# Re: The Big Bang

Michael Lorrey (retroman@together.net)
Wed, 12 Nov 1997 10:53:14 -0500

Anders Sandberg wrote:
>
> Greg Butler <gsb1997@ix.netcom.com> writes:
>
> > On Tuesday, November 04, 1997 8:54 AM, Anders Sandberg
> > > No. The amount of radiation you get from a star decreases with the
> > > square of the distance. But the number of stars grows with the square
> > > of the distance, so the total radiation from a given distance is
> > > approximately constant.
> >
> > I don't understand how one can compare one infinite number
> > with another (such as an infinite volume of space and an infinite
> > number of stars). If there is an infinite space per star, it seems
> > possible that it won't get filled. One infinite amount does not
> > necessarily equal another. For instance, how big is an infinite
> > amount of space squared? Is it bigger than a "regular" infinite
> > amount? Greg
>
> Ok, wait a moment while I don my impressive mathematicians robe
> (complete with chalk dust)... :-)
>
> Infinities are tricky, but if you know the rules and are a bit careful
> you can handle them with no danger. Yes, there are differently "sized"
> infinities, some much "larger" than others. But in this case it is
> much simpler.
>
> In the star example, let's say there are around 1 star per cubic
> lightyear. That means that on a sphere of radius r there will be
> around 4 Pi r^2 stars. The intensity of a star at distance r is L/r^2
> where L is a constant (I assume all stars are equally bright now, for
> simplicitys sake). This means that the 4 pi r^2 stars on the sphere
> around us will shine with a total power of 4 Pi L - the two r^2 terms
> cancel each other. So the total luminosity from a spherical shell
> around us is independent of it's radius. This means that if we get one
> unit of light from all stars 1 lightyear away, we will get one unit
> from the stars 2 lightyears away, one from all stars 3 lightyears
> away, and so on. As you sum up the contributions from all the shells
> the sum get's larger and larger. And if you assume an infinite
> universe you will have to sum up the contributions of ever more remote
> stars, and the total luminosity will become larger than any finite
> luminosity.

However, unlike pure numbers, in terms of "real" photons, you can't have
a fraction of one from a source. You either have none or one or more
integers of photons. Fractional photons only exist in a virtual state,
as in the ZPF. Given this, you need to treat the light from each star
relative to distance like its on a halflife type scale. THis should
limit the total luminosity of a infinite universe, shouldn't it?
Additionally, consider the filtering effects of non-luminous matter. For
example, the actual core of our galaxy is, from our point of view, one
of the dimmest parts of the milky way, because that region also has the
most dust clouds, but it also has the highest density of stars, but it
isn't bright, is it? The existence of dark matter can also be used in a
steady infinite universe to explain the dark sky.

```--
TANSTAAFL!!!
Michael Lorrey
------------------------------------------------------------
mailto:retroman@together.net	Inventor of the Lorrey Drive
MikeySoft: Graphic Design/Animation/Publishing/Engineering
------------------------------------------------------------
How many fnords did you see before breakfast today?
```