Pioneer 10 looks like red shift, not blue
John C. Polasek
question of red shift or blue shift remains. I try to show that the
phenomenon is a red shift, acceleration away from the sun. This is a
matter that should be cleared up.
It seems pretty clear from close analysis of [1] Anderson's et al
paper Oct. 98 and [2] Turyshev Mar. 99 that the Pioneer acceleration
should be defined as a red shift, not blue, a very important
difference. The claimed blue shift would have the acceleration toward
the sun for Pioneer 10 and P11 which were on opposite sides of the
sun.
However, just the opposite (red shift) seems to be the case as a
result of detecting some questionable algebra where a sign was
switched.
Ref. [2] defines the Doppler frequency shift or beat as the received
frequency vs standard:
(1) dF = F - F0
Clearly if the craft is leaving the sun, rdot>0, and dF<0, the usual
redshift condition causing a lower received frequency. Conversely,
while nearing the sun, the opposites are true but in either case, dR
and rdot are of opposite polarity so
(2) dF = -rdot*F/c defines the relationship. For acceleration r",
(3) dF/dt = -r"F/c
It is repeatedly stated in [1] and [2] that the observed frequency
drift over a 20 year period was always negative with the value:
(4) dF/dt = -6x10^-9Hz/sec (-1.5 Hz in 8 years)
It is at this point that a polarity-switch occurred when following the
statement defined the (negative!) clock acceleration of dF/Fdt as
(5) -A_t = -2.x10^-18s/s^2 instead of
(6) A_t = -2.x10^-18s/s^2
so that by incorporating two negatives we get the incorrect
(8X) A_p = A_t*c with the result that a negative A_t incorrectly
forces a negative acceleration of -8.5e-8cm/s^. The true value should
be positive:
(8) A_p = -A_t*c = +8.5 x 10^-8 cm/s^2, positive acceleration not
negative. There seems little latitude for interpretation.
There is a curious footnote in [2] pg.3 that says dF "is positive for
a spacecraft receding from the tracking station (red shift), negative
for approaching" (Note: both interpretations are wrong) "just the
opposite of the usual convention". The last remark does nothing to
clarify the situation, but it seems algebraically impossible to
interpret the negatively increasing dF/dt as an indication of negative
acceleration.
This topic was discussed before but I bring it up again, so we can
finally settle it. It is natural to "want" some kind of sun gravity,
and certainly not sun repulsion, but a hard look at the algebra raises
serious questions. In my and others' analysis, the sun has no part in
the phenomenon and there's no need to "make nice" with the sun.
John Polasek
http://www.dualspace.net
Thomas Smid
I think the problem is merely one of unfortunate and misleading
notation rather than a genuine error:
if you look at footnote #38 in Anderson et al. 2002
(http://arxiv.org/abs/gr-qc/0104064 ) it is clear that
the basic frequency difference is actually defined as f0-f rather than
f-f0 (where f0 is the transmitted frequency and f the received
frequency). Note that this frequency difference is positive for a
redshift (f<f0).
Now the (two-way) difference between the observed frequency difference
f0-f and the theoretical frequency difference f0-f' is then related to
the excess accelaration 'a' by
(1) (f0-f)-(f0-f') = -2*f0*t/c*a
or
(2) f'-f = -2*f0*t/c*a
The acceleration 'a' is defined such that it is positive if directed
towards the sun, so in this case the right hand side is negative and
hence it is required that f>f' on the left hand side, which corresponds
to an excess blue-shift (which is consistent with the assumption that
the acceleration is towards the sun). In Eq.(15) in the above mentioned
paper they write f-f' rather than f'-f, so this may be a typo or
(probably more likely) just an unfortunate notation as they actually
were referring to (1) rather than (2) (i.e. to df-df' rather than
f'-f).
It is rather annoying that at this level they can't even manage to
produce a consistent notation throughout the paper, but I think the
problem here is just this and has nothing to do with a sign switch in
the data analyis itself.
There are in my opinion actually more important inconsistencies in
statements made in these papers, which suggest that the theoretical
modelling of the data is actually anything but clear-cut. At the
beginning of chapter 2.1 of the Turyshev paper
(http://xxx.sf.nchc.gov.tw/abs/gr-qc/9903024 ) they give the
acceleration due to radiation pressure at 20 AU as 5*10^-8 cm/sec^2 ( a
value which is also mentioned in the Anderson paper), but towards the
bottom of the same chapter it suddenly says that 'at distances >10-15
AU it (the radiation pressure) produces an acceleration that is much
less then 8*10^-8 cm/sec^2'. Now, 5 is certainly not much less then 8,
and taking also into account that the statistical error of the data is
about 2*10^-8 cm/sec^2 (as mentioned in the Anderson paper), this means
that errors in the modelling of the radiation pressure force could
completely alter the data. Additionally, as I have pointed out on my
webpage http://www.physicsmyths.org.uk/pioneer.htm , signal propagation
effects (due to an incorrect application of the principle of the
invariance of the speed of light) would also lead to a contribution of
the same order of magnitude. A combination of all these factors could
well explain the data.
Thomas
frank...@yahoo.com
John C. Polasek wrote:
> The Pioneer 10 anomaly is a problem that is far from settled, but the
> question of red shift or blue shift remains. I try to show that the
> phenomenon is a red shift, acceleration away from the sun. This is a
> matter that should be cleared up.
I think you make an excellent and very significant claim the the
Pioneer 10 anomaly is actually a red-shift instead of the usually
claimed blue shift. I was wondering if you have compared the Pioneer
red-shift with the red-shift associated with the expanding universe.
If we are observing red-shift, then there could be 1 of 2 causes.
Either the spacecraft is accelerating away faster than we expect (the
normally accepted explanation) or there is something intrinsic about
space which is causing the signal to become red-shifted as it travels
through space and Pioneer 10 is exactly where it is supposed to be.
This might lend some credibility to the "tired light" hypothesis if the
rates of red-shift for Pioneer 10 just happened to line up with the
red-shift observed from distant stars. Such a correlation would be
evidence that the red shift is not caused by an expanding universe, but
rather, some unexplained intrinsic property of space.
I had previously thought about how the Pioneer 10 anomoly might be used
to help test the tired light hypothesis, but I had initialy ruled it
out since the anomaly appeared to be a blue-shift instead of the
expected red-shift. But your reasoning puts it back as a red-shift
which makes the test viable.
If anybody has a comparison of Pioneer 10 red-shift compared to
galactic red-shift caused by the expansion of the universe, that would
be interesting to see.
fhuredshift
George Dishman
<frank...@yahoo.com> wrote in message
news:1129959719.6...@z14g2000cwz.googlegroups.com...
John is wrong it is a blue shift. Look at the graph showing the
sum of the anomaly and the solar radiation pressure. Solar
radiation pushes the craft away from the sun adding a red shift,
but the total of that and the anomaly goes through zero showing
the anomaly is a blue shift. This was also confirmed by Craig
Markwardt who re-analysed the data from scratch and found the
same result, the frequency is higher than expected.
> If anybody has a comparison of Pioneer 10 red-shift compared to
> galactic red-shift caused by the expansion of the universe, that would
> be interesting to see.
This was done to death in the group a few years ago. Numerically
the acceleration appears close to the Hubble value.
a_p ~ c H
However, tired light says the red shift depends on distance so the
key question is the rate at which the distance is increasing. The
formula is then
a_tl = 2 v H
where v is the speed of the craft and the factor of 2 is because
the shift applies on both uplink and downlink.
Bottom line is that the anomaly is about four orders of magnitude
larger than tired light as well as being in the wrong direction.
George
Hero.van...@gmx.de
George Dishman wrote:
> <frank...@yahoo.com> wrote in message
> news:1129959719.6...@z14g2000cwz.googlegroups.com...
> >
> > John C. Polasek wrote:
> >> The Pioneer 10 anomaly is a problem that is far from settled, but the
> >> question of red shift or blue shift remains. I try to show that the
> >> phenomenon is a red shift, acceleration away from the sun. This is a
> >> matter that should be cleared up.
> sum of the anomaly and the solar radiation pressure. Solar
> radiation pushes the craft away from the sun adding a red shift,
> but the total of that and the anomaly goes through zero showing
> the anomaly is a blue shift. This was also confirmed by Craig
> Markwardt who re-analysed the data from scratch and found the
> same result, the frequency is higher than expected.
>
> > If anybody has a comparison of Pioneer 10 red-shift compared to
> > galactic red-shift caused by the expansion of the universe, that would
> > be interesting to see.
>
Blue shift- that means Pioneer is coming back ?
Hero
Dishman
Hero.van...@gmx.de wrote:
> George Dishman wrote:
> > ... the anomaly is a blue shift. This was also confirmed by Craig
> > Markwardt who re-analysed the data from scratch and found the
> > same result, the frequency is higher than expected.
> Blue shift- that means Pioneer is coming back ?
The anomaly is a blue shift. It suggests the craft is being
slowed with constant acceleration. The cumulative loss of
speed was around 100 mm/s over 8 years.
Overall, the shift is mostly red. The craft is leaving the
Solar system at about 12 km/s but the Earth's speed in orbit
is around 30 km/s so when we are moving towards it the shift
is blue (for about 19 weeks, up to 18 km/s) and when Earth is
moving the other way it is red (up to 42 km/s).
George
OmegaNumber
> Blue shift- that means Pioneer is coming back ?
> Hero
Not quite :j
Spud
cma...@yahoo.com
No, it's a tiny unexpected blue shift added to the much bigger expected
red shift. So the net result is still a red shift, the Pioneers are not
coming back. The blue shift is said to be "unmodeled".
Chris
Hero.van...@gmx.de
> Hero.van...@gmx.de wrote:
> > George Dishman wrote:
> > > ... the anomaly is a blue shift.
>
> The anomaly is a blue shift. It suggests the craft is being
> slowed with constant acceleration. The cumulative loss of
> speed was around 100 mm/s over 8 years.
>
> Overall, the shift is mostly red. The craft is leaving the
> Solar system at about 12 km/s but the Earth's speed in orbit
> is around 30 km/s so when we are moving towards it the shift
> is blue (for about 19 weeks, up to 18 km/s) and when Earth is
> moving the other way it is red (up to 42 km/s).
>
of pioneer, we observe red-shift. But it is a little bit less, what
could be calculated from the continuation of an undisturbed hyperbolic
orbit.
But, we have this right angle movement twice a year, with a distance of
14 light-minutes = two astronomical units in between, and that means a
lot of radiation from sun in between. This should shift the frequency
too (isn't it?) and what is observed in this aspect?
Thanks
Hero
PS. Posting with google-usenet claims, that my message will appear in
the unmoderated groups momentarily and in
sci.physics.research may be later.
But already twice it didn't appear in the unmoderated groups, so now i
took sci.physics reasearch out of the adress.
Jan Panteltje
Hero.van...@gmx.de wrote in
<1130247705.9...@z14g2000cwz.googlegroups.com>:
>But, we have this right angle movement twice a year, with a distance of
>14 light-minutes = two astronomical units in between, and that means a
>lot of radiation from sun in between. This should shift the frequency
>too (isn't it?)
One EM frequency does not affect an other.
One radio station at a specific frequency will NOT change frequency if
an other one comes on line, or noise is present.
_________________________________________
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Hero.van...@gmx.de
angles moving with respect to the movement of pioneer one observes a
red-shift. But not as much as could be expected from the formulas of a
hyperbolic orbit.
But observing it at right angles, this opportunity comes twice a year
and there's a difference between these two, it is a length of ca. 14
lightminutes= 2 astronimical units in between and this is filled with
lots of radiation from the sun. This should show up in the observation
of shift, shouldn't it ?
Thanks
Hero
Thomas Smid
> John is wrong it is a blue shift. Look at the graph showing the
> sum of the anomaly and the solar radiation pressure. Solar
> radiation pushes the craft away from the sun adding a red shift,
> but the total of that and the anomaly goes through zero showing
> the anomaly is a blue shift.
George, you are missing John's point. If the sign of the anomalous
force has been switched by mistake, then it would of course in reality
have the same sign as the radiation pressure force.
Not that I think this is the case though. As already mentioned in my
post above, the definitions in the papers (i.e. Eq.(15) and foonote #38
in Anderson et al. 2002 (http://arxiv.org/abs/gr-qc/0104064 ) are
simply ambiguous and do strictly speaking neither allow the conclusion
of a red shift nor a blue shift. Of course it has to be one of those,
and judging from the more basic definition (footnote #38) it should be
a blue shift. This is also confirmed by a statement on page 10 of this
paper (top left): "the JPL Doppler records are not frequency
measurements. Rather, they are digitally counted measurements of the
Doppler phase difference between the transmitted and received S-Band
frequencies, divided by the count time". In other words, they are
actually not comparing the measured and theoretical frequencies
(because they don't actually have the former) but they are comparing
the measured and theoretical red shifts. Eq.(15) in Anderson's paper is
therefore written incorrectly: it should not be [f(obs)-f(model)] but
[df(obs)-df(model] where df(obs)=f0-f(obs) and df(model)=f0-f(model)
(with f0 being the transmitted frequency and f(obs) and f(model) the
actual and theoretical received frequency respectively). This means
that if [df(obs)-df(model] is negative (as they say), the red shift
df(obs) is red shifted but the frequency f(obs) itself is blue shifted.
Another remark regarding the radition pressure though. In the above
quoted paper they mention also data with regard to the Galileo and
Ulysses spacecraft which seem to yield very similar results regarding
the anomalous acceleration. However, in these cases they mention that
this effect is strongly correlated to the radiation pressure (with
correlation coefficients of 0.99 and 0.89 respectively), which very
much suggests that the radiation pressure has been incorrectly
modelled. Given this fact, I find it quite astonishing that this point
is not examined in more detail for the Pioneer data. At least for
distance up to 40 AU or so the radiation pressure should still be
significant according to their own figures, so at least part of the
discrepancy could be removed by re-modelling the radiation pressure
force.
Thomas
Andrew P.
graph of positive solar pressure As, together with negative unmodeled
acceleration Ap, and their sum vs. distance in AU.
At first it looks compelling that the two accelerations are opposite in
polarity but it appears you could also draw Fig. 6 with positive Ap and no
harm done. After all it is foregone that As is positive, and whoever drew
this very likely had their mind made up for negative Ap.
Neither is there any big conclusion to be drawn when the graph shows the two
forces providentially totalling to a null at 12-13 AU, since real analysis
did not begin till As became negligible at 20+ AU, because (zero) As would
have no effect either on +Ap or on -Ap.
What would be convincing would be to determine whether both functions were
added/subtracted into the computer equations while retaining the same
contrary sense. The drawing alone is subject to "adjustment" and in my
opinion, the red/blue resolution is still open.
John P.
<Hero.van...@gmx.de> wrote in message
news:1130098828.1...@g44g2000cwa.googlegroups.com...
George Dishman
"Andrew P." <and...@att.net> wrote in message
news:BSs7f.498850$5N3....@bgtnsc05-news.ops.worldnet.att.net...
> George, thank you for pointing out Fig. 6 in gr-qc/0104064 v5 that is a
> graph of positive solar pressure As, together with negative unmodeled
> acceleration Ap, and their sum vs. distance in AU.
>
> At first it looks compelling that the two accelerations are opposite in
> polarity but it appears you could also draw Fig. 6 with positive Ap and no
> harm done. After all it is foregone that As is positive, and whoever drew
> this very likely had their mind made up for negative Ap.
>
> Neither is there any big conclusion to be drawn when the graph shows the
> two forces providentially totalling to a null at 12-13 AU, since real
> analysis did not begin till As became negligible at 20+ AU, because (zero)
> As would have no effect either on +Ap or on -Ap.
>
> What would be convincing would be to determine whether both functions were
> added/subtracted into the computer equations while retaining the same
> contrary sense. The drawing alone is subject to "adjustment" and in my
> opinion, the red/blue resolution is still open.
You can also look at the text immediately preceding the graph:
"Here, v_0 is the reference frequency, the factor 2 is
because we use two- and three-way data [36]. v_model
is the modeled velocity of the spacecraft due to the
gravitational and other large forces discussed in
Section IV. (This velocity is outwards and hence
produces a red shift.) We have already included the
sign showing that aP is inward. (Therefore, a_P produces
a slight blue shift on top of the larger red shift.)"
Or the heading of figure 7 which says "Acceleration Directed
Toward the Sun.", a statement also made in the first sentence
of the abstract.
You can also look at Table II, the Error Budget on page 43 where
the reaction force of the radio beam which points at Earth is
included as +1.10 and provides the majority of the bias of +0.90.
That has the effect of increasing the anomaly through eqn (53) on
the previous page from 7.84 to 8.74. If the anomaly were a red
shift, it would need to be negative and reduce a_p compared to
a_p(exper). You might need to think about that one a little ;-)
The direction is also obvious when you consider that a lot of
work has been done on the possibility that the anomaly is caused
by heat reflected off the back og the HGA into deep space.
George
George Dishman
<Hero.van...@gmx.de> wrote in message
news:1130228790.9...@g44g2000cwa.googlegroups.com...
>
> Dishman wrote:
>> Hero.van...@gmx.de wrote:
>> > George Dishman wrote:
>> > > ... the anomaly is a blue shift. This was also confirmed by Craig
>> > > Markwardt who re-analysed the data from scratch and found the
>> > > same result, the frequency is higher than expected.
>>
>> > Blue shift- that means Pioneer is coming back ?
>>
>> The anomaly is a blue shift. It suggests the craft is being
>> slowed with constant acceleration. The cumulative loss of
>> speed was around 100 mm/s over 8 years.
>>
>> Overall, the shift is mostly red. The craft is leaving the
>> Solar system at about 12 km/s but the Earth's speed in orbit
>> is around 30 km/s so when we are moving towards it the shift
>> is blue (for about 19 weeks, up to 18 km/s) and when Earth is
>> moving the other way it is red (up to 42 km/s).
>>
>> George
> Thanks George, i've got it. So, when observed, when earth is at right
> angles moving with respect to the movement of pioneer one observes a
> red-shift.
Not quite. This diagram shows the sun S and the craft P and
Earth at various positions in its orbit, a through f. (You
will need to use a fixed-pitch font like Courier to see it
properly.)
d
e S c P -> 12km/s
f b
a
-> 30km/s
Pioneer should be much father to the right but it would be off
your screen.
Pioneer is moving at 12km/s almost directly away from the Sun.
When Earth is at point a, it is moving towards Pioneer at about
30km/s so is catching it up at 18km/s and that motion produces
a blue shift. At point d it is moving away from the craft so
the two motions give a total rate of separation of 42km/s and
we get a large red shift. At points c and e, the Earth's motion
is transverse so we see only the red shift due to the craft's
speed of 12km/s. At points f and b, the motion of the Earth is
about 12km/s towards the craft so the total is no shift.
On top of all that, there is a tiny, inexplicable blue shift
at all times suggesting something is slowing the craft. Drag
from dust would produce the right effect but we know the dust
is too thin to explain the size of the anomaly.
> But not as much as could be expected from the formulas of a
> hyperbolic orbit.
> But observing it at right angles, this opportunity comes twice a year
> and there's a difference between these two, it is a length of ca. 14
> lightminutes= 2 astronimical units in between and this is filled with
> lots of radiation from the sun. This should show up in the observation
> of shift, shouldn't it ?
Radiation from the sun pushes the craft at all times so adds a
little red shift, but that was smaller than the anomaly for the
period in question, it was so far from the Sun.
George
Andrew P.
<Hero.van...@gmx.de> wrote in message
news:1130098828.1...@g44g2000cwa.googlegroups.com...
"To George Dishman:
You mention the Table II Error budget on pg. 43, but it seems to shore up my
red shift/recession argument.
The radio beam force b.xp is mentioned at eq. 33 page 32. mentioning that
the antenna power reaction b.xp is "away from the Sun". It's logical.
On pg. 42 @ Eq. (53,54) they evaluate b.xp = 1.1e-8cms2 and "It makes the
Pioneer effect larger". Therefore Ap and b.xp are of the same sign, implying
outward acceleration as far as I can see.
In addition, they state on pg. 42 that Aheat = -0.55e-8cms2 for heat off
the back, indicating inward acceleration and using a proper negative sign.
Doesn't all of that indicate that Ap is outward acceleration and therefore,
increasing red shift?
How can this be misinterpreted?"
JP
Craig Markwardt
Thomas Smid <thoma...@gmail.com> writes:
> George Dishman wrote:
>
> > John is wrong it is a blue shift. Look at the graph showing the
> > sum of the anomaly and the solar radiation pressure. Solar
> > radiation pushes the craft away from the sun adding a red shift,
> > but the total of that and the anomaly goes through zero showing
> > the anomaly is a blue shift.
>
> George, you are missing John's point. If the sign of the anomalous
> force has been switched by mistake, then it would of course in reality
> have the same sign as the radiation pressure force.
The sign of the anomalous acceleration has not been switched by
mistake. I performed an independent confirmation of the Anderson et
al results, all using the "usual" frequency conventions. My results
matched Anderson et al's results quite closely.
The anomalous acceleration is consistent with all these interpretations:
* a slightly higher received frequency than expected;
* a slightly "bluer" frequency shift than expected;
* a slightly faster receding craft than expected.
There is no room for a directional sign error. The doppler shifts due
to planetary motion+rotation, spacecraft rotation, etc. are very large
(hundreds of thousands of Hertz), and a sign error would have been
obvious.
> Not that I think this is the case though. As already mentioned in my
> post above, the definitions in the papers (i.e. Eq.(15) and foonote #38
> in Anderson et al. 2002 (http://arxiv.org/abs/gr-qc/0104064 ) are
> simply ambiguous and do strictly speaking neither allow the conclusion
> of a red shift nor a blue shift. Of course it has to be one of those,
> and judging from the more basic definition (footnote #38) it should be
> a blue shift. This is also confirmed by a statement on page 10 of this
> paper (top left): "the JPL Doppler records are not frequency
> measurements. Rather, they are digitally counted measurements of the
> Doppler phase difference between the transmitted and received S-Band
> frequencies, divided by the count time". In other words, they are
> actually not comparing the measured and theoretical frequencies
> (because they don't actually have the former) but they are comparing
> the measured and theoretical red shifts. Eq.(15) in Anderson's paper is
> therefore written incorrectly: it should not be [f(obs)-f(model)] but
> [df(obs)-df(model] where df(obs)=f0-f(obs) and df(model)=f0-f(model)
> (with f0 being the transmitted frequency and f(obs) and f(model) the
> actual and theoretical received frequency respectively). This means
> that if [df(obs)-df(model] is negative (as they say), the red shift
> df(obs) is red shifted but the frequency f(obs) itself is blue shifted.
Your "in other words" interpretation is incorrect. The observable is
the number of accumulated carrier cycles, which when differentiated
gives the measured frequency.
Your interpretation of the frequency conventions is erroneous. Please
see:
http://groups.google.com/group/sci.physics/msg/9b1b52545dab08fd?hl=en&
and the manual DSN 810-005.
> Another remark regarding the radition pressure though. In the above
> quoted paper they mention also data with regard to the Galileo and
> Ulysses spacecraft which seem to yield very similar results regarding
> the anomalous acceleration. However, in these cases they mention that
> this effect is strongly correlated to the radiation pressure (with
> correlation coefficients of 0.99 and 0.89 respectively), which very
> much suggests that the radiation pressure has been incorrectly
> modelled. Given this fact, I find it quite astonishing that this point
> is not examined in more detail for the Pioneer data. At least for
> distance up to 40 AU or so the radiation pressure should still be
> significant according to their own figures, so at least part of the
> discrepancy could be removed by re-modelling the radiation pressure
> force.
However, the anomalous acceleration remains constant, while the
radiation pressure decreases as 1/r^2 ~ 1/t^2. The anomalous
acceleration remains even after the solar radiation pressure declines
to insignificance.
Correlated parameters do not mean that the radiation pressure has been
incorrectly modeled. Rather, it is a statistical statement about how
well the data can distiguish between two fitted parameters in a given
model.
It sounds like you need to do a little more research before making
unsubstantiated claims.
CM
Jonathan Silverlight
<geo...@briar.demon.co.uk> writes
>
>The direction is also obvious when you consider that a lot of
>work has been done on the possibility that the anomaly is caused
>by heat reflected off the back og the HGA into deep space.
>
That's not a very good argument :-)
The argument is that because the direction is toward the sun, the cause
may be heat reflected from the back of the HGA, not vice versa.
--
Boycott Yahoo!
Remove spam and invalid from address to reply.
George Dishman
"Jonathan Silverlight" <jsilve...@spam.merseia.fsnet.co.uk.invalid> wrote
in message news:m2iZ+jKi...@merseia.fsnet.co.uk...
> In message <djm7ik$cp$1...@news.freedom2surf.net>, George Dishman
> <geo...@briar.demon.co.uk> writes
>>
>>The direction is also obvious when you consider that a lot of
>>work has been done on the possibility that the anomaly is caused
>>by heat reflected off the back og the HGA into deep space.
>>
>
> That's not a very good argument :-)
> The argument is that because the direction is toward the sun, the cause
> may be heat reflected from the back of the HGA, not vice versa.
My argument is that, if the anomalous frequency shift
was a reduction, Anderson et al would have pointed
out that heat reflected from the back of the antenna
would have the opposite effect to what they observed
and saved everyone the effort, including themselves.
The fact that they take this seriously and have studied
samples for differential deterioration means it is a
credible explanation.
George
N:dlzc D:aol T:com (dlzc)
"Craig Markwardt" <crai...@REMOVEcow.physics.wisc.edu> wrote in
message news:onbr1dn...@cow.physics.wisc.edu...
..
> The anomalous acceleration is consistent with all these
> interpretations:
> * a slightly higher received frequency than expected;
> * a slightly "bluer" frequency shift than expected;
> * a slightly faster receding craft than expected.
Did you mean to say the last point here? A blue (actually less
red) shift would be a slightly slower receding craft than
expected, wouldn't it?
David A. Smith
Thomas Smid
> Thomas Smid <thoma...@gmail.com> writes:
>
> > George Dishman wrote:
> >
> > > John is wrong it is a blue shift. Look at the graph showing the
> > > sum of the anomaly and the solar radiation pressure. Solar
> > > radiation pushes the craft away from the sun adding a red shift,
> > > but the total of that and the anomaly goes through zero showing
> > > the anomaly is a blue shift.
> >
> > George, you are missing John's point. If the sign of the anomalous
> > force has been switched by mistake, then it would of course in reality
> > have the same sign as the radiation pressure force.
>
>
> The sign of the anomalous acceleration has not been switched by
> mistake. I performed an independent confirmation of the Anderson et
> al results, all using the "usual" frequency conventions. My results
> matched Anderson et al's results quite closely.
>
> The anomalous acceleration is consistent with all these interpretations:
> * a slightly higher received frequency than expected;
> * a slightly "bluer" frequency shift than expected;
> * a slightly faster receding craft than expected.
A faster receding craft would correspond to a frequency red shift not a
blue shift. This holds independently of any convention.
The frequencies are not open to interpretations or conventions. As you
mentioned yourself in the referenced thread, they are by definition
always positive. Hence if you have the difference [f(obs)-f(model)]
and find that this is negative, you would have to conclude that
f(obs)<f(model) i.e. the observed frequency would have an excessive
redshift. Therefore, if you claim there is a blueshift, Eq.(15) in
Anderson's paper must be incorrect. It should read [f(model)-f(obs)] or
alternatively [df(obs)-df(model)] (using the DSN conventions for df).
>
> > Another remark regarding the radition pressure though. In the above
> > quoted paper they mention also data with regard to the Galileo and
> > Ulysses spacecraft which seem to yield very similar results regarding
> > the anomalous acceleration. However, in these cases they mention that
> > this effect is strongly correlated to the radiation pressure (with
> > correlation coefficients of 0.99 and 0.89 respectively), which very
> > much suggests that the radiation pressure has been incorrectly
> > modelled. Given this fact, I find it quite astonishing that this point
> > is not examined in more detail for the Pioneer data. At least for
> > distance up to 40 AU or so the radiation pressure should still be
> > significant according to their own figures, so at least part of the
> > discrepancy could be removed by re-modelling the radiation pressure
> > force.
>
> However, the anomalous acceleration remains constant, while the
> radiation pressure decreases as 1/r^2 ~ 1/t^2. The anomalous
> acceleration remains even after the solar radiation pressure declines
> to insignificance.
>
> Correlated parameters do not mean that the radiation pressure has been
> incorrectly modeled. Rather, it is a statistical statement about how
> well the data can distiguish between two fitted parameters in a given
> model.
This is a rather strange interpretation of statistical correlations.
Usually you consider statistical correlations exactly in order to
examine the possibility of physical correlations. If you find almost a
100% statistical correlation and still deny any physical connection,
you should have very good reasons to to so. However, I could not find
any explanation in Anderson's paper as to where this correlation should
come from if not from the radiation pressure term in their model. What
they should have done is to re-model the radiation pressure force such
that the anomalous acceleration for the Galileo and Ulysses data
disappears, and then use this modified model also on the Pioneer data.
I am not saying that for the latter the resultant anomalous
acceleration would also be zero, but it should change significantly and
hence call for different explanations (a non-constant acceleration
would for instance make the interpretation as a recoil effect due to
outgassing etc. very much unlikely and may also invalidate other
speculative theories in this respect).
Thomas
George Dishman
>"George Dishman" <geo...@briar.demon.co.uk> wrote in message
>news:djm7ik$cp$1...@news.freedom2surf.net...
>>
>> You can also look at Table II, the Error Budget on page 43 where
>> the reaction force of the radio beam which points at Earth is
>> included as +1.10 and provides the majority of the bias of +0.90.
>> That has the effect of increasing the anomaly through eqn (53) on
>> the previous page from 7.84 to 8.74. If the anomaly were a red
>> shift, it would need to be negative and reduce a_p compared to
>> a_p(exper). You might need to think about that one a little ;-)
"Andrew P." <and...@att.net> wrote in message
news:chB7f.501279$5N3.1...@bgtnsc05-news.ops.worldnet.att.net...
>
> "To George Dishman:
> You mention the Table II Error budget on pg. 43, but it seems to shore up
> my red shift/recession argument.
>
> The radio beam force b.xp is mentioned at eq. 33 page 32. mentioning that
> the antenna power reaction b.xp is "away from the Sun". It's logical.
Yes, since the beam points towards us.
> On pg. 42 @ Eq. (53,54) they evaluate b.xp = 1.1e-8cms2 and "It makes the
> Pioneer effect larger". Therefore Ap and b.xp are of the same sign,
> implying outward acceleration as far as I can see.
>
> In addition, they state on pg. 42 that Aheat = -0.55e-8cms2 for heat off
> the back, indicating inward acceleration and using a proper negative sign.
>
> Doesn't all of that indicate that Ap is outward acceleration and
> therefore, increasing red shift?
Not quite, it is more subtle. In units of 10^-8 cm/s^2,
the radio beam produces +1.10 and the heat -0.55 as you
say. The total bias including other effects is +0.90.
Since we know the radio beam produces a red shift and
it is has a positive value while the reflected heat
produces a blue shift and has a negative value, the
positive total of +0.90 must also represent a red shift.
The measured value ("experimental") has a magnitude of
7.84. If that was a red shift, they would then say that
part of that could be explained by the bias and reduce
it by 0.90 to get a value for the anomaly of 6.94.
Instead they add the 0.90 to increase the value to 8.74
because the anomaly first has to cancel out the red shift
from the radio beam and then there is an additional 7.84
which is measurable.
Basically, they say "It makes the Pioneer effect larger"
because the anomalous blue shift of 8.74 includes both
the actual measured blue shift of 7.84 and enough extra
to overcome the known red shift of 0.90 from the beam
and other biases.
> How can this be misinterpreted?"
Very easily, hence the wink when I said "You might need
to think about that one a little."
George
mrdualspace
It is simple to prove that Pioneers' acceleration would be +8.74 e-10
m/s^2 or AWAY from the sun for a red shift, not a blue shift.
All the necessary facts can be found on just pages 2 and 3 in
gr-qc/99003024 (Turyshev et al).
First, the footnote on pg. 3 affirms that they (JPL and DSN) use the
standard and proper convention for Doppler frequency shift:
df = f - f0
Secondly, also on page 3, primary data is cited showing that the
received frequency has a constant negative drift with time:
df/dt = -6.152 e-9Hz/s = fdot (tuned from -6)
We can get the proper relationship between fdot and acceleration by
taking the derivative of their (correct) Eq. 1 on page 2:
fdot = -f0*Ap/c f0 = 2.11Ghz
so Ap = -c*fdot/f0 = +8.74x10^-10 m/s^2
Therefore, the supposed Pioneer acceleration is away from the sun,
contrary to popular interpretation. It should be construed as a red
shift phenomenon. This conclusion is based on a sufficient few
irrefutable facts, both experimental and kinematic.
The declaration (footnote pg. 3) that the red or blue shifts will be
interpreted inversely should be ignored and is not needed in an
independent analysis.
John Polasek
cma...@yahoo.com
about the conclusions. But I find it hard to think about the enigma,
because of all the technical details that are explicitly or implicitly
mentioned. If you had to pose the Pioneer problem concisely in a
textbook so that students could start thinking about it, how would you
formulate it?
Chris
George Dishman
"mrdualspace" <jpol...@cfl.rr.com> wrote in message
news:1130377703....@f14g2000cwb.googlegroups.com...
>
> It is simple to prove that Pioneers' acceleration would be +8.74 e-10
> m/s^2 or AWAY from the sun for a red shift, not a blue shift.
Yes, there is no doubt that motion away from the Sun
would produce a reduced frequency which could properly
be described as a red shift, and I don't think anyone
contributing to this thread would disagree.
From http://www.arxiv.org/abs/gr-qc/0104064, at the
bottom of page 19, you will also find the following
statement confirming what you say:
"v_model is the modeled velocity of the spacecraft due to
the gravitational and other large forces discussed in
Section IV. (This velocity is outwards and hence produces
a red shift.)"
That is immediately followed by:
"We have already included the sign showing that a_P is
inward. (Therefore, a_P produces a slight blue shift
on top of the larger red shift.)"
So there you have it in black and white, the anomaly looks
like an acceleration which is "inward" while the velocity
of the craft is "outward". The anomaly is seen as a blue
shift while the speed produces a larger red shift.
> All the necessary facts can be found on just pages 2 and 3 in
> gr-qc/99003024 (Turyshev et al).
One zero too many: http://www.arxiv.org/abs/gr-qc/9903024
> First, the footnote on pg. 3 affirms that they (JPL and DSN) use the
> standard and proper convention for Doppler frequency shift:
>
> df = f - f0
That note caused a lot of confusion because df = f - f_0
should be negative for a red shift since f < f_0 but the
DSN convention is that it is positive for a red shift. The
equation should have been df = f_0 - f. The note states
specifically this is "just the opposite of the usual
convention".
This error was corrected in note [38] of the more recent
paper http://www.arxiv.org/abs/gr-qc/0104064 :
"The JPL and DSN convention for Doppler frequency shift
is (dv)DSN = v_0 ? v, where v is the measured frequency
and v_0 is the reference frequency. It is positive for
a spacecraft receding from the tracking station (red
shift), and negative for a spacecraft approaching the
station (blue shift), just the opposite of the usual
convention."
> Secondly, also on page 3, primary data is cited showing that the
> received frequency has a constant negative drift with time:
>
> df/dt = -6.152 e-9Hz/s = fdot (tuned from -6)
And as the footnote stated "negative for a spacecraft
approaching the station (blue shift)".
> We can get the proper relationship between fdot and acceleration by
> taking the derivative of their (correct) Eq. 1 on page 2:
>
> fdot = -f0*Ap/c f0 = 2.11Ghz
>
> so Ap = -c*fdot/f0 = +8.74x10^-10 m/s^2
>
> Therefore, the supposed Pioneer acceleration is away from the sun,
> contrary to popular interpretation. It should be construed as a red
> shift phenomenon.
No, the footnote specifically states that negative numbers
imply a blue shift and "approaching the station".
> This conclusion is based on a sufficient few
> irrefutable facts, both experimental and kinematic.
> The declaration (footnote pg. 3) that the red or blue shifts will be
> interpreted inversely should be ignored and is not needed in an
> independent analysis.
On the contrary, it is precisely because the DSN uses a
convention which is the opposite of the normal that they
added that comment. It's a shame they got the equation
the wrong way round but the more recent paper fixes that.
George
Craig Markwardt
Sorry, a typo.
* a slightly slower receding craft than expected.
Craig
--
--------------------------------------------------------------------------
Craig B. Markwardt, Ph.D. EMAIL: crai...@REMOVEcow.physics.wisc.edu
Astrophysics, IDL, Finance, Derivatives | Remove "net" for better response
--------------------------------------------------------------------------
John C. Polasek
first principles. I refuse to be drawn into the quagmire caused by
line 2 of the footnote on page 3 of
http://www.arxiv.org/abs/gr-qc/0104064.
Line 1 is fine. Df = f - f0 as everyone will agree. It gives the proper
sense to the experimental finding of fdot = -6e-9 Hz/sec.
The 2nd line of the footnote is simply a gratuitous notice of the
writer's intention to upend the usual velocity/frequency/redshift
relationships for whatever reason. This transgression can easily be
sidestepped since, using just first principles, we can differentiate
Eq. 1 and solve for Ap. These 2 steps are compulsory and also
mathematically immaculate.
The result, as I showed, is that the Pioneers do not accelerate
"toward the sun" but away from the sun. Ap = +8.4e-10ms2 which
indicates a red shift. The musings of line 2 only give rise to
bewilderment.
To say that a later paper "fixes" the frequency thing; it can't.
Standards are standards. The JPL and DSN programs must see wide
application in every sort of project and it would be a stretch of logic
to think they would they perversely choose to use an upside down
convention that would need explanation at every instance, when it is so
easy to do it the right way.
My theory shows there is no actual acceleration, but is a cosmic time
phenomenon. I will present a modified version explaining the Pioneer
anomaly on my website in light of certainty of this finding.
John Polasek
http://www.dualspace.net
George Dishman
"John C. Polasek" <jpol...@cfl.rr.com> wrote in message
news:1130443425.8...@g49g2000cwa.googlegroups.com...
> George, I tried to point out that the whole thing can be solved from
> first principles. I refuse to be drawn into the quagmire caused by
> line 2 of the footnote on page 3 of
> http://www.arxiv.org/abs/gr-qc/0104064.
> Line 1 is fine. Df = f - f0 as everyone will agree.
No John, nobody who knows the DSN convention will agree
with that. The Deep Space Network uses df = f_0 - f, the
opposite of what you expect. See reference 38 in that
paper.
> It gives the proper
> sense to the experimental finding of fdot = -6e-9 Hz/sec.
>
> The 2nd line of the footnote is simply a gratuitous notice of the
> writer's intention to upend the usual velocity/frequency/redshift
> relationships for whatever reason.
The reason is that the DSN software which has been in use
for decades reports the results that way. It isn't talking
of an intention but a fait acomplis which they carry across
into their paper rather than cause confusion by inverting
the DSN standard.
> This transgression can easily be
> sidestepped since, using just first principles, we can differentiate
> Eq. 1 and solve for Ap. These 2 steps are compulsory and also
> mathematically immaculate.
Differentiating doesn't help since the negative sign also
negates the derivative.
> The result, as I showed, is that the Pioneers do not accelerate
> "toward the sun" but away from the sun. Ap = +8.4e-10ms2 which
> indicates a red shift. The musings of line 2 only give rise to
> bewilderment.
>
> To say that a later paper "fixes" the frequency thing; it can't.
> Standards are standards.
The DSN standard is df = f_0 - f, not df = f - f_0. That
is what their software works on and that same software is
common to many missions. They weren't going to introduce
a special convention for Pioneer just because the data was
publicised outside the usual community.
> The JPL and DSN programs must see wide
> application in every sort of project and it would be a stretch of logic
> to think they would they perversely choose to use an upside down
> convention that would need explanation at every instance, when it is so
> easy to do it the right way.
Well for whatever reason that is the way it is. The DSN
documentation makes that clear. I think it is logical since
they are usually interested in craft moving away from the
Earth so a negative frequency shift indicates positive speed
away from Earth.
> My theory shows there is no actual acceleration, but is a cosmic time
> phenomenon. I will present a modified version explaining the Pioneer
> anomaly on my website in light of certainty of this finding.
There are several ways one can make certain including those I
have listed but these statements in the paper may be simplest:
"This velocity is outwards and hence produces a red shift.
... Therefore, a_P produces a slight blue shift on top of
the larger red shift."
George
John C. Polasek
Pioneer, there is an absolutely brilliant Powerpoint show at
giants.stanford.edu/~pln/Texas/Turyshev_Pioneer.pdf. It took great
talent to produce such a nice display.
John Polasek
George Dishman
"John C. Polasek" <jpol...@cfl.rr.com> wrote in message
news:1130469013.1...@g49g2000cwa.googlegroups.com...
Note that on pages 8 and 19, Slava highlights in red
that the direction of the anomaly is towards the Sun.
George
John C. Polasek
As usual we try stick to first principles and avoid the entanglements
previously noted. It seems clear that Doppler frequency is primary data
and conclusions about polarity of acceleration are a result of
intermediate calculation subject to error
Refer to arXiv:general relativity-qc/99003024 v2 9 Mar. 1999, Turyshev
et al. This is one of several Pioneer papers that largely repeat each
other in certain sections and mostly use unsigned numbers, but this
paper is a bit more specific than the others. On page 3, Sec. 2.2 are
described CHASMP results that are independent of JPL's with its
unique and troublesome Doppler signature. Note especially that CHASMP
takes care to specify normal Doppler arithmetic, to wit:
"Without using the apparent acceleration, CHASMP shows a steady
frequency drift of -6 x 10^-9 Hz/s, or 1.5 Hz over 8 years (one way
only). The drift in the Doppler residuals (observed minus computed
data) is clear, definite and cannot be removed without the added
[unsigned] acceleration Ap."
We will show that the added acceleration Ap must be positive.
First of all, with (orthodox) drift as defined above, there are these
relationships from first principles:
df = -f0*V/c = kV where k = -f0/c
df/dt = k*Ap = -6x10^-9 Hz/s = fdot
>From this, the solution for Ap is a positive value:
Ap = fdot/k
Ap = +8.7 x 10^-8 cm/ss
making Ap a positive number, contrary to the prevailing view.
Secondly, as I interpret it, the model seeks to duplicate the observed
data by accumulating all known effects, after which a comparison
finally leaves an unknown. Using k to convert V to frequency, we can
represent the balancing act as follows, first leaving an unknown X as
reported by CHASMP:
Obs(Knowns +X)k - Model(Knowns)k = fdot = -6x10^-9 Hz/s
X*k = fdot, the observed negative drift.
The missing element in the model is
X = fdot/k = -6e-9/(-f0/c) = +8.74 x 10^-8 cm/ss
Again we have agreement that a positive value for Ap must be added to
the model to achieve a null. (Repeatedly in these papers, the addition
Ap has been unsigned).
It seems clear that the range rate acceleration for the Pioneers is
away from the sun, not toward. Of course such a reversal would have a
major on efforts to deduce the cause of the anomaly.
John Polasek
George Dishman
> Here is a further effort to convince you that Ap is a positive number.
> As usual we try stick to first principles and avoid the entanglements
> previously noted.
Then you will get it wrong again. The DSN uses the opposite
convention from everyone else as you are told in the footnote
to which you referred. There are no previous "entanglements",
you are simply ignoring the standard which the DSN has used
for many years.
> It seems clear that Doppler frequency is primary data
> and conclusions about polarity of acceleration are a result of
> intermediate calculation subject to error
> Refer to arXiv:general relativity-qc/99003024 v2 9 Mar. 1999, Turyshev
> et al.
I would prefer that you refer instead to the more recent
paper:
http://www.arxiv.org/abs/gr-qc/0104064
As I have pointed out, the 1999 paper had an error in the
key equation for the Doppler shift which negates the sign.
> This is one of several Pioneer papers that largely repeat each
> other in certain sections and mostly use unsigned numbers, but this
> paper is a bit more specific than the others. On page 3, Sec. 2.2 are
> described CHASMP results that are independent of JPL's with its
> unique and troublesome Doppler signature. Note especially that CHASMP
> takes care to specify normal Doppler arithmetic, to wit:
>
> "Without using the apparent acceleration, CHASMP shows a steady
> frequency drift of -6 x 10^-9 Hz/s, or 1.5 Hz over 8 years (one way
Right, the negative sign indicates a blue shift as stated in the
footnote "It is ... negative for a spacecraft approaching the
station (blue shift), just the opposite of the usual convention."
> only). The drift in the Doppler residuals (observed minus computed
> data) is clear, definite and cannot be removed without the added
> [unsigned] acceleration Ap."
> We will show that the added acceleration Ap must be positive.
>
> First of all, with (orthodox) drift as defined above,
If by "orthodox" you mean df = f_received - f_transmitted
then that is the wrong way round. The DSN uses the opposite
convention, df = f_transmitted - f_received
<snip more based on your erroneous doppler formula>
> It seems clear that the range rate acceleration for the Pioneers is
> away from the sun, not toward. Of course such a reversal would have a
> major on efforts to deduce the cause of the anomaly.
It would, however, you have reversed the sign all the
way through by again neglecting the fact that the DSN
convention is the opposite of what you consider to be
orthodox. I can only again repeat the words of the
authors on page 20:
"This velocity is outwards and hence produces a red
shift. ... Therefore, a_P produces a slight blue
shift on top of the larger red shift."
They then say quite clearly:
"By DSN convention [38], the first of Eqs. (15) is
[dv_obs ? dv_model]_usual = ?[dv_obs ? dv_model]_DSN.
Note the negative sign in front of the second square
bracket. The hyperlink [38] takes you to the explanatory
note:
"[38] The JPL and DSN convention for Doppler frequency
shift is (dv)_DSN = v0 ? v, where v is the measured
frequency and v0 is the reference frequency. It is
positive for a spacecraft receding from the tracking
station (red shift), and negative for a spacecraft
approaching the station (blue shift), just the
opposite of the usual convention, (dv)_usual = v?v_0.
In consequence, the velocity shift, dV = V?V_0, has
the same sign as (dv)_DSN but the opposite sign to
(dv)_usual. Unless otherwise stated, we will use the
DSN frequency shift convention in this paper. We
thank Matthew Edwards for asking us about this."
As long as you continue to ignore the fact that the DSN
protocol reverses the sign, you will continue to get the
acceleration in the wrong direction, contrary to what the
authors say in the abstract and emphasised in red twice
in the later presentation.
George
John C. Polasek
<geo...@briar.demon.co.uk> wrote:
George: (I am having to send this by google.groups as
sci.physics.research has conspired to cut me off at the server. I
cannot post one message in this thread via my news-...@dfl.rr.com) I
have discussed this with Khavine but he denies it.)
>"John C. Polasek" <jpol...@cfl.rr.com> wrote in message
>news:1130520901.8...@g49g2000cwa.googlegroups.com...
>> Here is a further effort to convince you that Ap is a positive number.
>> As usual we try stick to first principles and avoid the entanglements
>> previously noted.
>
>Then you will get it wrong again. The DSN uses the opposite
>convention from everyone else as you are told in the footnote
>to which you referred. There are no previous "entanglements",
>you are simply ignoring the standard which the DSN has used
>for many years.
Don't you understand that I am bypassing DSN for the independent
analysis of CHASMP and am ignoring the upside down Doppler?
>>
>> It seems clear that Doppler frequency is primary data
>> and conclusions about polarity of acceleration are a result of
>> intermediate calculation subject to error
>> Refer to arXiv:general relativity-qc/99003024 v2 9 Mar. 1999, Turyshev
>> et al.
>
>I would prefer that you refer instead to the more recent
>paper:
>
> http://www.arxiv.org/abs/gr-qc/0104064
>
>As I have pointed out, the 1999 paper had an error in the
>key equation for the Doppler shift which negates the sign.
You failed to name which key equation in 0104064 makes the correction,
but if you mean (16) where
At = Ap/c (16)
of course this is wrong fundamentally, needing a minus sign, and in
fact contradicts (15)* in the same paper which can be differentiated to
yield
At = -Ap/c where At = df/fdt
and so we witness the introduction of the "big fib". There is no way on
earth that At and Ap can have the same sign.
Are you touting this as the correction (you weren't specific). There is
no evidence that CHASMP uses the same perverted Doppler arithmetic even
though it's built into DSN and as I pointed out, no evidence that they
had to comply with it. Why do you stick with DSN when we have clear
evidence in plain English from CHASMP?
*(identical to Eq. (1) in 9903024 v2).
>> This (99) is one of several Pioneer papers that largely repeat each
>> other in certain sections and mostly use unsigned numbers, but this
>> paper is a bit more specific than the others. On page 3, Sec. 2.2 are
>> described CHASMP results that are independent of JPL's with its
>> unique and troublesome Doppler signature. Note especially that CHASMP
>> takes care to specify normal Doppler arithmetic, to wit:
>>
>> "Without using the apparent acceleration, CHASMP shows a steady
>> frequency drift of -6 x 10^-9 Hz/s, or 1.5 Hz over 8 years (one way
>
>Right, the negative sign indicates a blue shift as stated in the
>footnote "It is ... negative for a spacecraft approaching the
>station (blue shift), just the opposite of the usual convention."
and using the usual convention which I am assuming CHASMP does, then
the negative sign means recession.
>> only). The drift in the Doppler residuals
>>(observed minus computed data)
Please note CHASMP's : "OBSERVED MINUS COMPUTED"
>>is clear, definite and cannot be removed without the added
>> [unsigned] acceleration Ap."
I take them at their word that from the raw data, taking observed minus
computed, they came up with a negative drift, and from that one fact
the rest can be deduced. It's only of clinical interest that DSN
chooses to go upside down. This is not DSN, it's CHASMP
>> We will show that the added acceleration Ap must be positive.
>>
>> First of all, with (orthodox) drift as defined above,
>
>If by "orthodox" you mean df = f_received - f_transmitted
>then that is the wrong way round. The DSN uses the opposite
>convention, df = f_transmitted - f_received
I am aware of DSN's idiosynchasies. They make nothing but trouble.
><snip more based on your erroneous doppler formula>
Quite audacious to snip. That's not your style is it?
I am restoring my legitimate equations. Please tell me where the errors
are and I'll try to fix them.
First of all, with (orthodox CHASMP) drift as defined above, there are
these relationships from first principles:
df = -f0*V/c = kV where k = -f0/c
df/dt = k*Ap = -6x10^-9 Hz/s = fdot
>From this, the solution for Ap is a positive value:
Ap = fdot/k
Ap = +8.7 x 10^-8 cm/ss
making Ap a positive number, contrary to the prevailing view.
Secondly, as I interpret it, the model seeks to duplicate the observed
data by accumulating all known effects, after which a comparison
finally leaves an unknown. Using k to convert V to frequency, we can
represent the balancing act as follows, first leaving an unknown X as
reported by CHASMP:
Obs(Knowns +X)k - Model(Knowns)k = fdot = -6x10^-9 Hz/s
X*k = fdot, the observed negative drift.
The missing element in the model is
X = fdot/k = -6e-9/(-f0/c) = +8.74 x 10^-8 cm/ss
Again we have agreement that a positive value for Ap must be added to
the model to achieve a null and match the observed. (Repeatedly in
these papers, the addition Ap has been unsigned). If they meant Ap to
be the usual negative Ap, there'd be no balance.
>> It seems clear that the range rate acceleration for the Pioneers is
>> away from the sun, not toward. Of course such a reversal would have a
>> major on efforts to deduce the cause of the anomaly.
>
>It would, however, you have reversed the sign all the
>way through by again neglecting the fact that the DSN
>convention is the opposite of what you consider to be
>orthodox.
Can I make myself any plainer? When you go with a lie like the DSN
convention, then it's no wonder there is little algebra but much
arithmetic in the papers. I get the strong impression that the party
line was "It's some kind of gravity; there are no repulsive forces.
Make it come true".
>I can only again repeat the words of the
>authors on page 20:
That is all that happens in these papers, is that each one is copy of
the previous one.
> "This velocity is outwards and hence produces a red
> shift. ... Therefore, a_P produces a slight blue
> shift on top of the larger red shift."
>
>They then say quite clearly:
>
> "By DSN convention [38], the first of Eqs. (15) is
> [dv_obs ? dv_model]_usual = ?[dv_obs ? dv_model]_DSN.
BUT BUT we're not doing DSN, so now it's my turn to snip. If you
haven't grasped that, then I'll try to clarify. (Hint I'm dealing with
CHASMP, the "OBSERVED MINUS COMPUTED" people.
SNIP
DSN is corrupt and Eq. 16 At = Ap/c is lie. There is no way to work
with that kind of material and come out right. I agree though, stick
with DSN and get blue shift.
>George
>
John Polasek
Craig Markwardt
I'm not sure this is a simple textbook problem. The simple part is
that the apparent anomaly is a constant acceleration, directed toward
the sun.
The complex part is that the anomaly involves modeling: modeling of
the gravitation of the sun and planets; the solar radiation pressure;
of the spacecraft environmnent; an understanding of earth rotation;
and of electromagnetic propagation in the solar system. Thankfully
all of those things are very very well understood, in part because of
careful observations over hundreds of years, and sensitive new
measurement techniques that have been developed over the past few
decades.
Ultimately we can account for all the motions of the spacecraft (a
speed of approximately 12 km/s), except for a small residual of a few
*millimeters* per second. We can say what this residual is *not* due
to: not solar wind, not dark matter, not swept up dust, not gravity of
the Kuiper belt. Perhaps it is some new physics. More likely it is
something like the modeling of the thermal/power environment of the
spacecraft. Unfortunately, spacecraft radiometric and thermal
modeling is is not a freshman physics textbook problem. :-(
Best luck,
George Dishman
> On Fri, 28 Oct 2005 21:59:09 +0100, "George Dishman"
> <geo...@briar.demon.co.uk> wrote:
>
> George: (I am having to send this by google.groups as
> sci.physics.research has conspired to cut me off at the server. I
> cannot post one message in this thread via my news-...@dfl.rr.com) I
> have discussed this with Khavine but he denies it.)
As I said by email, I believe that's the way the
system works. It would appear on all only after
approval.
>>"John C. Polasek" <jpol...@cfl.rr.com> wrote in message
>>news:1130520901.8...@g49g2000cwa.googlegroups.com...
>>> Here is a further effort to convince you that Ap is a positive number.
>>> As usual we try stick to first principles and avoid the entanglements
>>> previously noted.
>>
>>Then you will get it wrong again. The DSN uses the opposite
>>convention from everyone else as you are told in the footnote
>>to which you referred. There are no previous "entanglements",
>>you are simply ignoring the standard which the DSN has used
>>for many years.
>
> Don't you understand that I am bypassing DSN for the independent
> analysis of CHASMP and am ignoring the upside down Doppler?
Yes I understand that. That's why you get the
acceleration upside down too.
>>> It seems clear that Doppler frequency is primary data
>>> and conclusions about polarity of acceleration are a result of
>>> intermediate calculation subject to error
>>> Refer to arXiv:general relativity-qc/99003024 v2 9 Mar. 1999, Turyshev
>>> et al.
>>
>>I would prefer that you refer instead to the more recent
>>paper:
>>
>> http://www.arxiv.org/abs/gr-qc/0104064
>>
>>As I have pointed out, the 1999 paper had an error in the
>>key equation for the Doppler shift which negates the sign.
>
> You failed to name which key equation in 0104064 makes the correction,
I have said twice now, reference [38], and quoted it both times.
The 1999 paper said in the footnote df = f - f_0
The current paper corrects that to df = f_0 - f
> but if you mean (16) where
> At = Ap/c (16)
>
> of course this is wrong fundamentally, needing a minus sign,
It is correct, the negation having been incorporated in
the definition of df.
> and in
> fact contradicts (15)* in the same paper which can be differentiated to
> yield
> At = -Ap/c where At = df/fdt
> and so we witness the introduction of the "big fib". There is no way on
> earth that At and Ap can have the same sign.
There is when one has already been negated.
> Are you touting this as the correction (you weren't specific).
No, the correction is endnote [38] of the 2001 paper compared to
the footnote on the page 3 of the 1999 document.
"[38] The JPL and DSN convention for Doppler frequency
shift is (dv)_DSN = v0 ? v, where v is the measured
frequency and v0 is the reference frequency. It is
positive for a spacecraft receding from the tracking
station (red shift), and negative for a spacecraft
approaching the station (blue shift), just the
opposite of the usual convention, (dv)_usual = v?v_0.
In consequence, the velocity shift, dV = V?V_0, has
the same sign as (dv)_DSN but the opposite sign to
(dv)_usual. Unless otherwise stated, we will use the
DSN frequency shift convention in this paper. We
thank Matthew Edwards for asking us about this."
> There is
> no evidence that CHASMP uses the same perverted Doppler arithmetic even
> though it's built into DSN and as I pointed out, no evidence that they
> had to comply with it.
It is the CHASMP and ODP programs that started this
because they follow the protocol stated for the
interface files which has this inverted Doppler
convention.
> Why do you stick with DSN when we have clear
> evidence in plain English from CHASMP?
Because the authors wrote
"Unless otherwise stated, we will use the DSN frequency
shift convention in this paper."
> *(identical to Eq. (1) in 9903024 v2).
>
>>> This (99) is one of several Pioneer papers that largely repeat each
>>> other in certain sections and mostly use unsigned numbers, but this
>>> paper is a bit more specific than the others. On page 3, Sec. 2.2 are
>>> described CHASMP results that are independent of JPL's with its
>>> unique and troublesome Doppler signature. Note especially that CHASMP
>>> takes care to specify normal Doppler arithmetic, to wit:
>>>
>>> "Without using the apparent acceleration, CHASMP shows a steady
>>> frequency drift of -6 x 10^-9 Hz/s, or 1.5 Hz over 8 years (one way
>>
>>Right, the negative sign indicates a blue shift as stated in the
>>footnote "It is ... negative for a spacecraft approaching the
>>station (blue shift), just the opposite of the usual convention."
>
> and using the usual convention which I am assuming CHASMP does,
CHASMP uses the DSN convention, it is DSN software.
> then
> the negative sign means recession.
>
>>> only). The drift in the Doppler residuals
>>>(observed minus computed data)
>
> Please note CHASMP's : "OBSERVED MINUS COMPUTED"
And "OBSERVED" means f_0 - f
For that reason COMPUTED also has to have the same
convention and the difference between them is negated
as a result.
>>>is clear, definite and cannot be removed without the added
>>> [unsigned] acceleration Ap."
>
> I take them at their word that from the raw data, taking observed minus
> computed, they came up with a negative drift,
So do I, and they say "It is positive for a spacecraft
receding from the tracking station (red shift), and
negative for a spacecraft approaching the station (blue
shift), just the opposite of the usual convention."
> and from that one fact
> the rest can be deduced. It's only of clinical interest that DSN
> chooses to go upside down. This is not DSN, it's CHASMP
CHASMP and ODP are DSN tools which were written to
the DSN protocol.
In fact that's why the paper is written the way it
is, CHASMP and ODP given these inverted outputs and
they didn't want to start manually inverting
everything because everyone at the DSN and JPL has
been using that inverted convention on every mission
they manage. The confusion that would introduce would
have been far worse.
>>> We will show that the added acceleration Ap must be positive.
>>>
>>> First of all, with (orthodox) drift as defined above,
>>
>>If by "orthodox" you mean df = f_received - f_transmitted
>>then that is the wrong way round. The DSN uses the opposite
>>convention, df = f_transmitted - f_received
> I am aware of DSN's idiosynchasies. They make nothing but trouble.
Indeed, but you cannot ignore or you get the wrong answer.
>><snip more based on your erroneous doppler formula>
> Quite audacious to snip. That's not your style is it?
Not often, but I agree with what you wrote. Your
derivation is fine but it starts with negated values
so gives a negated result. I was just trying to focus
on where the difference between us lies.
> I am restoring my legitimate equations. Please tell me where the errors
> are and I'll try to fix them.
>
> First of all, with (orthodox CHASMP)
That's it, the one and only error. You assumed CHASMP
used the conventional definition when in fact it uses
the opposite.
The party line at the time was that they had no idea.
>>I can only again repeat the words of the
>>authors on page 20:
>
> That is all that happens in these papers, is that each one is copy of
> the previous one.
>
>> "This velocity is outwards and hence produces a red
>> shift. ... Therefore, a_P produces a slight blue
>> shift on top of the larger red shift."
>>
>>They then say quite clearly:
>>
>> "By DSN convention [38], the first of Eqs. (15) is
>> [dv_obs ? dv_model]_usual = ?[dv_obs ? dv_model]_DSN.
>
> BUT BUT we're not doing DSN, so now it's my turn to snip. If you
> haven't grasped that, then I'll try to clarify. (Hint I'm dealing with
> CHASMP, the "OBSERVED MINUS COMPUTED" people.
I won't snip but I will repeat, endnote [38] tells
you that OBSERVED = f_0 - f, the opposite if the
usual convention, and that they are using that DSN
convention throughout the paper.
> SNIP
>
> DSN is corrupt and Eq. 16 At = Ap/c is lie. There is no way to work
> with that kind of material and come out right. I agree though, stick
> with DSN and get blue shift.
Let me just highlight that because we are in agreement
about this:
> I agree though, stick
> with DSN and get blue shift.
CHASMP was written to the DSN standard and the authors
say they are sticking with the DSN convention throughout
the paper.
George
Hero.van...@gmx.de
> On Fri, 28 Oct 2005 21:59:09 +0100, "George Dishman"
> <geo...@briar.demon.co.uk> wrote:
>
> George: (I am having to send this by google.groups as
> sci.physics.research has conspired to cut me off at the server. I
> cannot post one message in this thread via my news-...@dfl.rr.com) I
> have discussed this with Khavine but he denies it.)
sci.physics.research in the adress, otherwise it might not appear at
all.
Also Google claims otherwise.
Mr.Khavine wrote to me:
"Please note that, since the article was posted to a moderated group
and
was not approved, it will not appear in ANY newsgroup. If you want to
post it to any unmoderated newsgroup, you must post it again, avoiding
any moderated newsgroups."
Hero
John C. Polasek
has it right: df = f - f0. But no one, not even the Almighty, can
legitimately follow it with the statement that blue shift is red shift
and red shift is blue shift and get away with just calling it a
convention.
Let's dignify it by calling it what it is: a transform. Transforms help
with the intermediate calculations.
But then, at the end, we are compelled to perform the inverse transform
to convert "toward the sun" to "away from the sun" and blue shift into
red shift.
John Polasek
George Dishman
> We can shorten this all up very easily.
Yes we could, you could just read what the authors have written.
> The first line of the footnote
> has it right: df = f - f0.
No John, the footnote had it wrong. They subsequently corrected
that error. See reference [38] in the current paper.
"[38] The JPL and DSN convention for Doppler frequency shift
is (dv)_DSN = v0 - v, where v is the measured frequency
and v0 is the reference frequency. It is positive for a
spacecraft receding from the tracking station (red shift),
and negative for a spacecraft approaching the station
(blue shift), just the opposite of the usual convention,
(dv)_usual = v - v_0. In consequence, the velocity shift,
dV = V - V_0, has the same sign as (dv)_DSN but the
opposite sign to (dv)_usual. Unless otherwise stated, we
will use theDSN frequency shift convention in this paper.
We thank Matthew Edwards for asking us about this."
> But no one, not even the Almighty, can
> legitimately follow it with the statement that blue shift is red shift
> and red shift is blue shift and get away with just calling it a
> convention.
They are not doing that though, the craft is being accelerated
towards the Sun which increases the frequency and they call
that "blue shift", you as you or I would. The only unusual
feature is that the DSN software uses negative numbers to
indicate blue shift and positive for red shift.
> Let's dignify it by calling it what it is: a transform. Transforms help
> with the intermediate calculations.
It is a convention, or more accurately an interface specification.
It tells the software writers who produce programs like CHASMP and
ODP what format they will find the data in when the read DSN files
and in what format they are required to write data back to those
files so that the programs communicate successfully with each other.
Call it a transform if you wish.
> But then, at the end, we are compelled to perform the inverse transform
> to convert "toward the sun" to "away from the sun" and blue shift into
> red shift.
We know the descriptions of red and blue have already been transformed
because of what they say at the top of page 19 about the velocity of
the craft:
"(This velocity is outwards and hence produces a red shift.)
We have already included the sign showing that a_P is inward.
(Therefore, a_P produces a slight blue shift on top of the
larger red shift.)"
We know the craft is moving away from the Sun and motion away
from the Sun would reduce the frequency which is conventionally
described as a red shift, so if you want to call that a transform,
they have already applied it.
In fact the whole thing is very simple, we know the craft is
receding and the describe that as producing a red shift while
the anomaly "a_P produces a slight blue shift on top of the
larger red shift." The anomaly is therefore in the opposite
direction to the overall craft velocity.
George
Jeff Root
> The only unusual feature is that the DSN software uses
> negative numbers to indicate blue shift and positive for
> red shift.
And there isn't anything strange about it, since positive
numbers mean the spacecraft is moving forward, away from
the origin, and negative numbers mean it is falling back,
toward the origin.
-- Jeff, in Minneapolis
Craig Markwardt
Thomas Smid <thoma...@gmail.com> writes:
> Craig Markwardt wrote:
> > Thomas Smid <thoma...@gmail.com> writes:
> >
> > > George Dishman wrote:
> > >
> > > > John is wrong it is a blue shift. Look at the graph showing the
> > > > sum of the anomaly and the solar radiation pressure. Solar
> > > > radiation pushes the craft away from the sun adding a red shift,
> > > > but the total of that and the anomaly goes through zero showing
> > > > the anomaly is a blue shift.
> > >
> > > George, you are missing John's point. If the sign of the anomalous
> > > force has been switched by mistake, then it would of course in reality
> > > have the same sign as the radiation pressure force.
> >
> >
> > The sign of the anomalous acceleration has not been switched by
> > mistake. I performed an independent confirmation of the Anderson et
> > al results, all using the "usual" frequency conventions. My results
> > matched Anderson et al's results quite closely.
> >
> > The anomalous acceleration is consistent with all these interpretations:
> > * a slightly higher received frequency than expected;
> > * a slightly "bluer" frequency shift than expected;
> > * a slightly faster receding craft than expected.
>
> A faster receding craft would correspond to a frequency red shift not a
> blue shift. This holds independently of any convention.
This was a typo on my part. Corrected:
* a slightly higher received frequency than expected;
* a slightly "bluer" frequency shift than expected;
* a slightly slower receding craft than expected.
> > There is no room for a directional sign error. The doppler shifts due
> > to planetary motion+rotation, spacecraft rotation, etc. are very large
> > (hundreds of thousands of Hertz), and a sign error would have been
> > obvious.
Note, no response.
Uh, whatever. In fact, the "DSN" convention for frequency differences
is the opposite of the "usual" convention, as in
[nu_obs - nu_model]_DSN = [nu_model - nu_obs]_usual
so there isn't really the contradiction that you so deeply seem to need.
And in fact, regardless of the text in Anderson et al, my own
independent analysis also showed that nu_obs > nu_model, i.e. a
slightly "bluer" shift than expected.
> you should have very good reasons to to so. ...
In this case, the Anderson authors take the conservative approach,
which is to say that since the "anomaly" parameter is so highly
correlated with the solar radiation pressure coefficient, it would be
improper to conclude that the anomaly is real.
So, yes, it is quite possible that the radiation pressure is not
modeled perfectly. Let's say that there is a <25% error in the
modeling of that quantity for spacecraft < 5 AU, e.g. Ulysses. That
would conservatively assume that *all* of the anomaly was due to
radiation pressure since the radiation pressure force is >40 x 10^{-8}
cm/s^2 for R < 5 AU (see Anderson et al Fig 6).
Now let's consider the Pioneer spacecraft, which is beyond 40 AU
starting in the year 1987. At that position, the radiation pressure
is < 1 x 10^{-8} cm/s^2. Our very conservative 25% modelling error
gives an upper limit of < 0.25 x 10^{-8} cm/s^2. Insignificant.
In reality, the radiation pressure modeling is better than 25%.
Thus, your "guilt by association" argument doesn't work. The
radiation pressure modeling at 1-5 AU for Ulysses or Galileo, even in
the worst case, is not really relevant to the Pioneer 10 environment.
[ Also, Ulysses and Galileo are not spin stabilized, and have many
more maneuvers, which contribute to the uncertainty in the presence of
the anomaly. ]
> ... However, I could not find
> any explanation in Anderson's paper as to where this correlation should
> come from if not from the radiation pressure term in their model. What
> they should have done is to re-model the radiation pressure force such
> that the anomalous acceleration for the Galileo and Ulysses data
> disappears, and then use this modified model also on the Pioneer data.
> I am not saying that for the latter the resultant anomalous
> acceleration would also be zero, but it should change significantly and
> hence call for different explanations
It probably would not change significantly (see above).
CM
Richard Saam
Craig and others
Would you consider the following "simple" model
based on historically accepted/verified physical concepts
as would be presented in a freshman physics textbook
arrives at an observed Pioneer spacecraft deceleration rate of:
8E-8 cm/sec2
_
Pose the problem in conventional hydrodynamic sense
E = F d
F = E / d
= m v^2 / d
let d be expressed as volume / Area ratio
d = volume / area
therefore:
F = m v^2 / d
= m v^2 area / volume
let
rho = medium density = m / volume
area = spacecraft frontal area
therefore
F = rho area v^2
now assume the possibility that
v = c
(space vacuum "unknown" energy is available to the Pioneer spacecraft
as mc^2 which greatly exceeds mv2)
and
rho = conventionally accepted space vacuum density
= 2 H2 / (8 pi G)
where:
H = Hubble constant
G = Newton's gravitational constant
a = F/M = anomalous deceleration
Where:
M is the Pioneer spacecraft mass.
deceleration "a" approximately equal to 8E-8 cm/sec2
for Pioneer physical design parameters
Why not start with the basics (newton's 2nd law)
onto which greater complexities map onto.
Richard Saam
Craig Markwardt
Richard Saam <rds...@att.net> writes:
> Would you consider the following "simple" model
> based on historically accepted/verified physical concepts
> as would be presented in a freshman physics textbook
> arrives at an observed Pioneer spacecraft deceleration rate of:
>
> F = rho area v^2
>
> now assume the possibility that
>
> v = c
>
> (space vacuum "unknown" energy is available to the Pioneer spacecraft
> as mc^2 which greatly exceeds mv2)
>
> and
>
> rho = conventionally accepted space vacuum density
> = 2 H2 / (8 pi G)
>
> where:
> H = Hubble constant
> G = Newton's gravitational constant
>
> a = F/M = anomalous deceleration
The approximate value of "rho" is 8.1E-30 g/cm^3 (using Ho = 72 km/s/Mpc).
The value of "a" in your expression,
a = F/M = rho * area * c^2 / M
= (8.1E-30)*(5.9E4)*(3.0E10)^2 / (2.5E5) ; (all values in c.g.s.)
= 1.7E-9 cm^2/s
So whether or not your "derivation" makes physical sense, which it
really does not, the numerology is not even right.
CM
Richard Saam
Craig
Your right in the numerology.
1.7E-9 cm/sec^2 is about 45 times
the observed deceleration value of 8E-8 cm/sec^2
which is fairly close considering the "powers of ten" nature in the universe.
In the context of a hydrodynamic problem
F = rho area c^2
= (8.1E-30)*(5.9E4)*(3.0E10)^2
= 4.3E-4 dyne
assumes kinetic impact of medium onto object moving through it.
This is generally a high Reynold's number
(inertial forces/viscous forces) case
With the 45 factor, a predominately viscous interaction between moving object
and medium would be considered. (A low Reynold's number case)
Stokes' Law would be indicated.
F(viscous) = 45 (4.3E-4)
= 1.9E-2 dyne
F(viscous) = 3 pi mu c D (for a sphere if diameter D)
D = sqrt(5.9E4*4/pi)
= 275 cm
mu = F(viscous) / (3 pi c D)
= 1.9E-2/(3*pi*(3.0E10)*(275))
= 2.4E-16 g/(cm sec)
mu = momentum / area
mu / x = rho c
x is interaction distance of the space craft with medium
x = mu / (rho c)
= (2.4E-16)/((8.1E-30)*(3.0E10))
= 9.9E2 cm
This is in the ball park of space craft dimensions
or at least the anticipated "fluid medium" disturbance
surrounding the spacecraft.
All this is assuming a sphere.
I am sure you realize that if conducting a hydrodynamic experiment,
one would not choose the Pioneer space craft shape.
A much more defined curved volumetric shape would be in order.
This type of work is conducted in hydraulics labs.
Work at very low Reynold's numbers is of much interest.
Any future space craft studying the deceleration anomaly
should take this potential hydrodynamic effect into consideration.
Richard Saam
sean
Richard Saam wrote:
> Craig Markwardt wrote:
Can the too high observed speed of stars rotation in galaxies be
explained by the pioneer anomaly? If it is .0000075 (or whatever) cms2
inwards acceleration for pioneer would the same value applied to the
mass of the rotating stars be enough to hold them in from being flung
off ?
sean
Craig Markwardt
No evidence supports such a speculation. The pioneer anomaly is a
constant *acceleration*. The galaxy velocity problem is a constant
*speed*. If one were to invoke a cloud of "dark matter" at the center
of the solar system to explain the pioneer anomaly, then the gravity
from the same matter would alter the orbits of the planets in a
detectable way, and no such planetary effect is found.
CM
frank...@yahoo.com
> > If anybody has a comparison of Pioneer 10 red-shift compared to
> > galactic red-shift caused by the expansion of the universe, that would
> > be interesting to see.
>
> This was done to death in the group a few years ago. Numerically
> the acceleration appears close to the Hubble value.
>
> a_p ~ c H
>
> However, tired light says the red shift depends on distance so the
> key question is the rate at which the distance is increasing. The
> formula is then
>
> a_tl = 2 v H
>
> where v is the speed of the craft and the factor of 2 is because
> the shift applies on both uplink and downlink.
Can you explain this 2nd formula a bit further - I don't understand the
significance. If you have an expression for the P10 acceleration, then
this must include the effects of the craft's speed etc. which might be
expressed by your 2nd formula. It's like you're saying it matches, but
then it doesn't match.
I find it very significant that the P10 acceleration is close to the
Hubble value - I don't think this could be coincidence. This shows the
P10 acceleration is not an acceleration at all, but rather part of the
same natural phenomenon that makes all stars appear to be red-shifted.
I think there is sufficient doubt in the calculations to suggest that a
red-shift is possible and that the value matches the Hubble value adds
more support to this interpretation. Such an interpretation would
simplify the explanations of red-shift and eliminate the need to think
that the universe was born in a mysterious big bang. I think it is far
more likely that the universe is infinite, uniform, and has always
existed in its current form, rather than pop into existence out of know
where.
George Dishman
<frank...@yahoo.com> wrote in message
news:1131260733....@g43g2000cwa.googlegroups.com...
> George Dishman wrote:
>> > If anybody has a comparison of Pioneer 10 red-shift compared to
>> > galactic red-shift caused by the expansion of the universe, that would
>> > be interesting to see.
>>
>> This was done to death in the group a few years ago. Numerically
>> the acceleration appears close to the Hubble value.
>>
>> a_p ~ c H
>>
>> However, tired light says the red shift depends on distance so the
>> key question is the rate at which the distance is increasing. The
>> formula is then
>>
>> a_tl = 2 v H
>>
>> where v is the speed of the craft and the factor of 2 is because
>> the shift applies on both uplink and downlink.
>
> Can you explain this 2nd formula a bit further - I don't understand the
> significance.
Sure. Tired light says that a photon loses energy as it travels
through space. The loss is exponential over large distances (of
the order of giga-parsecs) but can be approximated as linear
within the Solar system. The distance the signal is travelling
controls the amount of frequency loss and that distance is
increasing at a rate determined by the speed of the craft. The
published value is an acceleration which is rate-of-change of
frequency shift, so it is proportional to rate-of-change of
distance, which is of course the speed of the craft.
> If you have an expression for the P10 acceleration, then
> this must include the effects of the craft's speed etc. which might be
> expressed by your 2nd formula. It's like you're saying it matches, but
> then it doesn't match.
The second equation is an expression for the amount of red shift
that would be produced by Tired Light given the speed of the
craft.
By comparison, the first equation is a numerical 'coincidence'
between the values.
> I find it very significant that the P10 acceleration is close to the
> Hubble value - I don't think this could be coincidence.
Indeed, it seems unlikely, but although the numbers coincide,
nobody has yet come up with a credible mechanism by which the
Hubble Constant could produce the observed effect. Tired Light
gives a prediction several thousand times too small and in the
wrong sense (as a phenomenon, it is ruled out by measurements
of Type Ia supernovae but that's another matter, Pioneer is to
insensistive to detect it).
> This shows the
> P10 acceleration is not an acceleration at all, but rather part of the
> same natural phenomenon that makes all stars appear to be red-shifted.
It may possibly be some natural phenomenon but the signal
from Pioneer is blue shifted, that's one reason why it has
been so hard to explain.
> I think there is sufficient doubt in the calculations to suggest that a
> red-shift is possible
You might consider the paper ambiguous, though I can't see
how, but there is no doubt about what is observed, the
authors of both papers have stated categorically that it
is a blue shift, the frequency is higher than predicted.
> and that the value matches the Hubble value adds
> more support to this interpretation. Such an interpretation would
> simplify the explanations of red-shift and eliminate the need to think
> that the universe was born in a mysterious big bang. I think it is far
> more likely that the universe is infinite, uniform, and has always
> existed in its current form, rather than pop into existence out of know
> where.
Ah, you are another one who perfers to re-write reality to match
your beliefs. I thought you were asking a serious question.
George
sean
I wasnt thinking about dark matter specifically but using that as one
possibility of a new physics to explain the anomaly I wonder if in fact
a cloud of dark matter could explain the anomaly. I would have thought
a cloud of dark matter would not give a constant accelleration but
rather one that diminished the farther Pioneeer was from the center of
the solar system.
I think only an evenly distributed amount of dark matter across the
universe would give a constant acceleration seen in the anomaly?
But seeing as gravity is decribed as accelleration then the anomaly
being constant would be effectively increasing the inward pull of
gravity on any planet at any orbit radius. Making the inward pull on
any one planet as...sun gravity + anomaly.
For the solar system, this slight increase in `gravity` on the planet
would mean that a planet would have to have a slightly faster orbital
speed to stay in its particular orbit to compensate for the increased
inward force of sun g + anomaly. Either that or have a larger weight
than currently calculated.So maybe planets do show an observable effect
from the anomaly. Its just that its disguised because calculations have
attributed it to an underestimate of planets mass.
For galaxies, we get the same thing happening due to the anomaly.
Either the orbital speeds have to be greater or the stars weights have
to be greater to counter the additional inward acceleration of the
anomaly. One would also get the effect of the observed orbital speed
being more diverged on the increased side from that predicted, the
farther out from the core on looks. This would be because the cores
gravity decreases as one gets farther from the core but the anomaly
stays constant. Im sure this is whats observed.
I also wonder if it (anomaly)can explain mercurys orbit , until now
only done so by GR?
Sean
Richard Saam
Sean
Keeping to my theme
deceleration = Cd rho c^2 / M
The apparent high observed speed of outer star rotation in galaxies
is due to forced redistribution with R of orbital velocities with time
due to disproportionate change in R of inner faster moving stars
relative to R of outer slower orbiting stars
in accordance with orbitial velocity (v) relationship:
v^2 = G M / R
This is a residual effect left over from the time
when galaxies were primarily dust or small particles
when the formula:
deceleration = Cd rho c^2 / M
rho = conventionally accepted space vacuum density
= 2 H2 / (8 pi G)
M = object mass
Cd = drag coefficient
was primarily applicable
and is still applicable
in as much as galaxies still consist of
small particles (< a few meters)
gravitationally coupled to large objects such as stars.
Thus the spirals appearing to reach for the center.
Richard Saam
Richard Saam
>
>>I find it very significant that the P10 acceleration is close to the
>>Hubble value - I don't think this could be coincidence.
>
>
> Indeed, it seems unlikely, but although the numbers coincide,
> nobody has yet come up with a credible mechanism by which the
> Hubble Constant could produce the observed effect.
George:
There is some logic to the following:
deceleration = Cd rho c^2 / M
rho = conventionally accepted space vacuum density
= 2 H^2 / (8 pi G)
M = mass of the object
Cd = drag coefficient
or
deceleration = Cd (2 H^2 / (8 pi G)) c^2 / M
deceleration = (Cd (2 H / (8 pi G)) c / M) c H
deceleration = c H
where (Cd (2 H / (8 pi G)) c / M) about equal to one(1) for Pioneer 10
Richard Saam
Dishman
> George Dishman wrote:
>
> >
> >>I find it very significant that the P10 acceleration is close to the
> >>Hubble value - I don't think this could be coincidence.
> >
> >
> > Indeed, it seems unlikely, but although the numbers coincide,
> > nobody has yet come up with a credible mechanism by which the
> > Hubble Constant could produce the observed effect.
>
> George:
>
> There is some logic to the following:
>
> deceleration = Cd rho c^2 / M
Well I would say acceleration is:
a = f / M
f = dp/dt
The mass in the volume swept in time dt is
dm = rho A v dt
where v is the relative speed of the material and A is
the cross-sectional area in the direction of flow. The
momentum is
dp = v dm
Hence:
a = rho A v^2 / M
The drag then depends on the relative speed of the
material as well as its density. Both speed and
direction are important. Further you are assuming
the material is moving at the speed of light yet
imparts finite momentum.
Richard, I know you aren't suggesting the following
but let me go over some alternatives to illustrate a
point.
The solar wind is too thin anyway but would not
slow the craft, it is moving outwards at around
400km/s so would cause an acceleration _away_
from the Sun for both Pioneer craft.
Dust in the region where the anomaly is seen is I
believe of two components, the interplanetary dust
which is mostly orbiting the Sun so would appear to
be almost motionless compared to Pioneer's 12km/s,
perhaps giving a small lateral acceleration, and an
interstellar component which is moving through the
solar system with a common direction of flow. The
former would explain the apparent acceleraton
_towards_ the Sun for both craft while the latter
would produce acceleration in different directions for
the two craft since they are on opposite sides of the
system.
> rho = conventionally accepted space vacuum density
> = 2 H^2 / (8 pi G)
Your suggestion seems to relate to some general
vacuum effect, but that would also have operated
on the material which produced the CMBR. Since
we have a motion (from the dipole) of about 620km/s
relative to that material, I would expect there to be
a common direction for the acceleration of the two
craft and I would have expected the speed of
620km/s to be used in your formula, not the speed
of light.
>From the fact that you dn't comment on the direction
of acceleration of the craft, I infer you are assuming
this unspecified material is at rest relative to the
Sun, but in that case you should be using 12km/s
instead of the speed of light.
Can you clarify any of this.
George
Richard Saam
George
Acknowledging your derivation presented above.
a = rho A v2 / M
Then
F = M a = rho A v^2
= (m/vol)Av^2
= (A/vol)mv^2
= (1/x)mv^2
= (1/x)E
or
Fx = E
The assumption is that Energy imparted
to Pioneer spacecraft and other objects
over distance x
is mc^2 and not mv^2.
mc^2 would be the total energy
available within the space vacuum
with density
rho = 2 H^2 / (8 pi G)
over distance x.
It does not matter how fast the object goes through x.
The energy mv^2 contribute
of dust and other particles
is calculable but insignificant.
then
F = M a = (1/x)E
= (1/x)mc^2
= (A/vol)mc^2
= (m/vol)Ac^2
= rho A c^2
and
a = rho A c^2 / M
A coefficient Cd may be added
for the general hydrodynamic case
a = Cd rho A c^2 / M
This means that the decceleration is constant
for any object velocity (v << c) or direction
(not tied to any particular reference frame)
which appears to be the observed case.
The Pioneer 10 and 11 opposing directions
to CMBR 620km/s is irrelevant in this concept.
The Pioneer velocities 12km/s are also irrelevant.
The deceleration is constant for all frames of reference
as observed.
Richard Saam
Dishman
Richard Saam wrote:
> Dishman wrote:
> > Richard Saam wrote:
> >
> >>George Dishman wrote:
> >>
>
> Then
>
> F = M a = rho A v^2
> = (m/vol)Av^2
> = (A/vol)mv^2
> = (1/x)mv^2
> = (1/x)E
>
> or
>
> Fx = E
Good old "work = force * distance".
> The assumption is that Energy imparted
> to Pioneer spacecraft and other objects
> over distance x is mc^2 and not mv^2.
It isn't energy that we are interested in though,
it is the change of momentum of the craft that
needs to be explained. Simply adding energy
would merely warm the craft, or you can think
of it as the kinetic energy lost by the craft in
moving against the force.
George
Richard Saam
George
In terms of momentum change
a relativistic approach may be in order:
Integrate[0,x] F dx = m Integrate[0,v] v dv / sqrt(1 - v^2/c^2)
F x = - m c^2 sqrt(1 - v^2/c^2)
= - m c^2 for v << c
then
F = M a = -(1/x)mc2
= -(A/vol)mc2
= -(m/vol)Ac2
= -rho A c2
and
a = -rho A c2 / M
A coefficient Cd may be added
for the general hydrodynamic case
a = -Cd rho A c2 / M
Dishman
Richard Saam wrote:
> Dishman wrote:
> > Richard Saam wrote:
<standard derivation snipped>
> >>Acknowledging your derivation presented above.
> >>
> >>a = rho A v^2 / M
> >>
> >>Then
> >>
> >>F = M a = rho A v^2
> >> = (m/vol)Av^2
> >> = (A/vol)mv^2
> >> = (1/x)mv^2
> >> = (1/x)E
> >>
> >>or
> >>
> >>Fx = E
> >
> > Good old "work = force * distance".
> >
> >>The assumption is that Energy imparted
> >>to Pioneer spacecraft and other objects
> >>over distance x is mc^2 and not mv^2.
> >
> > It isn't energy that we are interested in though,
> > it is the change of momentum of the craft that
> > needs to be explained. Simply adding energy
> > would merely warm the craft, or you can think
> > of it as the kinetic energy lost by the craft in
> > moving against the force.
>
> In terms of momentum change
> a relativistic approach may be in order:
>
> Integrate[0,x] F dx = m Integrate[0,v] v dv / sqrt(1 - v^2/c^2)
>
> F x = - m c^2 sqrt(1 - v^2/c^2)
>
> = - m c^2 for v << c
Not really Richard, if you look at each side, you will
see you are just equating energies. The equation is
still scalar while we are looking for the momentum
vector. You have no direction information in this.
There seems to be no physics behind your equations.
Anderson et al already acknowledge the coincidence
of the numbers so I don't se that you are adding
anything to what has been published.
> then
>
> F = M a = -(1/x)mc2
> = -(A/vol)mc2
> = -(m/vol)Ac2
> = -rho A c2
> and
>
> a = -rho A c2 / M
>
> A coefficient Cd may be added
> for the general hydrodynamic case
>
> a = -Cd rho A c2 / M
>
> This means that the decceleration is constant
> for any object velocity (v << c) or direction
> (not tied to any particular reference frame)
But that's the problem, acceleration is a
vector so your equations should _predict_
its direction.
> which appears to be the observed case.
> The Pioneer 10 and 11 opposing directions
> to CMBR 620km/s is irrelevant in this concept.
> The Pioneer velocities 12km/s are also irrelevant.
> The deceleration is constant for all frames of reference
> as observed.
The magnitude is observed to be constant over
time but the direction is not constant, the craft
are being accelerated in almost (but not quite)
opposite directions and your approach of using
energy doesn't address the direction issue at all.
Sorry Richard, but I don't see that this leads
anywhere.
best regards
George
Richard Saam
The most recent Anderson et al report
http://arxiv.org/abs/gr-qc/0510081
is a continuation of prior assertions about the Pioneer anomaly.
Anderson et al appear to still indicate the deceleration numbers are real.
What do you mean by:
"Anderson et al already acknowledge the coincidence
of the numbers" ????????
For clarification:
The deceleration is constant for all frames of reference
including spin
and in direction opposite to velocity vector.
From our position on earth,
we observe constant spacecraft deceleration in direction opposite to the
velocity vector we observe.
(p 10 & 11 spacecraft heading away from us in opposite directions but having
same deceleration opposite to velocity vector)
If one could position oneself any place in the solar system
the spacecraft would be observed to decelerate in the same constant manner
a = -Cd rho A c2 / M
in direction opposite to observed velocity vector whatever the magnitude of that
velocity vector.
It is acknowledged that E is a state variable
We are talking about Work = Force through distance.
Work = Integrate[0,x] F dx
In terms of momentum change
a relativistic approach is in order:
F dx = m v dv / sqrt(1 - v2/c2)
m = constant
and fixed by Heisenberg Uncertainty principle
m delta v delta x = h / 4pi
Assume a change in velocity over linear path establishing direction.
Work = Integrate[0,delta x] F dx
= m Integrate[0,delta v] v dv / sqrt(1 - v2/c2)
Work = F delta x = - m c^2 sqrt(1 - v^2/c^2)
= - m c^2 for v << c
then
F = M a = -(1/delta x)mc^2
= -(A/vol)mc^2
= -(m/vol)Ac^2
= -rho A c^2
and
a = -rho A c^2 / M
vector a is defined by linear path: delta x
A coefficient 'Cd' may be added
for the general hydrodynamic case
a = -Cd rho A c^2 / M
This means that the decceleration is constant
for any object velocity (v << c) or direction
(not tied to any particular reference frame)
which appears to be the observed case.
The Pioneer 10 and 11 opposing directions
to CMBR 620km/s is irrelevant in this concept.
The Pioneer velocities 12km/s are also irrelevant.
The deceleration is constant for all frames of reference
as observed.
I agree that there is no new physics here.
It just Basic 101 physics.
Nothing mysterious.
Very Basic.
And most importantly, it fits the data.
http://arxiv.org/abs/physics/9905007
Awaiting future space flights to confirm.
Best regards,
Richard Saam
George Dishman
"Richard Saam" <rds...@att.net> wrote in message
news:i4ocf.66882$zb5....@bgtnsc04-news.ops.worldnet.att.net...
They say clearly that the anomaly has been confirmed by
three separate analyses. From the middle of page 2 of
the above document:
"By now several studies of the Pioneer Doppler
navigational data have demonstrated that the
anomaly is unambiguously present in the
Pioneer 10 and 11 data."
However, there remain two quite different possibilities,
that the anomaly is indicative of a force acting on the
craft which is changing its speed and consequently its
distance from the Sun at any given time, or that it is
a direct effect on the measured frequency such as a
quadratic drift in the DSN station clocks which would of
course have no effect on the distance to the craft.
> What do you mean by:
>
> "Anderson et al already acknowledge the coincidence
> of the numbers" ????????
From the last paragraph of section 2.1 of the above paper:
"Finally we note that, motivated by the numerical
coincidence a_P ~ c H_0, where c the speed of light
and H_0 is the Hubble constant at the present time,
there have been many attempts to explain the anomaly
in terms of the expansion of the Universe."
Note their words, "numerical coincidence". The value of
a_P happens to be very close to the product of c and H_0,
but there is no known mechanism that can explain why they
should be connected.
> For clarification:
> The deceleration is constant for all frames of reference
> including spin
> and in direction opposite to velocity vector.
> From our position on earth,
> we observe constant spacecraft deceleration in direction opposite to the
> velocity vector we observe.
> (p 10 & 11 spacecraft heading away from us in opposite directions but
> having same deceleration opposite to velocity vector)
> If one could position oneself any place in the solar system
> the spacecraft would be observed to decelerate in the same constant manner
Yes, that is what is observed.
> a = -Cd rho A c2 / M
I have pointed this out already but you are repeating
hence so must I. If you are suggesting that there is some
sort of drag effect going on, which is implied by your
inclusion of the Cd term, then you must use v^2, not c^2
otherwise you get a finite acceleration when the speed is
zero and you cannot identify a direction for the vector.
A drag term must eventually reduce the speed of the object
to zero in the medium which is causing the drag which
means that it must be proportional to v for sufficiently
small values of v, even if it limits asymptotically to some
constant value at higher speeds.
Your use of the (m c^2) in deriving this suggests you are
trying to use energy instead of momentum but think how that
would apply if the responsible agent were dust. You would
be taking the bound energy in the dust particles and using
that value as if there was perfect conversion of matter to
energy going on. That doesn't make any sense.
> in direction opposite to observed velocity vector whatever the magnitude
> of that velocity vector.
>
> It is acknowledged that E is a state variable
>
> We are talking about Work = Force through distance.
You appear to be, I am not. The anomaly is not an energy
problem as I have said before, the question is where the
_momentum_ comes from.
<snip repeat of prevous equations>
> I agree that there is no new physics here.
> It just Basic 101 physics.
No it isn't Richard, all you are writing is a bunch of
equations with no physics to justify them. Basic 101
physics would describe this as a drag, there is no
viscosity between interplanetary dust particles that
I can see and and particle hit by the craft either gets
swept up (inelastic) or would ricochet off so the
equation I gave and you acknowledged is appropriate:
a = rho A v^2 / M
> Nothing mysterious.
Nothing mysterious about drag 101 certainly. What is
mysterious is the mechanism you are proposing that
justifies the use of c^2 instead of v^2 and implies the
direction of the acceleration. So far you don't seem to
have any suggestion, hence my view that you are just
restating the coincidence of the values without adding
any explanation.
> Very Basic.
>
> And most importantly, it fits the data.
>
> http://arxiv.org/abs/physics/9905007
>
> Awaiting future space flights to confirm.
Confirm what? You haven't offered any explanation to be
checked yet.
George
sean
> hence so must I. If you are suggesting that there is some
> sort of drag effect going on, which is implied by your
> inclusion of the Cd term, then you must use v^2, not c^2
> otherwise you get a finite acceleration when the speed is
> zero and you cannot identify a direction for the vector.
> A drag term must eventually reduce the speed of the object
> to zero in the medium which is causing the drag which
> means that it must be proportional to v for sufficiently
> small values of v, even if it limits asymptotically to some
> constant value at higher speeds.
>
> Your use of the (m c^2) in deriving this suggests you are
> trying to use energy instead of momentum but think how that
> would apply if the responsible agent were dust. You would
> be taking the bound energy in the dust particles and using
> that value as if there was perfect conversion of matter to
> energy going on. That doesn't make any sense.
Could you clarify this a bit for me. Are you suggesting that the
observed acceleration towards the sun on Pioneer isnt constant but is
decreasing ever so slightly?
George Dishman
No, something quite different. Suppose I throw a golf ball
at you. It might hurt but the momentum won't knock you
off your feet. A golf ball has a mass of 45g so by E=mc^2
it has about 1 megaton of energy. Apply that as kinetic
energy to an object of mass 100kg using E=1/2 m v^2
and you get a speed of 9000km/s. That doesn't mean that,
if I hand you a golf ball, you will hurtle off into the
sunset at Mach 26, but that is how Richard's calculations
work. As he says, both equations are basic physics but
that doesn't mean he is using them in a meaningful way.
The whole point of interest in the Pioneer anomaly is that
the mechanism by which the acceleration is produced is
unknown and Richard is not addressing that at all.
George
Richard Saam
Yes, I have thought about this.
The object would never go to zero
because of Heisenberg Uncertainty principle
and at that point no acceleration.
One would have to talk in terms
of probabilities expressed in quantum theory.
Work = Integrate[0,x] F dx
In terms of momentum change
a relativistic approach is in order:
F dx = m v dv / sqrt(1 - v2/c2)
m = constant
and fixed by Heisenberg Uncertainty principle
m delta v delta x = h / 4pi
Assume a change in velocity over linear path establishing direction.
Work = Integrate[0,delta x] F dx
= m Integrate[0,delta v] v dv / sqrt(1 - v^2/c^2)
Work = F delta x = - m c^2 sqrt(1 - (delta v)^2/c^2)
But there would always be residual movement
within the uncertainty principle
expressed in the probabilities of quantum theory.
Within the space medium itself, I calculate
delta v at .0015 cm/sec
delta x at 22 cm
Work approximately equal to - m c^2 for v << c
>
> Your use of the (m c^2) in deriving this suggests you are
> trying to use energy instead of momentum but think how that
> would apply if the responsible agent were dust. You would
> be taking the bound energy in the dust particles and using
> that value as if there was perfect conversion of matter to
> energy going on. That doesn't make any sense.
That is exactly what I am assuming.
but it is not the "bound energy in the dust particles".
It is the energy content of the space vacuum.
There is a direct conversion of energy (mc^2) going on
between this space vacuum, quintesence,
aether, dark matter, dark energy or whatever
numerically quantified as space density 'rho'
and the space craft or other object.
rho = conventionally accepted space vacuum density
= 2 H^2 / (8 pi G)
= 6.4E-30 g/cm^3
The dust particles are also interacting
with the spacecraft as mv^2
but at a much smaller energy level.
Given the assumption that all of the energy of the space
(other than dust particles)
in which the object is passing through is available to the object
then the physics concludes:
a = -Cd rho A c2 / M
Also an answer to Anderson et al 's coincidence notation is provided:
a = -Cd rho A c2 / M
= -Cd (2 H^2 / (8 pi G)) A c^2 / M
= -Cd ((2 H / (8 pi G)) A c / M) (c H)
where:
Cd (2 H / (8 pi G)) A c / M = 1
and
a = - cH
for Pioneer 10 & 11
My question is:
After the object passes through the space volume
and "consumes" the energy content in that space volume
as reflected in object decelerating,
does the space energy content regenerate itself
after the object passing???
Does it regenerate in place
or does the rest of the space medium
fill it in like a fluid filling a vessel???
Richard Saam
Richard Saam
George
I think I am addressing it very well
within the scope of elementary physics
to which all further insights must map
taking my lead from established
theoretical/experimental knowledge base
compiled in aircraft and ship design.
a = -Cd rho A v2 / M
(no question about direction of acceleration vector 'a'
a = -Cd rho A c^2 / M
You know enough about this development
not to characterize the calculations
in terms of golf balls moving at Mach 26.
As you know, we are talking
about densities of 6E-30 g/cm^3.
The mc^2 energy associated with this tenuous
medium decelerates an object the size of Pioneer 10 at 8E-8 cm/sec^2.
We are talking faint whispers here, not Hiroshima bombs.
Your assertion that the acceleration mechanism
is unknown seems to imply inquiry is unwarranted.
The basic elementary physics approach
as presented provides some insight.
The deceleration concept remains mostly unknown
especially in terms of the space medium nature.
At least the elementary procedure opens up the field
for more questions.
Richard Saam
Craig Markwardt
"sean" <jaymo...@hotmail.com> writes:
> Craig Markwardt wrote:
> > "sean" <jaymo...@hotmail.com> writes:
> > > Richard Saam wrote:
> > > > Craig Markwardt wrote:
> > >
> > > Can the too high observed speed of stars rotation in galaxies be
> > > explained by the pioneer anomaly? If it is .0000075 (or whatever) cms2
> > > inwards acceleration for pioneer would the same value applied to the
> > > mass of the rotating stars be enough to hold them in from being flung
> > > off ?
> >
> > No evidence supports such a speculation. The pioneer anomaly is a
> > constant *acceleration*. The galaxy velocity problem is a constant
> > *speed*. If one were to invoke a cloud of "dark matter" at the center
> > of the solar system to explain the pioneer anomaly, then the gravity
> > from the same matter would alter the orbits of the planets in a
> > detectable way, and no such planetary effect is found.
>
> I wasnt thinking about dark matter specifically but using that as one
> possibility of a new physics to explain the anomaly I wonder if in fact
> a cloud of dark matter could explain the anomaly. I would have thought
> a cloud of dark matter would not give a constant accelleration but
> rather one that diminished the farther Pioneeer was from the center of
> the solar system.
The acceleration profile of a cloud of dark matter would depend on the
density distribution of the cloud. It could be fashioned to make a
constant acceleration quite readily (i.e. a cumulative mass
distribution that goes as radius to the inverse square power).
> I think only an evenly distributed amount of dark matter across the
> universe would give a constant acceleration seen in the anomaly?
Not really. See above.
> But seeing as gravity is decribed as accelleration then the anomaly
> being constant would be effectively increasing the inward pull of
> gravity on any planet at any orbit radius. Making the inward pull on
> any one planet as...sun gravity + anomaly.
> For the solar system, this slight increase in `gravity` on the planet
> would mean that a planet would have to have a slightly faster orbital
> speed to stay in its particular orbit to compensate for the increased
> inward force of sun g + anomaly. Either that or have a larger weight
> than currently calculated.So maybe planets do show an observable effect
> from the anomaly. Its just that its disguised because calculations have
> attributed it to an underestimate of planets mass.
Umm, no. Gravitational acceleration of a body is independent of the
mass of the body; it depends on the mass of the sun and any "dark
matter" interior to the orbit. This is basic Newton's laws. If there
were a cloud of dark matter, the mass of the sun derived from
different planets would be different. And yet, all the planetary
orbits are consistent with a single mass for the sun.
> For galaxies, we get the same thing happening due to the anomaly.
> Either the orbital speeds have to be greater or the stars weights have
> to be greater to counter the additional inward acceleration of the
> anomaly. One would also get the effect of the observed orbital speed
> being more diverged on the increased side from that predicted, the
> farther out from the core on looks. This would be because the cores
> gravity decreases as one gets farther from the core but the anomaly
> stays constant. Im sure this is whats observed.
Again, no. What is observed is a constant velocity profile in
galactic disks. Simple algebra and Newton's lawas would show that a
constant acceleration does not make a constant velocity profile.
(i.e. v = sqrt(GM/r + a r) where a would be an anomalous acceleration,
and r is the distance; note that v is not independent of r in that
formula).
> I also wonder if it (anomaly)can explain mercurys orbit , until now
> only done so by GR?
Finally, no. The precession of mercury's perihelion requires an
addition of a 1/r^3 term in the gravitational effective acceleration,
not a constant. Relativistic precession has been detected in
asteroids and binary pulsars as well. It would be a strange kind of
dark matter that mimicked all those properties.
CM
George Dishman
> George Dishman wrote:
> >
> > at you. It might hurt but the momentum won't knock you
> > off your feet. A golf ball has a mass of 45g so by E=mc^2
> > it has about 1 megaton of energy. Apply that as kinetic
> > energy to an object of mass 100kg using E=1/2 m v^2
> > and you get a speed of 9000km/s. That doesn't mean that,
> > if I hand you a golf ball, you will hurtle off into the
> > sunset at Mach 26, but that is how Richard's calculations
> > work. As he says, both equations are basic physics but
> > that doesn't mean he is using them in a meaningful way.
> >
> > The whole point of interest in the Pioneer anomaly is that
> > the mechanism by which the acceleration is produced is
> > unknown and Richard is not addressing that at all.
>
> George
>
> within the scope of elementary physics
> to which all further insights must map
> taking my lead from established
> theoretical/experimental knowledge base
> compiled in aircraft and ship design.
>
> a = -Cd rho A v2 / M
> (no question about direction of acceleration vector 'a'
>
> a = -Cd rho A c^2 / M
>
> You know enough about this development
> not to characterize the calculations
> in terms of golf balls moving at Mach 26.
> As you know, we are talking
> about densities of 6E-30 g/cm^3.
> The mc^2 energy associated with this tenuous
> medium decelerates an object the size of Pioneer 10 at 8E-8 cm/sec^2.
> We are talking faint whispers here, not Hiroshima bombs.
Only the magnitude differs, the principle remains the
same. The problem is illustrated by the golf ball
exageration, if I hand you a golf ball, essentially
at rest, and you then clap your hands to convert it
to energy and fly off at Mach 26, you have violated
conservation of momentum. Since energy is a scalar
there is no associated direction so there is nothing
to determine the direction of your flight.
> Your assertion that the acceleration mechanism
> is unknown seems to imply inquiry is unwarranted.
I don't see how that follows. If I said the mechanism
was known then that could imply further consideration
was pointless, but saying it is unknown invites more
study IMO.
> The basic elementary physics approach
> as presented provides some insight.
"basic elementary physics" would conserve momentum
and would use v^2 instead of c^2 in the equation.
George
George Dishman
> George Dishman wrote:
> > "Richard Saam" <rds...@att.net> wrote in message
> > news:i4ocf.66882$zb5....@bgtnsc04-news.ops.worldnet.att.net...
> >
I presume when you say "The object would never
go to zero because of ...", you mean "... go to
zero SPEED because of...". Looking back, you
said in similar vein
> >>From our position on earth,we observe constant
> >>spacecraft deceleration in direction opposite to
> >>the velocity vector we observe.
Now consider two astronauts near the craft, one
moving radially out from the Sun at 11km/s and
the other at 13km/s. The craft is moving at 12km/s.
The first astronaut sees the craft moving outwards
at 1km/s with the acceleration in the opposite
direction. The second astronaut sees the craft
moving towards the Sun at 1km/s with the
acceleration in the _same_ direction. Basic
physics 101, velocity is frame dependent.
Both would see a non-zero acceleration based
on your explanation (but in different directions)
while an astronaut moving at 12km/s, the same
speed as the craft should see none since the
speed he measures is zero.
> One would have to talk in terms
> of probabilities expressed in quantum theory.
>
> Work = Integrate[0,x] F dx
>
> In terms of momentum change
> a relativistic approach is in order:
Sorry, in terms of acceleration, a vector approach
is in order. Physics 101 Richard.
> > Your use of the (m c^2) in deriving this suggests you are
> > trying to use energy instead of momentum but think how that
> > would apply if the responsible agent were dust. You would
> > be taking the bound energy in the dust particles and using
> > that value as if there was perfect conversion of matter to
> > energy going on. That doesn't make any sense.
>
> That is exactly what I am assuming.
> but it is not the "bound energy in the dust particles".
> It is the energy content of the space vacuum.
> There is a direct conversion of energy (mc^2) going on
> between this space vacuum, quintesence,
> aether, dark matter, dark energy or whatever
> numerically quantified as space density 'rho'
> and the space craft or other object.
That only conserves energy, not momentum. You are
ignoring the most important aspect. Energy is a
scalar so your approach is incapable of predicting
the direction of the acceleration, unlike the dust
case where the acceleration is in the direction of
the dust relative to the craft, conservation of
momentum resolves the problem.
George
Richard Saam
George:
We are talking "work" here.
From my elementary text:
Work = Ek2 - Ek1
That is, the work of the resultant force exerted on a particle equals the change
in kinetic energy of the particle.
Work = mv1^2/2 - mv2^2/2
Work = mc^2 - 0
What you are saying is that the Hiroshima bomb did no work.
The city devastation and that delayed impulse
to the departing Enola Gay indicate otherwise.
Substitute Enola Gay with Pioneer 10
and with appropriate scaling
witness the proposed deceleration effect.
There is no reason to indicate that momentum
is not conserved in the mc^2 process.
Richard
George Dishman
No we aren't Richard, we are talking momentum. The
power involved is tiny as you say, and the RTGs can
produce kW, many orders of magnitude more than is
required to explain the anomaly if there was some
reaction mass available. Even using pure radiation
thrust, it only takes 63W of anisotropically radiated
heat to explain a_P so "work" has never been a problem.
> From my elementary text:
>
> Work = Ek2 - Ek1
>
> That is, the work of the resultant force exerted on a particle equals the
> change in kinetic energy of the particle.
>
> Work = mv1^2/2 - mv2^2/2
v2 and v1 are the speed of the particle before and
after the impulse is applied respectively. This is
the Newtonian formula applicable only if v1 << c
and v2 << c.
> Work = mc^2 - 0
What you are saying there is that the speed of the
particle has changed from zero the speed of light
but your are using the Newtonian formula for kinetic
energy which is unusable for v=c. If your particles
are moving at an appreciable fraction of c then you
need to use the relativistic energy formula.
> What you are saying is that the Hiroshima bomb did no work.
No I'm not saying that at all. The bomb heated the
air which expanded away from that point in _all_
directions thud conserving momentum. As the air was
pushed up, the Earth would be pushed down but overall
the centre of momentum of the planet was unaffected,
momentum was conserved.
What I am saying is that just accounting for energy
isn't of use unless you can explain how the momentum
is conserved and explain the mechanism that is
responsible for both. It is the _mechanism_ that is
unknown.
> The city devastation and that delayed impulse
> to the departing Enola Gay indicate otherwise.
The plane was pushed by the shock wave from the
expansion of the air. All the bomb did was heat
the surroundings at ground zero.
> Substitute Enola Gay with Pioneer 10
> and with appropriate scaling
> witness the proposed deceleration effect.
There is no identified reaction mass available in
the case of Pioneer, no air to expand. You could
reasonably argue that the energy would heat the
craft but you cannot produce motion from that
without addressing the problem of momentum.
> There is no reason to indicate that momentum
> is not conserved in the mc^2 process.
"mc^2" isn't a process, but that aside, your use of
the equation to change kinetic energy does violate
conservation of momentum. I am saying you need not
only to calculate the energy but also the momentum.
Tell me how your proposed mechanism will conserve
it and you will have something worth discussing.
George
Richard Saam
> I am saying you need not
> only to calculate the energy but also the momentum.
> Tell me how your proposed mechanism will conserve
> it and you will have something worth discussing.
>
> George
>
>
George:
I have used the hydrodynamic model
F = Cd rho area c^2
to correlate with Pioneer Spacecraft data
In order to do this, the value of Cd is on the order of 50.
In the hydrodynamic model this would indicate
a laminar flow regime or a viscous effect
or in other words enlisting the Stokes' law analog:
F = 3pi viscosity c D
where:
D = space craft diameter
viscosity = momentum transfered per area
momentum = mc
area = space vacuum unit cell area
m = mass of space vacuum in that space vacuum unit cell.
This momentum is transfered to passing spacecraft.
Does this fall into something you would consider discussing further?
Richard
George Dishman
> George Dishman wrote:
>
>> I am saying you need not
>> only to calculate the energy but also the momentum.
>> Tell me how your proposed mechanism will conserve
>> it and you will have something worth discussing.
>>
>> George
>>
>>
>
> George:
>
> I have used the hydrodynamic model
>
> F = Cd rho area c^2
That is incorrect, the c^2 should be v^2, the
speed of the viscous substance relative to
the craft. If you are saying the material
hits the craft at the speed of light then
I need to look at it quite differently.
> to correlate with Pioneer Spacecraft data
>
> In order to do this, the value of Cd is on the order of 50.
>
> In the hydrodynamic model this would indicate
> a laminar flow regime or a viscous effect
> or in other words enlisting the Stokes' law analog:
>
> F = 3pi viscosity c D
>
> where:
>
> D = space craft diameter
>
> viscosity = momentum transfered per area
>
> momentum = mc
>
> area = space vacuum unit cell area
>
> m = mass of space vacuum in that space vacuum unit cell.
>
> This momentum is transfered to passing spacecraft.
>
> Does this fall into something you would consider discussing further?
Yes, you said "This momentum is transfered to [the]
passing spacecraft." The first question is transferred
FROM what?
The second question is how would you characterise the
motion of that substance relative to the Sun, moving
at the speed of light, orbiting it, or what?
George
sean
`a` is observed to be constant regardless of pioneers distance from the
sun then wouldnt the formula be v = sqrt(GM/r + a)?
Anyways, if I could follow the galaxy speed point a bit more now you`ve
supplied more information.
(Ive based my following argument on the assumption that in your post
you mean:
G= gravity constant
M= total galaxy mass
and your term "constant velocity profile" means all stars in galaxy
both inner and outer are observed to orbit at the same speed.)
If the formula v = sqrt(GM/r) supposedly works for the solar system but
not the galaxy, then how about this line of thought: What if the
galaxies mass distribution is different from what is predicted. For
instance the solar systems mass is presumably 99% in the sun which
focuses the total mass of the system into the very centre of all the
planetary orbits which in turn gives a good match to the formulas
predictions.
However galaxies do seem to have a different orbiting disc mass/
central coremass ratio then the solar systems planet mass /sun mass
ratio.
If one were to *speculate* that the central core was far less heavy
then expected and the mass of the galaxy was distributed more evenly
throughout the entire disc then it is possible to end up with a
velocity calculaytion using Nerwtons formula that gives a constant
velocity profile to stars orbiting speeds.
Just imagine if the core had 1/5 of the galaxy mass and the disc 4/5.
Then split the disc in half bisected by the core and each `side` has
2/5 of the mass. This model would mean that a star orbiting at the edge
of the disc would have all of the mass pulling at it from the core
direction divided by so the formula v = sqrt(GM/r) would apply.
But a star orbiting as close to the central core would have a different
gravitational pull effected on it. Because in fact 2/5 of the galaxy
mass which Ive attributed to each 1/2 disc section would be pulling at
the star from one side whilst the core and the other 1/2 disc
comprising of 3/5 of the mass would be pulling from the other side.
In effect then the 2/5 and 3/5 cancel each other out and leave only an
excess of 1/5 gravitaional pull on a central orbitting star from the
core direction. In other words a central star has only 1/5 the galaxies
mass pulling on it whereas the outer star has 5/5 of the galaxies mass
pulling on it. However as the radius of the outer star is larger than
the inner star then when it is divided by radius r the result balances
out
And the formula of Newtons for galaxies would thus be written , as
follows: v = sqrt(GM/R)/r where R is the total radius of the galaxy.
If Ive got my maths right the resulting formula should give a constant
velocity to orbiting stars regardless of r.
> > I also wonder if it (anomaly)can explain mercurys orbit , until now
> > only done so by GR?
>
> Finally, no. The precession of mercury's perihelion requires an
> addition of a 1/r^3 term in the gravitational effective acceleration,
> not a constant. Relativistic precession has been detected in
> asteroids and binary pulsars as well. It would be a strange kind of
> dark matter that mimicked all those properties.
If I understand you correctly, you are adding in a small constant
(1/r^3) to the g acceleration acting on mercury. Isnt that the same as
what I said above in that by adding in an additional small constant
acceleration `a` (pioneer anomaly) to the g acceleration one gets the
observed precession?
I added a constant `a` and you added a constant 1/r^3.
Sean
Richard Saam
George:
I tried to address all of your questions below:
(note '~' means approximate)
The following overall presentation is presented in terms
of answering questions related to a particular space density model
and how momentum is transfered from it to Pioneer 10 & 11 spacecraft
and other space objects:
Given a parity geometry (with dimensiona A and B)
which necessarily fulfills determinate identity:
| | | |
| -B 0 0 | | 0 0 -B |
| | | |
| | | |
| | | |
| 0 B -B | = - | 0 B -B |
| ------ ------ | | ------ ------ |
| sqrt(3) sqrt(3) | | sqrt(3) sqrt(3) |
| | | |
| -A 0 A | | -A 0 A |
| --- --- | | --- --- |
| 2 2 | | 2 2 |
| | | |
and this geometry fills space lattice
with hexagonal cells each of volume (2 sqrt(3) A B^2)
and having property that lattice = reciprical lattice.
Now each cell contains a charge pair e+ e-
in conjunction with cell dimensions A B
defining a permittivity and permeability
such that:
v2 = c2 / (permittivity permeability)
v is necessarily + or -
and
(+permittivity +permeability) = (-permittivity -permeability)
now time(t) is defined as interval to traverse cell:
t = 2B/v
t is necessarily + or -
Further given the Heisenberg Uncertainty
delta_p delta_x => hb/2 and delta_E delta_t => hb/2
The parity geometry
satisfies following condition:
delta_p delta_x = delta_E delta_t => hb/2
where:
delta_x = 2B
delta_v = v
delta_t = t
m = constant and always +
delta_p = m delta_v
delta_E = m(delta_v)^2/2 (always +)
delta_t = delta_x/delta_v
hb = h/(2*pi)
the B/A must have a specific ratio (2.38)
and a constant +mass(m) value of 110 x electron mass(me) to make this work
although the absolute values of A and B may be of any value
as limited by design where v <<<< c.
and modified Eddington relation (fine structure) holds:
(m/me)2 = 3 (hb c / e^2) (7 (A / B)2 + 12 (B / A)2 + 19)
Do virtual particles arise or conjure
to satisfy this very well defined
geometrical symmetrical situation?
The superconductivity phenomenon
wherein defined CPT theorem conditions may be related
in a particulary non obvious way
to atomic molecular structure
wherein either the + or - state
is "pinned" in space
such that supercurrents are observed.
In terms of modeling Pioneer 10 & 11 deceleration, the following is observed:
Space craft deceleration is opposite the direction of Pioneer motion and that
would include any velocity (x,y,z cartesian or polar coordinate) component vector.
The space lattice could move through interstellar space
and into the solar system at CMBR velocity (640 m/sec)
and not affect the Pioneer deceleration model.
Conventional momentum transfer from a fluid
to a body moving through it would be:
Integrate[0,x] F dx = m Integrate[0,v] v dv
Work = Fx = mv^2/2
In terms of momentum transfer to spacecraft
from the space lattice described above,
a relativistic approach is used:
Integrate[0,x] F dx = m Integrate[0,v] v dv / sqrt(1 - v2/c2)
Work = F x = - m c^2 sqrt(1 - v^2/c^2)
= 0 for v = c
~ - m c^2 for v << c
(note: this is consistent with particle accelerator experiments)
then
F = M a ~ -(1/x)mc^2
~ -(A/vol)mc^2
~ -(m/vol)Ac^2
~ -rho A c^2
and
a ~ -rho A c^2 / M
A coefficient Cd may be added
for the general hydrodynamic case
And then these decelerations modeled by:
F = M a ~ -Cd Area rho c^2
M = mass of space craft
Area = space craft cross sectional area
rho = space vacuum density using critical density
= 3 H2 / (8 pi G) = 9.6E-30 g/cc
Assuming universe in inflationary mode,
rho could be some fraction (.2 - .9) of the critical density
assume fraction .67
H= Hubble Constant 2.31E-18/sec
or 71.23 km/sec/million parsec
Cd = cooefficient which conventionally varies
as Reynold's number (inertial/viscous forces)
= 24/Reynold's number for Stokes' viscous (momentum/area) case
= near one for the inertial case
(note: space craft would never decelerate
to zero speed because of Heisenberg Uncertainty)
(note: Cd is on the order of 50 for the Pioneer spacecraft
indicating a laminar or viscous regime)
Pioneer 10 & 11 do vary in velocity (helio or geocentric) (dr/dt)
and have the same deceleration.
(8.74 ± 1.33) x 10^-8 cm/sec2
Based on a graphical method:
Heliocentric Trajectories for Selected Spacecraft, Planets, and Comets, NSSDC
Goddard National Spaceflight Center
http://nssdc.gsfc.nasa.gov/space/helios/heli.html
the spacecraft’s velocities were in a range of about 9% (28,600 – 27,500 mph for
Pioneer 10 “3Jan87 - 22July88” and 26,145 – 26,339 mph for Pioneer 11 ”5Jan87 -
01Oct90”)
But although velocities differ, the variance still falls within the observed
common deceleration:
(8.74 ± 1.33) x 10^-8 cm/sec2
The Primary data set may be viewed at:
Study of the Pioneer Anomaly:
A Problem Set Slava G. Turyshev,
Michael Martin Nieto,
and John D. Anderson
(Dated: February 24, 2005)
http://xxx.lanl.gov/abs/physics/0502132
Summary Motion data:
Pioneer 10 about 28,000 mph 1,250,000 cm/sec
Pioneer 11 about 26,000 mph 1,160,000 cm/sec
with deceleration for both at (8.74 ± 1.33) x 10^-8 cm/sec2
and the pioneer spacecraft rotational spin rates
Pioneer 10 about 4 rpm (2,581 cm/sec tip speed)
Pioneer 11 about 7 rpm (4,517 cm/sec tip speed)
with deceleration for both at .0067 rpm/year
Applying the model:
F = M a ~ -Cd Area rho c^2
in the torque mode
Torque = (spin axis Moment of Inertia) * (change in rotation rate)
(spin axis Moment of Inertia = 5.88E9 g cm2)
assuming psuedo paddles in the rotating Pioneer space craft
http://xxx.lanl.gov/abs/physics/9905007 page 21
provides calculations indicating that the common Pioneer 10 & 11 spin
deceleration data can be modeled.
This would seem to indicate that the Hypothesis:
Space craft deceleration is opposite the direction of Pioneer motion and that
would include any velocity (x,y,z cartesian or polar coordinate) component vector.
When the model is applied to objects rotating
about the sun for 5,000,000,000 years
a size cut off reflecting present day Asteroid size is achieved
(all smaller objects would have decelerated an fell into the sun or planets).
Conclusions:
1. Dark Matter and Dark Energy are the same thing and are represented by a
Trisine Elastic Space CPT lattice which is composed of virtual particles
adjacent to Heisenberg Uncertainty
2. Dark Matter and Dark Energy are related to Cosmic Microwave Background
Radiation (CMBR) through very large space dielectric (permittivity) and
permeability constants.
3. A very large Trisine dielectric constant and CPT equivalence makes the Dark
Matter/Energy invisible to electromagnetic radiation.
4. A velocity independent (v<<c) drag force acts on all objects passing through
the Trisine Space CPT lattice and models the translational and rotational
deceleration of Pioneer 10 & 11 space craft.
5. The Asteroid Belt marks the interface between solar wind and Trisine, the
radial extent an indication of the dynamics of the energetic interplay of these
two fields
6. The Trisine Model predicts no Dust in the Kuiper Belt. The Kuiper Belt is
made of large Asteroid like objects.
7. The observed Matter Clumping in Galaxies is due to magnetic or electrical
energetics in these particular areas at such a level as to destroy the Coherent
Trisine.
8. The apparent high observed speed of outer star rotation in galaxies is due to
forced redistribution with R of orbital velocities with time due to
disproportionate change in R of inner faster moving stars relative to R of outer
slower orbiting stars in accordance with orbital velocity (v) relationship:
This is a residual effect left over from the time when galaxies were primarily
dust or small particles and is still applicable in as much as galaxies still
consist of small particles (< a few meters) gravitationally coupled to large
objects such as stars.
9. The observed Universe Expansion is due to Trisine Space CPT lattice internal
pressure.
10. When the Trisine is scaled to molecular dimensions, superconductor
parameters such as critical fields, penetration depths, coherence lengths,
Cooper CPT Charge conjugated pair densities are modeled.
11. A superconductor is consider electrically neutral with balanced conjugated
charge as per CPT theorem. Observed current flow is due to one CPT time frame
being pinned relative to observer.
George Dishman
>
> George:
>
> I tried to address all of your questions below:
It looks more like you just cut and paste
previous text.
> (note '~' means approximate)
>
> The following overall presentation is presented in terms
> of answering questions related to a particular space density model
> and how momentum is transfered from it to Pioneer 10 & 11 spacecraft
> and other space objects:
>
> Given a parity geometry (with dimensiona A and B)
> which necessarily fulfills determinate identity:
>
> | | | |
> | -B 0 0 | | 0 0 -B |
> | | | |
> | | | |
> | | | |
> | 0 B -B | = - | 0 B -B |
> | ------ ------ | | ------ ------ |
> | sqrt(3) sqrt(3) | | sqrt(3) sqrt(3) |
> | | | |
> | -A 0 A | | -A 0 A |
> | --- --- | | --- --- |
> | 2 2 | | 2 2 |
> | | | |
>
>
> and this geometry fills space lattice
> with hexagonal cells each of volume (2 sqrt(3) A B^2)
> and having property that lattice = reciprical lattice.
> Now each cell contains a charge pair e+ e-
> in conjunction with cell dimensions A B
> defining a permittivity and permeability
> such that:
>
> v2 = c2 / (permittivity permeability)
You haven't defined v2 or c2 and neither
appears above anyway.
> v is necessarily + or -
> and
> (+permittivity +permeability) = (-permittivity -permeability)
>
> now time(t) is defined as interval to traverse cell:
Time for what to traverse the cell, the craft?
> t = 2B/v
>
> t is necessarily + or -
t is positive.
Nothing in that is of the slightest relevance
that I can see.
> In terms of modeling Pioneer 10 & 11 deceleration, the following is
> observed:
>
> Space craft deceleration is opposite the direction of Pioneer motion and
> that would include any velocity (x,y,z cartesian or polar coordinate)
> component vector.
No, it is only opposite in heliocentric coordinates
as I explained before. In a coordinate scheme with
origin moving radially out at 13km/s, the velocity
and acceleration of the craft are in the same
direction.
> The space lattice could move through interstellar space
> and into the solar system at CMBR velocity (640 m/sec)
> and not affect the Pioneer deceleration model.
That should be "km/s", not "m/s". It makes a big
difference as 640m/s is small compared to the
craft speed of 12km/s but the 620km/s of the CMBR
is far larger.
If this "lattice" is what causes the drag, then
the 'v' in your equation should be the vector
sum of 640km/s and the heliocentric craft speeds,
not the speed of light as you used. The result
will be that one craft will (probably) be
accelerated towards the Sun but the other will
be accelerated away from it.
> Conventional momentum transfer from a fluid
> to a body moving through it would be:
>
> Integrate[0,x] F dx = m Integrate[0,v] v dv
> Work = Fx = mv^2/2
Wrong, conventional momentum transfer is given by
a = rho A v^2 / M
> In terms of momentum transfer to spacecraft
> from the space lattice described above,
> a relativistic approach is used:
>
> Integrate[0,x] F dx = m Integrate[0,v] v dv / sqrt(1 - v2/c2)
This is just the same error you made before
Richard, that is the equation for energy, not
momentum and sticking in a gamma term doesn't
make it a "relativistic approach". For that,
you would need to use the relativistic momentum,
not the relativistic energy.
I don't see any point in going endlessly in circles.
George
sean
Correction
For a constant velocity profile the formula should be
v = sqrt{GM(r/R)/r}
Craig Markwardt
Nope. I'm assuming circular orbits here for simplicity, which give
the centripetal acceleration as v^2 / r.
Thus, if the acceleration is
accel = (gravitational) + (anomalous)
= (GM/r^2) + (a)
and you solve for v by multiplying by r on both sides, you get my formula.
> Anyways, if I could follow the galaxy speed point a bit more now you`ve
> supplied more information.
> (Ive based my following argument on the assumption that in your post
> you mean:
> G= gravity constant
> M= total galaxy mass
> and your term "constant velocity profile" means all stars in galaxy
> both inner and outer are observed to orbit at the same speed.)
>
> If the formula v = sqrt(GM/r) supposedly works for the solar system but
> not the galaxy, then how about this line of thought: What if the
> galaxies mass distribution is different from what is predicted. For
> instance the solar systems mass is presumably 99% in the sun which
> focuses the total mass of the system into the very centre of all the
> planetary orbits which in turn gives a good match to the formulas
> predictions.
> However galaxies do seem to have a different orbiting disc mass/
> central coremass ratio then the solar systems planet mass /sun mass
> ratio.
...
So this is exactly the theory of dark matter. A distribution of
matter which does not follow the distribution of luminous matter like
stars and dust. Dark matter is not very scientifically satisfying.
However, one can postulate some behaviors for it and then see if those
behaviors are consistent with the known universe... which is what is
being done.
> > > I also wonder if it (anomaly)can explain mercurys orbit , until now
> > > only done so by GR?
> >
> > Finally, no. The precession of mercury's perihelion requires an
> > addition of a 1/r^3 term in the gravitational effective acceleration,
> > not a constant. Relativistic precession has been detected in
> > asteroids and binary pulsars as well. It would be a strange kind of
> > dark matter that mimicked all those properties.
> If I understand you correctly, you are adding in a small constant
> (1/r^3) to the g acceleration acting on mercury. Isnt that the same as
> what I said above in that by adding in an additional small constant
> acceleration `a` (pioneer anomaly) to the g acceleration one gets the
> observed precession?
> I added a constant `a` and you added a constant 1/r^3.
No, r is a radius, so it varies with position, and is therefore not
constant.
While it is conceivable that one could construct a strange
distribution of mass in the solar system that might mimic the Mercury
perihelion effect, it's not clear how that same distribution would
affect the perihelion of the asteroid Icarus, or the bending of light
in the solar system, or Shapiro time delay. Or how it would affect
the orbits of the outer planets which would be sensitive to the extra
mass enclosed within their orbits. In short, by fine tuning dark
matter distribution to get one effect (Mercury perihelion advance),
one ruins the other high precision tests in the solar system.
CM
Richard Saam
>
>>In terms of momentum transfer to spacecraft
>>from the space lattice described above,
>>a relativistic approach is used:
>>
>>Integrate[0,x] F dx = m Integrate[0,v] v dv / sqrt(1 - v2/c2)
>
>
> This is just the same error you made before
> Richard, that is the equation for energy, not
> momentum and sticking in a gamma term doesn't
> make it a "relativistic approach". For that,
> you would need to use the relativistic momentum,
> not the relativistic energy.
The momentum concept is very clear,
based on well established concept of effective or relativistic mass
m / sqrt(1 - v2/c2)
out of favor in current physics, but still valid,
and applicable in this case for computing space medium momentum
to be transfered to spacecraft or object:
(m / sqrt(1 - v2/c2)) v
or
m v / sqrt(1 - v2/c2)
integrated across de Broglie momentum of space lattice
as imparted work to passing space craft or other object
inelastically
where 'm' is constant?
>
> I don't see any point in going endlessly in circles.
>
> George
To all:
George has made some minor corrections
and suggested clarifications to the presentation
which are included below:
If anyone else wishes to comment, I would be interested.
(note '~' means approximate)
The following overall presentation is presented in terms
of answering questions related to a particular space density model
which is related to the phenomenon of superconductivity
and how momentum is transferred from it to Pioneer 10 & 11 spacecraft
and other space objects
I have tried to be complete here, but
http://xxx.lanl.gov/abs/physics/9905007
with numerical references to CMBR temp 2.711 degree Kelvin
is the final reference:
Define a space lattice within the context of the
Charge - Conjugation
Parity - Change
Time - Reversal
Theorem
Given a parity geometry (with dimensions A and B)
which necessarily fulfills determinate identity:
| | | |
| -B 0 0 | | 0 0 -B |
| | | |
| | | |
| | | |
| 0 B -B | = - | 0 B -B |
| ------ ------ | | ------ ------ |
| sqrt(3) sqrt(3) | | sqrt(3) sqrt(3) |
| | | |
| -A 0 A | | -A 0 A |
| --- --- | | --- --- |
| 2 2 | | 2 2 |
| | | |
and this geometry fills space lattice
with hexagonal cells each of volume (2 sqrt(3) A B2)
and having property that lattice = reciprocal lattice.
Now each cell contains a charge pair e+ e-
in conjunction with cell dimensions A B
defining a permittivity and permeability
such that:
v = hb pi / (m B) (space lattice de Broglie velocity)
v^2 = c^2 / (permittivity permeability)
v is necessarily + or -
and
(+permittivity +permeability) = (-permittivity -permeability)
now time(t) is defined as interval to traverse space lattice cell:
t = 2B/v
t is necessarily + or -
Further given the Heisenberg Uncertainty
delta_p delta_x => hb/2 and delta_E delta_t => hb/2
The parity geometry
satisfies following condition:
delta_p delta_x = delta_E delta_t => hb/2
where:
delta_x = 2B
delta_v = v
delta_t = t
m = constant and always +
delta_p = m delta_v
delta_E = m(delta_v)^2/2 (always +)
delta_t = delta_x/delta_v
hb = h/(2*pi)
the B/A must have a specific ratio (2.38)
and a constant +mass(m) value of 110 x electron mass(me) to make this work
although the absolute values of A and B may be of any value
as limited by design where v <<<< c.
and modified Eddington relation (fine structure) holds:
(m/me)2 = 3 (hb c / e2) (7 (A / B)2 + 12 (B / A)2 + 19)
Do virtual particles arise or conjure
to satisfy this very well defined
geometrical symmetrical situation?
The superconductivity phenomenon
wherein defined CPT theorem conditions may be related
in a particularly non obvious way
to atomic molecular structure
wherein either the + or - state
is "pinned" in space
such that supercurrents are observed.
In terms of modeling Pioneer 10 & 11 deceleration, the following is observed:
Space craft deceleration is opposite the direction of Pioneer motion and that
would include any velocity (x,y,z cartesian or polar coordinate) component
vector. This is evidenced by constant observed Pioneer deceleration
(sun Pioneer radial velocity of 11 km/sec)
even though observed from the earth with a 29.80 km/sec sun orbital component.
Confirmation:
http://arxiv.org/abs/gr-qc/0104064
page 26, figure 17
Annual Pioneer deceleration sinusoidal amplitude = 1.6E-8 cm/sec^2
or 1.6/8.74 or about 20% of anomalous deceleration.
In other words:
Any two objects moving towards each other
or away from each other would observe
the same deceleration
(which would be an addition of their individual decelerations
as proportional to their individual mass to cross-sectional area ratios).
Example:
If the Pioneer spacecraft turned around
headed towards the earth
the same numerical (including sign) deceleration
would be observed by doppler shift
and indistinguishable doppler shift deceleration
from out going Pioneer SpaceCraft.
The space lattice could move through interstellar space
and into the solar system at CMBR velocity (640 km/sec)
and at any angle
and not affect the Pioneer deceleration model
as long as object velocity<<c.
Conventional momentum transfer from a fluid
to a body moving through it would be:
Integrate[0,x] F dx = m Integrate[0,v] v dv
Work = Fx = mv^2/2
which is equivalent to:
a = rho A v^2 / M
where v is object velocity.
In terms of momentum transfer to spacecraft
from the space lattice described above,
a relativistic momentum approach is used.
This is equivalent to integrating the de Broglie momentum
across the space lattice cell making available
that momentum to the passing Pioneer space craft.
Integrate[0,x] F dx = m Integrate[0,v] v dv / sqrt(1 - v^2/c^2)
now v is de Broglie velocity of elastic space lattice.
use: http://integrals.wolfram.com/
Work = F x = - m c^2 sqrt(1 - v^2/c^2)
= 0 for v = c
~ - m c^2 for v << c (confirm: photon has zero mass)
(note: this is consistent with particle accelerator experiments)
then
F = M a ~ -(1/x)mc^2
~ -(A/vol)mc^2
~ -(m/vol)Ac^2
~ -rho A c^2
and
a ~ -rho A c^2 / M
A coefficient Cd may be added
for the general hydrodynamic case
And then these decelerations modeled by:
F = M a ~ -Cd Area rho c^2
M = mass of space craft
Area = space craft cross sectional area
rho = space vacuum density using critical density
= 3 H^2 / (8 pi G) = 9.6E-30 g/cc
Assuming universe in inflationary mode,
rho could be some fraction (.2 - .9) of the critical density
assume fraction .67
H= Hubble Constant 2.31E-18/sec
or 71.23 km/sec/million parsec
Cd = cooefficient which conventionally varies
as Reynold's number (inertial/viscous forces)
= 24/Reynold's number for Stokes' viscous (momentum/area) case
= near one for the inertial case
(note: space craft would never decelerate
to zero speed because of Heisenberg Uncertainty)
(note: Cd is on the order of 50 for the Pioneer spacecraft
indicating a laminar or viscous regime)
Pioneer 10 & 11 do vary in velocity (helio or geocentric) (dr/dt)
and have the same deceleration.
(8.74 ± 1.33) x 10^-8 cm/sec^2
Based on a graphical method:
Heliocentric Trajectories for Selected Spacecraft, Planets, and Comets, NSSDC
Goddard National Spaceflight Center
http://nssdc.gsfc.nasa.gov/space/helios/heli.html
the spacecraft’s velocities were in a range of about 9% (28,600 – 27,500 mph for
Pioneer 10 “3Jan87 - 22July88” and 26,145 – 26,339 mph for Pioneer 11 ”5Jan87 -
01Oct90”)
But although velocities differ, the variance still falls within the observed
common deceleration:
(8.74 ± 1.33) x 10^-8 cm/sec^2
The Primary data set may be viewed at:
Study of the Pioneer Anomaly:
A Problem Set Slava G. Turyshev,
Michael Martin Nieto,
and John D. Anderson
(Dated: February 24, 2005)
http://xxx.lanl.gov/abs/physics/0502132
Summary Motion data:
Pioneer 10 about 28,000 mph 1,250,000 cm/sec
Pioneer 11 about 26,000 mph 1,160,000 cm/sec
with deceleration for both at (8.74 ± 1.33) x 10^-8 cm/sec2
and the pioneer spacecraft rotational spin rates
Pioneer 10 about 4 rpm (2,581 cm/sec tip speed)
Pioneer 11 about 7 rpm (4,517 cm/sec tip speed)
with deceleration for both at .0067 rpm/year
Applying the model:
F = M a ~ -Cd Area rho c^2
in the torque mode
Torque = (spin axis Moment of Inertia) * (change in rotation rate)
(spin axis Moment of Inertia = 5.88E9 g cm2)
assuming pseudo paddles in the rotating Pioneer space craft
http://xxx.lanl.gov/abs/physics/9905007 page 21
provides calculations indicating that the common Pioneer 10 & 11 spin
deceleration data can be modeled.
This would seem to indicate that the Hypothesis:
Space craft deceleration is opposite the direction of Pioneer motion and that
would include any velocity (x,y,z cartesian or polar coordinate) component vector.
When the model is applied to objects rotating
about the sun for 5,000,000,000 years
a size cut off reflecting present day Asteroid size is achieved
(all smaller objects would have decelerated an fell into the sun or planets).
Conclusions:
1. Dark Matter and Dark Energy are the same thing and are represented by a
Elastic Space CPT lattice which is composed of virtual particles adjacent to
Heisenberg Uncertainty
2. Dark Matter and Dark Energy are related to Cosmic Microwave Background
Radiation (CMBR) through very large space dielectric (permittivity) and
permeability constants.
3. A very large Elastic Space CPT lattice dielectric constant and CPT
equivalence makes the Dark Matter/Energy invisible to electromagnetic radiation.
4. A velocity independent (v<<c) drag force acts on all objects passing through
the Elastic Space CPT lattice and models the translational and rotational
deceleration of Pioneer 10 & 11 space craft.
5. The Asteroid Belt marks the interface between solar wind and Elastic Space
CPT lattice, the radial extent an indication of the dynamics of the energetic
interplay of these two fields
6. The Elastic Space CPT lattice Model predicts no Dust in the Kuiper Belt. The
Kuiper Belt is made of large Asteroid like objects.
7. The observed Matter Clumping in Galaxies is due to magnetic or electrical
energetics in these particular areas at such a level as to destroy the Coherent
Elastic Space CPT lattice.
8. The apparent high observed speed of outer star rotation in galaxies is due to
forced redistribution with R of orbital velocities with time due to
disproportionate change in R of inner faster moving stars relative to R of outer
slower orbiting stars in accordance with orbital velocity (v) relationship:
This is a residual effect left over from the time when galaxies were primarily
dust or small particles and is still applicable in as much as galaxies still
consist of small particles (< a few meters) gravitationally coupled to large
objects such as stars.
9. The observed Universe Expansion is due to Elastic Space CPT lattice internal
pressure.
10. When the Elastic Space CPT lattice is scaled to molecular dimensions,
sean
Craig Markwardt wrote:
> Thus, if the acceleration is
> accel = (gravitational) + (anomalous)
> = (GM/r^2) + (a)
> and you solve for v by multiplying by r on both sides, you get my formula.
Hi
Sorry Craig . I think I inadvertently mislead you in an earlier post
into making you think I was attributing both `a` and the galaxy
rotation to dark matter . I wasnt, and my apologies on that.
The point I was trying to make regarding a and the perehilion was; That
for the centripedal acceleration one has to factor in r... Obviously.
Because its strength varies with r when calculating the suns effect on
earth depending on earths distance from the sun . No problem there. But
with `a`, seeing as its not a centripedal acceleration dependent on
distance from the sun or at least it doesnt appear to be and seeing as
its observed to be constant, regardless of r, then one shouldnt
multiply it by r in the velocity equation above ? Otherwise the
equation assumes that `a` changes with r.
Sorry my mistake, I wasnt suggesting that the constant velocity
profile was attributed to extra dark matter. I was suggesting that the
assumed mass distribution of the visible mass of galaxies is incorrect.
To support this argument I offer the formula...
v = sqrt{GM(r/R)/r}...to see if a different mass distribution of
visible mass can explain the constant velocity profile. And it does.
Or at least I was asking you if you would mind having a look at my
formula ...
v = sqrt{GM(r/R)/r} to see if you could show that it does or doesnt
predict a constant velocity profile. As far as I can tell , the formula
does predict a constant velocity profile. And thats observed, so the
two do seem to match.
> > If I understand you correctly, you are adding in a small constant
> > (1/r^3) to the g acceleration acting on mercury. Isnt that the same as
> > what I said above in that by adding in an additional small constant
> > acceleration `a` (pioneer anomaly) to the g acceleration one gets the
> > observed precession?
> > I added a constant `a` and you added a constant 1/r^3.
>
> No, r is a radius, so it varies with position, and is therefore not
> constant.
Yes, obviously 1/r^3 r varies with r, or its distance from the sun over
the orbital period. But year to year thats roughly constant isnt it?
And year to year so is `a`. So technically they are both constant over
the full term of the year/orbit. But of course within the year 1/r^3 r
varies.
In fact so would `a` as its relative strength to mercuries orbital
position and velocity would vary.
So I would have thought that ultimately, both 1/r^3 r and `a` are
constant over full orbital periods.
Sean
George Dishman
>
> The momentum concept is very clear,
> based on well established concept of effective or relativistic mass
> m / sqrt(1 - v2/c2)
> out of favor in current physics, but still valid,
> and applicable in this case for computing space medium momentum
> to be transfered to spacecraft or object:
>
> (m / sqrt(1 - v2/c2)) v
>
> or
>
> m v / sqrt(1 - v2/c2)
Since v << c, the gamma factor is negligible. The momentum is
close to m * v but v is the vector sum of the 12km/s of the
craft and the CMBR dipole of 620 km/s as you said before. The
higher value predominates so we can determine the direction
your idea predicts for the acceleration of both craft, it
should be aligned with the CMBR dipole.
Well I'm afraid it isn't, it is towards the Sun for both craft
so I'm sorry Richard, it doesn't work.
George
Craig Markwardt
"sean" <jaymo...@hotmail.com> writes:
> Craig Markwardt wrote:
>
> > Thus, if the acceleration is
> > accel = (gravitational) + (anomalous)
> > = (GM/r^2) + (a)
> > and you solve for v by multiplying by r on both sides, you get my formula.
>
> Hi
> Sorry Craig . I think I inadvertently mislead you in an earlier post
> into making you think I was attributing both `a` and the galaxy
> rotation to dark matter . I wasnt, and my apologies on that.
> The point I was trying to make regarding a and the perehilion was; That
> for the centripedal acceleration one has to factor in r... Obviously.
> Because its strength varies with r when calculating the suns effect on
> earth depending on earths distance from the sun . No problem there. But
> with `a`, seeing as its not a centripedal acceleration dependent on
> distance from the sun or at least it doesnt appear to be and seeing as
> its observed to be constant, regardless of r, then one shouldnt
> multiply it by r in the velocity equation above ? Otherwise the
> equation assumes that `a` changes with r.
Your question doesn't make sense. My above equation is correct for
circular orbits. I showed how it was gotten by basic algebra. I
explicitly discuss "multiplying by r," but this is so basic I'm not
sure it's worth discussing.
But this is dark matter! One has to invoke some distribution of mass
other than the visible ones to explain galaxy rotation curves.
> To support this argument I offer the formula...
> v = sqrt{GM(r/R)/r}...to see if a different mass distribution of
> visible mass can explain the constant velocity profile. And it does.
> Or at least I was asking you if you would mind having a look at my
> formula ...
There is no physics in your formula so it is largely irrelevant. You
invented something that is constant. I could have just as easily said
v = CONSTANT. But that doesn't contain any physics or understanding
of the system, and hence provides no predictive power.
> > > If I understand you correctly, you are adding in a small constant
> > > (1/r^3) to the g acceleration acting on mercury. Isnt that the same as
> > > what I said above in that by adding in an additional small constant
> > > acceleration `a` (pioneer anomaly) to the g acceleration one gets the
> > > observed precession?
> > > I added a constant `a` and you added a constant 1/r^3.
> >
> > No, r is a radius, so it varies with position, and is therefore not
> > constant.
> Yes, obviously 1/r^3 r varies with r, or its distance from the sun over
> the orbital period. But year to year thats roughly constant isnt it?
> And year to year so is `a`. So technically they are both constant over
> the full term of the year/orbit. But of course within the year 1/r^3 r
> varies.
*How* the force law varies in space or time is important in
determining any net effect. 1/r^3 and constant radial accelerations
obviously have very different forms, and so they produce very
different secular effects. I would describe none of these effects as
"constant."
> In fact so would `a` as its relative strength to mercuries orbital
> position and velocity would vary.
> So I would have thought that ultimately, both 1/r^3 r and `a` are
> constant over full orbital periods.
Whether or not one can average a function over one period, does *not*
say anything about whether it is constant or not.
CM
Richard Saam
> "Richard Saam" <rds...@att.net> wrote in message
> news:Q00ff.116767$zb5....@bgtnsc04-news.ops.worldnet.att.net...
>
>>The momentum concept is very clear,
>>based on well established concept of effective or relativistic mass
>>m / sqrt(1 - v2/c2)
>>out of favor in current physics, but still valid,
>>and applicable in this case for computing space medium momentum
>>to be transfered to spacecraft or object:
>>
>>(m / sqrt(1 - v2/c2)) v
>>
>>or
>>
>>m v / sqrt(1 - v2/c2)
>
>
> Since v << c, the gamma factor is negligible.
>The momentum is
> close to m * v but v is the vector sum of the 12km/s of the
> craft and the CMBR dipole of 620 km/s as you said before.
These velocity magnitudes and any others (0<v<<c)
are irrelevant to observed space craft or object deceleration.
The
> higher value predominates so we can determine the direction
> your idea predicts for the acceleration of both craft, it
> should be aligned with the CMBR dipole.
Alignment with the CMBR dipole has nothing to do with deceleration effect.
>
> Well I'm afraid it isn't, it is towards the Sun for both craft
Best evidence and model prediction would be that it is towards the observer
wherever that may be.
> so I'm sorry Richard, it doesn't work.
>
> George
>
>
George
I am not about to go against
Gottfried Wilhelm Leibnitz (1646 - 1716)
or
Sir Isaac Newton (1643 - 1727)
in their origin of infinitesimal calculus
with some hand waving phrases such as "close to"
and "afraid it isn't".
The fact is that:
Work = Integrate[0,x] F dx = m Integrate[0,v] v dv / sqrt(1 - v2/c2)
where:
v is de Broglie velocity of elastic space lattice cell.
v << c
and
x is the elastic space lattice cell dimension.
such that uncertainty principle maintained
m v x = h / (4 pi)
(not just any mass will do)
(m is found to be equivalent to 110 electron mass)
(v is .0015 cm/sec)
(x is 3.5 cm)
(There could conceivably be other values of 'm' unknown at this time
conforming to some other lattice geometry than indicated)
use: http://integrals.wolfram.com/
therefore:
Work = F x = - m c2 sqrt(1 - v2/c2)
= 0 for v = c (confirm: photon has zero mass)
~ - m c2 for v << c
(note: this is consistent with particle accelerator experiments)
then
F = M a ~ -(1/x)mc2
~ -(A/vol)mc2
~ -(m/vol)Ac2
~ -rho A c2
and
a ~ -rho A c2 / M
A coefficient Cd may be added
for the general hydrodynamic case
And then these decelerations modeled by:
F = M a ~ -Cd Area rho c2
M = mass of space craft
Area = space craft cross sectional area
rho = space vacuum density using critical density
= 3 H2 / (8 pi G) = 9.6E-30 g/cc
Assuming universe in inflationary mode,
rho could be some fraction (.2 - .9) of the critical density
assume fraction .67
H= Hubble Constant 2.31E-18/sec
or 71.23 km/sec/million parsec
Cd = cooefficient which conventionally varies
as Reynold's number (inertial/viscous forces)
= 24/Reynold's number for Stokes' viscous (momentum/area) case
= near one for the inertial case
(note: space craft would never decelerate
to zero speed because of Heisenberg Uncertainty)
(note: Cd is on the order of 50 for the Pioneer spacecraft
indicating a laminar or viscous regime)
The pioneer sun referenced velocity of 12km/s
or earth (around sun) orbital velocity of 29.8 km/sec
or the CMBR dipole of 620 km/s
or any other velocity 'v' magnitude (0<v<<c)
do not contribute to the observed Pioneer deceleration
(as observed).
There are a lot of interesting questions to be discussed
in the context of the above (this is just the start).
For one, how to establish the size of the space vacuum de Broglie cell.
Relate the CMBR to the space vacuum de Broglie cell.
Can all of this be scaled to a laboratory experiment on earth?
Richard
George Dishman
> George Dishman wrote:
>> "Richard Saam" <rds...@att.net> wrote in message
>> news:Q00ff.116767$zb5....@bgtnsc04-news.ops.worldnet.att.net...
>>
>>>The momentum concept is very clear,
>>>based on well established concept of effective or relativistic mass
>>>m / sqrt(1 - v2/c2)
>>>out of favor in current physics, but still valid,
>>>and applicable in this case for computing space medium momentum
>>>to be transfered to spacecraft or object:
>>>
>>>(m / sqrt(1 - v2/c2)) v
>>>
>>>or
>>>
>>>m v / sqrt(1 - v2/c2)
>>
>>
>> Since v << c, the gamma factor is negligible.
>
> but not zero and worthy of analysis
The uncertainty in the craft measurements is
going to be vastly larger.
> >The momentum is
>> close to m * v but v is the vector sum of the 12km/s of the
> m is not just any mass as the conventional case that you are suggesting.
It is the amount by which the mass of the craft increases
due to whatever it is you are sweeping up.
>> craft and the CMBR dipole of 620 km/s as you said before.
> I never said vector sum (12 + 620)
No but you should have. You said:
"Richard Saam" <rds...@att.net> wrote in message
news:gf1ef.98646$zb5....@bgtnsc04-news.ops.worldnet.att.net...
>
> The space lattice could move through interstellar space
> and into the solar system at CMBR velocity (640 m/sec)
> and not affect the Pioneer deceleration model.
If the heliocentric speed of the "space lattice" is
640km/s and the heliocentric speed of the craft is
12km/s then the speed of the latice relative to the
craft is the vector sum of those and that is the
speed that determines the momentum in
p = m * v / sqrt(1 - (v/c)^2
(The figure I have is 620km/s but that's an aside.)
> These velocity magnitudes and any others (0<v<<c)
> are irrelevant to observed space craft or object deceleration.
IOn the contrary, they are fundamental if you are saying
that is the heliocentric speed of your "space lattice"
ad drag from that lattice is the cause of the anomaly.
>> The
>> higher value predominates so we can determine the direction
>> your idea predicts for the acceleration of both craft, it
>> should be aligned with the CMBR dipole.
> Not so. The model does not predict this.
If drag from your lattice is your suggested mechanism
then it does.
> Alignment with the CMBR dipole has nothing to do with deceleration effect.
>>
>> Well I'm afraid it isn't, it is towards the Sun for both craft
> Deceleration resolution has not been resolved between sun and earth.
> Best evidence and model prediction would be that it is towards the
> observer
> wherever that may be.
Right, but frag from a "space lattice" would not be so
aligned unless it was close to being at rest relative
to the Sun. Im that case the speed in the momentum
equation becomes close to the 12km/s of the craft and
the direction close to opposing the velocity.
>> so I'm sorry Richard, it doesn't work.
> Your conclusion is based on limited information on problem scope.
Possibly, but I should have said enough for you to
understand what you need to clarify.
> I am not about to go against
> Gottfried Wilhelm Leibnitz (1646 - 1716)
> or
> Sir Isaac Newton (1643 - 1727)
> in their origin of infinitesimal calculus
> with some hand waving phrases such as "close to"
"close to" just means I do not intend to look up
the exact headings of the Pioneers or the alignment
of the CMBR dipole to do the exact calculations.
When one number is 50 times larger than the other,
the vector sum will be "close to" the larger. Since
that is in the direction of the dipole for both craft,
your "space lattice" does not explain the nearly
opposed directions of the anomaly for the two craft.
> and "afraid it isn't".
>
> The fact is that:
>
> Work = Integrate[0,x] F dx = m Integrate[0,v] v dv / sqrt(1 - v2/c2)
As I have pointed out several times, work, or energy,
is not the problem, it is mechanism for the change of
momentum that needs to be explained. This equation
isn't wrong, but only the one you give at the top of
this post is relevant to my comments.
George
Richard Saam
>>
>>Work = Integrate[0,x] F dx = m Integrate[0,v] v dv / sqrt(1 - v2/c2)
>
>
> As I have pointed out several times, work, or energy,
> is not the problem, it is mechanism for the change of
> momentum that needs to be explained. This equation
> isn't wrong, but only the one you give at the top of
> this post is relevant to my comments.
>
> George
>
>
Work = Integrate[0,x] F dx = - m Integrate[0,v] v dv
+ m Integrate[0,v] v dv / sqrt(1 - v^2/c^2)
Therefore:
Work = Fx = -mv^2/2 - mc^2 * sqrt(1 - v^2/c^2)
Therefore:
Fx = -mv^2/2 - mc^2 * sqrt(1 - v^2/c^2)
Therefore:
F = -(m /x)(v^2/2 + c^2 * sqrt(1 - v^2/c^2))
Given:
1/x = (Swept Volume) / (Pioneer Cross-sectional Area)
= Volume / Area
and
rho = m / Volume = 110*electron mass / Volume
where: A & B are space lattice dimensions
A = 9.29 cm and B = 27.11 cm
Volume = 2 sqrt(3) A B^2 = de Broglie Cell Volume
vd = de Broglie velocity = hb pi / (m B) = .0015 cm/sec
v = n vd (n = vd multiples composing v)
space lattice (quantum mechanical medium is composed
of conjoining cells with no voids.
Therefore:
F = -rho Area (v^2/2 + c^2 * sqrt(1 - v^2/c^2))
Inspection would indicate that as (0<v<<c):
F ~ -rho Area c^2
and specifically for:
sun referenced Pioneer velocity of 12km/s
or earth (around sun) orbital velocity of 29.8 km/sec
or sun (around milky way center) rotation velocity of 400 km/sec
or the CMBR dipole of 620 km/s
or any other velocity 'v' magnitude (0<v<<c)
do not contribute to the observed Pioneer deceleration
(as observed)
(noting that c ~ 300,000 km/sec).
if deceleration or slowing down is observed in one space lattice cell
the same will be observed in a series of space lattice cells
and independent of v or n vd
The observed deceleration will be
in the line of observer to object
in accordance with a = F/M
where M is the Mass of the object.
This deceleration or slowing rate
will be numerically the same whether object
is moving away from the observer or towards the observer.
Conceptually, it is like a disc sweeping out a cylinder
through a quantum mechanical medium of virtual particles
in equilibrium with CMBR
having the characteristics of a Bose Einstein Condensate
(hb^2 / 2m) (pi / B)^2 = m v^2 / 2
It does not matter how fast the disc is moving.
The Work performed in doing is:
m Integrate[0,v] v dv / sqrt(1 - v^2/c^2)
which approaches mc^2
There is nothing more to give (mc^2) per distance or per volume,
so resulting 'work' has to be nearly constant with velocity 'v' << c.
If the observer and pioneer/object
where in line with CMBR 640 km/sec dipole
and having + or - relative velocity 'v'
the same pioneer/object (deceleration or slowing down)
would be observed as if
observer and pioneer line was orthogonal
to CMBR 640 km/sec dipole.
The Force and (deceleration or slowing rate)
is always the same for a given F/M.
(M = Mass of Pioneer or object)
Richard
George Dishman
>>>
>>>Work = Integrate[0,x] F dx = m Integrate[0,v] v dv / sqrt(1 - v2/c2)
>>
>>
>> As I have pointed out several times, work, or energy,
>> is not the problem, it is mechanism for the change of
>
> Work = Integrate[0,x] F dx = - m Integrate[0,v] v dv
> + m Integrate[0,v] v dv / sqrt(1 - v^2/c^2)
>
As I said, energy is not of interest. If all you
can do is keep repeating this, I see no point in
continuing.
George
Richard Saam
> "Richard Saam" <rds...@att.net> wrote in message
> news:78rgf.143265$zb5....@bgtnsc04-news.ops.worldnet.att.net...
>
>>George Dishman wrote:
>>
>>>"Richard Saam" <rds...@att.net> wrote in message
>>>news:Szsff.68627$qk4....@bgtnsc05-news.ops.worldnet.att.net...
>>>
>>>>Work = Integrate[0,x] F dx = m Integrate[0,v] v dv / sqrt(1 - v2/c2)
>>>
>>>
>>>As I have pointed out several times, work, or energy,
>>>is not the problem, it is mechanism for the change of
>>>momentum that needs to be explained. ...
>>
>>Work = Integrate[0,x] F dx = - m Integrate[0,v] v dv
>> + m Integrate[0,v] v dv / sqrt(1 - v^2/c^2)
>>
>>Therefore: ...
>
>
> As I said, energy is not of interest. If all you
> can do is keep repeating this, I see no point in
> continuing.
>
> George
>
>
If you do not understand
the Work Energy Principle
and relationship to Momentum
I guess not.
Basic Physics 101
Richard
Richard Saam
> "Richard Saam" <rds...@att.net> wrote in message
news:78rgf.143265$zb5....@bgtnsc04-news.ops.worldnet.att.net...
>
>> George Dishman wrote:
>>
>>> "Richard Saam" <rds...@att.net> wrote in message
news:Szsff.68627$qk4....@bgtnsc05-news.ops.worldnet.att.net...
>>>
>>>> Work = Integrate[0,x] F dx = m Integrate[0,v] v dv / sqrt(1 - v2/c2)
>>>
>>>
>>>
>>> As I have pointed out several times, work, or energy,
>>> is not the problem, it is mechanism for the change of
>>> momentum that needs to be explained. ...
>>
>>
>> Work = Integrate[0,x] F dx = - m Integrate[0,v] v dv
>> + m Integrate[0,v] v dv / sqrt(1 - v2/c2)
>>
>> Therefore: ...
>
>
>
> As I said, energy is not of interest. If all you
> can do is keep repeating this, I see no point in
> continuing.
>
> George
>
>
George
If you do not understand
the Work Energy Principle or Theorem
and relationship to Momentum
I guess not.
Basic Physics 101
Newton's Second Law
F = m dv/dt
differential identity
dv/dt = v dv/dx
Therefore:
F = m v dv/dx
F dx = m v dv
integrate and then
Work = Ek2 - Ek1
Statement of Work Energy Principle or Theorem
"The work of the resultant force
exerted on a particle
equals the change in kinetic energy
of the particle"
In terms of relativistic mass
F dx = m (1 - v^/c^2)^(-1) v dv
where m is constant
and m v x = h / (4 pi) (Heisenberg's Uncertainty)
Integrate, then work in sweeping across 'x'
Work = -mc^2
Richard
sean
Craig Markwardt wrote:
> But this is dark matter! One has to invoke some distribution of mass
> other than the visible ones to explain galaxy rotation curves.
If a galaxies visible mass (visible stars and gas) is distributed
evenly across its radius profile and this in turn gives a constant
velocity proofile using newtons formula then why invoke dark matter?
Its not needed.
> There is no physics in your formula so it is largely irrelevant. You
> invented something that is constant.
I didnt add a constant. You `invented` that. I only added a variable
`R` (radius of galaxy) to your formula. And R couldnt be constant as
all galaxies have different radii.
>I could have just as easily said
> v = CONSTANT. But that doesn't contain any physics or understanding
> of the system, and hence provides no predictive power.
>
Another mistake of yours here. The formula does have predictive power.
It predicts a constant velocity profile and that is observed.
Your problem is that when you use v = sqrt(GM/r) you incorrectly
assume that GM is constant regardless of r. Thats wrong because its
obvious that GM decreases where r decreases. If GM stayed constant
regardless of r one would have to assume that ALL the mass or at least
99.5 % of it was located in the core.
> *How* the force law varies in space or time is important in
> determining any net effect. 1/r^3 and constant radial accelerations
> obviously have very different forms, and so they produce very
> different secular effects. I would describe none of these effects as
> "constant."
Yes it does vary, but then again thats irrelevent to the point
I make. Both `a` and 1/r^3 are on similar orders of magnitude *each*
orbit, so it doesnt matter if one isnt constant.
Sean
Craig Markwardt
"sean" <jaymo...@hotmail.com> writes:
> Craig Markwardt wrote:
>
> > But this is dark matter! One has to invoke some distribution of mass
> > other than the visible ones to explain galaxy rotation curves.
>
> If a galaxies visible mass (visible stars and gas) is distributed
> evenly across its radius profile and this in turn gives a constant
> velocity proofile using newtons formula then why invoke dark matter?
> Its not needed.
Because (a) a uniform disk of matter does not produce a constant
velocity profile, and (b) the amount of visible matter (stars, dust)
is not enough to explain the magnitude of the observed velocity.
> > There is no physics in your formula so it is largely irrelevant. You
> > invented something that is constant.
>
> I didnt add a constant. You `invented` that. I only added a variable
> `R` (radius of galaxy) to your formula. And R couldnt be constant as
> all galaxies have different radii.
Actually, the discussion which you deleted was about whether the
"Pioneer anomaly," a constant acceleration, could account for galactic
rotation curves. My point was to show that a constant acceleration
produces a non-constant velocity profile and hence does perform as you
had supposed.
Then you speculated that v = sqrt(GM)/r, then v = sqrt(GM(r/R)/r).
The main point is that there was no real basis advanced for either of
these arbitrary formulae, and neither of these produces a constant
radial acceleration like the Pioneer "anomaly."
In your second formula velocity is trivially constant with respect to
r. How could that be? Let's say that the mass distribution is such
that the mass enclosed at a radius r is an effective mass M' = M(r/R).
In which case one gets the standard formula v = sqrt(GM'/r), but with
a revised mass distribution. This effectively the model of dark
matter. Why? Because the visible mass in stars, gas and dust do not
follow this mass distribution. There must be some other undetectable
mass. Your supposition *is* dark matter!
In addition, your supposed mass distribution, M' ~ r, is incosistent
with your *other* supposition above that the mass distribution is
"even" (uniform), which produces M' ~ r^2. Your two suppositions lead
to a contradiction, which is not good.
> >I could have just as easily said
> > v = CONSTANT. But that doesn't contain any physics or understanding
> > of the system, and hence provides no predictive power.
> >
> Another mistake of yours here. The formula does have predictive power.
> It predicts a constant velocity profile and that is observed.
> Your problem is that when you use v = sqrt(GM/r) you incorrectly
> assume that GM is constant regardless of r. Thats wrong because its
> obvious that GM decreases where r decreases. If GM stayed constant
> regardless of r one would have to assume that ALL the mass or at least
> 99.5 % of it was located in the core.
See above.
> > *How* the force law varies in space or time is important in
> > determining any net effect. 1/r^3 and constant radial accelerations
> > obviously have very different forms, and so they produce very
> > different secular effects. I would describe none of these effects as
> > "constant."
>
> Yes it does vary, but then again thats irrelevent to the point
> I make. Both `a` and 1/r^3 are on similar orders of magnitude *each*
> orbit, so it doesnt matter if one isnt constant.
I note that you did not substantiate your claim. In fact, constant
and 1/r^3 force laws produce quite different secular results.
Consider that the orbit of Mercury which varies by a factor of 1.5 in
radius from the sun, or a variation of (1.5)^3 ~ 4 in force, whereas
the "anomaly" is constant. For other bodies, like the asteroid
Icarus, the variations are even more extreme, and the test of GR is
even stronger.
And let's not ignore the fact that even at the *mean* orbital radius
of Mercury, the acceleration due to GR is more than 4 times the
Pioneer "anomaly." Such a discrepancy could have easily been
detected, but has not been.
And finally, there are many tests of GR which are too numerous to list
here, but have been cited before. Why you want to replace a
well-tested theory with a totally unexplained "anomaly" is the real
puzzle.
CM
George Dishman
> George Dishman wrote:
>
> > As I said, energy is not of interest. If all you
> > can do is keep repeating this, I see no point in
> > continuing.
>
> the Work Energy Principle or Theorem
> and relationship to Momentum
> I guess not.
I understand them quite well thanks.
> Basic Physics 101
>
> Newton's Second Law
>
> F = m dv/dt
Sorry Richard, that's wrong. If your going to do
"Basic Physics 101" you need to get it right. Force
is defined by:
F = dp/dt
where p is the momentum. For v << c we can approximate
the momentum by the non-relativistic p = m v hence
F = m dv/dt + v dm/dt
Note that the second term may not be negligible where
you have a moving object being slowed by drag. However,
let's gloss over that and see where this goes.
Note importantly that F and v are vectors.
> differential identity
>
> dv/dt = v dv/dx
>
> Therefore:
>
> F = m v dv/dx
Noting again that F, v and dv/dx are all vectors.
What about dv/dx and dv/dz ? For Physics 101 you
need to write the complete formula and then for
example you could say the trajectory is in the
x,y plane and set z to zero but you can only
ignore one since the path is hyperbolic.
> F dx = m v dv
Noting again that F, v and dv/dx are all vectors
and you are again neglecting y and z components.
> integrate and then
>
> Work = Ek2 - Ek1
Hold on there Richard, force * distance is the
work done only when the distance is measured in
the direction of force. You have forgotten that
most of your quantities are vectors.
> Statement of Work Energy Principle or Theorem
>
> "The work of the resultant force
> exerted on a particle
> equals the change in kinetic energy
> of the particle"
>
> In terms of relativistic mass
>
> F dx = m (1 - v^/c^2)^(-1) v dv
And since you assumed above that v << c, we can
neglect the gamma term. If you don't like that,
start again but use the relativistic formula for
momentum from the beginning.
> where m is constant
You started by neglecting the dm/dt term.
> and m v x = h / (4 pi) (Heisenberg's Uncertainty)
Oh dear, no. The Uncertainty Principle applies to
the _uncertainties_ in the measurements, not the
measurements themselves. Here mv is the momentum
and x is one coordinate of the location (previously
you used x as the distance moved in the x direction)
but it is the product of the uncertainties in the
measurements that is constrained. Also note that
it is an inequality:
dx * dp > h_bar / 2
That says we cannot measure both the location and
momentum to unlimited accuracy.
http://en.wikipedia.org/wiki/Uncertainty_principle
In the case of Pioneer, practical limitations mean
we are nowhere near that limit, the uncertainty in
the location alone is thousands of km.
At the end of the day Richard, all you are showing
is that applying a force to a mass changes its
momentum and kinetic energy which as you say is
basic physics, but what you haven't attempted to do
is state the mechanism that is applying the force
in the case of the Pioneer craft. As I have said
several times, the kinetic energy in that case is
tiny compared to the RAG waste heat so energy is
not of interest, it is the _mechanism_ by which the
force is applied that is being sought. If you are
claiming it is drag then the dm/dt factor is
fundamental and you have simply ignored it.
George
sean
Craig Markwardt wrote:
> Because (a) a uniform disk of matter does not produce a constant
> velocity profile,
I think I wasnt clear enough with my description as your expression
`uniform disk of matter` doesnt seem to fit . Especially considering
that when M is multiplied by (r/R) and then inserted into your formula
it does give a mathematical constant velocity profile.
Please see my explanation below if you want a clear picture of what
I mean by M(r/R)
>and (b) the amount of visible matter (stars, dust)
> is not enough to explain the magnitude of the observed velocity.
I dont see how we can accurately calculate the `mass` of any galaxy
except by observing its internal velocities?
Theres one thing you suggest above that I was wondering if you
could clarify. Youve said previously that all galaxies have constant
velocity profiles . But are you also saying above that in larger
galaxies stars all spin at faster speeds than in the smaller galaxies ?
Or do you mean that all sizes of galaxies spin at the same speed at
constant velocity profiles but its just that the velocity profile
is different from that predicted?
> Then you speculated that v = sqrt(GM)/r, then v = sqrt(GM(r/R)/r).
> The main point is that there was no real basis advanced for either of
> these arbitrary formulae, and neither of these produces a constant
> radial acceleration like the Pioneer "anomaly."
Yes I did mention already that I diverged from the anomaly
with this latest argumnent on rotation curves. Im not reffering to
the anomaly here now . Only below with mercury. And I now accept your
argument below about ruling out the anomaly> In your second formula
velocity is trivially constant with respect to
> r. How could that be? Let's say that the mass distribution is such
> that the mass enclosed at a radius r is an effective mass M' = M(r/R).
> In which case one gets the standard formula v = sqrt(GM'/r), but with
> a revised mass distribution. This effectively the model of dark
> matter. Why? Because the visible mass in stars, gas and dust do not
> follow this mass distribution. There must be some other undetectable
> mass. Your supposition *is* dark matter!
What I question is how do you know the mass distribution
does not follow this mass distribution. As I mentioned above, the only
accurate way to determine mass distribution is by observing rotation
speeds.
Incidentally I have tried a computor simulation of this distribution
to see if it produces a galaxy shape and it does. Very well.
Ill send you a quicktime output if you want.
Another argument agaisnt current modeling is that your original
formula v=sqrt(GM/r) only works if 99% of the mass is in the core.
Visually this is not corroborated at all by what we see. A visual
assessment of galaxy shape to me conveys the impression that no more
than
a 1/4 of the mass is in the core. Defitely nowhere near 99%.
So that rules out current mass distribution models and means
that v=sqrt(GM/r) is not an acceptable formula for determining
galaxy mass and internal velocities. For the solar system , yes , but
thats because 99% of the solar system mass is in the sun?
> In addition, your supposed mass distribution, M' ~ r, is incosistent
> with your *other* supposition above that the mass distribution is
> "even" (uniform), which produces M' ~ r^2. Your two suppositions lead
> to a contradiction, which is not good.
I dont think Ive described the distribution well enough.
I would say that my use of `even` is at best more like the first not
the
second expression you use above. It is definitely not described by a
cube root function. Is M(r/R) the same as M ~ r^3 ?
I dont think so. For that matter is M(r/R) the same as
M ~ r? I dont think so either as r/R is not the same as r.
If you could read all the explanation below I think it should
help clarify things. Ive made it as short as possible.
If a galaxy is 10 light years wide and is divided into 5 concentric
rings (one includes the core), each 1 light year wide.
<............10 wide...........>
5 4 3 2 1(core)1* 2 3 4 5**
The core is in the middle.
If you then say that 1/5 of the mass is assigned to each
ring so that for instance between 2-3 lightyears from the core
there is 1/5 of the galaxies mass. And in the core within the
1 light year radius also 1/5 of the mass.etc etc.
This doesnt mean that the density is uniform.In the outer
rings,although the ring area is greater the stars are
concentrated in 2 spiral arms. And conversely in the core
where the area of the `ring` is smaller, the core protudes
either side of the disc in its profile.
I`ve modelled this in a 3-d computer simulation and a
distribution of mass in this manner produces a typical
galaxy shape. I can send you a quicktime showing this.
Anyways this means that GM must be calculated for each
different r star using GM(r/R)
Why ? Because for instance a star `*` (see illustration)
at 1r has 3/5 of the galaxy mass pulling at it from the left
and 2/5 from the opposite direction. This can only mean
that the pull balances out and only 1/5 mass (1/5GM)is
felt by the star EVEN though its near the core. A star
at the rim `**` at 5r on the other hand has all 5/5 of the
mass pulling at it inwards and thus it feels the pull
of 5/5 (5/5 GM)
However once you calculated GM for the two stars
seperately and put these values into your formula..
v=sqrt (GM)/r
You can see that mathematically the two produce the
same velocity even though one star is near the core
and one is near the rim.Because the 5/5GM star at
5r gets divided by 5 whereas the 1/5GM star at 1 r
only gets divided by 1.
> And let's not ignore the fact that even at the *mean* orbital radius
> of Mercury, the acceleration due to GR is more than 4 times the
> Pioneer "anomaly." Such a discrepancy could have easily been
> detected, but has not been.
If 1/R^3 is 4 * 7.5 10-8cm/s^2 or
30 * 10-8cm/s^2 then
I accept this point as conclusive against the anomaly.
Sean
Hero.van...@gmx.de
sean wrote:
> ....mass .. galaxy
I do not understand much of Your discussion. But talking about mass and
galaxies, are You hinting at the gravity pull of the other stars of our
milky way - pure in Newton physics. Just like a sledge pulled by a
horse and a mouse, You hardly will notice the pull of the mouse, but
when You replace horse by dog and dog by cat and so forth suddenly one
will notice the pull of the mouse, especially if the pull is into a
different direction from the second animal.
Or for a more beautifull picture of this read Arthur C. Clarke story
about a chinese vase and in different texts a more detailed picture
about gravity funnels of several stars adding up.
Does the gravity of other stars explain part of the minmal acceleration
on top of the reduction, which is normal, when drifting on a
hyperbolic route away from our sun?
Regards
Hero
Craig Markwardt
"sean" <jaymo...@hotmail.com> writes:
> Craig Markwardt wrote:
>
> > Because (a) a uniform disk of matter does not produce a constant
> > velocity profile,
>
> I think I wasnt clear enough with my description as your expression
> `uniform disk of matter` doesnt seem to fit . Especially considering
> that when M is multiplied by (r/R) and then inserted into your formula
> it does give a mathematical constant velocity profile.
> Please see my explanation below if you want a clear picture of what
> I mean by M(r/R)
Your argument is essentially that of the people requiring dark matter.
In order to make (nearly) constant galactic rotation velocity
profiles, the mass enclosed at a radius r must be proportional to r.
HOWEVER, we can "measure" the amount of mass by looking a the massive
things which emit radiation: stars (optical & IR light); dust (IR);
hot plasma (X-rays); cold gas (radio). The conclusions from those
studies are that the cumulative measureable mass does *not* increase
linearly with r. Nor is there enough mass to make the rotation speeds
of ~200 km/s... which leads to the conclusion that there is something
masive which does *not* emit radiation: dark matter!
> >and (b) the amount of visible matter (stars, dust)
> > is not enough to explain the magnitude of the observed velocity.
>
> I dont see how we can accurately calculate the `mass` of any galaxy
> except by observing its internal velocities?
See above.
> Theres one thing you suggest above that I was wondering if you
> could clarify. Youve said previously that all galaxies have constant
> velocity profiles .
Nearly constant, but you can look this up for yourself if you wish.
> ... But are you also saying above that in larger
> galaxies stars all spin at faster speeds than in the smaller galaxies ?
I did not say that. Some have faster rotation curves than others, but
I didn't say anything specific about "large" galaxies.
> Or do you mean that all sizes of galaxies spin at the same speed at
> constant velocity profiles but its just that the velocity profile
> is different from that predicted?
A large number of galactic rotation curves have constant portions
which fall in the range 200 - 300 km/s.
...
> > r. How could that be? Let's say that the mass distribution is such
> > that the mass enclosed at a radius r is an effective mass M' = M(r/R).
> > In which case one gets the standard formula v = sqrt(GM'/r), but with
> > a revised mass distribution. This effectively the model of dark
> > matter. Why? Because the visible mass in stars, gas and dust do not
> > follow this mass distribution. There must be some other undetectable
> > mass. Your supposition *is* dark matter!
>
> What I question is how do you know the mass distribution
> does not follow this mass distribution. As I mentioned above, the only
> accurate way to determine mass distribution is by observing rotation
> speeds.
No. See above. It is the cross checking of several different methods
which leads to the requirement for some kind of dark matter.
> Incidentally I have tried a computor simulation of this distribution
> to see if it produces a galaxy shape and it does. Very well.
I'm not disagreeing that this distribution works well. It's the
functional form of the cumulative mass distribution that a galaxy must
have to make a constant velocity profile.
But what you are doing is saying, if I multiply "M" by some arbitrary
function then I get a good result! That is only mildly interesting.
What *is* interesting is the scientific implications of that arbitrary
manipulation, which is that there must be some kind of massive but
dark matter to account for it.
> > In addition, your supposed mass distribution, M' ~ r, is incosistent
> > with your *other* supposition above that the mass distribution is
> > "even" (uniform), which produces M' ~ r^2. Your two suppositions lead
> > to a contradiction, which is not good.
>
> I dont think Ive described the distribution well enough.
> I would say that my use of `even` is at best more like the first not
> the
> second expression you use above. It is definitely not described by a
> cube root function. Is M(r/R) the same as M ~ r^3 ?
???? If you are being serious, you betray your lack of algebraic
skills necessary for your endeavor.
> I dont think so. For that matter is M(r/R) the same as
> M ~ r? I dont think so either as r/R is not the same as r.
Please. "M ~ r" is common notation to mean "M proportional to r",
which it clearly is for your "formula." Your inclusion of an
arbitrary constant R is irrelevant to the discussion.
> If you could read all the explanation below I think it should
> help clarify things. Ive made it as short as possible.
While your discussion is clear, you are still arguing for dark matter!
Visible (detectable) matter does not follow the distribution you
describe!
It's kind of silly to proceed. It is fairly clear that you are doing
arbitrary mathematical manipulations without concern for what actual
observations have to say. You are in effect arguing for dark matter
without even realizing it.
CM
Hero.van...@gmx.de
sean
Craig Markwardt wrote:
> Your argument is essentially that of the people requiring dark matter.
> In order to make (nearly) constant galactic rotation velocity
> profiles, the mass enclosed at a radius r must be proportional to r.
>
> HOWEVER, we can "measure" the amount of mass by looking a the massive
> things which emit radiation: stars (optical & IR light); dust (IR);
> hot plasma (X-rays); cold gas (radio). The conclusions from those
> studies are that the cumulative measureable mass does *not* increase
> linearly with r. Nor is there enough mass to make the rotation speeds
> of ~200 km/s... which leads to the conclusion that there is something
> masive which does *not* emit radiation: dark matter!
All these estimates are made using unproven theoretical assumptions
of how galaxies form and how stars form. These are estimates based
on impossible to prove theoretical assumptions. In fact I argue that
observed rotation curves are proof that galaxy and star formation
theories are incorrect. Anyways, why not measure galaxies mass on
rotation speeds only? Thats how we do it for the solar system.
> A large number of galactic rotation curves have constant portions
> which fall in the range 200 - 300 km/s.
If all galaxies have roughly constant
rotation curves regardless of size and distance from a assumed denser
core Then it must follow that the rotation speeds are not
dependent directly on the mass of the system. If they were,
larger heavier galaxies would spin faster. So obviously your
statement that visible mass cannot account for the faster rotation
speeds is shown to be invalid by the fact that more mass
is observed to not supply faster roation speeds. WITH or
without dark matter.
> No. See above. It is the cross checking of several different methods
> which leads to the requirement for some kind of dark matter.
It seems unscientific that in this `cross checking` process the
only measurement that is truly accurate ( observed speeds) is the
only one that is ignored.
> ???? If you are being serious, you betray your lack of algebraic
> skills necessary for your endeavor.
>
> > I dont think so. For that matter is M(r/R) the same as
> > M ~ r? I dont think so either as r/R is not the same as r.
>
> Please. "M ~ r" is common notation to mean "M proportional to r",
> which it clearly is for your "formula." Your inclusion of an
> arbitrary constant R is irrelevant to the discussion.
Why is algebra neccesary for understanding this discussion on
rotation curves? Its an inferior tool compared to what
our brains 3-d visualizing potential offers.
In fact I came up with the same mass distribution
that you agree dark matter uses without ANY algebra used at all.
I only had to visualize and analyse it in my head in a couple of
minutes.
Not only can I visualize and `measure` anything algebra does,
I can do so with an added time element
You say 1/r^3 is what old einstein worked out
using relativity. He didnt get that from his GR, he tried
out a variety of formulas on top of newtons to see which gave
the results he needed. 1/r^3 is a fairly basic mathemnatical
expression and was probably one of the first he tried.
If Newton had had the available data he would have known that
his formula wasnt sufficient to account for what he saw.
What if he had measured gravity with modern technology?
He would have seen the need for an extra expression
in his formula where gravity also has a falloff described by
1/r^3 and added that in his principia. You dont need any
GR to figure that out.
Sean
sean
galaxies, are You hinting at the gravity pull of the other stars of our
milky way - pure in Newton physics. Just like a sledge pulled by a
horse and a mouse, You hardly will notice the pull of the mouse, but
when You replace horse by dog and dog by cat and so forth suddenly one
will notice the pull of the mouse, especially if the pull is into a
different direction from the second animal.
Dont worry about not understanding our discussion . You seem to
have grasped it perfectly.
Or for a more beautifull picture of this read Arthur C. Clarke story
about a chinese vase and in different texts a more detailed picture
about gravity funnels of several stars adding up.
Does the gravity of other stars explain part of the minmal acceleration
on top of the reduction, which is normal, when drifting on a
hyperbolic route away from our sun?
Possibly. I havent heard of this minimal extra acceleration
you speak of but it seems reasonable to say that the rest of the galaxy
is pulling us away from the sun by a small amount. The only thing is
that it would have to be a directional pull towards the milky way
although having said that maybe as we are inbedded in an `Arm`
and our local stars are closer, the overall effect is
one of a pull away to everywhere in our`sky` from local stars.
You should find out more about this.
Sean
Richard Saam
but for constant m, the uncertainties exist within the de Broglie condition
see below.
Here mv is the momentum
> and x is one coordinate of the location (previously
> you used x as the distance moved in the x direction)
> but it is the product of the uncertainties in the
> measurements that is constrained. Also note that
> it is an inequality:
>
> dx * dp > h_bar / 2
do not forget the equal sign possibility
dx * dp >= h_bar / 2
dx * dp >= h / (4pi)
>
> That says we cannot measure both the location and
> momentum to unlimited accuracy.
>
> http://en.wikipedia.org/wiki/Uncertainty_principle
>
> In the case of Pioneer, practical limitations mean
> we are nowhere near that limit, the uncertainty in
> the location alone is thousands of km.
No No, see below
>
> At the end of the day Richard, all you are showing
> is that applying a force to a mass changes its
> momentum and kinetic energy which as you say is
> basic physics, but what you haven't attempted to do
> is state the mechanism that is applying the force
> in the case of the Pioneer craft. As I have said
> several times, the kinetic energy in that case is
> tiny compared to the RAG waste heat so energy is
> not of interest, it is the _mechanism_ by which the
> force is applied that is being sought. If you are
> claiming it is drag then the dm/dt factor is
> fundamental and you have simply ignored it.
No it is not ignored, see clarification below.
>
> George
>
>
George
Clarification:
Apparently you forgot that "m"
is not the mass of the spacecraft
but is 110 * electron mass
and represents the mass in the space density of 6E-30 g/cc.
and available as work energy 6E-30*c^2 erg/cc or 5.4E-9 erg/cc.
Above all "m" is constant
or more specifically rest mass.
Newton's Second Law
F = dp/dt
Newton's Relativistic Second Law
F = (m v (1 - v^2/c^2)^(-1) / dt
m = constant
One dimensional assumption
therefore
F = m (v (1 - v^2/c^2)^(-1) / dt
differential identity
dt = dx/dv
work - energy theorem
F dx = m (1 - v^2/c^2)^(-1) v dv
m = constant (reminder)
Heisenberg Uncertainty
Dp * Dx >= h/(4pi)
m * Dv * Dx >= h/(4pi)
m >= h/(4pi) * 1/(Dv * Dx)
de Broglie relation
p * x = h
m * v * x = h
m = h * 1/(v * x)
Therefore:
m >= m
but this cannot be
(constant mass cannot be greater than itself)
therefore
h * 1/(v * x) = h/(4pi) * 1/(Dv * Dx)
v * x = 4pi * Dv * Dx
both Uncertainty and de Broglie condition are satisfied.
Uncertainty is dimensionally within the de Broglie condition.
Back to the work - energy theorem
F dx = m (1 - v^2/c^2)^(-1) v dv
Integrate across one space lattice de Broglie cell
F x = -m c^2 (1 - v^2/c^2) + Constant
F x = -mc^2 where Constant = 0 as 0 to v & 0 to x
Given:
1/x = (Swept Volume) / (Pioneer Cross-sectional Area)
= Volume / Area
and
rho = m / Volume = 110*electron mass / Volume
where: A & B are space lattice dimensions
A = 9.29 cm and B = 27.11 cm
Volume = 2 sqrt(3) A B2 = de Broglie Cell Volume
vd = de Broglie velocity = hb pi / (m B) = .0015 cm/sec
v = n vd (n = vd multiples composing v)
space lattice (quantum mechanical medium is composed
of conjoining cells with no voids.
Therefore:
F = -rho Area (v2/2 + c2 * sqrt(1 - v2/c2))
Inspection would indicate that as (0<v<<c):
F ~ -rho Area c2
and specifically for:
sun referenced Pioneer velocity of 12km/s
or earth (around sun) orbital velocity of 29.8 km/sec
or sun (around milky way center) rotation velocity of 400 km/sec
or the CMBR dipole of 620 km/s
or any other velocity 'v' magnitude (0<v<<c)
do not contribute to the observed Pioneer deceleration
(as observed)
(noting that c ~ 300,000 km/sec).
if deceleration or slowing down is observed in one space lattice cell
the same will be observed in a series of space lattice cells
and independent of v or n vd
The observed deceleration will be
in the line of observer to object
in accordance with a = F/M
where M is the Mass of the object.
This deceleration or slowing rate
will be numerically the same whether object
is moving away from the observer or towards the observer.
Conceptually, it is like a disc sweeping out a cylinder
through a quantum mechanical medium of virtual particles
in equilibrium with CMBR
having the characteristics of a Bose Einstein Condensate
(hb2 / 2m) (pi / B)2 = m v2 / 2
It does not matter how fast the disc is moving.
The Work performed in doing is:
m Integrate[0,v] v dv / sqrt(1 - v2/c2)
which approaches mc2
There is nothing more to give (mc2) per distance or per volume,
so resulting 'work' has to be nearly constant with velocity 'v' << c.
If the observer and pioneer/object
where in line with CMBR 640 km/sec dipole
and having + or - relative velocity 'v'
the same pioneer/object (deceleration or slowing down)
would be observed as if
observer and pioneer line was orthogonal
to CMBR 640 km/sec dipole.
The Force and (deceleration or slowing rate)
is always the same for a given F/M.
(M = Mass of Pioneer or object)
At the end of the day,
the above hypothesis falls within available experimental data.
Verification will depend on further space flights.
Richard
Hero.van...@gmx.de
Hero wrote:
> > Does the gravity of other stars explain part of the minmal acceleration
> > on top of the reduction, which is normal, when drifting on a
> > hyperbolic route away from our sun?
Sean wrote :
> Possibly. I havent heard of this minimal extra acceleration
> you speak of
Part of the shift in pioneer 10's electromagnetic and pulse responses,
mentioned in the title of this thread.
> but it seems reasonable to say that the rest of the galaxy
> is pulling us away from the sun by a small amount. The only thing is
> that it would have to be a directional pull towards the milky way
> although having said that maybe as we are inbedded in an `Arm`
> and our local stars are closer, the overall effect is
still directional, and not, what You say
> one of a pull away to everywhere in our`sky` from local stars.
> You should find out more about this.
Sean, let's try a picture of this: air bubbles in water. The
temperature of the air and the water should be the same, but still the
molecules of the airbubble are smaller and of less mass than the
molecules of the surrounding water. So there is inside the molecule
speed/velocity ( i don't mean the direction, just the amount of)
greater than in the water, which is resisting and reflecting because
the water molecules have more mass. Now the pressure in our oceans is
less at the surface and high in the depth. These differences in this
skalar field result in the movement of the bubble in direction to the
surface.
The universe is a skalar field of unevenly distributed matter in
motion and the differences are leading to a change of motion and
different distribution of matter. As there's no center, we started with
ourselves. Our gravity, which let the apples fall from trees, is
according to Newton not the pull of earth, but the difference of the
pull of earth to the rest. Proceed to earth and sun. We now know sun
is moving around the center of milky way with about 30 km/sec and with
the local group in a speed of 250 km/sec towards direction Hercules
stars.
A two body motion is a conical path around their common centre. Now
with more bodies spaced apart with great distances, like sun and
pioneer 10 and following voyagers and pluto, sedna and some comets we
should observe a four or more bodies path (which we can calculate for
the solar system bodies for a certain amount of time) but also a
difference of pull on sun and on pioneer from the rest of the universe
according to the distances and masses, part of it due to milky way,
part of it related to the motion to Hercules, and may be more.
And actually with the motion to Hercules we should think of a helical
spiral movement of sun and the planets ( why is in the word "helix" the
word for sun= hel ?)
A conical path has two centers, for planets and sun one is their center
of mass with iso-planes of equal pull (the pull directed normal to the
iso-planes) and the other center can be explained with the directed
speed/velocity of the bodies moving through these planes. The more
these two centers are spaced apart, the more the path will be distorted
by unequal pull from "outside".
> You should find out more about this.
> Sean
Actually i think, that physicists and astronomers should correct my
picture, especially those, who can measure with such high precision in
instruments and in their thinking.
Regards
Hero
PS What's really intriguing me is the my question in "Just a
millimeter".
George Dishman
> George Dishman wrote:
>> "Richard Saam" <rds...@att.net> wrote in message
>> news:ENEhf.90886$qk4....@bgtnsc05-news.ops.worldnet.att.net...
>>
True, my mistake.
> dx * dp >= h / (4pi)
>
>>
>> That says we cannot measure both the location and
>> momentum to unlimited accuracy.
>>
>> http://en.wikipedia.org/wiki/Uncertainty_principle
>>
>> In the case of Pioneer, practical limitations mean
>> we are nowhere near that limit, the uncertainty in
>> the location alone is thousands of km.
>
> No No, see below
>>
>> At the end of the day Richard, all you are showing
>> is that applying a force to a mass changes its
>> momentum and kinetic energy which as you say is
>> basic physics, but what you haven't attempted to do
>> is state the mechanism that is applying the force
>> in the case of the Pioneer craft. As I have said
>> several times, the kinetic energy in that case is
>> tiny compared to the RAG waste heat so energy is
>> not of interest, it is the _mechanism_ by which the
>> force is applied that is being sought. If you are
>> claiming it is drag then the dm/dt factor is
>> fundamental and you have simply ignored it.
>
> No it is not ignored, see clarification below.
>
> George
>
> Clarification:
>
> Apparently you forgot that "m"
> is not the mass of the spacecraft
> but is 110 * electron mass
No Richard, apparently you forgot we were discussing
"Basic Physics 101". The above equations are entirely
general and you have yet to apply them to anything. On
that basis, they _should_ include the v dm/dt term to
be correct.
> and represents the mass in the space density of 6E-30 g/cc.
> and available as work energy 6E-30*c^2 erg/cc or 5.4E-9 erg/cc.
>
> Above all "m" is constant
> or more specifically rest mass.
M _might_ be constant once you apply the equations to
a specific case. Later you say of your hypothesis:
> Conceptually, it is like a disc sweeping out a cylinder
> through a quantum mechanical medium of virtual particles
> in equilibrium with CMBR
> having the characteristics of a Bose Einstein Condensate
In that case, m is the total mass of the craft
plus the "virtual particles" which have been
swept up. Of course the latter is negligible in
comparison so we can treat m as constant but for
"Basic Physics 101", you cannot neglect the term.
> Newton's Second Law
>
> F = dp/dt
>
> Newton's Relativistic Second Law
>
> F = (m v (1 - v^2/c^2)^(-1) / dt
>
> m = constant
>
> One dimensional assumption
No Richard, F, v and p are all vectors and in the
case of the Pioneer craft they are not aligned.
> therefore
>
> F = m (v (1 - v^2/c^2)^(-1) / dt
>
> differential identity
>
> dt = dx/dv
Where "x" is the distance moved by m in time dt.
> work - energy theorem
>
> F dx = m (1 - v^2/c^2)^(-1) v dv
>
> m = constant (reminder)
>
> Heisenberg Uncertainty
>
> Dp * Dx >= h/(4pi)
>
> m * Dv * Dx >= h/(4pi)
>
> m >= h/(4pi) * 1/(Dv * Dx)
>
> de Broglie relation
>
> p * x = h
p * lambda = h where lambda is the wavelength of the
particle.
> m * v * x = h
Wrong, you have confused the wavelength of the particle
with the distance it moves in some time. The next part
rested on this and is therefore invalid so I'll snip it
to save space. However, For the sake of argument, I'll
assume your conclusion in the rest of the reply, that
each cell produces an amount of slowing which is
independent of the speed of the craft through it.
> and specifically for:
> sun referenced Pioneer velocity of 12km/s
> or earth (around sun) orbital velocity of 29.8 km/sec
> or sun (around milky way center) rotation velocity of 400 km/sec
> or the CMBR dipole of 620 km/s
> or any other velocity 'v' magnitude (0<v<<c)
> do not contribute to the observed Pioneer deceleration
> (as observed)
> (noting that c ~ 300,000 km/sec).
>
> if deceleration or slowing down is observed in one space lattice cell
> the same will be observed in a series of space lattice cells
> and independent of v or n vd
Each cell would contribute the same amount of slowing so
the net effect will depend on the rate at which cells are
traversed therefore it should be proportional to v. That
isn't necessarily in conflict with observation since the
speed of the Pioneers was fairly constant over the period
of the analysis even relative to the sun, and relative to
the CMBR, the variation is even smaller.
> The observed deceleration will be
> in the line of observer to object
> in accordance with a = F/M
> where M is the Mass of the object.
No, the deceleration should be in the same direction as
the _vector_ F, that's "Physics 101" Richard
> This deceleration or slowing rate
> will be numerically the same whether object
> is moving away from the observer or towards the observer.
>
> Conceptually, it is like a disc sweeping out a cylinder
> through a quantum mechanical medium of virtual particles
> in equilibrium with CMBR
> having the characteristics of a Bose Einstein Condensate
This is the same basic mistake I pointed out some time
ago, if the particles are in equilibrium with CMBR, the
deceleration shold be aligned with and opposite to the
direction of motion of the craft in a local frame in
which there is no CMBR dipole. The milky way is moving
towards Hydra at ~ 600km/s so the deceleration of both
craft should be opposite to that direction.
> If the observer and pioneer/object
> where in line with CMBR 640 km/sec dipole
> and having + or - relative velocity 'v'
> the same pioneer/object (deceleration or slowing down)
> would be observed as if
> observer and pioneer line was orthogonal
> to CMBR 640 km/sec dipole.
No but what you say next the correct:
> The Force and (deceleration or slowing rate)
> is always the same for a given F/M.
> (M = Mass of Pioneer or object)
Exactly, F is the same for both craft but you are
forgetting that F is a vector. The acceleration should
be in the same direction as F and should therefore be
in the same direction (away from Hydra) for both craft.
The observed acceleration is roughly towards the Sun
for both craft and they are on opposite sides of the
Sun.
> At the end of the day,
> the above hypothesis falls within available experimental data.
Not for the direction, which is what I pointed out right
at the beginning.
George
Craig Markwardt
"sean" <jaymo...@hotmail.com> writes:
> Craig Markwardt wrote:
>
> > Your argument is essentially that of the people requiring dark matter.
> > In order to make (nearly) constant galactic rotation velocity
> > profiles, the mass enclosed at a radius r must be proportional to r.
> >
> > HOWEVER, we can "measure" the amount of mass by looking a the massive
> > things which emit radiation: stars (optical & IR light); dust (IR);
> > hot plasma (X-rays); cold gas (radio). The conclusions from those
> > studies are that the cumulative measureable mass does *not* increase
> > linearly with r. Nor is there enough mass to make the rotation speeds
> > of ~200 km/s... which leads to the conclusion that there is something
> > masive which does *not* emit radiation: dark matter!
>
> All these estimates are made using unproven theoretical assumptions
> of how galaxies form and how stars form. These are estimates based
Really? Which estimates precisely are you talking about? How do you
know which estimates are impossible to test? And how are "how
galaxies form and how stars form" relevant to measuring the quantity
of stars? In short, because you haven't substantiated your claim, it
it is irrelevant. And beyond that, stars were only one of the mass
indicators I mentioned. Most galactic rotation curves are derived
from 21 cm radio observations of cold hydrogen, which is easy to
measure and extends far beyond detectable stars.
> ... In fact I argue that
> observed rotation curves are proof that galaxy and star formation
> theories are incorrect. Anyways, why not measure galaxies mass on
> rotation speeds only? Thats how we do it for the solar system.
Why limit ourselves to only rotation speeds? There are so many ways
to measure a galaxy, why limit ourselves to just one? That is what
science is about: making theories and testing them. We can make a
theory about how massive galaxies are, and then test that theory by
other means. You basically propose to merely pretend that a dark
matter problem does not exist by ignoring certain measurements.
Of course, until now we've not been discussing the other evidence for
dark matter, such as the microwave background, supernova measurements,
Einstein lensing by clusters, and so on. We can't pretend this
problem away.
> > A large number of galactic rotation curves have constant portions
> > which fall in the range 200 - 300 km/s.
>
> If all galaxies have roughly constant
> rotation curves regardless of size and distance from a assumed denser
> core Then it must follow that the rotation speeds are not
> dependent directly on the mass of the system. If they were,
> larger heavier galaxies would spin faster. So obviously your
> statement that visible mass cannot account for the faster rotation
> speeds is shown to be invalid by the fact that more mass
> is observed to not supply faster roation speeds. WITH or
> without dark matter.
I did not make a statement about the relationship to the total mass of
the galaxy, and thus your claim is irrelevant. In fact, there are
also a significant number of galaxies with rotation speeds in the
50-200 km/s range. (cf. Brownstein & Moffat, astro-ph/0506370).
> > No. See above. It is the cross checking of several different methods
> > which leads to the requirement for some kind of dark matter.
> It seems unscientific that in this `cross checking` process the
> only measurement that is truly accurate ( observed speeds) is the
> only one that is ignored.
Ignored? Who is ignoring it? And in what precise manner are imaging
measurements of starlight, hot gas, dust, and cold hydrogen gas not
accurate? Your criticism is embedded in a web of unsubstantiated
statements. In fact, why are *you* proposing to ignore the imaging
measurements? What kind of science is that?
> > ???? If you are being serious, you betray your lack of algebraic
> > skills necessary for your endeavor.
> >
> > > I dont think so. For that matter is M(r/R) the same as
> > > M ~ r? I dont think so either as r/R is not the same as r.
> >
> > Please. "M ~ r" is common notation to mean "M proportional to r",
> > which it clearly is for your "formula." Your inclusion of an
> > arbitrary constant R is irrelevant to the discussion.
> Why is algebra neccesary for understanding this discussion on
> rotation curves? Its an inferior tool compared to what
> our brains 3-d visualizing potential offers.
What your brain can or cannot do is irrelevant. Astronomy is advanced
enough that it makes actual quantitative measurements. Therefore any
models will need to make quantitative (i.e. mathematical) predictions.
"3-d visualizing," while pretty, is irrelevant to the question of the
quantitative observations and models at hand.
...
> You say 1/r^3 is what old einstein worked out
> using relativity. He didnt get that from his GR, he tried
> out a variety of formulas on top of newtons to see which gave
> the results he needed. 1/r^3 is a fairly basic mathemnatical
> expression and was probably one of the first he tried.
Really? This is another unsubstantiated claim from you. In fact, the
approximate expression for the acceleration due to GR,
a = GM/r^2 (1 + GM/(3rc^2)) is an expansion of the more-correct
full GR equations. The second term is proportional to 1/r^3, but the
constant of proportionality is completely determined. It's not like
anybody had to try different functional forms or guess. Thus, your
claims are erroneous.
> If Newton had had the available data he would have known that
> his formula wasnt sufficient to account for what he saw.
> What if he had measured gravity with modern technology?
> He would have seen the need for an extra expression
> in his formula where gravity also has a falloff described by
> 1/r^3 and added that in his principia. You dont need any
> GR to figure that out.
Sorry, where are you going with this? Gravity is very much a 1/r^2
force, with a small additive ~ 1/r^3 correction due to GR (see above).
When you keep making such erroneous and unsubstantiated claims, who do
you expect to take you seriously?
CM
sean
Hero.van...@gmx.de wrote:
> Sean wrote :
> > Possibly. I havent heard of this minimal extra acceleration
> > you speak of
>
> Part of the shift in pioneer 10's electromagnetic and pulse responses,
> mentioned in the title of this thread.
Sorry yes of course.I got carried away with another seperate
explanation for the galaxy rotation speeds that didnt use the anomaly.
My conclusions on that so far is that if the rotation speeds are
roughly constant (200-300km/s) regardless of galaxy size then that
small variation in speed must come from different galaxies having
different densities of visible matter. More visble matter per unit
volume gives that galaxy a slightly higher rotation speed profile. And
of course as Ive expalined already within that galaxy the mass of
visible matter can be calculated by V=sqrt(GM{r/R})/r. Obviously to
date we`ve underestimated the mass of visible matter in galaxies. Maybe
a correlation can be found between age of galaxies and observed
rotation speeds . That is, if older galaxies are thought to be heavier
or lighter than younger galaxies there might be a correlation between
observed rotation speeds and age in my model.
I think I get what you are suggesting here. You suggest the anomaly is
the pull of all the local group etc like our milky way and hercules.
This is a weaker gravitational force on top of the stronger gravity we
feel from sun. Its hard to model what direction this bias would be in
but the only problem that I can see from this idea is that in that case
the anomaly wouldnt appear to be sun centered. It would be an
acceleration pointing elsewhere. To me the anomaly is either sun
centered (in which case not from the scalar field of the universe) or
just possibly... an acceleration that opposes the direction of movement
. (Pioneer could move at a tangent to the sun and still be subjected to
the same constant accleration opposing its direction of movement.)> >
PS
What's really intriguing me is the my question in "Just a
> millimeter".
Im not sure where this question is?
Sean
sean
Craig Markwardt wrote:
> Really? Which estimates precisely are you talking about? How do you
> know which estimates are impossible to test? And how are "how
> galaxies form and how stars form" relevant to measuring the quantity
> of stars? In short, because you haven't substantiated your claim, it
> it is irrelevant. And beyond that, stars were only one of the mass
> indicators I mentioned. Most galactic rotation curves are derived
> from 21 cm radio observations of cold hydrogen, which is easy to
> measure and extends far beyond detectable stars.
You are being evasive here. You said "other methods" were available
to my observed velocities method. I assumed you meant Other
methods. Now I realize you are presenting the same method that
I cite. Observed velocities.
And that method shows conclusively that the visible mass is
more than predicted.
> Why limit ourselves to only rotation speeds? There are so many ways
> to measure a galaxy, why limit ourselves to just one? That is what
> science is about: making theories and testing them. We can make a
> theory about how massive galaxies are, and then test that theory by
> other means. You basically propose to merely pretend that a dark
> matter problem does not exist by ignoring certain measurements.
Im not ignoring dark matter! It didnt exist until we
found out that accurate measurements didnt match the
predictions. Technically, they made up dark matter *after*
we had the visible mass of galaxies confirmed by accurate velocity
observations. Not before
> I did not make a statement about the relationship to the total mass of
> the galaxy, and thus your claim is irrelevant. In fact, there are
> also a significant number of galaxies with rotation speeds in the
> 50-200 km/s range. (cf. Brownstein & Moffat, astro-ph/0506370).
Sorry I misunderstood you. I had asked if you meant "that all sizes
of galaxies spin at the same speed at constant velocity profiles?"
and you replied "..A large number of galactic rotation curves have
constant portions which fall in the range 200 - 300 km/s."
To me that sounded like a `yes`. Obviously now I know that you meant
`No`.
Anyways Ive had a initial cruise through the paper and found out
the answer to my question is that yes, galaxy rotation speeds are
dependent on mass and by the looks of it in table 1, I can
devine that the larger the galaxy the faster the rotation speeds.
Not only that I`ve found that rotation velocities are NOT constant
within each galaxy contrary to your claim here in this
thread that they are. Note ngc 4010 has speeds from 40- 140 km/s
and our milky way has 190-250 km/s range. In fact they seem to
be all over the place from galaxy to galaxy.
I note that you say repeatedly that rotation
velocities can only be explained by extra dark matter.
Yet the very paper you cite has in its abstracts opening first line
the very clear statement that in fact they can explain
observed rotation curves without invoking dark matter!
> Sorry, where are you going with this? Gravity is very much a 1/r^2
> force, with a small additive ~ 1/r^3 correction due to GR (see above).
>
> When you keep making such erroneous and unsubstantiated claims, who do
> you expect to take you seriously?
What is erroneous or unsubstantiated about pointing out that Newton was
probably not aware that there is a very small extra gravitational
acceleration
pulling on mass that fell off with r but not at r^2. Are you suggesting
in fact he did know and never could understand why his formula
couldnt explain mercurys preccesion? My apologies if he did know
Sean
Richard Saam
Here is something to contemplate
and perhaps you have some insight.
When you indicate "direction"
perhaps you are indicating the direction of
longitudinal and transverse forces
as indicated as possibly archaic
but still valid concepts
presented in my physics 101 text?
Now for general relativistic newtonian force
within the context of work energy theorem:
F = dp/dt
Constant mass m
Work Energy theorem
F = m v/dt = m v dv/dx
de Broglie cell:
x*y*z
defined by de Broglie condition:
m * v_x * x = h
m * v_y * y = h
m * v_z * z = h
Work - Energy on object moving through de Broglie space cell in x direction:
General form:
F dx = m v dv = (m*v*(1-v^2/c^2)^(n)) dv
integration:
F x = work = energy = - mc^2 *(1-v^2/c^2)^(n+1) /(2*(n+1))
for n = -1/2 (transverse condition)
F x = work = energy = - 0 * mc^2 for 0<v<<c
Transverse forces in y & z direction cancel
with zero force on object moving
in x direction through space de Broglie cell.
for n = -3/2 (longitudinal condition)
F x = work = energy = + mc^2 for 0<v<<c
but in general (for all directional conditions)
F x = work = energy = - mc^2 *(1-v^2/c^2)^(n+1) /(2*(n+1))
= - mc^2 /(2*(n+1)) for 0<v<<c
for n = rational numbers -infinity to +infinity
and n <> integers -1,0,1,2,3,4,5,6,7.......
and
F x = work = energy = mv^2/2 for 0<v<<c
for n = integers 0,1,2,3,4,5,6,7.......
and
F x = work = energy = undefined for 0<v<<c
for n = -1
Subscripts may be off, but that is the general idea.
Use:
http://integrals.wolfram.com/
as tool.
Richard