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Gravitational redshift
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wgilmour@i-zoom.net
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Joined: 22 Nov 2005
Posts: 12

PostPosted: Mon Jul 17, 2006 3:44 pm    Post subject: Gravitational redshift Reply with quote

This is an interesting speculation regarding a thought experiment and
its subsequent gravitational redshift consequences.

Lets say you are standing at the center of the earth down a long hollow
shaft and send a beam of light back up the shaft towards the surface.
This beam of light will be gravitationally redshifted according to

Zg = 4.19 G d r^2/c^2 (approx equation)

Where;
G= gravitational constant, d=density of earth, r= radius traveled, c=
speed of light.

This is the standard gravitational potential well given in most texts.

It should be noted that Gausses theorem comes into play here, in that a
consequence of the theorem shows that the mass (i.e. density) above
radius r does not come into play in the calculation.
It could be zero mass or extremely large mass outside radius r, it
doesn't care, all effects cancel.

Now for a given r, there is only one variable, density (d).
This means that we can vary d to any extreme, either higher or lower,
and we just observe either a higher or lower redshift.
Redshift can only approach zero when d approaches zero.

So far pretty standard.
Now here is the interesting part.
Lets vary d (density) to say 10^-28 gms/cm^3.
Do we still get a redshift?
Yes of course we do, very small for sure but there nonetheless.
But what have we done here? [some may see it already].
10^-28gms/cm^3 is the density of the universe!
The background density. [i.e. Critical density].

What we have done here is, instead of the photon climbing out of a
potential well of the earth density, it is climbing out of a potential
well of the universe density, which is the same density in which the
sphere is imbedded [remember density inside r can be the same as
outside r, but Gausses theorem makes the outside part irrelevant
anyway].

Remember this argument may be constructed at any point within the
universe and should hold true.

We seem to have added an additional Gravitational component to observed
redshift. (with consequences for QSOs)

Thus Zo = Zd + Zg
Where Zo = observed redshift, Zd = cosmological redshift (normally
explained as a dopler shift due expansion), and Zg = gravitational
redshift due to the mass of the universe.

While there is no absolute center to the universe, from an
observational point of view every point in the universe is the center,
and when it comes to measurements this is all that matters.
One could argue that every point in the universe is at the center of
its own visible horizon, and a photon upon being created finds itself
at the center of a gravitational potential well consisting of the mass
of the universe.


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WG
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Igor Khavkine
science forum Guru


Joined: 01 May 2005
Posts: 607

PostPosted: Mon Jul 17, 2006 11:30 pm    Post subject: Re: Gravitational redshift Reply with quote

WG wrote:

Quote:
Thus Zo = Zd + Zg
Where Zo = observed redshift, Zd = cosmological redshift (normally
explained as a dopler shift due expansion), and Zg = gravitational
redshift due to the mass of the universe.

If by

Quote:
10^-28gms/cm^3 is the density of the universe!
The background density. [i.e. Critical density].

you mean the "dark energy" density attributed to a cosmological
constant, then what you've done is a back-of-the-envelope calculate to
estimate its impact on the redshift. It is possible to do a much more
accurate calculation from cosmological solutions of Einstein's
Equations (the equations of general relarivity). Once such a solution
is obtained from a model (such as the Friedman-Robertson-Walker model
with a non-zero cosmological constant), the correct way to obtain the
redshift is the find the so-called "null geodesics". An analysis of the
null geodesics will give a redshift that automatically takes all the
cosmological contributions present in the model into account. The
actual calculation of the redshift sometimes can be done exactly,
sometimes with an analytical approximation (as you did), or even
numerically.

The important point is that there is a systematic way of taking all
known contributions to the redshift into account. And it is used
routinely in any serious analysis of, say, the CMB data.

Hope this helps.

Igor
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wgilmour@i-zoom.net
science forum beginner


Joined: 22 Nov 2005
Posts: 12

PostPosted: Mon Jul 17, 2006 11:30 pm    Post subject: Re: Gravitational redshift Reply with quote

A more succinct way to state this problem would be;

If you take a sphere of density d and imbed it in a background density
of d1, where d=d1, would you still get a redshift?
[i.e. does redshift approach 0 when d approaches d1, or d approaches 0
(it has to be 0)]

The answer has to be yes, you still get a redshift for the following 2
reasons.

1. From Zg = 4.19 G d r^2/c^2
For a given r, Zg can only approach 0 when d approaches 0.

2. From Gausses theorem, the matter [i.e. density] outside r
does not come in to play, the theorem makes it irrelevant,
it could be 0,
extremely large or anything in between including d.


WG wrote:
Quote:
This is an interesting speculation regarding a thought experiment and
its subsequent gravitational redshift consequences.

Lets say you are standing at the center of the earth down a long hollow
shaft and send a beam of light back up the shaft towards the surface.
This beam of light will be gravitationally redshifted according to

Zg = 4.19 G d r^2/c^2 (approx equation)

Where;
G= gravitational constant, d=density of earth, r= radius traveled, c=
speed of light.

This is the standard gravitational potential well given in most texts.

It should be noted that Gausses theorem comes into play here, in that a
consequence of the theorem shows that the mass (i.e. density) above
radius r does not come into play in the calculation.
It could be zero mass or extremely large mass outside radius r, it
doesn't care, all effects cancel.

Now for a given r, there is only one variable, density (d).
This means that we can vary d to any extreme, either higher or lower,
and we just observe either a higher or lower redshift.
Redshift can only approach zero when d approaches zero.

So far pretty standard.
Now here is the interesting part.
Lets vary d (density) to say 10^-28 gms/cm^3.
Do we still get a redshift?
Yes of course we do, very small for sure but there nonetheless.
But what have we done here? [some may see it already].
10^-28gms/cm^3 is the density of the universe!
The background density. [i.e. Critical density].

What we have done here is, instead of the photon climbing out of a
potential well of the earth density, it is climbing out of a potential
well of the universe density, which is the same density in which the
sphere is imbedded [remember density inside r can be the same as
outside r, but Gausses theorem makes the outside part irrelevant
anyway].

Remember this argument may be constructed at any point within the
universe and should hold true.

We seem to have added an additional Gravitational component to observed
redshift. (with consequences for QSOs)

Thus Zo = Zd + Zg
Where Zo = observed redshift, Zd = cosmological redshift (normally
explained as a dopler shift due expansion), and Zg = gravitational
redshift due to the mass of the universe.

While there is no absolute center to the universe, from an
observational point of view every point in the universe is the center,
and when it comes to measurements this is all that matters.
One could argue that every point in the universe is at the center of
its own visible horizon, and a photon upon being created finds itself
at the center of a gravitational potential well consisting of the mass
of the universe.


Reply to group or private email.
WG
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Sue...
science forum Guru


Joined: 08 May 2005
Posts: 2684

PostPosted: Tue Jul 18, 2006 5:48 pm    Post subject: Re: Gravitational redshift Reply with quote

WG wrote:

<< This is the standard gravitational potential well given in most
texts.>>

Gravitational potential (greatest at the planet's surface) is not
the same as gravitational potential *energy*. 'Most texts' are
far too cavalier in this distinction.

Equation 14 is an unambiguous expression in
this reagard:
http://xxx.lanl.gov/abs/gr-qc/9606079

Cosmology aside, Okun explains why most expressions
in many texts, based on potential energy have to be
dismissed as violations of causality:
http://arxiv.org/abs/physics/9907017

The importance of the distinction can also be
seen, comparing the titles of Pound-Rebka and
Pound-Snider experiments. 'apparent weight of photon'
is changed to 'effect on nuclear resonance' in the
later paper.

Sue...
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Thomas.Palm
science forum beginner


Joined: 10 Dec 2005
Posts: 3

PostPosted: Tue Jul 18, 2006 5:48 pm    Post subject: Re: Gravitational redshift Reply with quote

WG <wgilmour@i-zoom.net> wrote in news:1153180335.343568.262620
@i42g2000cwa.googlegroups.com:

Quote:
A more succinct way to state this problem would be;

If you take a sphere of density d and imbed it in a background density
of d1, where d=d1, would you still get a redshift?
[i.e. does redshift approach 0 when d approaches d1, or d approaches 0
(it has to be 0)]

The answer has to be yes, you still get a redshift for the following 2
reasons.

1. From Zg = 4.19 G d r^2/c^2
For a given r, Zg can only approach 0 when d approaches 0.

2. From Gausses theorem, the matter [i.e. density] outside r
does not come in to play, the theorem makes it irrelevant,
it could be 0,
extremely large or anything in between including d.

To make things a bit more tricky: if you get a redshift sending light away
from the center of that sphere, surely you must get a blueshift if you send
it the other way, towards the center. But if the density is in reality
constant you can create spheres with any center you wish so that light that
travels away from the center of one travels toward the center of a
displaced sphere. Will it be red- or blueshifted?
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wgilmour@i-zoom.net
science forum beginner


Joined: 22 Nov 2005
Posts: 12

PostPosted: Wed Jul 19, 2006 6:35 am    Post subject: Re: Gravitational redshift Reply with quote

Thomas Palm wrote:
Quote:
WG <wgilmour@i-zoom.net> wrote in news:1153180335.343568.262620
@i42g2000cwa.googlegroups.com:

A more succinct way to state this problem would be;

If you take a sphere of density d and imbed it in a background density
of d1, where d=d1, would you still get a redshift?
[i.e. does redshift approach 0 when d approaches d1, or d approaches 0
(it has to be 0)]

The answer has to be yes, you still get a redshift for the following 2
reasons.

1. From Zg = 4.19 G d r^2/c^2
For a given r, Zg can only approach 0 when d approaches 0.

2. From Gausses theorem, the matter [i.e. density] outside r
does not come in to play, the theorem makes it irrelevant,
it could be 0,
extremely large or anything in between including d.

To make things a bit more tricky: if you get a redshift sending light away
from the center of that sphere, surely you must get a blueshift if you send
it the other way, towards the center. But if the density is in reality
constant you can create spheres with any center you wish so that light that
travels away from the center of one travels toward the center of a
displaced sphere. Will it be red- or blueshifted?


Yes, I had previously thought of this argument and it is a good one.
The only thing I can suggest is that the symmetry of the argument may
be broken here and the 2 cases different..

In order to get a blueshift you need an observer sitting in the middle
of his visible horizon [i.e. potential well] and the photon has to be
created somewhere closer to the edge of this well and travel inward. In
this case the photon would clearly feel that there is more mass in the
direction of the observer, and thus blueshift occurs. In the photons
reference frame however there is not more mass in one direction since
it sits at the middle of its own visible horizon. [read well], also the
gravitational forces act on the moving photon not the observer so this
should be the only consideration.

This may be unsatisfactory but it's the best I can do for now.
In any case this still does not resolve my initial argument.

This idea does suggest one other point.
Why is the value of C what it is?
What if the minimum kinetic energy of a photon be exactly equal to the
minimum required to raise it out of the potential well of the universe?
This is a Mach's principle type of argument which states that the
properties of matter [inertial] are a result of the gravitating mass of
the universe as a whole.
This sets the value of C according to the value of the mass of the
universe out to the visible horizon.

For the complete argument please visit;
http://home.i-zoom.net/~wgilmour/Machs%20Principle.html
It also suggests a VLS theory.
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