Re: Cosmic acceleration rediscovered

From: greywolf42 (mingstb_at_marssim-ss.com)
Date: 01/05/05


Date: Wed, 05 Jan 2005 20:40:19 GMT

George Dishman <george@briar.demon.co.uk> wrote in message
news:crevs1$4f7$1@news.freedom2surf.net...
>
> "greywolf42" <mingstb@marssim-ss.com> wrote in message
> news:2jjCd.16800$XZ.10055@news.flashnewsgroups.com...
> > George Dishman <george@briar.demon.co.uk> wrote in message
> > news:crbthk$8mv$1@news.freedom2surf.net...
> >>

{snip abandoned attempt at description}

> We are talking past each other so let me try another
> way. Consider a area of 1m^2 at a distance of 1000AU
> from the Sun. In one second an energy E passes through
> that square. That will deposit a small fraction dE
> into the aether.

Delta E. Yes.

> Now consider an area of 1m^2 at a distance of 2000AU.
> The energy passing through in one second will be
> almost E/4 but very slightly less due to tired light.

Yes.

> Over this distance though the tired light loss will
> be negligible compared to the r^-2 loss. The energy
> deposited into the aether here will therefore be no
> more than dE/4 and in fact slightly less.

Yes.

> If the temperature of the aether at 1000AU is T, that
> at 2000AU should be very close to (but slightly less
> than) T/sqrt(2) since the power from a black body
> radiator is proportional to T^4.

Oops, here is your error.

The aether temperature is not merely a function of the local addition of
energy from starlight degradation. (Temperature is a function of E, not of
dE.) Starlight degradation is a miniscule contribution to the pre-existing
aether energy density. After all, the aether temperature (energy content)
is what drives gravitation. In fact, gravitation is a competing effect --
lowering the local aether temperature a tiny amount, in the vicinity of the
star.

> >> I don't think so. It still seems to me that dE in
> >> the equation you posted will follow an inverse square
> >> from each (point) source of E.
> >
> > As demonstrated above, you are incorrect. A tired light model will
> > always be (slightly) below a pure inverse square model.
>
> Yes, that's what I'm saying, though I had ignored the
> (slightly) part as it is negligible over short ranges
> and it adds to the effect anyway. You are arguing that
> the temperature is uniform aren't you?

Very close to uniform. Not uniform. Normal gravitation and light energy
are tiny variations on the pre-existing local energy density. (The sun's
surface gravitation is about 1 part in 10^8 of the maximum gravitational
acceleration.)

--
greywolf42
ubi dubium ibi libertas
{remove planet for return e-mail}


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