Gravitational Doppler

On Fri, 04 Aug 2006 06:48:27 GMT, Odysseus
wrote:

In article ,
Lester Zick wrote:

snip

Rocket thrust as a constant force in this context is what I would call
a second order nonlinear force whereas friction is what I would term a
first order linear force. In other words an ordinary retro rocket
causes an absolute acceleration while friction brakes can only cause a
relative acceleration (deceleration in my own vernacular) in direct
proportion to underlying velocity. Hence constant retro rocket
acceleration will not vary directly as a function of velocity while
friction brake deceleration will.

Why should the latter be the case? Air resistance is a function of
velocity (quadratic IIANM), but a caliper brake exerting a constant
force on a wheel will produce a constant frictional force and therefore
a constant deceleration, to a first approximation at least, until the
speed becomes very low and the transition between sliding and static
friction takes place.

Are you suggesting "air resistance" and "friction braking" are
different? They look pretty much the same to me. I'm not sure I
understand the point you're trying to make. You ask "why should the
latter be the case?" without suggesting any alternative that I can
see.

Were the braking proportional to velocity, the train would never quite
come to a halt; instead its speed would approach zero asymptotically,
like the amount of radioactive material in a sample, decaying
exponentially. Cf. also Newton's law of cooling.

Sure. Friction braking is asymptotic until sufficiently slowing is
effected to bring elastic electrostatic forces into play which stop
motion.

Lester Zick
~v~~

On 4 Aug 2006 13:03:52 -0400, (Steve Willner)
wrote:

In article ,
Lester Zick writes:
In the case of friction if we ignore slippage the magnitude of the
deceleration (counterceleration?) can only be a function of velocity:
proportionately stronger at greater velocities, zero at zero velocity,
and reversing direction at velocities in opposite directions.

I don't know what you mean by "ignore slippage,"

"Ignoring slippage" just means assuming friction is maintained evenly
throughout. There are occasions when friction is not maintained evenly
throughout as when brake pads are worn unevenly or the brakepad has
oil residue.

but mechanical
friction is generally greatest at low speeds (just above zero) and
nearly constant independent of speed at higher speeds.

At very low speeds you have a mixture of first order friction effects
and second order elastic molecular forces in operation. When second
order elastic molecular forces are very slight in comparison to first
order friction effects the net force exerted by the latter depends on
how much friction is generated and that depends in turn on velocity.

Of course its
direction is always opposite the direction of motion.

Of course. That marks it as a first order force.

It's not hard to think of other examples where force depends on
velocity. Air resistance and magnetic force on a moving charge are
two that come to mind.

Air resistance is certainly a first order force but magnetic force is
a second order quadratic effect in that the interaction of magnetic
fields is positive and inverse square. I think what you might mean is
that a magnetic field is a linear function of velocity.But interaction
of the fields is inverse square if the fields are present. However for
what it's worth I'd prefer not to argue magnetic effects because their
underlying mechanics are so complicated in dynamic terms that they
don't admit of readlily apparent analysis and illustrations.

Lester Zick
~v~~

[sci.physics.relativity snipped yet again.]

In article ,
Lester Zick writes:
That marks it as a first order force.

"First order force" is not a term I've ever heard used by
physicists.

SWIt's not hard to think of other examples where force depends on
SWvelocity. Air resistance and magnetic force on a moving charge are
SWtwo that come to mind.

Air resistance is certainly a first order force but magnetic force is
a second order quadratic effect in that the interaction of magnetic
fields is positive and inverse square.

Please read again what I wrote. Magnetic force on a moving charge is
linearly proportional to the charge velocity and perpendicular to its
direction.

I think what you might mean is
that a magnetic field is a linear function of velocity.

I was assuming the magnetic field to be independent of the velocity
of the charge, as will be the case if the electric charge is
sufficiently small.

But interaction
of the fields is inverse square if the fields are present.

No, it isn't. In the case I mentioned, force is linear with magnetic
field strength and with electric charge. How the magnetic field
varies with position wasn't specified; that depends on how the field
was generated.

what it's worth I'd prefer not to argue magnetic effects because their
underlying mechanics are so complicated in dynamic terms that they
don't admit of readlily apparent analysis and illustrations.

Force on a single moving charge is much simpler than mechanical
friction (about which you seem to have very strange ideas) or star
cluster dynamics. However, the topic is usually presented in the
second semester of college physics, so we can ignore it for present
purposes. Just take it as an example that force can depend on
velocity and need not be in the same direction.

By the way, air resistance depends on velocity in quite a complex
way, but for typical everyday objects and everyday speeds, saying
it's proportional to the square of velocity is a good approximation.
The approximation breaks down for very small objects or very low
speeds ("small Reynolds number") and for speeds approaching the speed
of sound ("large Mach number").

--
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
email may be sent to your ISP.)

Steve Willner wrote:

air resistance depends on velocity in quite a complex
way, but for typical everyday objects and everyday speeds, saying
it's proportional to the square of velocity is a good approximation.
The approximation breaks down for very small objects or very low
speeds ("small Reynolds number") and for speeds approaching the speed
of sound ("large Mach number").


As you must know

F = Mass a = Cd * (Area) * (m/volume) * v^2 /2

or

a = Cd * (Area/Mass) * (m/volume) * v^2 /2

where

Area = Pioneer crossection

v is pioneer speed relative to sun

m/volume = density of space medium (dust or other)

As has been pointed out by many others

deceleration 'a' is extremely small
by many orders of magnitude
when compared to observed 8.37E-8 cm/sec^2

For completness

Cd = f(Reynold's Number)

At high Reynold's Number,

Cd ~ 1

F~ k v^2

At low Reynold's Number (Re)

Cd = 24/Re

F~ k v

but let us consider that the force is not a drag
but a thrust based on the Potential energy
of the space medium as calculated as follows:

rho = m/volume
= 3 H^2 / (8 pi G) = 6.38E-30 g/cm^3

where

H = Hubble Constant
G = Newton Gravity Constant


a ~ 2*Area*rho*c2 / M
~ 2*(58,965 cm2)*(6.38E-30 g/cm3)*(3E10 cm/sec)^2 /(241,000 g)
~ 2.8E10-9 cm/sec2 for Pioneer spacecraft

This value is a little low compared to actual value of 8E-8 cm/sec2
but it makes the thrust alternative worthy of further investigation
particularily when the factor 8.37E-8/2.8E-9 or 29
could be interpreted as a reasonable Cd factor.


Richard

Lester Zick wrote:
On 4 Aug 2006 13:03:52 -0400, (Steve Willner)
wrote:

In article ,
Lester Zick writes:
In the case of friction if we ignore slippage the magnitude of the
deceleration (counterceleration?) can only be a function of velocity:
proportionately stronger at greater velocities, zero at zero velocity,
and reversing direction at velocities in opposite directions.

I don't know what you mean by "ignore slippage,"

"Ignoring slippage" just means assuming friction is maintained evenly
throughout. There are occasions when friction is not maintained evenly
throughout as when brake pads are worn unevenly or the brakepad has
oil residue.

but mechanical
friction is generally greatest at low speeds (just above zero) and
nearly constant independent of speed at higher speeds.

At very low speeds you have a mixture of first order friction effects
and second order elastic molecular forces in operation. When second
order elastic molecular forces are very slight in comparison to first
order friction effects the net force exerted by the latter depends on
how much friction is generated and that depends in turn on velocity.

Of course its
direction is always opposite the direction of motion.

Of course. That marks it as a first order force.

It's not hard to think of other examples where force depends on
velocity. Air resistance and magnetic force on a moving charge are
two that come to mind.

Air resistance is certainly a first order force but magnetic force is
a second order quadratic effect in that the interaction of magnetic
fields is positive and inverse square. I think what you might mean is
that a magnetic field is a linear function of velocity.But interaction
of the fields is inverse square if the fields are present. However for
what it's worth I'd prefer not to argue magnetic effects because their
underlying mechanics are so complicated in dynamic terms that they
don't admit of readlily apparent analysis and illustrations.

Lester Zick
~v~~

xxein: Given your post, you have no idea of a physic or how to obtain
one.

In article ,
Lester Zick wrote:

On Fri, 04 Aug 2006 06:48:27 GMT, Odysseus
wrote:

In article ,
Lester Zick wrote:

snip

Rocket thrust as a constant force in this context is what I would call
a second order nonlinear force whereas friction is what I would term a
first order linear force. In other words an ordinary retro rocket
causes an absolute acceleration while friction brakes can only cause a
relative acceleration (deceleration in my own vernacular) in direct
proportion to underlying velocity. Hence constant retro rocket
acceleration will not vary directly as a function of velocity while
friction brake deceleration will.

Why should the latter be the case? Air resistance is a function of
velocity (quadratic IIANM), but a caliper brake exerting a constant
force on a wheel will produce a constant frictional force and therefore
a constant deceleration, to a first approximation at least, until the
speed becomes very low and the transition between sliding and static
friction takes place.

Are you suggesting "air resistance" and "friction braking" are
different? They look pretty much the same to me. I'm not sure I
understand the point you're trying to make. You ask "why should the
latter be the case?" without suggesting any alternative that I can
see.

I certainly am; they're not at all the same. Sliding friction between
two solid surfaces is independent of speed; it's just the product of the
normal force with the coefficient of friction. Where the normal force is
provided by the weight of the object on a level surface (as for the
train, if its wheels stop turning and slide on the rails) the
deceleration is just k*g, where k is the relevant coefficient of
friction (here for steel on steel) and g is the acceleration of gravity.
No velocity parameter required.

Air resistance is very different, the force being strongly dependent on
the speed. This has to do with the effects of pressure and turbulence,
which are quite complicated. At very low speeds or in a viscous fluid,
drag is roughly proportional to the velocity, but at high speeds in air
it's a function of velocity squared.

Were the braking proportional to velocity, the train would never quite
come to a halt; instead its speed would approach zero asymptotically,
like the amount of radioactive material in a sample, decaying
exponentially. Cf. also Newton's law of cooling.

Sure. Friction braking is asymptotic until sufficiently slowing is
effected to bring elastic electrostatic forces into play which stop
motion.

It is not in fact asymptotic, but linear; simple experiments, requiring
little in the way of apparatus, will readily demonstrate that. I
performed several such in my high-school and first-year physics classes.

--
Odysseus

On 4 Aug 2006 00:45:09 -0700, "George Dishman"
wrote:


Lester Zick wrote:
You know, George, you've raised such an interesting issue with your
first point that I think I'd like to reply to it separately now and
try the other points later. Frankly after considering the issue for a
while I think the first point holds the key to the misunderstanding
between us on the whole problem. - LZ

Yes, I think it works. Once you understand the nature
of the problem, possible causes can follow. I'll trim
the background until we get this straightened out. It
means I'm going to be a bit repetetive, sorry.

On 3 Aug 2006 01:20:51 -0700, "George Dishman"
wrote:
Lester Zick wrote:
...
Well I'm not so sure this doesn't create a misunderstanding even where
the convention is understood. You could describe your braking force in
analogous terms but you wouldn't expect putting the brakes on at rest
to cause any motion even though you could characterize the effect as
an acceleration at V=0.

No, but you would expect an acceleration at V=0
if the train used retro-rockets instead.

Very good counter argument. However I think you'll find that there is
a distinct and recognizable difference despite what you say below.
Sure you could use retro rockets instead of friction brakes. But the
ratio wouldn't be a linear function of speed unless you devise a
system to decrease the rocket thrust in proportion to the decrease in
speed. In which case of course one wouldn't expect any acceleration at
V=0.

George, it seems I've been winging it a little too carelessly. My
calculations only imply that v/V is the same as V/c for 2004 and do
not imply v/V is constant. Obviously I've been sloppy in my comments
but this still stikes me as a remarkable coincidence and I'm convinced
gravitational doppler is what underlies the phenomenon, at least for
2004, whether or not I can link it all up. So I'll have to rethink the
problem. For the time being I'll just have to let the issue rest. But
I certainly appreciate your interest and comments.

That is true but the anomaly isn't known to be
a linear function of speed, it is a constant
acceleration as far as we can tell. The accuracy
of the measurement is about +/-15% but the
speed only varied by 3.1%, mostly due to
slowing by the Sun's gravity.

Rocket thrust as a constant force in this context is what I would call
a second order nonlinear force whereas friction is what I would term a
first order linear force. In other words an ordinary retro rocket
causes an absolute acceleration while friction brakes can only cause a
relative acceleration (deceleration in my own vernacular) in direct
proportion to underlying velocity. Hence constant retro rocket
acceleration will not vary directly as a function of velocity while
friction brake deceleration will.

The key thing you need to realise is that there
is no measurable variation in the anomaly, it
looks exactly like the retro-rockets.

As an aside, the frictional scenario in this case
would typically be a drag and the formula for that
has the acceleration dependent on the square of
the speed. My terminology would be that this would
be second order, something proportional to speed
would be first order and constant acceleration would
be zeroth order based on the value of N where in
general:

a_P = a_0 + k * v^N

However it could also depend on other factors such
as range.

As I understand actual measurement of the Pioneer anomaly the ratio of
the anomaly to the underlying speed is constant.

No, we only have one point so whether it is related
to speed in any way or is independent is unknown.

And this is a sure
giveaway that it's a linear anomaly (either friction, gravitational
doppler, or some other comparable linear effect) rather than a second
order nonlinear anomaly like escaping gas. (We might conclude there is
an exponential decay in the force exerted by an escaping gas as the
source of the gas is depleted and pressure falls. However the decrease
will almost certainly not be in constant proportion to the underlying
velocity.)

That is something that was looked for but any
change of acceleration is less than the accuracy
of the measurement.

In the case of friction if we ignore slippage the magnitude of the
deceleration (counterceleration?) can only be a function of velocity:
proportionately stronger at greater velocities, zero at zero velocity,
and reversing direction at velocities in opposite directions.

Probably, but we do not see any variation of
acceleration, it is constant.

So I think we can definitely say the source of the Pioneer anomaly
cannot be a nonlinear second order force and must be some linear first
order effect, either friction, gravitational doppler, or another
comparable linear effect.

No, we do not have enough information to decide
whether there is a variation or not, we can only
say that none is detected. That isn't entirely
useless as the range varied by 50% so it allows
us to rule out any explanation that varies with
range, in particular inverse square effects.

Imagine watching a train on a short straight
track between two nearby stations and using a
radar gun to measure the speed. The train is
travelling at a constant 50 mph. Then it starts
to slow. You measure a constant (negative)
acceleration over some time during which the
speed of the train falls from 50.00 mph to
48.47 mph. Then the batteries go flat in your
radar gun. Can you tell whether the train was
using friction brakes or retro-rockets? That
is the situation we are in.

Sure. And I strongly suspect the answer is that yes one can tell the
difference if one can accurately track the ratio of acceleration to
underlying velocity.

Maybe, but the speed didn't change enough to
be detectable at the accuracy they achieved
with the initial data set. That may change now
that large amounts of extra data have been
recovered but perhaps not.

As an informative excercise, you might like to
work out what the speed will be at one light
year range, or at infinity ignoring extra-solar
influences and the anomaly. Just use the
effect of the Sun.

For even in that short period the use of a
nonlinear second order force such as a retro rocket will produce a
varying ratio between decreasing velocity and retro acceleration. With
a linear first order force the decreasing change in velocity will
remain a function of and constant with respect to underlying velocity.

For a change of speed of 3.1%, you get a
3.1% change for linear and a 6.2% change if
the acceleration depends on the square (drag).
The accuracy of the measured value is +/-15%.

:-(

I haven't seen values or error estimates
for da/dr or da/dv which would be interesting.

In any event I hope this makes sense to you and others. As a practical
matter your suggestion of retro rockets versus friction brakes puts
the whole problem into such clear perspective that I can't really
imagine a better way to get at the crux of what I'm talking about.

Thanks. When you understand the problem
produced by the limited accuracy, I think you
will see why you cannot assume it is related
to speed but the 50% variation in range does
allow us to rule out anything related to r^2.

To put this in context, remember a lot of work
has been done looking at the possibility that
the RTGs radiate more away from the Sun than
towards it. A retro-rocket with a thrust equivalent
to the radiation pressure of a 63W light bulb
exactly matches the anomaly. The other early
possibility of a gas leak also has a retro-rocket
characteristic. A ot of people consider these
to be the most likely explanations and they
certainly cannot be ruled out at this stage.

George

Lester Zick
~v~~

"Lester Zick" wrote in message
...
....
George, it seems I've been winging it a little too carelessly. My
calculations only imply that v/V is the same as V/c for 2004 and do
not imply v/V is constant.

In that case all the previous discussion is
really moot. However I hope you now have a
lot more information at your disposal.

Obviously I've been sloppy in my comments
but this still stikes me as a remarkable coincidence and I'm convinced
gravitational doppler is what underlies the phenomenon, at least for
2004, whether or not I can link it all up.

Well the key I think is the formula for your
"gravitational doppler". If it is a multiplier
onto the normal gravitational acceleration as
it appeared earlier in the thread then the
cruial test is how the anomaly varies with
range. At constant speed, the anomaly from
gravitational doppler would vary as the
inverse square of the distance from the Sun
but the observed anomaly is constant,
independent of range. That seems to rule it
out but correct me if I'm assuming too much.

So I'll have to rethink the
problem. For the time being I'll just have to let the issue rest. But
I certainly appreciate your interest and comments.

That's OK, I've been scratching this itch for
5 years now ;-)

Good luck with the hunt.

George

On Sat, 05 Aug 2006 08:27:11 GMT, Odysseus
wrote:

In article ,
Lester Zick wrote:

On Fri, 04 Aug 2006 06:48:27 GMT, Odysseus
wrote:

In article ,
Lester Zick wrote:

snip

Rocket thrust as a constant force in this context is what I would call
a second order nonlinear force whereas friction is what I would term a
first order linear force. In other words an ordinary retro rocket
causes an absolute acceleration while friction brakes can only cause a
relative acceleration (deceleration in my own vernacular) in direct
proportion to underlying velocity. Hence constant retro rocket
acceleration will not vary directly as a function of velocity while
friction brake deceleration will.

Why should the latter be the case? Air resistance is a function of
velocity (quadratic IIANM), but a caliper brake exerting a constant
force on a wheel will produce a constant frictional force and therefore
a constant deceleration, to a first approximation at least, until the
speed becomes very low and the transition between sliding and static
friction takes place.

Are you suggesting "air resistance" and "friction braking" are
different? They look pretty much the same to me. I'm not sure I
understand the point you're trying to make. You ask "why should the
latter be the case?" without suggesting any alternative that I can
see.

I certainly am; they're not at all the same. Sliding friction between
two solid surfaces is independent of speed; it's just the product of the
normal force with the coefficient of friction.

Here you say sliding friction between two solid surfaces is
independent of speed but below you indicate the effect is linear. I
don't mind being corrected on the subject but now I'm confused. In any
event I acknowledged to George that my reading on the Pioneer anomaly
is incorrect so I'm not sure the issue matters much at the moment.

Where the normal force is
provided by the weight of the object on a level surface (as for the
train, if its wheels stop turning and slide on the rails) the
deceleration is just k*g, where k is the relevant coefficient of
friction (here for steel on steel) and g is the acceleration of gravity.
No velocity parameter required.

Air resistance is very different, the force being strongly dependent on
the speed. This has to do with the effects of pressure and turbulence,
which are quite complicated. At very low speeds or in a viscous fluid,
drag is roughly proportional to the velocity, but at high speeds in air
it's a function of velocity squared.

Sure. My mistake. With air we're dealing with a compressible fluid.

Were the braking proportional to velocity, the train would never quite
come to a halt; instead its speed would approach zero asymptotically,
like the amount of radioactive material in a sample, decaying
exponentially. Cf. also Newton's law of cooling.

Sure. Friction braking is asymptotic until sufficiently slowing is
effected to bring elastic electrostatic forces into play which stop
motion.

It is not in fact asymptotic, but linear; simple experiments, requiring
little in the way of apparatus, will readily demonstrate that. I
performed several such in my high-school and first-year physics classes.

Hmm. Linear as a function of velocity? Wouldn't the make the approach
to v=0 asymptotic?

Lester Zick
~v~~

On 4 Aug 2006 17:42:03 -0400, (Steve Willner)
wrote:

[sci.physics.relativity snipped yet again.]

Yeah, Steve, look your messages are coming through with no groups at
all. I don't much care if you want to remove or add a group. Except
for sci.physics these aren't the groups in my original alist. Everyone
around here just seems to make up their own rules and modify alists to
suit themselves. In any even if you want replies you'll just have to
make sure to keep some groups in the alist or the reply won't post.


In article ,
Lester Zick writes:
That marks it as a first order force.

"First order force" is not a term I've ever heard used by
physicists.

Well I seem to have ****ed up royally with my interpretation of the
subject at the moment. So until I can justify my treatment of the
Pioneer anomaly in mechanical terms beyond a correct calculation just
for 2004 there isn't much point to going into a lot of detail on the
subject. Suffice it to say that by "nonlinear" "quadratic" or "second
order" force I'm referring to classical elastic inverse square forces
and by "linear" or "first order" forces I'm referring to things like
friction and longitudinal doppler. I'm not sure there is a recognized
treatment of the subject. My opinion is that first order forces are
always negative and quadratic forces always positive according to
direction.

SWIt's not hard to think of other examples where force depends on
SWvelocity. Air resistance and magnetic force on a moving charge are
SWtwo that come to mind.

Air resistance is certainly a first order force but magnetic force is
a second order quadratic effect in that the interaction of magnetic
fields is positive and inverse square.

Please read again what I wrote. Magnetic force on a moving charge is
linearly proportional to the charge velocity and perpendicular to its
direction.

And please read again what I wrote which was directly responsive to
the issue you raised.

I think what you might mean is
that a magnetic field is a linear function of velocity.

I was assuming the magnetic field to be independent of the velocity
of the charge, as will be the case if the electric charge is
sufficiently small.

Huh?

But interaction
of the fields is inverse square if the fields are present.

No, it isn't. In the case I mentioned, force is linear with magnetic
field strength and with electric charge. How the magnetic field
varies with position wasn't specified; that depends on how the field
was generated.

Yeah well I addressed the question and explained why I'd rather not go
into it at present.

what it's worth I'd prefer not to argue magnetic effects because their
underlying mechanics are so complicated in dynamic terms that they
don't admit of readlily apparent analysis and illustrations.

Force on a single moving charge is much simpler than mechanical
friction (about which you seem to have very strange ideas) or star
cluster dynamics. However, the topic is usually presented in the
second semester of college physics, so we can ignore it for present
purposes. Just take it as an example that force can depend on
velocity and need not be in the same direction.

I don't consider magnetism to be an independent force and until
someone can show me a magnetic monopole I see no reason to consider
it such. And if you think my ideas on first order linear forces and
star cluster dynamics are strange you ain't seen nothin yet.

By the way, air resistance depends on velocity in quite a complex
way, but for typical everyday objects and everyday speeds, saying
it's proportional to the square of velocity is a good approximation.
The approximation breaks down for very small objects or very low
speeds ("small Reynolds number") and for speeds approaching the speed
of sound ("large Mach number").

I apologized to Odysseus for not realizing that air is a compressible
fluid.

Lester Zick
~v~~