# How do you think of the Cosmic Event Horizon?

1. Oct 25, 2013

### marcus

If you click on Jorries calculator
http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7/LightCone.html
you see in the default table that immediately comes up (without your doing anything) that the present-day CEH is 16.5 billion ly. What does that mean to you? How do you picture what is happening around that distance?

More precisely 16.472 but let's round it to 16.5.

Here's how I picture what that distance means. If some light from a distant galaxy is heading in our direction today it will eventually reach us only if it is already within the CEH. It will get here eventually only if has already made it within the magic circle and is within 16.5 Gly of us.

That light may have come from much farther away. It might have been emitted a long time ago from a galaxy which today is, I don't know, 40 or 50 billion lightyears from here. In fact it could even be losing ground at present: aimed at us but being dragged back by expansion. But it will still make it eventually if it is today within 16.472. The CEH is a kind of threshold.

Let's say the light is now at 16.47 Gly. Well the current Hubble radius, the distance that today is expanding at c, is 14.4 (as you can see if you clicked the link--it's one of the things the calculator gives you up front.)
So 16.47/14.4 = 1.14.
That light is in space that is receding at 1.14 c and it is traveling towards us at c, so it's net recession is 14% of the speed of light. But it will eventually get here! because it is within the Cosmic Event Horizon distance.

2. Oct 25, 2013

### marcus

If you didn't understand how that's possible and still had some question about it, even after you clicked on the link and looked at the table, then of course ask. But also you could look at the R column of the table, which gives the Hubble radius, showing its growth history. That's the key to it. R reaches out to light that is traveling towards us and as soon as the light is within Hubble radius range it begins to make headway.
http://www.einsteins-theory-of-relativity-4engineers.com/LightCone7/LightCone.html

3. Oct 25, 2013

### Tanelorn

Thanks marcus, that is interesting, and the expansion is accelerating I believe.
Dark energy is the name given to that which causes the expansion.
I still dont understand where this energy could come from?
What in the Universe (or perhaps even outside it) could be providing this energy and perhaps increasing this energy to cause accelerated expansion?
Is the dark energy being used up as it drives the expansion of space? Is work being done as space expands?
I am guessing it is not energy as we would normally understand it?

Last edited: Oct 26, 2013
4. Oct 26, 2013

### marcus

That would be my guess too, Tanelorn.
It may be that "dark energy" was an unfortunate choice of jargon.

when Einstein introduced the new constant in his GR equation he didn't call it energy. He gave it the units of curvature (a reciprocal area) not an energy density. And called it "cosmological constant"

A small inherent baseline curvature. A very slight constant "vacuum curvature" the same throughout all time and space that the other curvature caused by stuff is added on top of. That was how the quantity started and how it was considered for some 80 years, until it got hyped up with the "energy" jargon. Renamed perhaps for public relations reasons. Sounds more exciting.

But so far the observations are still consistent with the idea that it is simply a small constant curvature. We do not know for a fact that this very small constant baseline curvature actually arises from anything we would normally call an "energy". So far as we know it's just the "cosmological constant" that appears naturally in Einstein GR equation, but which for the first 80 years most people thought was exactly zero. They thought that without any good scientific proof, a nonzero value just had not been detected, and then in 1998 it was found that it's NOT exactly zero after all. Just very small

It's late here and I'm off to bed. I'll try to reply more completely tomorrow.

Last edited: Oct 26, 2013
5. Oct 26, 2013

### marcus

Hi Tanelorn, I think you are right to express skepticism about whether the cosmo constant (aka "dark energy") is actually an energy as normally understood. And right to point out there are puzzling questions around this.

Let's just call the constant Lambda. It's a term that occurs naturally both in the GR equation and in the simplified Friedman form of GR equation that cosmologists use.

Until 1998 people thought Λ = 0, but they had known since 1930s or 1940s about expansion. Expansion is something that does not require "dark energy" or Lambda or anything like that.
Lambda is only responsible for a very modest fraction of the expansion that has occurred so far.
That's WHY most people thought it was zero. The observations covering the past history are almost perfectly consistent with there not being any Lambda. So it wasn't needed to explain the observations (up to those made in 1998) and people thought it was zero.

So I would rephrase what you said like this:

Dark energy is the [hype] name given to that [slight intrinsic vacuum curvature] which caused the expansion [to start accelerating slightly around year 7 billion].

I would guess that it might have some explanation down at the microscopic level of quantum geometry in the insufficiently explored regions of physics where geometry and matter share a common foundation. Why should the vacuum have this nearly undetectable inherent curvature that the changing evolving curvature we are used to comes down on top of and almost completely hides?

It's a real puzzle! I see no point to putting a preconceived spin on it---jumping aboard some speculative notion that it is due to this or that---to some putative "energy". Some people like to do that, I don't. It's probably less confusing to reserve judgment and wait until we known a little more.

Anyway let's look at the history! I love looking at the tables the "Lightcone" calculator makes!
Here's a table showing EXPANSION SPEED OF A SAMPLE GENERIC DISTANCE as it changes over time. The reason it SLOWS AT FIRST is because other effects dwarfed the very slight intrinsic tendency to expand represented by the curvature constant Lambda. The speed only begins to increase after around year seven billion. Acceleration is almost imperceptible at first and is still very gradual.

Last edited: Oct 26, 2013
6. Oct 26, 2013

### marcus

Let's look at the expansion speed history of a sample distance.
Let's start in year 545 million around the time the first galaxies were forming, and distances were 1/10 their present size and run the history out into future when they will be 2.5 times present size.

For the first few billion years the growth speed was slowing and then around year 7.6 billion it started to pick up. This is shown in the rightmost column with the "a-prime" label. a is the scale factor and a'(t) is the time-derivative of the scale factor. That's just notation. The essential thing is it is the growth speed of a generic distance expressed as multiples of the speed of light.

The other cosmological distances would be acting in concert with this generic one, but proportional to their size. Bigger size means proportionally faster, smaller means proportionally slower.

$${\scriptsize\begin{array}{|c|c|c|c|c|c|}\hline R_{0} (Gly) & R_{\infty} (Gly) & S_{eq} & H_{0} & \Omega_\Lambda & \Omega_m\\ \hline 14.4&17.3&3400&67.9&0.693&0.307\\ \hline \end{array}}$$ $${\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline a=1/S&S&T (Gy)&R (Gly)&D_{now} (Gly)&D_{then}(Gly)&a'R_{0} (c) \\ \hline 0.100&10.000&0.5454&0.82&30.684&3.068&1.76\\ \hline 0.118&8.483&0.6982&1.05&29.279&3.451&1.62\\ \hline 0.139&7.197&0.8937&1.34&27.755&3.857&1.49\\ \hline 0.164&6.105&1.1436&1.71&26.103&4.275&1.38\\ \hline 0.193&5.179&1.4626&2.19&24.313&4.694&1.27\\ \hline 0.228&4.394&1.8692&2.78&22.377&5.093&1.18\\ \hline 0.268&3.728&2.3863&3.53&20.289&5.443&1.09\\ \hline 0.316&3.162&3.0412&4.46&18.045&5.706&1.02\\ \hline 0.373&2.683&3.8653&5.59&15.650&5.834&0.96\\ \hline 0.439&2.276&4.8927&6.93&13.116&5.763&0.91\\ \hline 0.518&1.931&6.1556&8.45&10.473&5.424&0.88\\ \hline 0.611&1.638&7.6782&10.08&7.769&4.743&0.87\\ \hline 0.720&1.389&9.4693&11.69&5.069&3.648&0.89\\ \hline 0.848&1.179&11.5164&13.17&2.450&2.079&0.93\\ \hline 1.000&1.000&13.7872&14.40&0.000&0.000&1.00\\ \hline 1.179&0.848&16.2374&15.35&2.272&2.678&1.11\\ \hline 1.336&0.748&18.1957&15.89&3.833&5.121&1.21\\ \hline 1.514&0.660&20.2130&16.29&5.252&7.953&1.34\\ \hline 1.717&0.583&22.2740&16.59&6.530&11.210&1.49\\ \hline 1.946&0.514&24.3666&16.80&7.676&14.936&1.67\\ \hline 2.206&0.453&26.4820&16.95&8.698&19.183&1.87\\ \hline 2.500&0.400&28.6133&17.06&9.606&24.014&2.11\\ \hline \end{array}}$$

7. Oct 26, 2013

### Haelfix

Ok, I sense some confusion here and in several other posts related to the same concept. There are a number of terminological points. I've noticed you have invented the word 'cosmological curvature constant' before. This is not a good idea for the following reason. Consider the Friedmann equations, reworked into the usual form where we can explicitly see the contributions to the Hubble rate from various different physical concepts.
$$H^{2}\left(z \right )= H_{0}^{2}\left [ \Omega _{\Lambda }\left ( 1+z \right )^{0}+ \Omega _{k}\left ( 1+z \right )^{2}+\Omega _{M}\left ( 1+z \right )^{3}+\Omega _{R}\left(1+z\right )^{4}\right]$$

The first term is from the cosmological constant, the second term arises from the curvature of space, the third from the contribution from matter and the fourth from the contribution from radiation. It is important to note that the second term and the first term are NOT RELATED. Indeed, they scale differently, and in fact observationally it is determined that omega curvature is very close to being zero, eg to being flat. The reason the word choice is not wise is that students would get confused that the "cosmological curvature constant" might be related to the curvature parameter or equation in some way.

Ok, now another point.. The cosmological constant IS an energy DENSITY if you believe in Einsteins equation. Full stop, period!

$$R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R + \Lambda^{^{cc}} g_{\mu\nu} = \frac{8\pi G}{c^4}T_{\mu\nu} + \Lambda^{^{vacuum}} g_{\mu\nu}$$
Where,
$$\Lambda ^{vacuum}=\frac{8\pi G}{c^{4}}\rho _{vacuum}$$
Let me write Einsteins equations suggestively, anticipating a misconception. (I have included a cosmological constant term on the right hand side naively appearing from contractions of the stress energy tensor that might perhaps arise from the quantum vacuum (or even from CLASSICAL contributions, like perhaps what might arise from spontaneous symmetry breaking)

The last term on the right hand side, has the same units as the last term on the left hand side. This is what Einstein's equations does, it relates terms related to geometry, with terms related to matter, shear and pressure (stress energy). It does not make sense to say that you believe Einsteins equations and deny that these two terms form exact identities. Nor does it make sense to appeal to observation to distinguish the two concepts, since they are exactly equal. No possible observation could ever distinguish between the two unless it also simultaneously refuted GR. So the point is, there is no way, even in principle, to untangle one concept from the other unless you went beyond Einstein (and in some sense this is ridiculous, b/c any putative quantum theory would necessarily need to be almost exactly GR in the infrared).

Now, what is true, is that we could in principle generalize to a larger class of possibilities, that goes by the name dark energy (these possibilities include time varying cosmological constants, cc's that are given by phantom fields, moduli, relics from inflation, and many other possibilities, some arising from geometry, and some from matter). In this form, the cosmological constant would then be the simplest form of dark energy, with w exactly equal to minus 1. So this is then given by replacing term 1 in the above Hubble rate equation with the following
$$\left (1+z\right )^{3(1+w)}$$
Where w is parametrized by the following phenomenological relation
$$w\left (z\right )=w_{0}+w_{a}\frac{z}{z+1}$$
Indeed these types of relationships have just been measured and is the source of a controversy (where a departure from the usual cosmological constant has just been claimed to be observed) see eg arXiv:1310.3828
But the point is, even if w is exactly equal to -1, and is the usual cosmological constant as expected (as standard model physics predicts) it is still represented by an energy density (or a negative pressure if you treat the FRW as modeling a perfect dust solution).

8. Oct 26, 2013

### marcus

That paper and its companion paper http://arxiv.org/pdf/1310.3824v1.pdf are certainly worth watching. If their result is confirmed by further study, showing that the equation of state w is NOT equal to -1, that would be a good reason to attribute the cosmological constant curvature quantity to some kind of ENERGY!

I believe there is divided expert opinion on that. It obviously depends on how you write the equations. My understanding is that Einstein originally included the Lambda constant as a reciprocal area---a curvature constant---on the lefthand side of his equation. That seems like a natural way to treat it, at least to some people :big grin:

The two papers by Scolnic et al that you point to (or at least you give a link to one) are really interesting because they go against the trend I've been seeing in the literature to consider that w is probably -1 and that the observational data is CONSISTENT with LambdaCDM model. In other words consistent with Lambda as cosmological constant.

Because the papers take a kind of contrarian direction and are interesting, I think I should give the basic info about them: link, title, abstract...
http://arxiv.org/abs/1310.3824
Systematic Uncertainties Associated with the Cosmological Analysis of the First Pan-STARRS1 Type Ia Supernova Sample
Scolnic et al (including Adam Riess!)
We probe the systematic uncertainties from 112 Type Ia supernovae (SNIa) in the Pan-STARRS1 (PS1) sample along with 201 SN Ia from a combination of low-redshift surveys. The companion paper by Rest et al. (2013) describes the photometric measurements and cosmological inferences from the PS1 sample...
...Assuming flatness in our analysis of only SNe measurements, we find
$w = -1.015^{+0.319}_{-0.201} (Stat)^{+0.164}_{-0.122}(Sys.)$
...
[and they go on, combining their data with a selection of various other studies thereby reducing the uncertainties]
http://arxiv.org/abs/1310.3828
Cosmological Constraints from Measurements of Type Ia Supernovae discovered during the first 1.5 years of the Pan-STARRS1 Survey
A. Rest, D. Scolnic, and others including Adam Riess again!
...The value of w is inconsistent with the cosmological constant value of -1 at the 2.4 sigma level. ... If we include WMAP9 CMB constraints instead of those from Planck, we find $w = -1.142^{+0.076}_{-0.087}$, which diminishes the discord to <2 sigma. We cannot conclude whether the tension with flat CDM is a feature of dark energy, new physics, or a combination of chance and systematic errors. The full Pan-STARRS1 supernova sample will be 3 times as large as this initial sample, which should provide more conclusive results.
==================
So stay tuned!

Some signs have appeared that w might NOT equal -1, in which case the observations can NOT be most simply explained by positing a Lambda constant.

If this initial finding is born out that would be extremely interesting. It would suggest some type of "dark energy" as the simplest explanation.

But that is a substantial "if".

Last edited: Oct 26, 2013
9. Oct 26, 2013

### Tanelorn

Thanks Marcus and Haelfix. I learned that this subject has very great depth!

So to summarize we are not sure if it is really energy or if work is being done in the classical sense.

10. Oct 27, 2013

### Haelfix

Hmm, there is certainly no controversy on what the units are for the cosmological constant. Again look at the equation.

$$R_{\mu\nu} - \frac{1}{2} g_{\mu\nu} R + \Lambda^{^{cc}} g_{\mu\nu} = 8 \pi T_{\mu\nu}$$

Working in units where C = G = 1 (Planck units), one see's that the last term on the left hand side must be dimensionally the same as the Ricci scalar right next to it, which has units of inverse squared length.

But those *are* units of energy density in Planck units. To see this, lets restore the C's and G, which gives the conversion factor that multiplies the left hand side like so:

$$\frac{c^{4}}{8\pi G}\rightarrow \frac{\left [ \frac{l^{4}}{t^{4}} \right ]}{\left [ \frac{l^{3}}{\left ( kg \right ) \left ( t^{2} \right )} \right ]} \rightarrow \frac{\left ( l \right )\left ( kg \right )}{\left (t^{2} \right )}\rightarrow \frac{J}{l}$$

so you end up with a J/L^3 which is an energy density.