Some Thoughts About Cosmological Horizons and Thermodynamics

Well, I was looking at the super simple model described in my last post. This model has flat space with a positive Cosmological Constant and no matter or energy. So basically I'm thinking about de Sitter space.

Just to recap a bit about that post, start with the Friedmann Equations

Now assume that matter and radiation are negligible and only the Cosmological term remains. Further, assume that space is flat.

Now the Friedmann Equations read as follows.

The two equations are consistent with each other and the first order equation is like a "first integral" of the second order one. The only difference is that the first equation requires a positive Cosmological Constant, while the second one does not. To satisfy both equations, in this simplest of models, the Cosmological Constant must be positive.

So we get a solution for a(t) which looks like

We can have an exponentially expanding solution, an exponentially contracting solution, or a linear combination of the two like a hyperbolic cosine solution.

Let's look at the metric in the x-t plane, for the expanding case. For simplicity, I will take units for which c=1.

Here I am calling x the Co-Moving coordinate, even though there is nothing to move in this model which is devoid of Matter or Regular Energy. Proper distance is then defined by

In order to investigate the Horizon, I need to work out the trajectories of light rays. Using the metric above things proceed in the usual way. To find the Horizon, I am looking for a light ray which reaches x=0 at t=infinity. The calculation goes as follows.

If I choose the other sign the solution is just flipped across the t-axis. Here is a graphical representation.

It is clear that I will never receive a signal from any events which occur in the shaded region, even if I wait till the end of time. So this is my true Cosmological Horizon. In terms of proper distance, this becomes...

So, in this case, my Horizon remains fixed at constant proper distance L=1/H = c/H. In other words, according to the Hubble Law, the Horizon is at exactly that proper distance at which Galaxies recede at the speed of light. So my Hubble Sphere and my Horizon, are the same thing, at least in this simple model.

This is not the case when boring ordinary Matter and or Energy are included in the mix. You can learn a lot about a much more complicated LCDM model in the Doctoral Dissertation of Prof. Tamara Davis, which was also published, in part, as an article in Scientific American.

In Prof. Davis's dissertation, she uses several different coordinate systems to make certain points clear. In particular, she uses Co-Moving coordinate vs so-called Conformal Time and in this coordinate system light rays are rendered as straight lines. Conformal Time is defined as follows...

So in the simple case under consideration here, this becomes

Notice that the Conformal Time is defined on the finite interval [0,1/H], as opposed to Hubble Time, t, which is on the interval [0, infinity).

That is

(I have chosen to begin Hubble Time at t=0. I could have started at
t = - infinity, but the results I am looking for would be essentially the same. So let's go with this.)

And so, my entire existence in this veil of tears must sadly end at the event T=1/H, x=0, which is the end of Conformal Time at the origin of Co-Moving coordinates.

Light rays travel in straight lines with slope equal to +-1, so I can make a diagram like this.

It is a simple matter to check that the line T+x=1/H is, in fact, the same as the Horizon/Hubble Sphere previously mentioned.

A few observations can be made from looking at the diagram.

1) Events outside the backwards light cone can never signal us. This makes evident our true Cosmological Horizon.

2) Events occurring near the Horizon, but inside the light cone, can be seen only near the end of Conformal Time.

3) Galaxies will cross the Horizon in finite Conformal Time, but we will not see this until the bitter end of days

4) We can reach, and indeed cross, the Horizon in our primitive, clunker, no warp drive, spaceship and do it in finite Conformal Time. But again, home base will neither see this nor get our report until the end of the world.

In general, no matter what the form of the scale factor a, light rays travel on straight lines at 45-degree slope when viewed in Conformal Time vs Co-Moving Coordinates. The calculation is as follows.

I mentioned above that for this model, with an exponential scale factor, the Conformal Time coordinate lives on a finite interval. This is not the case where the Cosmological Constant is zero and either Matter or Regular Energy dominate. In those cases, the scale factor, a, goes like a power of t and not an exponential.

 In the Radiation-Dominated case, for example, Conformal Time looks like this.

So there is no end to Conformal Time and the backward light cone of an observer at x=0 fills all Spacetime as T approaches infinity. The Spacetime diagram looks like this.

The Matter-Dominated case goes the same way. There is no Horizon when the Cosmological Term is absent, only an ever expanding Hubble Sphere.

Let's calculate the acceleration of a typical galaxy/test particle caught up in the flow of cosmic expansion. Assume the Galaxy is originally at Co-Moving Coordinate x0. At Hubble Time t, the Galaxy is still at Co-Moving Coordinate x0 but it's proper distance, velocity and acceleration are given by the following.

I want to get the acceleration of a Galaxy which is near the Horizon.

Notice how stuff accelerates towards Horizon like it is being pulled by "gravity". I guess this is the anti-gravity of the mysterious dark energy. I had a hard time with this because Newtonian Gravity says if you have a spherical hollow in a uniform material, say spherical shell or infinite medium, the force on a particle inside the hollow cancels and nobody accelerates.  But that's not what's going on here. It is the flow of cosmic expansion which gives us this acceleration or "force".

So far this has been pretty tame. Warning: The following is a very weak heuristic argument if even that. Take with a grain of salt.

Let's calculate the temperature of the horizon. If a photon is added to the Horizon an amount of Entropy is also added.

Let's choose the photon wavelength so that the smallest possible Entropy is added. That would be one e bit of information or one k (Boltzmann's Constant) of Entropy. The photon should have the smallest possible Energy, in other words, the longest possible wavelength. I choose the wavelength to be the size of the observable universe! (Cool huh? I can just do that because it's my blog. Who knows? Let's go with this for a bit.)

And that gives the temperature of my Horizon. OK. That agrees within a geometric factor with the usual result for the de Sitter temperature.

Actually, this is disingenuous, If I assume hc/lmax = kT, where lmax is the "largest possible wavelength available", whatever that might mean, then assuming lmax= R is already equivalent to stating the temperature goes like 1/R. So the argument is circular. I disavow this line of reasoning completely. I refute my own argument!

But hey. Let's go with this just a bit further. If I substitute the above result for the acceleration at the Horizon, I can express the temperature in terms of that acceleration, in a manner similar to the Unruh-Davies temperature experienced by an accelerated observer.

The usual Unruh-Davies result is

It is my hope that this heuristic "argument", as messed up as it is, can somehow shed some light on this subject matter for someone. All this is similar to my other post about Black Hole temperature, the basic idea being cribbed from Susskind's book, "The Black Hole Wars". The argument is a crazy one, which, none the less, gives more or less the right answer.

As usual, this post has not been subject to peer review. And so, dear reader, I ask you to please educate me concerning any flaws or failings in this post. I did reach out to someone who is specialized in this area, asking for comments. But to date, I have received no reply.

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Estimating the Cosmological Constant

I was talking to an old friend on the phone the other day when he asked a question about the Cosmological Constant.  I said that you can estimate the value of the Cosmological Constant if you know the size of the Observable  Universe. This was an idea which had occurred to me while watching one of Prof. Leonard Susskind's awesome youtube lectures on Cosmology. So I was then determined to show exactly how that works out.

Start with the Friedmann Equations.

Now assume that matter and radiation are negligible and only the Cosmological term remains. Further assume that space is flat.

Now the Friedmann Equations read as follows.

The two equations are consistent with each other and the first order equation is like a "first integral" of the second order one.

Recalling that

gives a simple expression for the Cosmological Constant in terms of the Hubble parameter.

or

If we are now truly at a point in time where the cosmological term dominates, as appears to be the conventional wisdom, then we can get an estimate for

(Sincere apologies for the large lambda since we all know it is supposed to be very small)

by plugging in a value for H something like

, which is more or less the current value of the Hubble Parameter.

This gives the following.

Converting to Planck units gives

Cool. This is the usually quoted  value for the Cosmological Constant in Planck units.

So what about the size of the Observable Universe? Well, we are not exactly in a case of Cosmological Term only. But humor me for a moment and assume that we are near to that case, with Cosmological Term dominating. Consider our Hubble Sphere, the radius of which is the distance at which galaxies recede from us at the speed of light. (In the case of Cosmological Term Only our Hubble Sphere is our true Cosmological Event Horizon and so I will use the terms interchangeably. )

If R denotes the radius of our Hubble Sphere then

And so, as I told my friend, the Cosmological Constant is small because our Observable Universe is big.

There is a bit more. Under the assumption of flat space, the surface area of our Hubble Sphere is

And so the Cosmological Constant is inversely proportional to the surface area of our Hubble Sphere or Cosmological Event Horizon.

But also the surface area of our Cosmological Horizon SHOULD BE proportional to it's entropy.

Under this assumption the Cosmological Constant, can be related to the entropy (S), or information content (S/k), of our Cosmological Horizon.

Also, as usual, the Cosmological Constant can be expressed in the form of an energy density, the density of "dark energy".

So

The density of dark energy would then also be inversely proportional to the entropy of the horizon.

In this calculation I have taken the density of dark energy to be measured in kg/m3 and followed units carefully at each stage in order to see that they work out correctly.

Just for hahas let's plug in some numbers and estimate the density of dark energy.

so

The accepted value of the density is

So my simple calculation is off by a factor of 5. Hey, that's the best I can come up with having such limited resources as only Purina Cat Chow, dry erase markers, and some whiteboard. In my defense I will point out that it's off by a factor of 5 in 27 orders of magnitude.

Well I am not much for error analysis but this seems like more than just sloppy arithmetic. Possibly the assumption that only the Cosmological Term remains in Friedmann's Equations is a bit too stringent.

Thank you for your kind consideration of my blitherings. In as much as this post has been written by a large orange cat and is not peer reviewed, I ask that you challenge these ideas if you find them lacking. As usual, please use the comment section to let me know if you think I am all wet here.

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Some Notes on Black Hole Thermodynamics

I noticed, while reading The Black Hole War, by Leonard Susskind, that he presents Jacob Bekenstein's simple argument about Black Hole Entropy. On page 150 he lays out the basic idea in numbers, but encourages the interested reader to work it out with equations. Astonishingly, this very easy to do.

Prof. Bekenstein argued that, to add one (e) bit to the black hole, one k of entropy, one should drop in a photon with a wavelength just the right size to resolve the black hole, but not smaller, which would provide more detail than needed. In other words, hit the hole with a photon who's wavelength is comparable to it's radius, just right sized to “see” the Black Hole.

From there the argument proceeds with a very few lines to result that the surface area increases by a constant amount, proportional to the square of the Planck Length.

Here

So adding one bit increases the surface area by a constant amount, which is on the order of a square Planck Length and therefore the Black Hole Entropy is roughly it's surface area in square Planck Lengths.

All this depends on the relationship between the radius and the mass of the Black Hole,

,and without this relationship the argument does not work.

Then on page 169 Prof. Susskind makes the offhand comment that Bekenstein had actually calculated the black hole temperature without noticing, since the temperature is found from the amount of energy needed to add one bit of information (using the conversion factor k). In other words, from the above,

,which compares nicely with Hawking's result

All this was a real eye opener for pod pod and well worth the price of Prof. Susskind's book.

Pod pod also found some online lectures by Prof. Don Marolf which address this topic from a more technical point of view. The lectures can be found, along with a lot of other great stuff, including, a self contained introduction to Quantum Field Theory by Prof. Tony Zee, calculation of the temperature of an observers cosmological horizon in the case of de Sitter Space by Prof. Maulik Parikh, an introduction to Particle Physics by Prof. Alan Martin, and Cosmology talks by Prof. George Ellis, all at the African Summer Theory Institute. Oh yeah, plus a bunch of String Theory stuff too! The site is actually pretty cool.

As usual, pod pod invites reader comments on this post and especially if you think I am all wet. Thanks in advance for your scathing critiques.

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