Pioneer : Anomaly Still Anonymous

"jpolasek" writes:

Craig Markwardt wrote:
John C. Polasek writes:

On 07 Jul 2006 01:26:00 -0400, Craig Markwardt
wrote:


John C. Polasek writes:
...
Well of course everything is done currently in the station and always
rechecked for conformance to 2.292Ghz. The model must likewise have
that same 2.292 imprimatur. And it must use it some way to produce the
total frequency fmod = f0 - vf0/c, of which the 2d term is
negligible. f0 is added to the converted velocity stream in the model
(f87 in my diagram).

But meanwhile the real clock frequencies have theoretically moved on
up by f0Ht with no one able to prove otherwise, right from clocks that
are currently certified as 2.292. NIST nor any one else can decide.
The declaration of 133Cs being 9,192,631,770 cycles to equal 1
second, with c *declared* as 299,792,458 m/s merely locks the
wavelength of the Cesium transition to 0.0326xxx m.
...

However, the Anderson et al paper checked for such a "drifting clock"
model, and it failed to fit the full data set. Therefore your
speculation is not substantiated.

CM
My premise is by no means disproved in arXiv:gr-qc/0104064. It has not
been properly tested there as I will show.

Eq. 60 purportedly seeks to find whether "drifting clocks" could
...

However, equation 60 of Anderson et al is not the only drifting clock
model considered. As George points out, equation 62 is essentially
your model, and is also inconsistent with the full data set. Thus,
your discussion of equation 60 is irrelevant.

You must have missed my note to George where I admitted it should be
Eq. 62 but your "Drifting Clocks" remark steered me to Eq. 60 .

At Eq. 62 the report says "This model adds constant frequency drift to
the reference S-band frequency. This model also fits Doppler well, but
again fits range poorly". That is an impossible conclusion.

This can be clarified by looking at the flowgraph on my website. It
seems hardly possible that a constant frequency drift added at
"f-adjust" downstream, could "fit range poorly" because range is
"upstream" at the left side of the diagram. In other words, the model
is self-sufficient at deriving velocity and range, but gets into
trouble when it tries to synthesize frequency in the right hand half.

What "seems hardly possible" to you is irrelevant. In fact, your flow
chart has nothing to do with the ranging observations to the other
spacecraft (note: the RANGE technique is different from DOPPLER).
Anderson et al's results have everything to do with the fact that *if*
the station frequency standards had been drifting, this would also
produce a temporal drift which would be inconsistent with ranging
data. The authors considered your hypothesis and can reject it based
on the observations.

[ And it's worth mentioning that if frequency standards were gradually
drifting linearly, then the accumulated phase error would build up
quadratically with time, and be detectable in the orbits of planets
and sattelites... but such a discrepancy is not detected. ]




Furthermore, as has been pointed out several times already, the
station clock is used as a reference frequency for both the
transmitter and the receiver. If frequencies were globally drifting
with time, a bias would be added at the transmit stage, but subtracted
at the receive/downconvert stage. Thus, your theory is erroneous.

The workmanship at the station is considered flawless.
But as I have already pointed out, the system does not have sufficient
resolving power to detect the anomaly in a normal round trip period. It
is only by tracking over the years that it became evident.
....

I note that you did not respond to the truly relevant criticism:
"If frequencies were globally drifting with time, a bias would be
added at the transmit stage, but subtracted at the
receive/downconvert stage."
Thus, the only "anomaly" that could be produced is during the time
between transmit and receive, i.e. the round trip light travel time.
And that is what the Anderson authors reject based on the full
doppler+ranging data set.

CM

On Sun, 9 Jul 2006 22:51:01 +0100, "George Dishman"
wrote:


"jpolasek" wrote in message
roups.com...

In keeping with your theory, ..

Not in keeping with any theory, in keeping with
the known method by which the hardware records
the data and the known method which Craig used
in the software

.. suppose that 1000 roundtrip tests had
been recorded over the 8 year period. How would you assemble these 1000
data sets into the linear graph shown? Did they each contain a 0.0015
Hz excess, ...

No. There would be 1000 points on the graph. The
points at the start of the graph would contain
almost no excess, those after 4 years would have
a 1.5Hz excess _each_ while those at the end of
the eight year period would _each_ have a 3Hz
excess.

Each excess is calculated by comparing the
received frequency with that transmitted less
than a day earlier and removing the Doppler due
to the speed of the craft etc. and the numbers
are the total for the two legs of the trip.

No, that's not how the anomaly is computed in gr-qc 0104064 . Look at
Eq. 15 pg. 18:
Anomaly = fobs(t) - fmodel(t) = - f0*2Ap*t/c
where t is the time since initialization, not the signal loop time.
See Figs. 6 & 7 covering 0-60 AU which is many years.
They do not, as you said, "compare the received frequency with that
transmitted less than a day earlier" but only with the model's version
of the frequency, which stays the same over the years.

As shown in Craig's paper, the noise on each
measurement is around 0.0042Hz, quite small
compared to the measurement of 3Hz at the end.

... and then by some logic, it was decided to box-car these
together till in the aggregate you get 1.5Hz? Not likely, but it seems
the only modus to support the arguments that you and George have been
advancing.

George

John Polasek

George Dishman wrote:
"jpolasek" wrote in message
oups.com...

In keeping with your theory, ..

Not in keeping with any theory, in keeping with
the known method by which the hardware records
the data and the known method which Craig used
in the software

.. suppose that 1000 roundtrip tests had
been recorded over the 8 year period. How would you assemble these 1000
data sets into the linear graph shown? Did they each contain a 0.0015
Hz excess, ...

No. There would be 1000 points on the graph. The
points at the start of the graph would contain
almost no excess, those after 4 years would have
a 1.5Hz excess _each_ while those at the end of
the eight year period would _each_ have a 3Hz
excess.

Each excess is calculated by comparing the
received frequency with that transmitted less
than a day earlier and removing the Doppler due
to the speed of the craft etc. and the numbers
are the total for the two legs of the trip.

?Received vs transmitted?
No, that's not how the anomaly is computed in gr-qc 0104064 . Look at
Eq. 15 pg. 18:
Anomaly = fobs(t) - fmodel(t) = - f0*2Ap*t/c
where t is the time since initialization, not the signal loop time. See
Figs. 6 & 7 covering 0-60 AU which is many years.

They clearly do not, as you said, "compare the received frequency with
that transmitted less than a day earlier" but only with the *model's*
version of the received frequency, which stays the same over the years.

As shown in Craig's paper, the noise on each
measurement is around 0.0042Hz, quite small
compared to the measurement of 3Hz at the end.

... and then by some logic, it was decided to box-car these
together till in the aggregate you get 1.5Hz? Not likely, but it seems
the only modus to support the arguments that you and George have been
advancing.

George

John Polasek

John C. Polasek wrote:
On Sun, 9 Jul 2006 22:51:01 +0100, "George Dishman"
wrote:


"jpolasek" wrote in message
roups.com...

In keeping with your theory, ..

Not in keeping with any theory, in keeping with
the known method by which the hardware records
the data and the known method which Craig used
in the software

.. suppose that 1000 roundtrip tests had
been recorded over the 8 year period. How would you assemble these 1000
data sets into the linear graph shown? Did they each contain a 0.0015
Hz excess, ...

No. There would be 1000 points on the graph. The
points at the start of the graph would contain
almost no excess, those after 4 years would have
a 1.5Hz excess _each_ while those at the end of
the eight year period would _each_ have a 3Hz
excess.

Each excess is calculated by comparing the
received frequency with that transmitted less
than a day earlier and removing the Doppler due
to the speed of the craft etc. and the numbers
are the total for the two legs of the trip.

No, that's not how the anomaly is computed in gr-qc 0104064 . Look at
Eq. 15 pg. 18:
Anomaly = fobs(t) - fmodel(t) = - f0*2Ap*t/c

Note that a_P already appears in that equation. The
words preceding the equation explain:

"The observed, two-way anomalous effect can be
expressed to first order in v/c as:"

In other words, after the value of a_P has been
extracted by the actual processing which I am
explaining, Eqn 15 is a way of explaining what
the anomaly would look like if the uplink were
at a fixed frequency. It is not a description of the
actual data processing method because the uplink
isn't fixed.

where t is the time since initialization, not the signal loop time.

They are saying that to first approximation the difference
between the measured doppler shift and the predicted
shift is a linear function of time since the start of the
analysis.

I am saying the actual shift is the difference between the
uplink and downlink frequencies. Both are true.

See Figs. 6 & 7 covering 0-60 AU which is many years.
They do not, as you said, "compare the received frequency with that
transmitted less than a day earlier"

Sorry John but that IS what is done for the analysis, there
is no alternative since the actual transmit frequency has
step changes for engineering purposes. Once the speed
has been determined from those two values, it can of
course be applied to a hypothetical exact signal such as
2.92GHz, which is the nature of Eqn 15, but that is not
part of the actual analysis.

After all that is said, the point remains, Eqn 15 says
that after 8 years the actual received frequency at any
time exceeds the expected value from the model by
3Hz (round trip) so each point on the graph is 3Hz,
not the tiny amount you mentioned.

George

John C. Polasek wrote:
On 09 Jul 2006 00:48:53 -0400, Craig Markwardt
wrote:


John C. Polasek writes:


On 07 Jul 2006 01:26:00 -0400, Craig Markwardt
wrote:


John C. Polasek writes:
...

Well of course everything is done currently in the station and always
rechecked for conformance to 2.292Ghz. The model must likewise have
that same 2.292 imprimatur. And it must use it some way to produce the
total frequency fmod = f0 - vf0/c, of which the 2d term is
negligible. f0 is added to the converted velocity stream in the model
(f87 in my diagram).

But meanwhile the real clock frequencies have theoretically moved on
up by f0Ht with no one able to prove otherwise, right from clocks that
are currently certified as 2.292. NIST nor any one else can decide.
The declaration of 133Cs being 9,192,631,770 cycles to equal 1
second, with c *declared* as 299,792,458 m/s merely locks the
wavelength of the Cesium transition to 0.0326xxx m.

...

However, the Anderson et al paper checked for such a "drifting clock"
model, and it failed to fit the full data set. Therefore your
speculation is not substantiated.

CM

My premise is by no means disproved in arXiv:gr-qc/0104064. It has not
been properly tested there as I will show.

Eq. 60 purportedly seeks to find whether "drifting clocks" could

..
However, equation 60 of Anderson et al is not the only drifting clock
model considered. As George points out, equation 62 is essentially
your model, and is also inconsistent with the full data set. Thus,
your discussion of equation 60 is irrelevant.


You must have missed my note to George where I admitted it should be
Eq. 62 but your "Drifting Clocks" remark steered me to Eq. 60 .

At Eq. 62 the report says "This model adds constant frequency drift to
the reference S-band frequency. This model also fits Doppler well, but
again fits range poorly". That is an impossible conclusion.

This can be clarified by looking at the flowgraph on my website. It
seems hardly possible that a constant frequency drift added at
"f-adjust" downstream, could "fit range poorly" because range is
"upstream" at the left side of the diagram. In other words, the model
is self-sufficient at deriving velocity and range, but gets into
trouble when it tries to synthesize frequency in the right hand half.


Furthermore, as has been pointed out several times already, the
station clock is used as a reference frequency for both the
transmitter and the receiver. If frequencies were globally drifting
with time, a bias would be added at the transmit stage, but subtracted
at the receive/downconvert stage. Thus, your theory is erroneous.


The workmanship at the station is considered flawless.
But as I have already pointed out, the system does not have sufficient
resolving power to detect the anomaly in a normal round trip period.
It is only by tracking over the years that it became evident.

In keeping with your theory, suppose that 1000 roundtrip tests had
been recorded over the 8 year period. How would you assemble these
1000 data sets into the linear graph shown? Did they each contain a
0.0015 Hz excess, and then by some logic, it was decided to box-car
these together till in the aggregate you get 1.5Hz? Not likely, but it
seems the only modus to support the arguments that you and George
have been advancing.

There is nothing phenomenological here. The anomaly comes from the
understandable practice of assigning a constant clock frequency to the
model while real clocks were advancing, albeit at a glacial rate.


CM


John Polasek
http://www.dualspace.net

The statement:

"real clocks were advancing, albeit at a glacial rate"

implies that all clocks increased montonically over time
"all clocks" would include Big Ben, London
my quartz wrist watch
and every quantum transition time (frequency)
including the one governing the station clock
as well as the 6E23 clocks per mole
of our known universe.

Is that what your are implying?

Richard

On 10 Jul 2006 01:23:31 -0700, "George Dishman"
wrote:


John C. Polasek wrote:
On Sun, 9 Jul 2006 22:51:01 +0100, "George Dishman"
wrote:


"jpolasek" wrote in message
roups.com...

In keeping with your theory, ..

Not in keeping with any theory, in keeping with
the known method by which the hardware records
the data and the known method which Craig used
in the software

.. suppose that 1000 roundtrip tests had
been recorded over the 8 year period. How would you assemble these 1000
data sets into the linear graph shown? Did they each contain a 0.0015
Hz excess, ...

No. There would be 1000 points on the graph. The
points at the start of the graph would contain
almost no excess, those after 4 years would have
a 1.5Hz excess _each_ while those at the end of
the eight year period would _each_ have a 3Hz
excess.

Each excess is calculated by comparing the
received frequency with that transmitted less
than a day earlier and removing the Doppler due
to the speed of the craft etc. and the numbers
are the total for the two legs of the trip.

No, that's not how the anomaly is computed in gr-qc 0104064 . Look at
Eq. 15 pg. 18:
Anomaly = fobs(t) - fmodel(t) = - f0*2Ap*t/c

Note that a_P already appears in that equation. The
words preceding the equation explain:

"The observed, two-way anomalous effect can be
expressed to first order in v/c as:"

Eq. 15b says
fmod = f0(1-2v/c)
Just as well I can write Eq. 15c which says
fobs = f0(1-2v/c)
Then Eq. 15a subtracts the two and comes up with not 0, but
-F0*2Apt/c
which, since there is no proved Ap, can be more correctly be written
as
-H*f0*2 where H = Ap/c
It's all algebra with not one jot of physics.
But to include my thesis of all clocks increasing at H we have
fobs - fmod = v0(1+Ht)(1-2v/c) - f0(1-2v/c) comes down to
fobs - fmod = fo(1-2v/c)Ht ~ f0Ht
where whole frequencies are used.
As I had said. The v/c dynamics are plainly negligible (0.0004x).
If you want to compare Dopplers it would look like (but it's not Eq.
15)
(fobs-f0) - (fmod-f0) = fobs - fmod would equal zero if f0
was constant everywhere. But with an advancing station clock we have
instead
(fobs-f0[1+Ht]) - (fmod-f0) = f0Ht
This version of Doppler comparison works out but Eq. 15 doesn't use
it.

In other words, after the value of a_P has been
extracted by the actual processing which I am
explaining, Eqn 15 is a way of explaining what
the anomaly would look like if the uplink were
at a fixed frequency. It is not a description of the
actual data processing method because the uplink
isn't fixed.

where t is the time since initialization, not the signal loop time.

They are saying that to first approximation the difference
between the measured doppler shift and the predicted
shift is a linear function of time since the start of the
analysis.

I am saying the actual shift is the difference between the
uplink and downlink frequencies. Both are true.

See Figs. 6 & 7 covering 0-60 AU which is many years.
They do not, as you said, "compare the received frequency with that
transmitted less than a day earlier"

Sorry John but that IS what is done for the analysis, there
is no alternative since the actual transmit frequency has
step changes for engineering purposes. Once the speed
has been determined from those two values, it can of
course be applied to a hypothetical exact signal such as
2.92GHz, which is the nature of Eqn 15, but that is not
part of the actual analysis.

You're obfuscating; we know there are two gearshifts in frequency that
have nothing to do with this problem.

After all that is said, the point remains, Eqn 15 says
that after 8 years the actual received frequency at any
time exceeds the expected value from the model by
3Hz (round trip) so each point on the graph is 3Hz,
not the tiny amount you mentioned.

I'm giving 3 Hz, but not in each turnaround.

George
John P

John C. Polasek wrote:
On 10 Jul 2006 01:23:31 -0700, "George Dishman"
wrote:


John C. Polasek wrote:
On Sun, 9 Jul 2006 22:51:01 +0100, "George Dishman"
wrote:


"jpolasek" wrote in message
roups.com...

In keeping with your theory, ..

Not in keeping with any theory, in keeping with
the known method by which the hardware records
the data and the known method which Craig used
in the software

.. suppose that 1000 roundtrip tests had
been recorded over the 8 year period. How would you assemble these 1000
data sets into the linear graph shown? Did they each contain a 0.0015
Hz excess, ...

No. There would be 1000 points on the graph. The
points at the start of the graph would contain
almost no excess, those after 4 years would have
a 1.5Hz excess _each_ while those at the end of
the eight year period would _each_ have a 3Hz
excess.

Each excess is calculated by comparing the
received frequency with that transmitted less
than a day earlier and removing the Doppler due
to the speed of the craft etc. and the numbers
are the total for the two legs of the trip.

No, that's not how the anomaly is computed in gr-qc 0104064 . Look at
Eq. 15 pg. 18:
Anomaly = fobs(t) - fmodel(t) = - f0*2Ap*t/c

Note that a_P already appears in that equation. The
words preceding the equation explain:

"The observed, two-way anomalous effect can be
expressed to first order in v/c as:"

Eq. 15b says
fmod = f0(1-2v/c)
Just as well I can write Eq. 15c which says
fobs = f0(1-2v/c)
Then Eq. 15a subtracts the two and comes up with not 0, but
-F0*2Apt/c
which, since there is no proved Ap, can be more correctly be written
as
-H*f0*2 where H = Ap/c
It's all algebra with not one jot of physics.

Exactly, it is all algebra. Figure 8 is produced by
plotting individual points as I explained above. It
is fairly easy to see the the result is close to a
straight line and Eqn 15 is then the algebraic
expression of the general form of such a line.

To understand what is being plotted however, you
need to consider how a value for Doppler shift is
produced, and the only way to do that when the
uplink changes from day to day is to find the
difference in frequency between the Tx and Rx
values. That is what was done.

....
Sorry John but that IS what is done for the analysis, there
is no alternative since the actual transmit frequency has
step changes for engineering purposes. Once the speed
has been determined from those two values, it can of
course be applied to a hypothetical exact signal such as
2.92GHz, which is the nature of Eqn 15, but that is not
part of the actual analysis.

You're obfuscating; we know there are two gearshifts in frequency that
have nothing to do with this problem.

What we know is that the hardware records the transmit
frequency and that is the value which must be used as
the reference against which the receive frequency is
compared to obtain each plotted point on Figure 8.

After all that is said, the point remains, Eqn 15 says
that after 8 years the actual received frequency at any
time exceeds the expected value from the model by
3Hz (round trip) so each point on the graph is 3Hz,
not the tiny amount you mentioned.

I'm giving 3 Hz, but not in each turnaround.

Then your document is worthless because 3Hz per
sample is what the analysis produces at the end
of the period. Ask Craig, he has done the actual
work so can confirm from first hand knowledge. My
work (unpublished) has been based on the residuals
remaining from his analysis so is second hand but
required me to understand in detail how they were
derived. Your hypothesis is one that should be
considered but to match the frequency results, you
need to increase your clock drift rate by a factor of
about 10000. Doing that would give you a process
that matched the Doppler observations and the test
would then come done to a range comparison which
is a much more complex task given the absence of
direct range measurements.

George

Craig Markwardt wrote:
"Aleksandr Timofeev" writes:


Craig Markwardt wrote:
"Aleksandr Timofeev" writes:

Craig Markwardt wrote:
[snip]
Dear Craig Markwardt, please, look at:

http://groups.google.com/group/sci.p...e=source&hl=en

My arguments and interpretation of anomaly of "Pioneers" can change
your approach to interpretation of the problem.

The principle of equivalence has been tested quite narrow tolerances
in the solar system already (Williams et al 1996), so your supposition
will probably not be fruitful.

CM

References
Williams, Newhall & Dickey 1996, Phys Rev D, 53, 6730

I do not believe in Williams, Newhall & Dickey interpretation of
experimental data absolutely.

Please, try to explain a difference between measurements of values
of masses of Jove and Saturn by two different measurement methods
of masses of planets in two epoch distant from each other by interval
of time per 25 years:

Your reference to Jupiter and Saturn is irrelevant, since the
equivalence test in Williams Newhall & Dickey (1996), regarded the
independent accelerations of the *earth and moon* toward the sun. The
accelerations match to with one part in 5 x 10^{-13}, even though the
two bodies have a factor ~80 difference in mass.

Furthermore, your calculation,
Jove was 317.735 in 1970 and has become 317.89 in 1990
Saturn was 95.147 in 1970 and has become 95.168 in 1990
is both in error and does not account for measurement uncertainties.

Assuming measurement uncertainties in the final IAU 1976 values,
one obtains,
IAU 1976 IERS (2003)
M_jupiter / M_earth = 317.89(1) 317.8942(1)
M_saturn / M_earth = 95.17(1) 95.185(2)

Whether you can point out the method of measurement of values of
planetary masses?
( Accordingly for IAU 1976 and for IERS (2003))?

http://groups.google.com/group/sci.a...e=source&hl=en

Jim Cobban wrote:

"As pointed out by this discussion, the methodology of determination of
the masses of solar system objects is significantly different between
the two eras.

Prior to the space age all that could be determined was the angular
position of solar system objects."


In short, the values are consistent to with a few sigma in the
uncertainties, and thus your claim reagarding mass differences is
irrelevant.

CM

Reference
McCarthy & Petit, IERS Conventions, IERS Technical Note No. 32
http://tai.bipm.org/iers/conv2003/conv2003.html
(number in parenthesis is estimated uncertainty in final digit)

IAU (1976)
M_moon / M_earth = 0.01230002
M_sun / (M_earth + M_moon) = 328900.5(1)
M_sun / M_jupiter = 1047.355(1)
M_sun / M_saturn = 3498.5(2)

M_jupiter / M_earth = 317.89(1)
M_saturn / M_earth = 95.17(1)


IERS (2003)
M_moon / M_earth = 0.0123000383(1)
M_sun / (M_earth + M_moon) = 328900.561400(1)
M_sun / M_jupiter = 1047.3486(8)
M_sun / M_saturn = 3497.898(18)

M_jupiter / M_earth = 317.8942(1)
M_saturn / M_earth = 95.185(2)

The spin network model in loop quantum gravity could explain the
sunward acceleration.

uri wrote:
The spin network model in loop quantum gravity could explain the
sunward acceleration.

Please explain how it can produce a gravitational
effect on the craft with a similar effect on larger
bodies.

Can you derive the value of a_P with that theory?

George