thought expt

"B Gilmour" wrote in message
...
There is a simple thought experiment posted at the following, which
suggests
that there may be a gravitational component to cosmological redshift which
only becomes significant at extreme QSO distances.
Ideas as to why it is incorrect would be appreciated.

http://home.i-zoom.net/~wgilmour/Redshift.html

Interesting. It seems to me though that there is a flaw; if, wherever a
photon is created, it sees itself as in the centre of a gravitational well,
then wherever it moves to it will continue to see itself in the centre of
the gravitational well since, wherever it moves to, it is still at the
centre of the universe (from its own POV). Therefore it never climbs out of
the well, so this won't cause a redshift.

Ian

"B Gilmour" wrote in message
...
There is a simple thought experiment posted at the following, which
suggests
that there may be a gravitational component to cosmological redshift which
only becomes significant at extreme QSO distances.
Ideas as to why it is incorrect would be appreciated.

http://home.i-zoom.net/~wgilmour/Redshift.html


Same answer I gave you in sci.physics:

Every point in the universe is the center. You
cannot change average gravitational potential
simply by moving.

On the other hand, since the universe is expanding,
we do expect a red shift to occur as a light ray
propagates over large distances.

There is a simple thought experiment posted at the following, which suggests
that there may be a gravitational component to cosmological redshift which
only becomes significant at extreme QSO distances.
Ideas as to why it is incorrect would be appreciated.

http://home.i-zoom.net/~wgilmour/Redshift.html

"Jaxtraw" wrote in message
...
"B Gilmour" wrote in message
...
There is a simple thought experiment posted at the following, which
suggests
that there may be a gravitational component to cosmological redshift
which
only becomes significant at extreme QSO distances.
Ideas as to why it is incorrect would be appreciated.
http://home.i-zoom.net/~wgilmour/Redshift.html

Interesting. It seems to me though that there is a flaw; if, wherever a
photon is created, it sees itself as in the centre of a gravitational
well,
then wherever it moves to it will continue to see itself in the centre of
the gravitational well since, wherever it moves to, it is still at the
centre of the universe (from its own POV). Therefore it never climbs out
of
the well, so this won't cause a redshift.
Ian

Yes I have thought of this argument before and it's a good one.
However in the climbing out from the center of the earth analogy there
clearly is a redshift which is measurable! And I am attempting to argue that
the two situations are identical.

Another poster in another forum states

Boundary conditions. There's no boundary to the universe,
so there is no center from which to measure the distance

This is true, when we speak of the universe as a whole, but surrounding each
point within the universe there is a visible horizon which is determined by
the value of C. There is of course more universe beyond this boundary but
this can never be communicated with. I am suggesting that this horizon
imposes a boundary condition [artificial if you will, but there none the
less] which makes the analogy between the earth example and universe analogy
valid. I tried to emphasis that due to Gausses law, the value of R [the
radius of the earth or radius to the visible horizon] is always unknown and
unknowable, yet in the climbing from the center of the earth analogy we do
indeed get a gravitational redshift even though there would appear to be no
known boundary at this point or at least one that can never be known. [A
photon climbing from the center can never know where the surface is before
it gets there.]

traveled towards any boundary beyond which there is no
mass.

And again I would suggest any point beyond the visible horizon would
represent this no mass area, since any mass there has yet to communicate its
presence.

Cheers

"B Gilmour" wrote in message
...

"Jaxtraw" wrote in message
...
"B Gilmour" wrote in message
...
There is a simple thought experiment posted at the following, which
suggests
that there may be a gravitational component to cosmological redshift
which
only becomes significant at extreme QSO distances.
Ideas as to why it is incorrect would be appreciated.
http://home.i-zoom.net/~wgilmour/Redshift.html

Interesting. It seems to me though that there is a flaw; if, wherever a
photon is created, it sees itself as in the centre of a gravitational
well,
then wherever it moves to it will continue to see itself in the centre
of
the gravitational well since, wherever it moves to, it is still at the
centre of the universe (from its own POV). Therefore it never climbs out
of
the well, so this won't cause a redshift.
Ian

Yes I have thought of this argument before and it's a good one.
However in the climbing out from the center of the earth analogy there
clearly is a redshift which is measurable! And I am attempting to argue
that
the two situations are identical.

Another poster in another forum states

Boundary conditions. There's no boundary to the universe,
so there is no center from which to measure the distance

This is true, when we speak of the universe as a whole, but surrounding
each
point within the universe there is a visible horizon which is determined
by
the value of C. There is of course more universe beyond this boundary but
this can never be communicated with. I am suggesting that this horizon
imposes a boundary condition [artificial if you will, but there none the
less] which makes the analogy between the earth example and universe
analogy
valid. I tried to emphasis that due to Gausses law, the value of R [the
radius of the earth or radius to the visible horizon] is always unknown
and
unknowable, yet in the climbing from the center of the earth analogy we do
indeed get a gravitational redshift even though there would appear to be
no
known boundary at this point or at least one that can never be known. [A
photon climbing from the center can never know where the surface is before
it gets there.]

traveled towards any boundary beyond which there is no
mass.

And again I would suggest any point beyond the visible horizon would
represent this no mass area, since any mass there has yet to communicate
its
presence.


I see your point; having thought about this though, I think there is one
difference. A photon climbing up from the centre of the earth is approaching
the boundary (earth's surface). In the universe, the boundary constantly
recedes ahead of the photon and approaches it from behind; i.e. the boundary
remains centred on the photon. In front of it, new parts of the universe are
coming into view; behind it part of the universe is disappearing. IOW the
boundary remains centred on the photon.

Umm. I think.

That bit about the behind boundary doesn't seem right, but I'm not sure why.
Dang, if only I knew what I'm talking about :'(

Ian

* "B Gilmour" schriebt:
There is a simple thought experiment posted at the following, which suggests
that there may be a gravitational component to cosmological redshift which
only becomes significant at extreme QSO distances.
Ideas as to why it is incorrect would be appreciated.

http://home.i-zoom.net/~wgilmour/Redshift.html

The idea of keeping a sphere of possible gravitational influence
fixed while a photon goes merrily on its way relative to it, is not
valid. A photon going from A to B in a homogenous universe loses
energy relative to the sphere with center at point A. However, it
gains exactly the same amount relative to the sphere with center at
point B. And so it is for all points on the path. Every point A on
the path has a corresponding (cancelling) point B; 0 net effect.

--
A: Because it messes up the order in which people normally read text.
Q: Why is top-posting such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

"Alf P. Steinbach" wrote in message
...
* "B Gilmour" schriebt:
There is a simple thought experiment posted at the following, which
suggests
that there may be a gravitational component to cosmological redshift
which
only becomes significant at extreme QSO distances.
Ideas as to why it is incorrect would be appreciated.

http://home.i-zoom.net/~wgilmour/Redshift.html

The idea of keeping a sphere of possible gravitational influence
fixed while a photon goes merrily on its way relative to it, is not
valid. A photon going from A to B in a homogenous universe loses
energy relative to the sphere with center at point A. However, it
gains exactly the same amount relative to the sphere with center at
point B. And so it is for all points on the path. Every point A on
the path has a corresponding (cancelling) point B; 0 net effect.

This would certainly be true for a "static" homogeneous universe,
but in an expanding universe density decreases with respect to time.
and I would argue that since the value of C is not infinite, photons travel
from past to future from an observers frame of reference (certainly not in
there own). So in reality when a photon is created at your point A the
density is P1, and then it is observed at point B where density is P2, and
P2P1. Whenever a photon travels from a dense to a less dense medium you get
a gravitational redshift proportional to the density difference P1 & P2.
This leads to the same conclusion which I have tried to point out at
http://home.i-zoom.net/~wgilmour/Den...ifference.html

"B Gilmour" wrote in message
...
There is a simple thought experiment posted at the following, which
suggests
that there may be a gravitational component to cosmological redshift which
only becomes significant at extreme QSO distances.
Ideas as to why it is incorrect would be appreciated.

http://home.i-zoom.net/~wgilmour/Redshift.html

For those of you who have read the page.

1] would you agree that a photon will be redshifted upon traveling from the
center ot the earth.
[of course it will be yes]
2] would you agree we should be able to vary both variables to any value we
want and redshift should still be observed. [variables being, -density of
medium, and radius]
[ the answer should be yes]
3] would you maintain that the redshift somehow disappears when these
two variables reach Pc (critical density) and the radius of the visible
horizon, but would still be there just prior to these values.

"B Gilmour" wrote in message
...

"Alf P. Steinbach" wrote in message
...

The idea of keeping a sphere of possible gravitational influence
fixed while a photon goes merrily on its way relative to it, is not
valid. A photon going from A to B in a homogenous universe loses
energy relative to the sphere with center at point A. However, it
gains exactly the same amount relative to the sphere with center at
point B. And so it is for all points on the path. Every point A on
the path has a corresponding (cancelling) point B; 0 net effect.

This would certainly be true for a "static" homogeneous universe,
but in an expanding universe density decreases with respect to time.

And the relative potential between two locations will not
change, as the density decreases uniformly. The effect
becomes indistinguishible from the cosmological redshift
due to expansion.

"B Gilmour" wrote in message
...
There is a simple thought experiment posted at the following, which
suggests
that there may be a gravitational component to cosmological redshift which
only becomes significant at extreme QSO distances.
Ideas as to why it is incorrect would be appreciated.

http://home.i-zoom.net/~wgilmour/Redshift.html


I think there's a problem because in your first experiment (man within the
earth) you are clearly moving away from a special position and creating an
asymmetry (more local mass below you than above you).

When you change this to the second experiment (photon within the earth), you
have this same special position to start with, and the same resulting
asymmetry. In both of these cases, you are correct that there would be a
measurable effect.

However, when you try and generalise it to the case of any photon anywhere
in the universe, then there is no special starting position and no asymmetry
can be developed that would lead to a generic red shift.

Do you agree?
Owen