Another attempt by the cranks to fool people
Mike
even deny the existence of well-documented affects by calling to their
help some crooks in Equador.
Tom Roberts wrote:
> Mike wrote:
> > If you and Roberts have any doubts theat a coriolis force for example
> > is a real force all you have to do is watch the water go down the sink
> > drain. Actually its rotation direction is different in the north and
> > south hemisphere,
>
> This is an old wives' tale, aided and abetted by impoverished
> entrepreneurs living near the equator who use it to extract money from
> tourists.
>
> In fact, for a typical drain, careful observations show it requires
> several hours of rest after filling before the water has settled down
> enough so its rotation direction is random when drained. Indeed,
> computations of Coriolis force for such a drain show it to be far too
> small to account for the observed direction -- asymmetries in the drain
> and/or initial conditions of the water are _FAR_ more important. There
> was an article in AJP a few years ago on this.
>
> The entrepreneurs can easily control which way the drain
> flows by the way they fill the container. So they walk
> 100 yards north of the equator and show one direction, and
> walk 100 yards south of the equator and show the other.
> They are in _complete_ control of the water's direction.
> Amazed tourists, gullible as anywhere else, pay up.
Here you go Roberts:
"...Any one of these factors is usually more than enough to overwhelm
the small contribution of the Coriolis effect in your kitchen sink or
bathtub. Research in the 1960s showed that if you do carefully
eliminate these factors, the Coriolis effect can be observed [1,2]."
1. Shapiro, 1962, Bath Tub Vortex, Nature, v 196, pp 1080-81 (Northern
Hemisphere)
2. Trefethen, et.al., 1965, The Bath Tub Vortex in the Southern
Hemisphere, Nature, v 207, pp 1084-85
http://www.physics.ohio-state.edu/~dvandom/Edu/newcor.html
I totally amazed of the logic some people use and their intelligence
level. I got no respect for self-proclaimed experts who cannot
understand what they read.
Mike
Hayek
Mike wrote:
> I totally amazed of the logic some people use and
> their intelligence level. I got no respect for
> self-proclaimed experts who cannot understand what
> they read.
Be glad !
That way, you can beat them anytime.... :-)
First, it is shocking, but then, you learn to turn
circles around them.
Uwe Hayek.
> Mike
>
--
Die wirtschaftliche Freiheit hat keine Sicherheit ohne
Politische Freiheit, und die politische findet ihre
Sicherheit nur in die wirtschaftlichen Freiheit." --
Eugen Richter
This is the bitterest pain among men, to have much
knowledge but no power.
Herodotus (484 BC - 430 BC), The Histories of Herodotus
IDIOCY - Never underestimate the power of stupid
people in large groups.
Randy Poe
Mike wrote:
> Another attempt by the cranks of sci.physics, sci.physics.relativity to
> even deny the existence of well-documented affects by calling to their
> help some crooks in Equador.
What in the world are you talking about?
So given that there are conmen in Ecuador who can make the
water swirl any direction they want, you think this proves
coriolis force is significant for small tubs?
Or what? What is your point exactly? Are you saying there's
something wrong with the calculations in that link?
> http://www.physics.ohio-state.edu/~dvandom/Edu/newcor.html
>
> I totally amazed of the logic some people use and their intelligence
> level. I got no respect for self-proclaimed experts who cannot
> understand what they read.
The link says Coriolis doesn't explain the direction of water going
down a sink, just the same as Tom Roberts said.
What do YOU think the article meant by this:
"Water in the sink doesn't go far enough to trigger a
noticeable north/south deflection."
or this:
"In any case, don't blame it on the Coriolis effect unless
your sink is the size of a small ocean."
?
What point do YOU think the article is making regarding
Coriolis and sinks that "self-proclaimed experts" are missing?
- Randy
Dirk Van de moortel
"Mike"
aka Bill Smith
aka Eleatis
aka Undeniable
<ele...@yahoo.gr> wrote in message news:1144956645....@v46g2000cwv.googlegroups.com...
A and B have a vessel with water and a plugged hole.
A turns 180 degrees clockwise and removes the plug.
B turns 180 degrees antclockwise and removes the plug.
A and B have a vortices in different directions everywhere
on the planet. Even on the equator.
That's how fakers manage to make a modest living.
Dirk Vdm
Martin Hogbin
"Mike" <ele...@yahoo.gr> wrote in message news:1144956645....@v46g2000cwv.googlegroups.com...
> Another attempt by the cranks of sci.physics, sci.physics.relativity to
> even deny the existence of well-documented affects by calling to their
> help some crooks in Equador.
What effects?
>
<Snip>>
> I totally amazed of the logic some people use and their intelligence
> level. I got no respect for self-proclaimed experts who cannot
> understand what they read.
If there is any misunderstanding, it is by you.
Martin Hogbin
Mike
1. Shapiro, 1962, Bath Tub Vortex, Nature, v 196, pp 1080-81 (Northern
Hemisphere)
2. Trefethen, et.al., 1965, The Bath Tub Vortex in the Southern
Hemisphere, Nature, v 207, pp 1084-85
Did you read the references or you got no interest in reading and
learning but only in perpetuating your misconceptions?
Mike
>
> Martin Hogbin
Mike
1. Shapiro, 1962, Bath Tub Vortex, Nature, v 196, pp 1080-81 (Northern
Hemisphere)
2. Trefethen, et.al., 1965, The Bath Tub Vortex in the Southern
Hemisphere, Nature, v 207, pp 1084-85
The point is that when people say that about the sink they refer to the
effect of coriolis force and something that can be notices under the
proper conditions whereas they surely do not refer to crooks who insist
they can do that under every condition.
Some, took advantage of the crook attempt to fool people and in turn
try to fool everybody that the effect does not exist using of course an
informal fallacy of the type:
if a crook sayd that then it must be false.
Under carefullt controlled laboratory conditions, tghe coriolis effect
on water going down a drain is noticable, this is my point. It is
surely noticable on larger scales on the rotation of cyclons.
My point and dispute with Roberts and other here is the Coriolis (and
centrifugal) is a real force and not ficticious as they call it. They
want to declare it ficticious because some in the past called it so.
But the contradictions here is that these people believe that there is
not preferred reference frame and since coriolis arises in rotating
reference frames calling it ficticious because it is not needed in
inertial frames is taking a prefered stand on what frame you will
consider as the basis of your physics.
Given also the fact that the idea of a purely inertial FoR is just a
mathematical abstraction whereas the coriolis effect is a measured one,
it seems to me that these people who insist it is ficticious are
cranks, simply because they denay experimental evidence in favor of
abstract and absurd models of reality.
Mike
>
> - Randy
Martin Hogbin
"Mike" <ele...@yahoo.gr> wrote in message news:1144999017.7...@i40g2000cwc.googlegroups.com...
You have completely missed the point. No one denies
that the Coriolis force exists. In very carefully designed
experiments with water, with extreme precautions
taken to minimise the rotation caused by filling, it may well
be possible to detect the Coriolis force. But, with any
ordinary bath tub or sink the Coriolis effect is completely
swamped by more practical effects.
As a student, I won small sums of money from my
fellow scholars by filling all the basins in my hall of
residence and taking bets on the direction of rotation.
If you want an example of where the Coriolis force is a
very useful concept just think of large-scale weather systems.
These are completely dominated by the Coriolis force.
Martin Hogbin
Randy Poe
I'll have to wait until I can get to a library with issues of Nature
going that far back. But for now I'll take the word of the
article which cited them, which says that it is possible to
eliminate the other factors under carefully controlled conditions
and to actually see the very slow coriolis effect on sinks.
> The point is that when people say that about the sink they refer to the
> effect of coriolis force and something that can be notices under the
> proper conditions whereas they surely do not refer to crooks who insist
> they can do that under every condition.
You've missed the point. The effect is very small for sinks.
The swirling we all see in our sinks is not Coriolis. The people
showing swirling in different directions on the equator are
con men.
> Some, took advantage of the crook attempt to fool people and in turn
> try to fool everybody that the effect does not exist using of course an
> informal fallacy of the type:
> if a crook sayd that then it must be false.
No. Did you read the article you cited?
It isn't saying the effect doesn't exist. We all know it exists.
We all agree it exists. The question is the magnitude of the
effect.
>From the online article:
"So, fast things moving over great distances can be significantly
affected by the Coriolis effect. But what about the sink?...
In a kitchen sink, of course, speeds and time scales are much smaller
than hours and miles...Quantitatively, putting these numbers into
Equation 1
results in an estimated change in rotation of ...less than an
arc-second (1/3600th of a degree) per second."
So we're all saying, including the article, that the reason the
swirling in a sink isn't Coriolis is that Coriolis would only
cause swirling of an arc-second per second. You wouldn't
even notice it without sensitive measuring equipment and
carefully controlled conditions to eliminate the much larger
swirling effect.
> Under carefullt controlled laboratory conditions, tghe coriolis effect
> on water going down a drain is noticable, this is my point.
Nobody disagrees. But it's an ARC SECOND PER SECOND.
It looks like unswirling water to the naked eye. The swirling
you see in your sink is not this effect.
> It is
> surely noticable on larger scales on the rotation of cyclons.
Of course it is. As the article says, those are much larger
distances and much higher velocities.
- Randy
Randy Poe
Mike wrote:
> My point and dispute with Roberts and other here is the Coriolis (and
> centrifugal) is a real force and not ficticious as they call it.
It seems to be the word "fictitious" that is causing you to
misinterpret everything you're reading on this subject.
Every physicist agrees there is a Coriolis effect in rotating
systems like the earth's surface, and that it is responsible
for such things as rotating weather systems (e.g. hurricanes)
and the fact that artillery shells follow a sideways curving path.
> They
> want to declare it ficticious because some in the past called it so.
No, they call it a fictitious "force" because it is not really a force,
but an artifact of rotating systems. It is a real effect due to the
conservation of momentum. Things want to fly in straight lines.
When you are in a rotating system, those paths look curved.
It is not a force making them look curved, but your rotation.
Not only do physicists believe in this effect, there are actual
equations for the magnitude of the effect.
> But the contradictions here is that these people believe that there is
> not preferred reference frame and since coriolis arises in rotating
> reference frames calling it ficticious because it is not needed in
> inertial frames is taking a prefered stand on what frame you will
> consider as the basis of your physics.
You're confused again. The physics of all inertial reference frames
is the same. The physics is different in accelerated frames, such
as rotating frames. Again, every physicist will agree on this.
There's no preferred INERTIAL frame. If you're in a closed box and
non-accelerated, you can't tell if you're moving or not.
But if you're rotating, you can. There IS a difference between rotating
and non-rotating frames.
- Randy
Mike
So you and Randy agree the Coriolis effect is real. I suppose the same
holds for the centrifugal effect. So now let us take it from here.
Calling these forces "ficticious" must be justified in the context of
what ficticious means and not in some other context, like for instance
a posteriori derivation from a certain model.
My argument is similar to that brought up by Newton. These forces are
real. The fact that we cannot explain how they arise in non-inertial
reference frames does not immediately justify calling them ficticious.
I cannot see also how they can arise but I will not go as far as
arguing that these forces do not really exist but they are the result
of the motions in the frame bacause this does not explain anything.
IMO nothing justifies calling these foces ficticious, there is no
compeling argument for that. Maybe the word "inertial forces" some
people have used is a better term. This term has been used extensively
in the past but lately some Relativists, including Roberts insist on
calling them ficticious.
I would be ready to discuss such possibility if the concept of force is
declared as ficticious all together. But we know forces do exist, if
someone has a different idea tell us why they do not and how physics
textbooks should be rewritten without them, especially statics and
dynamics, not kinematics of course.
Mike
Mike
No I am not confused. I was arguing with Roberts who stated that the
objective of GR was to eliminate these forces so that physics in ALL
moving reference frames is the same. GR attempts to apply the strong
POR version under which physics laws are the same in all moving
reference frames, at least locally.
See my reply to Martin below. If you agrree that centrifugal and
coriolis effects are real then you should procees and justify why
centrifugal and coriolis forces should be called ficticious, unless you
deny the concept of force all together and we take it from there.
Mike
Mike
See the post above by Roberts and replies to get an idea how this
controversy started.
Mike
Randy Poe
Mike wrote:
> Martin Hogbin wrote:
> So you and Randy agree the Coriolis effect is real. I suppose the same
> holds for the centrifugal effect. So now let us take it from here.
Yes, they're real effects of conservation of momentum in
accelerated frames of reference.
> Calling these forces "ficticious" must be justified in the context of
> what ficticious means and not in some other context, like for instance
> a posteriori derivation from a certain model.
It means they are not truly forces but artifacts of a coordinate
transformation. There isn't really an acceleration corresponding
to these "forces".
> My argument is similar to that brought up by Newton. These forces are
> real. The fact that we cannot explain how they arise in non-inertial
> reference frames does not immediately justify calling them ficticious.
We know precisely how they arise, that's how we can derive
equations for them. If I am driving in a straight line but
you are in a circle, you will conclude a "force" is making me
"curve". I can tell you precisely where that illusion is coming
from and calculate the precise amount.
Same thing with centrifugal and Coriolis "forces".
- Randy
Mike
Randy Poe wrote:
> Mike wrote:
> > Martin Hogbin wrote:
> > So you and Randy agree the Coriolis effect is real. I suppose the same
> > holds for the centrifugal effect. So now let us take it from here.
>
> Yes, they're real effects of conservation of momentum in
> accelerated frames of reference.
Momentum is not conserved in accelerated frames.
>
> > Calling these forces "ficticious" must be justified in the context of
> > what ficticious means and not in some other context, like for instance
> > a posteriori derivation from a certain model.
>
> It means they are not truly forces but artifacts of a coordinate
> transformation. There isn't really an acceleration corresponding
> to these "forces".
>
> > My argument is similar to that brought up by Newton. These forces are
> > real. The fact that we cannot explain how they arise in non-inertial
> > reference frames does not immediately justify calling them ficticious.
>
> We know precisely how they arise, that's how we can derive
> equations for them. If I am driving in a straight line but
> you are in a circle, you will conclude a "force" is making me
> "curve". I can tell you precisely where that illusion is coming
> from and calculate the precise amount.
Do you know the informal fallacy of "false analogy"? Either you want to
have a serious discussion or you want to turn this into a play.
Mike
Randy Poe
Mike wrote:
> Randy Poe wrote:
> > Mike wrote:
> > > Martin Hogbin wrote:
> > > So you and Randy agree the Coriolis effect is real. I suppose the same
> > > holds for the centrifugal effect. So now let us take it from here.
> >
> > Yes, they're real effects of conservation of momentum in
> > accelerated frames of reference.
>
>
> Momentum is not conserved in accelerated frames.
That's the point.
Take a situation in which momentum is conserved and the
entire situation is described by conservation of momentum.
Translate all the same motions into a rotating frame of
reference.
The very same dynamics which are completely explainable
by conservation of momentum now has, when described
in rotating coordinates, a centrifugal and coriolis term.
> > > Calling these forces "ficticious" must be justified in the context of
> > > what ficticious means and not in some other context, like for instance
> > > a posteriori derivation from a certain model.
> >
> > It means they are not truly forces but artifacts of a coordinate
> > transformation. There isn't really an acceleration corresponding
> > to these "forces".
> >
> > > My argument is similar to that brought up by Newton. These forces are
> > > real. The fact that we cannot explain how they arise in non-inertial
> > > reference frames does not immediately justify calling them ficticious.
> >
> > We know precisely how they arise, that's how we can derive
> > equations for them. If I am driving in a straight line but
> > you are in a circle, you will conclude a "force" is making me
> > "curve". I can tell you precisely where that illusion is coming
> > from and calculate the precise amount.
>
> Do you know the informal fallacy of "false analogy"? Either you want to
> have a serious discussion or you want to turn this into a play.
It's not only not a "false analogy", it wasn't even intended to
be an analogy. It's a direct description. An object moving in
a constant plane, like a Foucault pendulum or an artillery shell,
will be seen by the person in the curving path to have a
curving component. That person will conclude there is
a "force" (the Coriolis "force") causing the pendulum or the
shell to "curve".
- Randy
Mike
No, you are using the opposite example, it is a false analogy. We are
not talking about what an observer in a rotating frame, for example,
concludes for bodies moving in inertial frames. This has not been an
issue in the discussion. We have only talkied about what an observer in
a non-inertial reference frame measures, forget about the inertial
observer.
Example: a spring-mass system attached at the center of a rotating
platform. The observer on the platform sees the mass at rest but a
force is measured by the spring, which is equal to the centripetal
force. The observer then based on that measurement must conclude that
there is also another force, equal and opposite to the centrifugal
force acting on the mass to keep it at rest. No mention to the outside
at all. Any mention, like you do, is a red herring.
At this point, if you deny the real existence of the centrifugal force
and call it ficticious you got a problem. Newton's first law holds
because a real force is balanced by a ficticious force.
The other choice is to extend Newton's law, the way I have shown to
Roberts, using esentially D'Alembert's law, SUM(F) = dp/dt. Now,
centrifugal forces have real existence in one frame but absent when
motion is observed from inertial frame. This is a problem also but it
does not compare real to ficticious.
Maybe there are other choices but what GR tries to do is not a
reasonable choice because it evades a solution to the problem.
Try the thought experiment that you are enlosed in a rotating space
capsule and you have no access to the ouside. You measure the force and
you conclude there is another force acting for you to be at rest. You
got no excuse for calling that force ficticious or a result of
non-conservation of momentum.
Mike
>
> - Randy
Mike
Sorry I meant to say:
The observer then based on that measurement must conclude that
there is also another force, equal and opposite to the centripetal
force acting......
Mike
Martin Hogbin
"Mike" <ele...@yahoo.gr> wrote in message news:1145024284.8...@z34g2000cwc.googlegroups.com...
>
> Martin Hogbin wrote:
> > "Mike" <ele...@yahoo.gr> wrote in message news:1144999017.7...@i40g2000cwc.googlegroups.com...
> > >
> > > Martin Hogbin wrote:
> > > > "Mike" <ele...@yahoo.gr> wrote in message news:1144956645....@v46g2000cwv.googlegroups.com...
> >
> > that the Coriolis force exists. In very carefully designed
> > experiments with water, with extreme precautions
> > taken to minimise the rotation caused by filling, it may well
> > be possible to detect the Coriolis force. But, with any
> > ordinary bath tub or sink the Coriolis effect is completely
> > swamped by more practical effects.
> >
> > As a student, I won small sums of money from my
> > fellow scholars by filling all the basins in my hall of
> > residence and taking bets on the direction of rotation.
> >
> > If you want an example of where the Coriolis force is a
> > very useful concept just think of large-scale weather systems.
> > These are completely dominated by the Coriolis force.
> >
> > Martin Hogbin
>
> So you and Randy agree the Coriolis effect is real. I suppose the same
> holds for the centrifugal effect.
Yes, if you expect Newton's laws to apply in a rotating reference
frame.
> So now let us take it from here.
> Calling these forces "ficticious" must be justified in the context of
> what ficticious means and not in some other context, like for instance
> a posteriori derivation from a certain model.
As I said in a different thread, the name 'fictitious' is not meant to
be taken too literally, just like with irrational and imaginary numbers.
> My argument is similar to that brought up by Newton. These forces are
> real. The fact that we cannot explain how they arise in non-inertial
> reference frames does not immediately justify calling them ficticious.
We can, very easily, explain how the Coriolis and centrifugal forces
arise in a rotating reference frame.
The case of large-sale weather sytems is an excellent example
of where the the concept of Coriolis force is a good one. We
have a very natural reference frame (the Earth) which for many
purposes (a game of pool for example) we can consider as
inertial. It is therefore very natural to expect Newtonian
physics to apply to this frame. However, we know that the Earth
is rotating and for weather systems this is very important. The
easiest way to deal with this is to use normal Newtonian physics
but to add in the Coriolis force.
You could do the analysis in an inertial (non-rotating) frame
but this would be horribly complicated.
> I cannot see also how they can arise...
Perhaps you should study some physics then.
> ...but I will not go as far as
> arguing that these forces do not really exist but they are the result
> of the motions in the frame bacause this does not explain anything.
Coriolis and centrifigal forces are perfectly well explained by
classical mechanics. Read a book on the subject.
> IMO nothing justifies calling these foces ficticious, there is no
> compeling argument for that. Maybe the word "inertial forces" some
> people have used is a better term. This term has been used extensively
> in the past but lately some Relativists, including Roberts insist on
> calling them ficticious.
It is just a name. More important is to understand how the forces arise
and when to apply them. You do not seem very clear on this..
Martin Hogbin
Randy Poe
Mike wrote:
> Randy Poe wrote:
> > > > We know precisely how they arise, that's how we can derive
> > > > equations for them. If I am driving in a straight line but
> > > > you are in a circle, you will conclude a "force" is making me
> > > > "curve". I can tell you precisely where that illusion is coming
> > > > from and calculate the precise amount.
> > >
> > > Do you know the informal fallacy of "false analogy"? Either you want to
> > > have a serious discussion or you want to turn this into a play.
> >
> > It's not only not a "false analogy", it wasn't even intended to
> > be an analogy. It's a direct description. An object moving in
> > a constant plane, like a Foucault pendulum or an artillery shell,
> > will be seen by the person in the curving path to have a
> > curving component. That person will conclude there is
> > a "force" (the Coriolis "force") causing the pendulum or the
> > shell to "curve".
>
> No, you are using the opposite example, it is a false analogy. We are
> not talking about what an observer in a rotating frame, for example,
> concludes for bodies moving in inertial frames. This has not been an
> issue in the discussion. We have only talkied about what an observer in
> a non-inertial reference frame measures, forget about the inertial
> observer.
You asked the question "where do Coriolis and centrifugal forces
come from?" You act like nobody knows the answer.
That is not a deep mystery. The answer is: "they come
from taking a situation described by Newton's Laws
and describing it in non-inertial coordinates."
> Example: a spring-mass system attached at the center of a rotating
> platform. The observer on the platform sees the mass at rest but a
> force is measured by the spring, which is equal to the centripetal
> force.
If the observer is not allowed to know he is rotating,
then he certainly would not say "there is a centripetal
force".
You either do or don't know that your frame is rotating
with respect to an inertial frame. Choose one. Do we
know?
> The observer then based on that measurement must conclude that
> there is also another force, equal and opposite to the centrifugal
> force acting on the mass to keep it at rest.
No, the observer is not obligated to draw this conclusion.
He is also allowed to draw the conclusion that he is
in a rotating frame and he is seeing an effect of
that rotation. He certainly wouldn't say "there is a
force balancing my centripetal force" if he didn't know
there was a centripetal force.
In fact, it is because of experiments like this that we
say rotation is absolute, and there are experiments you
can do in rotating closed boxes to tell you are rotating
and at what rate, with respect to what axis.
Perhaps you meant to say something like this: an
observer is in a rotating frame but he doesn't know
he is rotating. By using a spring, he learns that
there is a mysterious force radiating outward
from one particular point. He learns this by
seeing how far and in which direction the spring
stretches when he attaches it to various places.
He calls this originating point "the center".
Plotting the results, he finds F = k*m*r where r is
the distance from the center and m is the mass
at the end of the spring and k is a constant.
He calls this force "centrifugal force" and says it
must be added to the list of fundamental forces
(gravitation, electrostatic, weak nuclear, strong
nuclear).
Another observer in the same small universe
has a different hypothesis. He says their universe
is rotating, and k = w^2 where w = angular rotation
rate. His hypothesis does not require an addition
to the list of fundamental forces. It explains the
mysterious "force" in terms of Newton's laws. There
is no force pushing out. There is only the tendency
of things to follow a straight line. He hypothesizes
that things sitting "still" in their universe are actually
not following straight lines.
Both these observers are non-inertial. The first one
is drawing the conclusion you say any observer
MUST conclude. The second one has a competing
hypothesis. He is not obligated to draw the conclusion
you insist on.
- Randy
mme...@cars3.uchicago.edu
>
>Mike wrote:
>> Martin Hogbin wrote:
>> So you and Randy agree the Coriolis effect is real. I suppose the same
>> holds for the centrifugal effect. So now let us take it from here.
>
>Yes, they're real effects of conservation of momentum in
>accelerated frames of reference.
>
>> Calling these forces "ficticious" must be justified in the context of
>> what ficticious means and not in some other context, like for instance
>> a posteriori derivation from a certain model.
>
>It means they are not truly forces but artifacts of a coordinate
>transformation. There isn't really an acceleration corresponding
>to these "forces".
Careful here. Relative to the accelerating reference frame, where
you've these forces, there most certainly is an acceleration
corresponding to said forces.
>
>> My argument is similar to that brought up by Newton. These forces are
>> real. The fact that we cannot explain how they arise in non-inertial
>> reference frames does not immediately justify calling them ficticious.
>
>We know precisely how they arise, that's how we can derive
>equations for them. If I am driving in a straight line but
>you are in a circle, you will conclude a "force" is making me
>"curve". I can tell you precisely where that illusion is coming
>from and calculate the precise amount.
>
It is not an illusion. Repeat, *it is not an ilusion*. It is a
perfectly real effect resulting from a specific choice of a reference
frame.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
Mike
Because I have done so carefully, I dispute the "modern" explanations
offered and especially the "illusion" one offered by Roberts and Randy.
>
> > IMO nothing justifies calling these foces ficticious, there is no
> > compeling argument for that. Maybe the word "inertial forces" some
> > people have used is a better term. This term has been used extensively
> > in the past but lately some Relativists, including Roberts insist on
> > calling them ficticious.
>
> It is just a name. More important is to understand how the forces arise
> and when to apply them. You do not seem very clear on this..
Neither do you. You also seem to be dancing around.
Mike
>
> Martin Hogbin
Mike
Act? I certainly not dance.
>
> That is not a deep mystery. The answer is: "they come
> from taking a situation described by Newton's Laws
> and describing it in non-inertial coordinates."
You act now like you do not understand the particular issue discussed.
That was not the issue I repeated so many times. The issue was whether
these forces can be called "ficticious" as Roberts and other
Relativists do.
>
> > Example: a spring-mass system attached at the center of a rotating
> > platform. The observer on the platform sees the mass at rest but a
> > force is measured by the spring, which is equal to the centripetal
> > force.
>
> If the observer is not allowed to know he is rotating,
> then he certainly would not say "there is a centripetal
> force".
>
> You either do or don't know that your frame is rotating
> with respect to an inertial frame. Choose one. Do we
> know?
>
> > The observer then based on that measurement must conclude that
> > there is also another force, equal and opposite to the centrifugal
> > force acting on the mass to keep it at rest.
>
> No, the observer is not obligated to draw this conclusion.
> He is also allowed to draw the conclusion that he is
> in a rotating frame and he is seeing an effect of
> that rotation. He certainly wouldn't say "there is a
> force balancing my centripetal force" if he didn't know
> there was a centripetal force.
> In fact, it is because of experiments like this that we
> say rotation is absolute, and there are experiments you
> can do in rotating closed boxes to tell you are rotating
> and at what rate, with respect to what axis.
>
That's the point.
> Perhaps you meant to say something like this: an
> observer is in a rotating frame but he doesn't know
> he is rotating. By using a spring, he learns that
> there is a mysterious force radiating outward
> from one particular point. He learns this by
> seeing how far and in which direction the spring
> stretches when he attaches it to various places.
> He calls this originating point "the center".
> Plotting the results, he finds F = k*m*r where r is
> the distance from the center and m is the mass
> at the end of the spring and k is a constant.
>
You making progress.
> He calls this force "centrifugal force" and says it
> must be added to the list of fundamental forces
> (gravitation, electrostatic, weak nuclear, strong
> nuclear).
>
Of course, he is disspointed when fo some reason his rotation stops.
I think you also like dancing around like Martin.
> Another observer in the same small universe
> has a different hypothesis. He says their universe
> is rotating, and k = w^2 where w = angular rotation
> rate. His hypothesis does not require an addition
> to the list of fundamental forces. It explains the
> mysterious "force" in terms of Newton's laws. There
> is no force pushing out. There is only the tendency
> of things to follow a straight line. He hypothesizes
> that things sitting "still" in their universe are actually
> not following straight lines.
Which universe? No such thing was mentioned.
>
> Both these observers are non-inertial. The first one
> is drawing the conclusion you say any observer
> MUST conclude. The second one has a competing
> hypothesis. He is not obligated to draw the conclusion
> you insist on.
The first is faced with a problem. The second sounds to me like a crank
(framing hypotheses). Which one do you prefer?
Again, the question whether these forces are real or ficticious was not
answered. If they are real, there is a problem. if they are ficticious,
there is a problem. Newton's answer was that they arise because of a
wrong choice of reference frame, that is motion must only be analyzed
from inertial frames. This means that all moving reference frames are
not equivalent and Relativity (GR) is wrong. If we reject Newton, we
must explain how these real effects arise. GRists have take the route
of calling them ficticious because they have failed to incorporate the
hypothesis you mentioned (Mach's) in their theory. Whatever they fail
to explain they pronounce it ficticious or paradox or whatever.
Centrifugal and Coriolis effects are real, there is no doubt about it.
The inability to offer laws of motion that explain how these effects
arise while presering their validity in all moving reference frame is a
failure of GR and it is an obvious failure since it converges to
Newton's laws at the weak field limit. To conceal this failure
Relativists has chosen the route of calling these effects "illusions".
That is not good science.
Mike
>
> - Randy
Daryl McCullough
>"Randy Poe" <poespa...@yahoo.com> writes:
>>We know precisely how they arise, that's how we can derive
>>equations for them. If I am driving in a straight line but
>>you are in a circle, you will conclude a "force" is making me
>>"curve". I can tell you precisely where that illusion is coming
>>from and calculate the precise amount.
>>
>It is not an illusion. Repeat, *it is not an illusion*. It is a
>perfectly real effect resulting from a specific choice of a reference
>frame.
Hmm. I'm not sure what "real effect" means here, but let's
make this concrete:
Suppose I observe a ball's motion and plot its position x versus
t. I get a straight line, so I conclude (via Newton's laws) that
there is no force acting on the ball (at least not in the x-direction).
Now, I change coordinates to X = arctan(x) and plot X versus t.
In the new plot, I *don't* get a straight line, but I get some
curved line. Is that a "real effect"? Does it indicate the presence
of a force? No, it's obviously an artifact of my particular
choice of coordinates.
Now suppose I'm on a rotating platform, and observe a ball's
motion. I plot x versus t, and I don't get a straight line.
Is that due to the presence of a force, or is it an artifact of
my particular choice of coordinates?
It can be hard to tell which is the case, but what's *not*
true is that because I'm on a rotating platform, I must take
into account centrifugal forces. Whether the plot of x versus
t is a straight line or not depends on what *coordinate* system
I choose. Just because I'm on a rotating platform doesn't
mean that I have to use one particular coordinate system
over another. I'll use whatever coordinate system is convenient,
but there are two issues involved in convenience: What is convenient
to *measure* with, and what is convenient to *work* with.
--
Daryl McCullough
Ithaca, NY
--
NewsGuy.Com 30Gb $9.95 Carry Forward and On Demand Bandwidth
Dirk Van de moortel
"Mike"
aka Undeniable
aka Eleatis
aka Bill Smith
<ele...@yahoo.gr> wrote in message news:1145047611....@v46g2000cwv.googlegroups.com...
[snip]
>
> The first is faced with a problem. The second sounds to me like a crank
> (framing hypotheses). Which one do you prefer?
This one:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/RattenFingure.html
Definitely.
Dirk Vdm
Martin Hogbin
"Mike" <ele...@yahoo.gr> wrote in message news:1145046293.9...@i39g2000cwa.googlegroups.com...
>
> Martin Hogbin wrote:
> >
> > Coriolis and centrifigal forces are perfectly well explained by
> > classical mechanics. Read a book on the subject.
>
> Because I have done so carefully, I dispute the "modern" explanations
Perhaps you could enlighten me. What is the 'modern' explanation
for Coriolis force and what is the old one?
>and especially the "illusion" one offered by Roberts and Randy.
Neither has offered such an explanation/
You do not seem very clear on this..
>
> Neither do you. You also seem to be dancing around.
No relation to Don Shead are you?
Martin Hogbin
Martin Hogbin
"Dirk Van de moortel" <dirkvand...@ThankS-NO-SperM.hotmail.com> wrote in message
news:Odz%f.370897$FS4.10...@phobos.telenet-ops.be...
>
> "Mike"
> aka Bill Smith
> aka Eleatis
> aka Undeniable
No relation of Don Shead?
Martin Hogbin
mme...@cars3.uchicago.edu
>mme...@cars3.uchicago.edu says...
>>"Randy Poe" <poespa...@yahoo.com> writes:
>
>>>equations for them. If I am driving in a straight line but
>>>you are in a circle, you will conclude a "force" is making me
>>>"curve". I can tell you precisely where that illusion is coming
>>>from and calculate the precise amount.
>>>
>>frame.
>
>make this concrete:
>
>Suppose I observe a ball's motion and plot its position x versus
>t. I get a straight line, so I conclude (via Newton's laws) that
>there is no force acting on the ball (at least not in the x-direction).
>
>Now, I change coordinates to X = arctan(x) and plot X versus t.
>In the new plot, I *don't* get a straight line, but I get some
>curved line. Is that a "real effect"? Does it indicate the presence
>of a force? No, it's obviously an artifact of my particular
>choice of coordinates.
We're not talking about changing coordinates, we're talking about
measurements of postions, velocities and accelerations. That's plain
kinematics. From this point on, the interpretation dpends on the
definition of a force. If you define force as, say, "transfer of
momentum resulting from an interaction between physical entities, an
interaction which can be expressed, in the ultimate account, by an
exchange of bosons between said physical entities", then no, the above
is not a real force. But, I do not recall force being strictly defined
this way. Mind you, I'm not saying that this is not a legitimate
definition (to the extent that the word "legitimate" applies to
defintions at all) just that it has never been mad universal. If, on
the other hand, you define force as anything causing change of
meomentum than yes, the above is force. You may say that "the change
of momentum is only apparent since there is no change relative to an
inertial frame, but this implies another defintion, namely that "a
valid mesaurement is a measurement performed relative to an inertial
frame". Again, says who?
Consider this simple example (which I used numerous times). In the
equation
a*x^2 + b*x + c = 0
the middle (linear) term can always be eliminated by a proper choice
of coordinates. So, does this mean that the linear term has no
meaning and no validity? No, nonsense. All this means is that I may
choose to eliminate it or to keep it, according to what is convenient.
Same thing goes for the inertial (or, if you prefer, "ficticious")
forces. Bottom line, when evaluating the equations of motion relative
to a non-inertial reference frame you get extra terms. Whether you
tack them on the F side or the ma side, that's a matter of
convenience. But they have real effects with respect to observations
performed in the non-inertial frame. That's all.
>
>Now suppose I'm on a rotating platform, and observe a ball's
>motion. I plot x versus t, and I don't get a straight line.
>Is that due to the presence of a force, or is it an artifact of
>my particular choice of coordinates?
>
>It can be hard to tell which is the case, but what's *not*
>true is that because I'm on a rotating platform, I must take
>into account centrifugal forces.
Of course you don't have to do it.
> Whether the plot of x versus
>t is a straight line or not depends on what *coordinate* system
>I choose. Just because I'm on a rotating platform doesn't
>mean that I have to use one particular coordinate system
>over another. I'll use whatever coordinate system is convenient,
>but there are two issues involved in convenience: What is convenient
>to *measure* with, and what is convenient to *work* with.
>
Sure.
Daryl McCullough
>stevend...@yahoo.com (Daryl McCullough) writes:
>>Suppose I observe a ball's motion and plot its position x versus
>>t. I get a straight line, so I conclude (via Newton's laws) that
>>there is no force acting on the ball (at least not in the x-direction).
>>
>>Now, I change coordinates to X = arctan(x) and plot X versus t.
>>In the new plot, I *don't* get a straight line, but I get some
>>curved line. Is that a "real effect"? Does it indicate the presence
>>of a force? No, it's obviously an artifact of my particular
>>choice of coordinates.
>
>We're not talking about changing coordinates,
It doesn't matter whether I *changed* coordinates or not. If
I had some strange kind of ruler that happened to measure X
instead of x, then plotting the trajectory of the ball would
yield a curved line, rather than a straight line. That doesn't
mean that there is any force involved, necessarily, it just
means that my coordinate system is non-Cartesian.
It's not likely that I'll ever have rulers that happen
to measure arctan(x), but it isn't hard to come up with
rulers that measure X = x cos(wt) + y sin(wt) and
Y = - x sin(wt) + y cos(wt).
>we're talking about measurements of postions,
>velocities and accelerations. That's plain kinematics.
What does it mean to measure positions, velocities,
and accelerations? I know how to measure those things
in a particular *coordinate* system, but how do you
measure them in a coordinate-independent way?
>From this point on, the interpretation dpends on the
>definition of a force. If you define force as, say, "transfer of
>momentum resulting from an interaction between physical entities, an
>interaction which can be expressed, in the ultimate account, by an
>exchange of bosons between said physical entities", then no, the above
>is not a real force. But, I do not recall force being strictly defined
>this way.
What is the definition of force, then? Or if we are assuming
Newton's law, F=ma, then we can ask: what is the definition
of "acceleration"?
What I would say is that acceleration is a vector (in classical
physics, anyway) defined by
a = (d/dt) v
where v is the velocity vector. In terms of components, we
can write v = v^i e_i where e_i is the ith basis vector. Then
we have
a = (d/dt v^i) e_i + v^i (d/dt e_i)
The fact that d/dt v^i = 0 does *not* imply that a=0 unless
we know that our basis vectors e_i are time-independent.
>If, on the other hand, you define force as anything causing change of
>meomentum than yes, the above is force.
That's what I'm disputing. How are you proposing to measure
momentum? How are you operationally deciding whether momentum
is changing are not?
>You may say that "the change of momentum is only apparent
>since there is no change relative to an inertial frame,
Regardless of whether you use inertial frames or not, what
does it *mean* to say that momentum is changing? How do
you operationally measure momentum?
>but this implies another defintion, namely that "a
>valid mesaurement is a measurement performed relative
>to an inertial frame". Again, says who?
Well, I certainly didn't say that. You can measure things
in whatever coordinate system you like, but not all coordinate
systems are Cartesian.
Mike
Delta(p) = p(final) - p(initial)
p = mv
v = ds/dt
All you need is to measure position using a ruler and a clock and then
take the slope. We are talking here physics 101. Measuring change in
momentum wrt a FoR is the easiest thing to do.
>
> >but this implies another defintion, namely that "a
> >valid mesaurement is a measurement performed relative
> >to an inertial frame". Again, says who?
>
> Well, I certainly didn't say that. You can measure things
> in whatever coordinate system you like, but not all coordinate
> systems are Cartesian.
So what? I think you are confused about his, maybe because you dropped
in late in the discussion. We are not talking here about motion as
observed from another frame. We are talking about measurements made in
a particular frame, in this case non-inertial, and inferences that can
be made from those measurements about forces. Of course, as I pointed
out before and Meron put it in the correct context, it all depends on
the definition of force. But assuming no particular definition of force
a priori and then talking about ficticious forces is non-sense physics
to me. And that's the main point of this IMO.
Thus, I sense that classical mechanics and Relativity have a different
conception of force and this is where I wanted to bring the discussion
to. If this is the case, any convergence of GR to Newtonian laws of
motion and gravitation at weak field limits is suspicious, to say the
least. The Newtonian definition is clear: force causes change in state
of motion and equals the rate of change of momentum. Without force
acting the state of rest or uniform linear motion is maintained.
Mike
Mike
Dirk Van de moortel wrote:
> "Mike"
> aka Bill Smith
> aka Eleatis
> aka Undeniable
> > Another attempt by the cranks of sci.physics, sci.physics.relativity to
> > even deny the existence of well-documented affects by calling to their
> > help some crooks in Equador.
> >
> >
> > > Mike wrote:
> > > > If you and Roberts have any doubts theat a coriolis force for example
> > > > is a real force all you have to do is watch the water go down the sink
> > > > drain. Actually its rotation direction is different in the north and
> > > > south hemisphere,
> > >
> > > This is an old wives' tale, aided and abetted by impoverished
> > > entrepreneurs living near the equator who use it to extract money from
> > > tourists.
> > >
> > > In fact, for a typical drain, careful observations show it requires
> > > several hours of rest after filling before the water has settled down
> > > enough so its rotation direction is random when drained. Indeed,
> > > computations of Coriolis force for such a drain show it to be far too
> > > small to account for the observed direction -- asymmetries in the drain
> > > and/or initial conditions of the water are _FAR_ more important. There
> > > was an article in AJP a few years ago on this.
> > >
> > > The entrepreneurs can easily control which way the drain
> > > flows by the way they fill the container. So they walk
> > > 100 yards north of the equator and show one direction, and
> > > walk 100 yards south of the equator and show the other.
> > > They are in _complete_ control of the water's direction.
> > > Amazed tourists, gullible as anywhere else, pay up.
> >
> > Here you go Roberts:
> >
> >
> > "...Any one of these factors is usually more than enough to overwhelm
> > the small contribution of the Coriolis effect in your kitchen sink or
> > bathtub. Research in the 1960s showed that if you do carefully
> > eliminate these factors, the Coriolis effect can be observed [1,2]."
> >
> > 1. Shapiro, 1962, Bath Tub Vortex, Nature, v 196, pp 1080-81 (Northern
> > Hemisphere)
> > 2. Trefethen, et.al., 1965, The Bath Tub Vortex in the Southern
> > Hemisphere, Nature, v 207, pp 1084-85
> >
> >
> > I totally amazed of the logic some people use and their intelligence
> > level. I got no respect for self-proclaimed experts who cannot
> > understand what they read.
>
> A turns 180 degrees clockwise and removes the plug.
> B turns 180 degrees antclockwise and removes the plug.
> A and B have a vortices in different directions everywhere
> on the planet. Even on the equator.
> That's how fakers manage to make a modest living.
I see you have a good understanding of how crooks operate. No wonder.
I suppose for you all Rolex are fake just because some crooks sell them
on the web.
Why don't you take a hike because we are having a serious discussion
here you cannot participate.
Mike
>
> Dirk Vdm
mme...@cars3.uchicago.edu
>mme...@cars3.uchicago.edu says...
>
>>>Suppose I observe a ball's motion and plot its position x versus
>>>t. I get a straight line, so I conclude (via Newton's laws) that
>>>there is no force acting on the ball (at least not in the x-direction).
>>>
>>>Now, I change coordinates to X = arctan(x) and plot X versus t.
>>>In the new plot, I *don't* get a straight line, but I get some
>>>curved line. Is that a "real effect"? Does it indicate the presence
>>>of a force? No, it's obviously an artifact of my particular
>>>choice of coordinates.
>>
>>We're not talking about changing coordinates,
>
>I had some strange kind of ruler that happened to measure X
>instead of x, then plotting the trajectory of the ball would
>yield a curved line, rather than a straight line. That doesn't
>mean that there is any force involved, necessarily, it just
>means that my coordinate system is non-Cartesian.
>
>It's not likely that I'll ever have rulers that happen
>to measure arctan(x), but it isn't hard to come up with
>rulers that measure X = x cos(wt) + y sin(wt) and
>Y = - x sin(wt) + y cos(wt).
>
>
>>velocities and accelerations. That's plain kinematics.
>
>and accelerations? I know how to measure those things
>in a particular *coordinate* system, but how do you
>measure them in a coordinate-independent way?
You don't use coordinates for measurement, you *assign* coordinates
based on measurement. What you use for measurement are standard
yardsticks and standard clocks, or the modern day equivalents.
>
>>From this point on, the interpretation dpends on the
>>definition of a force. If you define force as, say, "transfer of
>>momentum resulting from an interaction between physical entities, an
>>interaction which can be expressed, in the ultimate account, by an
>>exchange of bosons between said physical entities", then no, the above
>>is not a real force. But, I do not recall force being strictly defined
>>this way.
>
>What is the definition of force, then? Or if we are assuming
>Newton's law, F=ma, then we can ask: what is the definition
>of "acceleration"?
No, since acceleration is a kinematic quantity, defined geometrically.
There is nothing in its definition that assumes anything "causing"
the acceleration. All you rely upon is the ability to measure
displacements and time intervals.
>
>What I would say is that acceleration is a vector (in classical
>physics, anyway) defined by
>
> a = (d/dt) v
>
Yes.
>where v is the velocity vector. In terms of components, we
>can write v = v^i e_i where e_i is the ith basis vector. Then
>we have
>
> a = (d/dt v^i) e_i + v^i (d/dt e_i)
>
>The fact that d/dt v^i = 0 does *not* imply that a=0 unless
>we know that our basis vectors e_i are time-independent.
>
Indeed. Now, how do you know that your basis vectors are time
independent?
>>If, on the other hand, you define force as anything causing change of
>>meomentum than yes, the above is force.
>
>That's what I'm disputing. How are you proposing to measure
>momentum? How are you operationally deciding whether momentum
>is changing are not?
Staying for the moment within Newtonian mechanics of massive bodies
I'll just take the low-brow p = mv.
>
>>You may say that "the change of momentum is only apparent
>>since there is no change relative to an inertial frame,
>
>Regardless of whether you use inertial frames or not, what
>does it *mean* to say that momentum is changing? How do
>you operationally measure momentum?
I measure velocity and use the above.
>
>>but this implies another defintion, namely that "a
>>valid mesaurement is a measurement performed relative
>>to an inertial frame". Again, says who?
>
>Well, I certainly didn't say that. You can measure things
>in whatever coordinate system you like, but not all coordinate
>systems are Cartesian.
Indeed, and I didn't claim they're. Also, the coordinate system may
be Cartesian, yet the reference frame may be non-inertial (in
Newtonian mechanics). However, we're under no obligation to limit
ourselves only to a specific set of reference frames.
Daryl McCullough
>stevend...@yahoo.com (Daryl McCullough) writes:
>>What does it mean to measure positions, velocities,
>>and accelerations? I know how to measure those things
>>in a particular *coordinate* system, but how do you
>>measure them in a coordinate-independent way?
>
>You don't use coordinates for measurement, you *assign*
>coordinates based on measurement.
That doesn't seem right to me. Measurements by themselves
don't give you coordinates, you have to do some extra set-up,
figure out what your basis vectors are.
You can draw a set X of nonintersecting lines on the ground
and label them x=0, x=1, etc. Then you can draw a set Y of
lines that intersect all the lines in the first set, and label
them y=0, y=1, etc. Then you can use those two sets of lines
to record positions of objects. So you don't actually need
measurements prior to setting up a coordinate system, you
can just start off with arbitrary curvilinear coordinates.
>>What is the definition of force, then? Or if we are assuming
>>Newton's law, F=ma, then we can ask: what is the definition
>>of "acceleration"?
>
>No, since acceleration is a kinematic quantity, defined geometrically.
I'm not sure what you mean by saying it's a kinematic quantity,
or what you mean by saying it is defined geometrically. If you
mean the vectorial relation
a = (d/dt) v
where v is the velocity vector, then under that definition, if
the acceleration is zero in one coordinate system, it is zero
in *every* coordinate system. (Vectors are coordinate-independent).
If you want to say that the path of a thrown ball is unaccelerated
in an inertial frame, but is accelerated in a rotating frame, then
you are *not* using the geometric notion of acceleration.
>There is nothing in its definition that assumes anything "causing"
>the acceleration. All you rely upon is the ability to measure
>displacements and time intervals.
What is your definition of acceleration? Do you just
mean the triple of quantities
(d/dt)^2 x
(d/dt)^2 y
(d/dt)^2 z
You can certainly measure those quantities, but they aren't *geometric*,
they are dependent on your choice of coordinate system for x, y, and z.
Those three quantities don't tell you the acceleration unless your
coordinates are inertial and Cartesian.
>>What I would say is that acceleration is a vector (in classical
>>physics, anyway) defined by
>>
>> a = (d/dt) v
>>
>Yes.
>
>>where v is the velocity vector. In terms of components, we
>>can write v = v^i e_i where e_i is the ith basis vector. Then
>>we have
>>
>> a = (d/dt v^i) e_i + v^i (d/dt e_i)
>>
>>The fact that d/dt v^i = 0 does *not* imply that a=0 unless
>>we know that our basis vectors e_i are time-independent.
>>
>Indeed. Now, how do you know that your basis vectors are time
>independent?
Well, I think it's theory-dependent. If we define G^i_jk implicitly
by
@/@x^j e_k = G^i_jk e_i
and define Q^i_j by
@/@t e_j = Q^i_j e_i
then mathematically, F=ma becomes, in components,
F^i = m [(d/dt v^i) + Q^i_j v^j + G^i_jk v^j v^k]
But the quantities Q^i_j and G^i_jk are not directly measurable, if
all we know is F=ma. We can only directly measure (d/dt v^i), and
thus the combination
F^i - m [Q^i_j v^j + G^i_jk v^j v^k]
How do we know which parts of (d/dt v^i) are due to F and
which are due to the Qs and Gs? Well, Newton had an answer...
If we assume Newton's *third* law (equal and opposite forces),
as well, then we have a big constraint on F. That is, if we assume that
whenever a force F acts on an object, that object exerts a force -F on
something else, then we can definitively figure out which terms are
acceleration terms, because there is no equal and opposite forces
associated with them.
Think about sitting on a rotating platform. You throw a baseball.
In your coordinate system, it goes spinning off on some curved
path. Is that due to a force, is it due to the Qs and Gs? Well,
if it were a force, then by Newton's third law, the baseball would
be exerting an equal and opposite force on something else. But
nothing else is affected at all by the baseball's curved trajectory.
So it's not a real force, it's an acceleration term.
>>That's what I'm disputing. How are you proposing to measure
>>momentum? How are you operationally deciding whether momentum
>>is changing are not?
>
>Staying for the moment within Newtonian mechanics of massive bodies
>I'll just take the low-brow p = mv.
Yes, but how do you measure whether the vector v is changing?
You can certainly measure whether the components v^i are changing,
but a vector is more than its components.
>>Regardless of whether you use inertial frames or not, what
>>does it *mean* to say that momentum is changing? How do
>>you operationally measure momentum?
>
>I measure velocity and use the above.
How do you measure velocity? I believe you can measure the
*components* of velocity d/dt x^i. Is that what you mean?
>>Well, I certainly didn't say that. You can measure things
>>in whatever coordinate system you like, but not all coordinate
>>systems are Cartesian.
>
>Indeed, and I didn't claim they're.
It seems to me that you can't say that acceleration is given
by
a^i = dv^i/dt
unless you know that your coordinate system is Cartesian. So
if you don't know that your coordinates are Cartesian, then
how do you measure acceleration?
>Also, the coordinate system may
>be Cartesian, yet the reference frame may be non-inertial (in
>Newtonian mechanics).
I'm lumping non-Cartesian and non=inertial together. The main
issue is whether the quantities Q and G defined above are zero.
If they are zero, then you have an inertial Cartesian coordinate
system. If they *aren't* zero, then how do you measure them? If
you can't measure them, then how do you measure acceleration?
mme...@cars3.uchicago.edu
>mme...@cars3.uchicago.edu says...
>
>>stevend...@yahoo.com (Daryl McCullough) writes:
>
>>>What does it mean to measure positions, velocities,
>>>and accelerations? I know how to measure those things
>>>in a particular *coordinate* system, but how do you
>>>measure them in a coordinate-independent way?
>>
>>You don't use coordinates for measurement, you *assign*
>>coordinates based on measurement.
>
>don't give you coordinates, you have to do some extra set-up,
>figure out what your basis vectors are.
I would rather say "decide what your basis vectors are".
>
>You can draw a set X of nonintersecting lines on the ground
>and label them x=0, x=1, etc. Then you can draw a set Y of
>lines that intersect all the lines in the first set, and label
>them y=0, y=1, etc. Then you can use those two sets of lines
>to record positions of objects. So you don't actually need
>measurements prior to setting up a coordinate system, you
>can just start off with arbitrary curvilinear coordinates.
Ah, but unless you either drew them equidistant or established some
rule allowing you to translate coordinate reading to distances, these
coordinates do you no good. So, there other other concepts, prior to
talking about coordinates. The concepts of length, of measuring rods
which carry a constant property of "length" regardles of their
location and orientation, of length comparison, of parallel
displacements etc. In short, geometry. And once you get this you
recognize that coordinates are a matter of convenience, not necessity.
People performed geometrical measurements (and pretty precise ones, at
that) for thousands of years prior to the appearance of the coordinate
concept.
Moreover, it is not through the use of coordinates that most such
measurements are performed. If I want to track, say, the motion of an
object on a 2d plane, I don't mark coordinates all over the plane and
run after the object writing down every moment which coordinate lines
it is currently crossing. That would be rather clumsy. No, what I
would to is to first establish a baseline (arbitrarily chosen but
one the length of which is precisely measured), and second locate, say,
two theodolites at both ends of the baseline. Once this is done then,
at any given moment, measurement of the angles between the lines of
sight to the object form both theodelites, and the baseline, uniquely
locate the object in the plane (with the assumption of a specific
geometry, yes, something needs to be assumed). As such location can
be represented as "so many measuring rod's length along the baseline,
so many in the direction normal to the baseline", then coordinates
follow from the measurement as an afterthought. But we can calculate
displacement between two consecutive locations without ever invoking
coordinates. This is how surveyors actually work. And this is how
astronomers always worked.
Coordinates are very useful as a way to codify the results of such
measurements as the above, in a wya which is generally clearer, more
uniform and more ameanable to theoretical treatment. They do not,
however, contain any information beyond this present in the oroiginal
measurement and they most certainly are not a necessary precursor to
the performance of such measurement.
>
>>>What is the definition of force, then? Or if we are assuming
>>>Newton's law, F=ma, then we can ask: what is the definition
>>>of "acceleration"?
>>
>>No, since acceleration is a kinematic quantity, defined geometrically.
>
>I'm not sure what you mean by saying it's a kinematic quantity,
>or what you mean by saying it is defined geometrically. If you
>mean the vectorial relation
>
> a = (d/dt) v
>
>where v is the velocity vector, then under that definition, if
>the acceleration is zero in one coordinate system, it is zero
>in *every* coordinate system. (Vectors are coordinate-independent).
You're thinking too much like a theoretician. I'm an experimentalist.
So, consider the following (note, in order to keep it simple I'm not
even getting into rotations): We've a moving train. Within the train
we establish a system as the one described above, two theodolites at
the ends of a baseline (lets say that the baseline is parallel to the
train'd direction of motion). On the ground, parallel to the rails,
we establish an identical system. We use both systems to track an
object, sayan earring in the left ear of a motionless passenger (lets
say she fell asleep). Now I perform a series of measurements of the
location of this earring, repeated in constant time intervals.
Relative to the train's baseline I'm getting a location which can be
described as a vector loc_tr, with a component parallel to the
baseline and a component orthogonal to it. This location doesn't
change with time. Relative to the ground baseline I'm gettin another
location, loc_gr, which again can be described as a vector, using the
*same* base. This location changes with time.
So, we've:
1) v_tr = (d/dt) loc_tr = 0
2) v_gr = (d/dt) loc_gr != 0
A vector is coordinate independent, but not reference frame
independent.
You can now repeat the same with velocities, using an accelerating
train. Note that we didn't even bring in the added complexity
resulting when the base vectors themselves change with time. Could
easily add it by having the train following a curve, but there is no
need for it at this stage.
>
>If you want to say that the path of a thrown ball is unaccelerated
>in an inertial frame, but is accelerated in a rotating frame, then
>you are *not* using the geometric notion of acceleration.
>
I'm most certainly using the geometrical notion of acceleration. See
above. I'm measuring locations using geometry alone, in constant time
intervals. I'm finding (approximate) velocities as ratios of
displacements (changes of locations) to changes of time. I'm then
finding (approximate) accelerations by takine the ratios of changes of
velocities to changes in time. By making the time intervals shorter
and shorter I'm making the approximate values approach the true
momentary values, closer and closer. At no point in this process are
any dynamical notions being used. Especially, the issue of "what is
causing the acceleration" is not being considered.
>>There is nothing in its definition that assumes anything "causing"
>>the acceleration. All you rely upon is the ability to measure
>>displacements and time intervals.
>
>What is your definition of acceleration? Do you just
>mean the triple of quantities
>
> (d/dt)^2 x
> (d/dt)^2 y
> (d/dt)^2 z
>
>You can certainly measure those quantities, but they aren't *geometric*,
>they are dependent on your choice of coordinate system for x, y, and z.
>Those three quantities don't tell you the acceleration unless your
>coordinates are inertial and Cartesian.
See above. Acceleration is (d/dt)^2 r where r is the location vector.
Whether r is represented using an orthogonal basis or any other basis,
this is utterly irrelevant. And that's all there is to the definition
of acceleration. The definition most certainly ***does not*** include
the requirement that the reference frame is inertial or that the
coordinate system used within said reference frame is Cartesian.
Heck, it doesn't even require the base vectors used to be the same in
every location. Consider using polar coordinates with the base
vectors a_r and a_phi. All that is required is that the base vectors
used in any location do not change with time *relative* to the same
reference frame we use to measure or calculate acceleration. Mind
you, they may very well change with time relative to another reference
frame.
Side note: If the base vectors change with location, the
differentiation is more complex than plain (coordinate by coordinate
differentiation". We need to concept of connection. That's a
technicality (albit an important one).
>
>>>What I would say is that acceleration is a vector (in classical
>>>physics, anyway) defined by
>>>
>>> a = (d/dt) v
>>>
>>Yes.
>>
>>>where v is the velocity vector. In terms of components, we
>>>can write v = v^i e_i where e_i is the ith basis vector. Then
>>>we have
>>>
>>> a = (d/dt v^i) e_i + v^i (d/dt e_i)
>>>
>>>The fact that d/dt v^i = 0 does *not* imply that a=0 unless
>>>we know that our basis vectors e_i are time-independent.
>>>
>>Indeed. Now, how do you know that your basis vectors are time
>>independent?
>
>Well, I think it's theory-dependent. If we define G^i_jk implicitly
>by
>
> @/@x^j e_k = G^i_jk e_i
>
>and define Q^i_j by
>
> @/@t e_j = Q^i_j e_i
>
>then mathematically, F=ma becomes, in components,
>
> F^i = m [(d/dt v^i) + Q^i_j v^j + G^i_jk v^j v^k]
>
>But the quantities Q^i_j and G^i_jk are not directly measurable, if
>all we know is F=ma. We can only directly measure (d/dt v^i), and
>thus the combination
>
> F^i - m [Q^i_j v^j + G^i_jk v^j v^k]
>
>How do we know which parts of (d/dt v^i) are due to F and
>which are due to the Qs and Gs? Well, Newton had an answer...
The problem in all the above is, again, that you think as theorist.
Acceleration is not jsut a theoretical concept it is a perfectly
measurable one. See above.
>
>If we assume Newton's *third* law (equal and opposite forces),
>as well, then we have a big constraint on F. That is, if we assume that
>whenever a force F acts on an object, that object exerts a force -F on
>something else, then we can definitively figure out which terms are
>acceleration terms, because there is no equal and opposite forces
>associated with them.
>
>Think about sitting on a rotating platform. You throw a baseball.
>In your coordinate system, it goes spinning off on some curved
>path. Is that due to a force, is it due to the Qs and Gs? Well,
>if it were a force, then by Newton's third law, the baseball would
>be exerting an equal and opposite force on something else. But
>nothing else is affected at all by the baseball's curved trajectory.
>So it's not a real force, it's an acceleration term.
Which is jsut semantics, to say "it can be made to disappear by using
an inertial reference frame. That was never in dispute. Yet, it is
perfectly real when describing motion relative to the rotating frame.
>
>>>That's what I'm disputing. How are you proposing to measure
>>>momentum? How are you operationally deciding whether momentum
>>>is changing are not?
>>
>>Staying for the moment within Newtonian mechanics of massive bodies
>>I'll just take the low-brow p = mv.
>
>Yes, but how do you measure whether the vector v is changing?
>You can certainly measure whether the components v^i are changing,
>but a vector is more than its components.
Again, see above. Try to think about measurements, not just
formalism.
>
>>>Regardless of whether you use inertial frames or not, what
>>>does it *mean* to say that momentum is changing? How do
>>>you operationally measure momentum?
>>
>>I measure velocity and use the above.
>
>How do you measure velocity? I believe you can measure the
>*components* of velocity d/dt x^i. Is that what you mean?
>
No, see above.
>>>Well, I certainly didn't say that. You can measure things
>>>in whatever coordinate system you like, but not all coordinate
>>>systems are Cartesian.
>>
>>Indeed, and I didn't claim they're.
>
>It seems to me that you can't say that acceleration is given
>by
>
> a^i = dv^i/dt
>
Sigh. Try not to attribute to me things I didn't say.
>unless you know that your coordinate system is Cartesian. So
>if you don't know that your coordinates are Cartesian, then
>how do you measure acceleration?
See above.
>
>>Also, the coordinate system may
>>be Cartesian, yet the reference frame may be non-inertial (in
>>Newtonian mechanics).
>
>I'm lumping non-Cartesian and non=inertial together. The main
>issue is whether the quantities Q and G defined above are zero.
>If they are zero, then you have an inertial Cartesian coordinate
>system. If they *aren't* zero, then how do you measure them? If
>you can't measure them, then how do you measure acceleration.
And for the last time, see above. I gave you an operational
procedure, and nowhere does this procedure asume anything about
inertiality or a Cartesian coordinate system. It does require various
geometrical concepts, to be sure, including the parallel transport
concept if the coordinate system isn't Cartesian. But that's all.
Timo Nieminen
> A vector is coordinate independent, but not reference frame
> independent.
... which is why transformations under which vectors are independent of
the transformation (eg rotations/translations for 3D Euclidean vectors,
Lorentz transformations for 4-vectors, etc) are _special_.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
mme...@cars3.uchicago.edu
>On Sat, 15 Apr 2006 mme...@cars3.uchicago.edu wrote:
>
>> independent.
>
>the transformation (eg rotations/translations for 3D Euclidean vectors,
>Lorentz transformations for 4-vectors, etc) are _special_.
>
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1145061987....@i40g2000cwc.googlegroups.com...
>
> Dirk Van de moortel wrote:
[snip]
> > A and B have a vessel with water and a plugged hole.
> > A turns 180 degrees clockwise and removes the plug.
> > B turns 180 degrees antclockwise and removes the plug.
> > A and B have a vortices in different directions everywhere
> > on the planet. Even on the equator.
> > That's how fakers manage to make a modest living.
>
>
>
> I see you have a good understanding of how crooks operate. No wonder.
>
> I suppose for you all Rolex are fake just because some crooks sell them
> on the web.
>
> Why don't you take a hike because we are having a serious discussion
> here you cannot participate.
You can't read my messages, retard.
You killfiled me:
29-Jan-2006:
http://groups.google.com/group/sci.physics.relativity/msg/18246d2052b...
| You are now "plonked" for good idiot. Anyway, you contribute nothing in
| these ng's other than psychosis and stupidity. I stick with it now and
| you "stock" your merde some place else.
1-Sep-2005:
http://groups.google.com/group/sci.math/msg/1b0256e5a1ad48b0
| Gee, Cantor, Goedel, Peano and the others would be proud of you Dirt.
| Especially if they knew that besides that you also understand the
| square root.
|
| Plonk
|
| Mike
5-Feb-2005:
http://groups.google.com/group/sci.physics/msg/986fdd22eebc33b9
| You got nothing to say, proven fact. You spend your whole misearable
| days maintaining a site where you store your impotence and psychotic
| behavior.
|
| I had enough with you schizo, a killfile will do it and then a clean of
| the Dirt you left around making pooppies all over the place.
|
| Mike
25-Sep-2004:
http://groups.google.com/group/sci.physics/msg/e5cfae098ed3d0a1
| Hi Dirk, Are you still working on a definition of sqrt? You and Alex
| can make a nice team. He thinks there is no time so you got all the
| time in the world to think about sqrt.
|
| Plonk Dirk
|
| Mike
Dirk Vdm
Martin Hogbin
"Mike" <ele...@yahoo.gr> wrote in message news:1145046293.9...@i39g2000cwa.googlegroups.com...
>
>
> Neither do you. You also seem to be dancing around.
If you are working in an inertial (non-rotating) frame:
You use Newton's laws.
There is no centrifugal or Coriolis force.
(Note: some older elementary text books did use the words
'centrifugal force ' in inertial frames. This was wrong then
and it is wrong now).
If you are working in a rotating frame:
You can use Newton's laws in exactly the same way as in an inertial frame.
You must invoke two additional forces, Coriolis and centrifugal.
That is it. The subject has been well understood for centuries.
Martin Hogbin
PD
Says Newton's 3rd law, which implicitly demands an agent to the force.
In particular, a force is defined as being the interaction between two
objects. It is this relationship that allows the third law to apply
universally. A coriolis or centrifugal "force" is completely consistent
with the notion of force more or less defined by the 2nd law alone, but
fails completely to satisfy the 3rd law.
Mike is pointing out -- I believe -- the rather narrow perspective of
the principle of equivalence, which in a way "senses" a force in the
same way the 2nd law defines it. Mach's principle -- whatever it means
-- appeals to the 3rd law, asking "what is the agent?"
I believe the modern view of force implicitly bears the notion of an
interaction between objects. Any action that produces an acceleration
but which cannot be attributed to an interaction is therefore missing
something essential and is considered a pseudoforce.
PD
PD
PD
I disagree. Newtonian laws of motion include the 3rd law in the
definition of forces. Please describe how a coriolis or a centripetal
force satisfies the 3rd law.
PD
PD
And when you do this, you find in certain situations that you are faced
with problem with several possible solutions:
1. Momentum as calculated as you suggest is not always conserved (the
current situation being case in point).
2. Momentum is in fact conserved, and we are further to assume that the
only way that momentum is changed is through the presence of a force: F
= dp/dt. We therefore conclude that there is a force present, though
for the life of us we're not sure what the agent of that force is.
3. Momentum is in fact conserved; however, we allow that there are
other means by which a momentum contribution can be added to a system
other than a "real" force.
It's not immediately apparent which of these should be chosen without
appeal to some independent criteria, such as additional criteria that a
force must satisfy other than F=dp/dt.
PD
joe_ave...@yahoo.com
I'm trying to follow your discussion with great interest. As a graduate
PhilSci student long time ago, my thesis dealt with the subject of
force. Some of you might not know that there were a few alternatives to
the Newtonian definition of force, before and after Newton. One such
definition was offered by Leibniz, equating (active) force to mv^2 and
(passive) force to mv. Newton's definition survived because it fitted
well with his law of gravitation.
Now, in a peper published last year in a second class rated journal --
this does not bother me at all -- another definition of force was
presented. Basically, that force is equal to the rate of change of
kinetic energy, also known as instantenuous power. I found this highly
interesting since defining force in such a way makes it a scalar
quantity.
In the paper I mentioned there is also a reference to the problem of
ficticious forces and how this new definition of force -- I call it new
because I have not seen it before -- makes the corresponding laws of
motion -- also presented in the paper -- form invariant in all frames
of reference, inertial or non-inertial, for freely moving particles or
particles in uniform motion, linear or curvilinear. This is a straight
forward result but neverthless very interesting. You can find the
paper at the following web link:
http://www.doaj.org/abstract?id=119444&toc=y/
Joe Avery
Tom Roberts
> [... I agree with what he said]
> If you want an example of where the Coriolis force is a
> very useful concept just think of large-scale weather systems.
> These are completely dominated by the Coriolis force.
Not quite "completely dominated" -- in the U.S.A. there are occasional
counter-rotating weather systems. But even so, "Coriolis force" is
definitely needed to model them (they are much less stable than most
normally rotating ones, and often disappear before reaching the east
coast). This happens perhaps once or twice a year, compared to hundreds
of normally rotating ones; TV weathermen always comment on this when
they pass through.
The real world is much more complex than theoretical physics can
accommodate in detail.
Tom Roberts tjro...@lucent.com
joe_ave...@yahoo.com
I'm not sure about that. Newton, during his famous debate with Leibniz
claimed these forces can only be referenced with respect to an absolute
space and, as such, the reaction to these forces is to be found in that
unobservable domain. Just because there is no immediate and direct
observation of a reaction this does not mean it is not present. This
is a subject of the realist/anti-realist debate, a metaphysical issue
rather than an issue of experimental physics.
>
> Mike is pointing out -- I believe -- the rather narrow perspective of
> the principle of equivalence, which in a way "senses" a force in the
> same way the 2nd law defines it. Mach's principle -- whatever it means
> -- appeals to the 3rd law, asking "what is the agent?"
If I recall correctly, force is attributed to the agent. Mach attempted
to offer a different explanation for the results of the bucket
experiment performed by Newton with the intent to show that centrifugal
forces must be referenced with respect to an absolute space. As far as
I know, Mach's idea is still considered a metaphysical hypothesis.
>
> I believe the modern view of force implicitly bears the notion of an
> interaction between objects. Any action that produces an acceleration
> but which cannot be attributed to an interaction is therefore missing
> something essential and is considered a pseudoforce.
Be careful here. "Cannot be attributed" does not mean "is not
attributed". The alternative would involve searching for such
interaction, something that was pursued only by a few philosophers and
experimentalists. I studied physics before I pursued PhilSci. One must
be very careful of the boundary: where physics stops and metaphysics
starts. Actually, Mach was a victim of such boundary crossing. In
physics, if you stick to what can be measured with clocks and rulers
you end up with enough information to draw your conclusions in a safe
way.
Joe Avery
>
> PD
>
> PD
joe_ave...@yahoo.com
Good point as I noted in another post. One answer is that we do not
know. Just because the answer is not known, one cannot draw
automatically the conclusion that Centrifugal --- I think you meant to
say --- and Coriolis forces are pseudoforces.
Centrifugal forces can be considered reactions to Centripetal forces. I
do not know of any experiment to generate Centrifugal forces in the
absence of Centripetal forces.
The Coriolis case is much more complicated.
Joe Avery
joe_ave...@yahoo.com
Mike wrote:
> Randy Poe wrote:
> > > Another attempt by the cranks of sci.physics, sci.physics.relativity to
> > > even deny the existence of well-documented affects by calling to their
> > > help some crooks in Equador.
> >
> >
> > So given that there are conmen in Ecuador who can make the
> > water swirl any direction they want, you think this proves
> > coriolis force is significant for small tubs?
> >
> > Or what? What is your point exactly? Are you saying there's
> > something wrong with the calculations in that link?
> >
> > > http://www.physics.ohio-state.edu/~dvandom/Edu/newcor.html
> > >
> > > I totally amazed of the logic some people use and their intelligence
> > > level. I got no respect for self-proclaimed experts who cannot
> > > understand what they read.
> >
> > down a sink, just the same as Tom Roberts said.
> >
> > What do YOU think the article meant by this:
> > "Water in the sink doesn't go far enough to trigger a
> > noticeable north/south deflection."
> > or this:
> > "In any case, don't blame it on the Coriolis effect unless
> > your sink is the size of a small ocean."
> > ?
> >
> > What point do YOU think the article is making regarding
> > Coriolis and sinks that "self-proclaimed experts" are missing?
>
> 1. Shapiro, 1962, Bath Tub Vortex, Nature, v 196, pp 1080-81 (Northern
> Hemisphere)
> 2. Trefethen, et.al., 1965, The Bath Tub Vortex in the Southern
> Hemisphere, Nature, v 207, pp 1084-85
>
> effect of coriolis force and something that can be notices under the
> proper conditions whereas they surely do not refer to crooks who insist
> they can do that under every condition.
>
> Some, took advantage of the crook attempt to fool people and in turn
> try to fool everybody that the effect does not exist using of course an
> informal fallacy of the type:
> if a crook sayd that then it must be false.
>
> Under carefullt controlled laboratory conditions, tghe coriolis effect
> on water going down a drain is noticable, this is my point. It is
> surely noticable on larger scales on the rotation of cyclons.
>
> My point and dispute with Roberts and other here is the Coriolis (and
> centrifugal) is a real force and not ficticious as they call it. They
> want to declare it ficticious because some in the past called it so.
> But the contradictions here is that these people believe that there is
> not preferred reference frame and since coriolis arises in rotating
> reference frames calling it ficticious because it is not needed in
> inertial frames is taking a prefered stand on what frame you will
> consider as the basis of your physics.
>
> Given also the fact that the idea of a purely inertial FoR is just a
> mathematical abstraction whereas the coriolis effect is a measured one,
> it seems to me that these people who insist it is ficticious are
> cranks, simply because they denay experimental evidence in favor of
> abstract and absurd models of reality.
Be careful here. What is real is not an easy thing to answer. Often, it
depends on your objectives. You can alway measure the velocity of a
shadow but no shadow, no matter how fast is moving, can carry any
useful information faster than the speed of light.
I'm still not sure whether the same thing can be both real and
ficticious depending on the circumstances. I think this is an
interesting question. In the past, I adopted much of the anti-realist
views. As I grow older, I tend to question them. I now think Einstein
was correct when he disputed the Copenhagen hypothesis. But at the
same time, this is just a belief of mine against the overwhelming
experimental evidence for quantum mechanics.
Joe Avery
>
> Mike
>
>
>
>
> >
> > - Randy
Tom Roberts
> On Sat, 15 Apr 2006 mme...@cars3.uchicago.edu wrote:
>> A vector is coordinate independent, but not reference frame
>> independent.
>
> ... which is why transformations under which vectors are independent of
> the transformation (eg rotations/translations for 3D Euclidean vectors,
> Lorentz transformations for 4-vectors, etc) are _special_.
Right, as far as you went. Specifically, in Newtonian mechanics
3-momentum is invariant under a rotation but not a Galilean boost. In
SR, 4-momentum is invariant under a rotation or Lorentz boost but not an
acceleration. In GR, 4-momentum is invariant under any coordinate
transform whatsoever -- indeed in GR there are no "special" coordinate
transformations at all.
Coordinates are pure constructs of the human mind, and therefore cannot
possibly affect physical phenomena -- but of course we often describe
phenomena using coordinates, so the _descriptions_ depend on them even
though the underlying phenomena themselves do not.
In Newtonian mechanics, the _description_ of phenomena using rotating
coordinates must include "centrifugal and Coriolis forces", but the
underlying phenomena cannot possibly depend on those "forces". That's
why we call them "fictitious" [#]. In SR it gets too complicated to
attempt to use rotating coordinates and AFAIK nobody has laid out how to
do it in any comprehensive manner (and there exist few if any rotating
systems in which applying SR would be fruitful, because either NM would
do or GR would be required). In GR this is all irrelevant, because it
does not matter which coordinates you use at all.
[#] Like literary fiction they are pure constructs of the
human mind. This is _precisely_ the right term for them.
Modern physics incorporates this inherently, because all quantities in
our fundamental theories of physics are tensors, and are thus
independent of coordinates. "Centrifugal, Coriolis, and gravitational
forces" are all simply components of the connection when projected onto
certain coordinate systems, but neither the connection nor its
projections appear in these theories in any way. None of those "forces"
affect the _tensors_.
Tom Roberts tjro...@lucent.com
Daryl McCullough
>>You can draw a set X of nonintersecting lines on the ground
>>and label them x=0, x=1, etc. Then you can draw a set Y of
>>lines that intersect all the lines in the first set, and label
>>them y=0, y=1, etc. Then you can use those two sets of lines
>>to record positions of objects. So you don't actually need
>>measurements prior to setting up a coordinate system, you
>>can just start off with arbitrary curvilinear coordinates.
>
>Ah, but unless you either drew them equidistant or established some
>rule allowing you to translate coordinate reading to distances, these
>coordinates do you no good.
On the contrary, the lesson of General Relativity is that
physics can be done using *any* coordinates whatsoever. The
exact form of your laws will depend on the coordinates, but
in very constrained way: only through the connection coefficients,
which can be empirically determined.
>So, there other other concepts, prior to
>talking about coordinates. The concepts of length, of measuring rods
>which carry a constant property of "length" regardles of their
>location and orientation, of length comparison, of parallel
>displacements etc. In short, geometry.
I know all that, but it seemed to me that the use of "centrifugal
force" as a force is antithetical to the geometric view. It seems
to me that centrifugal force is a coordinate-dependent effect, and
in the geometric view, vectors are independent of coordinates.
Actually, maybe the distinction you are making is between
non-Cartesian and non-inertial. I tend to lump them together, since
(trained in relativity) I think of time as just another dimension.
But it is certainly consistent to take a geometric view of the
*spatial* coordinates, and still take a parameter view of time.
>You're thinking too much like a theoretician. I'm an experimentalist.
>So, consider the following (note, in order to keep it simple I'm not
>even getting into rotations): We've a moving train. Within the train
>we establish a system as the one described above, two theodolites at
>the ends of a baseline (lets say that the baseline is parallel to the
>train'd direction of motion). On the ground, parallel to the rails,
>we establish an identical system. We use both systems to track an
>object, sayan earring in the left ear of a motionless passenger (lets
>say she fell asleep). Now I perform a series of measurements of the
>location of this earring, repeated in constant time intervals.
>Relative to the train's baseline I'm getting a location which can be
>described as a vector loc_tr, with a component parallel to the
>baseline and a component orthogonal to it. This location doesn't
>change with time. Relative to the ground baseline I'm gettin another
>location, loc_gr, which again can be described as a vector, using the
>*same* base. This location changes with time.
>
>So, we've:
>
>1) v_tr = (d/dt) loc_tr = 0
>2) v_gr = (d/dt) loc_gr != 0
>
>A vector is coordinate independent, but not reference frame
>independent.
Right. I keep forgetting that, because I've been too thoroughly
"indoctrinated" in the 4-D spacetime view, which treats space
and time equivalently. In the 4-D view, vectors are both
coordinate-independent and reference frame independent.
I'm not talking special relativity, specifically. You can cast
Galilean spacetime into a 4-D view, as well. It's not something
that would have occurred to Newton, perhaps, but it is extremely
convenient even when doing Newtonian gravity, because it unifies
effects due to non-Cartesian coordinates and non-inertial
coordinates.
But you're right, in a 3D view, vectors are not reference frame
independent.
The geometric space-time view is this (let's just use 1 spatial
dimension, for simplicity): Let P(t) be the spacetime location
of the object at time t. The velocity vector V associated with
P has coordinates (V^0,V^1) such that for any scalar field Phi
(function of space and time)
d/dt Phi(P(t)) = V^0 @/@x^0 Phi + V^1 @/@x^1 Phi
For Newtonian spacetime, we can just always take x^0 to be t,
and x^1 = x, and V^0 = 1. So (V^0,V^1) = (1,v) where v is
the ordinary velocity of the object.
To compute the components of velocity in a different frame
with coordinates x'^0, x'^1 (where again, we'll let x'^0 = t,
x'^1 = x')
you just use:
V'^0 = 1
V'^1 = @x'/@t + v @x'/@x
So, if the relationship between the coordinates is
x' = x - ut
then
@x'/@t = -u
@x'/@x = 1
So in the new coordinate system, the velocity has coordinates
(V'^0, V'^1) = (1, v-u)
It works out fine. You don't need to assume Einstein's
theory to unify space and time, Galilean relativity can
be unified just as well.
>You can now repeat the same with velocities, using an accelerating
>train. Note that we didn't even bring in the added complexity
>resulting when the base vectors themselves change with time.
In the space-time view, a switch to an accelerating reference
frame *does* mean using time-dependent basis vectors. If the
initial basis vectors are e_i, and the new basis vectors are
e'_j, then we have
e'_j = @x^k/@x'^j e_k
For the transformation
x = x' + 1/2 at'^2
t = t'
we have
e'_0 = @x/@t' e_1 + @t/@t' e_0
= at' e_1 + e_0
e'_1 = @x/@x' e_1 + @t/@x' e_0
= e_1
So the new basis vectors *are* time-dependent. e'_0 is time-dependent
(but e'_1 is not).
>>If you want to say that the path of a thrown ball is unaccelerated
>>in an inertial frame, but is accelerated in a rotating frame, then
>>you are *not* using the geometric notion of acceleration.
>>
>I'm most certainly using the geometrical notion of acceleration. See
>above.
Yes, you are right, using 3-D geometry. It isn't using *spacetime*
geometry.
The thing that's a little weird about your "geometric" notion of
acceleration is that it does *not* consider the centrifugal force
due to using polar coordinates to be a real force, but it *does*
consider the centrifugal force due to using a rotating frame to
be a real force, even though they have the same (mathematical)
origin.
>>What is your definition of acceleration? Do you just
>>mean the triple of quantities
>>
>> (d/dt)^2 x
>> (d/dt)^2 y
>> (d/dt)^2 z
>>
>>You can certainly measure those quantities, but they aren't *geometric*,
>>they are dependent on your choice of coordinate system for x, y, and z.
>>Those three quantities don't tell you the acceleration unless your
>>coordinates are inertial and Cartesian.
>
>See above. Acceleration is (d/dt)^2 r where r is the location vector.
Okay. However, if instead you took the spacetime view that
acceleration is the time derivative of the spacetime velocity vector,
then you *don't* get the acceleration to be zero for an object at
rest in a rotating frame.
[stuff deleted]
>>But the quantities Q^i_j and G^i_jk are not directly measurable, if
>>all we know is F=ma. We can only directly measure (d/dt v^i), and
>>thus the combination
>>
>> F^i - m [Q^i_j v^j + G^i_jk v^j v^k]
>>
>>How do we know which parts of (d/dt v^i) are due to F and
>>which are due to the Qs and Gs? Well, Newton had an answer...
>
>The problem in all the above is, again, that you think as theorist.
F = ma is a physical *theory*. If all you are interested in is
measuring things, you don't need the concept of force. The
concepts of Newtonian physics do *not* generalize to noninertial
frames if you try to view "centrifugal force" as a force.
>>Think about sitting on a rotating platform. You throw a baseball.
>>In your coordinate system, it goes spinning off on some curved
>>path. Is that due to a force, is it due to the Qs and Gs? Well,
>>if it were a force, then by Newton's third law, the baseball would
>>be exerting an equal and opposite force on something else. But
>>nothing else is affected at all by the baseball's curved trajectory.
>>So it's not a real force, it's an acceleration term.
>
>Which is jsut semantics,
No, it's a matter of whether you are using Newton's laws of motion,
or not. You *cannot* view centrifugal force as a "force" and keep
all of Newton's laws of motion.
Yes, you can always make F = ma true even in a noninertial
frame by *defining* force to be a/m. However, that definition
of "force" doesn't in general obey the other law of motion,
that there are equal and opposite forces.
>>It seems to me that you can't say that acceleration is given
>>by
>>
>> a^i = dv^i/dt
>>
>Sigh. Try not to attribute to me things I didn't say.
Okay, I'm sorry. I see that you have non-Cartesian coordinates covered.
In my definitions, the acceleration looks like
a^i = d/dt v^i + Q^i_j v^j + G^i_jk v^j v^k
in coordinates. Choosing Cartesian coordinates makes G vanish.
Choosing inertial coordinates makes Q vanish.
I mistakenly thought that your approach was treating the
G-terms as if they were forces, rather than acceleration terms. You
aren't doing that. However, you *are* treating the Q-terms as
forces, rather than acceleration.
The Q terms and the G terms have the same origin, in the
spacetime geometric view.
>>I'm lumping non-Cartesian and non=inertial together. The main
>>issue is whether the quantities Q and G defined above are zero.
>>If they are zero, then you have an inertial Cartesian coordinate
>>system. If they *aren't* zero, then how do you measure them? If
>>you can't measure them, then how do you measure acceleration.
>
>And for the last time, see above.
No, you haven't provided an operational procedure for measuring
the terms Q. You are basically lumping them in with the force
terms.
>I gave you an operational procedure, and nowhere does this
>procedure assume anything about inertiality or a Cartesian
>coordinate system.
I don't think that's true. I think you are ignoring the
distinction between inertial and noninertial coordinates.
Once again, the full expression for the acceleration is
this:
a = a^j e_j
where
a^j = d/dt v^i + Q^i_j v^j + G^i_jk v^j v^k
You are providing an operational way to measure the
combination
a_partial = e_i [d/dt v^i + G^i_jk v^j v^k]
(without the Q terms). I would claim that that is
*not* the full acceleration, but that's a matter of
semantics, I suppose.
PD
Yes, of course: centrifugal. My bad.
>
> Centrifugal forces can be considered reactions to Centripetal forces.
I disagree. This is a common physics misconception. In an action-pair,
if the action is the force that A exerts on B, the reaction is the
equal and opposite force that B exerts on A. The reaction force is NOT
an equal and opposite force that acts on B.
PD
Mike
You are confusing reference frames. I repeat to you once more a fact
you seem you do not want to accept: Newton's law apply without
modification only in inertial reference frames. This holds for the
second law and it must also hold for the third law. In the inertial
frame, the reaction to the centripetal force is the centrifugal force.
In a non-indertial reference frame, there is no issue of
action-reaction since these two forces must cancel each other to have
the state of rest. There is no net action in a non-inertial reference
frame and this was the point all along.
If you keep jumping around reference frames you will never get it
straight.
Mike
Mike
I try to stay away from philosophical issues in this as much as
possible. To me real is that can be measured using clocks and rulers.
Since coriolis and centrifugal accelerations can be measured, they are
real effects. If one then accepts that F= dp/dt then the forces are
real. That is all.
Mike
Mike
Dirk Van de moortel wrote:
> "Mike" <ele...@yahoo.gr> wrote in message news:1145061987....@i40g2000cwc.googlegroups.com...
> >
> > Dirk Van de moortel wrote:
>
> [snip]
>
> > > A and B have a vessel with water and a plugged hole.
> > > A turns 180 degrees clockwise and removes the plug.
> > > B turns 180 degrees antclockwise and removes the plug.
> > > A and B have a vortices in different directions everywhere
> > > on the planet. Even on the equator.
> > > That's how fakers manage to make a modest living.
> >
> >
> >
> > I see you have a good understanding of how crooks operate. No wonder.
> >
> > I suppose for you all Rolex are fake just because some crooks sell them
> > on the web.
> >
> > Why don't you take a hike because we are having a serious discussion
> > here you cannot participate.
>
> You can't read my messages, retard.
> You killfiled me:
Don't cry crank. Take you medication and shut up if you cannot
participate in this discussion in a productive manner.
Mike
PD
That is incorrect and precisely the misconception to which I referred.
In the case of swinging a bucket around in a circle with your hand, if
the inward (centripetal) force on the bucket is the one exerted by your
hand, then the reaction force referred to in Newton's third law is the
force that the bucket exerts on the hand. It is *not* an outward
(centrifugal) force exerted on the bucket.
Newton's 3rd law refers to forces exerted on each *object* of an
interacting pair, not to two forces acting on the *same* object in
opposite directions.
This misconception is heartily stomped on in just about every textbook
from the 9th grade on up.
PD
Dirk Van de moortel
"Mike"
aka Bill Smit
aka Eleatis
aka Undeniable
<ele...@yahoo.gr> wrote in message news:1145124024.1...@g10g2000cwb.googlegroups.com...
>
> PD wrote:
> > joe_ave...@yahoo.com wrote:
> > > Centrifugal forces can be considered reactions to Centripetal forces.
> >
> > I disagree. This is a common physics misconception. In an action-pair,
> > if the action is the force that A exerts on B, the reaction is the
> > equal and opposite force that B exerts on A. The reaction force is NOT
> > an equal and opposite force that acts on B.
>
> You are confusing reference frames. I repeat to you once more a fact
> you seem you do not want to accept: Newton's law apply without
> modification only in inertial reference frames. This holds for the
> second law and it must also hold for the third law. In the inertial
> frame, the reaction to the centripetal force is the centrifugal force.
In the inertial frame, there is only the centripetal
force that makes the object go round in a circle.
If there was a reaction centrifugal force, the object
would not move in a circle to begin with.
> In a non-indertial reference frame, there is no issue of
> action-reaction since these two forces must cancel each other to have
> the state of rest. There is no net action in a non-inertial reference
> frame and this was the point all along.
The point was all along that you are incapable of
admitting your mistakes.
>
> If you keep jumping around reference frames you will never get it
> straight.
If you keep digging, you'll end up choking in your
droppings.
Dirk Vdm
Martin Hogbin
"Tom Roberts" <tjro...@lucent.com> wrote in message news:sE80g.1992$Lm5....@newssvr12.news.prodigy.com...
> Martin Hogbin wrote:
> > [... I agree with what he said]
> > If you want an example of where the Coriolis force is a
> > very useful concept just think of large-scale weather systems.
> > These are completely dominated by the Coriolis force.
>
> Not quite "completely dominated" -- in the U.S.A. there are occasional
> counter-rotating weather systems. But even so, "Coriolis force" is
> definitely needed to model them (they are much less stable than most
> normally rotating ones, and often disappear before reaching the east
> coast). This happens perhaps once or twice a year, compared to hundreds
> of normally rotating ones; TV weathermen always comment on this when
> they pass through.
Interesting. Are these relatively local phenomena?
Martin Hogbin
Mike
Yes indeed. So when someone offers you a blue banana be alert. Most
probable it is painted.
http://plato.stanford.edu/entries/qualia-knowledge/
Mike
>
>
> Tom Roberts tjro...@lucent.com
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1145124348.9...@e56g2000cwe.googlegroups.com...
>
> Dirk Van de moortel wrote:
> > "Mike" <ele...@yahoo.gr> wrote in message news:1145061987....@i40g2000cwc.googlegroups.com...
> > >
> > > Dirk Van de moortel wrote:
> >
> > [snip]
> >
> > > > A and B have a vessel with water and a plugged hole.
> > > > A turns 180 degrees clockwise and removes the plug.
> > > > B turns 180 degrees antclockwise and removes the plug.
> > > > A and B have a vortices in different directions everywhere
> > > > on the planet. Even on the equator.
> > > > That's how fakers manage to make a modest living.
> > >
> > >
> > >
> > > I see you have a good understanding of how crooks operate. No wonder.
> > >
> > > I suppose for you all Rolex are fake just because some crooks sell them
> > > on the web.
> > >
> > > Why don't you take a hike because we are having a serious discussion
> > > here you cannot participate.
> >
> > You can't read my messages, retard.
> > You killfiled me:
>
> Don't cry crank. Take you medication and shut up if you cannot
> participate in this discussion in a productive manner.
The only productive way to handle a dog that continues
to shit on the kitchen floor is by rubbing its nose in it:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/BrainHoles.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/IdiotsAndrocles.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/RattenFingure.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/StockWithIt.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/EleatisStyle.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/OfCourseBozzo.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Bourbaki.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Psychotic.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Learned.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Playground.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Dirt.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Imbecile.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/HiPsycho.html
Dirk Vdm
Mike
Dirk Van de moortel wrote:
> aka Bill Smit
> aka Eleatis
> aka Undeniable
> <ele...@yahoo.gr> wrote in message news:1145124024.1...@g10g2000cwb.googlegroups.com...
> >
> > PD wrote:
> > > joe_ave...@yahoo.com wrote:
>
> > > > Centrifugal forces can be considered reactions to Centripetal forces.
> > >
> > > I disagree. This is a common physics misconception. In an action-pair,
> > > if the action is the force that A exerts on B, the reaction is the
> > > equal and opposite force that B exerts on A. The reaction force is NOT
> > > an equal and opposite force that acts on B.
> >
> > You are confusing reference frames. I repeat to you once more a fact
> > you seem you do not want to accept: Newton's law apply without
> > modification only in inertial reference frames. This holds for the
> > second law and it must also hold for the third law. In the inertial
> > frame, the reaction to the centripetal force is the centrifugal force.
>
> In the inertial frame, there is only the centripetal
> force that makes the object go round in a circle.
> If there was a reaction centrifugal force, the object
> would not move in a circle to begin with.
Idiot you do not even understand Newton's laws and you want to post in
sci.physics groups. To every action there is an equal and opposite
reaction.
When a stone is attached to a rope and revolving above your empty head,
the reaction to the centripetal force is at your head moron. It tries
to take your hand out of the center of the revolution, that is the
reason it is called centri-fugal.
Idiot
Mike
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1145124241.1...@u72g2000cwu.googlegroups.com...
You should stay away from physicis as much as possible:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Physics101.html
Dirk Vdm
Mike
Did I say that? Either you do not understand or like Dirt von der
Vomit you are incapable of basic physics.
>
> Newton's 3rd law refers to forces exerted on each *object* of an
> interacting pair, not to two forces acting on the *same* object in
> opposite directions.
DId I say that? Where did you read it? The reaction to the centripetal
force acts on the agent who generates the centripetal force. On your
hand that is.
>
> This misconception is heartily stomped on in just about every textbook
> from the 9th grade on up.
You try to convert your [probably intentional] misunderstandings into
alleged misconceptions at no avail.
Mike
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1145125439.4...@t31g2000cwb.googlegroups.com...
[snip]
> > > > > Centrifugal forces can be considered reactions to Centripetal forces.
> > > >
> > > > I disagree. This is a common physics misconception. In an action-pair,
> > > > if the action is the force that A exerts on B, the reaction is the
> > > > equal and opposite force that B exerts on A. The reaction force is NOT
> > > > an equal and opposite force that acts on B.
> > >
> > > You are confusing reference frames. I repeat to you once more a fact
> > > you seem you do not want to accept: Newton's law apply without
> > > modification only in inertial reference frames. This holds for the
> > > second law and it must also hold for the third law. In the inertial
> > > frame, the reaction to the centripetal force is the centrifugal force.
> >
> > That is incorrect and precisely the misconception to which I referred.
> > In the case of swinging a bucket around in a circle with your hand, if
> > the inward (centripetal) force on the bucket is the one exerted by your
> > hand, then the reaction force referred to in Newton's third law is the
> > force that the bucket exerts on the hand. It is *not* an outward
> > (centrifugal) force exerted on the bucket.
>
>
> Did I say that? Either you do not understand or like Dirt von der
> Vomit you are incapable of basic physics.
Keep digging, pig ;-)
I'm sure you'll manage to get a followup to this one:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Physics101.html
Dirk Vdm
Mike
Hello crank. Are you busy trying to understand Newton's third law?
Mike
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1145125085.3...@i39g2000cwa.googlegroups.com...
Yes, just like I predicted, you did it again:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Physics101bis.html
Grrrrreat :-)
Dirk Vdm
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1145125602.4...@t31g2000cwb.googlegroups.com...
Nope, just trying to make you understand that you should stay
away from physics as far as you piggibly can:
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Physics101.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Physics101bis.html
I just love you :-)
Dirk Vdm
Mike
How many times I have said that the reaction is not exerted on the
bucket but at the hand revolving the bucket?
I think you read what you want to read.
>
> Newton's 3rd law refers to forces exerted on each *object* of an
> interacting pair, not to two forces acting on the *same* object in
> opposite directions.
>
> This misconception is heartily stomped on in just about every textbook
> from the 9th grade on up.
Also, at the same grade on up, people are supposed to be able to read
and understand what they read.
Mike
Mike
paranoid.
Mike
> Dirk Vdm
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1145126037....@e56g2000cwe.googlegroups.com...
Even when they are "hidding behing their own ratten fingure"?
Dirk Vdm
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1145126167.9...@j33g2000cwa.googlegroups.com...
No no no.... I'm highly amused by the way you always dig
yourself in - Just like Schoenfeld.
It would be so easy to say "Ah yes, of course, sorry, I was
mistaken". But no... there is that Deep Hole to be Dug...
You are a master at it. Congratulations.
Dirk Vdm
mme...@cars3.uchicago.edu
>
>mme...@cars3.uchicago.edu wrote:
>> In article <e1p2g...@drn.newsguy.com>, stevend...@yahoo.com (Daryl McCullough) writes:
>> >mme...@cars3.uchicago.edu says...
>> >>"Randy Poe" <poespa...@yahoo.com> writes:
>> >
>> >>>We know precisely how they arise, that's how we can derive
>> >>>equations for them. If I am driving in a straight line but
>> >>>you are in a circle, you will conclude a "force" is making me
>> >>>"curve". I can tell you precisely where that illusion is coming
>> >>>from and calculate the precise amount.
>> >>>
>> >>It is not an illusion. Repeat, *it is not an illusion*. It is a
>> >>perfectly real effect resulting from a specific choice of a reference
>> >>frame.
>> >
>> >Hmm. I'm not sure what "real effect" means here, but let's
>> >make this concrete:
>> >
>> >t. I get a straight line, so I conclude (via Newton's laws) that
>> >there is no force acting on the ball (at least not in the x-direction).
>> >
>> >Now, I change coordinates to X = arctan(x) and plot X versus t.
>> >In the new plot, I *don't* get a straight line, but I get some
>> >curved line. Is that a "real effect"? Does it indicate the presence
>> >of a force? No, it's obviously an artifact of my particular
>> >choice of coordinates.
>>
>> measurements of postions, velocities and accelerations. That's plain
>> definition of a force. If you define force as, say, "transfer of
>> momentum resulting from an interaction between physical entities, an
>> interaction which can be expressed, in the ultimate account, by an
>> exchange of bosons between said physical entities", then no, the above
>> is not a real force. But, I do not recall force being strictly defined
>> definition (to the extent that the word "legitimate" applies to
>> the other hand, you define force as anything causing change of
>> of momentum is only apparent since there is no change relative to an
>> valid mesaurement is a measurement performed relative to an inertial
>> frame". Again, says who?
>>
>
Yes, *in an inertial frame*.
>In particular, a force is defined as being the interaction between two
>objects.
Again, yes, *in an inertial frame*.
> It is this relationship that allows the third law to apply universally.
No, not universally. Just *in an inertial frame*.
In no way does any of this say "only measurements performed relative
to inertial frames are valid". There is absolutely no reason for
physics to limit itself so. What it does say is "Note what is it that
you're measuring". To wit, the first law gives us a default state of
motion of an object, the second tells us that the state of motion can
be changed by an interaction (more than this, it tells you how will it
change) and the third tells us (as you noted) that the interaction *has*
to be with other objects (not necessarily a single other object).
And all of this is true *as long* as the reference frame used is
inertial.
Now, would all of this has been topped by "and if it is not inertial,
you cannot say anything", that it would've been of rather limited
usefullness. Fortunately, this is not the case. We can transform
from one frame to another, including to non-inertial frames, thus we
can do physics even in non-inertial frames. Only, there the rules
change somewhat. The 1st law is no longer true and same goes for the
third. The second, however, can still be applied as long as proper
correction terms are included. That's all.
> A coriolis or centrifugal "force" is completely consistent
>with the notion of force more or less defined by the 2nd law alone, but
>fails completely to satisfy the 3rd law.
>
Indeed. It applies in a non-inertial frame where the 3rd law is no
longer valid.
>Mike is pointing out -- I believe -- the rather narrow perspective of
>the principle of equivalence, which in a way "senses" a force in the
>same way the 2nd law defines it. Mach's principle -- whatever it means
>-- appeals to the 3rd law, asking "what is the agent?"
>
>I believe the modern view of force implicitly bears the notion of an
>interaction between objects.
I wouldn't quite say so. I would say that the above is true in the
context of "fundamental forces" but not in the context of equations of
motion.
> Any action that produces an acceleration
>but which cannot be attributed to an interaction is therefore missing
>something essential and is considered a pseudoforce.
Sure, no problem. This doesn't negate the fact that, when describing
motion relative to a non-inertial reference frame, such pseudoforce has
perfectly real effects, where by "real" I mean "observable". If one wants
to limit the term "real" only to those things which are completely
independent of any reference frame then sure, this is legitimate, but
it limits the scope of "real" to a rather small class of entities. I
don't see the need for this.
Mati Meron | "When you argue with a fool,
me...@cars.uchicago.edu | chances are he is doing just the same"
mme...@cars3.uchicago.edu
>
>mme...@cars3.uchicago.edu wrote:
>> >mme...@cars3.uchicago.edu says...
>> >
>> >>>Suppose I observe a ball's motion and plot its position x versus
>> >>>t. I get a straight line, so I conclude (via Newton's laws) that
>> >>>there is no force acting on the ball (at least not in the x-direction).
>> >>>
>> >>>Now, I change coordinates to X = arctan(x) and plot X versus t.
>> >>>In the new plot, I *don't* get a straight line, but I get some
>> >>>curved line. Is that a "real effect"? Does it indicate the presence
>> >>>of a force? No, it's obviously an artifact of my particular
>> >>>choice of coordinates.
>> >>
>> >>We're not talking about changing coordinates,
>> >
>> >I had some strange kind of ruler that happened to measure X
>> >instead of x, then plotting the trajectory of the ball would
>> >yield a curved line, rather than a straight line. That doesn't
>> >mean that there is any force involved, necessarily, it just
>> >means that my coordinate system is non-Cartesian.
>> >
>> >It's not likely that I'll ever have rulers that happen
>> >to measure arctan(x), but it isn't hard to come up with
>> >rulers that measure X = x cos(wt) + y sin(wt) and
>> >Y = - x sin(wt) + y cos(wt).
>> >
>> >
>> >>velocities and accelerations. That's plain kinematics.
>> >
>> >and accelerations? I know how to measure those things
>> >in a particular *coordinate* system, but how do you
>> >measure them in a coordinate-independent way?
>>
>> based on measurement. What you use for measurement are standard
>> yardsticks and standard clocks, or the modern day equivalents.
>> >
>> >>From this point on, the interpretation dpends on the
>> >>definition of a force. If you define force as, say, "transfer of
>> >>momentum resulting from an interaction between physical entities, an
>> >>interaction which can be expressed, in the ultimate account, by an
>> >>exchange of bosons between said physical entities", then no, the above
>> >>is not a real force. But, I do not recall force being strictly defined
>> >>this way.
>> >
>> >Newton's law, F=ma, then we can ask: what is the definition
>> >of "acceleration"?
>>
>> There is nothing in its definition that assumes anything "causing"
>> the acceleration. All you rely upon is the ability to measure
>> displacements and time intervals.
>> >
>> >What I would say is that acceleration is a vector (in classical
>> >physics, anyway) defined by
>> >
>> > a = (d/dt) v
>> >
>>
>> >where v is the velocity vector. In terms of components, we
>> >can write v = v^i e_i where e_i is the ith basis vector. Then
>> >we have
>> >
>> > a = (d/dt v^i) e_i + v^i (d/dt e_i)
>> >
>> >The fact that d/dt v^i = 0 does *not* imply that a=0 unless
>> >we know that our basis vectors e_i are time-independent.
>> >
>> independent?
>>
>> >>If, on the other hand, you define force as anything causing change of
>> >>meomentum than yes, the above is force.
>> >
>> >momentum? How are you operationally deciding whether momentum
>> >is changing are not?
>>
>> I'll just take the low-brow p = mv.
>> >
>> >>You may say that "the change of momentum is only apparent
>> >>since there is no change relative to an inertial frame,
>> >
>> >does it *mean* to say that momentum is changing? How do
>> >you operationally measure momentum?
>>
>
>And when you do this, you find in certain situations that you are faced
>with problem with several possible solutions:
>1. Momentum as calculated as you suggest is not always conserved (the
>current situation being case in point).
Aha. Momentum isn't always conserved. It is only conserved when
there is no external force acting.
>2. Momentum is in fact conserved, and we are further to assume that the
>only way that momentum is changed is through the presence of a force: F
>= dp/dt. We therefore conclude that there is a force present, though
>for the life of us we're not sure what the agent of that force is.
In the context of the second law, what matters is that we can say what
the force is. It *does not* matter what the agent of the force is.
>3. Momentum is in fact conserved; however, we allow that there are
>other means by which a momentum contribution can be added to a system
>other than a "real" force.
We opt for (2).
>
>It's not immediately apparent which of these should be chosen without
>appeal to some independent criteria, such as additional criteria that a
>force must satisfy other than F=dp/dt.
>
This only matters if you insist that there has to be an agent.
mme...@cars3.uchicago.edu
>
>mme...@cars3.uchicago.edu wrote:
>> >
>> >Mike wrote:
>> >> Martin Hogbin wrote:
>> >> So you and Randy agree the Coriolis effect is real. I suppose the same
>> >> holds for the centrifugal effect. So now let us take it from here.
>> >
>> >Yes, they're real effects of conservation of momentum in
>> >accelerated frames of reference.
>> >
>> >> Calling these forces "ficticious" must be justified in the context of
>> >> what ficticious means and not in some other context, like for instance
>> >> a posteriori derivation from a certain model.
>> >
>> >It means they are not truly forces but artifacts of a coordinate
>> >transformation. There isn't really an acceleration corresponding
>> >to these "forces".
>>
>> Careful here. Relative to the accelerating reference frame, where
>> you've these forces, there most certainly is an acceleration
>> corresponding to said forces.
>>
>> >
>> >> My argument is similar to that brought up by Newton. These forces are
>> >> real. The fact that we cannot explain how they arise in non-inertial
>> >> reference frames does not immediately justify calling them ficticious.
>> >
>> >We know precisely how they arise, that's how we can derive
>> >equations for them. If I am driving in a straight line but
>> >you are in a circle, you will conclude a "force" is making me
>> >"curve". I can tell you precisely where that illusion is coming
>> >from and calculate the precise amount.
>> >
>> perfectly real effect resulting from a specific choice of a reference
>> frame.
>>
>> me...@cars.uchicago.edu | chances are he is doing just the same"
>
>
>PhilSci student long time ago, my thesis dealt with the subject of
>force. Some of you might not know that there were a few alternatives to
>the Newtonian definition of force, before and after Newton. One such
>definition was offered by Leibniz, equating (active) force to mv^2 and
>(passive) force to mv. Newton's definition survived because it fitted
>well with his law of gravitation.
>
No, not only because of this. It fits well with everything else as
well.
>Now, in a peper published last year in a second class rated journal --
>this does not bother me at all -- another definition of force was
>presented. Basically, that force is equal to the rate of change of
>kinetic energy, also known as instantenuous power. I found this highly
>interesting since defining force in such a way makes it a scalar
>quantity.
>
And why this should be a plus? We know that same interaction but
acting in different directions will result in different motions.
Unless, of course, you consider all motions with same kinetic energy
the same. But then you've just lost lots of information.
>In the paper I mentioned there is also a reference to the problem of
>ficticious forces and how this new definition of force -- I call it new
>because I have not seen it before -- makes the corresponding laws of
>motion -- also presented in the paper -- form invariant in all frames
>of reference, inertial or non-inertial, for freely moving particles or
>particles in uniform motion, linear or curvilinear.
Sounds a big claim. Will have to see it to believe. Anyway, two
things worth mentioning:
1) There is no "problem of ficticious forces". The only problems
here are semantic.
2) While the notion of force was fundamental in the early formulation
of mechanics, currently it is rather secondary.
This is a straight
>forward result but neverthless very interesting. You can find the
>paper at the following web link:
>
>http://www.doaj.org/abstract?id=119444&toc=y/
>
>
>Joe Avery
mme...@cars3.uchicago.edu
>mme...@cars3.uchicago.edu says...
>>stevend...@yahoo.com (Daryl McCullough) writes:
>
>>>and label them x=0, x=1, etc. Then you can draw a set Y of
>>>lines that intersect all the lines in the first set, and label
>>>them y=0, y=1, etc. Then you can use those two sets of lines
>>>to record positions of objects. So you don't actually need
>>>measurements prior to setting up a coordinate system, you
>>>can just start off with arbitrary curvilinear coordinates.
>>
>>Ah, but unless you either drew them equidistant or established some
>>rule allowing you to translate coordinate reading to distances, these
>>coordinates do you no good.
>
>On the contrary, the lesson of General Relativity is that
>physics can be done using *any* coordinates whatsoever. The
>exact form of your laws will depend on the coordinates, but
>in very constrained way: only through the connection coefficients,
>which can be empirically determined.
Yes, see above regarding "...established some rule...etc."
>
>>So, there other other concepts, prior to
>>talking about coordinates. The concepts of length, of measuring rods
>>which carry a constant property of "length" regardles of their
>>location and orientation, of length comparison, of parallel
>>displacements etc. In short, geometry.
>
>I know all that, but it seemed to me that the use of "centrifugal
>force" as a force is antithetical to the geometric view. It seems
>to me that centrifugal force is a coordinate-dependent effect, and
>in the geometric view, vectors are independent of coordinates.
It is not "coordinate dependent" but "reference frame dependent". You
appear to use "coordinates' and "reference frames" interchangeably,
but they're not quite the same in the current context. So ..
>
>Actually, maybe the distinction you are making is between
>non-Cartesian and non-inertial.
Exactly.
> I tend to lump them together, since
>(trained in relativity) I think of time as just another dimension.
>
The current discussion is in the context of Newtonian mechanics. In
the context of GR we've no issue of "pseudoforces", it disappears (or,
as Tom Roberts noted, is relagated to the connections).
Yes, for sure. I think we were talking past each other, here, because
we used different contexts.
Yes, that's neat. And, indeed, doing this eliminates the distinction
between reference frames and coordinate systems.
>
>>You can now repeat the same with velocities, using an accelerating
>>train. Note that we didn't even bring in the added complexity
>>resulting when the base vectors themselves change with time.
>
>In the space-time view, a switch to an accelerating reference
>frame *does* mean using time-dependent basis vectors.
Of course, it has to.
> If the
>initial basis vectors are e_i, and the new basis vectors are
>e'_j, then we have
>
> e'_j = @x^k/@x'^j e_k
>
>For the transformation
>
> x = x' + 1/2 at'^2
> t = t'
>
>we have
>
> e'_0 = @x/@t' e_1 + @t/@t' e_0
> = at' e_1 + e_0
>
> e'_1 = @x/@x' e_1 + @t/@x' e_0
> = e_1
>
>So the new basis vectors *are* time-dependent. e'_0 is time-dependent
>(but e'_1 is not).
>
>>>If you want to say that the path of a thrown ball is unaccelerated
>>>in an inertial frame, but is accelerated in a rotating frame, then
>>>you are *not* using the geometric notion of acceleration.
>>>
>>I'm most certainly using the geometrical notion of acceleration. See
>>above.
>
>Yes, you are right, using 3-D geometry. It isn't using *spacetime*
>geometry.
>
Agreed. As I said, different context.
>The thing that's a little weird about your "geometric" notion of
>acceleration is that it does *not* consider the centrifugal force
>due to using polar coordinates to be a real force, but it *does*
>consider the centrifugal force due to using a rotating frame to
>be a real force, even though they have the same (mathematical)
>origin.
That's inherent in the Newtonian context where coordinates are just a
*static* labeling of locations and time is a parameter, separate from
space. Obviously, the physics is independent of labeling.
>
>>>What is your definition of acceleration? Do you just
>>>mean the triple of quantities
>>>
>>> (d/dt)^2 x
>>> (d/dt)^2 y
>>> (d/dt)^2 z
>>>
>>>You can certainly measure those quantities, but they aren't *geometric*,
>>>they are dependent on your choice of coordinate system for x, y, and z.
>>>Those three quantities don't tell you the acceleration unless your
>>>coordinates are inertial and Cartesian.
>>
>>See above. Acceleration is (d/dt)^2 r where r is the location vector.
>
>Okay. However, if instead you took the spacetime view that
>acceleration is the time derivative of the spacetime velocity vector,
>then you *don't* get the acceleration to be zero for an object at
>rest in a rotating frame.
Agreed. So, again, context. And I should mention that by "time
derivative", in the above, I mean the, what's the term (it has been a
long time) "covariant derivative" I think. Meaning, parallel
transporting v(t+dt) to the location of v for the purpose of
subtraction.
>
>[stuff deleted]
>
>>>But the quantities Q^i_j and G^i_jk are not directly measurable, if
>>>all we know is F=ma. We can only directly measure (d/dt v^i), and
>>>thus the combination
>>>
>>> F^i - m [Q^i_j v^j + G^i_jk v^j v^k]
>>>
>>>How do we know which parts of (d/dt v^i) are due to F and
>>>which are due to the Qs and Gs? Well, Newton had an answer...
>>
>>The problem in all the above is, again, that you think as theorist.
>
>F = ma is a physical *theory*. If all you are interested in is
>measuring things, you don't need the concept of force. The
>concepts of Newtonian physics do *not* generalize to noninertial
>frames if you try to view "centrifugal force" as a force.
>
Depends which ones. Obviously, the 3rd law doesn't stand as stated
(though one make take refuge in the metaphysical "the reaction is
acting on the universe as a whole"). So what? I can write the
Lagrangian for, say, an object suspended in the cargo bay of the space
shuttle. Then I can switch to a reference frame comoving with the
shuttle, with, say, the z axis directed at the center of the Earth.
So, we've a reference frame the center of which moves in a circular
(for convenience) orbit around Earth and, in addition, the axes rotate
(with the same period). And, I can then take this Lagrangian and
solve the equations of motion for an object with some initial location
and velocity relative to the center of the shuttle. The equations
will have terms corresponding to centrifugal and coriolis forces, so?
Which Newtonian concepts got lost here?
>>>Think about sitting on a rotating platform. You throw a baseball.
>>>In your coordinate system, it goes spinning off on some curved
>>>path. Is that due to a force, is it due to the Qs and Gs? Well,
>>>if it were a force, then by Newton's third law, the baseball would
>>>be exerting an equal and opposite force on something else. But
>>>nothing else is affected at all by the baseball's curved trajectory.
>>>So it's not a real force, it's an acceleration term.
>>
>>Which is jsut semantics,
>
>No, it's a matter of whether you are using Newton's laws of motion,
>or not. You *cannot* view centrifugal force as a "force" and keep
>all of Newton's laws of motion.
The only thing I don't keep is the 3rd, which is not relevant to the
equations of motion.
>
>Yes, you can always make F = ma true even in a noninertial
>frame by *defining* force to be a/m. However, that definition
>of "force" doesn't in general obey the other law of motion,
>that there are equal and opposite forces.
>
Indeed, but as I said, this is not relevant to the equations of
motion. As long as we limit the scope of the second law to inertial
frames, we're fine.
We have again an issue of context here (though a different context than
before). Obviously, we cannot keep both the 2nd and the 3rd law
intact when working in a non-inertial frame. So, which one we care
about more? As long as what we're doing is writing and solving
equations of motion, it is the 2nd that matters, and in this context
"inertial forces" act same as any other forces.
>>>It seems to me that you can't say that acceleration is given
>>>by
>>>
>>> a^i = dv^i/dt
>>>
>>Sigh. Try not to attribute to me things I didn't say.
>
>Okay, I'm sorry. I see that you have non-Cartesian coordinates covered.
>In my definitions, the acceleration looks like
>
> a^i = d/dt v^i + Q^i_j v^j + G^i_jk v^j v^k
>
>in coordinates. Choosing Cartesian coordinates makes G vanish.
>Choosing inertial coordinates makes Q vanish.
Yes.
>
>I mistakenly thought that your approach was treating the
>G-terms as if they were forces, rather than acceleration terms. You
>aren't doing that. However, you *are* treating the Q-terms as
>forces, rather than acceleration.
>
Which, in the context mentioned above, is really a matter of
semantics. Schematically, if you have a reference frame accelerating
at a_0, then you get
F = m(a + a_0)
which you can also write as
F - ma_0 = ma
In the first case you treat a_0 as acceleration correction, in the
second as a force correction. It matters not how you call it.
>The Q terms and the G terms have the same origin, in the
>spacetime geometric view.
Certainly. And if we take the spacetime view cosnsitently all the
way, to GR, we don't have to discuss inertial forces at all since the
issue disappears.
>
>>>I'm lumping non-Cartesian and non=inertial together. The main
>>>issue is whether the quantities Q and G defined above are zero.
>>>If they are zero, then you have an inertial Cartesian coordinate
>>>system. If they *aren't* zero, then how do you measure them? If
>>>you can't measure them, then how do you measure acceleration.
>>
>>And for the last time, see above.
>
>No, you haven't provided an operational procedure for measuring
>the terms Q. You are basically lumping them in with the force
>terms.
Yes.
>
>>I gave you an operational procedure, and nowhere does this
>>procedure assume anything about inertiality or a Cartesian
>>coordinate system.
>
>I don't think that's true. I think you are ignoring the
>distinction between inertial and noninertial coordinates.
>
I'm not even using coordinates, in the procedure. I'm measuring
changes of observation angles, that's all.
>Once again, the full expression for the acceleration is
>this:
>
> a = a^j e_j
>
>where
>
> a^j = d/dt v^i + Q^i_j v^j + G^i_jk v^j v^k
>
>You are providing an operational way to measure the
>combination
>
> a_partial = e_i [d/dt v^i + G^i_jk v^j v^k]
>
>(without the Q terms). I would claim that that is
>*not* the full acceleration, but that's a matter of
>semantics, I suppose.
Well, yes, exactly. It is acceleration relative to the frame used.
That's what is within scope in this context.
Timo Nieminen
> In SR it gets too complicated to attempt to use
> rotating coordinates and AFAIK nobody has laid out how to do it in any
> comprehensive manner (and there exist few if any rotating systems in which
> applying SR would be fruitful, because either NM would do or GR would be
> required).
No, rotating coordinate systems can be useful and feasible in SR.
Specifically, for the analysis of the interaction of electromagnetic waves
and dielectrics in uniform rotation. For example, scattering by rotating
objects, Sagnac interferometer. Van Bladel devotes a chapter to rotating
systems in his Relativity and Engineering.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
Martin Hogbin
"Mike" <ele...@yahoo.gr> wrote in message news:1145126167.9...@j33g2000cwa.googlegroups.com...
Despite your protestations that you are only interested
in discussing physics rather than trading insults you
have failed to reply to my latest post in this thread.
Martin Hogbin.
Daryl McCullough
>>No, it's a matter of whether you are using Newton's laws of motion,
>>or not. You *cannot* view centrifugal force as a "force" and keep
>>all of Newton's laws of motion.
>
>The only thing I don't keep is the 3rd, which is not relevant to the
>equations of motion.
Well, you don't need Newton at all in order to compute
trajectories: If you know the position, the velocity, and
the acceleration at one time, then you can integrate to
get the position at a later time. So what is Newton doing
for you, in this case?
--
Daryl McCullough
Ithaca, NY
--
NewsGuy.Com 30Gb $9.95 Carry Forward and On Demand Bandwidth
mme...@cars3.uchicago.edu
>mme...@cars3.uchicago.edu says...
>
>>>or not. You *cannot* view centrifugal force as a "force" and keep
>>>all of Newton's laws of motion.
>>
>>The only thing I don't keep is the 3rd, which is not relevant to the
>>equations of motion.
>
>trajectories: If you know the position, the velocity, and
>the acceleration at one time, then you can integrate to
>get the position at a later time. So what is Newton doing
>for you, in this case?
>
the acceleration at all times. How do you know it if not from the
equations of motion? And, where do the equations of motion come from?
Daryl McCullough
>You need to know position and velocity at one time. You need to know
>the acceleration at all times. How do you know it if not from the
>equations of motion? And, where do the equations of motion come from?
Applying Newton's theory in an inertial reference frame, and then
performing a coordinate transformation?
mme...@cars3.uchicago.edu
>mme...@cars3.uchicago.edu says...
>
>>the acceleration at all times. How do you know it if not from the
>>equations of motion? And, where do the equations of motion come from?
>
>performing a coordinate transformation?
>
non-inertial reference frame. Details depend on convenience. In any
case, you do need Newton's theory to set the equations.
Tom Roberts
> Centrifugal forces can be considered reactions to Centripetal forces.
No, they cannot.
When swinging a stone by a string, the force on my hand is radially
outward at radius zero, so the "centrifugal force" is zero. The force on
my hand is quite clearly the tension in the string, not "centrifugal
force". The force on the rock is also the tension on the string.
"Centrifugal force" is a _fiction_ specifically invented so one could
analyze that situation in a frame co-rotating with the rock and make
believe Newton's laws hold in the rotating frame, hence "centrifugal
force" is needed to cancel the tension of the string on the rock so the
rock remains motionless in this rotating frame. Similarly it must go to
zero at the center because there is no force that needs canceling there.
Tom Roberts tjro...@lucent.com
Tom Roberts
> Since coriolis and centrifugal accelerations can be measured, they are
> real effects.
Sure. Nobody denies that if one uses rotating coordinates such
accelerations will be observed. But there is no corresponding force
other than the _fictitious_ ones due to choice of a rotating frame.
In particular, while you can measure these accelerations in your
rotating frame, you cannot _measure_ the "centrifugal or Coriolis force".
Tom Roberts tjro...@lucent.com
Tom Roberts
> On Sat, 15 Apr 2006, Tom Roberts wrote:
>> In SR it gets too complicated to attempt to use rotating coordinates
>> and AFAIK nobody has laid out how to do it in any comprehensive manner
>
> Specifically, for the analysis of the interaction of electromagnetic
> waves and dielectrics in uniform rotation. For example, scattering by
> rotating objects, Sagnac interferometer. Van Bladel devotes a chapter to
> rotating systems in his Relativity and Engineering.
Interesting! Thanks.
Tom Roberts tjro...@lucent.com
PD
Right. Now, in an inertial frame it is always possible to enlarge the
system enough where that external force becomes internal to the system,
because you've made the interacting pair both part of the system. In a
rotating frame, it's a little trickier to do this. This is the first
clue that something is amiss.
>
> >2. Momentum is in fact conserved, and we are further to assume that the
> >only way that momentum is changed is through the presence of a force: F
> >= dp/dt. We therefore conclude that there is a force present, though
> >for the life of us we're not sure what the agent of that force is.
>
> In the context of the second law, what matters is that we can say what
> the force is. It *does not* matter what the agent of the force is.
Well, I for one would be hard-pressed to identify the force without
knowing what the agent is. Unless of course you want to label it
according to the kind of acceleration it produces, e.g. centrifugal or
coriolis.
>
> >3. Momentum is in fact conserved; however, we allow that there are
> >other means by which a momentum contribution can be added to a system
> >other than a "real" force.
>
> We opt for (2).
> >
> >It's not immediately apparent which of these should be chosen without
> >appeal to some independent criteria, such as additional criteria that a
> >force must satisfy other than F=dp/dt.
> >
> This only matters if you insist that there has to be an agent.
Which is what I opt for.
PD
I understand the point.
>
> >In particular, a force is defined as being the interaction between two
> >objects.
>
> Again, yes, *in an inertial frame*.
>
> > It is this relationship that allows the third law to apply universally.
>
> No, not universally. Just *in an inertial frame*.
>
> In no way does any of this say "only measurements performed relative
> to inertial frames are valid". There is absolutely no reason for
> physics to limit itself so. What it does say is "Note what is it that
> you're measuring". To wit, the first law gives us a default state of
> motion of an object, the second tells us that the state of motion can
> be changed by an interaction (more than this, it tells you how will it
> change) and the third tells us (as you noted) that the interaction *has*
> to be with other objects (not necessarily a single other object).
> And all of this is true *as long* as the reference frame used is
> inertial.
The problem I have with this is that the "note what it is that you are
measuring" begs the question. If one can *distinguish* between a
coriolis "force" and a force that is due to some charge-induced field,
for example, then you are already saying, "Yes, of course, I know that
this force is not due to another partner" or "I can tell from the
situation that this is not a force like that one is." In essence, you
are dividing forces (as acceleration-makers) into two camps but giving
them the same label. I choose to label the two camps differently, into
forces (with interaction partners) and pseudoforces (without). I see no
reason not to do this.
>
> Now, would all of this has been topped by "and if it is not inertial,
> you cannot say anything", that it would've been of rather limited
> usefullness. Fortunately, this is not the case. We can transform
> from one frame to another, including to non-inertial frames, thus we
> can do physics even in non-inertial frames. Only, there the rules
> change somewhat. The 1st law is no longer true and same goes for the
> third. The second, however, can still be applied as long as proper
> correction terms are included. That's all.
And the crux is that I use Newton's three laws to define a force. You
are using only the second.
Mike
Tom Roberts wrote:
> joe_ave...@yahoo.com wrote:
> > Centrifugal forces can be considered reactions to Centripetal forces.
>
> No, they cannot.
>
> When swinging a stone by a string, the force on my hand is radially
> outward at radius zero, so the "centrifugal force" is zero. The force on
> my hand is quite clearly the tension in the string, not "centrifugal
> force". The force on the rock is also the tension on the string.
You are wrong.
>
> "Centrifugal force" is a _fiction_ specifically invented so one could
> analyze that situation in a frame co-rotating with the rock and make
> believe Newton's laws hold in the rotating frame, hence "centrifugal
> force" is needed to cancel the tension of the string on the rock so the
> rock remains motionless in this rotating frame. Similarly it must go to
> zero at the center because there is no force that needs canceling there.
>
Again wrong. A simple force analysis reveals the opposite you claim.
Mike
>
> Tom Roberts tjro...@lucent.com
Mike
Tom Roberts wrote:
> Mike wrote:
> > Since coriolis and centrifugal accelerations can be measured, they are
> > real effects.
>
> Sure. Nobody denies that if one uses rotating coordinates such
> accelerations will be observed. But there is no corresponding force
> other than the _fictitious_ ones due to choice of a rotating frame.
Can you ask yourself how in the world a "ficticious" force gives rise
to a real, measurable effect?
Then, do you imply inertial reference frames are priviledged frames? As
Newton did? And real forces can be referenced only wrt these frames?
Any way you try to answer it seems you fall into a trap of some type, a
result of the inconsistencies present in Relativity theory.
>
> In particular, while you can measure these accelerations in your
> rotating frame, you cannot _measure_ the "centrifugal or Coriolis force".
I am amazed by this statement. Do you deny the definition of force as
being equal to the time rate of change of momentum?
Do you further deny that every effect, in any moving reference frame,
must have a proximate cause or you believe in magic causes and
action_at_a_distance?
In summary:
do you think:
1. inertial frames are priviledged?
2. There are real measurable effects without a real cause in their
frame of ref?
3. the definition of force F =dp/dt is wrong?
Please answer these basic questions that arise from your answers so we
can get an idea if we have a common ground for a discussion.
Mike
>
>
> Tom Roberts tjro...@lucent.com
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1145192944.6...@i39g2000cwa.googlegroups.com...
Three in a row:
"Physics 101 re-invented again once more":
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Physics101ter.html
Mike, we love you - keep digging!
Dirk Vdm
Mike
You see martin it is very hard with this idiot around to maintain a
level of conversation. We have started an interesting discussion and he
is intruding with insults. I concede it is my mistake I reply to him. I
never insult gentlemen like you although I may have a tense debate with
them.
Mike
>
> Martin Hogbin.
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1145193711....@j33g2000cwa.googlegroups.com...
It is a miracle to me why people like Martin, Tom, Randy,
and PD still seem to treat an obvious obnoxious little shit
like you as a real person.
But please don't mistake my remarks as insults towards
your person. I mean, really, how can one possible insult a
piece of garbage?
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Physics101.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Physics101bis.html
http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Physics101ter.html
Dirk Vdm
Mike
> universally. A coriolis or centrifugal "force" is completely consistent
> with the notion of force more or less defined by the 2nd law alone, but
> fails completely to satisfy the 3rd law.
It does not, you can always consider the reaction to these forces as
part of an absolute space, as Newton insisted.
>
> Mike is pointing out -- I believe -- the rather narrow perspective of
> the principle of equivalence, which in a way "senses" a force in the
> same way the 2nd law defines it. Mach's principle -- whatever it means
> -- appeals to the 3rd law, asking "what is the agent?"
The appeal you are talking about is not evident from Mach's principle.
Although this can be a narrow interpretation of it, the principle
implies much more than that.
>
> I believe the modern view of force implicitly bears the notion of an
> interaction between objects. Any action that produces an acceleration
> but which cannot be attributed to an interaction is therefore missing
> something essential and is considered a pseudoforce.
This is at last a good point but I can see a problem with it. What are
those actions (rather effects of some actions) that have a pseudoforce
as an agent? You see, if dynamics can be modeled completely as
interactions between objects, then you already preclude reactions that
have pseudo actions.
The only way for your approach to work is to accept axiomatically
pseudo actions. As you realize, this is a problem, a big one. Your
stopped at the problem, but it is there where one must start to tackle
the issue.
Mike
>
> PD
>
> PD
Mike
For Relativists there is. Actually, much of the foundation of the
theory was based on the claim that Newtonian physics needs ficticious
forces to work in non-inertial FoR and thus covariance should be
introduced to eliminate these forces. Of course, you are left with real
effects with no cause in this way.
>
> 2) While the notion of force was fundamental in the early formulation
> of mechanics, currently it is rather secondary.
What is fundamental currently? There is no convincing alternative to
F=dp/dt being a fundamental concept. I am not convinced the Action
Principle is fundamental to Mechanics. Are you?
Mike
Mike
Dirk Van de moortel wrote:
If you are a grown up responsible person you must understand when
legitimate debate ends and harassment starts, an activity that is
punishable by law. You keep harassing me in every post I reply to other
people and this constitutes a repeat offense I am kindly asking you to
terminate immediately or face the consequences.
Mike
Mike
Dirk Van de moortel wrote:
> "Mike" <ele...@yahoo.gr> wrote in message news:1145193711....@j33g2000cwa.googlegroups.com...
> >
> > Martin Hogbin wrote:
> > > "Mike" <ele...@yahoo.gr> wrote in message news:1145126167.9...@j33g2000cwa.googlegroups.com...
> > > >
> > > > Dirk Van de moortel wrote:
> > > > > Yes, just like I predicted, you did it again:
> > > > > http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Physics101bis.html
> > > > > Grrrrreat :-)
> > > > >
> > > > You should seek for professional medical help idiot. You are highly
> > > > paranoid.
> > > >
> > >
> > > Despite your protestations that you are only interested
> > > in discussing physics rather than trading insults you
> > > have failed to reply to my latest post in this thread.
> >
> > You see martin it is very hard with this idiot around to maintain a
> > level of conversation. We have started an interesting discussion and he
> > is intruding with insults. I concede it is my mistake I reply to him. I
> > never insult gentlemen like you although I may have a tense debate with
> > them.
>
> It is a miracle to me why people like Martin, Tom, Randy,
> and PD still seem to treat an obvious obnoxious little shit
> like you as a real person.
I really do not understand where you hate comes from, unless you are a
very sick person.
If you are a grown up responsible person you must understand when
legitimate debate ends and harassment starts, an activity that is
punishable by law. You keep harassing me in every post I reply to other
people and this constitutes a repeat offense I am kindly asking you to
terminate immediately or face the consequences.
Mike
Dirk Van de moortel
"Mike"
aka Eleatis
aka Undeniable
aka Bill Smith
<ele...@yahoo.gr> wrote in message news:1145195644....@z34g2000cwc.googlegroups.com...
Trolls cannot be harassed in the technical sense.
>, an activity that is
> punishable by law. You keep harassing me in every post I reply to other
> people and this constitutes a repeat offense I am kindly asking you to
> terminate immediately or face the consequences.
Try me, troll - I've got plenty to show ;-)
Here's a few to warm up.
29-Jan-2006:
http://groups.google.com/group/sci.physics.relativity/msg/18246d2052b351eb
| You are now "plonked" for good idiot. Anyway, you contribute nothing in
| these ng's other than psychosis and stupidity. I stick with it now and
| you "stock" your merde some place else.
1-Sep-2005:
http://groups.google.com/group/sci.math/msg/1b0256e5a1ad48b0
| Gee, Cantor, Goedel, Peano and the others would be proud of you Dirt.
| Especially if they knew that besides that you also understand the
| square root.
|
| Plonk
|
| Mike
5-Feb-2005:
http://groups.google.com/group/sci.physics/msg/986fdd22eebc33b9
| You got nothing to say, proven fact. You spend your whole misearable
| days maintaining a site where you store your impotence and psychotic
| behavior.
|
| I had enough with you schizo, a killfile will do it and then a clean of
| the Dirt you left around making pooppies all over the place.
|
| Mike
25-Sep-2004:
http://groups.google.com/group/sci.physics/msg/e5cfae098ed3d0a1
| Hi Dirk, Are you still working on a definition of sqrt? You and Alex
| can make a nice team. He thinks there is no time so you got all the
| time in the world to think about sqrt.
|
| Plonk Dirk
|
| Mike
Dirk Van de moortel
"Mike" <ele...@yahoo.gr> wrote in message news:1145195908.9...@t31g2000cwb.googlegroups.com...
>
> Dirk Van de moortel wrote:
> > "Mike" <ele...@yahoo.gr> wrote in message news:1145193711....@j33g2000cwa.googlegroups.com...
> > >
> > > Martin Hogbin wrote:
> > > > "Mike" <ele...@yahoo.gr> wrote in message news:1145126167.9...@j33g2000cwa.googlegroups.com...
> > > > >
> > > > > Dirk Van de moortel wrote:
> > > > > > Yes, just like I predicted, you did it again:
> > > > > > http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Physics101bis.html
> > > > > > Grrrrreat :-)
> > > > > >
> > > > > You should seek for professional medical help idiot. You are highly
> > > > > paranoid.
> > > > >
> > > >
> > > > Despite your protestations that you are only interested
> > > > in discussing physics rather than trading insults you
> > > > have failed to reply to my latest post in this thread.
> > >
> > > You see martin it is very hard with this idiot around to maintain a
> > > level of conversation. We have started an interesting discussion and he
> > > is intruding with insults. I concede it is my mistake I reply to him. I
> > > never insult gentlemen like you although I may have a tense debate with
> > > them.
> >
> > It is a miracle to me why people like Martin, Tom, Randy,
> > and PD still seem to treat an obvious obnoxious little shit
> > like you as a real person.
>
> I really do not understand where you hate comes from, unless you are a
> very sick person.
You don't understand, Mike - I don't hate you.
I LOVE you - really - honestly.
You can't beat Androcles, but're almost the best. Really :-)
Dirk Vdm
Martin Hogbin
"Mike" <ele...@yahoo.gr> wrote in message news:1145193711....@j33g2000cwa.googlegroups.com...
>
> Martin Hogbin wrote:
> > "Mike" <ele...@yahoo.gr> wrote in message news:1145126167.9...@j33g2000cwa.googlegroups.com...
> > >
> > > Dirk Van de moortel wrote:
> > > > Yes, just like I predicted, you did it again:
> > > > http://users.telenet.be/vdmoortel/dirk/Physics/Fumbles/Physics101bis.html
> > > > Grrrrreat :-)
> > > >
> > > You should seek for professional medical help idiot. You are highly
> > > paranoid.
> > >
> >
> > Despite your protestations that you are only interested
> > in discussing physics rather than trading insults you
> > have failed to reply to my latest post in this thread.
>
> You see martin it is very hard with this idiot around to maintain a
> level of conversation.
Not at all. Simply reply to my post.
>We have started an interesting discussion and he
> is intruding with insults.
Dirk has not intruded into my discussions wioth you.
You have simply dropped out.
Martin Hogbin
PD
Mike wrote:
> Tom Roberts wrote:
> > joe_ave...@yahoo.com wrote:
> > > Centrifugal forces can be considered reactions to Centripetal forces.
> >
> > No, they cannot.
> >
> > When swinging a stone by a string, the force on my hand is radially
> > outward at radius zero, so the "centrifugal force" is zero. The force on
> > my hand is quite clearly the tension in the string, not "centrifugal
> > force". The force on the rock is also the tension on the string.
>
> You are wrong.
No, he's quite right. And if you look in a basic physics text, this is
precisely what you'll see.
PD
mme...@cars3.uchicago.edu
>
>mme...@cars3.uchicago.edu wrote:
>> >
>> >mme...@cars3.uchicago.edu wrote:
>> >> >mme...@cars3.uchicago.edu says...
>> >> >
>> >> >>>Suppose I observe a ball's motion and plot its position x versus
>> >> >>>t. I get a straight line, so I conclude (via Newton's laws) that
>> >> >>>there is no force acting on the ball (at least not in the x-direction).
>> >> >>>
>> >> >>>Now, I change coordinates to X = arctan(x) and plot X versus t.
>> >> >>>In the new plot, I *don't* get a straight line, but I get some
>> >> >>>curved line. Is that a "real effect"? Does it indicate the presence
>> >> >>>of a force? No, it's obviously an artifact of my particular
>> >> >>>choice of coordinates.
>> >> >>
>> >> >>We're not talking about changing coordinates,
>> >> >
>> >> >I had some strange kind of ruler that happened to measure X
>> >> >instead of x, then plotting the trajectory of the ball would
>> >> >yield a curved line, rather than a straight line. That doesn't
>> >> >mean that there is any force involved, necessarily, it just
>> >> >means that my coordinate system is non-Cartesian.
>> >> >
>> >> >It's not likely that I'll ever have rulers that happen
>> >> >to measure arctan(x), but it isn't hard to come up with
>> >> >rulers that measure X = x cos(wt) + y sin(wt) and
>> >> >Y = - x sin(wt) + y cos(wt).
>> >> >
>> >> >
>> >> >>velocities and accelerations. That's plain kinematics.
>> >> >
>> >> >and accelerations? I know how to measure those things
>> >> >in a particular *coordinate* system, but how do you
>> >> >measure them in a coordinate-independent way?
>> >>
>> >> You don't use coordinates for measurement, you *assign* coordinates
>> >> based on measurement. What you use for measurement are standard
>> >> yardsticks and standard clocks, or the modern day equivalents.
>> >> >
>> >> >>definition of a force. If you define force as, say, "transfer of
>> >> >>momentum resulting from an interaction between physical entities, an
>> >> >>interaction which can be expressed, in the ultimate account, by an
>> >> >>exchange of bosons between said physical entities", then no, the above
>> >> >>is not a real force. But, I do not recall force being strictly defined
>> >> >>this way.
>> >> >
>> >> >Newton's law, F=ma, then we can ask: what is the definition
>> >> >of "acceleration"?
>> >>
>> >> No, since acceleration is a kinematic quantity, defined geometrically.
>> >> There is nothing in its definition that assumes anything "causing"
>> >> the acceleration. All you rely upon is the ability to measure
>> >> displacements and time intervals.
>> >> >
>> >> >What I would say is that acceleration is a vector (in classical
>> >> >physics, anyway) defined by
>> >> >
>> >> > a = (d/dt) v
>> >> >
>> >> Yes.
>> >>
>> >> >where v is the velocity vector. In terms of components, we
>> >> >can write v = v^i e_i where e_i is the ith basis vector. Then
>> >> >we have
>> >> >
>> >> > a = (d/dt v^i) e_i + v^i (d/dt e_i)
>> >> >
>> >> >The fact that d/dt v^i = 0 does *not* imply that a=0 unless
>> >> >we know that our basis vectors e_i are time-independent.
>> >> >
>> >> Indeed. Now, how do you know that your basis vectors are time
>> >> independent?
>> >>
>> >> >>meomentum than yes, the above is force.
>> >> >
>> >> >momentum? How are you operationally deciding whether momentum
>> >> >is changing are not?
>> >>
>> >> Staying for the moment within Newtonian mechanics of massive bodies
>> >> I'll just take the low-brow p = mv.
>> >> >
>> >> >>since there is no change relative to an inertial frame,
>> >> >
>> >> >does it *mean* to say that momentum is changing? How do
>> >> >you operationally measure momentum?
>> >>
>> >> I measure velocity and use the above.
>> >
>> >And when you do this, you find in certain situations that you are faced
>> >with problem with several possible solutions:
>> >1. Momentum as calculated as you suggest is not always conserved (the
>> >current situation being case in point).
>>
>> Aha. Momentum isn't always conserved. It is only conserved when
>> there is no external force acting.
>
>Right. Now, in an inertial frame it is always possible to enlarge the
>system enough where that external force becomes internal to the system,
>because you've made the interacting pair both part of the system. In a
>rotating frame, it's a little trickier to do this. This is the first
>clue that something is amiss.
>
Also, nobody (to my knowledge at least) claimed that in Newtonian
mechanics there is no difference between inertial and non inertial
frames. Would that have been the case, we wouldn't need the "inertial
frame" concept at all.
>>
>> >2. Momentum is in fact conserved, and we are further to assume that the
>> >only way that momentum is changed is through the presence of a force: F
>> >= dp/dt. We therefore conclude that there is a force present, though
>> >for the life of us we're not sure what the agent of that force is.
>>
>> In the context of the second law, what matters is that we can say what
>> the force is. It *does not* matter what the agent of the force is.
>
>Well, I for one would be hard-pressed to identify the force without
>knowing what the agent is. Unless of course you want to label it
>according to the kind of acceleration it produces, e.g. centrifugal or
>coriolis.
But of course. That's all it is. A term with the dimensions of force
which appears in the equations of motion when they're written relative
to a rotating reference frame and since it appears often we may as
well have a name for it. No mystical connotations whatsoever. I'll
notice in passing that Landau (in his "Mechanics") had no qualms about
referring to these terms as "inertial forces".
>
>>
>> >3. Momentum is in fact conserved; however, we allow that there are
>> >other means by which a momentum contribution can be added to a system
>> >other than a "real" force.
>>
>> We opt for (2).
>> >
>> >It's not immediately apparent which of these should be chosen without
>> >appeal to some independent criteria, such as additional criteria that a
>> >force must satisfy other than F=dp/dt.
>> >
>> This only matters if you insist that there has to be an agent.
>
>Which is what I opt for.
>
Feel free, but that's overloading the meaning of "force". And you're
still left with these terms in the equations and it is still
convenient to have names for them.
mme...@cars3.uchicago.edu
>
>mme...@cars3.uchicago.edu wrote:
>> >
>> >mme...@cars3.uchicago.edu wrote:
>> >> >mme...@cars3.uchicago.edu says...
>> >> >
>> >> >>>We know precisely how they arise, that's how we can derive
>> >> >>>equations for them. If I am driving in a straight line but
>> >> >>>you are in a circle, you will conclude a "force" is making me
>> >> >>>"curve". I can tell you precisely where that illusion is coming
>> >> >>>from and calculate the precise amount.
>> >> >>>
>> >> >>perfectly real effect resulting from a specific choice of a reference
>> >> >>frame.
>> >> >
>> >> >make this concrete:
>> >> >
>> >> >t. I get a straight line, so I conclude (via Newton's laws) that
>> >> >there is no force acting on the ball (at least not in the x-direction).
>> >> >
>> >> >Now, I change coordinates to X = arctan(x) and plot X versus t.
>> >> >In the new plot, I *don't* get a straight line, but I get some
>> >> >curved line. Is that a "real effect"? Does it indicate the presence
>> >> >of a force? No, it's obviously an artifact of my particular
>> >> >choice of coordinates.
>> >>
>> >> measurements of postions, velocities and accelerations. That's plain
>> >> definition of a force. If you define force as, say, "transfer of
>> >> momentum resulting from an interaction between physical entities, an
>> >> interaction which can be expressed, in the ultimate account, by an
>> >> exchange of bosons between said physical entities", then no, the above
>> >> is not a real force. But, I do not recall force being strictly defined
>> >> definition (to the extent that the word "legitimate" applies to
>> >> the other hand, you define force as anything causing change of
>> >> of momentum is only apparent since there is no change relative to an
>> >> valid mesaurement is a measurement performed relative to an inertial
>> >> frame". Again, says who?
>> >>
>> >
>> >Says Newton's 3rd law, which implicitly demands an agent to the force.
>>
>
>I understand the point.
>
>>
>> >objects.
>>
>>
>>
>>
>> In no way does any of this say "only measurements performed relative
>> to inertial frames are valid". There is absolutely no reason for
>> physics to limit itself so. What it does say is "Note what is it that
>> you're measuring". To wit, the first law gives us a default state of
>> motion of an object, the second tells us that the state of motion can
>> be changed by an interaction (more than this, it tells you how will it
>> change) and the third tells us (as you noted) that the interaction *has*
>> to be with other objects (not necessarily a single other object).
>> And all of this is true *as long* as the reference frame used is
>> inertial.
>
>The problem I have with this is that the "note what it is that you are
>measuring" begs the question. If one can *distinguish* between a
>coriolis "force" and a force that is due to some charge-induced field,
>for example, then you are already saying, "Yes, of course, I know that
>this force is not due to another partner" or "I can tell from the
>situation that this is not a force like that one is." In essence, you
>are dividing forces (as acceleration-makers) into two camps but giving
>them the same label. I choose to label the two camps differently, into
>forces (with interaction partners) and pseudoforces (without). I see no
>reason not to do this.
I have no problem with calling them pseudoforces, or "inertial forces"
(the term Landau used) or any other convenient term> And I certainly
agree that they should be named in a way which distinguihes them from
the physical forces, those with an interaction partner. I only have a
problem with trying to pretend that "these forces are a fiction, they
don't really exist". They certainly exist in the sense of appearing
in the equations of motion relative to an accelerated frame and having
observable effects relative to said frames. That's all there is to
it.
>
>>
>> Now, would all of this has been topped by "and if it is not inertial,
>> you cannot say anything", that it would've been of rather limited
>> usefullness. Fortunately, this is not the case. We can transform
>> from one frame to another, including to non-inertial frames, thus we
>> can do physics even in non-inertial frames. Only, there the rules
>> change somewhat. The 1st law is no longer true and same goes for the
>> third. The second, however, can still be applied as long as proper
>> correction terms are included. That's all.
>
>And the crux is that I use Newton's three laws to define a force. You
>are using only the second.
>
In the context of equations of motion, I only use the second since
only the second is relevant there. In other contexts (for example, in
the context of analyzing interactions), I can do differently. Context
matters.
Herman Trivilino
> Mike wrote:
>> Tom Roberts wrote:
>> > joe_ave...@yahoo.com wrote:
>> > > Centrifugal forces can be considered reactions to Centripetal forces.
>> >
>> > No, they cannot.
>> >
>> > When swinging a stone by a string, the force on my hand is radially
>> > outward at radius zero, so the "centrifugal force" is zero. The force
>> > on
>> > my hand is quite clearly the tension in the string, not "centrifugal
>> > force". The force on the rock is also the tension on the string.
>>
>> You are wrong.
>
> No, he's quite right. And if you look in a basic physics text, this is
> precisely what you'll see.
Take a look at the connection between the string and the rock. The string
exerts a centripetal force on the rock, and the rock exerts a centrifugal
force on the string. They form a Third Law pair.
The confusion arises when a Freshman claims there is a centrifugal force on
the rock. You have to wait until you're a Junior before you can do that!
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Mike
Herman Trivilino wrote:
> "PD" <TheDrap...@gmail.com> wrote ...
>
> > Mike wrote:
> >> Tom Roberts wrote:
> >> > joe_ave...@yahoo.com wrote:
> >> > > Centrifugal forces can be considered reactions to Centripetal forces.
> >> >
> >> > No, they cannot.
> >> >
> >> > When swinging a stone by a string, the force on my hand is radially
> >> > outward at radius zero, so the "centrifugal force" is zero. The force
> >> > on
> >> > my hand is quite clearly the tension in the string, not "centrifugal
> >> > force". The force on the rock is also the tension on the string.
> >>
> >> You are wrong.
> >
> > No, he's quite right. And if you look in a basic physics text, this is
> > precisely what you'll see.
>
> Take a look at the connection between the string and the rock. The string
> exerts a centripetal force on the rock, and the rock exerts a centrifugal
> force on the string. They form a Third Law pair.
Precisely. For whatever action that is trying to move something towards
a center (centripetal) there is an equal an opposite reaction that
tries to take it away from that center (centrifugal). Plain and
beautiful newtonian mechanics.
>
> The confusion arises when a Freshman claims there is a centrifugal force on
> the rock. You have to wait until you're a Junior before you can do that!
At least. freshman have the bad habbit of jumping around reference
frames while solving a problem.
Mike