Re: Can energy conservation be derived from Newton's motion laws only?
From: Igor Khavkine (igor.kh_at_gmail.com)
Date: 01/06/05
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Date: Thu, 6 Jan 2005 22:10:11 +0000 (UTC)
Strong_Field wrote:
> "Newton's laws" and "Energy conservation" appears to be two different names
> for the same thing. In any derivation, there is the first statement X, and
> there is the last stament Y. If your derivation starts from X and ends with
> the expression Y, then the expression Y must be hidden in the expression X.
> If Y is not hidden in X and you could find Y starting from X you would be
> violating conservation laws. You would be creating something from nothing.
> Equations themselves obey the rules of thermodynamics. An equation is
> nothing more than a definiton of conservation rules. Otherwise you cannot
> express the laws of thermodynamics with physics equations. So, in my
> opinion, what physicists call a "derivation" is nothing more than an
> algebraic simplification process. To claim otherwise would be against the
> laws of physics. Corrections, comments welcome.
The above paragraph makes very little sense. From what I understand, you
are trying to apply something very vague to something very precise. It
is unlikely that you can apply thermodynamics to mathematical logic
unless propositional statements suddenly turn into an ideal gas.
And, as I've already mentioned, one does not discriminate against
derivations or proofs, whether one is a physicist, mathematician,
chemist, psychologist, bus driver, etc. One only discriminates against
assumptions.
> To a disinterested observer who does not know the conventions and
traditions
> of physics but is familiar with mathematical notation, what you call
"Newton
> 's second law" is simply a definition. "Newton's second
law" makes the two
> symbols "F" and "a" synonyms. There is no reason to write the
superluous
> symbol "m" which will cancel at the next step. (Or if you wish we
can set it
> to unity.)
>
> F =3D=3D a is in itself nothing more than saying that in your equations
you may
> write "F" or if you think it looks better you may write "a"
instead. There
> is no physics content in F =3D=3D a.
This is not true. Both F and a have independent meaning and can be
measured
independently. The latter with a ruler and a stop watch and the former
with a standard, like a spring of a certain stiffness. Equating them
gives
a non-trivial statement about how interactions between objects affect
their
motion. Also, the mass does not cancel from the equation unless we have
only gravitational forces. But since it's a dimensionful quantity, you
could
set its numerical value to be 1 in the appropriate units.
> > > We can write the first derivative in delta notation as the
increment in
> > > distance x divided by the increment in time t: delta x / delta t.
> >
> > Just keep in mind that this is only an approximation. The exact
result
> > is recovered only when delta t goes to zero.
>
> What do you mean by exact? To me the distance the moon moves in delta
t =3D
> 1/1000 seconds is an exact observation. Delta t is infinitely distant
from
> 0, but it is exact. What I mean is that no new insight is gained by
> considering the case t-->0. In practice you choose the smallest unit
you can
> measure.
You must keep in mind that Newtonian mechanics is a simplified model of
the real world. In this model time, space, and other quantities are
continuous and even smooth. That is why one can speak of derivatives
and
how they can be evaluated exactly. There is no garantee that numbers
from
a theoretical calculation will predict exactly numbers obtained from
experiment, only that they'll be close.
> > > But
> > > the distance moved in the first increment of time equals the
> > > acceleration. Therefore, we can write
> > >
> > > delta x/delta t =3D [-grad V].
> >
> > This is completely incorrect. What you should have written is
> >
> > delta (delta x/delta t) / delta t =3D -grad V.
> >
> What I wrote is elementary physics. I am saying that velocity in the
first
> unit of time equals acceleration. You are saying this is completely
wrong.
If you believe this to be elementary physics, you clearly do not
understand
it. In words, your statement is "velocity is equal to minus the
gradient
of the potential," keeping in mind that your "=3D" sign actually means
"approximately equal when delta t is small". The correct statement
is "acceleration is equal to minus the gradient of the potential". If
you
really want to stick to the case where a particle starts accelerating
from rest, then you could write something like
v(at time delta t) =3D (-grad V)*(delta t),
which reads "if a particle starts from rest (velocity=3D0 at time 0),
after
a time delta t its velocity is (-grad V)*(delta t)". The devil is in
the details, as you can see.
>
> > grad V is a vector attached to every point in space...
>
> First, thanks for this explanation. Instead of considering the most
general
> case, let's look at the specific case of orbits. I believe grad V
could be
> applied to orbital situation. There is no doubt that the method of
studying
> motion with the gradient vector point of view may be useful in some
> situations. But in this case, it appears to me, the complications
that ensue
> by introducing this notation is unnecessarily high for physics
returns we
> get. Take the case of circular orbits. You divide the circle into
points.
> These points are spaced with equal intervals. You attach an arrow to
each
> point. Each arrow is perpendicular to the radius of the orbit. Each
arrow
> has the same length.... All this cluttering of the figure is done in
order
> to say that the orbit has uniform motion. What is gained here by
talking
> about grad V? It seems superfluous to me.
Once again, I am lost in what you are trying to say. Force is a vector,
it is applied to a particle when it's at a given position, say x. This
vector at position x could for example come from the gradient of a
scalar potential V, namely grad V. I fail to see what orbits,
especially
circular ones have to do with this. Yes, a circular orbit is a
trajectory
of a particle in a certain potential, the Newtonian gravitational
potential.
No, your example does not say anything about superfluousness of grad V,
whatever you mean by that.
> > Perhaps
> > what you are trying to get at is that the definition of a
conservative
> > force is taylored for the proof conservation of energy.
>
> Yes. Something like that. What does it mean when you assume "only
> conservative forces?" As I understand it, by tradition physicists
prefer to
> call "acceleration" "force." Substituting "acceleration"
in "only
> conservative force" I get:
No.
> "Only conservative [acceleration]."
???
> "Only Conservative accelerations" may be physicists' way of
saying "let's
> consider motions where net acceleration is zero." The trivial case
of which
> is the uniform motion. From a na=EFve point of view, it appears that
your
> derivation amounts to writing "acceleration" in two different
notations,
> then setting, acceleration to zero, and calling this situation
"gradient
> force." I am not sure this is correct though.
This is correct in no shape or form.
> > This is true,
> > as the name "conservative" implies. Further justification can only
go
> > to experimental observation, and we do observe that energy is
conserved
> > in simple mechanical systems.
>
> This seems not to be relevant to the derivation question. The
question is
> not about the observation of energy conservation but about the
derivation of
> it from the Newtonian laws of motion.
This question has been answered. Newton's laws + assumption of
conservative
forces give conservation of enerty. Not it is not a vaccuous statement.
Both, Newton's laws and assumption of conservative force are justified
by observing experiments; nature is the highest authority.
Igor
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