🚀 go-pugleaf

"Eric Gisse" <jowr.pi@gmail.com> wrote in message
news:84c8c8de-d68c-4319-998a-475f9783b160@q39g2000hsf.googlegroups.com...
> Max Keon wrote:
>> Eric Gisse wrote:
>>> Max Keon wrote:
>>>> Eric Gisse wrote:
>>>>> The CMBR has been detected many times over the past 40 years.
>>>>
>>>> Yes, it has. But what is detected today requires that spectral
>>>> line frequencies generated when the CMBR was first transmitted
>>>> be exactly what they are when generated in today's environment.
>>>> That's hardly likely when the mass of the entire universe was
>>>> within, at most, 10000000 * 2 light years of everything.
>>>>
>>>> Can't you see that?
>>>
>>> Your argument would be more meaningful if you had a calculation
>>> supporting it. Especially if the argument wasn't that we cannot
>>> possibly be observing what has been routinely observed.
---

>> Every natural oscillator is affected by every bit of matter in
>> the universe. If the mass of the currently visible universe is
>> i.e. 5e+56 kilograms, and all of that mass resided in a thin
>> shell at the visible extremes, at a 13 billion light year radius,
>> or 1.23e+26 meters, clock running speed is  t' = t/(G*M/(c^2*r))
>>  t' = 3.32e-4 of what it would be in a universe where only the
>> clock exists.

> When you learn to stop pulling equations out of your ass, let me
> know and this can continue.

I chose to use the equation in the above format because
 t' = t*(G*M/r))  generates numbers large enough to frighten you
away. We can do it your way if you like. t' is of course the
clock slowing rate per (whatever time unit you want).

You will no doubt recognize the time slowing difference between
the Earth surface and the GPS according to that equation.
A clock at 6378000 meter radius loses 6.937e-10 sec/sec.
A clock at 26570000 meter radius loses 1.665e-10 sec/sec.
The difference is 5.272e-10 seconds per second or 4.5422e-05
seconds per sidereal day. You're happy with that?

The actual clock rates are per  t' = t + t*(G*M/(r*c^2))
Each second on the Earth's surface is 1.0000000006937 seconds
according to a clock that's unaffected by Earth's gravity, and
is 1.0000000001665 seconds at a GPS radius.

Consider an Earth diameter and Earth mass sphere where all of
the matter is spread uniformly throughout a thin outer shell. One
second for a clock at the center would be
 t' = t + t*(G*M/(r*c^2)) = 1.0000000006937 unaffected clock
seconds. That will be the case regardless of where the clock is
placed within the shell. You can do the calculations.

But the clock that's considered free from the effects of Earth's
gravity is still affected by the matter in the universe. It's
never free to run unaffected by gravity. And that is what
demolishes the big bang theory.

Now consider a visible universe diameter sphere, 13 billion
light years or 1.23e+26 meter radius, with all of the matter,
i.e. 5e+56 kg, contained in a thin outer shell.
 t' = t + t*(G*M/(r*c^2)) = 3014. A truly free clock ticks 3014
seconds in one second according to the universe housed clock.

Then take the clock back to the early universe, 300000 years
after the big bang. Allowing for as much as a 20 times light
speed initial expansion, gives a shell radius of 5.68e+22 meters.
The mass of the universe remains constant. The clock now ticks 1
second per 6523866 free clock seconds. That's 2165 times slower
than today's clock.

The same applies for the spectral line generating oscillators at
the time. They are affected in exactly the same way as the atomic
clock oscillator. Spectral lines are redshifted ENORMOUSLY before
the expansion is even considered.

_Spectral line generating oscillator rates and clock rates
increase as the universe evolves_.

Here's how the zero origin universe accommodates the CMBR,
 http://members.optusnet.com.au/maxkeon/cmbr.html
You should be able to recognize the real universe by now.

-----

Max Keon