Elementary Physics, Mechanics, and Acoustics
This appendix derives some basic results from the field of
physics, particularly
mechanics and
acoustics,
which are referenced elsewhere in this book, or which are known to be
needed in practical synthesis models.

Newton's Laws of Motion
Perhaps the most heavily used equation in
physics is
Newton's second
law of motion:
That is, when a
force is applied to a
mass, the mass experiences an
acceleration proportional to the applied force. Denoting the mass by

, force at time

by

, and acceleration by
we have
 |
(B.1) |
In this formulation, the applied force

is considered positive
in the direction of positive mass-position

. The force

and acceleration

are, in general,
vectors in
three-dimensional space

. In other words, force and
acceleration are generally vector-valued functions of time

. The
mass

is a
scalar quantity, and can be considered a measure of the
inertia of the physical system (see §
B.1.1 below).
Newton's three laws of motion may be stated as follows:
- Every object in a state of uniform motion will remain in that
state of motion unless an external force acts on it.
- Force equals mass times acceleration [
].
- For every action there is an equal and opposite reaction.
The first law, also called the
law of inertia, was pioneered by
Galileo. This was quite a conceptual leap because it was not possible
in Galileo's time to observe a moving object without at least some
frictional forces dragging against the motion. In fact, for over a
thousand years before Galileo, educated individuals believed
Aristotle's formulation that, wherever there is motion, there is an
external force producing that motion.
The second law,

, actually implies the first law, since
when

(no applied force), the acceleration

is zero,
implying a constant
velocity 
. (The velocity is simply the
integral with respect to time of

.)
Newton's third law implies
conservation of momentum
[
137]. It can also be seen as following from the
second law: When one object ``pushes'' a second object at some
(massless) point of contact using an applied force, there must be an
equal and opposite force from the second object that cancels the
applied force. Otherwise, there would be a nonzero net force on a
massless point which, by the second law, would accelerate the point of
contact by an infinite amount.
In summary,
Newton's laws boil down to

. An enormous quantity
of physical science has been developed by applying this
simple
B.1 mathematical law to different physical
situations.
Mass is an intrinsic property of matter.
From
Newton's second law,

, we have that the amount of
force required to accelerate an object, by a given amount, is
proportional to its mass. Thus, the mass of an object quantifies its
inertia--its resistance to a change in
velocity.
We can measure the mass of an object by measuring the
gravitational force between it and another known mass,
as described in the next section. This is a special case of measuring
its acceleration in response to a known force. Whatever the force

,
the mass

is given by

divided by the resulting acceleration

, again by Newton's second law

.
The usual mathematical model for an
ideal mass is a dimensionless
point at some location in space. While no real objects are
dimensionless, they can often be treated mathematically as
dimensionless points located at their
center of mass, or
centroid (§
B.4.1).
The
physical state of a mass

at time

consists of its
position 
and
velocity

in 3D space.
The amount of mass itself,

, is regarded as a fixed parameter that
does not change. In other words, the
state

of a
physical system typically changes over time, while any
parameters of the system, such as mass

, remain fixed over
time (unless otherwise specified).
Gravitational Force
We are all familiar with the
force of gravity. It is a
fundamental observed property of our universe that any two
masses

and

experience an
attracting force 
given by the
formula
 |
(B.2) |
where

is the distance between the
centroids of the masses

and

at time

, and

is the
gravitation
constant.
B.2
The law of gravitation Eq.

(
B.2) can be accepted as an
experimental fact which defines the concept of a
force.
B.3 The giant conceptual leap taken by
Newton was that the law of
gravitation is
universal--applying to celestial bodies as well as objects on
earth. When a mass is ``dropped'' and allowed to ``fall'' in a
gravitational field, it is observed to experience a
uniform acceleration proportional to its mass. Newton's second
law of motion (§
B.1) quantifies this result.
Consider an ideal
spring suspending a
mass from a rigid ceiling, as
depicted in Fig.
B.1. Assume the mass is at rest,
and that its distance

from the ceiling is fixed.
Figure B.1:
Mass hung by a spring from a rigid ceiling.
 |
If

denotes the mass of the earth, and

is the distance of mass

's center from the earth's
center of mass, then the downward
force
on the mass

is given by Eq.

(
B.2) as
where
is called the
acceleration due to gravity. Changes in

due
to the motion of the mass are assumed negligible relative to the
radius of the earth (about

miles), and so

is treated as a
constant for most practical purposes near the earth's surface. We see
that if we double the mass

, we double the force

pulling on
the spring. It is an
experimental fact that typical springs
exhibit a
displacement 
that is approximately proportional to the
applied force

for a wide range of applied forces. This is
Hooke's law for ideal springs:
(Hooke's Law) |
(B.3) |
where

is the
displacement of the spring from its
natural length. We call

the
spring constant, or
stiffness of the spring. In terms of our previous notation, we have
where

is the length of the spring with no mass attached.
Note that the force on the spring in Fig.
B.1 is
gravitational force. Equal and opposite to the force of gravity is
the
spring force exerted upward by the spring on the mass
(which is not moving). We know that the spring force is equal and
opposite to the gravitational force because the mass would otherwise
be accelerated by the net force.
B.4 Therefore, like gravity, a
displaced spring can be regarded as a definition of an applied force.
That is, whenever you have to think of an applied force, you can
always consider it as being delivered by the end of some ideal spring
attached to some external physical system.
Note, by the way, that normal interaction forces when objects touch
arise from the
Coulomb force (electrostatic force, or repulsion
of like charges) between electron orbitals. This electrostatic force
obeys an ``
inverse square law'' like gravity, and therefore also
behaves like an ideal spring for small displacements.
B.5
The specific value of

depends on the physical units adopted as
well as the ``stiffness'' of the spring. What is most important in
this definition of force is that a doubling of spring displacement
doubles the force. That is, the spring force is a
linear
function of spring displacement (compression or stretching).
Applying Newton's Laws of Motion
As a simple example, consider a
mass 
driven along a frictionless
surface by an ideal
spring 
, as shown in Fig.
B.2.
Assume that the mass position

corresponds to the spring at rest,
i.e., not stretched or compressed. The
force necessary to compress the
spring by a distance

is given by
Hooke's law (§
B.1.3):
This force is balanced at all times by the
inertial force

of
the mass

,
i.e. 
,
yielding
B.6
 |
(B.4) |
where we have defined

as the initial
displacement of the mass
along

. This is a
differential equation whose solution
gives the equation of motion of the mass-spring junction for all
time:
B.7
 |
(B.5) |
where

denotes the
frequency of
oscillation in radians per second. More generally, the complete
space of solutions to Eq.

(
B.4), corresponding to all possible
initial displacements

and initial velocities

, is the
set of all
sinusoidal oscillations at frequency

:
The amplitude of oscillation

and phase offset

are
determined by the
initial conditions,
i.e., the initial position

and initial
velocity

of the mass (its
initial
state) when we ``let it go'' or ``push it off'' at time

.
Work = Force times Distance = Energy
Work is defined as
force times distance. Work is a measure
of the
energy expended in applying a force to move an
object.
B.8
The work required to compress a
spring 
through a
displacement of
meters, starting from rest, is then
 |
(B.6) |
Work can also be negative. For example, when uncompressing an
ideal spring, the (positive) work done by the spring on its moving end
support can be interpreted also as saying that the end support
performs negative work on the spring as it allows the spring to
uncompress. When negative work is performed, the driving system is
always accepting energy from the driven system. This is all simply
accounting. Physically, one normally considers the
driver as
the agent performing the positive work,
i.e., the one expending energy
to move the driven object. Thus, when allowing a spring to
uncompress, we consider the spring as performing (positive) work on
whatever is attached to its moving end.
During a
sinusoidal mass-spring oscillation, as derived in
§
B.1.4, each
period of the oscillation can be divided into
equal sections during which either the
mass performs work on the spring,
or vice versa.
Gravity, spring forces, and electrostatic forces are examples of
conservative forces. Conservative forces have the property
that the work required to move an object from point

to point

,
either with or against the force, depends only on the locations of
points

and

in space, not on the path taken from

to

.
Potential Energy
in a Spring
When compressing an ideal spring,
work is performed, and this
work is
stored in the spring in the form of what we call
potential energy. Equation (
B.6) above gives the quantitative
formula for the potential energy

stored in an ideal spring
after it has been compressed
meters from rest.
Kinetic energy is energy associated with
motion. For
example, when a
spring uncompresses and accelerates a mass, as in the
configuration of Fig.
B.2, work is performed on the mass
by the spring, and we say that the
potential energy of the spring is
converted to
kinetic energy of the mass.
Suppose in Fig.
B.2 we have an initial spring compression
by
meters at time

, and the mass
velocity is zero at

. Then from the equation of motion Eq.

(
B.5), we can calculate
when the spring returns to rest (

). This first happens at the
first zero of

, which is time

. At this time, the velocity,
given by the time-derivative of Eq.

(
B.5),
can be evaluated at

to yield the mass velocity
![$ v[(\pi/2)/\omega_0] = -A\omega_0 = -A\sqrt{k/m}$](http://www.dsprelated.com/josimages_new/pasp/img2652.png)
, which is when all
potential energy from the spring has been converted to kinetic energy
in the mass. The square of this value is
and we see that if we multiply

by

, we get
which is the initial potential energy stored in the spring. We
require this result. Therefore, the kinetic energy of a mass must be
given by
in order that the kinetic energy of the mass when spring compression
is zero equals the original potential energy in the spring when the
kinetic energy of the mass was zero. In the next section we derive
this result in a more general way.
From
Newton's second law,

(introduced in Eq.

(
B.1)),
we can use d'Alembert's idea of
virtual work to derive the
formula for the kinetic energy of a mass given its speed

.
Let

denote a small (infinitesimal)
displacement of the mass in
the

direction. Then we have, using the
calculus of differentials,
Thus, by Newton's second law, a differential of work

applied to a
mass

by
force 
through distance

boosts the kinetic energy
of the mass by

. The kinetic energy of a mass moving at
speed

is then given by the integral of all such differential
boosts from 0 to

:
where

denotes the
kinetic energy of mass

traveling at speed

.
The quantity

is classically called the
virtual work
associated with force

, and

a
virtual displacement
[
544].
Summarizing the previous sections, we say that a compressed
spring
holds a
potential energy equal to the
work required to
compress the spring from rest to its current
displacement. If a
compressed spring is allowed to expand by pushing a
mass, as in the
system of Fig.
B.2, the potential energy in the spring
is converted to
kinetic energy in the moving mass.
We can draw some inferences from the oscillatory motion of the
mass-spring system written in Eq.

(
B.5):
- From a global point of view, we see that
energy is conserved, since the oscillation never decays.
- At the peaks of the displacement
(when
is either
or
), all energy is in the form of potential energy,
i.e., the spring is either maximally compressed or stretched, and the mass
is momentarily stopped as it is changing direction.
- At the zero-crossings of
, the spring is momentarily
relaxed, thereby holding no potential energy; at these instants, all
energy is in the form of kinetic energy, stored in the motion of the mass.
- Since total energy is conserved (§B.2.5), the kinetic
energy of the mass at the displacement zero-crossings is exactly the
amount needed to stretch the spring to displacement
(or compress
it to
) before the mass stops and changes direction. At all
times, the total energy
is equal to the sum of the potential
energy
stored in the spring, and the kinetic energy
stored in the mass:
Regarding the last point, the potential energy,

was
defined as the work required to
displace the spring by
meters,
where work was defined in Eq.

(
B.6). The kinetic energy of a mass

moving at speed

was found to be

. The constance of the potential plus
kinetic energy at all times in the mass-spring oscillator is easily
obtained from its equation of motion using the trigonometric identity

(see Problem
3).
Energy Conservation
It is a remarkable property of our universe that
energy is
conserved under all circumstances. There are no known exceptions to
the conservation of energy, even when relativistic and quantum effects
are considered.
B.9
Energy may be defined as the ability to do
work, where work may be defined as
force times distance
(§
B.2).
Energy Conservation in the
Mass-Spring System
Recall that
Newton's second law applied to a
mass-
spring system, as in
§
B.1.4, yields
which led to the
differential equation obeyed by the mass-spring system:
Multiplying through by

gives
Thus, Newton's second law and
Hooke's law imply conservation of energy
in the mass-spring system of Fig.
B.2.
Momentum
The
momentum of a
mass 
is usually defined by
where

is the
velocity of the mass. Note that momentum is a vector
quantity in general, so we should more clearly write
where

and

are
vectors in 3D space. We will
return to vector momentum in §
B.4.1 below.
Like energy,
momentum is
conserved in physical systems. For
example, when two
masses collide and recoil from each other, the total
momentum after the collision equals that before the collision. Since
the momentum is a three-dimensional vector in Euclidean space,
momentum conservation provides three simultaneous equations, in
general.
B.10
Rigid-Body Dynamics
Below are selected topics from
rigid-body dynamics, a subtopic
of
classical mechanics involving the use of
Newton's laws of
motion to solve for the motion of
rigid bodies moving in 1D,
2D, or 3D space.
B.11 We may think
of a rigid body as a
distributed mass, that is, a mass that has
length, area, and/or volume rather than occupying only a single point
in space. Rigid body models have application in
stiff strings
(modeling them as disks of mass interconnect by ideal
springs), rigid
bridges,
resonator braces, and so on.
We have already used Newton's

to formulate mathematical dynamic
models for the ideal
point-mass (§
B.1.1),
spring
(§
B.1.3), and a simple
mass-spring system (§
B.1.4).
Since many physical systems can be modeled as assemblies of masses and
(normally damped) springs, we are pretty far along already. However,
when the springs interconnecting our point-masses are very stiff, we
may approximate them as rigid to simplify our simulations. Thus,
rigid bodies can be considered mass-spring systems in which the
springs are so stiff that they can be treated as rigid massless rods
(infinite spring-constants

, in the notation of §
B.1.3).
So, what is new about
distributed masses, as opposed to the
point-masses considered previously? As we will see, the main new
ingredient is
rotational dynamics. The
total momentum
of a rigid body (distributed mass) moving through space will be
described as the
sum of the
linear momentum of its
center of mass (§
B.4.1 below) plus the
angular
momentum about its center of mass (§
B.4.13 below).
The
center of mass (or
centroid) of a rigid body is
found by averaging the spatial points of the body

weighted by the mass

of those points:
B.12
Thus, the center of mass is the
mass-weighted average location
of the object. For a continuous mass distribution totaling up to

,
we can write
where the volume integral is taken over a volume

of 3D space that
includes the rigid body, and

denotes the mass contained within the differential volume
element

located at the point

, with

denoting the
mass density at the point

. The total mass
is
A nice property of the center of mass is that
gravity acts on a
far-away object as if all its mass were concentrated at its center of
mass. For this reason, the center of mass is often called the
center of gravity.
Consider a system of

point-
masses 
, each traveling with
vector
velocity

, and not necessarily rigidly attached to
each other. Then the total
momentum of the system is
where

denotes the total mass, and

is the velocity
of the center of mass.
Thus, the
momentum

of any collection of masses

(including rigid bodies) equals the total mass

times the
velocity of the center-of-mass.
The previous result might be surprising since we said at the outset
that we were going to
decompose the total
momentum into a sum
of linear plus angular momentum. Instead, we found that the total
momentum is simply that of the
center of mass, which means any angular
momentum that might have been present just went away. (The center of
mass is just a point that cannot rotate in a measurable way.) Angular
momentum does not contribute to
linear momentum, so it provides three
new ``degrees of freedom'' (three new energy storage dimensions, in 3D
space) that are ``missed'' when considering only linear momentum.
To obtain the desired decomposition of momentum into linear plus
angular momentum, we will choose a fixed reference point in space
(usually the center of mass) and then, with respect to that reference
point, decompose an arbitrary mass-particle travel direction into the
sum of two mutually
orthogonal vector components: one will be the
vector component pointing
radially with respect to the
fixed
point (for the ``linear momentum'' component), and the other will be
the vector component pointing
tangentially with respect to the
fixed point (for the ``angular momentum''), as shown in
Fig.
B.3. When the reference point is the center of mass, the
resultant radial
force component gives us the force on the center of
mass, which creates
linear momentum, while the net tangential
component (times distance from the center-of-mass) give us a resultant
torque about the reference point, which creates
angular
momentum. As we saw above, because the tangential force component
does not contribute to linear momentum, we can simply sum the external
force vectors and get the same result as summing their radial
components. These topics will be discussed further below, after some
elementary preliminaries.
Figure:
Rigid body having center of mass at
, experiencing an external force
that can be expressed as
the sum of radial and tangential components
. The
radial component
accelerates the center-of-mass, while the
tangential component
causes rotation about the center of
mass.
![\includegraphics[width=1.5in]{eps/rigidbody}](http://www.dsprelated.com/josimages_new/pasp/img2710.png) |
The
translational kinetic energy of a collection of
masses

is given by
where

is the total mass, and

denotes the speed of the
center-of-mass. We have

, where

is
the
velocity of the center of mass.
More generally, the
total energy of a collection of masses
(including distributed and/or rigidly interconnected point-masses) can
be expressed as the
sum of the
translational and
rotational kinetic energies [
270, p.
98].
Figure B.4:
Point-mass
rotating in a
circle of radius
with tangential speed
, where
denotes the angular velocity in rad/s.
![\includegraphics[width=1.2in]{eps/masscircle}](http://www.dsprelated.com/josimages_new/pasp/img2715.png) |
The
rotational kinetic energy of a rigid assembly of masses (or
mass distribution) is the sum of the rotational kinetic energies of
the component masses. Therefore, consider a point-mass

rotating
B.13 in a circular orbit
of radius

and angular
velocity 
(radians per second), as
shown in Fig.
B.4. To make it a closed system, we can
imagine an effectively infinite mass at the origin

. Then the
speed of the mass along the circle is

, and its kinetic
energy is

. Since this is what we want
for the
rotational kinetic energy of the system, it is
convenient to define it in terms of angular velocity

in
radians per second. Thus, we write
 |
(B.7) |
where
 |
(B.8) |
is called the
mass moment of inertia.
The
mass moment of inertia 
(or simply
moment of
inertia), plays the role of
mass in rotational dynamics, as
we saw in
Eq.

(
B.7) above.
The mass moment of inertia of a rigid body, relative to a given axis
of rotation, is given by a weighted sum over its mass, with each
mass-point weighted by the square of its distance from the rotation
axis. Compare this with the
center of mass (§
B.4.1) in which each
mass-point is weighted by its vector location in space (and divided by
the total mass).
Equation (
B.8) above gives the moment of inertia for a single point-mass

rotating a distance

from the axis to be

. Therefore,
for a rigid collection of point-masses

,

,
B.14 the
moment of inertia about a given axis of rotation is obtained by adding
the component moments of inertia:
 |
(B.9) |
where

is the distance from the axis of rotation to the

th
mass.
For a continuous mass distribution, the moment of inertia is given by
integrating the contribution of each differential mass element:
 |
(B.10) |
where

is the distance from the axis of rotation to the mass
element

.
In terms of the
density

of a continuous mass
distribution, we can write
where

denotes the mass density (kg/m

) at the
point

, and

denotes a differential volume element
located at

.
Circular Disk Rotating in Its Own Plane
For example, the
moment of inertia for a uniform circular disk of
total
mass 
and radius

, rotating in its own plane about a
rotation axis piercing its center, is given by
Circular Disk Rotating About Its Diameter
The
moment of inertia for the same circular disk rotating about an
axis in the plane of the disk, passing through its center, is given by
Thus, the uniform disk's moment of inertia in its own plane is twice
that about its diameter.
In general, for any 2D distribution of
mass, the
moment of inertia
about an axis
orthogonal to the plane of the mass equals the sum of
the moments of inertia about any two mutually orthogonal axes in the
plane of the mass intersecting the first axis. To see this, consider
an arbitrary mass element

having rectilinear coordinates

in the plane of the mass. (All three coordinate axes
intersect at a point in the mass-distribution plane.) Then its moment
of inertia about the axis orthogonal to the mass plane is

while its moment of inertia about coordinate axes
within the mass-plane are respectively

and

.
This, the
perpendicular axis theorem is an immediate consequence of
the
Pythagorean theorem for right triangles.
Let

denote the
moment of inertia for a rotation axis passing
through the
center of mass, and let

denote
the moment of inertia for a rotation axis parallel to the first
but a distance

away from it. Then the
parallel axis theorem
says that
where

denotes the total
mass. Thus, the added inertia due to
displacement by
meters away from the centroidal axis is equal to
that of a
point mass 
rotating a distance

from the
center of rotation.
Stretch Rule
Note that the
moment of inertia does not change when
masses are moved
along a vector parallel to the axis of rotation (see,
e.g.,
Eq.

(
B.9)). Thus, any rigid body may be ``stretched'' or
``squeezed'' parallel to the rotation axis without changing its moment
of inertia. This is known as the
stretch rule, and it can be
used to simplify
geometry when finding the moment of inertia.
For example, we saw in §
B.4.4 that the moment of inertia
of a point-mass

a distance

from the axis of rotation is given
by

. By the stretch rule, the same applies to an ideal
rod of mass

parallel to and distance

from the axis of
rotation.
Note that mass can be also be ``stretched'' along the circle of
rotation without changing the moment of inertia for the mass about
that axis. Thus, the point mass

can be stretched out to form a
mass
ring at radius

about the axis of rotation without
changing its moment of inertia about that axis. Similarly, the ideal
rod of the previous paragraph can be stretched tangentially to form a
cylinder of radius

and mass

, with its axis of symmetry
coincident with the axis of rotation. In all of these examples, the
moment of inertia is

about the axis of rotation.
The
area moment of inertia is the second moment of an area

around a given axis:
where

denotes a differential element of the area (summing to

), and

denotes its distance from the axis of rotation.
Comparing with the definition of
mass moment of inertia in
§
B.4.4 above, we see that mass is replaced by area in
the area moment of inertia.
In a planar mass distribution with total mass

uniformly
distributed over an area

(
i.e., a constant mass density of

), the mass moment of inertia

is given by the area
moment of inertia

times mass-density

:
For a planar distribution of
mass rotating about some axis in the
plane of the mass, the
radius of gyration is the distance from
the axis that all mass can be concentrated to obtain the same mass
moment of inertia. Thus, the radius of gyration is the ``equivalent
distance'' of the mass from the axis of rotation. In this context,
gyration can be defined as
rotation of a planar region
about some axis lying in the plane.
For a bar cross-section with area

, the radius of gyration is given by
 |
(B.11) |
where

is the
area moment of inertia (§
B.4.8)
of the cross-section about a given axis of rotation lying in the plane
of the cross-section (usually passing through its
centroid):
where

denotes the distance of the differential area element

from the axis of gyration.
Rectangular Cross-Section
For a rectangular cross-section of height

and width

, area

, the
area moment of inertia about the horizontal midline is
given by
The
radius of gyration about this axis is then
Similarly, the
radius of gyration about a vertical axis passing
through the center of the cross-section is

.
The radius of gyration can be thought of as the ``effective radius''
of the
mass distribution with respect to its inertial response to
rotation (``gyration'') about the chosen axis.
Most cross-sectional shapes (
e.g., rectangular), have at least two
radii of gyration. A circular cross-section has only one, and its
radius of gyration is equal to half its radius, as shown in the next
section.
Circular Cross-Section
For a circular cross-section of radius

, Eq.

(
B.11) tells us
that the squared
radius of gyration about any line passing through the
center of the cross-section is given by
Using the elementrary trig identity

, we readily
derive
The first two terms of this expression contribute zero to the integral
from 0 to

, while the last term contributes

,
yielding
Thus, the
radius of gyration about any midline of a circular cross-section of
radius

is
For a circular
tube in which the
mass of the cross-section lies
within a circular
annulus having inner radius

and outer
radius

, the radius of gyration is given by
 |
(B.12) |
Two Masses Connected by a Rod
Figure B.5:
Two ideal point-masses
connected by an ideal, rigid, massless rod of length
.
![\includegraphics[width=1.5in]{eps/massrodmass}](http://www.dsprelated.com/josimages_new/pasp/img2758.png) |
As an introduction to the decomposition of rigid-body motion into
translational and
rotational components, consider the
simple system shown in Fig.
B.5. The excitation
force
density
B.15 
can be
applied anywhere between

and

along the connecting rod.
We will deliver a vertical
impulse of
momentum to the mass on the
right, and show, among other observations, that the total
kinetic
energy is split equally into (1) the
rotational kinetic energy
about the
center of mass, and (2) the
translational kinetic
energy of the total mass, treated as being located at the center of
mass. This is accomplished by defining a
new frame of
reference (
i.e., a moving coordinate system) that has its origin at
the center of mass.
First, note that the driving-point
impedance (§
7.1)
``seen'' by the driving force

varies as a function of

.
At

, The excitation

sees a ``point mass''

, and no
rotation is excited by the force (by symmetry). At

, on the
other hand, the excitation

only sees mass

at time
0, because the vertical motion of either point-mass initially only
rotates the other point-mass via the massless connecting rod. Thus,
an observation we can make right away is that the
driving point
impedance seen by
depends on the striking point
and,
away from
, it depends on time
as well.
To avoid dealing with a time-varying driving-point impedance, we will
use an impulsive force input at time

. Since momentum is the
time-integral of force (

), our
excitation will be a
unit momentum transferred to the two-mass
system at time 0.
First, consider

. That is, we apply an
upward unit-
force impulse at time 0 in the middle of the rod. The
total
momentum delivered in the neighborhood of

and

is
obtained by integrating the applied force density with respect to time
and position:
This unit momentum is transferred to the two
masses 
. By symmetry,
we have

. We can also refer to

as the
velocity of the center of mass, again obvious by symmetry.
Continuing to refer to Fig.
B.5, we have
Thus, after time zero, each mass is traveling upward at speed

, and there is no rotation about the center of mass at

.
The
kinetic energy of the system after time zero is
Note that we can also compute

in terms of the total mass

and the velocity of the center of mass

:
Now let

. That is, we apply an
impulse
of vertical
momentum to the
mass on the right at time 0.
In this case, the unit of vertical momentum is transferred entirely to
the mass on the right, so that
which is twice as fast as before. Just after time zero, we have

,

, and, because the massless rod remains rigid,

.
Note that the
velocity of the
center-of-mass 
is the
same as it was when we hit the midpoint of the rod. This is an
important general equivalence: The sum of all external
force vectors
acting on a rigid body can be applied as a single resultant force
vector to the total mass concentrated at the center of mass to find
the linear (translational) motion produced. (Recall from §
B.4.1
that such a sum is the same as the sum of all radially acting external
force components, since the tangential components contribute only to
rotation and not to translation.)
All of the
kinetic energy is in the mass on the right just after time
zero:
 |
(B.13) |
However, after time zero, things get more complicated, because the
mass on the left gets dragged into a rotation about the center of
mass.
To simplify ongoing analysis, we can define a
body-fixed frame
of referenceB.16 having its origin at the center of mass. Let

denote a velocity in this frame. Since the velocity of the center of
mass is

, we can convert any velocity

in the
body-fixed frame to a velocity

in the original frame by adding

to it,
viz.,
The mass velocities in the body-fixed frame are now
and of course

.
In the body-fixed frame, all kinetic energy is
rotational about
the origin. Recall (Eq.

(
B.9)) that the
moment of inertia for this
system, with respect to the center of mass at

, is
Thus, the
rotational kinetic energy (§
B.4.3) is found to
be
This is
half of the kinetic energy we computed in the original
``space-fixed'' frame (Eq.

(
B.13) above). The other half is in the
translational kinetic energy not seen in the body-fixed frame.
As we saw in §
B.4.2 above, we can easily calculate the
translational kinetic energy as that of the total mass

traveling at the center-of-mass velocity

:
Adding this translational kinetic energy to the rotational kinetic
energy in the body-fixed frame yields the total kinetic energy, as it
must.
In summary, we defined a moving body-fixed frame having its origin at
the center-of-mass, and the total kinetic energy was computed to be
in agreement with the more complicated (after time zero) space-fixed
analysis in Eq.

(
B.13).
It is important to note that, after time zero, both the
linear
momentum of the center-of-mass (

), and the
angular momentum in the body-fixed frame
(

) remain
constant over time.
B.17 In the original space-fixed
frame, on the other hand, there is a complex transfer of momentum back
and forth between the masses after time zero.
Similarly, the translational kinetic energy of the total mass, treated
as being concentrated at its center-of-mass, and the rotational
kinetic energy in the body-fixed frame, are both constant after time
zero, while in the space-fixed frame, kinetic energy transfers back
and forth between the two masses. At all times, however, the total
kinetic energy is the same in both formulations.
When working with rotations, it is convenient to define the
angular-velocity vector as a vector

pointing
along the
axis of rotation. There are two directions we could
choose from, so we pick the one corresponding to the
right-hand
rule,
i.e., when the fingers of the right hand curl in the direction
of the rotation, the thumb points in the direction of the angular
velocity vector.
B.18 The
length

should obviously equal the angular
velocity

. It is convenient also to work with a unit-length
variant

.
As introduced in Eq.

(
B.8) above, the
mass moment of inertia is
given by

where

is the distance from the (instantaneous)
axis of rotation to the mass

located at

. In
terms of the angular-velocity vector

, we can write this as
(see Fig.
B.6)
where
denotes the
orthogonal projection of

onto

(or

)
[
451]. Thus, we can project the mass position

onto the
angular-velocity vector

and subtract to get the component of

that is
orthogonal to

, and the length of that difference
vector is the distance to the rotation axis

, as shown in
Fig.
B.6.
Figure:
Mass position vector
and its
orthogonal projection
onto the angular
velocity vector
for purposes of finding the distance
of
the mass
from the axis of rotation
.
![\includegraphics[width=1.5in]{eps/pxov}](http://www.dsprelated.com/josimages_new/pasp/img2806.png) |
Using the
vector cross product (defined in the next section),
we will show (in §
B.4.17) that

can be written more succinctly as
The
vector cross product (or simply
vector product, as
opposed to the
scalar product (which is also called the
dot product, or
inner product)) is commonly used in
vector calculus--a basic mathematical toolset used in
mechanics [
270,
258],
acoustics [
349], electromagnetism [
356], quantum
mechanics, and more. It can be defined symbolically in the form of
a
matrix determinant:
B.19
where

denote the
unit vectors in

. The
cross-product is a vector in 3D that is
orthogonal to the plane
spanned by

and

, and is oriented positively according to
the
right-hand rule.
B.20
The second and third lines of Eq.

(
B.15) make it clear that

. This is one example of a host of identities that
one learns in vector calculus and its applications.
It is a straightforward exercise to show that the
cross-product
magnitude is equal to the product of the vector lengths times the sine
of the angle between them:
B.21
 |
(B.16) |
where
with
(Recall that the
vector cosine of the angle between two vectors
is given by their
inner product divided by the product of their
norms
[
451].)
To derive Eq.

(
B.16), let's begin with the cross-product in
matrix
form as

using the first matrix form in the
third line of the cross-product definition in Eq.

(
B.15) above. Then
where
![$ \mathbf{E}=[\underline{e}_1,\underline{e}_2,\underline{e}_3]$](http://www.dsprelated.com/josimages_new/pasp/img2823.png)
denotes the identity matrix in

,

denotes the
orthogonal-
projection matrix onto

[
451],

denotes the projection matrix onto
the orthogonal
complement of

,

denotes the component of

orthogonal to

, and we used the fact that
orthogonal projection matrices

are
idempotent (
i.e.,

) and
symmetric (when real, as we have here) when we replaced

by

above. Finally,
note that the length of

is

, where

is the angle
between the 1D subspaces spanned by

and

in the plane
including both vectors. Thus,
which establishes the desired result:
Moreover, this proof gives an appealing geometric interpretation of
the
vector cross-product

as having magnitude given by
the product of

times the norm of the difference
between

and the
orthogonal projection of

onto

(

)
or vice versa
(

).
In this geometric picture it is clear that the cross-product magnitude
is maximized when the vectors are orthogonal, and it is zero when the
vectors are collinear. It is ``length times orthogonal length.''
The direction of the cross-product vector is then taken to be
orthogonal to both

and

according to the right-hand
rule. This
orthogonality can be checked by verifying that

. The right-hand-rule parity can be checked by
rotating the space so that
![$ \underline{x}'=[ \vert\vert\,\underline{x}\,\vert\vert ,0,0]^T$](http://www.dsprelated.com/josimages_new/pasp/img2838.png)
and
![$ \underline{y}'=[ \vert\vert\,\underline{y}\,\vert\vert \cos(\theta), \vert\vert\,\underline{y}\,\vert\vert \sin(\theta),0]^T$](http://www.dsprelated.com/josimages_new/pasp/img2839.png)
in
which case

. Thus, the cross
product points ``up'' relative to the

plane for

and ``down'' for

.
In Eq.

(
B.14) above, the
mass moment of inertia was expressed
in terms of
orthogonal projection as

, where

. In terms of the
vector cross
product, we can now express it as
where

is the distance from the rotation
axis out to the point

(which equals the length of the vector

).
Referring again to Fig.
B.4, we can write the
tangential velocity vector

as a
vector cross product of
the
angular-velocity vector

(§
B.4.11) and the position
vector

:
 |
(B.17) |
To see this, let's first check its direction and then its magnitude.
By the right-hand rule,

points up out of the page in
Fig.
B.4. Crossing that with

, again by the right-hand
rule, produces a tangential velocity vector

pointing as shown
in the figure. So, the direction is correct. Now, the magnitude:
Since

and

are mutually
orthogonal, the angle between them
is

, so that, by Eq.

(
B.16),
as desired.
The
angular momentum of a
mass 
rotating in a circle of
radius

with
angular velocity 
(rad/s), is defined by
where

denotes the
mass moment of inertia of the rigid
body (§
B.4.4).
Recall (§
B.3) that the
momentum of a
mass 
traveling
with
velocity 
in a straight line is given by
while the
angular momentum of a point-mass

rotating along a circle
of radius

at

rad/s is given by
where

.
The
tangential speed of the mass along the circle of radius

is given by
Expressing the angular momentum

in terms of

gives
 |
(B.18) |
Thus, the angular momentum

is

times the linear momentum

.
Linear momentum can be viewed as a renormalized special case of
angular momentum in which the radius of rotation goes to infinity.
Like
linear momentum, angular
momentum is fundamentally a vector in

. The definition of the previous section suffices when the
direction does not change, in which case we can focus only on its
magnitude

.
More generally, let

denote the 3-space coordinates
of a point-
mass 
, and let

denote its
velocity
in

. Then the
instantaneous angular momentum vector
of the mass relative to the origin (not necessarily rotating about a
fixed axis) is given by
 |
(B.19) |
where

denotes the
vector cross product, discussed in
§
B.4.12 above. The identity

was discussed
at Eq.

(
B.17).
For the special case in which

is
orthogonal to

, as in Fig.
B.4, we have that

points, by the right-hand rule, in the direction of the
angular
velocity vector

(up out of the page), which is
satisfying. Furthermore, its magnitude is given by
which agrees with the
scalar case.
In the more general case of an arbitrary mass velocity vector

, we know from §
B.4.12 that the magnitude of

equals the product of the distance from the axis
of rotation to the mass,
i.e.,

, times the length of
the component of

that is orthogonal to

,
i.e.,

, as needed.
It can be shown that vector angular momentum, as defined, is
conserved.
B.22 For
example, in an orbit, such as that of the moon around the earth, or
that of Halley's comet around the sun, the orbiting object speeds up
as it comes closer to the object it is orbiting. (See
Kepler's laws
of planetary motion.) Similarly, a spinning ice-skater spins faster
when pulling in arms to reduce the
moment of inertia about the spin
axis. The conservation of angular momentum can be shown to result
from the principle of least action and the isotrophy of space
[
270, p. 18].
The two
cross-products in Eq.

(
B.19) can be written out with the help
of the
vector analysis identity
B.23
This (or a direct calculation) yields, starting with Eq.

(
B.19),
where
with

, and

, for

. That is,
 |
|
|
(B.21) |
The
matrix

is the Cartesian representation of the
mass
moment of inertia tensor, which will be explored further in
§
B.4.15 below.
The vector
angular momentum of a rigid body is obtained by summing the
angular
momentum of its constituent mass particles. Thus,
Since

factors out of the sum, we see that the mass moment of
inertia tensor for a rigid body is given by the sum of the mass
moment
of inertia tensors for each of its component mass particles.
In summary, the angular momentum vector

is given by the mass
moment of inertia tensor

times the
angular-velocity vector

representing the axis of rotation.
Note that the angular momentum vector

does
not in general
point in the same direction as the
angular-velocity vector

. We
saw above that it does in the special case of a point mass traveling
orthogonal to its position vector. In general,

and

point
in the same direction whenever

is an
eigenvector of

, as will be discussed further below (§
B.4.16). In this
case, the rigid body is said to be
dynamically balanced.
B.24
As derived in the previous section, the
moment of inertia
tensor, in 3D Cartesian coordinates, is a three-by-three
matrix

that can be multiplied by any
angular-velocity vector to
produce the corresponding
angular momentum vector for either a point
mass or a rigid mass distribution. Note that the origin of the
angular-velocity vector

is always fixed at

in the space
(typically located at the
center of mass). Therefore, the
moment of
inertia tensor

is defined relative to that origin.
The moment of inertia tensor can similarly be used to compute the
mass moment of inertia for any normalized angular
velocity
vector

as
 |
(B.22) |
Since rotational energy is defined as

(see
Eq.

(
B.7)), multiplying Eq.

(
B.22) by

gives the following
expression for the rotational
kinetic energy in terms of the moment of
inertia tensor:
 |
(B.23) |
We can show Eq.

(
B.22) starting from Eq.

(
B.14). For a
point-mass

located at

, we have
where again

denotes the three-by-three identity matrix, and
 |
(B.24) |
which agrees with Eq.

(
B.20). Thus we have derived the moment of
inertia

in terms of the moment of inertia tensor

and the
normalized angular velocity

for a point-mass

at

.
For a collection of

masses

located at

, we
simply sum over their masses to add up the moments of inertia:
Finally, for a continuous mass distribution, we integrate as usual:
where

is the total mass.
Consider a
mass 
at
![$ \underline{x}=[x,0,0]^T$](http://www.dsprelated.com/josimages_new/pasp/img2890.png)
. Then the mass
moment of inertia
tensor is
For the
angular-velocity vector
![$ \underline{\omega}=[\omega,0,0]^T$](http://www.dsprelated.com/josimages_new/pasp/img2892.png)
, we obtain the
moment of
inertia
This makes sense because the axis of rotation passes through the point
mass, so the moment of inertia
should be zero about that
axis. On the other hand, if we look at
![$ \underline{\omega}=[0,1,0]^T$](http://www.dsprelated.com/josimages_new/pasp/img2894.png)
, we get
which is what we expected.
Now let the
mass 
be located at
![$ \underline{x}=[1,1,0]^T$](http://www.dsprelated.com/josimages_new/pasp/img2896.png)
so that
We expect
![$ \underline{\omega}=[1,1,0]$](http://www.dsprelated.com/josimages_new/pasp/img2898.png)
to yield zero for the
moment of inertia, and
sure enough

. Similarly, the vector
angular
momentum is zero, since

.
For
![$ \underline{\omega}=[1,0,0]^T$](http://www.dsprelated.com/josimages_new/pasp/img2901.png)
, the result is
which makes sense because the distance from the axis

to

is one. The same result is obtained for rotation about

.
For

, however, the result is

, as
expect.
This all makes sense, but what about those

off-diagonal terms in

? Consider the vector
angular momentum (§
B.4.14):
We see that the off-diagonal terms

correspond to a
coupling of rotation about

with rotation about

.
That is, there is a component of
moment-of-inertia 
that is
contributed (or subtracted, as we saw above for
![$ \underline{\omega}=[1,1,0]^T$](http://www.dsprelated.com/josimages_new/pasp/img2910.png)
) when
both 
and

are nonzero. These cross-terms can be eliminated by
diagonalizing the
matrix [
449],
B.25as discussed further in the next section.
Principal Axes of Rotation
A
principal axis of rotation (or
principal direction) is
an
eigenvector of the
mass moment of inertia tensor (introduced
in the previous section) defined relative to some point (typically the
center of mass). The corresponding
eigenvalues are called the
principal moments of inertia.
Because the moment of
inertia tensor is defined relative to the point

in the space, the principal axes all pass through that point
(usually the center of mass).
As derived above (§
B.4.14), the
angular momentum vector is given by
the moment of inertia tensor times the
angular-velocity vector:
If

is an eigenvector of

, then we have
where the (
scalar) eigenvalue

is called a
principal moment of
inertia. If we set the rigid body assocated with

rotating
about the axis

, then

is the mass moment of inertia of
the body for that rotation. As will become clear below, there are
always three mutually
orthogonal principal axes of rotation, and three
corresponding principal moments of inertia (in 3D space, of course).
From the form of the
moment of inertia tensor introduced in Eq.

(
B.24)
it is clear that

is
symmetric. Moreover, for any normalized
angular-velocity vector

we have
since

is unit length, and projecting it onto any other vector
can only shorten it or leave it unchanged. That is,

, with equality occurring for

for any nonzero

. Zooming out,
of course we expect any
moment of inertia 
for a positive
mass 
to be nonnegative. Thus,

is
symmetric
nonnegative definite. If furthermore

and

are not
collinear,
i.e., if there is any nonzero angle between them, then

is
positive definite (and

). As is well known in
linear algebra [
329], real, symmetric, positive-definite
matrices have
orthogonal eigenvectors and
real, positive
eigenvalues. In this context, the orthogonal eigenvectors are
called the
principal axes of rotation. Each corresponding
eigenvalue is the moment of inertia about that principal axis--the
corresponding principal moment of inertia. When
angular velocity
vectors

are expressed as a
linear combination of the principal
axes, there are no cross-terms in the moment of
inertia tensor--no
so-called
products of inertia.
The three principal axes are
unique when the eigenvalues of

(principal moments of inertia) are
distinct. They are
not unique when there are repeated eigenvalues, as in the example
above of a disk rotating about any of its diameters
(§
B.4.4). In that example, one principal
axis, the one corresponding to eigenvalue

, was

(
i.e.,
orthogonal to the disk and passing through its center), while any two
orthogonal diameters in the plane of the disk may be chosen as the
other two principal axes (corresponding to the repeated eigenvalue

).
Symmetry of the rigid body about any axis

(passing through the
origin) means that

is a principal direction. Such a symmetric
body may be constructed, for example, as a
solid of
revolution.
B.26In rotational dynamics, this case is known as the
symmetric top
[
270]. Note that the
center of mass will lie
somewhere along an axis of symmetry. The other two principal axes can
be arbitrarily chosen as a mutually orthogonal pair in the (circular)
plane orthogonal to the

axis, intersecting at the

axis. Because of the circular symmetry about

, the two
principal moments of inertia in that plane are equal. Thus the moment
of inertia tensor can be
diagonalized to look like
where

is the principal moment of inertia about

, and

is the (twice repeated) principal moment of inertia about the
two axes in the circular-symmetry plane. We saw in §
B.4.5
(
Perpendicular Axis theorem) that if the mass distribution is planar,
then

.
If a point-
mass is located at

and is rotating about an
axis-of-rotation

with
angular velocity 
, then the
distance from the rotation axis to the mass is

,
or, in terms of the
vector cross product,

. The tangential
velocity of the mass is
then

, so that the kinetic energy can be expressed as
(
cf. Eq.

(
B.23))
 |
(B.25) |
where
In a collection of

masses

having velocities

, we of
course sum the individual kinetic energies to get the total kinetic
energy.
Finally, we may also write the rotational kinetic energy as half the
inner product of the
angular-velocity vector and the
angular-momentum vector:
B.27
where the second form (introduced above in Eq.

(
B.7)) derives from
the vector-dot-product form by using Eq.

(
B.20) and Eq.

(
B.22)
to establish that

.
Figure B.7:
Application of torque
about the
origin given by a tangential force
on a lever arm of length
.
![\includegraphics[width=1.1in]{eps/torque}](http://www.dsprelated.com/josimages_new/pasp/img2937.png) |
When twisting things, the rotational force we apply about the center
is called a
torque (or
moment, or
moment of
force). Informally, we think of the torque as the
tangential
applied force 
times the
moment arm (length of the
lever arm)
 |
(B.26) |
as depicted in Fig.
B.7. The moment arm is the distance from the
applied force to the point being twisted. For example, in the case of
a wrench turning a bolt,

is the force applied at the end of the
wrench by one's hand,
orthogonal to the wrench, while the moment arm

is the length of the wrench. Doubling the length of the wrench
doubles the torque. This is an example of
leverage. When

is increased, a given twisting angle

is spread out over a
larger arc length

, thereby reducing the tangential force

required to assert a given torque

.
For more general applied forces

, we may compute the
tangential component

by
projecting

onto the
tangent direction. More precisely, the
torque 
about the
origin

applied at a point

may be defined by
 |
(B.27) |
where

is the applied force (at

) and

denotes the
cross product, introduced above in §
B.4.12.
Note that the torque vector

is orthogonal to both the lever
arm and the tangential-force direction. It thus points in the
direction of the
angular velocity vector (along the axis of rotation).
The torque magnitude is
where

denotes the angle from

to

. We can
interpret

as the length of the projection of

onto the tangent direction (the line orthogonal to

in the
direction of the force), so that we can write
where

, thus getting back
to Eq.

(
B.26).
Newton's Second Law
for Rotations
The rotational version of
Newton's law 
is
 |
(B.28) |
where

denotes the
angular
acceleration. As in the previous section,

is
torque
(tangential
force 
times a moment arm

), and

is the
mass moment of inertia. Thus, the net applied torque

equals the time derivative of
angular momentum 
, just as
force

equals the time-derivative of
linear momentum 
:
To show that Eq.

(
B.28) results from
Newton's second law

,
consider again a mass

rotating at a distance

from an axis
of rotation, as in §
B.4.3 above, and
let

denote a
tangential force on the mass, and

the corresponding tangential acceleration. Then we have, by Newton's
second law,
Multiplying both sides by

gives
where we used the definitions

and

.
Furthermore, the left-hand side is the definition of
torque 
.
Thus, we have derived
from Newton's second law

applied to the tangential force

and acceleration

of the mass

.
In summary, force equals the time-derivative of linear
momentum, and
torque equals the time-derivative of angular momentum. By Newton's
laws, the time-derivative of linear momentum is mass times
acceleration, and the time-derivative of angular momentum is the mass
moment of inertia times angular acceleration:
Equations of
Motion for Rigid Bodies
We are now ready to write down the general equations of motion for
rigid bodies in terms of

for the
center of mass and

for the rotation of the body about its center of
mass.
As discussed above, it is useful to decompose the motion of a rigid
body into
- (1)
- the linear velocity
of its center of mass, and
- (2)
- its angular velocity
about its center of mass.
The linear motion is governed by
Newton's second law

, where

is the total mass,

is the
velocity of the center-of-mass, and

is the sum of all external
forces on the rigid body. (Equivalently,

is the sum of the
radial force components pointing toward or away from the center of
mass.) Since this is so straightforward, essentially no harder than
dealing with a point mass, we will not consider it further.
The angular motion is governed the
rotational version of
Newton's second law introduced in §
B.4.19:
 |
(B.29) |
where

is the vector
torque defined in Eq.

(
B.27),

is
the
angular momentum,

is the mass
moment of inertia tensor,
and

is the
angular velocity of the rigid body about its center
of mass. Note that if the center of mass is moving, we are in a
moving coordinate system moving with the center of mass (see next
section). We may call

the
intrinsic momentum of the
rigid body,
i.e., that in a coordinate system moving with the center of
the mass. We will translate this to the non-moving coordinate system
in §
B.4.20 below.
The driving torque

is given by the
resultant moment of
the external forces, using Eq.

(
B.27) for each external force to
obtain its contribution to the total moment. In other words, the
external moments (tangential forces times moment arms) sum up for the
net torque just like the radial force components summed to produce the
net driving force on the center of mass.
Body-Fixed and Space-Fixed Frames of Reference
Rotation is always about some (instantaneous) axis of rotation that is
free to change over time. It is convenient to express rotations in a
coordinate system having its origin (

) located at the
center-of-mass of the rigid body (§
B.4.1), and its coordinate axes
aligned along the principal directions for the body (§
B.4.16).
This
body-fixed frame then moves within a stationary
space-fixed frame (or ``star frame'').
In Eq.

(
B.29) above, we wrote down Newton's second law for angular
motion in the
body-fixed frame,
i.e., the coordinate system
having its origin at the center of
mass. Furthermore, it is simplest
(

is diagonal) when its axes lie along principal directions
(§
B.4.16).
As an example of a local body-fixed coordinate system, consider a
spinning top. In the body-fixed frame, the ``vertical'' axis
coincides with the top's axis of rotation (spin). As the top loses
rotational
kinetic energy due to
friction, the top's rotation-axis
precesses around a circle, as observed in the space-fixed
frame. The other two body-fixed axes can be chosen as any two
mutually
orthogonal axes intersecting each other (and the spin axis)
at the center of mass, and lying in the plane orthogonal to the spin
axis. The space-fixed frame is of course that of the outside
observer's inertial frame
B.28in which the top is spinning.
Angular Motion
in the Space-Fixed Frame
Let's now consider angular motion in the presence of linear motion of
the
center of mass. In general, we have [
270]
where the sum is over all
mass particles in the rigid body, and

denotes the vector
linear momentum for each particle. That
is, the
angular momentum is given by the tangential component of the
linear
momentum times the associated moment arm. Using the
chain rule
for differentiation, we find
However,

, so that
which is the sum of moments of all external
forces.
Euler's Equations
for Rotations in the Body-Fixed Frame
Suppose now that the body-fixed frame is rotating in the space-fixed
frame with
angular velocity

. Then the total
torque on the rigid
body becomes [
270]
 |
(B.30) |
Similarly, the total external
forces on the
center of mass become
If the body-fixed frame is aligned with the principal axes of rotation
(§
B.4.16), then the
mass moment of inertia tensor is diagonal, say

diag

. In this frame, the
angular momentum is simply
so that the term

becomes (
cf. Eq.

(
B.15))
Substituting this result into Eq.

(
B.30), we obtain the following
equations of angular motion for an object rotating in the body-fixed
frame defined by its three principal axes of rotation:
These are call
Euler's
equations:B.29Since these equations are in the body-fixed frame,

is the mass
moment of inertia about principal axis

, and

is the
angular
velocity about principal axis

.
For a uniform sphere, the cross-terms disappear and the
moments of
inertia are all the same, leaving

, for

.
Since any three
orthogonal vectors can serve as
eigenvectors of the
moment of inertia tensor, we have that, for a uniform sphere, any
three orthogonal axes can be chosen as principal axes.
For a cylinder that is not spinning about its axis, we similarly
obtain two uncoupled equations

, for

, given

(no spin). Note, however, that if we replace the
circular cross-section of the cylinder by an
ellipse, then

and there is a coupling term that drives

(unless

happens to cancel it).
Properties of Elastic Solids
Young's modulus can be thought of as the
spring constant
for solids. Consider an ideal
rod (or bar) of length

and
cross-sectional area 
. Suppose we apply a
force 
to the face of
area

, causing a
displacement 
along the axis of the rod.
Then Young's modulus

is given by
where
For wood, Young's modulus

is on the order of

N/m

.
For aluminum, it is around

(a bit higher than glass which is near

), and structural steel has

[
180].
Recall (§
B.1.3) that
Hooke's Law defines a
spring
constant

as the applied
force 
divided by the spring
displacement 
, or

. An elastic solid can be viewed as a
bundle of ideal springs. Consider, for example, an
ideal
bar (a rectangular solid in which one dimension, usually its
longest, is designated its length

), and consider compression by

along the length dimension. The length of each spring in
the bundle is the length of the bar, so that each spring constant

must be inversely proportional to

; in particular, each doubling of
length

doubles the length of each ``spring'' in the bundle, and
therefore halves its stiffness. As a result, it is useful to
normalize displacement

by length

and use
relative
displacement

. We need displacement per unit length
because we have a constant spring compliance per unit length.
The
number of springs in parallel is proportional to the
cross-sectional area 
of the bar. Therefore, the force applied to
each spring is proportional to the total applied force

divided by
the cross-sectional area

. Thus,
Hooke's law for each spring in the
bundle can be written
where

is
Young's modulus.
We may say that Young's modulus is the Hooke's-law spring constant for
the spring made from a specifically cut section of the solid material,
cut to length 1 and cross-sectional area 1. The shape of the
cross-sectional area does not matter since all displacement is assumed
to be longitudinal in this model.
String Tension
The
tension of a
vibrating string is the
force 
used
to stretch it. It is therefore directed along the axis of the string.
A force

must be applied at the endpoint on the right, and a force

is applied at the endpoint on the left. Each point interior to
the string is pulled equally to the left and right,
i.e., the net force on an interior point is

. (A nonzero
force on a massless point would produce an infinite acceleration.)
If the
cross-sectional area of the string is

, then the tension is
given by the stress on the string times

.
Consider an elastic string under tension which is at rest along the

dimension. Let

,

, and

denote the unit vectors in
the

,

, and

directions, respectively. When a wave is
present, a point

originally at

along the string is
displaced to some point

specified by the
displacement
vector
Note that typical derivations of the
wave equation consider only the
displacement

in the

direction. This more general treatment
is adapted from [
122]. An alternative clear
development is given in [
391].
The displacement of a neighboring point originally at

along the string can be specified as
Let

denote string tension along

when the string is at rest, and

denote the vector tension at the point

in the present displaced
scenario under analysis. The net vector
force acting on the infinitesimal
string element between points

and

is given by the vector sum of
the force

at

and the force

at

, that is,

. If the string
has stiffness, the two forces will in general not be tangent to the string
at these points. The
mass of the infinitesimal string element is

,
where

denotes the mass per unit length of the string at rest. Applying
Newton's second law gives
 |
(B.31) |
where

has been canceled on both sides of the equation. Note
that no approximations have been made so far.
The next step is to express the force

in terms of the tension

of the string at rest, the elastic constant of the string, and
geometrical factors. The displaced string element

is the
vector
having magnitude
 |
(B.34) |
Let's now assume the string is perfectly flexible (zero stiffness) so
that the direction of the
force vector

is given by the unit
vector

tangent to the string. (To accommodate
stiffness, it would be necessary to include a force component at right
angles to the string which depends on the curvature and stiffness of
the string.) The magnitude of

at any position is the rest
tension

plus the incremental tension needed to stretch it the
fractional amount
If

is the constant
cross-sectional area of the string, and

is the
Young's modulus (stress/strain--the ``
spring constant''
for solids--see §
B.5.1), then
so that
![$\displaystyle \mathbf{K}= \left[K+ SY\left(\frac{ds}{dx} - 1\right)\right]\frac{d{\bf s}}{ds}$](http://www.dsprelated.com/josimages_new/pasp/img3016.png) |
(B.35) |
where no
geometrical limitations have yet been placed on the
magnitude of

and

, other than to prevent the
string from being stretched beyond its elastic limit.
The four equations (
B.31) through (
B.35) can be combined
into a single
vector wave equation that expresses the
propagation of waves on the string having three
displacement
components. This
differential equation is
nonlinear, so that
superposition no longer holds. Furthermore, the three
displacement components of the wave are
coupled together at all
points along the string, so that the
wave equation is no longer
separable into three independent
1D wave equations.
To obtain a linear, separable
wave equation, it is necessary to assume
that the
strains

,

, and

be
small compared with unity. This is the same assumption
(

) necessary to derive the
usual
wave equation for
transverse vibrations only in the

-

plane.
When (
B.35) is expanded into a
Taylor series in the strains,
and when only the first-order terms are retained, we obtain
 |
(B.36) |
This is the
linearized wave equation for the string, based only
on the assumptions of elasticity of the string, and strain magnitudes
much less than unity. Using this linearized equation for the force

, it is found that (
B.31) separates into the three wave
equations
where

is the
longitudinal wave velocity, and

is the
transverse wave velocity.
In summary, the two
transverse wave components and the longitudinal
component may be considered
independent (
i.e., ``superposition''
holds with respect to vibrations in these three dimensions of
vibration) provided powers higher than 1 of the strains (relative
displacement) can be neglected,
i.e.,

and
The physical forward
momentum carried by a
transverse wave
along a string is conveyed by a secondary
longitudinal wave
[
391].
A less simplified
wave equation which
supports
longitudinal wave momentum is given by [
391, Eqns. 38ab]
where

and

denote longitudinal and
transverse
displacement, respectively, and the commonly used ``dot'' and
``prime'' notation for partial derivatives has been introduced,
e.g.,
(See also Eq.

(
C.1).) We see that the term

in the first equation above
provides a mechanism for
transverse waves to ``drive'' the generation
of longitudinal waves. This coupling cannot be neglected if momentum
effects are desired.
Physically, the rising edge of a transverse wave generates a
longitudinal displacement in the direction of wave travel that
propagates ahead at a much higher speed (typically an order of
magnitude faster). The falling edge of the transverse wave then
cancels this forward displacement as it passes by. See
[
391] for further details (including computer
simulations).
Properties of Gases
Particle Velocity of a Gas
The
particle velocity of a gas flow at any point can be defined
as the average velocity (in
meters per second, m/s) of the air
molecules passing through a plane cutting
orthogonal to the flow. The
term ``velocity'' in this book, when referring to air, means
``particle velocity.''
It is common in acoustics to denote particle velocity by lower-case

.
The
volume velocity 
of a gas flow is defined as particle
velocity

times the
cross-sectional area 
of the flow, or
where

denotes position along the flow, and

denotes time in
seconds. Volume velocity is thus in physical units of volume per
second (m

/s).
When a flow is confined within an enclosed channel, as it is in an
acoustic tube,
volume velocity is conserved when the tube
changes cross-sectional area, assuming the density

remains
constant. This follows directly from conservation of
mass in a flow:
The total mass passing a given point

along the flow is given by
the mass density

times the integral of the volume volume
velocity at that point, or
As a simple example, consider a constant flow through two cylindrical
acoustic tube sections having cross-sectional areas

and

,
respectively. If the particle velocity in cylinder 1 is

, then
the particle velocity in cylinder 2 may be found by solving
for

.
It is common in the field of acoustics to denote volume velocity by an
upper-case

. Thus, for the two-cylinder acoustic tube example above,
we would define

and

, so that
would express the conservation of volume velocity from one tube
segment to the next.
According the
kinetic theory of ideal gases
[
180],
air pressure can be defined
as the
average momentum transfer per unit area per unit time
due to molecular collisions between a confined gas and its boundary.
Using Newton's second law, this pressure can be shown to be given by
one third of the average kinetic energy of molecules in the gas.
Here,

denotes the average squared particle
velocity in
the gas. (The constant

comes from the fact that we are
interested only in the kinetic energy directed along one dimension in
3D space.)
Proof: This is a classical result from the
kinetic theory of gases
[
180]. Let

be the total
mass of a gas
confined to a rectangular volume

, where

is the area of
one side and

the distance to the opposite side. Let

denote the average molecule velocity in the

direction. Then the
total net molecular momentum in the

direction is given by

. Suppose the momentum

is directed
against a face of area

. A rigid-wall
elastic collision by a mass

traveling into the wall at velocity

imparts a momentum of
magnitude

to the wall (because the momentum of the mass is
changed from

to

, and momentum is conserved).
The average momentum-transfer per unit area is therefore

at any instant in time. To obtain the definition of pressure, we need
only multiply by the average collision rate, which is given by

. That is, the average

-velocity divided by the
round-trip distance along the

dimension gives the collision rate
at either wall bounding the

dimension. Thus, we obtain
where

is the density of the gas in mass per unit volume.
The quantity

is the average kinetic energy density of
molecules in the gas along the

dimension. The total kinetic
energy density is

, where

is the average molecular
velocity magnitude of the gas. Since the gas pressure must be the
same in all directions, by symmetry, we must have

, so that
Bernoulli Equation
In an ideal
inviscid, incompressible flow, we have, by
conservation of energy,

constant
where
This basic energy conservation law was published in 1738 by Daniel
Bernoulli in his classic work
Hydrodynamica.
From §
B.7.3, we have that the
pressure of a gas is
proportional to the average
kinetic energy of the molecules making up
the gas. Therefore, when a gas flows at a constant height

, some
of its ``pressure kinetic energy'' must be given to the kinetic energy
of the flow as a whole. If the mean height of the flow changes, then
kinetic energy trades with
potential energy as well.
Bernoulli Effect
The
Bernoulli effect provides that, when a gas such as air
flows, its
pressure drops. This is the basis for how aircraft wings
work: The cross-sectional shape of the wing, called an
aerofoil
(or
airfoil),
forces air to follow a longer path over the top
of the wing, thereby speeding it up and creating a net upward force
called
lift.
Figure B.8:
Illustration of the Bernoulli effect in
an acoustic tube.
 |
Figure
B.8 illustrates the Bernoulli effect for the case
of a reservoir at constant pressure

(``mouth pressure'') driving
an acoustic tube. Any flow inside the ``mouth'' is neglected. Within the
acoustic channel, there is a flow with constant particle
velocity 
.
To conserve energy, the pressure within the acoustic channel must drop
down to

. That is, the flow
kinetic energy subtracts
from the pressure kinetic energy within the channel.
For a more detailed derivation of the Bernoulli effect, see,
e.g.,
[
179]. Further discussion of its relevance
in musical acoustics is given in [
144,
197].
Referring again to Fig.
B.8, the gas flow exiting the
acoustic tube is shown as forming a
jet. The jet ``carries its
own
pressure'' until it dissipates in some form, such as any
combination of the following:
Pressure recovery refers to the
conversion of flow
kinetic energy back to pressure kinetic energy. In
situations such as the one shown in Fig.
B.8,
the flow itself is
driven by
the pressure drop between the confined
reservoir (pressure

) and the outside air (pressure

). Therefore, any pressure recovery would erode the
pressure drop and hence the flow
velocity 
.
For a summary of more advanced aeroacoustics, including consideration
of vortices, see [
196]. In addition, basic textbooks on
fluid mechanics are relevant [
171].
Acoustic intensity may be defined by
where
For a
plane traveling wave, we have
where
is called the
wave impedance of air, and
Therefore, in a
plane wave,
Acoustic Energy Density
The two forms of energy in a wave are
kinetic and
potential. Denoting them at a particular time

and position

by

and

, respectively, we can write them in
terms of
velocity 
and
wave impedance 
as follows:
More specifically,

and

may be called the
acoustic kinetic
energy density and the
acoustic potential energy density, respectively.
At each point in a
plane wave, we have

(
pressure equals wave-
impedance times velocity), and so
where

denotes the acoustic
intensity (pressure times velocity) at time

and position

.
Thus, half of the acoustic intensity

in a plane wave is kinetic,
and the other half is potential:
B.30
Note that acoustic intensity

has units of
energy per unit
area per unit time while the acoustic energy density

has
units of
energy per unit volume.
Energy Decay through Lossy Boundaries
Since the acoustic energy density

is the energy per unit
volume in a 3D sound field, it follows that the total energy of the
field is given by integrating over the volume:
In reverberant rooms and other acoustic systems, the field energy
decays over time due to losses. Assuming the losses occur only at the
boundary of the volume, we can equate the rate of total-energy change
to the rate at which energy exits through the boundaries. In other
words, the energy lost by the volume

in time interval

must equal the acoustic
intensity

exiting the volume,
times

(approximating

as constant between times

and

):
The term

is the dot-product of the (vector)
intensity

with a unit-vector

chosen to be normal to the
surface at position

along the surface. Thus,

is
the component of the acoustic intensity

exiting the volume
normal to its surface. (The tangential component does not exit.)
Dividing through by

and taking a limit as

yields the following conservation law, originally published by
Kirchoff in 1867:
Thus, the rate of change of energy in an ideal acoustic volume

is
equal to the surface integral of the power crossing its boundary. A
more detailed derivation appears in [
349, p. 37].
Sabine's theory of acoustic energy decay in reverberant room
impulse
responses can be derived using this conservation relation as a
starting point.
The
ideal gas law can be written as
 |
(B.45) |
where
The alternate form

comes from the
statistical
mechanics derivation in which

is the number of gas molecules in
the volume, and

is
Boltzmann's constant. In this
formulation (the
kinetic theory of ideal gases), the
average kinetic
energy of the gas molecules is given by

. Thus,
temperature is proportional to average kinetic energy of the
gas molecules, where the kinetic energy of a molecule

with
translational speed

is given by

.
In an ideal gas, the molecules are like little rubber balls (or
rubbery assemblies of rubber balls) in a weightless vacuum, colliding
with each other and the walls elastically and losslessly (an ``ideal
rubber''). Electromagnetic
forces among the molecules are neglected,
other than the electron-orbital repulsion producing the
elastic
collisions; in other words, the molecules are treated as electrically
neutral far away. (Gases of ionized molecules are called
plasmas.)
The
mass 
of the gas in volume

is given by

, where

is
the
molar mass of the gass (about 29 g per mole for air). The
air density is thus

so that we can write
That is,
pressure 
is proportional to density

at constant
temperature

(with

being a constant).
We normally do not need to consider the (
nonlinear)
ideal gas law in
audio acoustics because it is usually
linearized about some
ambient pressure

. The physical pressure is then

, where

is the usual acoustic pressure-
wave variable. That is, we are
only concerned with small pressure
perturbations 
in typical
audio acoustics situations, so that, for example, variations in volume

and density

can be neglected. Notable exceptions include
brass instruments which can achieve nonlinear
sound-pressure regions,
especially near the mouthpiece [
198,
52].
Additionally, the aeroacoustics of air
jets is nonlinear
[
196,
530,
531,
532,
102,
101].
If air compression/expansion were
isothermal (constant
temperature 
), then, according to the
ideal gas law 
, the
pressure 
would simply be proportional to density

. It turns
out, however, that
heat diffusion is much slower than audio acoustic
vibrations. As a result, air compression/expansion is much closer to
isentropic (constant
entropy 
) in normal acoustic
situations. (An isentropic process is also called a
reversible
adiabatic process.) This means that when air is compressed by
shrinking its volume

, for example, not only does the pressure

increase (§
B.7.3), but the temperature

increases as
well (as quantified in the next section). In a constant-entropy
compression/expansion, temperature changes are not given time to
diffuse away to thermal equilibrium. Instead, they remain largely
frozen in place. Compressing air heats it up, and relaxing the
compression cools it back down.
The relative amount of compression/expansion energy that goes into
temperature 
versus
pressure 
can be characterized by the
heat capacity ratio
where

is the
specific heat (also called
heat
capacity) at constant pressure, while

is the specific heat at
constant volume. The
specific heat, in turn, is the amount of
heat required to raise the temperature of the gas by one degree. It
is derived in
statistical thermodynamics [
138]
that, for an
ideal gas, we have

, where

is the
ideal
gas constant (introduced in Eq.

(
B.45)). Thus,

for any
ideal gas. The extra heat absorption that occurs when heating a gas
at constant pressure is associated with the
work (§
B.2)
performed on the volume boundary (fore times distance = pressure times
area times distance) as it expands to keep pressure constant. Heating
a gas at constant volume involves increasing the
kinetic energy of the
molecules, while heating a gas at constant pressure involves both that
and pushing the boundary of the volume out. The reason not all
gases have the same

is that they have different
internal degrees of freedom, such as those associated with
spinning and vibrating internally. Each degree of freedom can store
energy.
In terms of

, we have
 |
(B.46) |
where

for dry air at normal temperatures. Thus,
if a volume of ideal gas is changed from

to

, the pressure
change is given by
and the temperature change is
These equations both follow from Eq.

(
B.46) and the
ideal gas law
Eq.

(
B.45).
The value

is typical for any
diatomic
gas.
B.31 Monatomic inert gases, on the other hand,
such as Helium, Neon, and Argon, have

.
Carbon
dioxide, which is
triatomic, has a
heat capacity ratio

. We see that more complex molecules have lower

values because they can store heat in more degrees of freedom.
In statistical thermodynamics [
175,
138],
it is derived that each molecular degree of freedom contributes

to the molar
heat capacity of an
ideal gas, where again

is the
ideal
gas constant.
An ideal
monatomic gas molecule (negligible spin) has only
three degrees of freedom: its
kinetic energy in the three spatial
dimensions. Therefore,

. This means we expect
a result that agrees well with experimental measurements
[
138].
For an ideal
diatomic gas molecule such as air, which can be
pictured as a ``bar bell'' configuration of two rubber balls, two
additional degrees of freedom are added, both associated with spinning
the molecule about an axis
orthogonal to the line connecting the
atoms, and piercing its
center of mass. There are two such
axes. Spinning about the connecting axis is neglected because the
moment of inertia is so much smaller in that case. Thus, for diatomic
gases such as dry air, we expect
as observed to a good degree of approximation at normal
temperatures.
At high temperatures, new degrees of freedom appear associated with
vibrations in the molecular bonds. (For example, the ``bar bell'' can
vibrate longitudinally.) However, such vibrations are ``frozen out''
at normal room temperatures, meaning that their (quantized) energy
levels are too high and spaced too far apart to be excited by room
temperature collisions [
138, p. 147].
B.32
The
speed of sound in a gas depends primarily on the
temperature, and can be estimated using the following formula
from the
kinetic theory of gases:
B.33
where, as discussed in the previous section, the
adiabatic gas
constant is

for dry air,

is the
ideal gas
constant for air in
meters-squared per second-squared per
degrees-
Kelvin-squared, and

is
absolute temperature in degrees
Kelvin (which equals degrees
Celsius + 273.15). For example, at zero
degrees Celsius (32 degrees
Fahrenheit), the speed of sound is
calculated to be 1085.1 feet per second. At 20 degrees Celsius, we
get 1124.1 feet per second.
Air Absorption
This section provides some further details regarding acoustic air
absorption [
318]. For a
plane wave, the decline of
acoustic
intensity as a function of
propagation distance

is given
by
where
Tables
B.1 and
B.2 (adapted from
[
314]) give some typical values for air.
Table B.1:
Attenuation constant
(in inverse
meters) at 20 degrees Celsius and standard atmospheric pressure
Relative |
Frequency in Hz |
Humidity |
1000 |
2000 |
3000 |
4000 |
40 |
0.0013 |
0.0037 |
0.0069 |
0.0242 |
50 |
0.0013 |
0.0027 |
0.0060 |
0.0207 |
60 |
0.0013 |
0.0027 |
0.0055 |
0.0169 |
70 |
0.0013 |
0.0027 |
0.0050 |
0.0145 |
|
Table B.2:
Attenuation in dB per kilometer at
20 degrees Celsius and standard atmospheric pressure.
Relative |
Frequency in Hz |
Humidity |
1000 |
2000 |
3000 |
4000 |
40 |
5.6 |
16 |
30 |
105 |
50 |
5.6 |
12 |
26 |
90 |
60 |
5.6 |
12 |
24 |
73 |
70 |
5.6 |
12 |
22 |
63 |
|
There is also a (weaker) dependence of air absorption on
temperature
[
183].
Theoretical models of energy loss in a gas are developed in Morse and
Ingard [
318, pp. 270-285]. Energy loss is caused by
viscosity,
thermal diffusion,
rotational
relaxation,
vibration relaxation, and
boundary losses
(losses due to
heat conduction and viscosity at a wall or other
acoustic boundary). Boundary losses normally dominate by several
orders of magnitude, but in resonant modes, which have
nodes along the
boundaries, interior losses dominate, especially for polyatomic gases
such as air.
B.34 For air having moderate amounts of water
vapor (

) and/or
carbon dioxide (

), the loss and dispersion
due to

and

vibration relaxation hysteresis becomes the
largest factor [
318, p. 300]. The vibration here
is that of the molecule itself, accumulated over the course of many
collisions with other molecules. In this context, a diatomic molecule
may be modeled as two
masses connected by an ideal
spring. Energy
stored in molecular vibration typically dominates over that stored in
molecular rotation, for polyatomic gas molecules [
318, p.
300]. Thus, vibration relaxation hysteresis is a loss
mechanism that converts wave energy into heat.
In a resonant mode, the attenuation per
wavelength due to vibration
relaxation is greatest when the
sinusoidal period (of the resonance)
is equal to

times the
time-constant for vibration-relaxation.
The relaxation time-constant for oxygen is on the order of one
millisecond. The presence of water vapor (or other impurities)
decreases the vibration relaxation time, yielding loss maxima at
frequencies above 1000 rad/sec. The energy loss approaches zero as
the frequency goes to infinity (wavelength to zero).
Under these conditions, the
speed of sound is approximately that of
dry air below the maximum-loss frequency, and somewhat higher above.
Thus, the humidity level changes the dispersion cross-over frequency
of the air in a resonant mode.
The
wave equation in 1D, 2D, or 3D may be written as
 |
(B.47) |
where, in 3D,

denotes the amplitude of the wave at time

and position

, and
denotes the
Laplacian operator in Euclidean coordinates. (In
general coordinates, it is often denoted by

.)
To investigate solutions of the
wave equation, as pursued in
§
B.8.3 below, it is useful to first develop some simple
expressions and notations for elementary waves in 2D and 3D.
Figure
B.9 shows a 2D

cross-section of a snapshot (in time)
of the
sinusoidal plane wave
for

and

, with

and

in the range

.
Figure:
Gray-scale density plot of the
cross-section of a sinusoidal plane wave
, at
with vector wavenumber
.
![\includegraphics[width=\twidth]{eps/planewave}](http://www.dsprelated.com/josimages_new/pasp/img3149.png) |
Figure
B.10 depicts a more mathematical schematic of a
sinusoidal plane wave traveling toward the upper-right of the figure.
The dotted lines indicate the
crests (peak amplitude location)
along the wave.
Figure:
Wave crests of the sinusoidal traveling
plane wave
, for some
fixed time
and
in the
plane.
 |
The direction of travel and
spatial frequency are indicated by the
vector wavenumber

, as discussed in in the following section.
Vector Wavenumber
Mathematically, a
sinusoidal plane wave, as in Fig.
B.9 or
Fig.
B.10, can be written as
 |
(B.48) |
where p(t,
x) is the
pressure at time

(seconds) and position

(3D Euclidean space). The amplitude

, phase

, and radian frequency

are ordinary
sinusoid parameters
[
451], and

is the
vector wavenumber:
where
-
(unit) vector of direction cosines
-
(scalar) wavenumber along travel
direction
Thus, the vector wavenumber

contains
- wavenumber along the travel direction in its magnitude
- travel direction in its orientation
Note that wavenumber units are
radians per meter (spatial
radian frequency).
To see that the vector wavenumber

has the claimed
properties, consider that the
orthogonal projection of any
vector

onto a vector collinear with

is given by

[
451].
B.35Thus,

is the component of

lying along the
direction of wave
propagation indicated by

. The
norm of this
component is

, since

is
unit-norm by construction. More generally,

is the
signed length (in meters) of the component of

along

.
This length times wavenumber

gives the spatial phase advance along
the wave, or,

.
For another point of view, consider the plane wave

,
which is the varying portion of the general plane-wave of
Eq.

(
B.48) at time

, with unit amplitude

and
zero phase

. The spatial phase
of this plane wave is given by
Recall that the general equation for a plane in 3D space is

constant
where

,

, and

are real constants, and

,

, and

are 3D spatial coordinates. Thus, the set of all points

yielding the same value

define a
plane of constant phase 
in

.
As we know from elementary vector
calculus, the direction of maximum
phase advance is given by the
gradient of the phase

:
This shows that the vector wavenumber

is equal to the gradient
of the phase

, so that

points in the direction of
maximum spatial-phase advance.
Since the wavenumber

is the
spatial frequency (in radians per
meter) along the direction of travel, we should be able to compute it
as the
directional derivative of

along

,
i.e.,
An explicit calculation yields
as needed.
Scattering of plane waves is discussed in §
C.8.1.
Since solving the
wave equation in 2D has all the essential features
of the 3D case, we will look at the 2D case in this section.
Specializing Eq.

(
B.47) to 2D, the
2D wave equation may
be written as
where
The 2D
wave equation is obeyed by
traveling sinusoidal plane
waves having any amplitude

, radian frequency

, phase

, and direction

:
where

denotes the vector-wavenumber,

denotes
the wavenumber (spatial radian frequency) of the wave along its
direction of travel, and

is a unit vector of direction
cosines. This is the
analytic-signal form of a sinusoidal
traveling plane wave, and we may define the real (physical)
signal as
the real part of the analytic signal, as usual [
451].
We see that the only constraint imposed by the wave equation on this
general
traveling-wave is the so-called
dispersion relation:
In particular, the wave can travel in any direction, with any
amplitude, frequency, and phase. The only constraint is that its
spatial frequency 
is tied to its temporal frequency

by
the dispersion relation.
B.36
The sum of two such waves traveling in opposite directions with the
same amplitude and frequency produces a
standing wave. For example,
if the waves are traveling parallel to the

axis, we have
 |
(B.49) |
which is a
standing wave along

.
We often wish to find solutions of the 2D
wave equation that obey
certain known
boundary conditions. An example is
transverse
waves on an ideal elastic membrane, rigidly clamped on its boundary to
form a rectangle with dimensions
meters.
Similar to the derivation of Eq.

(
B.49), we can
subtract
the second
sinusoidal traveling wave from the first to yield
which satisfies the zero-
displacement boundary condition along the

axis. If we restrict the wavenumber

to the set

, where

is any positive integer, then we also satisfy the boundary
condition along the line parallel to the

axis at

. Similar
standing waves along

will satisfy both boundary conditions along

and

.
Note that we can also use
products of horizontal and vertical
standing waves
because, when taking the partial derivative with respect to

, the
term

is simply part of the constant coefficient, and vice
versa.
To build solutions to the
wave equation that obey all of the boundary
conditions, we can form
linear combinations of the above standing-wave
products having zero displacement (``
nodes'') along all four boundary
lines:
 |
(B.50) |
where
By construction, all linear combinations of the form Eq.

(
B.50)
are solutions of the
wave equation that satisfy the zero boundary
conditions along the rectangle

-

-

. Since
sinusoids at
different frequencies are
orthogonal,
the solution building-blocks

are orthogonal under the
inner product
It remains to be shown that the set of functions

is
complete, that is, that they form a
basis for the set of
all solutions to the
wave equation satisfying the boundary
conditions. Given that, we can solve the problem of
arbitrary
initial conditions. That is, given any initial

over the
membrane (subject to the boundary conditions, of course), we can find
the amplitude of each excited mode by simple projection:
Showing completeness of the basis

in the desired solution
space is a special case (zero boundary conditions) of the problem of
showing that the 2D
Fourier series expansion is complete in the space
of all continuous rectangular surfaces.
The
Wikipedia page (as of 1/31/10) on the
Helmholtz equation
provides a nice ``entry point'' on the above topics and further
information.
3D Sound
The mathematics of 3D sound is quite elementary, as we will see below.
The hard part of the theory of practical systems typically lies in the
mathematical approximation to the ideal case. Examples include
Ambisonics [
158] and
wave field synthesis
[
49].
Consider a point source at position

. Then the
acoustic complex amplitude at position

is given by
where

denotes the complex amplitude one
meter from the
point source in any direction, and

denotes the wavenumber
(spatial radian frequency). Distributed acoustic sources are
handled as a superposition of point sources, so the point source is a
completely general building block for all types of sources in linear
acoustics.
The fundamental approximation problem in 3D sound is to approximate
the complex acoustic field at one or more listening points using a
finite set of

loudspeakers, which are often modeled as a point
source for each speaker.
Next Section: Digital Waveguide TheoryPrevious Section: A History of Enabling Ideas Leading to
Virtual Musical Instruments