Friiedman Fun Facts to know and tell

[jonmtkisco]

Hi Pervect,

I ran a "radiation only" scenario, and happily, the universe remains geometrically flat because its instantaneous expansion rate is always exactly equal to the escape velocity of the total radiation mass/energy. As with all of these flat-universe expansion curves, the expansion continues forever at an ever decreasing rate.

In that scenario of course, the number of photons doesn't decrease, but the energy-per-photon continues decreasing forever. Which raises a question, is there an absolute minimum energy-per-photon threshold, or can a photon possesses an energy that is infinitely close to zero, if its wavelength is stretched long enough?

Also, although I understand that the Friedmann equations mathematically treat the gravity added by radiation pressure as having no direct effect on geometric curvature, is there a straightforward "physical" explanation for why, in the abstract, one "flavor" of gravity directly affects geometric curvature while another concurrent "flavor" of gravity does not?

I guess that's a dumb question because it isn't really a case where there are two different "flavors" of gravity. The reality is simply that twice as much gravity (of any flavor) is needed to offset the simultaneous decline in total mass/energy, and thereby maintain flatness.

It strikes me that in a flat universe like ours, "there is no such thing" as free radiation that isn't constantly degenerating (losing energy). Free radiation is fundamentally unstable.

Jon

 

[jonmtkisco]

Hi Mr or Ms Hillman,

I must protest your abusive and non-responsive posts.

Jon

 

[Chris Hillman]

Jon, click on "UserCP" at the top of the display shown (probably) by your browser when you are looking at (almost) any PF page. Click on buddy/ignore lists and add me to your ignore list. After I post this I will do the same for you. This will ensure that henceforth we don't see each other's posts, which I trust will be agreeable to you.

 

[jonmtkisco]

Well I guess there's no point in responding to Chris because he's ignoring me now.

I'm interested in robust dialogue, not cutting it off. Obviously Chris has tremendous expertise and I'm quite interested in reading what he has to say. I want him to direct some of that towards the substance of my questions...

I'm going to end this thread now and pursue questions in separate threads.

Jon

 

[pervect]

OK, here is my final $.02 on the topic. I'm going to lock the thread after this as I mentioned previously, because I think that everything that needs to be said will have been said and the natives here on the forum are getting restless (plus I'm getting tired of it, too).

This also implies that I don't want to see new threads on this same topic for at least a couple of weeks, if not longer.

I would like to go on record as to encourage people such as jonmtkisco to brush off their rusty calculus skills and actually do some work. Doing work does take work, but it pays off in understanding. That's another reason I'm going to lock the thread, I'm hoping it will encourage more thought and less typing, and this thread is getting way too long.

With all the factors of c and G put back in we get the following symbolic solutions for a(t), rho(t) and P(t) with K=0

case1: no pressure (matter dominated universe)a(t) = a_0 \, t^{\frac{2}{3}}
rho(t) = \frac{1}{6 \pi G t^2}
P(t) = 0

these match up with MTW's equations (pg 735) except for the unit conversion factors.

To verify that these are correct, put them into F1 and F2 (the dynamic and initial value Friedmann equations), and show that they satisfy both equations.

Satisfying both F1 and F2 (the dynamic and initial value equations) is necessary to satisfy Einstein's field equations. Furthermore, it is both necessary and sufficient for F1 and F2 to both be simultaneously satisfied to ensure local energy conservation.

case 2: radiation dominated
<br /> a(t) = a_0 \, t^{\frac{1}{2}}
<br /> rho(t) = \frac{3}{32 \pi G t^2}<br />
<br /> P(t) = c^2 \frac{rho(t)}{3} = \frac{c^2}{32 \pi G t^2}<br />

Again, these can be verified by showing that they satisfy F1 and F2.

In geometric units, for the radiation dominated case, P = rho/3. In non-geometric units, this becomes P = c^2 rho / 3. This is the proper relationship between density and pressure for a universe of pure radiation. P = c^2 rho/3 is an example of what is called an "equation of state" and is the "equation of state" of radiation. As implied previously, the "equation of state" for cold matter is just P=0.

If Jon's spreadsheet gives values that are compatible with these symbolic expressions, great. If not, there is probably a problem with the spreadsheet.

If Jon's spreadsheet is giving these (correct) results, then Jon's problem is with the interpretation of his results, not with the spreadsheet.

I believe that Jon may be attempting to splice these two solutions together, and that may be the source of his difficulty. A smooth splice must have the following properties at a minimum:

a(t) must be continuous (no sudden jumps) at the transition
da/dt must be continuous (no sudden jumps) at the transition, because d^2a/dt^2 must exist

Such a transition is *not* a part of standard cosmology. The standard Friedmann cosmology assumes that there is no significant coupling between radiation and matter.

[add]
Initially, I thought there wasn't any smooth way to splice together a radiation dominated cosmology to a matter dominated cosmology with K=0, but I've changed my mind.

I'll note that such a solution is not a standard textbook solution. Standard textbook solutions assume little coupling between matter and radiation.

The splice I think I've found is:

a(t) = sqrt(t), t<1
a(t) = (t+37/27)^(2/3) t>1

a(t) and a'(t) are continuous at t=1, there is however a step function change in a'' at t=1. (I don't think this is a problem).

One should be able to work forwards from this using both Friedmann equations and the assumption that K=0 and \Lambda=0 to get a valid solution for rho(t) and P(t). Interestingly enough, rho(t) does not appear to be continuous either.