Re: How can I tell if F is a string or if it is a number?
- From: Pioneer1 <1pioneer1@xxxxxxxxx>
- Date: Thu, 29 May 2008 09:58:20 -0700 (PDT)
On May 19, 10:47 am, tc...@xxxxxxxxxxxxx wrote:
First I want to thank you for the positive encouragement at the end of
your post and for not dismissing this research without seeing
computations with Kepler's rule. I agree that computations must work.
But even though Kepler's rule is a simple proportion (R^3/T^2)
relating radius and period of the orbit, "Newton's equations of
motion" is not that simple. R^3/T^2 includes only two terms and it is
itself a rule. Newton's equations of motion is said to be tied to
"Newton's laws." I need to understand how Newton's laws exist in a
formula used for measurement of orbits.
Below are my replies:
It *does* enter the formulas, in Newton's formulation of the situation.
Maybe I am not using the word formula in its proper sense. Please
correct if there is another word for what I am trying to say. I
checked "formula" in Wikipedia and an example of a formula would be
computation of volume: V = 4/3 pi R^3. This is the type of formula I
have in mind. If someone decides to derive volume from Newton's laws
and derives it from force, I would still say that volume is
proportional to R^3 and volume is independent of force. In the orbital
case, the final formula is R^3/T^2 and this does not contain a term
for force because we eliminated it to obtain this formula. The orbit
defined by R^3/T^2 does not know that we wrote force terms. So I say
that R^3/T^2 is independent of force. I believe that you would agree
that volume is independent of force. Why is it that R^3/T^2 is
different than V = 4/3 pi R^3? They are both independent of force.
Should I be using another word than "formula?"
Yes, you *are* reasoning from just an example, namely the simple case of a
two-body problem.
But perturbations are additive. As Newton did, I can compute
perturbations between three bodies by computing perturbations between
each pair and add. If force does not enter two-body problem what is
the justification that it will enter three-body problem?
As I see it, the observational rule is for orbits, not for bodies. The
orbit is described by R and T, both are properties of the orbit, not
of a body. This did not fit Newton's beliefs so he defined orbits in
terms of mass (R^3/T^2) and forces (R/T^2) and (1/R^2) but because the
rule is for the orbit and not for force and mass both Newtonian terms
cancel.
Kepler's rules are good enough for such a simple case.
But this is about the absence of force in Newton's equations of
motion. Not about the precision of Kepler's rule. Without force and
mass Newton's equations of motion reduce to Kepler's rule.
There is no Kepler law for three bodies.
I believe that the rule R^3/T^2 is valid for the entire solar system,
not only for pairs of bodies. Adding computational sugar to R^3/T^2
does not make R^3/T^2 Newtonian. It just makes it easier to compute
with. The rule is still the same. For instance, to apply Kepler's rule
to a three-body problem I look at the perturbations in the orbits of
Saturn and Jupiter due to each other and the Sun. This is Newton's
proposition 13 in Book III. Newton computes R/T^2 = 1/R^2 for Sun-
Saturn, Jupiter-Saturn and Jupiter-Sun and gets 16, 81 and 156,609. I
am sure that you could compute the same with equational form of R^3 =
T^2 by using textbook methods. You will get the same results, except
that Newton's Sun-Saturn distance is too low (9) compared to modern
value (10.4). Newton uses Earth-Venus distance as his unit and that
complicates everything. We use AU and kilometers and seconds. Modern
methods too use Kepler's rule, there is no other rule. I would
appreciate if you can tell me what rule textbook version is using
different than R^3/T^2.
How do you propose
to account for three-body motion with appealing to the concept of force?
By applying Kepler's rule the way Newton has done in Principia. Force
terms cancel because we must eliminate them to obtain R^3/T^2. We
don't need the force terms if we are not using them in calculations. I
assume that your "with" was a typo and you meant "without" but to me
it makes perfect sense: "How do you propose to account for orbital
motion by using concept of force?" You cannot because you must
eliminate force terms.
And there is no other rule that Newton discovered.
Not true.
If it is not true that Newton did not discover a new rule, what is the
new rule that Newton discovered? Force is not a rule, it is a cause,
and because it is a cause, it does not enter formulas used in
measurements.
I think it would help your understanding greatly if you made a
serious attempt to analyze the motion of three bodies of roughly equal size.
You will then see that Kepler's laws don't work.
I am assuming simple circular orbits, so, all three of "Kepler's laws"
do not apply in this case. When I say "Kepler's rule" I mean the
proportionality R^3 = T^2, or Kepler's third law. This is the rule
Newton used by writing it as R/T^2 = 1/R^2.
They work in the solar
system only because the sun is so much more massive than any other body, so
you can approximate the situation by assuming that the sun is fixed.
The way I understand it, this is not about approximation. When we
eliminate force, force is gone, it is not approximated. Perturbations
in Saturn's orbit were observed in Newton's time and as I showed above
Newton computed the perturbations from Kepler's rule. So it is not
true that Kepler's rule works in the solar system "only because the
sun is so much more massive than any other body." The effect is
observable and it is computed by using Kepler's rule. This makes sense
to me, because to compute orbits we only need to know R and T. All
other names Newton invented for combinations of R and T can be
ignored. Mass is R^3/T^2. You may include it or not.
Newton's laws, however, work just fine.
"Newton's equations of motion" do not include "Newton's laws."
Calculations involve equations of motions, not laws. And it has been
proven that Newton did not solve the three-body problem. He faked it.
"Newton never managed to get the numbers to come out right, and
finally had to resort to the ancient Greek device of the epicycle to
complete his lunar theory."(From Dana Densmore, Newton's Principia:
The central argument. p. 509) This is my point. To this day three-body
problem is solved by geometric or trigonometric methods and the result
is attributed to Newton's laws. The three-body problem itself is a
problem created by the assumption of force. Without force there is no
three-body problem.
Your comments about "Keplerian proportion with units" are simply false.
Can you please give more detail on this? Why do you think a = GM/R^2
is a different rule than R/T^2 = 1/R^2? In other words why do you
think if I compute the period T of an orbit from its given radius R, I
would get two different results if I use the equational version and
proportional version? I believe that we would get the same results.
However, suppose I correct that error for you, and rephrase what you are
saying in a way that is not obviously false. There is a distinction in
the philosophy of science between "realism" and "anti-realism." . . .
Thanks for this example, it was thought provoking but I do not agree
that Newtonian forces
are merely a convenient computational device that we use to
predict our observations.
I object that forces "are merely a convenient computational device." I
believe that force terms are not used to compute orbits, they are
eliminated. Force is not computational, it is doctrinal. Observing
this fact I say that orbits are independent of force. The orbit sees
only R and T. We cannot have a formula which contains F, T and R and
use it to compute orbits. We can write F and then eliminate it but the
orbit does not know that.
It can be fruitful, as a thought experiment, to adopt realist and
anti-realist attitudes towards various concepts in physics, to see where that
leads you. Adopting an anti-realist attitude towards Newtonian force might
lead you to think of the principle of equivalence between gravitation and
acceleration and therefore to general relativity.
As I mentioned above, thanks for this positive encouragement and I
agree that such questioning of physics concepts at least will help
understanding them better. But I am not sure that my attitude is
reality or anti-reality. I don't know what it is called, maybe it is
rationality and an absolute belief in the authority of measurement and
observation and total denial of doctrine. If the formula we use to
measure a quantity does not include a term, I believe that the problem
is independent of that term. I believe this to be true absolutely, not
case by case. So if I use the formula Volume :: R^3 then I believe
that volume is independent of force, no matter what Newton says. If
Density :: 1/T^2 then I believe that density is independent of force,
no matter what Newtonian doctrine says. If R^3 and 1/T^2 are
independent of force then R^3 :: T^2 is independent of force too. Do
you believe in this principle? If you do, what makes you think that
there is force in R^3 :: T^2?
I also like to mention that the lack of force in orbital formulas is
not a thought experiment, it is an observation. That orbital formulas
must have force is doctrine. I go with observations.
I am not discouraging you from exploring these ideas.
However, regardless of your stance on realism versus anti-realism, you still
have to get your calculations right.
Yes, thanks again. I agree but to get the calculations right and
compare Kepler's rule with Newton's equations of motion I will have to
state clear definitions of both. I don't believe that
(R/T^2 = 1/R^2) + (units and constants) + (coordinates) + (calculus
notation)
turns (R/T^2 = 1/R^2) magically into a Newtonian rule.
.
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