Valentin
Voroshilov, Ph.D.
Physics
Department, Boston University,
A
Remark to “A Primer on
Work-Energy Relationships for Introductory Physics”
of
Carl E. Mungan, “The Physics Teacher”, Vol. 43, January 2005.
Any science is
based on definitions and laws. Physics is a science. Hence Physics is based on
definitions and laws. We can observe the history of Physics as the history of
developing of definitions and of opening of laws of Physics. However, I hardly
doubt on the necessity of inventing new definitions during teaching of the
Introductory Physics course. The correct using of the existing definitions is
one of the most important teacher’s work.
In the notated
publication, the author is playing with such categories as “part”, “particle”, “system”, “object”, “work”,
“center-of-mass work”, “particle work”, mechanical energy”, “kinetic energy”,
etc, using them in a specific author’s way. “The paper is intended to clarify
the central equations…” but instead of achieving this goal the relationships between
fundamental categories became more fogged only. I do not intend to discuss the
whole article here. But there is a part, where the author’s ideas lead just to
wrong physical conclusions.
The below is
the part of the text I want to talk about.
“(ii) The
kinetic energy of a system depends on how you partition it and thus one must
clearly specify not only the system but also its parts. For example, consider a
Frisbee® of mass m and
moment of inertia I at temperature T thrown on the conventional way so that
it spins with rotational speed ω
as it sails through the air with translational speed v.
If we take our system to be a single part, the Frisbee as a whole, then K = ½mv2; one might call this
the macroscopic view. On the other
hand, if we take this system to be composed of a large number of bits of
plastic that are small compared to the size of the Frisbee, but large compared
to molecular dimensions, then K =
½mv2
+ ½Iω2, which one might call the mesoscopic view. Finally, if one
resolves the Frisbee into its N
individual atoms, then K = ½mv2
+ ½Iω2 +³/2NkBT
(at high temperatures, typically at room temperature), where kB is the Boltzmann constant;
this is the microscopic view. While
this is often confusing initially to students, an appreciation of the fact that
kinetic energy can be “hidden” inside an object in this way is crucial to the
development of the concept of internal
energy.”
Let us first to
analyze the ideas hidden behind the text.
Accumulating
the logic of the author (“The kinetic energy of a system depends on how you
partition it”) we obtain that kinetic
energy of the moving and rotating rigid body depends on the mental
operations performed by the person whishing to calculate the kinetic energy.
It looks like the body reads the person’s mind and stores as much kinetic
energy as the person wants. I wonder what will happen with the Frisbee if three
different persons will be trying to calculate the three author’s kinetic
energies at the same time.
The idea of
dependence of a physical quantity on the ways of thinking of the quantity may
make study Physics really confusing. Physics, in this point of view, becomes
being the collection of peoples whishes, instead of being the collection of
laws of Nature. As one of the consequences of that view we obtain the rotating rigid body without rotational
kinetic energy (K = ½mv2)
just because we use “the macroscopic view”.
Why did it
happen?
Because of
misusing of the definitions.
Let’s get
things straightened. All we need to do to make that is rereading the common
textbook on General Physics, for example, of Douglas C. Giancoli (1984).
There is not a
specific physics definition for the word “a part”. But we’ll find that “a
particle” is “considered to be a mathematical point, and as such has no spatial
extent (no size)” (page 12). Hence, the correct definition of “a particle”
clearly shows that a particle dose not have any structure, it is an idealized
object the size of such defines as zero or approaches to zero during any
calculations. Talking about “a part” we have to fix the properties of the
“part”, i.e. is it “a particle” or just a small “rigid body” or maybe a small
body being able to change its size or form, etc? A “single part” is undefined
term in Physics, but saying “as a whole” means “the abject has a size but we do
not care about its structure”, however, it means “we do not consider body as a
particle”.
The Frisbee is
an extended object; hence to analyze its motion we must use all the physical
quantities significant to any rigid body, even if in some specific situations
the size of the object dose not play a significant role.
On the pages
183 – 185 (Douglas C. Giancoli,1984)we can find general solution for kinetic energy of a rigid body (rotating
about the axis fixed in direction); K
= ½mv2
+ ½Iω2, here v is
a velocity of the center of mass of the body, etc.; ½mv2
is the translational kinetic
energy of a rigid body; ½Iω2
is the rotational kinetic
energy of a rigid body. It is really important to
understand that we must use both of kinetic energies, i.e., translational and
rotational, to describe the behavior of a rigid body. In addition, we can see
that we have used definitions of three different
kinetic energies, but all of them have the same structure, i.e. what kind of kinetic energy of what*.
We can add at
least two more different kinetic energies:
- kinetic
energy of a particle ½mv2
(by a definition a particle dose not have
a size, hence, it dose not have a rotational kinetic energy or kinetic
energy of vibrations, hence, we do not have to specify what kind of kinetic
energy dose a particle have; but if somebody wants to chose a different
definition for “a particle” he has to reorganize the total system of definitions, which, probably, is possible to
make, but will confuse lots of other people);
- kinetic
energy of a system of particles, which is by a definition just an algebraic sum
of kinetic energies of the particles forming the system. Every time when we see
the term ½mv2 or we pronounce “kinetic energy” we
must be sure what are specific system we
talking about.
We have to see
the difference between “the object” and “the system”. Fixing the object of the
study dose not mean yet the fixing the system of the study. The object is the
something that attracts our attention. But by fixing the system we have automatically fixed all the
idealizations we have applied to the object (including its parts). We do not
work directly with objects in Physics; we work with specific idealizations,
which we choose according to our
(right or wrong) interpretation of the situation. By fixing the system
we fix the physical quantities crucial for describing the properties of the
system (however it dose not mean there do not exist other physical quantities
which could be used to describing the parent to the system object). To
calculate the values of the quantities we can use different mental operations,
including partition the system (considering subsystems), but the way of our
thinking of the system cannot have an influence on the result of calculation.
Nobody can
forbid dividing in our head (!) a rigid body on particles like atoms or any
other parts. But the way of thinking of
the rigid body cannot have an influence on the value of calculated physical
quantities. Rigid body always has rotational kinetic energy.
Thinking of the
Frisbee as of a rigid body (hence, having a size), we cannot consider it as of
a particle (saying “it spins with rotational speed ω” we automatically said “it is not a particle”). Thinking of the Frisbee as of a
particle, we cannot consider it as of a rigid body. What way of thinking should we chose depends on the specific
physics situation and problem we want to investigate, but it dose not depend on the way we
partition the Frisbee.
If we need just
to calculate the distance traveled, and if we can neglect the influence of air
(and any other objects in the Nature
except the Earth) we can consider the Frisbee as of a particle having all
the mass at the center of mass of the body (because the size of the body is
significantly smaller then the distance traveled, and there are not any
significant influences on any part of the object). It means we define our
system as a particle. But it dose not mean the Frisbee dose not have a
rotational kinetic energy or even internal energy; it means these energies do
exist, but we don’t care abut them.
On the other
hand, if we want to study the influence of the air on the motion of the
Frisbee, we must include in the consideration the rotational kinetic energy of
the Frisbee, because the part of it is transferring into the work of the air
friction force. It means we define our system as a rigid body (notice, that a
particle and a rigid body are different idealizations using to describe the
motion of the same real object, i.e. the Frisbee).
Saying “the
kinetic energy of the Frisbee as a whole is K = ½mv2”
is just wrong, it cannot be correct at
any conditions because using the term “a whole” automatically means consideration
the Frisbee as a rigid body. But we can definitely say “the translational
kinetic energy of the Frisbee as a whole is K
= ½mv2”
(and keep in mind that v is a velocity of the center of mass).
And we can say “in
the situation where, because of la-la-la,
the size and the rotational motion of the Frisbee do not have any significance,
the Frisbee can be considered as a
particle” (hence its kinetic energy is K
= ½mv2).
However, we have to understand, that our choice dose not depend on how we partition the system, it depends
on analyzing of the physical conditions
of the considering physical situation.
The last
author’s equation, that is K =
½mv2
+ ½Iω2 + ³/2NkBT,
looks fine. But correct explanation
of the equation by using the microscopic view
is beyond the matters of Introductory Physics.
To calculate
kinetic energy, we have to define first kinetic energy of what it is? Let us consider the object (i.e. the flying Frisbee)
as a group of atoms contained inside the physical boundaries of the Frisbee. In
this situation, by a definition, the
kinetic energy of the system is K = ΣKi, the summation is
going through all the atoms of the Frisbee. This form of kinetic energy of system dose not contains any information
on the phase of the parent object; it
can be a rigid body, but it can be a liquid or a gas. If we want to deal with a
rigid body, we must apply additional conditions on the system, like “the
distance between any two particles are fixed during the motion of the system”.
But we obviously cannot apply this condition to the atoms (plus, the form of the last term depends on the
assumptions we make on atoms, i.e. what kind of the idealization we use for
atoms, i.e. are they interacted particles
or oscillators, etc?). Direct
deriving the latter expression for K from an atomic view is more sophisticated
then it seems to be. We have to use the “old-fashion” approach, i.e. using the common definitions for mechanic
energy of rigid bogy and for the
internal energy of solid body, we obtain the final result as E = ½mv2 + ½Iω2 + U**, where E is the total energy of the body, ½mv2
is the translational kinetic energy of the rigid body, ½Iω2 is the rotational kinetic energy of the
rigid body, U is the
(thermodynamically calculated) total
average energy of the atoms of the body when
it is hold at rest and without rotation at the temperature of T. But this way do not have any
coherence with “ a microscopic view”. The Dulong-Petit rule, used by the
author, related directly to the total average energy of the atoms U, which is the internal energy of the
body. But if we want to come to kinetic energy K we need to have a specific talk on how do we use the expression for the total average energy to calculate the value of the average kinetic energy Uk = ³/2NkBT,
which is not so trivial (especially for Introductory Physics).
The easiest way
to explain the ³/2NkBT term would be treating the system as an
ideal gas, but in this case we definitely cannot extract the rotational kinetic
energy from the general expression for kinetic energy (we cannot use two
deferent idealizations, like an ideal gas and a rigid body, to the same object
at the same time).
I do not see a
necessity to analyze here the other author’s speculations (for example, he can
treat rotational kinetic energy of a rigid body as an internal energy of this
body, which is a new word in Physics, at least for Douglas C. Giancoli, see
page 375). I want to say only, when an author intends to discuss matters about
which “there exist differences of opinion among educators” there exist a high
probability to make a mistake. To avoid mistakes, like we saw in the notated
publication, the paper intended to clarify the matters has to be based on the
clear definitions using of the majority of physicist.
*P.S. It is
interesting subject to discuss, what do some everyday-using Physics terms
really mean? For example, pronouncing “a force” we mean (or at least must mean)
“a force exerted by that specific object
and acting on this specific object”;
pronouncing “a velocity” we mean (or at least must mean) “a velocity of this specific object with respect to that specific object”, etc.
** Strycly
speaking, the real Frisbee has always
all kinds of energy, i.e. translational and rotational kinetic energies,
internal energy U and even potential
energy P, so, the total energy
of the Frisbee is always E =
½mv2
+ ½Iω2 + U +P. However,
the behavior of any physical system is not defined by the value of its total energy, the behavior is defined by the change in the total energy. Physicists
know that an arbitrary constant can
be always included in the total energy of system without making any changes in
its behavior. Hence, when we deal with the Frisbee in the specific
circumstances, we need to consider each term of the total energy specifically
in point of view of possible reasons for the term being changed. If, for
example, the change in the rotational
kinetic energy and the internal energy can be neglected, hence the rotational
kinetic energy and the internal energy can be considered as a constant, and the
total energy of the Frisbee will be E
= ½mv2
+ P
+ constant. In this situation, the
dynamics of the Frisbee is just equal to
the dynamics of the particle of mass m in
the potential field P (that is
because we can use this idealization). But we cannot usually use this kind of
view on the dynamics problems while teaching Introductory Physics, because we
need to introduce first each term separately, but when we have it done, we are
done with the Introductory Mechanics as well.