Gravitational redshift Query

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"WG" wrote in message
...
Can someone here shed some light on why the gravitational potential well
equation gives a answer that is close to what is observed, but maybe it
should not be.

While there is no center of the universe, from an observational point of
view every point in the universe is at the center. We even have a name for
it, the Visible Horizon. The Visible Horizon is a sphere of mass surrounding
every point in the universe. While there may be mass beyond this sphere, it
is forever hidden since any effects, (light, gravity etc) can not have
reached us yet.

Each point in the universe has a different Visible Horizon that is displaced
by the distance between the two points, (i.e. there is a small wedge of mass
in each sphere whose effects cannot be felt by the other). It is this
displacement which introduces an asymmetry into the argument.
Now for you and me this displacement is negligible, but for distant QSOs the
asymmetry is significant.
I have posted a dia at http://tinyurl.com/3uzqkf .
Of course this a static picture and a dynamic one would look different,
[i.e. the photon travels, the radius expands], but the underlying asymmetry
must still remain.

Now if we treat the photon at its point of emission as traveling out from
the center of its Visible Horizon, or climbing out of a potential well it
will be gravitationally redshifted according to Zg=4.19GDr^2/C^2.

One might argue that if the photon is climbing out of a potential well at
point of emission it would be falling into the potential well of the
observer, and thus the effects cancel, but this is incorrect due to the
inherent asymmetry. The photon can only ever be climbing out, never falling
in due to the some mass being hidden (hatched area in dia).

Plugging in the values for observed values of Z, say Z = 1 thru 5 (I
believe up to 7 has been observed)
yields a Radius of the universe in the order of 10^26 meters, which is
curiously close to the radius calculated by other methods. Radius taken
from John Baez FAQ at http://math.ucr.edu/home/baez/distances.html .

Now this may be just a coincidence, but I sure hate it when its this close.

P.S. I am well aware I am using a Newtonian, flat, Euclidian, isotropic
calculation, and that a treatment under GR might be different.

Cheers

The hatched area is only hidden for the QSO, not for the photon.
Once it reaches A then its horizon is A's horizon, and so it is a
moving horizon. There is no asymmetry.

Can someone here shed some light on why the gravitational potential well equation gives a answer that is close to what is observed, but maybe it should not be.

While there is no center of the universe, from an observational point of view every point in the universe is at the center. We even have a name for it, the Visible Horizon. The Visible Horizon is a sphere of mass surrounding every point in the universe. While there may be mass beyond this sphere, it is forever hidden since any effects, (light, gravity etc) can not have reached us yet.

Each point in the universe has a different Visible Horizon that is displaced by the distance between the two points, (i.e. there is a small wedge of mass in each sphere whose effects cannot be felt by the other). It is this displacement which introduces an asymmetry into the argument.
Now for you and me this displacement is negligible, but for distant QSOs the asymmetry is significant.
I have posted a dia at http://tinyurl.com/3uzqkf .
Of course this a static picture and a dynamic one would look different, [i.e. the photon travels, the radius expands], but the underlying asymmetry must still remain.

Now if we treat the photon at its point of emission as traveling out from the center of its Visible Horizon, or climbing out of a potential well it will be gravitationally redshifted according to Zg=4.19GDr^2/C^2.

One might argue that if the photon is climbing out of a potential well at point of emission it would be falling into the potential well of the observer, and thus the effects cancel, but this is incorrect due to the inherent asymmetry. The photon can only ever be climbing out, never falling in due to the some mass being hidden (hatched area in dia).

Plugging in the values for observed values of Z, say Z = 1 thru 5 (I believe up to 7 has been observed)
yields a Radius of the universe in the order of 10^26 meters, which is curiously close to the radius calculated by other methods. Radius taken from John Baez FAQ at http://math.ucr.edu/home/baez/distances.html .

Now this may be just a coincidence, but I sure hate it when its this close.

P.S. I am well aware I am using a Newtonian, flat, Euclidian, isotropic calculation, and that a treatment under GR might be different.

Cheers

Dear WG:

On Apr 29, 12:12 pm, "WG" wrote:
Can someone here shed some light on why the
gravitational potential well equation gives a
answer that is close to what is observed, but
maybe it should not be.

While there is no center of the universe, from
an observational point of view every point in
the universe is at the center. We even have a
name for it, the Visible Horizon.

.... aka. the Rindler Horizon

The Visible Horizon is a sphere of mass
surrounding every point in the universe.

.... no "mass", "distance" beyond which we can never get light that is
newer than *now*.

While there may be mass beyond this sphere, it
is forever hidden since any effects, (light,
gravity etc) can not have reached us yet.

Each point in the universe has a different
Visible Horizon that is displaced by the
distance between the two points, (i.e. there is
a small wedge of mass in each sphere whose
effects cannot be felt by the other). It
is this displacement which introduces an
asymmetry into the argument.

What is the shape of this "wedge", in closed space? It is symmetric
about the axis joining the points...

Now for you and me this displacement is
negligible, but for distant QSOs the asymmetry
is significant. I have posted a dia at
http://tinyurl.com/3uzqkf .
Of course this a static picture and a dynamic
one would look different, [i.e. the photon
travels, the radius expands], but the
underlying asymmetry must still remain.

Now if we treat the photon at its point of
emission as traveling out from the center of
its Visible Horizon, or climbing out of a
potential well it will be gravitationally
redshifted according to Zg=4.19GDr^2/C^2.

No. Your assumptions of "flat space" is reasonable, so you must be
aware that this implies there is a uniform (if sparse) distribution of
mass in all directions. So the "potential well" has a gradient only
in time.

One might argue that if the photon is
climbing out of a potential well at point
of emission it would be falling into the
potential well of the observer, and thus the
effects cancel,

.... which is obviously not the problem ...

but this is incorrect due to the inherent
asymmetry. The photon can only ever be
climbing out, never falling in due to the
some mass being hidden (hatched area in dia).

Does not apply.

Plugging in the values for observed values
of Z, say Z = 1 thru 5 (I believe up to 7
has been observed)

CMBR, z = 1024 (give or take)

yields a Radius of the universe in the order
of 10^26 meters, which is curiously close to
the radius calculated by other methods.
Radius taken from John Baez FAQ at
http://math.ucr.edu/home/baez/distances.html.

Now this may be just a coincidence, but I
sure hate it when its this close.

David A. Smith

This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/

"WG" wrote in message
...

"Androcles" wrote in message
...


--
This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/

"WG" wrote in message
...
Can someone here shed some light on why the gravitational potential well
equation gives a answer that is close to what is observed, but maybe it
should not be.

While there is no center of the universe, from an observational point of
view every point in the universe is at the center. We even have a name for
it, the Visible Horizon. The Visible Horizon is a sphere of mass
surrounding
every point in the universe. While there may be mass beyond this sphere,
it
is forever hidden since any effects, (light, gravity etc) can not have
reached us yet.

Each point in the universe has a different Visible Horizon that is
displaced
by the distance between the two points, (i.e. there is a small wedge of
mass
in each sphere whose effects cannot be felt by the other). It is this
displacement which introduces an asymmetry into the argument.
Now for you and me this displacement is negligible, but for distant QSOs
the
asymmetry is significant.
I have posted a dia at http://tinyurl.com/3uzqkf .
Of course this a static picture and a dynamic one would look different,
[i.e. the photon travels, the radius expands], but the underlying
asymmetry
must still remain.

Now if we treat the photon at its point of emission as traveling out from
the center of its Visible Horizon, or climbing out of a potential well it
will be gravitationally redshifted according to Zg=4.19GDr^2/C^2.

One might argue that if the photon is climbing out of a potential well at
point of emission it would be falling into the potential well of the
observer, and thus the effects cancel, but this is incorrect due to the
inherent asymmetry. The photon can only ever be climbing out, never
falling
in due to the some mass being hidden (hatched area in dia).

Plugging in the values for observed values of Z, say Z = 1 thru 5 (I
believe up to 7 has been observed)
yields a Radius of the universe in the order of 10^26 meters, which is
curiously close to the radius calculated by other methods. Radius taken
from John Baez FAQ at http://math.ucr.edu/home/baez/distances.html .

Now this may be just a coincidence, but I sure hate it when its this
close.

P.S. I am well aware I am using a Newtonian, flat, Euclidian, isotropic
calculation, and that a treatment under GR might be different.

Cheers

The hatched area is only hidden for the QSO, not for the photon.
Once it reaches A then its horizon is A's horizon, and so it is a
moving horizon. There is no asymmetry.


| This is incorrect.... The visible horizon at point A is not a static
sphere, it is expanding
| (cosmologically) as well.
| So in the dia,,, you get a photon traveling from right to left,

(and getting slower and slower, hence the red shift...)

| chasing its own horizon which is expanding right to left,

If dx/dt = ax where a some ill-defined constant as claimed,
then d^2x/dt^2 = a and is non-zero, the QSO is accelerating away.
Since acceleration requires a force and gravity works in the
opposite direction you'll have to invent a dark force to account
for it. That's not physics, that pure bloody-minded mysticism.

Prove expansion.
Just because every other crackpot goes along with the first crank like
a load of silly sheep (including you) doesn't mean you are right.

What is more, this picture clearly shows the speed of light differs
depending on frequency:
http://antwrp.gsfc.nasa.gov/apod/ap070411.html

The arms of the galaxy appear displaced because IR takes
a different time to reach us than X-ray.
This popular blind faith in only one speed of light is total nonsense.
http://www.androcles01.pwp.blueyonde...rbit/Orbit.htm
http://www.androcles01.pwp.blueyonde...lgol/Algol.htm

Also disproved by Sagnac:
http://www.androcles01.pwp.blueyonde...nac/Sagnac.htm

"dlzc" wrote in message ...
Dear WG:

On Apr 29, 12:12 pm, "WG" wrote:
Can someone here shed some light on why the
gravitational potential well equation gives a
answer that is close to what is observed, but
maybe it should not be.

While there is no center of the universe, from
an observational point of view every point in
the universe is at the center. We even have a
name for it, the Visible Horizon.

... aka. the Rindler Horizon
Correct

The Visible Horizon is a sphere of mass
surrounding every point in the universe.

... no "mass", "distance" beyond which we can never get light that is
newer than *now*.

Yes it is usually stated as a distance, but it is the mass within this distance (radius) which causes gravitational effects,,, unless you assume gravity (gravitons under QM) or (Curvature if you assume GR) spreads its effect faster than C (i.e. instantaneously), in which case R- infinity and no grav redshift would occur. I believe that gravity propogating at C is currently accepted and thus the visible horizon with respect to gravity is finite.
Minor point.

While there may be mass beyond this sphere, it
is forever hidden since any effects, (light,
gravity etc) can not have reached us yet.

Each point in the universe has a different
Visible Horizon that is displaced by the
distance between the two points, (i.e. there is
a small wedge of mass in each sphere whose
effects cannot be felt by the other). It
is this displacement which introduces an
asymmetry into the argument.

What is the shape of this "wedge", in closed space? It is symmetric
about the axis joining the points...

????
See dia,,, it is certainly symmetric about the axis vertically which causes no effect (Gausses theorem alows you to reduce it to a point mass on the axis for calculation purposes but does not effect the argument at all), but it is definately assymetric horizontally since mass beyond the horizon of QSO is hidden (which would be the point on the left if you reduced it).

Now for you and me this displacement is
negligible, but for distant QSOs the asymmetry
is significant. I have posted a dia at
http://tinyurl.com/3uzqkf .
Of course this a static picture and a dynamic
one would look different, [i.e. the photon
travels, the radius expands], but the
underlying asymmetry must still remain.

Now if we treat the photon at its point of
emission as traveling out from the center of
its Visible Horizon, or climbing out of a
potential well it will be gravitationally
redshifted according to Zg=4.19GDr^2/C^2.

No. Your assumptions of "flat space" is reasonable, so you must be
aware that this implies there is a uniform (if sparse) distribution of
mass in all directions.

Correct, I am useing Critical density 10^-29 gms cc^3 which is indeed sparse, but does not extend to infinity for the purposes of Gravitational effects.
I stated isotropic and should have said homogenoeus as well.

So the "potential well" has a gradient only
in time.

Correct, but those photons gobble up distance when they have time.

One might argue that if the photon is
climbing out of a potential well at point
of emission it would be falling into the
potential well of the observer, and thus the
effects cancel,

... which is obviously not the problem ...

Yep,,,, we agree,,,does not occur

but this is incorrect due to the inherent
asymmetry. The photon can only ever be
climbing out, never falling in due to the
some mass being hidden (hatched area in dia).

Does not apply.]

Ahh, we finally disagree... I can only refer you back to the dia. as this is the crux of the argument.

Plugging in the values for observed values
of Z, say Z = 1 thru 5 (I believe up to 7
has been observed)

CMBR, z = 1024 (give or take)

Correct again but the CMBR is not exclusive to just the tiny volume within any given visible horizon (when compared to the entire universe). That is different horizons don't have different CMBRs. The CMBR extends to the entire universe even if we cannot see those parts. Thats why its Z is so high. But the common temp at 2.8 deg was established for all horizons at the inflationary ephoc.
The fact that we only see QSOs out to about Z=7 and not Z= 283 attests to this.
Yeah and I know... A Z of 283 would be too distant to be seen anyway, so it's a poor example.

yields a Radius of the universe in the order
of 10^26 meters, which is curiously close to
the radius calculated by other methods.
Radius taken from John Baez FAQ at
http://math.ucr.edu/home/baez/distances.html.

Now this may be just a coincidence, but I
sure hate it when its this close.

David A. Smith

--
This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/

"WG" wrote in message
...

"Androcles" wrote in message
...

| sadly I have seen your inane posts before!!
| I can only suggest you stay off the thread.

Wassup, don't like real data, prefer crackpot theories?
Sad cretin!

**** off.
*plonk*

"Androcles" wrote in message ...


--
This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/

"WG" wrote in message
...
Can someone here shed some light on why the gravitational potential well
equation gives a answer that is close to what is observed, but maybe it
should not be.

While there is no center of the universe, from an observational point of
view every point in the universe is at the center. We even have a name for
it, the Visible Horizon. The Visible Horizon is a sphere of mass surrounding
every point in the universe. While there may be mass beyond this sphere, it
is forever hidden since any effects, (light, gravity etc) can not have
reached us yet.

Each point in the universe has a different Visible Horizon that is displaced
by the distance between the two points, (i.e. there is a small wedge of mass
in each sphere whose effects cannot be felt by the other). It is this
displacement which introduces an asymmetry into the argument.
Now for you and me this displacement is negligible, but for distant QSOs the
asymmetry is significant.
I have posted a dia at http://tinyurl.com/3uzqkf .
Of course this a static picture and a dynamic one would look different,
[i.e. the photon travels, the radius expands], but the underlying asymmetry
must still remain.

Now if we treat the photon at its point of emission as traveling out from
the center of its Visible Horizon, or climbing out of a potential well it
will be gravitationally redshifted according to Zg=4.19GDr^2/C^2.

One might argue that if the photon is climbing out of a potential well at
point of emission it would be falling into the potential well of the
observer, and thus the effects cancel, but this is incorrect due to the
inherent asymmetry. The photon can only ever be climbing out, never falling
in due to the some mass being hidden (hatched area in dia).

Plugging in the values for observed values of Z, say Z = 1 thru 5 (I
believe up to 7 has been observed)
yields a Radius of the universe in the order of 10^26 meters, which is
curiously close to the radius calculated by other methods. Radius taken
from John Baez FAQ at http://math.ucr.edu/home/baez/distances.html .

Now this may be just a coincidence, but I sure hate it when its this close.

P.S. I am well aware I am using a Newtonian, flat, Euclidian, isotropic
calculation, and that a treatment under GR might be different.

Cheers

The hatched area is only hidden for the QSO, not for the photon.
Once it reaches A then its horizon is A's horizon, and so it is a
moving horizon. There is no asymmetry.


This is incorrect.... The visible horizon at point A is not a static sphere, it is expanding (cosmologically) as well.
So in the dia,,, you get a photon traveling from right to left, chasing its own horizon which is expanding right to left, but the Horizon around the point A is also expanding right to left. The hatched portion will always be hidden (because it has moved) even when the photon arrives at A. Asymmetry is preserved.

"Androcles" wrote in message ...

sadly I have seen your inane posts before!!
I can only suggest you stay off the thread.

Dear WG:

"WG" wrote in message
...
"dlzc" wrote in message
...
....
The Visible Horizon is a sphere of mass
surrounding every point in the universe.

... no "mass", "distance" beyond which we
can never get light that is newer than *now*.

Yes it is usually stated as a distance, but it
is the mass within this distance (radius)
which causes gravitational effects,,,

You have assumed, and space is consistent with, flat space. This
means you are not climbing out of a mass-based gravity well.
There is as much mass in front of you as behind you, for any
location (in general).

unless you assume gravity (gravitons under
QM) or (Curvature if you assume GR) spreads
its effect faster than C (i.e. instantaneously),
in which case R- infinity and no grav redshift
would occur.

The complexity is not required, and your point is incorrect.
Expansion red shift can simply be our "clocks speeding up" due to
relaxation of *global* curvature. All paths are into the future.

I believe that gravity propogating at C is
currently accepted and thus the visible
horizon with respect to gravity is finite.
Minor point.

Finite horizon is correct. Assuming you are mving away from mass
in flat space is incorrect.

....
Each point in the universe has a different
Visible Horizon that is displaced by the
distance between the two points, (i.e. there is
a small wedge of mass in each sphere whose
effects cannot be felt by the other). It
is this displacement which introduces an
asymmetry into the argument.

What is the shape of this "wedge", in closed
space? It is symmetric about the axis joining
the points...

????

It is your assertion...

See dia,,, it is certainly symmetric about the
axis vertically which causes no effect (Gausses
theorem alows you to reduce it to a point mass
on the axis for calculation purposes but does
not effect the argument at all), but it is definately
assymetric horizontally since mass beyond the
horizon of QSO is hidden (which would be the
point on the left if you reduced it).

Any mass in the Universe has had ~14 Gy to express its
contribution to curvature. Moving in any particular direction
will not change that "uniform distribution".

....
Now if we treat the photon at its point of
emission as traveling out from the center of
its Visible Horizon, or climbing out of a
potential well it will be gravitationally
redshifted according to Zg=4.19GDr^2/C^2.

No. Your assumptions of "flat space" is
reasonable, so you must be aware that this
implies there is a uniform (if sparse)
distribution of mass in all directions.

Correct, I am useing Critical density
10^-29 gms cc^3 which is indeed sparse,
but does not extend to infinity for the
purposes of Gravitational effects.

Yes, it does. Everywhere we look, we see CMBR, at an equal
distance. This stuff was more or less uniformly distributed, and
"gelled out" to become our neighbors. At no point are you moving
towards empty space.

I stated isotropic and should have said
homogenoeus as well.

Makes life simpler.

So the "potential well" has a gradient only
in time.

Correct, but those photons gobble up distance
when they have time.

And are not moving towards empty space. There is Universe in
every direction. A Universe with flat space.

One might argue that if the photon is
climbing out of a potential well at point
of emission it would be falling into the
potential well of the observer, and thus the
effects cancel,

... which is obviously not the problem ...

Yep,,,, we agree,,,does not occur

but this is incorrect due to the inherent
asymmetry. The photon can only ever be
climbing out, never falling in due to the
some mass being hidden (hatched area in dia).

Does not apply.]

Ahh, we finally disagree... I can only refer you
back to the dia. as this is the crux of the
argument.

A (picture is worth a) thousand words of wrong is still wrong.
There is an equal portion of mass in every direction. Space is
flat in every *direction*. The *only* gradient is in the time
axis.

Plugging in the values for observed values
of Z, say Z = 1 thru 5 (I believe up to 7
has been observed)

CMBR, z = 1024 (give or take)

Correct again but the CMBR is not exclusive
to just the tiny volume within any given visible
horizon (when compared to the entire universe).
That is different horizons don't have different
CMBRs. The CMBR extends to the entire
universe even if we cannot see those parts.
Thats why its Z is so high. But the common
temp at 2.8 deg was established for all horizons
at the inflationary ephoc.

Be careful... the CMBR was 9+ K a billion years ago, and other
hotter temperatures right back to when the plasma that emitted
the CMBR quenched.

The fact that we only see QSOs out to about
Z=7 and not Z= 283 attests to this. Yeah and
I know... A Z of 283 would be too distant to be
seen anyway, so it's a poor example.

This is "just" a technology problem. Whether or not structures
existed at Z=283 would be very interesting to see. What would a
star look and act like in a Universe that required it to lose
heat into a background of "2000K"?

David A. Smith

"Androcles" wrote in message
...


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This message is brought to you by Androcles
http://www.androcles01.pwp.blueyonder.co.uk/

"WG" wrote in message
...

"Androcles" wrote in message
...

| sadly I have seen your inane posts before!!
| I can only suggest you stay off the thread.

Wassup, don't like real data, prefer crackpot theories?
Sad cretin!

way to go , Anndro

Yup, they need good solid fibre filled stuff like the crap you have on
your website. Aint no srgument with stuff that smells like your crap 'n
dat.