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Gravitational Redshift as Measured by the Pound-Rebka-Snider Experiment

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Murat Ozer

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May 23, 2012, 4:41:42 PM5/23/12
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Hello,
I would like you to help me clarify an issue in the Pound ---Rebka-
Snider Experiment verifying the Gravitational Redshift (GR).
According to GR, photons traveling from a higher field region to
a lower field region (equivalently from the bottom to the top of a
tower) get redshifted, namely their frequency decreases (or their
wavelength increases). Thus, when the Mossbauer source is at the
bottom of the tower and the absorber is at the top, the frequencies
are related by
f_T = f_B(1 - gH/c^2), (1)

However, if we look at the details of the experiment this is what
happens: First, photons (gamma rays) are emitted with frequency
f_B by the source nuclei at the bottom of the tower (the energy of
these photons is 14.4 keV). According to the Mossbauer effect,
these photons cannot be absorbed by the absorber nuclei at the top
because their energy (or frequency) does not match the absorbtion
energy (or frequency). The absorbtion energy can be found as
follows: The excited and ground state energy levels of the nuclei
at the bottom have the energies

E*_B = Mc^2 + E_ph = (M + E_ph/c^2)c^2 = M*c^2, (2)
E_B = Mc^2, (3)

where E_ph = E*_B --- E_B is the emitted photon (gamma ray)
energy (14.4 keV) and M* = M + E_ph/c^2 and M are the masses
of the nuclei in the excited and ground states, respectively.
Also, the potential energy of the nuclei has been taken as zero
at the bottom of the tower. The excited and ground state energy
levels of the nuclei at the absorber have the energies

E*_T = M*c^2 + M*gH, (4)
E_T = Mc^2 + MgH, (5)

where g is the gravitational acceleration (strength of the field)
and H is the height difference between the top and bottom of
the tower.The deexcitation energy (energy of the photons emitted
as the absorber nuclei fall to the ground state from the excited
state) is

E*_ph = E*_T --- E_T = E_ph + (E_ph/c^2)gH (6)

Thus, the photons emitted by the source at the bottom with
energy E_ph need to have their energy increased to E*_ph
so that they can be absorbed by the absorber nuclei. The
energy shift dE, where

dE = E*_ph --- E_ph = (E_ph/c^2)gH = (E_ph/c)gH/c, (7)

is provided by the Doppler effect by moving the source up and
down with velocity
dv = gH/c (8)
My question is this: As is seen from Eq.(6), the frequency
w_T of the photons emitted by the absorber nuclei during
deexcitation is related to the frequency w_B = f_B as

w_T = w_B(1 + gH/c^2).(9)

This is the frequency _measured_ in the experiment and
it differs from the frequency _predicted_ by the theory.

How is the measured frequency reconciled with the predicted
one despite the sign difference between equations (1) and (9) ?

If the experiment, as described above, is repeated by putting the
source at the top and the absorber at the bottom, we still get a
sign difference between the predicted and measured frequencies.
As far as I understand, there is no mention of this rather
important point in the Pound-Rebka and Pound-Snider papers.

Thanks,
Murat Ozer

Tom Roberts

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May 24, 2012, 5:09:23 PM5/24/12
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On 5/23/12 5/23/12 3:41 PM, Murat Ozer wrote:
> Thus, when the Mossbauer source is at the
> bottom of the tower and the absorber is at the top, the frequencies
> are related by
> f_T = f_B(1 - gH/c^2), (1)
> [...]
> w_T = w_B(1 + gH/c^2).(9)
> This is the frequency _measured_ in the experiment and
> it differs from the frequency _predicted_ by the theory.
>
> How is the measured frequency reconciled with the predicted
> one despite the sign difference between equations (1) and (9) ?

Somewhere you made a sign error. There is no doubt that the frequency
measured at the top is lower (smaller) than that measured at the bottom.
We know this because the detector at the top must be moving upward, away
from the source, to maximize detection of the gammas; the Doppler shift
from such motion reduces the measured frequency of the gammas.

Tom Roberts

Murat Ozer

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May 25, 2012, 9:13:39 AM5/25/12
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On Friday, May 25, 2012 12:09:23 AM UTC+3, Tom Roberts wrote:
> On 5/23/12 5/23/12 3:41 PM, Murat Ozer wrote:
> > Thus, when the Mossbauer source is at the
> > bottom of the tower and the absorber is at the top, the frequencies
> > are related by
> > f_T = f_B(1 - gH/c^2), (1)
> > [...]
> > w_T = w_B(1 + gH/c^2).(9)
> > This is the frequency _measured_ in the experiment and
> > it differs from the frequency _predicted_ by the theory.
> >
> > How is the measured frequency reconciled with the predicted
> > one despite the sign difference between equations (1) and (9) ?
>
> Somewhere you made a sign error.

No, I did not make a sign error. The algebra is very simple.
Please show me the error if there is any.


>There is no doubt that the frequency
> measured at the top is lower (smaller) than that measured at the bottom.
> We know this because the detector at the top must be moving upward, away
> from the source, to maximize detection of the gammas; the Doppler shift
> from such motion reduces the measured frequency of the gammas.

The detector at the top is not moved at all. It's the source that is
moved.

Murat Ozer


>
> Tom Roberts


Anon E. Mouse

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May 25, 2012, 5:40:03 PM5/25/12
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Quoting wikipedia:

http://en.wikipedia.org/wiki/Pound%E2%80%93Rebka_experiment

"When an atom transits from an excited state to a base state, it emits
a photon with a specific frequency and energy. When the same atom in
its base state encounters a photon with that same frequency and
energy, it will absorb that photon and transit to the excited state."

Photon's have a negative momenta, they transfer excitation to the
absorber. This is the sign inversion you encountered in your
reasoning. The transposition of the ground and excited states.

Sincerely,

AAG

Rich L.

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May 27, 2012, 10:17:31 AM5/27/12
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Actually you got this backwards Tom, a rare event for you! The detector at the top is moving DOWNWARDS so that the gammas coming up from below are blue shifted again back to the absorbtion frequency of the upper detector. The key idea to realize is that this experiment does not measure the frequency of the gammas coming from below, but rather shifts the gamma frequency (via Doppler shift) to match that of the upper detector. The required velocity of the detector downward gives the red shift of the gammas coming up through the gravitational field.

Rich L.

Jonathan Thornburg [remove -animal to reply]

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May 28, 2012, 5:01:40 PM5/28/12
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Rich L. <ralivi...@sbcglobal.net> wrote:
> The detector at the top is moving DOWNWARDS so that the gammas
> coming up from below are blue shifted again back to the absorbtion
> frequency of the upper detector. The key idea to realize is that
> this experiment does not measure the frequency of the gammas coming
> from below, but rather shifts the gamma frequency (via Doppler
> shift) to match that of the upper detector. The required velocity
> of the detector downward gives the red shift of the gammas coming
> up through the gravitational field.

Actually, things were a bit more complicated than anyone has mentioned
in this thread so far.

Let's take a look at the actual Pound-Rebka paper:
Pound and Rebka
Physical Review Letters 4, 337 (1960)
http://prl.aps.org/abstract/PRL/v4/i7/p337_1

First, we see that the APS *still* keeps this paper behind a paywall,
52 years after it was published! :(

Beyond that, we see that to better control systematic errors, Pound &
Rebka actually alternated their experiment between two configurations:
(a) moving source at the top of the tower, sending gammas *down* to
stationary absorber (detector) at the bottom of the tower (this is
the configuration shown in figure 1 of their paper)
(b) moving source at the bottom of the tower, sending gammas *up* to
stationary absorber (detector) at the top of the tower

That is, (a) and (b) differ by swapping the two ends of the apparatus,
or equivalently, by flipping the sign of the gravitational force. In
all cases, Pound & Rebka moved the *source* (allowing the doppler shift
to compensate for the gravitational redshift/blueshift), and kept the
*absorber* (*detector*) stationary.

There's also a followup paper by Pound & Snider [I don't know if "Snider"
is the correct spelling, or a typo for "Snyder"] making more accurate
measurements with the same setup:

Pound & Snider
Physical Review Letters 13, 539 (1964)
http://prl.aps.org/abstract/PRL/v13/i18/p539_1

ciao,

--
-- "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
"Washing one's hands of the conflict between the powerful and the
powerless means to side with the powerful, not to be neutral."
-- quote by Freire / poster by Oxfam

Murat Ozer

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May 28, 2012, 9:25:00 PM5/28/12
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This just seems to be a sign convention.

>they transfer excitation to the absorber.

True, of course.

>This is the sign inversion you encountered in your
> reasoning. The transposition of the ground and excited states.

Yes, but why? Why is there a sign inversion between the deexcited
photons and the original incoming photons absorbed by the absorber?

Please expound on this.

Thanks,
Murat Ozer
>
> Sincerely,
>
> AAG

Murat Ozer

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May 28, 2012, 9:25:37 PM5/28/12
to
On Sunday, May 27, 2012 5:17:31 PM UTC+3, Rich L. wrote:
> On Thursday, May 24, 2012 4:09:23 PM UTC-5, Tom Roberts wrote:
> > On 5/23/12 5/23/12 3:41 PM, Murat Ozer wrote:
> > > Thus, when the Mossbauer source is at the
> > > bottom of the tower and the absorber is at the top, the frequencies
> > > are related by
> > > f_T = f_B(1 - gH/c^2), (1)
> > > [...]
> > > w_T = w_B(1 + gH/c^2).(9)
> > > This is the frequency _measured_ in the experiment and
> > > it differs from the frequency _predicted_ by the theory.
> > >
> > > How is the measured frequency reconciled with the predicted
> > > one despite the sign difference between equations (1) and (9) ?
> >
> > Somewhere you made a sign error. There is no doubt that the frequency
> > measured at the top is lower (smaller) than that measured at the bottom.
> > We know this because the detector at the top must be moving upward, away
> > from the source, to maximize detection of the gammas; the Doppler shift
> > from such motion reduces the measured frequency of the gammas.
> >
> > Tom Roberts
>
> Actually you got this backwards Tom, a rare event for you! The detector at >the top is moving DOWNWARDS so that the gammas coming up from below are blue >shifted again back to the absorbtion frequency of the upper detector.

The detector at the top is not moving. It's the source that's moving up and
down.

>The key idea to realize is that this experiment does not measure the frequency >of the gammas coming from below, but rather shifts the gamma frequency (via >Doppler shift) to match that of the upper detector.

The detector does not have a frequency. You probably mean the following:
According to the Einstein interpretation of the Gravitational Redshift
photons interact with the Gravitational Field as they travel upward against
the gravitational field and lose energy. Their frequency decreases as

f_T = f_B(1 - gH/c^2).

Since f_T is different from f_B fluorescence cannot be realized. Thus, the source is moved up and down so as to alter f_T to

f_T = f_B(1-gH/c^2)(1+ v/c) = f_B(1 - gH/c^2 + v/c) = f_B,

which yields v = gH/c. Pound, Rebka, and Snider measure this v in the
experiment and conclude from this that the fractional redshift is
- gH/c^2. However, Okun et.al. (Am.J.Phys,68(2000)115; arxiv,physics/
9907017) claim that the Einstein interpretation is wrong and the correct
interpretation is as given in my original post, namely photons do not
loose any energy as they climb up the gravitational field. They interact
with the absorber and the deexcitation frequency is given by

w_T = w_B(1 + gH/c^2) (blueshift)

But, they say, the incoming photons appear to be redshifted:

f_T = f_B(1 + gH/c^2).

this is the point that bothers me. How and why, according to what principle,
the incoming photons appear to be redshifted?

Thanks,
Murat Ozer

Ralph Hartley

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May 29, 2012, 4:26:15 PM5/29/12
to
On 05/28/2012 09:25 PM, Murat Ozer wrote:
> The detector does not have a frequency.

The detector used has a *very* narrow bandwidth. It can only absorb
photons with a particular frequency.

> You probably mean the following:
> According to the Einstein interpretation of the Gravitational Redshift
> photons interact with the Gravitational Field as they travel upward against
> the gravitational field and lose energy. Their frequency decreases as
>
> f_T = f_B(1 - gH/c^2).
>
> Since f_T is different from f_B fluorescence cannot be realized. Thus, the source is moved up and down so as to alter f_T to
>
> f_T = f_B(1-gH/c^2)(1+ v/c) = f_B(1 - gH/c^2 + v/c) = f_B,
>
> which yields v = gH/c. Pound, Rebka, and Snider measure this v in the
> experiment and conclude from this that the fractional redshift is
> - gH/c^2.

The photons are redshifted so the source needs to move upward (adding a
blueshift) to compensate.

> However, Okun et.al. (Am.J.Phys,68(2000)115; arxiv,physics/
> 9907017) claim that the Einstein interpretation is wrong and the correct
> interpretation is as given in my original post, namely photons do not
> loose any energy as they climb up the gravitational field. They interact
> with the absorber and the deexcitation frequency is given by
>
> w_T = w_B(1 + gH/c^2) (blueshift)

The frequency *required* for absorption is higher, so the source needs
to move upwards (adding a blueshift) to compensate.

Ralph Hartley

Anon E. Mouse

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May 30, 2012, 9:40:23 AM5/30/12
to

>
> > Photon's have a negative momenta,
>
> This just seems to be a sign convention.
>
> >they transfer excitation to the absorber.
>
> True, of course.
>
> >This is the sign inversion you encountered in your
> > reasoning. The transposition of the ground and excited states.
>
> Yes, but why? Why is there a sign inversion between the deexcited
> photons and the original incoming photons absorbed by the absorber?
>
> Please expound on this.
>
> Thanks,
> Murat Ozer

Initial - final = change
ground`- excited = positive moment
ground + photon = excited
photon = - momenta

If you accidentally transpose the initial and final states as you did
the sign error propagates creating inequalities instead of equalities.
Sign conventions have a real physical meaning and they keep everything
orderly as well.

AAG

news

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Sep 6, 2013, 5:59:05 AM9/6/13
to
Here's an exclamation that "works":
Let the free space frequency of the radiator = f0
When you put that radiator down in the gravity field,
it begins to radiate with reduced frequency
fB =f0*(1 - gH/c^2)
and then we find
f, as read at the top = fB a lower value
I'm saying the frequency does not change on the way up, rather that
it starts low due to gravitational red shift. The frequency is a
measure of energy, and it retains that energy on the way up,
where it compares 'low' compared to the upper filter.
It does not get tired on the way up but starts out with the initial
handicap of being generated in a gravitational field.
That equation is just an approximation for
sqrt(1-2Mg/rc^2)
John Polasek



On Mon, 28 May 2012 21:25:00 EDT, Murat Ozer <Murat....@gmail.com>
wrote:

J. J. Lodder

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Sep 6, 2013, 6:30:38 PM9/6/13
to
news <ne...@fx21.iad.highwinds-media.com> wrote:

> Here's an exclamation that "works":
> Let the free space frequency of the radiator = f0
> When you put that radiator down in the gravity field,
> it begins to radiate with reduced frequency
> fB =f0*(1 - gH/c^2)
> and then we find
> f, as read at the top = fB a lower value
> I'm saying the frequency does not change on the way up, rather that
> it starts low due to gravitational red shift. The frequency is a
> measure of energy, and it retains that energy on the way up,
> where it compares 'low' compared to the upper filter.
> It does not get tired on the way up but starts out with the initial
> handicap of being generated in a gravitational field.
> That equation is just an approximation for
> sqrt(1-2Mg/rc^2)
> John Polasek

A mistake.
In this quasi-Newtonian approximation
the photon's energy -does- change on the way up.
It's gravitational mass is \hbar\omega/c^2
so moving it up against a given g by dh
decreases it's energy by gdh\hbar\omega/c^2,
which is just the amount needed to give the red shift.

Quasi-Newtonically (that is, for small red shifts)
the gravitational redshift for free photons can be explained
as caused by work done against the gravitational potential.

OTOH, if you force the photon's frequency to remain the same
(by locking it in a massless perfectly reflecting box for example)
you can move it up at fixed energy and frequency.

But in that case -you- have to do the work to raise it,
and it is the work that you do on the photon that keeps
its frequency constant,

Jan

Ralph Hartley

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Sep 6, 2013, 6:31:02 PM9/6/13
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On 09/06/2013 05:59 AM, news wrote:
> Here's an exclamation that "works":
> Let the free space frequency of the radiator = f0
> When you put that radiator down in the gravity field,
> it begins to radiate with reduced frequency
> fB =f0*(1 - gH/c^2)
> and then we find
> f, as read at the top = fB a lower value
> I'm saying the frequency does not change on the way up, rather that
> it starts low due to gravitational red shift. The frequency is a
> measure of energy, and it retains that energy on the way up,
> where it compares 'low' compared to the upper filter.
> It does not get tired on the way up but starts out with the initial
> handicap of being generated in a gravitational field.

Well that's one way of looking at it, but not the only valid way. Since
frequency is frame dependent, there are many equivalent explanations.
Being equivalent, all are equally valid.

You can just as well say that the receivers frequency standard is
running fast due to gravitational blue shift, making the signal
frequency appear lower.

Or that the red shift occurs in transit. All are equivalent descriptions
of the same events.

Consider an inertial (free falling) reference frame, let the velocity of
the transmitter and receiver be 0 when the signal is sent. There is no
gravitational field in the frame, but the transmitter and receiver are
accelerating upwards. In this frame the explanation is a simple Doppler
effect. The receiver sees a red shift because it is moving when it
receives the signal, but the transmitter was stationary when it was
sent.

General relativity says that acceleration and gravitation are (locally)
equivalent. That's how we know that there is a gravitational red shift,
because it is required to agree with the free fall result.

Ralph Hartley

Tom Roberts

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Sep 8, 2013, 12:22:53 PM9/8/13
to
On 9/6/13 9/6/13 4:59 AM, news wrote:
> Here's an exclamation that "works":
> [...]

It may "work", but is probably not the best way to think about this. In
particular you have not related it to the very similar Doppler shift due
to relative motion between detector and source, or to the very similar
cosmological redshift.

GR models the gravitational redshift quite well, but in GR this is a
purely geometrical effect, and all types of Doppler shifts are seen to
be manifestations of orientation in 4-d spacetime. In particular, the
source, the detector, and the light traveling between them are all
unchanged (from their "free space" properties). But since the
orientations in spacetime of the source and detector can be different,
and the metric at their locations can be different, the detector can
measure a different frequency for the light than does the source (being
below implies blueshift, being above implies redshift, moving toward
implies blueshift, moving away implies redshift, expanding universe
implies redshift over cosmological distances, etc.).

In GR, the general way to compute the redshift/blueshift between source
and detector is:
A) Form the displacement 4-vector between successive wavecrests of the
source. This is necessarily parallel to the source's 4-velocity, and
is thus timelike. Its norm is clearly the time interval between
wavecrests that the source measures.
B) parallel transport that timelike 4-vector along the null geodesic
followed by the light ray to the detector.
C) compute the dot product of the parallel-transported 4-vector with the
4-velocity of the detector. The result is the time interval between
wavecrests that the detector measures.

In Minkowski spacetime this yields the same formula as SR for various
velocities of source and detector. It also applies to gravitational
redshift and cosmological redshift in GR -- they of course appear in the
metric used in the norm of (A), the parallel transport of (B), and the
dot product of (C). It also applies to combinations of these. It also
works in time-dependent situations such as the Shapiro time delay.

Note that in (B), parallel transport does not change the norm of the
4-vector being transported, so the light's intrinsic properties remain
unchanged (neither frequency nor wavelength are among those properties,
however). And as always, neither source nor detector properties depend
on their locations or motion. This is purely geometry.

Tom Roberts

news

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Sep 8, 2013, 12:23:14 PM9/8/13
to
On 06 Sep 2013 23:30:38 +0100 (BST), nos...@de-ster.demon.nl (J. J.
Lodder) wrote:

>news <ne...@fx21.iad.highwinds-media.com> wrote:
>
>> Here's an exclamation that "works":
>> Let the free space frequency of the radiator = f0
>> When you put that radiator down in the gravity field,
>> it begins to radiate with reduced frequency
>> fB =f0*(1 - gH/c^2)
>> and then we find
>> f, as read at the top = fB a lower value
>> I'm saying the frequency does not change on the way up, rather that
>> it starts low due to gravitational red shift. The frequency is a
>> measure of energy, and it retains that energy on the way up,
>> where it compares 'low' compared to the upper filter.
>> It does not get tired on the way up but starts out with the initial
>> handicap of being generated in a gravitational field.
>> That equation is just an approximation for
>> sqrt(1-2Mg/rc^2)
>> John Polasek
>
>A mistake.
>In this quasi-Newtonian approximation
>the photon's energy -does- change on the way up.
>It's gravitational mass is \hbar\omega/c^2
>so moving it up against a given g by dh
>decreases it's energy by gdh\hbar\omega/c^2,
>which is just the amount needed to give the red shift.
>
>Quasi-Newtonically (that is, for small red shifts)
>the gravitational redshift for free photons can be explained
>as caused by work done against the gravitational potential.
>
>OTOH, if you force the photon's frequency to remain the same
>(by locking it in a massless perfectly reflecting box for example)
>you can move it up at fixed energy and frequency.
>
>But in that case -you- have to do the work to raise it,
>and it is the work that you do on the photon that keeps
>its frequency constant,
>
>Jan
>
>> On Mon, 28 May 2012 21:25:00 EDT, Murat Ozer <Murat....@gmail.com>
>> wrote:
>>
SNIP

Let's start all over. You are talking about the putative gravitational
mass of a photon.this

I'm talking about an oscillator (as used in the actual experiment) that
has a frequency f0 in free space and that is brought down into a gravity
well, where I claim that its frequency is reduced already, just as given
by the relativistic formula for gravitational red shift. The oscillator
itself is slowed down.

You don't say how you got your photons down to the bottom from free
space, but according to your logic, they could just fall down into the
well and, by your theory, thereby acquire a blueshift, just as you claim
they lost energy climbing up the hill.

So, by your theory the photon would have to begin at normal frequency
down in the gravity well and lose frequency on the way up. Can you show
a way that the photon has normal frequency at the bottom of the gravity
well?

John Polasek

Tom Roberts

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Sep 8, 2013, 12:37:03 PM9/8/13
to
[[Mod. note -- I have manually rewrapped overly-long lines. -- jt]]

On 9/6/13 9/6/13 4:59 AM, news wrote:
> Here's an exclamation that "works":

news

unread,
Sep 8, 2013, 12:37:43 PM9/8/13
to
On 06 Sep 2013 23:30:38 +0100 (BST), nos...@de-ster.demon.nl (J. J.
Lodder) wrote:

What you described doesn't 'work'.
You have to consider all of the steps in the process.

You Just assumed that the free space frequency would not change when
the radiator was brought down into the gravity field, that it's still
f0, And only then could you make the case that it would climb up the
hill and lose energy and become red shifted.
But it's not right because in the first phase you must take it from
free space and let it fall in the gravity field to the bottom,
whereupon it would become blue shifted, by your photon logic.
Then it would arrive at the top with its original frequency due to the
red shift in its climb.
John Polasek

[[Mod. note -- 110 excessively-quoted lines snipped here. -- jt]]

J. J. Lodder

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Sep 10, 2013, 6:21:23 PM9/10/13
to
I assumed nothing of the kind, see above for what I actually wrote.

> But it's not right because in the first phase you must take it from
> free space and let it fall in the gravity field to the bottom,
> whereupon it would become blue shifted, by your photon logic.

See above again: Intergrating over dh, introducing \Phi
(the Newtonian poteltial) and expanding for \Phi/c^2 << 1
(Newtonian approximation) gives you that

E_photon + \Phi = constant = E_\infty.
(with suitably chosen integration constant)

> Then it would arrive at the top with its original frequency due to the
> red shift in its climb.

Yes, of course. It's conservative,

Jan

John Polasek

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Sep 12, 2013, 9:27:23 PM9/12/13
to
On Sunday, September 8, 2013 12:37:03 PM UTC-4, Tom Roberts wrote:
[[Mod. note --
1. The present article certainly comes very close to, and quite likely
goes beyond, the boundaries of our newsgroup charter's prohibition
against "excessively speculative" submissions.... but I'm allowing
it on the grounds that (I hope) it will spark interesting discussions.
2. Here I have snipped 53 lines of Tom Rober's very clear explanation
of how the Mossbauer effect is understood in the context of general
relativity; interesting readers can find this explanation in
Tom Roberts' posting.
-- jt]]

Not one iota of all this general relativity razzle-dazzle can be
measured or even realistically connected with the real world.

I have a different theory of gravity which for this case is easy
to understand: Atoms oscillate more slowly, and the speed of light
is similarly reduced because a gravity field represents a deprived
region. There is no time dilation. Therefore , in the well, L= c/f
retains its native value. The Mossbauer source at the bottom is
therefore "slugged" to the lower frequency and lower speed of light.

However, as it makes its way up to the detector, the speed of light
increases thereby stretching the 'native' wavelength on its way up,
while the frequency retains its lower value. This lower frequency
(and longer wavelength) at the top is recognized as red shift.

In the experiment, the Mossbauer radiator at the bottom was given
a sinusoidal excursion, with the highest radiation count occurring
at that part of the cycle when it is moving upward in order to
generate a blueshift. That blue-shifts the radiation to match the
narrow acceptance bandwidth of the Mossbauer Detector.

Notice something very unique here: I claim that down in the well
the frequency is lower but the wavelength is unchanged and that is
because the velocity of light is also lower there, therefore the
quotient L is constant. As wavelength moves up out of the well the
speed of light is increasing, thus stretching the wavelength, so
at the top we have both the normal frequency lower and the wavelength
longer. Any theory will have to reconcile all of these facets at
every stage in order to make everything "work". I believe every
other arrangement will fail to produce the red shift at top.

news

unread,
Sep 13, 2013, 5:05:32 AM9/13/13
to
On 10 Sep 2013 23:21:23 +0100 (BST), nos...@de-ster.demon.nl (J. J.
John P:
(I don't know what happens to my messages. Communicating over SPR
borders on spooky. The time delays are intolerable. I am resubmitting
this because it seems to have disappeared since September 10).

Conservative field makes it easy to explain, if we just make a round
trip with your photon. On a round-trip in a conservative field, the
integral will be zero.
In this case a photon starts with a frequency f and an energy E, up
top, and when you take it down to negative potential, its energy
increases as well as its frequency, according to your equation, so it
is blueshifted.
It will redshift on the way up, returning to its original value f and
E.
The experimental finding is that it should have a red shift up top.
but you get a null result instead.
I don't see any way out of that, except of course, the sequence I
described in which the radiating source itself is slowed down by
putting it in the gravity field. The frequency of the radiator will be
lower there, and this radiation will not change on the way up, so we
get a red shift, as required.
John Polasek

Jonathan Thornburg [remove -animal to reply]

unread,
Sep 13, 2013, 5:05:53 AM9/13/13
to
John Polasek <jack...@hotmail.com> wrote:
> Not one iota of all this general relativity razzle-dazzle can be
> measured or even realistically connected with the real world.

There are many situations (e.g., Shapiro time delay, studies of the
Earth-Moon orbital dynamics to millimeter precision, studies of binary
neutron star orbits) where general relativity's predictions
(a) deviate from those of Newtonian dynamics by *much* more than the
observational error bars, and
(b) are consistent with observations to well within those error bars.

If I understand you correctly, you're saying that you have a simpler
theory of gravity which agrees with all these observations, i.e.,
you're saying that your theory is observationally indistinguishable
from general relativity?

> I have a different theory of gravity which for this case is easy
> to understand: Atoms oscillate more slowly, and the speed of light
> is similarly reduced because a gravity field represents a deprived
> region. There is no time dilation. Therefore , in the well, L= c/f
> retains its native value. The Mossbauer source at the bottom is
> therefore "slugged" to the lower frequency and lower speed of light.

The (or a) key problem with these statements is that they lacks
specificity:

In the usual language of (post-1983) physics, the speed of light is a
defined constant (299 792 458 m/s). So... since you're saying the
speed of light varies, you must be using a different operational
definition of the speed of light. What is your definition? I.e., how
in your theory do we measure the speed of light (which you assert varies
depending on one's position in a gravitational field)?

Similarly, what's your operational definition of what it means to say
"Atoms oscillate more slowly"? More slowly with respect to *what*?
I.e., what's your definition of a frequency (or equivalently, time)
standard for this comparison?

Until we know all of these things, we can't even tell whether your
theory *is* different from general relativity, or whether it's simply
a different way to describe general relativity (cf. Schroedinger vs
Heisenberg descriptions of quantum mechanics).

We also can't tell whether your theory generalizes cleanly to stronger
gravitational fields (e.g., near the surface of a neutron star).

--
-- "Jonathan Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA
"There was of course no way of knowing whether you were being watched
at any given moment. How often, or on what system, the Thought Police
plugged in on any individual wire was guesswork. It was even conceivable
that they watched everybody all the time." -- George Orwell, "1984"

Phillip Helbig---undress to reply

unread,
Sep 13, 2013, 10:03:42 PM9/13/13
to
In article
<alpine.BSO.2.00.1...@cobalt.astro.indiana.edu>, "Jonathan
Thornburg [remove -animal to reply]" <jth...@astro.indiana-zebra.edu>
writes:

> In the usual language of (post-1983) physics, the speed of light is a
> defined constant (299 792 458 m/s). So... since you're saying the
> speed of light varies, you must be using a different operational
> definition of the speed of light. What is your definition? I.e., how
> in your theory do we measure the speed of light (which you assert varies
> depending on one's position in a gravitational field)?

While of course I agree that one needs some quantitative definition
describing concrete measurements, the "speed of light is constant, by
definition" argument is something of a red herring. People have
discussed theories with a variable speed of light. Physics didn't change
when the meter was re-defined in 1983. All that changed is that if one
wants to look for changes in the speed of light, one obviously can't use
the definition of a meter which is defined as the distance light travels
in a certain time. (I would say that rather than the speed of light
being defined as 299 792 458 m/s, rather the meter is defined as the
distance light travels in 1/(299 792 458) seconds, under the assumption
that the speed of light is constant. That doesn't mean that we can't
test this assumption, but it does mean that we need to avoid a circular
definition.)

news

unread,
Sep 13, 2013, 10:04:13 PM9/13/13
to
On 10 Sep 2013 23:21:23 +0100 (BST), nos...@de-ster.demon.nl (J. J.
Conservative field makes it easy to explain, if we just make a round
trip with your photon. On a round-trip in a conservative field,
integral will be zero.
In this case a photon has a frequency f and an energy E, up top, and

J. J. Lodder

unread,
Sep 13, 2013, 10:08:02 PM9/13/13
to
The simple explanation is that you are mistaken about it.
(by conservation of energy)
If the photon energy after a round trip through the field
were different from what it started out with
it would be possible in principle to build a perpetuum mobile.
(of the first kind)

This is obviuously absurd, so ...

Jan

John Polasek

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Sep 14, 2013, 2:08:06 AM9/14/13
to
On Friday, September 13, 2013 5:05:53 AM UTC-4, Jonathan Thornburg [remove -animal to reply] wrote:
> John Polasek <jack...@hotmail.com> wrote:
>
>> Not one iota of all this general relativity razzle-dazzle can be
>
>> measured or even realistically connected with the real world.
>
>
>
> There are many situations (e.g., Shapiro time delay, studies of the
>
> Earth-Moon orbital dynamics to millimeter precision, studies of binary
>
> neutron star orbits) where general relativity's predictions
>
> (a) deviate from those of Newtonian dynamics by *much* more than the
>
> observational error bars, and
>
> (b) are consistent with observations to well within those error bars.
>
>
>
> If I understand you correctly, you're saying that you have a simpler
>
> theory of gravity which agrees with all these observations, i.e.,
>
> you're saying that your theory is observationally indistinguishable
>
> from general relativity?
>
>
>
>> I have a different theory of gravity which for this case is easy
>
>> to understand: Atoms oscillate more slowly, and the speed of light
>
>> is similarly reduced because a gravity field represents a deprived
>
>> region. There is no time dilation. Therefore , in the well, L= c/f
>
>> retains its native value. The Mossbauer source at the bottom is
>
>> therefore "slugged" to the lower frequency and lower speed of light.

Jonathan, thank you for making a courteous request for details. Here's
what I can tell about my theory without handing you my book.

My theory is called dual space theory, the dual space being the source
for all the mass found in the universe, is taken from this "pair
space". Gravitation and electrical fields cannot exist in free empty
space, even though we write equations as if these forces could
exist. The subtractions of electrons from pair space to make our stars,
leaves positron star images in pairspace, and creates negative pressure
gradients around them for lack of support due to missing electrons.
There is a one-to-one correspondence between the electron objects we see
and the positron objects in pair-space. The gravity field is a lower
pressure "sink" surrounding the mass image, such that objects do not
"fall", but rather, succumb to the pressure gradient . My law of
gravity comes from Navier stokes:

del p = -rho*g p/rho = c^2/2
del = d/dr d/dr(c^2/2) = c*dc/dr = MG/r^2 = g
dc/dr = MG/r^2c Integrate to infinity for change in light speed:
Int dc/dr = MG/rc is change in lightspeed to infinity.

This works out to 634.78 m/s for the Sun from its surface, exactly as
determined by Breault in his experiment with the sodium D line. (See
description in "Gravitation" by MeisThoWheeler. The reduction at the
surface of the Earth is 0.208 m/s I think you can see what I derived
from scratch, deriving Brault's result easily. The basis of it all is
change of lightspeed in a gravitational field. Another bonus is that it
shows where the stars were created from, pair-space. Notice that it
also disposes of the problem of what happened to all the
anti-particles. That's easy, they are in one-to-one correspondence with
our particles. Furthermore, dc/dr = H0 in the Pioneer anomaly so
Ap = -H0*c. H0 = Ap/c =1/T T = 3.43e17s = 10.86Gyr.

It's all covered in my book which I wrote in 2004 but never published
"Dual Space Theory". I have discovered a whole lot more since then, so
some parts of the book are incorrect, but it is in revision.

J. J. Lodder

unread,
Sep 14, 2013, 11:49:22 AM9/14/13
to
John Polasek <jack...@hotmail.com> wrote:

> On Sunday, September 8, 2013 12:37:03 PM UTC-4, Tom Roberts wrote:
> [[Mod. note --
> 1. The present article certainly comes very close to, and quite likely
> goes beyond, the boundaries of our newsgroup charter's prohibition
> against "excessively speculative" submissions.... but I'm allowing
> it on the grounds that (I hope) it will spark interesting discussions.
> 2. Here I have snipped 53 lines of Tom Rober's very clear explanation
> of how the Mossbauer effect is understood in the context of general
> relativity; interesting readers can find this explanation in
> Tom Roberts' posting.
> -- jt]]
>
> Not one iota of all this general relativity razzle-dazzle can be
> measured or even realistically connected with the real world.

You might want to look at
<http://en.wikipedia.org/wiki/Tests_of_general_relativity>
for some real world connections.

> I have a different theory of gravity which for this case is easy
> to understand: Atoms oscillate more slowly, and the speed of light
> is similarly reduced because a gravity field represents a deprived
> region. There is no time dilation. Therefore , in the well, L= c/f
> retains its native value. The Mossbauer source at the bottom is
> therefore "slugged" to the lower frequency and lower speed of light.

Words. Do write down some equations.
Apart from that, you may notice that the gravitational red shift
does not depend on the full apparatus of general relativity.
It is a direct consequence of the equivalence principle.
(as noted by Einstein 1907)

So the obvious question is:
does this theory of yours conform to the equivalence principle?
If yes, you have said nothing that's worth noticing.
If no, you have a major problem dealing with
the experimental verifications of the equivalence principle.

> However, as it makes its way up to the detector, the speed of light
> increases thereby stretching the 'native' wavelength on its way up,
> while the frequency retains its lower value. This lower frequency
> (and longer wavelength) at the top is recognized as red shift.

Nothing but words again.
Do explain what length is, and what time is,
and how these are to be measured in a consistent way.

Finally, a word of advice.
You are talking here to an audience that knows what science is about.
Using value-laden terms such as 'general relativity razzle-dazzle'
may perhaps be good propaganda in a layman debate.
Here they decrease your credibility before you have even begun.

Just dont do it,

Jan

news

unread,
Sep 14, 2013, 11:49:43 AM9/14/13
to
On 10 Sep 2013 23:21:23 +0100 (BST), nos...@de-ster.demon.nl (J. J.
It's not enough just to push formulas around.
I am presenting a table of sequences and conditions that can act as a
test for any theory, which will reveal whether a theory is correct or
not. The following is my theory in action (RS=0.01) as the source
starts at the top, is taken to the bottom where it radiates, and
lastly we look at the photon as it arrives back at the top, red
shifted:
Top Down Back up
Fsrc 1.00 .99 ---
Fphotn 1.00 .99 remains .99 RS
L W/l 1.00 1.00 stretches 1.01 RS
c 1.00 .99 reverts to 1.00
In order to make this "work", we must assume that both the speed of
light and the frequency are reduced to 0.99.
It should be possible to enter numbers for any theory that would
explain the Pound Rebka experiment and reveal any inconsistencies.
John Polasek

John Polasek

news

unread,
Sep 17, 2013, 5:51:56 PM9/17/13
to
On Fri, 13 Sep 2013 22:08:02 EDT, nos...@de-ster.demon.nl (J. J.
Lodder) wrote:

snip
>The simple explanation is that you are mistaken about it.
>(by conservation of energy)
>If the photon energy after a round trip through the field
>were different from what it started out with
>it would be possible in principle to build a perpetuum mobile.
>(of the first kind)
>This is obviuously absurd, so ...
>
>Jan

You are reciting Einstein's argument as given in MTL "Gravitation"
pg 137: (z=.01). The logic is defective, predicting no red shift as
was found in the Pound-Rebka experiment.
He says
Let a mass m (E = 1.0) fall by a distance h so at the bottom he had
m--->m +gh (E = 1.01) extra energy,which he then converted to a photon
1.01 and let it radiate up.
It seemed logical that at the top he could now convert back from the
high energy photon (E = 1.01) to a mass m (E=1.0) plus some free
energy.
He had to reject the idea of free energy, so he argued there has to be
a red shift on the way up.
This gets rid of the spurious extra energy and gets him the red shift
he was shooting for on the way up.
But by starting with E = 1.0 and ending with E = 1.0, it would mean
that the Pound-Rebka experiment conducted in 1960, five years after
his death, would not show any redshift.
Everything seems to be in place, except that there is no possibility
of detectable red shift, since E1 = E2.

It's easy to concentrate on the up-leg and believe that a red shift
has occurred, while forgetting that its purpose is to neutralize a
previous blueshift. It appears to be impossible to reconcile the
observed red shift with a conservative field.

I think you will find that there is no other modification that can
make this "work".

It does work if you consider my theory that the radiator itself is
retarded when it is put down in a gravitational field as I described
before*. There the photon is red shifted at the bottom and stays red
shifted all the way to the top. Only the speed of the photon would be
affected by a gravitational field

Do you have any idea why I am identified as "news" instead of my
moniker John Polasek?

[Moderator's note: Posts to moderated newsgroups, however they
originate, ultimately get emailed to the moderation address. Your post
has the following RFC822 header:

From: "news" <ne...@fx22.iad.highwinds-media.com>

My guess is that this is set by whatever posting software you use. -P.H.]

John Polasek

*in the unstructured melange that is Google Groups

J. J. Lodder

unread,
Sep 17, 2013, 10:53:56 PM9/17/13
to
[[Mod. note -- I apologise for the long delay in processing this article,
which was originally submitted on 14.Sept.2011.
-- jt]]

news <ne...@fx07.iad.highwinds-media.com> wrote:

> On 10 Sep 2013 23:21:23 +0100 (BST), nos...@de-ster.demon.nl (J. J.
> Lodder) wrote:
[[Mod. note -- 73 excessively-quoted lines snipped here. -- jt]]
> Conservative field makes it easy to explain, if we just make a round
> trip with your photon. On a round-trip in a conservative field,
> integral will be zero.
> In this case a photon has a frequency f and an energy E, up top, and
> when you take it down to negative potential, its energy increases as
> well as its frequency, according to your equation, so it is
> blueshifted.
> It will redshift on the way up, returning to its original value f and
> E.
> The experimental finding is that it should have a red shift up top.
> but you get a null result instead.

Once again, give us some numbers.
By how much should the photon be red-shifted
after going down and back up? (according to your 'theory')

> I don't see any way out of that, except of course, the sequence I
> described in which the radiating source itself is slowed down by
> putting it in the gravity field. The frequency of the radiator will be
> lower there, and this radiation will not change on the way up, so we
> get a red shift, as required.

So conversely you should get a blue shift the other way round.
As it happens, that experiment has already been done,
in the lunar laser ranging experiments.
You shine a laser beam up at the moon,
and get photons back from the laser reflectors
left there by the Apollo crews.
As it happens the reflected intensity is very low,
one photon every few seconds, even with the most powerful lasers.

However, the reflected photons can be indentified
(among a much larger background of other light)
by passing them through a very narrow-band filter
at exactly the laser frequency.
So I guess your 'theory' stands falsified.

But -do- produce some equations and numbers,

Jan

Tom Roberts

unread,
Sep 17, 2013, 10:54:50 PM9/17/13
to
On 9/12/13 9/12/13 - 8:27 PM, John Polasek wrote:
> On Sunday, September 8, 2013 12:37:03 PM UTC-4, Tom Roberts wrote:
> [...]
> Not one iota of all this general relativity razzle-dazzle can be
> measured or even realistically connected with the real world.

It's not "razzle-dazzle", but is simply the way this is described and calculated
in the best theory of gravitation we have today. As you apparently don't
understand that, it seems unlikely you have anything of significance to
contribute [#]. Note also that GR _has_ been "connected with the real world" via
the usual method of science: being tested experimentally; the method I described
is among the best-tested aspects of GR.

You merely display your personal ignorance here.

[#] The history of physics shows no instance in which someone
has made an important contribution while being ignorant of
the then-current theories and experiments. Today for gravitation
that is GR and several dozen major experiments of which Pound-
Rebka-Snider is merely two.


> I have a different theory of gravity which for this case is easy
> to understand: Atoms oscillate more slowly, and the speed of light
> is similarly reduced because a gravity field represents a deprived
> region. [...]

Yes, I know; I read what you wrote earlier. But apparently you did not read what
I wrote: as I said, you have not related this to the very similar Doppler shift
due to relative motion between detector and source (which is essential to this
experiment), or to the very similar cosmological redshift. For those you must
postulate some other mechanism -- why do you think such similar processes have
different mechanisms?

In GR all these effects are related to the same phenomenon: differences of
orientation in spacetime.


> However, as it makes its way up to the detector, the speed of light
> increases thereby stretching the 'native' wavelength on its way up,
> while the frequency retains its lower value. This lower frequency
> (and longer wavelength) at the top is recognized as red shift.

Sure, I read what you wrote earlier. But this "increase in the speed of light"
is not observed experimentally [@]. As stated, your "theory" is refuted by
experiments. This might merely be in the way you stated it, but you seem to not
know the standard vocabulary of physics, or have much familiarity with the
relevant experimental record. There are MANY experiments in addition to
Pound-Rebka-Snider, and they ALL are relevant....

[@] Standards laboratories near sea level, and at rather
high altitudes (Boulder CO), all get the same value for
the vacuum speed of light, to significantly higher accuracy
than your claimed difference. The accuracy of the GPS
extends this to higher altitudes.


> Notice something very unique here: [...] I believe every
> other arrangement will fail to produce the red shift at top.

Your "theory" may account for Pound-Rebka-Snider QUALITATIVELY, but you have
given no QUANTITATIVE equations to apply it in other situations. I remind you
that physics is a quantitative science, and your verbal ramblings are woefully
insufficient.

And your "theory" is not "unique" at all: you are wrong in your last sentence --
for this case GR certainly does predict a redshift that is in excellent
agreement with measurements. (If GR did not get the right answer here, it would
have been abandoned long ago.)


Tom Roberts

John Polasek

unread,
Sep 17, 2013, 10:55:45 PM9/17/13
to
Jan: It appears you have not read my equations that were enclosed in my reply to Thornburg so I'll repeat them here

del p = -rho*g p/rho = c^2/2
del = d/dr d/dr(c^2/2) = c*dc/dr = MG/r^2 = g
dc/dr = MG/r^2c Integrate to infinity for change in light speed:
Int dc/dr = MG/rc is change in light speed to infinity.

dc/dr tells how the speed of light increases on its way up out of the well, which can be written as dc/dr = g/c. The speed change over the height h would be gh/c = 7e-7m/s
and the red shift would be given by
gh/c^2 = MGh/r^2c^2 = 2.45e-15.
My equations are real derivations from my dual space theory that agree with relativity numerically.
Notice the play-by-play derivation attributed to Einstein in MLT Gravitation on page 137, (assuming a red shift of 0.01).
"That a photon must be affected by a gravitational field -1911 etc. etc."
Briefly,
Let a particle of mass m fall by distance h, where it gains kinetic energy mgh at the bottom:
m---->m + mgh 1.0--->1.01 E = 1.01
where it annihilates into a photon of energy E of strength 1.01.
It is then explained how red shift on the way up would reduce E back to 1.00:
[which would result in no measurable red shift, since, starting at the top with m1.0 (=E1.0) and E1.01 at bottom, the final state is E1.0 again)].
"If the photon merely radiated back to the top without change, it could be converted by a suitable apparatus into another particle of rest mass m, plus an excess of energy mgh that costs nothing to produce. To avoid this contradiction Einstein saw that the photon must suffer a red shift. "
"The photon energy at top and bottom must be related by:
Ebot = Etop(1+gh/c^2)
Notice that this merely acknowledges a blueshift at the bottom, but to make it complete, he would need this equation:
Etop' = Etop
which his derivation seems to have proved.
He focuses on the work done on the way up, but in fact there are two steps for the photon, a blueshift followed by red shift, resulting in no measurable red shift, contradicting the experiment.
I'm sure he is making a mistake in trying to work with the photon alone without realizing that it is necessary to have a photon emitter and for it to suffer an impairment, as I described, when it is placed in a gravity well.
It looks like he's not predicting the red shift as measured by Pound. (This is morbid, but, did he not die in 1955, five years before Pound?).
Just sayin'.

John Polasek

news

unread,
Sep 19, 2013, 2:53:32 AM9/19/13
to
On Tue, 17 Sep 2013 22:53:56 EDT, nos...@de-ster.demon.nl (J. J.
Yes, they got a null off of the moon, from the round-trip of the
photons, but this *does not* apply to the Pound Rebka experiment in
which a red shift was detected.
In the PR experiment they took a piece of apparatus, a metal strip,
down to the basement.
>From there they let it radiate its Mossbauer frequency up to a similar
strip that acted as a sharp filter, nominally of the same identical
frequency by virtue of construction.
Redshift was detected only by virtue of the Doppler effect: The
emitter strip had to be moved upwardly to neutralize the red shift, to
maximize a Geiger counter output.

I repeat: according to my theory when the strip is brought down into
the gravity well, its resonant frequency f is reduced, as well as the
speed of light (by .01, say), both to 0.99, while their quotient,
which is the wavelength, remains at 1.00. This is because I designate
the gravitational field to be a deprived region, so both f and c
suffer.
The emitted photon, at f0.99, stays that way all the way up-the energy
and frequency is unaffected by the gravity field.
Wavelength at 1.00 expands to 1.01 at the top because the speed of
light changes from .99 to 1.00 on the way up.
There's a genuine red shift, frequency down and wavelength up. Thus
all the quantities, f, L, c are properly accounted for. I believe you
will find that this is the only way this experiment can "work".

It appears you have not read my equations that were enclosed in my
reply to Thornburg so I'll repeat them here I won't bother to try to
describe the rest of the theory.

My theory of gravity is based on the Navier Stokes equation:
del p = -rho*g

p/rho = c^2/2 (I have derived)
del = d/dr

d/dr(c^2/2) = c*dc/dr = MG/r^2 = g
c*dc/dr = MG/r^2
dc/dr = MG/r^2c = g/c
Increase in velocity of light in the tower height h is
Dc = gh/c = 7.194e-7m/s
Dc/c = 2.4e-15 the reported redshift
(The Doppler velocity Dc is produced by a sinusoidal amplitude of
loudspeaker of 1.8e-9 m or 35 Bohr radii at 10 Hz, (lowest possible
then??))

Integrate to infinity for change in light speed:
Int dc/dr = MG/rc = Dc is change in lightspeed to infinity.

This provides another validation:
For our Sun, the value Dc is 635 m/s, just as found by Brault using
the sodium D1 line, also described in "Gravitation"

My equations are real derivations from my dual space theory that
agree with relativity numerically.

John Polasek

John Polasek

unread,
Sep 23, 2013, 3:50:15 PM9/23/13
to
On Wednesday, May 23, 2012 4:41:42 PM UTC-4, Murat Ozer wrote:

> I would like you to help me clarify an issue in the Pound ---Rebka-
> Snider Experiment verifying the Gravitational Redshift (GR).

MO:
There is always the chance of misinterpretation if we do not watch our
algebra. The value of g = -9.8m/ss, and oftentimes it's simply taken as
a 9.8 m/ss which must have happened here. The frequency at the top is
lower. Also the algebraic potential U is a negative number (being the
work done from infinity with a force that's opposite to the motion).
John Polasek

[Moderator's note: WAY too much quoted text---which contained a blank
line between each quoted line---snipped and new text re-formatted to
have a sensible line length. -P.H.]

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