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From: tchow on 23 Apr 2008 13:34
In article <2beb583b-ae0e-4ebf-9f16-c6de8642867e(a)h1g2000prh.googlegroups.com>,
Pioneer1 <1pioneer1(a)gmail.com> wrote:
>On Apr 15, 12:08�pm, tc...(a)lsa.umich.edu wrote:
>
>> No. �F is not an abbreviation, not even a meaningless one.
>
>Okay. What is it? Does it have a meaning? If it does, its meaning is
>not coming from observations. Then, where does it come from?

F stands for the numerical value of the force, where "force" is a concept in
our physical theory of the world.

>> No. �ma is not an abbreviation for mR/T^2. �F is not an abbreviation
>> for ma.
>
>So why does it cancel? It's not a placeholder. It's not a temporary
>variable. It's not a label. It's supposed to be a "physical" quantity
>but it cancels out of formulas.

Canceling in formulas happens whenever you have more than one equation for an
unknown. Suppose you have two coins of unknown weight. By doing some
experiments you find that the difference in weight is 4 and the sum of the
weights is 10. So I can let X be the weight of one coin and let Y be the
weight of the other coin. I get the equations

X = Y + 4
X = 10 - Y

I can cancel out X to obtain the equation Y + 4 = 10 - Y, whence Y = 3
and X = 7. Does the fact that I have canceled out X mean that X is an
"abbreviation" for "Y + 4"? No. X is the weight of the coin. X = Y + 4
is not a definition, but a fact that you obtained from experiment.

Elimination of variables is just a mathematical technique for solving
equations. The fact that you can eliminate a variable doesn't mean that
the variable doesn't mean anything or is just an abbreviation; it just
means that you can make progress in *solving* the equations.

Perhaps you understand the coin-weighing example but still have trouble
making the leap to F = ma. So let's take another coin-weighing example.
Suppose I have a balance beam that has unequal length arms. One arm has
length A and the other arm has length B. If I put a weight X on one side
and a weight Y on the other side and they balance, then I have the "physical
law" AX = BY. Here we have an equation involving multiple variables. Does
this mean that "A" is an abbreviation for "BY/X"? No. A is the length of
the first arm. AX = BY is a generalization from many empirical observations.
It is not a mathematical identity, but a physical law.

What if I find some other physical laws that also involve A, X, B, and Y,
and that allow me to solve for these quantities by eliminating variables?
Would that mean that A suddenly becomes an abbreviation for "BY/X"? No.
A is the length of the first arm. Even if you manage to find some other
equation for A that allows you to eliminate A and solve for some other
unknowns, A continues to denote the length of the first arm.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
From: Patricia Shanahan on 24 Apr 2008 09:46
Pioneer1 wrote:
....
> No. I neither disagree or agree with Newtonian mechanics. To me,
> Newtonian mechanics is a meaningless generalization. I'm questioning
> one expression: F=ma. I cannot understand the types of the symbols in
> this expression.
....

I think you are making a mistake in even trying to understand one
equation in isolation from the theory of which it is part. The "F" in
"F=ma" carries meaning from static mechanics.

Patricia
From: Pioneer1 on 6 May 2008 07:50
On Apr 30, 8:37 pm, tc...(a)lsa.umich.edu wrote:

> This is perhaps the crucial point.  You could, perhaps, develop a theory
> of gravitation and mechanics without any reference to forces.  In a sense,
> that is what general relativity theory does.  

I understand that general relativity excludes forces in its realm. But
that doesn't help in this case. As you mentioned above Newton's force
is an experimental quantity routinely measured in Cavendish
experiments even by students. Physics admits the action-at-a-distance
Newtonian force as an experimental quantity and also general
relativity's no force explanation.

General relativity specifies a different mechanism for gravity but
that does not remove Newtonian action-at-a-distance force from
physics. But my point is that, the force term is eliminated and
therefore the orbit is independent of force. Since force is
eliminated, the fact that it is action-at-a-distance becomes
irrelevant.

> Forces are, as you say, not directly observed.  However, the way Newtonian
> theory works is that physicists *postulate* that there is something called a
> "force" that obeys certain laws.

Okay, I understand that Newton postulated something called "force"
that obeys Newton's laws and physicists start their derivation from
the assumption that something called force is needed to describe
orbits. But it is not true that orbital predictions are made by using
equations that contain force. The force terms are eliminated during
the derivation process:

1. Force is assumed
2. Force terms are eliminated
3. Orbits are computed with formulas that do not contain force terms.

From the above list I conclude that orbits are independent of force.
How do physicists conclude that orbits are described by the assumed
force after they eliminate the force from equations?

From the above I also conclude that force is not a quantity that is
observed indirectly, but it is a quantity that is eliminated, in other
words, it does not exist in orbital motion. To assume that force
remains in the equations after it is eliminated appears to me to be an
unjustified assumption that does not have a physical basis.
For instance, in the Newtonian description of orbits we assume that
force deflects rectilinear motion into curved orbits. If force is not
represented in the equations how does it do that? This seems like
magic to me. But first thing we learn in physics is that magic and
similar non-physical things are not part of physics.

> I think that once you accept the idea that one can hypothesize the existence
> of things that are not directly observed, in order to help develop a
> conceptually satisfactory model of how the world works, you will be more
> comfortable with "force."

In the case of orbits, there is a conceptually satisfactory model
describing orbits well without an assumption of force. Why is there a
need to assume a hidden cause not visible in the formulas? To me the
assumption of force adds nothing to the model and it is eliminated and
therefore is not needed.

> I don't understand why you think that "F_ma is proportional to R."
> F_ma is proportional to *acceleration*, which is a very different thing
> from the *distance between the objects*.

I expand acceleration a by writing it as a = R/T2 or F = m R/T2. Then
for a given period, force F is proportional to R. And this is true by
observation because in a sling, as Newton mentioned in Definition 5
while defining this type of force, for a given angle the force on the
string increases with R. This follows from radian motion theta = s/r
or for a given angle, s is proportional to r.

Newton then generalized this sling-type rotation to all orbits by
saying that "the same applies to all bodies that are made to move in
orbits." But planetary orbits do not obey this rule, planets do not
rotate by radian rule as a sling does, they revolve by Kepler's rule.
In other words, in planetary motion, for a given angle radius is not
respected as in radian motion, and radius and arc are not directly
proportional. Increasing the radius r decreases the arc s according to
Kepler's rule.

When physicists equate F_ma and F_GM they are saying that orbits
rotate the way slings rotate. This is not true. The sling rotation and
planetary revolutions are different. I realize that this was very
ingenious on Newton's part. He created this confusion to let his force
and his dynamical acceleration concept to be the rule also for
Keplerian orbits. But orbits are still Keplerian and not Newtonian and
dynamical. We must eliminate Newton's force to describe orbits with
Kepler's rule. Planetary orbits and a stone rotating on a sling are
not the same type of motion.

> You can, in fact, usually tell what y was, by substituting x = a+b back
> into one of your original equations.

Yes, that was my point. If you know the original equation you can
substitute y. In this case you do not know the original equations.
Similarly, the orbit does not know that physicists write F terms in
intermediary equations and then cancel them. For the orbit, there
exists no force because it was eliminated.

> Part of your problem, it seems to me, is that you don't see how to calculate
> the value of the force.  

This is true. I don't see how to compute the value of force by using
the same formula we used to compute the orbit because the formula we
used to compute the orbit does not include a term for force.

> After it's eliminated, and we solve for the motion,
> we never seem to care about it any more.  So wasn't it redundant in the first
> place?

Yes. This is how I see the problem. But not only we don't care about
force after we eliminate it, the orbit does not care about it either.
The orbit does not know that physicists wrote equations with force
terms in them.

> For the particular problems you're currently trying to solve, it may appear
> that the force term is redundant.  

But do you think it is redundant in the orbital motion? If it is
redundant, would it be correct to say that "orbits are independent of
force" because we eliminated the force term?

> However, Newton's laws apply much more
> generally, to other physical situations . . .

I understand this. But instead of considering Newton's laws as a
general framework, I'm just looking at the specific assumption of
force in the computation of orbital motion. The assumption is not
justified, as far as I understand it, because we assume it but then we
eliminate it, and compute orbits without it. If it's not used why
assume it?

> . . . and in other situations, you will
> be able to calculate the force.

I would like to make sure that this is the same force that enters
orbital calculations but no force term enter orbital calculations.

> For example, you can calculate the force
> if you have some way of determining the mass of the object in question
> as well as its acceleration.  

But there is no way to compute masses of celestial objects because
they don't appear in formulas alone. As far as I understand, I could
only compute force, a quantity that does not enter orbital
computations, from another quantity, the mass m, that does not enter
orbital computations. Given that neither is orbital, this computation
will not be relevant to orbit computations.

> Presumably you don't have any problems with
> measuring acceleration, which boils down to measuring distance and time?

I believe that in the case of orbits, the time in the a = R/T2 refers
to the period of the orbit. And I'm having, I think, the same problem
as force regarding acceleration as well. As far as I understand, in
physics, it is considered that a = R/T2 refers to two different
quantities, acceleration and R/T2. I believe that since only R and T
are measured, the live physical quantity is R/T2, not a different
quantity called acceleration. I would say that I am measuring the
ratio R/T2 and calling it acceleration.

> Do you have any problem with measuring mass, by using a balance pan and a
> "standard mass" like the one they keep locked up in Paris?

No problem. But this mass is not an orbital quantity because it is
eliminated. And in this case physicists say that the orbit is
independent of m because it is eliminated.

> If you can
> measure mass and measure acceleration then you can calculate the value
> of the force by multiplying the mass by the acceleration.

Okay. Say F_ma = X Newtons. This quantity is not a quantity that
enters in orbital computations because F_ma is eliminated.

There is a reason why F is eliminated from orbital computations:
Orbits are not rotational motion like a sling where R/T2 alone defines
the orbit. In definition 5 Newton called sling motion an "orbit" and
then generalized to all orbits. This way Newton confused the issue by
making acceleration a different property than the orbital quantity R/
T2. The fact that Newton calls R/T2 "acceleration" does not change the
fact that the orbital arc traveled in unit time is proportional to the
radius. Newton wants to project this sling "acceleration" to orbital
motion in order to say that orbits are formed by this force which
bends rectilinear inertial motion into curved orbits. But this does
not work.

The sling-type rotation is described by radian motion, theta =s/r or s
= r theta. Dividing by time gives velocity: s 1/t = r theta/t.
Dividing again by time: s 1/tt = r theta/tt is called acceleration. In
other words, however many times you divide by time will never change
the fact that the arc s is proportional to r for a given angle theta.
And the rotational motion will never become orbital motion. The
acceleration is true for sling-type rotational motion but not true for
revolutionary motion that obeys Kepler's rule.

The R/T2 hidden in acceleration a in F = ma is really half of Kepler's
rule, R/T2=1/R2. R/T2 alone works only for rotational motion. To get
orbital Keplerian motion we must eliminate Newton's force and obtain
Kepler's rule R/T2=1/R2. And this is no longer rotational radian
motion, now, R^1.5 is proportional to the period T. Or rotational
motion is R proportional to s and orbital motion is R^1.5 is
proportional to 1/s. Or in the planetary motion, the arc traveled per
unit time is no longer directly proportional to R. In orbital motion
radius is not respected. If you increase R, the arc s does not
increase directly as R, it decreases according to Kepler's rule. This
is the point of using Kepler's rule to describe orbits, instead of
radian motion.

But Newton confused the two type of motion, rotational and orbital,
and physics tradition still follows Newton and writes Newton's force
terms and then cancels them. Newton's projection from radian rotation
to orbital revolution is wrong. We don't need to write and cancel
Newton's force just because he did so. This is how I understand it.

Thanks again for all your help.
From: Patricia Shanahan on 6 May 2008 09:30
Pioneer1 wrote:
....
> I understand this. But instead of considering Newton's laws as a
> general framework, I'm just looking at the specific assumption of
> force in the computation of orbital motion. The assumption is not
> justified, as far as I understand it, because we assume it but then we
> eliminate it, and compute orbits without it. If it's not used why
> assume it?
....

That misses the key point of Newtonian physics, and indeed of much of
science. The main objective of the game is to find simple theories that
explain and predict a lot of observations. By that standard, Newtonian
physics is one of the most spectacularly successful theories ever developed.

You might be able to build a reasonably simple forceless theory for
orbital motion, but it would probably be at least as complex as
Newtonian physics and be much less useful for even closely related
questions, such as deciding how heavy components of a satellite need to
be to survive accelerations during launch.

Patricia
From: tchow on 6 May 2008 12:08
In article <0995457c-fbad-4159-ba55-bcd21eed0de7(a)u12g2000prd.googlegroups.com>,
Pioneer1 <1pioneer1(a)gmail.com> wrote:
>But it is not true that orbital predictions are made by using
>equations that contain force. The force terms are eliminated during
>the derivation process:

You're contradicting yourself. Something cannot be "eliminated" unless it is
there in the first place. The calculation *does* use equations that contain
force, and then eliminates them.

In a sense, you are right that if all you care about are orbits, then you get
by with Kepler's laws. But as Patricia Shanahan pointed out, physicists seek
to find laws that govern *all* physical phenomena. It's true that you can
view the world as a disjointed set of isolated, unrelated problems. Today
I'm interested in orbits. Tomorrow I may be interested in airplane design.
The next day I may be interested in semiconductors. I could treat each of
these problems separately, seeking to find the minimum number of assumptions
required to explain each situation, and ignoring the fact that each time I am
inventing a new set of assumptions that has nothing to do with the assumptions
I used yesterday for a different problem.

Alternatively, I could seek a few very general principles that apply in *all*
situations. Most scientists find this approach much more satisfying than the
piecemeal, ad hoc approach above.

Newton's law of gravitation is called the *universal* law of gravitation for
this reason. In a particular problem, say the calculation of a planetary
orbit, you might be able to get away without assuming forces. But to
conclude from that one example that forces are *in general* an unnecessary
theoretical device is to take a very narrow view of science. Before you
discard the notion of force, ask yourself if you could calculate the dynamics
of two colliding galaxies without assuming a theory of gravitation. Kepler's
laws won't be good enough any more. If a simple theory of gravitation
explains both galaxy collision and orbital motion, then it makes sense to
claim that gravity is at work in orbits, even if for some particularly simple
problems you don't need the general theory.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
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