Re: framerates

From: P.T. Breuer (ptb_at_oboe.it.uc3m.es)
Date: 05/08/04 

Date: Fri, 07 May 2004 23:30:18 GMT

In alt.os.linux Kadaitcha Man <nospam@kadaitcha.cx> wrote:
> P.T. Breuer wrote:
> > In alt.os.linux Kadaitcha Man <nospam@kadaitcha.cx> wrote:
> >> t = time
> >> c = the speed of light
> >> k = constant that describes the geometry of universe
> >> G = gravity constant
> >> p = pressure in universe
> >> R = length or diameter of universe <--- watch that one.
> >> r = the density of radial matter in universe
> >>
> >> A) 2(d^2R/dt^2)/R+[(dR/dt)/R]^2+kc^2/R^2 = -8(pi)Gp/c^2

What does this really say? One thing I noticed as I read it was that it
appeared to be dimensionless in terms of R, or rather the whole thing is in
seconds^-2. That's weird.

> >> B) [(dR/dt)/R]^2+kc^2/R^2 = 8(pi)GR/3)

This identifies two of the terms above as actually summing to O(R).

> > Some parens have gone wrong at right.
>
> Only one. It was done to see if anyone reading it was awake.
> Congratulations. You're the first pedant to spot it.

Nevertheless, the sum you do is literally incorrect, as noted BELOW.
The unbalenced paren merely makes me wonder as to whether there is
something missing that you intended to write.

> >> Where dR/dt and d^2R/dt^2 represent velocity and acceleration.
> >>
> >> Subtracting B) from A) yields:
> >>
> >> C) 2(d^2R/dt^2)/R = -8(pi)G(p+3p/c^2)/3
> >
> > well, almost. You get on the RHS -8(pi)Gp/c^2 - 8(pi)GR/3)
> > (sic) which would be, if I skip the final paren,
> >
> > -8(pi)G(p/c^2 +R/3)
> >
> > If you then pull out the 3 and switch the summands you get
> >
> > -8(pi)G(R+3p/c^2)/3
> >
> > So it looks as though you have a p <-> R problem.
>
> <snicker>

It requires a yes/no/maybe, not a "<snicker>".

For really small R, 3p dominates, probably in the negative direction,
and pushes R out with huge positive acceleration.

As R grows, however, your (B) brings the velocity close to the R^3/2
range. The acceleration therefore is in the R^1/2 range at that point.
At that point (C) would be saying something like O(1/R^1/2) = O(R+3p/c^2).
That's impossible if R continues growing and p doesn't, so
probably either R stops somewhere, or else R is mostly matched by
-3p/c^2 and both keep growing.

> See below.

I did. At first glance nothing relevant to that appears below ...
but I'm prepared to reread.

> >> The left of equation C) viz 2(d^2R/dt^2)/R is 2/R multiplied
> >> by universe radial acceleration.
> >
> > Yo!
> >
> >> So since:
> >>
> >> G is a force of attraction and is +ve
> >>
> >> THEN:
> >>
> >> C) shows the universe is ... what?
> >
> > Well, goshemi, is it speeding up or slowing down its radial expansion?
>
> That was your job to figure out, linuxfucktard.

If you think goshemi is an insult that requires responding to with an
insult back, you go right ahead and think that.

> > I guess that depends on the sign (and size, in my equation) of p. In
> > any case, in yours it just depends on the sign. In mine it seems to
> > say that when the universe was tiny and pressure was also negative,
> > then the universe underwent rapid radial acceleration. Inflation, I
> > suppose.
> >
> > Since then it appears to have been decelerating its expansion, I
> > suppose, but I haven't checked to see if R will grow faster positively
> > than p negatively ....
>
> You bombed out. At the point where the test was terminated, the universe is
> DECELERATING according to the equation. But...
>
> Since: R = length or diameter of universe
>
> For low values of R, p can be ignored, thus R EITHER

You mean for small R, p is even way smaller. However, where's the
reasoning? I don't see anything above that fixes a scale for R. As
I remarked, the only comparison evident is between p and R. The rest
is dimensionless for R.

As far as I can see, R can be absolutely huge. p just has to be even
bigger and negative to start off driving inflation. I suppose you're
going to say, "OK, let's call that scale the small one, then". OK.

> increases to infinity over time, OR R reaches a maximum
> value then decreases over time. So, the universe must

I didn't see a decrease as being obvious, though the driving
stresses in the equations clearly cross over somehere in the finite
case. Where's the shrink?

> necessarily be expanding or it must necessarily have
> expanded at some point in time past.

OK. (if I accept your idea that the thing will bounce)

> p can be ignored for low values of R because p is lower
> than r.

r does not appear in your equations. Therefore your reasoning is
invalid.

If p can be ignored for small R, then there would never have been ny
negative pressure to have driven inflation with. Maybe you mean "not
small R, but not too big R". Uhuh ... in the finite R case, p and R
must be commensurate at inflexion points.

> Current observations show that the universe is accelerating,
> thus Einsteinian general relativity holds by way of observation.
>
> So, by way of observation, we must have a low value of R. Since
> R = length or diameter of universe, the universe [*1] must be [*2].
>
> That is, we exist in an accelerating and finite universe.

Ultimately finite, you mean. Well, I'm too sleepy to think about it.
Maybe, probably. Gnight.

Peter