| Topic: |
Science > Physics |
| User: |
"Mike" |
| Date: |
13 Apr 2006 02:30:46 PM |
| Object: |
Another attempt by the cranks to fool people |
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.
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
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| User: "Dirk Van de moortel" |
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| Title: Re: Another attempt by the cranks to fool people |
15 Apr 2006 01:36:38 PM |
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"Mike" <eleatis@yahoo.gr> wrote in message news:1145126037.415734.23360@e56g2000cwe.googlegroups.com...
PD wrote:
Mike wrote:
PD wrote:
joe_avery_2005@yahoo.com wrote:
PD wrote:
Mike wrote:
Daryl McCullough wrote:
mmeron@cars3.uchicago.edu says...
stevendaryl3016@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?
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.
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
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.
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.
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.
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.
Even when they are "hidding behing their own ratten fingure"?
Dirk Vdm
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| User: "PD" |
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| Title: Re: Another attempt by the cranks to fool people |
15 Apr 2006 08:45:30 AM |
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wrote:
In article <e1p2g10vpf@drn.newsguy.com>, (Daryl McCullough) writes:
says...
"Randy Poe" <poespam-trap@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.
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?
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
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| User: "" |
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| Title: Re: Another attempt by the cranks to fool people |
15 Apr 2006 10:55:16 AM |
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PD wrote:
mmeron@cars3.uchicago.edu wrote:
In article <e1p2g10vpf@drn.newsguy.com>, (Daryl McCullough) writes:
mmeron@cars3.uchicago.edu says...
"Randy Poe" <poespam-trap@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.
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?
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.
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
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| User: "Mike" |
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| Title: Re: Another attempt by the cranks to fool people |
16 Apr 2006 08:38:38 AM |
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PD wrote:
mmeron@cars3.uchicago.edu wrote:
In article <e1p2g10vpf@drn.newsguy.com>, (Daryl McCullough) writes:
mmeron@cars3.uchicago.edu says...
"Randy Poe" <poespam-trap@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.
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?
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.
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
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| User: "" |
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| Title: Re: Another attempt by the cranks to fool people |
15 Apr 2006 01:42:48 PM |
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In article <1145108730.755242.293490@t31g2000cwb.googlegroups.com>, "PD" <TheDraperFamily@gmail.com> writes:
mmeron@cars3.uchicago.edu wrote:
In article <e1p2g10vpf@drn.newsguy.com>, (Daryl McCullough) writes:
mmeron@cars3.uchicago.edu says...
"Randy Poe" <poespam-trap@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.
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?
Says Newton's 3rd law, which implicitly demands an agent to the force.
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,
meron@cars.uchicago.edu | chances are he is doing just the same"
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| User: "PD" |
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| Title: Re: Another attempt by the cranks to fool people |
16 Apr 2006 05:28:54 AM |
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wrote:
In article <1145108730.755242.293490@t31g2000cwb.googlegroups.com>, "PD" <TheDraperFamily@gmail.com> writes:
wrote:
In article <e1p2g10vpf@drn.newsguy.com>, (Daryl McCullough) writes:
says...
"Randy Poe" <poespam-trap@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.
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?
Says Newton's 3rd law, which implicitly demands an agent to the force.
Yes, *in an inertial frame*.
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.
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,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "Mike" |
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| Title: Re: Another attempt by the cranks to fool people |
16 Apr 2006 05:01:34 PM |
|
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PD wrote:
mmeron@cars3.uchicago.edu wrote:
In article <1145108730.755242.293490@t31g2000cwb.googlegroups.com>, "PD" <TheDraperFamily@gmail.com> writes:
mmeron@cars3.uchicago.edu wrote:
In article <e1p2g10vpf@drn.newsguy.com>, (Daryl McCullough) writes:
mmeron@cars3.uchicago.edu says...
"Randy Poe" <poespam-trap@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.
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?
Says Newton's 3rd law, which implicitly demands an agent to the force.
Yes, *in an inertial frame*.
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.
Force is defined in the second law very clearly. The first law defines
the reference frames in which the second law applies. The third law is
a direct result of the law of conservation of momentum. Actually, only
the second law is necessary and the other two it is shown in several
textbooks how they are easily derived from it and conservation laws.
Remember that the definition of force comes before the laws are stated.
You do not use the laws that use the definition of force to define it,
that is a petittio principii.
But even if you were required to use all three laws to define a ''real"
force, it is not necessary for a force to be ficticious if you fail to
observe its reaction. You got no justification for such claims.
Mike
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,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "Herman Trivilino" |
|
| Title: Re: Another attempt by the cranks to fool people |
17 Apr 2006 07:23:19 PM |
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"Mike" <eleatis@yahoo.gr> wrote ...
Force is defined in the second law very clearly. The first law defines
the reference frames in which the second law applies. The third law is
a direct result of the law of conservation of momentum. Actually, only
the second law is necessary and the other two it is shown in several
textbooks how they are easily derived from it and conservation laws.
The First Law asserts that inertial reference frames are equivalent. This
is not implied by the Second Law. It's a common textbook error to assert
that the First Law is a consequence of the Second. It's been discussed in
the literature.
The Third Law can be used to derive Conservation of Momentum. But, as you
say, it can be done the other way around. In fact, it it is better to do
so, because in modern physics, where the field approach is used, the Third
Law doesn't hold. In other words, Conservation of Momentum implies the
Third Law, but only for contact forces.
----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----
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| User: "" |
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| Title: Re: Another attempt by the cranks to fool people |
16 Apr 2006 02:10:39 PM |
|
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In article <1145183334.838492.11500@i40g2000cwc.googlegroups.com>, "PD" <TheDraperFamily@gmail.com> writes:
mmeron@cars3.uchicago.edu wrote:
In article <1145108730.755242.293490@t31g2000cwb.googlegroups.com>, "PD" <TheDraperFamily@gmail.com> writes:
mmeron@cars3.uchicago.edu wrote:
In article <e1p2g10vpf@drn.newsguy.com>, (Daryl McCullough) writes:
mmeron@cars3.uchicago.edu says...
"Randy Poe" <poespam-trap@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.
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?
Says Newton's 3rd law, which implicitly demands an agent to the force.
Yes, *in an inertial frame*.
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.
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.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
|
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| User: "" |
|
| Title: Re: Another attempt by the cranks to fool people |
14 Apr 2006 07:48:47 PM |
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|
In article <e1pa8601n34@drn.newsguy.com>, (Daryl McCullough) writes:
mmeron@cars3.uchicago.edu says...
(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?
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.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
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| User: "Daryl McCullough" |
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| Title: Re: Another attempt by the cranks to fool people |
14 Apr 2006 09:39:14 PM |
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says...
stevendaryl3016@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?
--
Daryl McCullough
Ithaca, NY
--
NewsGuy.Com 30Gb $9.95 Carry Forward and On Demand Bandwidth
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| User: "PD" |
|
| Title: Re: Another attempt by the cranks to fool people |
15 Apr 2006 09:00:23 AM |
|
|
wrote:
In article <e1pa8601n34@drn.newsguy.com>, (Daryl McCullough) writes:
says...
(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?
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.
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
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.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
|
|
|
|
| User: "" |
|
| Title: Re: Another attempt by the cranks to fool people |
15 Apr 2006 01:49:11 PM |
|
|
In article <1145109623.678677.28390@j33g2000cwa.googlegroups.com>, "PD" <TheDraperFamily@gmail.com> writes:
mmeron@cars3.uchicago.edu wrote:
In article <e1pa8601n34@drn.newsguy.com>, (Daryl McCullough) writes:
mmeron@cars3.uchicago.edu says...
(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?
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.
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.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
|
|
|
| User: "PD" |
|
| Title: Re: Another attempt by the cranks to fool people |
16 Apr 2006 05:19:41 AM |
|
|
wrote:
In article <1145109623.678677.28390@j33g2000cwa.googlegroups.com>, "PD" <TheDraperFamily@gmail.com> writes:
wrote:
In article <e1pa8601n34@drn.newsguy.com>, (Daryl McCullough) writes:
says...
(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?
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.
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.
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.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
.
|
|
|
|
| User: "" |
|
| Title: Re: Another attempt by the cranks to fool people |
16 Apr 2006 01:56:55 PM |
|
|
In article <1145182781.325999.231540@i39g2000cwa.googlegroups.com>, "PD" <TheDraperFamily@gmail.com> writes:
mmeron@cars3.uchicago.edu wrote:
In article <1145109623.678677.28390@j33g2000cwa.googlegroups.com>, "PD" <TheDraperFamily@gmail.com> writes:
mmeron@cars3.uchicago.edu wrote:
In article <e1pa8601n34@drn.newsguy.com>, (Daryl McCullough) writes:
mmeron@cars3.uchicago.edu says...
(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?
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.
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.
Indeed. The frame is accelerating. Nobody claimed that it isn't.
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
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