Valentin Voroshilov,
Ph.D.
Physics Department,
Boston University,
www.beaplus.com
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.