Re: Expansion is wrong and its soooo freakin' obvious

On May 14, 11:09 pm, The Ghost In The Machine
<ew...@xxxxxxxxxxxxxxxxxxxxxxx> wrote:
In sci.physics, Michael Helland
<mobyd...@xxxxxxxxx>
wrote
on Tue, 13 May 2008 00:50:30 -0700 (PDT)
<929b7963-2286-4108-8c15-9f5db9a99...@xxxxxxxxxxxxxxxxxxxxxxxxxxx>:



On May 12, 10:47 pm, The Ghost In The Machine
<ew...@xxxxxxxxxxxxxxxxxxxxxxx> wrote:
In sci.physics, Michael Helland
<mobyd...@xxxxxxxxx>
wrote
on Mon, 12 May 2008 15:05:00 -0700 (PDT)
<af058eff-aa5e-4e12-9b12-4547c051b...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>:

On May 12, 2:20 pm, Eric Gisse <jowr...@xxxxxxxxx> wrote:
On May 12, 12:24 pm, Michael Helland <mobyd...@xxxxxxxxx> wrote:

What happened to "but eric, deceleration is the /same/ as
expansion!!!" ?

They match the same data.

Hell, what happened to the LAST TWO THREADS you made
about this subject but abandoned mid-stream?

Here - I'll refresh your memory.

http://groups.google.com/group/sci.physics/msg/6609337492105fdb?dmode......

How does my theory predict correct luminosity as a function of
distance?

Rewrite as a function of time

Expansion and deceleration both add time to the light's journey,
unlike tired light.

Extra time is redshift: f = 1/t

Observation: Hubble redshift

Explanation: The loss in frequency and energy is the natural
deceleration of light signals in steady space that fits the same curve
as constant light speed in an expanding space.

http://www.astro.ucla.edu/~wright/tiredlit.htm

Tired light is wrong.

It says light loses energy, but not velocity.

If there is no deceleration and no expansion, then no time is added in
conjunction with redshift.

That's why it fails the predictions.

It can't match the same curve that expansion and deceleration can.

Prediction 1: The next round of telescopes (2013?) discovers, once
again, galaxies far too old and distant for the big bang.

Why not make an actual prediction about the multipole moments in the
CMBR?

Because I don't understand multiple moments.

Prediction 2: We'll find that quasars are light signals from galaxies
near the end of the EM field's range, just before the very end of the
range which limps in as the CMB.

Idiot. Quasars are distributed over a long [but still far away] range
of distances. Plus the photon is massless - I told you this before,
but apparently you didn't bother reading it.

I accept that the photon is massless.

It can still have a finite range.

OK, I'll bite. What would the range of a photon be?

It holds to the inverse square law until Hubble redshift begins.

You'll have to clarify that. Both the inverse square law and
the Hubble redshift are continuous and have indefinite (some
might say infinite) range.


I'm saying that the inverse square law does NOT have an indefinite
range;

That the velocity, energy, and frequency begin to die out starting at
around 60 Mpc and it is observed as Hubble redshift.


When we observe redshift, we're literally observing the end of the EM
field, even if we aren't prepared to admit it.

And what does "the end of the EM field" really mean here?

Let's say I shine a light into the sky, and it travels at c away from
earth.

Then let's say it travels 60 Mpc, it starts to travel at less than c.

If you rewrite the Hubble parameter in the units of deceleration, then
you can derive the new velocity like so:

v = c - Ht

<snip>
It seems there is evidence that gamma rays may delay
faster than other photons.

Delay or decay? Also, how much faster?

4 minutes later, after 500 million light years:

http://www.physorg.com/news110480559.html


Prediction 3: We'll observe clusters that actually bleed right into
the CMB.

Idiot. The CMB is isotropic to parts per million level, has
inhomogeneities that have a Gaussian distribution [what does this
mean, Mike?] and has a blackbody spectrum [again, what does this man?]
and as such can /not/ possibly be from disparate sources like clusters.

The point is that beyond the range of the EM force are an infinite
amount of galaxies just too far for their light to reach us, and vice
versa.

An infinite number of galaxies aren't really that disparate.

This does get around the "sky as bright as the sun" problem
but I'd still like a formula for the range of a light beam.

Well, if E = hf, then maybe every megaparsec or so, E loses and h. So
E = hf - h, untill E is 0, then, blam.

Why that particular constant? Also, that would generate a readily
identifiable spectrum shift.

You're also not horribly clear on the concept of units, apparently.

I *might* buy into the notion that E' = E * exp(-kd) for some
k, were it not for the fact that no evidence at all exists
for this sort of thing.

Hubble redshift is evidence.

That's exactly what your formula predicts.


And yes, it does solve the night sky paradox. Thank you very much for
acknowledging that.

It's an obvious paradox; if one has an infinite dust-free Universe,
any extended view-ray eventually hits a star surface. Any singleton
dust clouds in such a Universe will eventually heat up to just
under the temperature of a star surface, as it is surrounded by
relatively hot stars. (It might take awhile.) Multiple dust
clouds won't fare much better.

Ergo, no one sleeps at night, as there's no night.

(In the real Universe, of course, there are two
compensating factors. First, it's of limited size; second,
for whatever reason the Hubble expansion cools the light.
So we get a nice black sky, instead -- at least in the
visual range.)


Or:

What we observe in outer space is just a drop of the Universe.

What we observe in outer space comes from light.

Light has a finite range, it begins red shifting and eventually stops.

Even though the Universe is infinite, not of all it is within our
observable drop and of that which is, not much is left in our visible
spectrum (thank you evolution).



Also, does said light beam's characteristics (frequency,
energy) change during its travels? If so, how? Also,
what would distinguish your theory from already discredited
"tired light" affairs?

Tired light tries to lose energy, but it never suggests the light gets
slower.

If one assumes E = 1/2 m v^2, diminishing E while holding
m constant requires that v decrease. Either the light particle
loses mass or speed; pick.


It loses speed.
.