🚀 go-pugleaf

"Eric Gisse" <jowr.pi@gmail.com> wrote in message
news:cf7fd5a6-005e-4d73-aef9-ba788194923d@s8g2000prg.googlegroups.com...
> Max Keon wrote:
>> Eric Gisse wrote:
>>> Max Keon wrote:

>>>> The results are the same as with GR. A clock fixed with the ECI
>>>> frame at an Earth surface radius loses 5.28e-10 seconds per
>>>> second of a GPS satellite radius clock. Multiply that by 86160
>>>>  = 4.55e-5 seconds per sidereal day. What's wrong with that?
>>>
>>> Are you unable to tell the difference between being a fair distance
>>> from a point mass to being immersed in an isotopic spray of matter?
>>>
>>>>  1+V/c^2 is the gravitational redshift factor, and that applies
>>>> to every electronic interaction, including clocks. V is GM/r. So
>>>> that becomes  1+(G*M/r)/c^2  which is the same as my equation
>>>>  t' = t+t*(G*M/r)/c^2  so long as "t" is specified as the time
>>>> unit, which is always 1.
>>>
>>> Your equation is wrong, as I already demonstrated.
>>
>> My equation gives the correct answer.

> Does it?
>
> Let's see the computation.

Firstly, all clocks are fixed with the ECI frame.

 1+(GM/r)/c^2: r=6378000 meters for the surface radius clock:
 1+(6.67e-11*5.97e24/6378000)/3e8^2 = 1.000000000694 seconds
per 1 second of a clock that's unaffected by Earth's gravity.

 1+(GM/r)/c^2: r=26570000 meters for the GPS radius clock:
 1+(6.67e-11*5.97e24/26570000)/3e8^2 = 1.0000000001665 seconds
per 1 second of a clock that's unaffected by Earth's gravity.

 1.000000000694 - 1.0000000001665 = 5.28e-10 difference.
The GPS radius clock is running 5.28e-10 seconds faster per
second than the lower clock. Multiply that by 86160 = 4.55e-5
seconds faster per sidereal day.

That's correct isn't it.

>>>> The way it was written may have been confusing to you, but it
>>>> wasn't wrong.
>>>
>>> Actually, it is. You are taking the weak field approximation to GR,
>>> and then you apply it incorrectly.
>>
>> But I get the right answer don't I.

> It has zero chance of getting the right answer for GPS since movement
> is not taken into account,

I wasn't including movement. The description was, "A clock fixed
with the ECI frame at an Earth surface radius loses 5.28e-10
seconds per second of a GPS satellite radius clock." I thought
it was clear that the two clocks were fixed with the ECI frame.

> and I see no comparison against Pound &
> Rebka.

If radioactive iron is chosen as the frequency generator in an
atomic clock, a GPS radius clock will run _slower_ than the
surface radius clock (all fixed with the ECI frame), by 4.55e-5
seconds per sidereal day. Not faster. The Pound and Rebka
experiment proves that gamma rays are generated in a vastly
different environment to that of lesser frequency rays.

>> At the Earth's center of mass the gravitational field is
>> isotropic, but the field hasn't gone anywhere. An isotropic
>> gravity doesn't remove its consequences. Do you think it logical
>> that a clock enroute to the Earth center from a distance above
>> the surface, will continue to slow until it enters the body of
>> the Earth, then rather rapidly speed up until it's running as
>> fast as would a clock in deep space, where Earth's gravity is of
>> virtually no consequence? Do you really imagine that Earth's
>> gravity is of no more consequence to electronic interactions at
>> its center than in deep space? I have a bridge that may interest
>> you.

> Woosh! The point goes over your head.
>
> The gravitational field of a point source at an extended distance is
> much different from the gravitational field of an extended
> distribution of matter. If you want to discuss how much redshift the
> CMBR has accumulated since its' time of emission, you need to use a
> model that is consistent with observation - namely, an extended
> isotropic distribution of matter.

I think it's you who missed the point here.

>> A clock tick rate won't further reduce once it has entered the
>> body of the Earth, nor will it speed up. What it was at the
>> surface will prevail right down to the center. Every electronic
>> interaction within the Earth remains redshifted as it would be
>> at the surface.
>>
>> Compressing the Earth to half its diameter doubles the previous
>> clock slowing rate at the surface, and that redshift rate
>> prevails throughout the body.
>>
>> So, shrinking the universe back to the era when it first became
>> transparent, when its temperature had reduced below 4000 K, is
>> going to produce an ENORMOUS gravitational redshift throughout
>> the body of the early universe. The CMBR started off with the
>> spectrum of a very cold blackbody radiator. Then it was
>> stretched.

> Let's see....you still don't have a computation supporting your
> assertions, nor do you have a reference justifying your 4000K
> assertion.

Around about May 2005, this is an excerpt from a post which was
posted on sci.astro.research , by Ulf Torkelsson, Chalmers
University of Technology, Sweden.

-    No, this is not what the big bang model says.  It says
- that a few hundred thousand years after the big bang the
- universe had become sufficiently cool, about 4000 K, that
- the electrons could combine with the atomic nuclei to
- form atoms.  At that time the universe becomes transparent
- to electromagnetic radiation.  Since then the universe
- has expanded by a factor 1300, and consequently the
- temperature of the universe has dropped to 4000/1300 = 3 K.

September 2006; The redshift of the CMBR, relative to now, when
it first became transparent was quoted as z=1079. That requires
a 3000 K universe at the time. 3000/1079 = 2.78 K today.

4000 K seems to be the correct threshold temperature where the
plasma breaks down though.

But take your pick.

>> Assuming that gravitational redshift is zero within the
>> isotropically distributed mass of the early universe will prove
>> to be fatal for the big bang theory.

> I never claimed that it was, I was merely trying to get you to think a
> little.

Perhaps it's time you put your thinking cap on. If gravitational
redshift is ANYTHING BUT ZERO the CMBR spectrum, at the time when
it was first released to travel the universe, would not have
been the same spectrum as for that temperature today. Any
redshift at all, means that time was running proportionally
slower back then, so an observer would still record the same
blackbody spectrum as today's observer. But the colder CMBR
spectrum relative to now is going to stretch to a lesser
temperature spectrum than that of a 2.73 K spectrum today.

>> This graph is according to your theories, showing no time loss
>> whatever for a clock at the Earth's center of mass.
>>  http://members.optusnet.com.au/maxkeon/erthclok.jpg
>> But the clock rate is very wrong according to mine.
 ---The above image is probably distorted
    if not viewed in full screen mode.

> DAMN you missed the point. See above.

The story has only just begun.

>> The curves were generated according to Gauss's law, as applied
>> to gravity. M was determined thus,
>>  (new M) = (last M) / (previous radius / current radius) ^3
>> Is that all too much for you?

> The irony of someone who doesn't know the difference between potential
> energy and force asking if I know Gauss' law is almost too much.
>
> BTW: That's not Gauss' law.

According to the Penguin science dictionary, "Gauss's law is also
used for the corresponding result for gravity: the total flux of
gravitational field at a closed surface is -4*pi*G times the
total mass enclosed by the surface."

The closed surface can be any imaginary shell within the Earth
at any radius about its center of mass. The total flux of the
gravity field for an imaginary shell inside the Earth alters
at a linear rate per shell radius because the mass M housed in
the shells alters at M*(r1^3/r2^3). r1 is the imaginary radius.
r2 is the Earth's radius. e.g. If the imaginary shell radius is
half Earth's radius, 3187000 meters,
 M' = M*(3187000^3 / 6374000^3) = 7.4625e23 kg housed in the
shell. For GM'/r^2: 6.67e-11 * 7.7625e23 / 3187000^2 = 4.9m/sec^2
which is half that for the true surface.

There's nothing wrong with that. The problem arises in the maths
for gravitational redshift, 1+V/c2, which is, 1+GM/r/c^2.
According to that formula, a clock within the Earth runs faster
as it's moved closer to the center of mass. But for that to be
so, the clock oscillator must somehow be directly linked to the
total flux of a gravitational field at a closed surface as well.

Your assertion that it is, is based entirely on what the maths
appears to dictate. But the pull of gravity and electronic
interactions are vastly different. Your assertion is in fact a
postulate because there's not one scrap of evidence that supports
it. The big problem for this postulate is that it can be
falsified.

If the entire mass of the Earth is of _direct_ consequence to
the interaction rate of any electronic interaction within the
Earth, wherever it is, that rate will remain as it is at the
surface. If the mass of the Earth remains constant and all of
its matter is compressed into a sphere of half its radius, the
time slowing at the new true surface will double, as it will
everywhere else throughout the body.

The entire mass of the early universe can be treated in the same
manner. It will all directly affect time rates. The average time
slowing everywhere is what it would be at the "radius of the
universe as gauged from the inside".

My claim is backed by the fact that it's a firm prediction of
the zero origin universe, where you've been lead down the garden
path by your maths.

Find your evidence.

>> Proving that clocks don't speed up as radius shrinks within the
>> Earth is relatively simple.
>>
>> One of two clocks synchronized at a depth below the Earth's
>> surface e.g. 6370000 meter radius, is shifted further into the
>> surface by 10000 meters, and they are both left running for some
>> time. The clock closest to the surface is then brought down to
>> the lower clock and compared.
>>
>> I say they will both display the same elapsed time.
>>
>> This is what you say.
>> For clocks at 6370000 and 6360000 radii:
>>  (new M) = 5.967e24 / (6370000 / 6360000) ^3 = 5.942e24 kg
>>
>>  (G*(last M)/r)/c^2 = 6.9457e-10 seconds per second.
>>  (G*(new M)/r)/c^2 =  6.924e-10 seconds per second
>>
>> The difference is 2.17e-12 seconds per second. Which is only
>> 6.8e-5 seconds per year. The difference is fairly minute, but
>> it should be detectable if it's there.
           (repaired)

-----

Max Keon