg

"which is of the order of the mass of the visible universe"

Then Carvalho uses a relation given by Weinberg (from his Gravitation and
Cosmology book):

$$m_{\pi }=\left( \frac{\hbar ^{2}H_{0}}{Gc}\right) ^{\frac{1}{3}}$$

He inserts that relation into $$\frac{c^{2}\hbar ^{2}}{G^{2}m_{\pi }^{3}}$$ to get:

$$\frac{c^{2}\hbar ^{2}}{G^{2}m_{\pi }^{3}}=\frac{c^{2}\hbar ^{2}}{% G^{2}\left( \left( \frac{\hbar ^{2}H_{0}}{Gc}\right) ^{\frac{1}{3}}\right) ^{3}}= \frac{c^{3}}{GH_{0}}$$

which is the same mass derived by the standard density approach.

He does a bit more math and concludes: "We have seen that, from a very simple argument, a new derivation of the mass of the universe is possible which involves only microscopic quantities. This strengthens the idea that there must exsist a deep relation between the microscopic and the macroscopic world within a complete and unified theory of the forces of nature."

Let me again try to get the point across.... these constants give the observable mass, not the whole mass. To get the whole mass, we would have to multiply the observable mass equation by yet another value, probably a scaled up version of one of these constants, meaning at least one of the constants is scaled to the observable.

"assume some simplicity in the symmetry ..." what??
If the universe generates these constants, why is it that observable does it, and not the whole? One of these constants, at least, is scaled to the observable.

9. Dec 7, 2014

### liometopum

Simon, one last thing before my day ends... "Infinite all along" ... That implies infinite time, if I read it correctly.

Let me share this idea I had a few years back: "If the universe is infinitely old, today can never happen"
Imagine an infinite time line, going both ways, past and future. If you put a point on it, representing today, then time could never progress to that point because that point is at an infinite distance from the past. You would have to arbitrarily place a 'starting point" on the line to get to today.

If you cannot progress back, you cannot progress to now. If that logic is correct, then the universe cannot be infinite in size, either.

10. Dec 7, 2014

### Staff: Mentor

No, it isn't, because "the past" is not a point on the time line. Any given point on the line is a finite distance from any other point. So no "infinite distance in time" is ever physically realized. It just happens that, for any point, there are points in time at arbitrarily large finite "distances in time" (in both directions) from that point. There's nothing logically wrong with that.

11. Dec 8, 2014

### Jorrie

AFAIK, the Hoyle mass refers to the energy contained in the present Hubble sphere (with RH = 1/H0), not the mass of the observable universe. The Hubble radius is roughly one third of the observable proper radius. It also assumes spatial flatness, for which the size of the 'whole universe' is indeterminably large.

One can play all sorts of games by assuming Ω to be slightly larger than unity, i.e. a marginally closed universe, for which a present proper radius can be calculated. Then one can, assuming that this whole lot presently resembles the observable universe, calculate a present energy content for this assumed universe. The answer is very sensitive to the degree of deviation from flatness of the spatial curvature, so it may not be very interesting, IMO.

12. Dec 8, 2014

### liometopum

Jorrie... you are correct, as the Carvalho paper uses the Hubble sphere. I will work on this more.
But the observation remains that we can calculate the 'observable' (define it as you wish) from these constants. I am not here to defend, but see I need to explain it. The goal is to get the opinion of others.

Regarding the infinity issue, I don't think you can place a point on an infinite timeline and call it "today". How can determine 'today', without reference? Placing another point as a reference leads to the same argument. It seems the only point of reference would be a beginning point.

13. Dec 8, 2014

### liometopum

Last edited: Dec 8, 2014
14. Dec 11, 2014

### yogi

The OP has raised a profound question - How is it that G, c and H can be combinable to predict the Hubble mass within the limits of experimental error - of even greater intrigue is the similarity of the results obtained using Planck's constant. While it is not to difficult to appreciate that H and c will be tied to Hubble properties, the involvement of G is not obvious. These and other mysteries like the ratio of GM/Rc^2 = 1 (where R is the Hubble scale and M the Hubble mass) seem to point to a holistic universe. Unfortunately we cannot discuss personal theories on these forums.

15. Dec 11, 2014

### yogi

Just realized - they are one in the same equation - Fred Hoyle's derivation differed by a factor of 2.

16. Dec 11, 2014

### yogi

Here is an interesting point - take the equation M = c^3/HG. If G is constant then since H is diminishing except in q = -1 accelerating universes, mass must be increasing as the universe ages

17. Dec 11, 2014

### Jorrie

Yes, if you take that equation at sight value. The Hubble volume is increasing and so does the energy therein.

18. Dec 12, 2014

### yogi

That would be logical for a decelerating universe where the Hubble scale is increasing faster than recessional velocity -

19. Dec 13, 2014

### Jorrie

Yes, only for the matter dominated phase, the amount of matter inside the Hubble sphere is increasing. If we include dark energy into the energy density mix, it is not so simple. The equation M = c^3/HG implies critical density and for as long as the Hubble sphere is growing, the "energy within" increases, despite accelerated expansion.

In this LightCone 7 graph, the Hubble radius is depicted by just R, with D_Hor the cosmic event horizon.

20. Dec 13, 2014

### liometopum

This is more of an exercise for me. Let us compare the universe when it was half the current age, to the current age.

Using Jorrie's cosmological calculator, and his default settings, we find that at redshift (z) =0.7764:

the Hubble constant was 105.5083 km/s/Mpc

the age of the universe was 6895.839068 My

the Hubble radius was 9269.416671 Mly

Omega matter = 0.7127337

Omega Lambda = 0.2870119

We will convert the Hubble constant, $105.5083$ km/s/Mpc to km/s/m

given $3.08567758\times 10^{22}\ {m}$ per megaparsec

we get

$\frac{105.5083\frac{\ {km}}{\ {s}}}{3.08567758\times 10^{22}\ {m}}% =\frac{3.4192911366974381037\times 10^{-18}}{\ {s}}=$the Hubble constant

When the universe was half its current age:

$\Omega =\frac{\rho }{\rho _{crit}}=\frac{\rho }{\frac{3H^{2}}{8\pi G}}$

$\rho =\frac{3H^{2}\Omega }{8\pi G}$

let us calculate the matter and vacuum densities for this time:

$\rho _{m}=\frac{3\left( \frac{3.4192911366974381037\times 10^{-18}}{\ {s}}\right) ^{2}\left( 0.7127337\right) }{8\pi \left( 6.67259\times 10^{-11}\ {m}^{3}\ {kg}% ^{-1}\ {s}^{-2}\right) }=\frac{1.4906868134080027291\times 10^{-26}}{% \ {m}^{3}}\ {kg}$

$\rho_{v}=\frac{3\left( \frac{3.4192911366974381037\times 10^{-18}}{\ {s}}\right) ^{2}\left( 0.2870119\right) }{8\pi \left( 6.67259\times 10^{-11}\ {m}^{3}\ {kg}% ^{-1}\ {s}^{-2}\right) }=\frac{6.0028711231302285621\times 10^{-27}}{% \ {m}^{3}}\ {kg}$

$\rho _{m}+\rho _{v}=\frac{1.4906868134080027291\times 10^{-26}}{\ {m}^{3}}\ {kg}+\frac{% 6.0028711231302285621\times 10^{-27}}{\ {m}^{3}}\ {kg}=\frac{2.0909739257210255853\times 10^{-26}}{% \ {m}^{3}}\ {kg}$

total density = $\frac{2.0909739257210255853\times 10^{-26}}{\ {m}^{3}}\ {kg}$

Hubble radius is $9269.416671\times 10^{6}\times 9.4605284\times 10^{15}% \ {m}=8.76935796674289564\times 10^{25}\ {m}$

$\frac{4}{3}\pi \left( 8.76935796674289564\times 10^{25}\ {m}\right) ^{3}=2.8248279687901783178\times 10^{78}\ {m}^{3}$

total mass is $\left( \frac{2.0909739257210255853\times 10^{-26}}{\ {m}^{3}}\ {kg}\right) \left( 2.8248279687901783178\times 10^{78}\ {m}^{3}\right)$

$=5.9066416273877498982\times 10^{52}\ {kg}$

----------------------------------------------

NOW (Current time)

the Hubble constant now is $67.9$ km/s/Mpc

$3.08567758\times 10^{22}\ {m}$ per megaparsec

$\frac{67.9\frac{\ {km}}{\ {s}}}{3.08567758\times 10^{22}\ {m}}% =\frac{2.2004891385962625428\times 10^{-18}}{\ {s}}$

current total mass:

$\rho _{m}=\frac{3\left( \ \frac{2.2004891385962625428\times 10^{-18}}{\ {s}}\right) ^{2}\left( 0.307\right) }{8\pi \left( 6.67259\times 10^{-11}\ {m}^{3}\ {kg}^{-1}% \ {s}^{-2}\right) }=\frac{2.592782943918842992\times 10^{-27}}{% \ {m}^{3}}\ {kg}$

$\rho _{v}=\frac{3\left( \frac{2.2004891385962625428\times 10^{-18}}{\ {s}}\ \right) ^{2}\left( 0.693\ \right) }{8\pi \left( 6.67259\times 10^{-11}\ {m}^{3}\ {kg}^{-1}% \ {s}^{-2}\right) }=\frac{6.0028659870149049491\times 10^{-27}}{% \ {m}^{3}}\ {kg}$

$\rho _{m}+\rho _{v}=\frac{2.6592782943918842992\times 10^{-27}}{\ {m}^{3}}\ {kg}+\frac{% 6.0028659870149049491\times 10^{-27}}{\ {m}^{3}}\ {kg}=\frac{8.6621442814067892483\times 10^{-27}}{% \ {m}^{3}}\ {kg}$

Hubble radius is $14403.53461\times 10^{6}\times 9.4605284\times 10^{15}% \ {m}=1.362650482 38287924\times 10^{26}\ {m}$

$\frac{4}{3}\pi \left( 1.36265048238287924\times 10^{26}\ {m}\right) ^{3}=1.0598442114935278101\times 10^{79}\ {m}^{3}$

total mass is $\left( \frac{8.662144281 4067892483\times 10^{-27}}{\ {m}^{3}}\ {kg}\right) \left( 1.0598442114935278101\times 10^{79}\ {m}^{3}\right)$

total mass is $=9.1805234757707496189\times 10^{52}\ {kg}$

--

The Hoyle mass is $\frac{c^{3}}{2GH}$

then: $\frac{c^{3}}{2GH}=5.9047510536371689098\times 10^{52}\ {kg}$

now: $\frac{c^{3}}{2GH}=\frac{\left( 2.99792458\times 10^{8}\ {m}% \ {s}^{-1}\right) ^{3}}{2\left( 6.67259\times 10^{-11}\ {m}^{3}\ {kg% }^{-1}\ {s}^{-2}\right) \left( \frac{2.2004891385962625428\times 10^{-18}}{% \ {s}}\right) }=9.1752613489317600659\times 10^{52}\ {kg}$
This matches the result just above.

so we get an increase in mass enclosed within the Hubble sphere with the passage of time.

------------------------------------

Compare to the Carvalho/Weinberg mass $\frac{c^{3}}{GH}$

now: $\frac{c^{3}}{GH}=\frac{\left( 2.99792458\times 10^{8}\ {m}\ {s}% ^{-1}\right) ^{3}}{\left( 6.67259\times 10^{-11}\ {m}^{3}\ {kg}^{-1}% \ {s}^{-2}\right) \left( \frac{2.2004891385962625428\times 10^{-18}}{% \ {s}}\right) }=1.8350522697863520132\times 10^{53}\ {kg}$

We find that the Hoyle mass, $\frac{c^{3}}{2GH}$ , which is derived from the product of critical density times the Hubble volume matches the standard cosmology model.

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Last edited: Dec 13, 2014