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John C. Polasek
science forum Guru
Joined: 30 Apr 2005
Posts: 321
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 Posted: Thu May 18, 2006 2:53 am Post subject:
Pioneer 10 Anomaly solved with expanding wavelength
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I have finally have got to my final solution to the Pioneer Anomaly,
and it is on my website as the first item, in red, as a pdf.
There have been two congresses already last year in Bremen and this
year in Glasgow with no results, using, naturally, relativity.
The solution is fairly (no, very) subtle and should be a relief to
those physicists who worried about range errors in 100,000km range. I
show how to modify the model to silence the nuisance "blue note".
I also sent this file to a number of scientists who were at last
year's congress, because I could not convince the arxiver's to print
it without an endorser (an expert who has written at least 2 articles
in 5 years or 5 articles in 2 years, I forget which).
Please check it out.
John Polasek
http://www.dualspace.net |
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dda1
science forum Guru
Joined: 06 Feb 2006
Posts: 762
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| Posted: Thu May 18, 2006 3:44 am Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
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jpolasek wrote:
| Quote: | I have finally have got to my final solution to the Pioneer Anomaly,
and it is on my website as the first item, in red, as a pdf.
There have been two congresses already last year in Bremen and this
year in Glasgow with no results, using, naturally, relativity.
The solution is fairly (no, very) subtle and should be a relief to
those physicists who worried about range errors in 100,000km range. I
show how to modify the model to silence the nuisance "blue note".
I also sent this file to a number of scientists who were at last
year's congress, because I could not convince the arxiver's to print
it without an endorser (an expert who has written at least 2 articles
in 5 years or 5 articles in 2 years, I forget which).
Please check it out.
John Polasek
http://www.dualspace.net
|
Duh, another kook. |
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John C. Polasek
science forum Guru
Joined: 30 Apr 2005
Posts: 321
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| Posted: Sun May 21, 2006 4:16 pm Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
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On 21 May 2006 04:59:04 -0500, Craig Markwardt
<craigmnet@REMOVEcow.physics.wisc.edu> wrote:
| Quote: |
John C. Polasek <jpolasek@cfl.rr.com> writes:
I have finally have got to my final solution to the Pioneer Anomaly,
and it is on my website as the first item, in red, as a pdf.
Note that there is a serious problem with this derivation, namely that
Polasek assumes that the spacecraft is traveling at the speed of
light, "c" (eqns. 4 & 5). The spacecraft is actually traveling at
speed ~12 km/s w.r.t the solar system center.
|
You have misread it entirely. First, Eq. 4 is a simple tutorial on
conversion in general of a kinematic velocity to a perceived
acceleration and c is not involved; it could be Vcraft or any velocity
conugate to the distance D. (we have 2: c and Vcraft)..
| Quote: |
The Hubble expansion rate is H = 72 km/s per Mpc of distance.
|
H is by no means established as 72, it is between 50 & 100
(established by 'standard candle')
and for our purposes, the anomaly provides us a way to determine H
much more accurately as 90.07 = Ap/c.
| Quote: | Since
the spacecraft distance is increasing at rate 12 km/s, the apparent
acceleration due to the Hubble flow is a_H = 72 km/s/Mpc * 12 km/s ~
3x10^{-12} cm/s^2. I.e. far too small by a factor of ~10^4.
|
Again, you must have skimmed the paper because it's what I point out
in Eq. 7a, stating that A_pcraft is 3.49e-14 m/ss, exactly equal to
your reduced value 3e-12cm/ss, 25,000 times lower than Ap.
A_pcraft when integrated over 7.5 years is the same velocity as Ap
integrated over the 2.63 hour flight time.
| Quote: | There is
no way for any "Hubble effect" to account for the Pioneer anomaly.
CM
|
In Eq. 5 I am stressing that the root phenomenon is that the wave, at
the moment of contact with the craft, is traveling at c, and lambda
is increasing at an as yet unestablished Hubble rate, which we derive
by setting
Ap = -Hc = -8.74e-10m/ss
in order to solve for H = 90.07km/smpc. This instant of contact with
expanding lambda is all the radar system knows about the phenomenon
and is the only information we have to work with. The team divided the
excess frequency by time of flight, don't forget.
Therefore, this event being the fundamental information kernel we
naturally integrate Ap over 2.63 hours Eq. 7c, to get the error
velocity. It is the exact same velocity as A_pcraft integrated over
7.5 years, namely -8.28e-6 m/s as per Eq. 7a.
Therefore, the velocity shown in the team's Fig. 1 is 25,000 times too
high because it is Ap mistakenly integrated over 7 years as in Eq. 7d.
I took considerable care to distinguish between the two velocities,
two accelerations and two periods of integration. The Pioneer
trepidation is overblown by a factor of c/V = 25,000 but I
show how to change the model by feeding back +HD to ward off the
annoying blue hum.
It is an easy mistake to interpret as per Fig. 1,, but at the same
time it was gratuitous to make the remark the "residuals...cannot be
removed without adding Ap", because as I have shown, it would have
been 25,000 times too high.
John Polasek
http://www.dualspace.net |
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John C. Polasek
science forum Guru
Joined: 30 Apr 2005
Posts: 321
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| Posted: Sun May 21, 2006 10:25 pm Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
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On 21 May 2006 16:01:56 -0500, Craig Markwardt
<craigmnet@REMOVEcow.physics.wisc.edu> wrote:
| Quote: |
John C. Polasek <jpolasek@cfl.rr.com> writes:
On 21 May 2006 04:59:04 -0500, Craig Markwardt
craigmnet@REMOVEcow.physics.wisc.edu> wrote:
John C. Polasek <jpolasek@cfl.rr.com> writes:
I have finally have got to my final solution to the Pioneer Anomaly,
and it is on my website as the first item, in red, as a pdf.
Note that there is a serious problem with this derivation, namely that
Polasek assumes that the spacecraft is traveling at the speed of
light, "c" (eqns. 4 & 5). The spacecraft is actually traveling at
speed ~12 km/s w.r.t the solar system center.
|
It later dawned on me that your first objection was that I did not
prove that Hubble could cause A_p and therefore failed. No, I showed
how you could obtain Ap from Hubble on a 2.6 hour scale, and then
showed that it was misapplied by being integrated over 7.5 years as is
manifest on the graph of Fig. 1.
| Quote: |
You have misread it entirely. First, Eq. 4 is a simple tutorial on
conversion in general of a kinematic velocity to a perceived
acceleration and c is not involved; it could be Vcraft or any velocity
conugate to the distance D. (we have 2: c and Vcraft)..
|
Let's get something straight first. I told you on page 1 that my
numerical examples would be bsed on Fig. 1where linear acceleration
integrated over linear time is plotted as velocity. This graph which
is repeated several places in the references covers a span of 7.5
years and so at 12km/s would span 19AU which is range D, for which the
signal time is 2.63 hours at velocity c. I took the trouble to
annotate the graph so there would be no mistaking it.
| Quote: |
Actually, the error occurs at equation 3. You define "velocity" as
dX/dt_e where t_e is apparently some arbitrary portion of the light
travel time. [*] However, velocity is actually defined as the
derivative w.r.t. some *coordinate* time, hence one of your errors.
|
How can you be so vapid? dX/dt_e is to distinguish time of emission-
from-the antenna at velocity c, from voyage time at velocity V_cr.
It's not "apparently some arbitrary blah blah blah", it's real time
the extent of which is limited to 2.63 hours because the time from
emission at the antenna to the target is 2.63 hours for 19AU.
The blatant error in the Pioneer experiment stems exactly from using
2.63 hour-data extrapolated over 7.5 years, giving a 25,000:1
magnification, that's why I affix "e" to the t in t_e.
| Quote: | The Hubble law, v_H = H0 d, is valid for bodies at a distance d. If
the distance of the body is changing with time at speed v, then one
can perhaps take the derivative of both sides w.r.t. *coordinate*
time, d(v_H)/dt = H0 d(d)/dt, H0 being a constant. Defining the left
hand side as an apparent acceleration, a_H, one finds, a_H = H0 v.
Note v, *not* c.
You really don't get it. I am saying that WL L = L0(1+Ht), and that |
the lengthening WL, L, unbeknownst to the model, would simulate a
shortening range by lower wave count, but better said mathematically
in Eq.3b as
dV =-HD = -Hct_e,
with t_e being the time since emission, t_e, till it strikes the
craft.
| Quote: | The Hubble expansion rate is H = 72 km/s per Mpc of distance.
H is by no means established as 72, it is between 50 & 100 ...
Apparently you haven't been aware of developments in cosmology over
the past decade (Freedman et al 2001; Spergel et al 2003). But
whether H0 is 50 or 100 or somewhere in between, the Hubble effect
still cannot explain the Pioneer anomaly since a factor of 2 is still
far too small.
I guess you still don't understand that I am not trying to show that |
Pioneer accelerated at the rate Ap. Ap is synthetic and it is all too
easy to be careless and apply it over the 7.5 year voyage time, which
is the crux of the problem, glaringly depicted in Fig. 1.
The latest cosmological developments also argue that there was a
sudden acceleration in expansion, which will prove to be poppycock,
besides which the "standard model" doesn't even have a cause for
expansion except dark energy or quintessence (which will go the way of
phlogiston).
| Quote: |
Since
the spacecraft distance is increasing at rate 12 km/s, the apparent
acceleration due to the Hubble flow is a_H = 72 km/s/Mpc * 12 km/s ~
3x10^{-12} cm/s^2. I.e. far too small by a factor of ~10^4.
Again, you must have skimmed the paper because it's what I point out
in Eq. 7a, stating that A_pcraft is 3.49e-14 m/ss, exactly equal to
your reduced value 3e-12cm/ss, 25,000 times lower than Ap.
A_pcraft when integrated over 7.5 years is the same velocity as Ap
integrated over the 2.63 hour flight time.
Irrelevant. Since you specially chose "2.63 hours" to be smaller than
7.5 years by the same factor that v is smaller than c,
|
you are hallucinating.
| Quote: | there is
nothing special about your subsequent results. It's just numerology.
Also note that 2.63 hours is not the one-way or round-trip light
travel time at any point during the analysis by either Anderson et al
or Markwardt.
|
I cannot believe the witless arguments. As I told you above, any
numerics refer to Fig. 1 which I annotated to forfend such talk. Look
at the x-axis and see if you can espy such as 2.63 hrs., 7.5 years,
19AU, 12 km/s. It is a fact that c/Vcraft = 25,000, which I call K.
| Quote: |
There is
no way for any "Hubble effect" to account for the Pioneer anomaly.
CM
In Eq. 5 I am stressing that the root phenomenon is that the wave, at
the moment of contact with the craft, is traveling at c, and lambda
is increasing at an as yet unestablished Hubble rate, which we derive
by setting
Ap = -Hc = -8.74e-10m/ss
in order to solve for H = 90.07km/smpc. This instant of contact with
expanding lambda is all the radar system knows about the phenomenon
and is the only information we have to work with. The team divided the
excess frequency by time of flight, don't forget.
I won't forget that your claim is incorrect. Neither of the "teams"
divided the frequency by the light travel time.
Golly moses, do I have to look that up again? It's right out of one of |
the standard papers on the topic. I started to include it as an image,
but forgot to do it. Oh, okay I found it, it's in refs. [14] and [9].
"The observed 2 way round trip anomalous effect....as
[fobs(t)-fmodel(t)]DSN=-2fP*t "
(or t_e would be suitable). To find fP is left as an exercise for the
student.
| Quote: |
Therefore, this event being the fundamental information kernel we
naturally integrate Ap over 2.63 hours Eq. 7c, to get the error
velocity. It is the exact same velocity as A_pcraft integrated over
7.5 years, namely -8.28e-6 m/s as per Eq. 7a.
This is nonsense. The apparent anomaly is very simply v = a_p * t,
where a_p is the apparent anomalous acceleration, and t is the
*coordinate* time. There is no integration over "2.63 hours." One
could easily take one half or one third of the data and still get
nearly the same result. The only thing special about 2.63 hours is
that you cherry picked it.
|
Think, man, think. And forget this coordinate time stuff, this is
straight Goldstein kinematics.
| Quote: | Since your premises are erroneous, your conclusions are irrelevant,
and there is no sense in proceeding.
CM
References
Freedman, W. L. et al. 2001, ApJ, 553, 47
Spergel, D. N., et al. 2003, ApJS, 148, 175
|
John Polasek
http://www.dualspace.net |
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John C. Polasek
science forum Guru
Joined: 30 Apr 2005
Posts: 321
|
| Posted: Mon May 22, 2006 4:04 pm Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
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On Sun, 21 May 2006 22:24:21 GMT, John C. Polasek
<jpolasek@cfl.rr.com> wrote:
| Quote: | On 21 May 2006 16:01:56 -0500, Craig Markwardt
craigmnet@REMOVEcow.physics.wisc.edu> wrote:
John C. Polasek <jpolasek@cfl.rr.com> writes:
On 21 May 2006 04:59:04 -0500, Craig Markwardt
craigmnet@REMOVEcow.physics.wisc.edu> wrote:
John C. Polasek <jpolasek@cfl.rr.com> writes:
I have finally have got to my final solution to the Pioneer Anomaly,
and it is on my website as the first item, in red, as a pdf.
Note that there is a serious problem with this derivation, namely that
snip
|
This is just a note to say I have revised my paper that I believe
makes it clearer in a number of areas after considering useful
comments by Craig Markwardt, to whom I also apologize for my
querulous expostulations vis a vis "cherry picking" , a practice which
apparently we both mutually hold in disrepute.
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John C. Polasek
science forum Guru
Joined: 30 Apr 2005
Posts: 321
|
| Posted: Wed May 24, 2006 3:02 pm Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
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On 24 May 2006 03:06:12 -0500, Craig Markwardt
<craigmnet@REMOVEcow.physics.wisc.edu> wrote:
| Quote: |
[ noting version "2" of the file. ]
John C. Polasek <jpolasek@cfl.rr.com> writes:
On 21 May 2006 16:01:56 -0500, Craig Markwardt
craigmnet@REMOVEcow.physics.wisc.edu> wrote:
John C. Polasek <jpolasek@cfl.rr.com> writes:
On 21 May 2006 04:59:04 -0500, Craig Markwardt
craigmnet@REMOVEcow.physics.wisc.edu> wrote:
John C. Polasek <jpolasek@cfl.rr.com> writes:
I have finally have got to my final solution to the Pioneer Anomaly,
and it is on my website as the first item, in red, as a pdf.
Note that there is a serious problem with this derivation, namely that
Polasek assumes that the spacecraft is traveling at the speed of
light, "c" (eqns. 4 & 5). The spacecraft is actually traveling at
speed ~12 km/s w.r.t the solar system center.
It later dawned on me that your first objection was that I did not
prove that Hubble could cause A_p and therefore failed. No, I showed
how you could obtain Ap from Hubble on a 2.6 hour scale, and then
showed that it was misapplied by being integrated over 7.5 years as is
manifest on the graph of Fig. 1.
However, the 2.6 hour time scale is still irrelevant. Fundamentally,
the Hubble effect is simply too small for a slow-moving spacecraft.
You have misread it entirely. First, Eq. 4 is a simple tutorial on
conversion in general of a kinematic velocity to a perceived
acceleration and c is not involved; it could be Vcraft or any velocity
conugate to the distance D. (we have 2: c and Vcraft)..
Let's get something straight first. I told you on page 1 that my
numerical examples would be bsed on Fig. 1where linear acceleration
integrated over linear time is plotted as velocity. This graph which
is repeated several places in the references covers a span of 7.5
years and so at 12km/s would span 19AU which is range D, for which the
signal time is 2.63 hours at velocity c. I took the trouble to
annotate the graph so there would be no mistaking it.
Whether you think things are "straight" or not is irrelevant. At no
point during the analysis period in Fig. 1 was the distance to the
spacecraft 19 AU, nor was the round trip light travel time 2.63 hours.
[ The mean distance was ~65 AU. Your 19 AU figure is the approx. *excess*
distance traveled during the analysis period. ]
This 19/65 AU business is easy to explain and this should clear up the |
whole matter: (have you seen v2 of the explanation?)
As you see, the plot was drawn AS IF things started happening at x = 0
AU, which you say is really 55 AU at which point there is shown zero
velocity residual.
This is easily remedied by adding a 55 AU extension to the left end of
the plot, so now it would extend from 0 to 75 AU or so. Now use the
same velocity slope from 0 to 75 AU and so the terminal residual
velocity plotted will now become ~20 cm/s x 75/19 = 78.9 cm/s instead
of 20.
Applying the usual arithmetic, with time extended by K = 75/19, we
find that at 12km/s it now takes 29.6 years and the signal time would
be 10.38 hours.
All of my computations apply as first presented.
| Quote: | That plot shows the velocity residuals when setting the anomalous
acceleration equal to zero. THERE IS NO "INTEGRATION" over 7.5 years
or 2.63 hours. The residuals are simply the difference between the
observed and model frequency. You can argue with me about this if you
wish... but then again, I actually did the analysis, and I know that
you are incorrect.
I recommend that you "get something straight" by considering that the
anomaly and its magnitude exist independent of the interpretation.
There is no freedom to dial down the amplitude by a factor of 25000.
Actually, the error occurs at equation 3. You define "velocity" as
dX/dt_e where t_e is apparently some arbitrary portion of the light
travel time. [*] However, velocity is actually defined as the
derivative w.r.t. some *coordinate* time, hence one of your errors.
How can you be so vapid? dX/dt_e is to distinguish time of emission-
from-the antenna at velocity c, from voyage time at velocity V_cr.
It's not "apparently some arbitrary blah blah blah", it's real time
the extent of which is limited to 2.63 hours because the time from
emission at the antenna to the target is 2.63 hours for 19AU.
The blatant error in the Pioneer experiment stems exactly from using
2.63 hour-data extrapolated over 7.5 years, giving a 25,000:1
magnification, that's why I affix "e" to the t in t_e.
I withdraw my comment about equation 3.
The Hubble law, v_H = H0 d, is valid for bodies at a distance d. If
the distance of the body is changing with time at speed v, then one
can perhaps take the derivative of both sides w.r.t. *coordinate*
time, d(v_H)/dt = H0 d(d)/dt, H0 being a constant. Defining the left
hand side as an apparent acceleration, a_H, one finds, a_H = H0 v.
Note v, *not* c.
There is another interpretation for H0 and that is fraction/time = |
100%/10Byrs, where H0 acts as the derivative operator d/dt as in H*L =
Ldot with H0 affixed.
I did get H0*V as "A_pcraft" lower by a factor of v/c. Ap is conjugate
to Ddot = c, and A_pcarft is conjugate to Ddto = Vcraft.
JP
| Quote: | You really don't get it. I am saying that WL L = L0(1+Ht), and that
the lengthening WL, L, unbeknownst to the model, would simulate a
shortening range by lower wave count, but better said mathematically
in Eq.3b as
dV =-HD = -Hct_e,
with t_e being the time since emission, t_e, till it strikes the
craft.
Equations 4 and 5 are still incorrect. You have erroneously confused
your "t_e" with "t". Velocity is a derivative with respect to
coordinate time = clock time = calendar time, *not* with respect to
light travel time, which results in your error of (v/c). As I show
above when the derivative is taken properly, a_H = H0 v.
The Hubble expansion rate is H = 72 km/s per Mpc of distance.
H is by no means established as 72, it is between 50 & 100 ...
Apparently you haven't been aware of developments in cosmology over
the past decade (Freedman et al 2001; Spergel et al 2003). But
whether H0 is 50 or 100 or somewhere in between, the Hubble effect
still cannot explain the Pioneer anomaly since a factor of 2 is still
far too small.
I guess you still don't understand that I am not trying to show that
Pioneer accelerated at the rate Ap. Ap is synthetic and it is all too
easy to be careless and apply it over the 7.5 year voyage time, which
is the crux of the problem, glaringly depicted in Fig. 1.
The latest cosmological developments also argue that there was a
sudden acceleration in expansion, which will prove to be poppycock,
besides which the "standard model" doesn't even have a cause for
expansion except dark energy or quintessence (which will go the way of
phlogiston).
Whatever. I'll reiterate that the Hubble constant measurements are
~72 km/s/Mpc.
No you can't do that. Do you realize what a wobbly standard the |
"standard candle" is? Some kind of Cepheids if I remember correctly
whose intensity is supposed to be constant.
JP
| Quote: |
Since
the spacecraft distance is increasing at rate 12 km/s, the apparent
acceleration due to the Hubble flow is a_H = 72 km/s/Mpc * 12 km/s ~
3x10^{-12} cm/s^2. I.e. far too small by a factor of ~10^4.
Again, you must have skimmed the paper because it's what I point out
in Eq. 7a, stating that A_pcraft is 3.49e-14 m/ss, exactly equal to
your reduced value 3e-12cm/ss, 25,000 times lower than Ap.
A_pcraft when integrated over 7.5 years is the same velocity as Ap
integrated over the 2.63 hour flight time.
Irrelevant. Since you specially chose "2.63 hours" to be smaller than
7.5 years by the same factor that v is smaller than c,
you are hallucinating.
there is
nothing special about your subsequent results. It's just numerology.
Also note that 2.63 hours is not the one-way or round-trip light
travel time at any point during the analysis by either Anderson et al
or Markwardt.
I cannot believe the witless arguments. As I told you above, any
numerics refer to Fig. 1 which I annotated to forfend such talk. Look
at the x-axis and see if you can espy such as 2.63 hrs., 7.5 years,
19AU, 12 km/s. It is a fact that c/Vcraft = 25,000, which I call K.
However, let's look at the values you indicate.
T ~ 7.5 years is the analysis interval (Markwardt 2002).
v ~ 12 km/s is the mean spacecraft speed
19 AU is *derived* = v * T
2.63 hours is *derived* = (v * T) / c = T * (v/c)
I.e. the "2.63 hour" time is *exactly* v/c times the analysis
interval. There is nothing interesting about that value, and it
certainly does not represent any kind of light travel time (one-way or
round trip) during the analysis interval that was plotted.
I think I answered that above, if you simply extend the graph. Don't |
forget the plot begins at 55AU with zero velocity residual, which
could not be the case.
JP
| Quote: |
There is
no way for any "Hubble effect" to account for the Pioneer anomaly.
Hubble only generates Ap/25,000. I said that. |
But the Hubble effect does account for the frequency discrepancy
logged by the team, and it was simply a natural mistake to take dV/dt
and multiply it by 7.5 years, but one which caused a lot of trouble.
JP
| Quote: | CM
In Eq. 5 I am stressing that the root phenomenon is that the wave, at
the moment of contact with the craft, is traveling at c, and lambda
is increasing at an as yet unestablished Hubble rate, which we derive
by setting
Ap = -Hc = -8.74e-10m/ss
in order to solve for H = 90.07km/smpc. This instant of contact with
expanding lambda is all the radar system knows about the phenomenon
and is the only information we have to work with. The team divided the
excess frequency by time of flight, don't forget.
I won't forget that your claim is incorrect. Neither of the "teams"
divided the frequency by the light travel time.
Golly moses, do I have to look that up again? It's right out of one of
the standard papers on the topic. I started to include it as an image,
but forgot to do it. Oh, okay I found it, it's in refs. [14] and [9].
"The observed 2 way round trip anomalous effect....as
[fobs(t)-fmodel(t)]DSN=-2fP*t "
(or t_e would be suitable). To find fP is left as an exercise for the
student.
Your quotation is correct, but "t_e" would *NOT* be suitable. The "t"
in the equation you cite is the coordinate time = clock time =
calendar time since the start of the analysis interval. You are
making the same error that Aladar Stolmar did, years ago.
At no time did anybody (except you and Stolmar) divide by the
frequency residuals by the "time of flight."
Therefore, this event being the fundamental information kernel we
naturally integrate Ap over 2.63 hours Eq. 7c, to get the error
velocity. It is the exact same velocity as A_pcraft integrated over
7.5 years, namely -8.28e-6 m/s as per Eq. 7a.
This is nonsense. The apparent anomaly is very simply v = a_p * t,
where a_p is the apparent anomalous acceleration, and t is the
*coordinate* time. There is no integration over "2.63 hours." One
could easily take one half or one third of the data and still get
nearly the same result. The only thing special about 2.63 hours is
that you cherry picked it.
Think, man, think. And forget this coordinate time stuff, this is
straight Goldstein kinematics.
Believe me, I've thought about it far more than you have. You need to
consider that not all "t"s are the same. The "t" in the equation
above is the standard clock time (time since the start of the analysis
interval in 1987). Call it coordinate time, calendar date, whatever,
but it is *not* the same as light travel time.
|
Again, when the Pioneer has traveled for N years and the team turns on
the radar and takes a reading from the first return M hours later,
they are both for the same distance V*Nyr or c*Mhrs. The one is a
proxy for the other. The correct Hubble analysis can only be done at
light speed and concentrating on that moment when the wave touches
the Pioneer and analyzing to see what the radar makes of this and it
turns out to be a false velocity equal to - H*D.
| Quote: | Since your premises are erroneous, your conclusions are irrelevant,
and there is no sense in proceeding.
CM
References
Freedman, W. L. et al. 2001, ApJ, 553, 47
Spergel, D. N., et al. 2003, ApJS, 148, 175
CM
It think you see that redrawn graph will make things clearer. The |
Hubble distance-shrink requires a bit of simple math. I didn't need
that graph-originally I used 70Au and 27.7 years, but numbers are
auditors.
John Polasek
http://www.dualspace.net |
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John C. Polasek
science forum Guru
Joined: 30 Apr 2005
Posts: 321
|
| Posted: Wed May 24, 2006 10:05 pm Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
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On Wed, 24 May 2006 15:01:17 GMT, John C. Polasek
<jpolasek@cfl.rr.com> wrote:
| Quote: | On 24 May 2006 03:06:12 -0500, Craig Markwardt
craigmnet@REMOVEcow.physics.wisc.edu> wrote:
SNIP for now
CM
It think you see that redrawn graph will make things clearer. The
Hubble distance-shrink requires a bit of simple math. I didn't need
that graph-originally I used 70Au and 27.7 years, but numbers are
auditors.
John Polasek
http://www.dualspace.net
|
In line with suggestions from Craig Markwardt, I have modified the
paper into Ver. 3 that is I believe more accepable, besides also
citing a reference that derives the expansion of the wavelength from
GR. The URL is
http://burro.astr.case.edu/Academics/Astr328/Notes/Redshift/redshift.html
It's a job to convey what is at the least, subtle, and at worst,
unbelievable, but I think now I've got it.
John Polasek
http://www.dualspace.net |
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George Dishman
science forum Guru
Joined: 08 May 2005
Posts: 963
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| Posted: Thu May 25, 2006 8:05 am Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
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John C. Polasek wrote:
....
| Quote: | Hubble only generates Ap/25,000. I said that.
But the Hubble effect does account for the frequency discrepancy
logged by the team, and it was simply a natural mistake to take dV/dt
and multiply it by 7.5 years, but one which caused a lot of trouble.
|
John, I think you are working under a misaprehension,
the team didn't make measurements over individual
transmissions and then extrapolate by 7.5 years as
you seem to be suggesting. The graph you are showing
is actual frequency residuals. They take the actual
increase in the residual and divide by 7.5 years to
find A_p, not the other way round. Thefirst paragraph of
your paper sums that up where you say:
"The Pioneer team found that after dividing the excess
frequency df by the signal flight time, (Fig. 1), they got
a quite definite and constant frequency acceleration
equal to
df' = 5.99*10^-9 Hz/s leading to
A_p = -8.74 * 10^-10 m/s^2 +/- 1.33 * 10^-10 m/s^2"
It would be more accurate to say this instead:
"The Pioneer team found that the change in excess
frequency between 1987 and 1994 was linear with
time at a best fit rate of
df' = 5.99*10^-9 Hz/s
equivalent to what would be produced via the
Doppler effect by a constant acceleration of
A_exp = 7.84 * 10^-10 m/s^2 +/-0.01 * 10^-10 m/s^2"
Note also that 5.99Hz/s is 7.84, not 8.74. The difference
of 0.90 is the result of the biases listed in table 2,
principally the radio beam reaction. All of these (other
than 2g) affect the motion of the craft rather than the
frequency of the beam so you need to treat them
differently.
George |
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Craig Markwardt
science forum addict
Joined: 16 Jun 2005
Posts: 66
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| Posted: Thu May 25, 2006 10:30 am Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
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"George Dishman" <george@briar.demon.co.uk> writes:
| Quote: | John C. Polasek wrote:
...
Hubble only generates Ap/25,000. I said that.
But the Hubble effect does account for the frequency discrepancy
logged by the team, and it was simply a natural mistake to take dV/dt
and multiply it by 7.5 years, but one which caused a lot of trouble.
John, I think you are working under a misaprehension,
the team didn't make measurements over individual
transmissions and then extrapolate by 7.5 years as
you seem to be suggesting. The graph you are showing
is actual frequency residuals. They take the actual
increase in the residual and divide by 7.5 years to
find A_p, not the other way round. Thefirst paragraph of
your paper sums that up where you say:
|
George, I tried explaining that to Polasek, but he ignored my comments
in that regard. Sigh.
Craig
--
--------------------------------------------------------------------------
Craig B. Markwardt, Ph.D. EMAIL: craigmnet@REMOVEcow.physics.wisc.edu
Astrophysics, IDL, Finance, Derivatives | Remove "net" for better response
-------------------------------------------------------------------------- |
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George Dishman
science forum Guru
Joined: 08 May 2005
Posts: 963
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| Posted: Thu May 25, 2006 12:20 pm Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
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|
Craig Markwardt wrote:
| Quote: | "George Dishman" <george@briar.demon.co.uk> writes:
John C. Polasek wrote:
...
Hubble only generates Ap/25,000. I said that.
But the Hubble effect does account for the frequency discrepancy
logged by the team, and it was simply a natural mistake to take dV/dt
and multiply it by 7.5 years, but one which caused a lot of trouble.
John, I think you are working under a misaprehension,
the team didn't make measurements over individual
transmissions and then extrapolate by 7.5 years as
you seem to be suggesting. The graph you are showing
is actual frequency residuals. They take the actual
increase in the residual and divide by 7.5 years to
find A_p, not the other way round. Thefirst paragraph of
your paper sums that up where you say:
George, I tried explaining that to Polasek, but he ignored my comments
in that regard. Sigh.
|
Hi Craig,
I can see where you pointed it out but from John's
reply, I don't think he appreciated what you were
saying. I've tried to avoid the other issues that have
come up as sidelines and just focus on that core
problem to see if that will help. Sometimes two
people saying the same thing in different ways
can emphasise the point.
best regards
George |
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John C. Polasek
science forum Guru
Joined: 30 Apr 2005
Posts: 321
|
| Posted: Thu May 25, 2006 3:20 pm Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
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On 25 May 2006 05:20:24 -0700, "George Dishman"
<george@briar.demon.co.uk> wrote:
| Quote: |
Craig Markwardt wrote:
"George Dishman" <george@briar.demon.co.uk> writes:
John C. Polasek wrote:
...
Hubble only generates Ap/25,000. I said that.
But the Hubble effect does account for the frequency discrepancy
logged by the team, and it was simply a natural mistake to take dV/dt
and multiply it by 7.5 years, but one which caused a lot of trouble.
John, I think you are working under a misaprehension,
the team didn't make measurements over individual
transmissions and then extrapolate by 7.5 years as
you seem to be suggesting. The graph you are showing
is actual frequency residuals. They take the actual
increase in the residual and divide by 7.5 years to
find A_p, not the other way round. Thefirst paragraph of
your paper sums that up where you say:
George, I tried explaining that to Polasek, but he ignored my comments
in that regard. Sigh.
Hi Craig,
I can see where you pointed it out but from John's
reply, I don't think he appreciated what you were
saying. I've tried to avoid the other issues that have
come up as sidelines and just focus on that core
problem to see if that will help. Sometimes two
people saying the same thing in different ways
can emphasise the point.
best regards
George
Take a look at http://www.dualspace.net/uploads/cosmoarxivnu3.pdf. |
I have edited the paper to clarify what Craig objected to. Now I
include as Figure 1 a transcript of the team's calculation of fP as
frequency difference over the 2-way round trip
fobs - fmod = -2*fP*t (2)
which clearly implies the signal round trip time.
That is the same basis I used and my argument still stands.
Take a look at that and see if there is still a conflict. Otherwise we
still have the spectre of undocumented forces.
John |
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George Dishman
science forum Guru
Joined: 08 May 2005
Posts: 963
|
| Posted: Thu May 25, 2006 3:46 pm Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
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|
John C. Polasek wrote:
| Quote: | On 25 May 2006 05:20:24 -0700, "George Dishman"
george@briar.demon.co.uk> wrote:
Take a look at http://www.dualspace.net/uploads/cosmoarxivnu3.pdf.
I have edited the paper to clarify what Craig objected to. Now I
include as Figure 1 a transcript of the team's calculation of fP as
frequency difference over the 2-way round trip
fobs - fmod = -2*fP*t (2)
which clearly implies the signal round trip time.
|
The value [fobs(t) - fmodel(t)] is the error in the
frequency measured in two-way mode. The model
was initialised in 1987 so at that time fmodel was
set equal to fobs. The discrepancy between the
measured frequency and the model prediction
by the end of 1994 was about 3Hz (two way). If
you divide 3Hz by ~8 years you get the figure they
quote of f'p ~ 6*10^-9Hz/s. In your equation above
therefore, the variable t represents the calendar
time since the model was initialised. In other
words, t is simply the horizontal axis of the graph
which you reproduce in your paper.
| Quote: | That is the same basis I used and my argument still stands.
Take a look at that and see if there is still a conflict. Otherwise we
still have the spectre of undocumented forces.
|
Yes, there is a conflict. I re-read your paper before
I posted because I was aware you were updating
it and might have corrected the description since I
last looked.
The point is that the value of a 3Hz discrepancy
by 1994 if the model is initialised in 1987 is a
direct measurement, not an integrated or
extrapolated number. All of the readings around
that time are about 3Hz higher than they should
be, and yes that does appear to require an as
yet undetermined force on the craft.
George |
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John C. Polasek
science forum Guru
Joined: 30 Apr 2005
Posts: 321
|
| Posted: Fri May 26, 2006 5:29 pm Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
|
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|
On 25 May 2006 08:46:10 -0700, "George Dishman"
<george@briar.demon.co.uk> wrote:
| Quote: |
John C. Polasek wrote:
On 25 May 2006 05:20:24 -0700, "George Dishman"
george@briar.demon.co.uk> wrote:
Take a look at http://www.dualspace.net/uploads/cosmoarxivnu3.pdf.
I have edited the paper to clarify what Craig objected to. Now I
include as Figure 1 a transcript of the team's calculation of fP as
frequency difference over the 2-way round trip
fobs - fmod = -2*fP*t (2)
which clearly implies the signal round trip time.
The value [fobs(t) - fmodel(t)] is the error in the
frequency measured in two-way mode. The model
was initialised in 1987 so at that time fmodel was
set equal to fobs. The discrepancy between the
measured frequency and the model prediction
by the end of 1994 was about 3Hz (two way). If
you divide 3Hz by ~8 years you get the figure they
quote of f'p ~ 6*10^-9Hz/s. In your equation above
therefore, the variable t represents the calendar
time since the model was initialised. In other
words, t is simply the horizontal axis of the graph
which you reproduce in your paper.
That is the same basis I used and my argument still stands.
Take a look at that and see if there is still a conflict. Otherwise we
still have the spectre of undocumented forces.
Yes, there is a conflict. I re-read your paper before
I posted because I was aware you were updating
it and might have corrected the description since I
last looked.
The point is that the value of a 3Hz discrepancy
by 1994 if the model is initialised in 1987 is a
direct measurement, not an integrated or
extrapolated number. All of the readings around
that time are about 3Hz higher than they should
be, and yes that does appear to require an as
yet undetermined force on the craft.
George
I have a new interpretation for the Pioneer anomaly which I have |
disstributed also to members of the Pioneer team. Your insistence that
the 7.5 year plot contained legitimate detected amplitudes and
George's remark about resetting the system in 1987 caused me to
rethink, as follows:
Gentlemen:
(I hope you appreciate the work I put into my Pioneer solution
"Pioneer 10 anomaly explained by Hubble expansion of wavelength" but I
have discarded it in favor of new findings below).
Previous theory was OK, but…
I finally have a new and hopefully final, solution for the Pioneer
anomaly problem. It is different from the one previously presented
that was based on wavelength expansion, which is still true, but is a
minor contributor to the extent of 1/25,000. l does expand at Hubble
rate after emission and contributes its snippet. The new solution
contains a major surprise.
Finding that the 7.5 year time plot could be true
At the insistence of Craig Markwardt and George Dishman I have
re-examined their statements that the plot in Fig. 2 was a plot of
real residuals, not algebraic extrapolations. But this meant that
Hubble had to be acting for the full 7.5 years, which seemingly would
be out of scale at the speed of light. For another thing, this in one
way seemed to imply the need to compare a clock’s rate with its own
rate 7.5 years ago, which seems an impossibility. Also, I was
convinced that one could only take frequency differences over
hours-long send-receive times (especially since Eq.(2) in Fig. 1
mentioned round trip times but did not specify that t was in calendar
years), to the exclusion of any 7.5 year comparison. (But I had also
misread fmodel = ftrans which was wrong).
Plot and frequency initialized in 1987
Then we discussed that Fig. 2 was a 25% slice of a 30-year voyage and
George made the useful remark that "The model was initialized in 1987
so that at that time fmodel was set equal to fobs".
The light dawns
That immediately struck a bell and I could see that the JPL model must
somehow contain within itself, evidence of a clock frozen in time with
a frequency that was true in 1987. But how could the digital program
hold the original clock rate? It cannot be done in the laboratory. We
know the program sets an increment Dt for use with its difference
equations to integrate Newtonian accelerations. At the same time, it
would later need a wavelength L with which to convert computed
velocities to actual frequency to be beat with the incoming fobs via:
f = -v/l where L = c/fobs
and f = F87 (constant)
In that way everything is locked together in the computer, including
F87 = fobs in 1987. It preserves the old clock frequency inside the
program enabling it to synthesize the original frequency. (I am only
quite sure that this train of logic is correct).
Hubble increase in clock f and c
Now we propose that all atomic clock rates increase at Hubble rate and
similarly that c increases at Hubble rate (H*c = Ap). These are
"proved" in my Dual Space text [12] (by proof I mean that there is a
convincing self-consistency in their derivation).
Therefore, using the previously derived value for Hubble:
H = -Ap/c = 2.9e-18 1/sec (= aP in Anderson, 1 Oct. 1998)
we call the fractional Hubble change over 7.5 years as eps:
eps = H*7.5 years = 6.9e-10
so that the increments in velocity and frequency over 7.5 years will
be:
Dv = eps*c = 20 cm/s
Df = eps*fdn = 1.5 Hz/s where fdn = 2.292 Ghz
thus matching the amplitudes in the plot of Fig. 2, by virtue of the
legacy value of frequency, F87. In short all the samplings can be
expressed as repeated comparisons with the original clock rate from
the computer:
Df_i = fobs_i - F87
linearly increasing in time since 1987.
The Big Surprise!
The above analysis also hides a big surprise. Since in any beat
calculation, we can use either of fobs or ftrans with little
difference (1 part in 25,000), we don’t need the Pioneer. This is not
a typographical error. All we are doing at the station is comparing
today’s station clock with its condition N years ago (as stored under
lock and key in the computer program)! All the peregrinations that
might occur on the 30-year voyage are therefore of no consequence.
Easy experiment to determine a fundamental cosmological constant:
Hubble H0
Here is an experiment that could be done with little expense, given
sufficient time, providing a wonderful opportunity to refine the value
of the Hubble constant. Only the exquisite statistics and longevity of
the Pioneer were able to bring about this finding.
I will be writing a formal paper to further cement the findings, and
while I am loath to abandon a quite ingenious train of thought, as
someone said (or will say) "Simplicity trumps ingenuity".
I thank you for your kind attention and especially George and Craig
for their persistence. I believe this solution is about as simple as
can possibly be.
John Polasek
http://www.dualspace.net
p.s. In justifying the H*f, H*c themes, you might want to visit my web
site and read paper #2, my explanation of gravity in which I affix a
new term to Newton’s equation:
cdc/dr = -g = MG/r^2 = c*H,
in which dc/dr is a kinematic coefficient (as explained in prior
Pioneer paper) combined with c, our (conjugate) velocity away from the
center of mass of the universe, the combination equating to cdc/dr =
-Ap. This says our universe has a slight acceleration away from the
center of mass, in a universe where time and space are separate. It is
an abstract from my book Dual Space, New Physics for a New Century. I
have not put it with Amazon since a perusal by several scholars has
elicited no comment. It is available at Principia Publishing Co., 1015
Maitland Center Commons, Suite 110, Maitland FL, 32751 or by email. |
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Craig Markwardt
science forum addict
Joined: 16 Jun 2005
Posts: 66
|
| Posted: Sun May 28, 2006 7:42 pm Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
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|
John C. Polasek <jpolasek@cfl.rr.com> writes:
....
| Quote: | Hubble increase in clock f and c
Now we propose that all atomic clock rates increase at Hubble rate and
similarly that c increases at Hubble rate (H*c = Ap). These are
"proved" in my Dual Space text [12] (by proof I mean that there is a
convincing self-consistency in their derivation).
.... |
I do *not* subscribe to your "proof," as it's worth noting that
Anderson et al found that a "drifting clock" model was rejected after
considering all of the data (Pioneers + other spacecraft).
| Quote: | The Big Surprise!
The above analysis also hides a big surprise. Since in any beat
calculation, we can use either of fobs or ftrans with little
difference (1 part in 25,000), we don¢t need the Pioneer. This is not
a typographical error. All we are doing at the station is comparing
today¢s station clock with its condition N years ago (as stored under
lock and key in the computer program)! All the peregrinations that
might occur on the 30-year voyage are therefore of no consequence.
|
That is incorrect. The Pioneer effect is based on the carrier
frequency uplinked to the spacecraft, which is then retransmitted to
the ground with a fixed turnaround frequency multiplier. Aside from
the round-trip light travel time, the Pioneer effect at a given time
is compared with references frequencies sent at that time, not in 1987.
The Pioneer effect is a discrepancy between the doppler measurements
and a model of the spacecraft motions. The special thing that
happened in 1987 was that the *model* was initialized then. I could
just have easily initialized the model in 1991 (or 1994), and then the
zero-crossing of the residuals would have occurred at that time
instead of 1987.
CM |
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John C. Polasek
science forum Guru
Joined: 30 Apr 2005
Posts: 321
|
| Posted: Sun May 28, 2006 9:13 pm Post subject:
Re: Pioneer 10 Anomaly solved with expanding wavelength
|
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|
On 28 May 2006 14:42:37 -0500, Craig Markwardt
<craigmnet@REMOVEcow.physics.wisc.edu> wrote:
| Quote: |
John C. Polasek <jpolasek@cfl.rr.com> writes:
...
Hubble increase in clock f and c
Now we propose that all atomic clock rates increase at Hubble rate and
similarly that c increases at Hubble rate (H*c = Ap). These are
"proved" in my Dual Space text [12] (by proof I mean that there is a
convincing self-consistency in their derivation).
...
I do *not* subscribe to your "proof," as it's worth noting that
Anderson et al found that a "drifting clock" model was rejected after
considering all of the data (Pioneers + other spacecraft).
The Big Surprise!
The above analysis also hides a big surprise. Since in any beat
calculation, we can use either of fobs or ftrans with little
difference (1 part in 25,000), we don¢t need the Pioneer. This is not
a typographical error. All we are doing at the station is comparing
today¢s station clock with its condition N years ago (as stored under
lock and key in the computer program)! All the peregrinations that
might occur on the 30-year voyage are therefore of no consequence.
That is incorrect. The Pioneer effect is based on the carrier
frequency uplinked to the spacecraft, which is then retransmitted to
the ground with a fixed turnaround frequency multiplier. Aside from
the round-trip light travel time, the Pioneer effect at a given time
is compared with references frequencies sent at that time, not in 1987.
JP |
Your statement just above, "the Pioneer effect at a...................
frequencies sent at that time, not in 1987" is not clear. In all
references the only comparison mentioned is between return frequency
and MODEL frequency, and never with transmitted frequency. In
dicussions of Ap and fP, reference is always made to Fig. 2 which is
clearly bench-marked at 1987.
I have already calculated a reduced version of Ap based on round trip
time and it's low by v/c. A full-value of Ap via Hubble can only occur
with the fractional increment of H0 x 7.5 years = 6.9e-10.
And I also pointed out that it's immaterial whether we use the
transmitted frequency vs the model, (and so having nothing to do with
the Pioneer), or the return signal. The magic comes in the fact that
the computer is able to deal in 1987 frequency, having no reason to
think it would increase at the very low Hubble rate. (At 2.292GHz it
takes 4.7 years to increase by 1 Hz). JP
| Quote: |
The Pioneer effect is a discrepancy between the doppler measurements
and a model of the spacecraft motions. The special thing that
happened in 1987 was that the *model* was initialized then. I could
just have easily initialized the model in 1991 (or 1994), and then the
zero-crossing of the residuals would have occurred at that time
instead of 1987.
CM
JP |
It's sufficient if the computer model was initialized in 1987 and then
left alone, while all clock frequencies increased at Hubble rate. Then
the difference with the model will rise to 1.5 Hz in the 7,t years, if
Hubble is taken as -Ap/c = 90kms/smpc. JP
John Polasek |
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