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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:

$\displaystyle \zbox {\mbox{\emph{Force = Mass $\times$\ Acceleration}}}
$

That is, when a force is applied to a mass, the mass experiences an acceleration proportional to the applied force. Denoting the mass by $ m$, force at time $ t$ by $ f(t)$, and acceleration by

$\displaystyle a(t)\isdefs {\ddot x}(t) \isdefs \frac{d^2 x(t)}{dt^2},
$

we have

$\displaystyle \zbox {f(t) = m\,a(t) = m\,{\ddot x}(t).} \protect$ (B.1)

In this formulation, the applied force $ f(t)$ is considered positive in the direction of positive mass-position $ x(t)$. The force $ f(t)$ and acceleration $ a(t)$ are, in general, vectors in three-dimensional space $ x\in{\bf R}^3$. In other words, force and acceleration are generally vector-valued functions of time $ t$. The mass $ m$ 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

Newton's three laws of motion may be stated as follows:

  1. Every object in a state of uniform motion will remain in that state of motion unless an external force acts on it.

  2. Force equals mass times acceleration [ $ f(t)=m\,a(t)$].

  3. 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, $ f(t)=m\,a(t)$, actually implies the first law, since when $ f(t)=0$ (no applied force), the acceleration $ a(t)$ is zero, implying a constant velocity $ v(t)$. (The velocity is simply the integral with respect to time of $ a(t)={\dot v}(t)$.)

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 $ f=ma$. An enormous quantity of physical science has been developed by applying this simpleB.1 mathematical law to different physical situations.


Mass

Mass is an intrinsic property of matter. From Newton's second law, $ f(t)=m\,a(t)$, 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 $ f$, the mass $ m$ is given by $ f$ divided by the resulting acceleration $ a$, again by Newton's second law $ f=ma$.

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 centroidB.4.1).

The physical state of a mass $ m$ at time $ t$ consists of its position $ x(t)$ and velocity $ {\dot x}(t)$ in 3D space. The amount of mass itself, $ m$, is regarded as a fixed parameter that does not change. In other words, the state $ (x,{\dot x})$ of a physical system typically changes over time, while any parameters of the system, such as mass $ m$, 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 $ m_1$ and $ m_2$ experience an attracting force $ f$ given by the formula

$\displaystyle f(t) = G\frac{m_1 m_2}{r^2(t)} \protect$ (B.2)

where $ r(t)$ is the distance between the centroids of the masses $ m_1$ and $ m_2$ at time $ t$, and $ G$ 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.


Hooke's Law

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 $ x(t)$ from the ceiling is fixed.

Figure B.1: Mass hung by a spring from a rigid ceiling.
\includegraphics{eps/springmass-gravity}

If $ M$ denotes the mass of the earth, and $ r$ is the distance of mass $ m$'s center from the earth's center of mass, then the downward force on the mass $ m$ is given by Eq.$ \,$(B.2) as

$\displaystyle f_m = g m,
$

where

$\displaystyle g = G\frac{M}{r^2}
$

is called the acceleration due to gravity. Changes in $ r$ due to the motion of the mass are assumed negligible relative to the radius of the earth (about $ 4000$ miles), and so $ g$ is treated as a constant for most practical purposes near the earth's surface. We see that if we double the mass $ m$, we double the force $ f_m$ pulling on the spring. It is an experimental fact that typical springs exhibit a displacement $ x_m$ that is approximately proportional to the applied force $ f_m$ for a wide range of applied forces. This is Hooke's law for ideal springs:

$\displaystyle \zbox {f(t) = k\,x_m(t),}$   (Hooke's Law)$\displaystyle \protect$ (B.3)

where $ x_m(t)$ is the displacement of the spring from its natural length. We call $ k$ the spring constant, or stiffness of the spring. In terms of our previous notation, we have

$\displaystyle x(t) = x_k + x_m(t),
$

where $ x_k$ 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 $ k$ 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

Figure B.2: Mass-spring system.
\includegraphics{eps/springmass-phy}

As a simple example, consider a mass $ m$ driven along a frictionless surface by an ideal spring $ k$, as shown in Fig.B.2. Assume that the mass position $ x=0$ corresponds to the spring at rest, i.e., not stretched or compressed. The force necessary to compress the spring by a distance $ x$ is given by Hooke's lawB.1.3):

$\displaystyle f_k(t) = -k\,x(t)
$

This force is balanced at all times by the inertial force $ f_m(x)=-m{\ddot x}$ of the mass $ m$, i.e. $ f_k+f_m=0$, yieldingB.6

$\displaystyle m{\ddot x}(t) + k\,x(t) = 0\, \quad \forall t\ge 0, \quad x(0)=A, \quad {\dot x}(0)=0, \protect$ (B.4)

where we have defined $ A$ as the initial displacement of the mass along $ x$. This is a differential equation whose solution gives the equation of motion of the mass-spring junction for all time:B.7

$\displaystyle x(t) = A\cos(\omega_0 t), \quad \forall t\ge 0, \protect$ (B.5)

where $ \omega_0\isdeftext \sqrt{k/m}$ 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 $ x(0)$ and initial velocities $ {\dot x}(0)$, is the set of all sinusoidal oscillations at frequency $ \omega_0$:

$\displaystyle x(t) = A\cos(\omega_0 t + \phi), \quad \forall A,\phi\in{\bf R}.
$

The amplitude of oscillation $ A$ and phase offset $ \phi$ are determined by the initial conditions, i.e., the initial position $ x(0)$ and initial velocity $ {\dot x}(0)$ of the mass (its initial state) when we ``let it go'' or ``push it off'' at time $ t=0$.


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 $ k$ through a displacement of $ x$ meters, starting from rest, is then

$\displaystyle W_k(x) = \int_0^x k\, \xi\, d\xi = \frac{1}{2} k\, x^2. \protect$ (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 $ a$ to point $ b$, either with or against the force, depends only on the locations of points $ a$ and $ b$ in space, not on the path taken from $ a$ to $ b$.

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 $ W_k(x)$ stored in an ideal spring after it has been compressed $ x$ meters from rest.


Kinetic Energy of a Mass

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 $ A$ meters at time $ t=0$, and the mass velocity is zero at $ t=0$. Then from the equation of motion Eq.$ \,$(B.5), we can calculate when the spring returns to rest ($ x(t)=0$). This first happens at the first zero of $ \cos(\omega_0t)$, which is time $ t=(\pi/2)/\omega_0=(\pi/2)\sqrt{m/k}$. At this time, the velocity, given by the time-derivative of Eq.$ \,$(B.5),

$\displaystyle v(t) = -A\omega_0\sin(\omega_0 t),
$

can be evaluated at $ t=(\pi/2)/\omega_0$ to yield the mass velocity $ v[(\pi/2)/\omega_0] = -A\omega_0 = -A\sqrt{k/m}$, which is when all potential energy from the spring has been converted to kinetic energy in the mass. The square of this value is

$\displaystyle v^2_{\mbox{max}} = A^2\omega_0^2 = A^2\frac{k}{m},
$

and we see that if we multiply $ v^2_{\mbox{max}}$ by $ m/2$, we get

$\displaystyle \frac{1}{2}m\,v^2_{\mbox{max}} = \frac{1}{2}k\,A^2,
$

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

$\displaystyle E_m(v) = \frac{1}{2}m\, v^2
$

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.


Mass Kinetic Energy from Virtual Work

From Newton's second law, $ f=ma=m{\ddot x}$ (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 $ v={\dot x}$. Let $ d x$ denote a small (infinitesimal) displacement of the mass in the $ x$ direction. Then we have, using the calculus of differentials,

\begin{eqnarray*}
f(t) &=& m\, {\ddot x}(t)\\
\,\,\Rightarrow\,\,\quad d W\isde...
...{1}{2}{\dot x}^2\right)\\
&=& d\left(\frac{1}{2}m\,v^2\right).
\end{eqnarray*}

Thus, by Newton's second law, a differential of work $ dW$ applied to a mass $ m$ by force $ f$ through distance $ d x$ boosts the kinetic energy of the mass by $ d(m\,v^2/2)$. The kinetic energy of a mass moving at speed $ v$ is then given by the integral of all such differential boosts from 0 to $ v$:

$\displaystyle E_m(v) = \int_0^v dW = \int_0^v d\left(\frac{1}{2}m \nu^2\right)
= \frac{1}{2}m v^2 = \frac{1}{2}m\,{\dot x}^2,
$

where $ E_m(v)$ denotes the kinetic energy of mass $ m$ traveling at speed $ v$.

The quantity $ dW=f\,dx$ is classically called the virtual work associated with force $ f$, and $ d x$ a virtual displacement [544].


Energy in the Mass-Spring Oscillator

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):

$\displaystyle x(t) = A\cos(\omega_0 t), \quad t\ge 0
$

  • From a global point of view, we see that energy is conserved, since the oscillation never decays.
  • At the peaks of the displacement $ x(t)$ (when $ \cos(\omega_0t)$ is either $ 1$ or $ -1$), 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 $ x(t)$, 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 $ -A$ (or compress it to $ +A$) before the mass stops and changes direction. At all times, the total energy $ E$ is equal to the sum of the potential energy $ E_k(t)$ stored in the spring, and the kinetic energy $ E_m(t)$ stored in the mass:

    $\displaystyle E = E_k(t) + E_m(t) =$   $\displaystyle \mbox{constant [$E_k(0)$]}$

Regarding the last point, the potential energy, $ E_k(t)=k\,x^2(t)/2$ was defined as the work required to displace the spring by $ x$ meters, where work was defined in Eq.$ \,$(B.6). The kinetic energy of a mass $ m$ moving at speed $ v$ was found to be $ E_m(t)=m\,v^2(t)/2=m\,{\dot x}^2/2$. 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 $ \cos^2(\theta)+\sin^2(\theta)=1$ (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 distanceB.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

$\displaystyle f_m(t) + f_k(t) = 0, \quad \forall t,
$

which led to the differential equation obeyed by the mass-spring system:

$\displaystyle m{\ddot x}(t) + k\,x(t) = 0 \quad \forall t
$

Multiplying through by $ {\dot x}(t)=v(t)$ gives

\begin{eqnarray*}
0
&=& m{\ddot x}(t){\dot x}(t) + k\,x(t){\dot x}(t)\\
&=& m\...
...{d}{dt} \left[ E_m(t) + E_k(t) \right]\\
&=& \frac{d}{dt} E(t).
\end{eqnarray*}

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 $ m$ is usually defined by

$\displaystyle p = m v,
$

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

$\displaystyle \underline{p}= m \underline{v},
$

where $ \underline{p}$ and $ \underline{v}$ are vectors in 3D space. We will return to vector momentum in §B.4.1 below.

Conservation of Momentum

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 $ f=ma$ to formulate mathematical dynamic models for the ideal point-massB.1.1), springB.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 $ k$, 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 massB.4.1 below) plus the angular momentum about its center of mass (§B.4.13 below).

Center of Mass

The center of mass (or centroid) of a rigid body is found by averaging the spatial points of the body $ \underline{x}_i\in{\bf R}^3$ weighted by the mass $ m_i$ of those points:B.12

$\displaystyle \underline{x}_c \isdefs \left. \sum_{i=1}^N m_i \underline{x}_i \right/ \sum_{i=1}^N m_i
$

Thus, the center of mass is the mass-weighted average location of the object. For a continuous mass distribution totaling up to $ M$, we can write

$\displaystyle \underline{x}_c \isdefs
\frac{1}{M}\int_V \underline{x}\, dm(\un...
...p \iiint_{\underline{x}\in V} \underline{x}\, \rho(\underline{x})\, dx\,dy\,dz
$

where the volume integral is taken over a volume $ V$ of 3D space that includes the rigid body, and $ dm(\underline{x}) = m(\underline{x})dV = m(\underline{x})\,
dx\,dy\,dz$ denotes the mass contained within the differential volume element $ dV$ located at the point $ \underline{x}\in{\bf R}^3$, with $ \rho(\underline{x})$ denoting the mass density at the point $ \underline{x}$. The total mass is

$\displaystyle M \eqsp \int_V dm(\underline{x}) \eqsp \int_V \rho(\underline{x})\, dV.
$

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.

Linear Momentum of the Center of Mass

Consider a system of $ N$ point-masses $ m_i$, each traveling with vector velocity $ \underline{v}_i$, and not necessarily rigidly attached to each other. Then the total momentum of the system is

$\displaystyle \underline{p}\eqsp \sum_{i=1}^N m_i \underline{v}_i
\eqsp \sum_{...
...ine{x}_i \right)
\eqsp M \frac{d}{dt} \underline{x}_c
\isdef M \underline{v}_c
$

where $ M=\sum m_i$ denotes the total mass, and $ \underline{v}_c$ is the velocity of the center of mass.

Thus, the momentum $ \underline{p}$ of any collection of masses $ m_i$ (including rigid bodies) equals the total mass $ M$ times the velocity of the center-of-mass.



Whoops, No Angular Momentum!

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 $ \underline{0}$, experiencing an external force $ \underline{f}$ that can be expressed as the sum of radial and tangential components $ \underline{f}=\underline{f}_r+\underline{f}_t$. The radial component $ \underline{f}_r$ accelerates the center-of-mass, while the tangential component $ \underline{f}_t$ causes rotation about the center of mass.
\includegraphics[width=1.5in]{eps/rigidbody}


Translational Kinetic Energy

The translational kinetic energy of a collection of masses $ m_i$ is given by

$\displaystyle E_K \eqsp \frac{1}{2} M v_c^2
$

where $ M=\sum_i m_i$ is the total mass, and $ v_c$ denotes the speed of the center-of-mass. We have $ v_c\isdeftext \left\Vert\,\underline{v}_c\,\right\Vert$, where $ \underline{v}_c$ 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].


Rotational Kinetic Energy

Figure B.4: Point-mass $ m$ rotating in a circle of radius $ R$ with tangential speed $ v=R\omega $, where $ \omega=\dot{\theta}$ denotes the angular velocity in rad/s.
\includegraphics[width=1.2in]{eps/masscircle}

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 $ m$ rotatingB.13 in a circular orbit of radius $ R$ and angular velocity $ \omega $ (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 $ \underline{0}$. Then the speed of the mass along the circle is $ v=R\omega $, and its kinetic energy is $ (1/2)mv^2=(1/2)mR^2\omega^2$. 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 $ \omega $ in radians per second. Thus, we write

$\displaystyle E_R \eqsp \frac{1}{2} I \omega^2, \protect$ (B.7)

where

$\displaystyle I \eqsp mR^2 \protect$ (B.8)

is called the mass moment of inertia.


Mass Moment of Inertia

The mass moment of inertia $ I$ (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 massB.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 $ m$ rotating a distance $ R$ from the axis to be $ I=mR^2$. Therefore, for a rigid collection of point-masses $ m_i$, $ i=1,\ldots,N$,B.14 the moment of inertia about a given axis of rotation is obtained by adding the component moments of inertia:

$\displaystyle I = m_1 R_1^2 + m_2 R_2^2 + \cdots + m_N R_N^2, \protect$ (B.9)

where $ R_i$ is the distance from the axis of rotation to the $ i$th mass.

For a continuous mass distribution, the moment of inertia is given by integrating the contribution of each differential mass element:

$\displaystyle I \eqsp \int_M R^2 dm,$ (B.10)

where $ R$ is the distance from the axis of rotation to the mass element $ dm$. In terms of the density $ \rho(\underline{x})$ of a continuous mass distribution, we can write

$\displaystyle I \,\mathrel{\mathop=}\,\int_V R^2(\underline{x})\,\rho(\underline{x})\,dV,
$

where $ \rho(\underline{x})$ denotes the mass density (kg/m$ \null^3$) at the point $ \underline{x}$, and $ dV=dx\,dy\,dz$ denotes a differential volume element located at $ \underline{x}\in{\bf R}^3$.

Circular Disk Rotating in Its Own Plane

For example, the moment of inertia for a uniform circular disk of total mass $ M$ and radius $ R$, rotating in its own plane about a rotation axis piercing its center, is given by

$\displaystyle I = \frac{M}{\pi R^2}\int_{-\pi}^\pi \int_0^R r^2\, r\,dr\,d\thet...
...c{2M}{R^2}\int_0^R r^3 dr
= \frac{2M}{R^2}\frac{1}{4} R^4
= \frac{1}{2} M R^2.
$


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

$\displaystyle I = \frac{M}{\pi R^2}\cdot 4\int_0^{\pi/2} \int_0^R [r\cos(\theta)]^2\, r\,dr\,d\theta
= \frac{1}{4}MR^2
$

Thus, the uniform disk's moment of inertia in its own plane is twice that about its diameter.


Perpendicular Axis Theorem

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 $ dm$ having rectilinear coordinates $ (x_1,x_2,0)$ 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 $ dm\,(x_1^2+x_2^2)$ while its moment of inertia about coordinate axes within the mass-plane are respectively $ dm\,x_1^2$ and $ dm\,x_2^2$. This, the perpendicular axis theorem is an immediate consequence of the Pythagorean theorem for right triangles.


Parallel Axis Theorem

Let $ I_0$ denote the moment of inertia for a rotation axis passing through the center of mass, and let $ I_d$ denote the moment of inertia for a rotation axis parallel to the first but a distance $ d$ away from it. Then the parallel axis theorem says that

$\displaystyle I_d = I_0 + Md^2
$

where $ M$ denotes the total mass. Thus, the added inertia due to displacement by $ d$ meters away from the centroidal axis is equal to that of a point mass $ M$ rotating a distance $ d$ 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 $ m$ a distance $ R$ from the axis of rotation is given by $ I=mR^2$. By the stretch rule, the same applies to an ideal rod of mass $ m$ parallel to and distance $ R$ 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 $ m$ can be stretched out to form a mass ring at radius $ R$ 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 $ R$ and mass $ m$, with its axis of symmetry coincident with the axis of rotation. In all of these examples, the moment of inertia is $ I=mR^2$ about the axis of rotation.


Area Moment of Inertia

The area moment of inertia is the second moment of an area $ A$ around a given axis:

$\displaystyle I_A = \int_A R^2(\underline{x}) dA
$

where $ dA$ denotes a differential element of the area (summing to $ A$), and $ R(\underline{x})$ 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 $ M$ uniformly distributed over an area $ A$ (i.e., a constant mass density of $ \rho=M/A$), the mass moment of inertia $ I_\rho$ is given by the area moment of inertia $ I_A$ times mass-density $ \rho$:

$\displaystyle I_\rho \isdefs \int_M R^2 dM \eqsp \int_A R^2 \rho\, dA \eqsp \rho I_A
$


Radius of Gyration

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 $ S$, the radius of gyration is given by

$\displaystyle R_g = \sqrt{\frac{I_S}{S}} \protect$ (B.11)

where $ I_S$ is the area moment of inertiaB.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):

$\displaystyle I_S = \int_S R^2 dS,
$

where $ R$ denotes the distance of the differential area element $ dS$ from the axis of gyration.

Rectangular Cross-Section

For a rectangular cross-section of height $ h$ and width $ w$, area $ S=hw$, the area moment of inertia about the horizontal midline is given by

$\displaystyle I_w
= w\int_{-h/2}^{h/2} y^2 dy
= w\left.\frac{1}{3}y^3\right\vert _{-h/2}^{h/2}
= \frac{Sh^2}{12}.
$

The radius of gyration about this axis is then

$\displaystyle R_g = \sqrt{\frac{I_w}{S}} = \sqrt{\frac{h^2}{12}} = \frac{h}{2\sqrt{3}}.
$

Similarly, the radius of gyration about a vertical axis passing through the center of the cross-section is $ R_g=w/(2\sqrt{3})$.

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 $ a$, Eq.$ \,$(B.11) tells us that the squared radius of gyration about any line passing through the center of the cross-section is given by

\begin{eqnarray*}
R_g^2 &=& \frac{1}{\pi a^2} \int_{-a}^a dx \int_{-\sqrt{a^2-x^...
...frac{4a^2}{3\pi} \int_{0}^{\frac{\pi}{2}} \cos^4(\theta)d\theta.
\end{eqnarray*}

Using the elementrary trig identity $ \cos(2\theta)=2\cos^2(\theta)-1$, we readily derive

$\displaystyle \cos^4(\theta) = \frac{1}{8}\cos(4\theta) + \frac{1}{2}\cos(2\theta) + \frac{3}{8}.
$

The first two terms of this expression contribute zero to the integral from 0 to $ \pi /2$, while the last term contributes $ 3\pi/16$, yielding

$\displaystyle R_g^2 = \frac{4a^2}{3\pi} \frac{3\pi}{16} = \frac{a^2}{4}.
$

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

$\displaystyle R_g = \frac{a}{2}.
$

For a circular tube in which the mass of the cross-section lies within a circular annulus having inner radius $ b$ and outer radius $ a$, the radius of gyration is given by

$\displaystyle R_g = \frac{\sqrt{a^2-b^2}}{2}. \protect$ (B.12)


Two Masses Connected by a Rod

Figure B.5: Two ideal point-masses $ m$ connected by an ideal, rigid, massless rod of length $ 2r$.
\includegraphics[width=1.5in]{eps/massrodmass}

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 densityB.15 $ f(t,x)$ can be applied anywhere between $ x=-r$ and $ x=r$ 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 impedance7.1) ``seen'' by the driving force $ f(t,x)dx$ varies as a function of $ x$. At $ x=0$, The excitation $ f(t,0)dx$ sees a ``point mass'' $ 2m$, and no rotation is excited by the force (by symmetry). At $ x=\pm R$, on the other hand, the excitation $ f(t,\pm R)dx$ only sees mass $ m$ 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 $ f(t,x)$ depends on the striking point $ x$ and, away from $ x=0$, it depends on time $ t$ as well.

To avoid dealing with a time-varying driving-point impedance, we will use an impulsive force input at time $ t=0$. Since momentum is the time-integral of force ( $ f=ma=m\dot{v}\,\,\Rightarrow\,\,mv=\int f\,dt$), our excitation will be a unit momentum transferred to the two-mass system at time 0.

Striking the Rod in the Middle

First, consider $ f(t,x)=\delta(t)\delta(x)$. 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 $ x=0$ and $ t=0$ is obtained by integrating the applied force density with respect to time and position:

$\displaystyle p \eqsp \iint f(t,x)\,dt\,dx \eqsp \iint \delta(t)\delta(x)\,dt\,dx \eqsp 1
$

This unit momentum is transferred to the two masses $ m$. By symmetry, we have $ v_{-r} = v_r = v_0$. We can also refer to $ v_0$ as the velocity of the center of mass, again obvious by symmetry. Continuing to refer to Fig.B.5, we have

$\displaystyle p \eqsp 1 \eqsp mv_{-r} + mv_r \eqsp (2m)v_0 \,\,\Rightarrow\,\,v_0 \eqsp
\frac{1}{2m}.
$

Thus, after time zero, each mass is traveling upward at speed $ v_0=1/(2m)$, and there is no rotation about the center of mass at $ x=0$.

The kinetic energy of the system after time zero is

$\displaystyle E_K \eqsp \frac{1}{2} mv_{-r}^2 + \frac{1}{2}mv_r^2 \eqsp
m\left(\frac{1}{2m}\right)^2 \eqsp \frac{1}{4m}.
$

Note that we can also compute $ E_K$ in terms of the total mass $ M=2m$ and the velocity of the center of mass $ v_0=1/(2m)$:

$\displaystyle E_K \eqsp \frac{1}{2} Mv_0^2 \eqsp \frac{1}{2}
(2m)\left(\frac{1}{2m}\right)^2 \eqsp \frac{1}{4m}
$


Striking One of the Masses

Now let $ f(t,x) = \delta(t)\delta(x-r)$. 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

$\displaystyle p \eqsp 1 \eqsp m v_r \,\,\Rightarrow\,\,v_r \eqsp \frac{1}{m},
$

which is twice as fast as before. Just after time zero, we have $ v_r=1/m$, $ v_{-r}=0$, and, because the massless rod remains rigid, $ v_0=1/(2m)$.

Note that the velocity of the center-of-mass $ 1/(2m)$ 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:

$\displaystyle E_K \eqsp \frac{1}{2}mv_r^2 \eqsp \frac{1}{2}m\left(\frac{1}{m}\right)^2 \eqsp \frac{1}{2m} \protect$ (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 $ v'$ denote a velocity in this frame. Since the velocity of the center of mass is $ v_0=1/(2m)$, we can convert any velocity $ v'$ in the body-fixed frame to a velocity $ v$ in the original frame by adding $ v_0$ to it, viz.,

$\displaystyle v \eqsp v' + v_0 \eqsp v' + \frac{1}{2m}.
$

The mass velocities in the body-fixed frame are now

$\displaystyle v'_{-r} \eqsp -\frac{1}{2m},\qquad\qquad v'_r \eqsp \frac{1}{2m},
$

and of course $ v'_0=0$.

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 $ x=0$, is

$\displaystyle I \eqsp m(-r)^2 + m r^2 \eqsp 2mr^2.
$

Thus, the rotational kinetic energyB.4.3) is found to be

$\displaystyle E'_R \eqsp \frac{1}{2}I\omega^2 \eqsp
\frac{1}{2}(2mr^2)\left(\frac{v'_r}{r}\right)^2
\eqsp mr^2\left(\frac{1}{2mr}\right)^2
\eqsp \frac{1}{4m}.
$

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 $ M=2m$ traveling at the center-of-mass velocity $ v_0=1/(2m)$:

$\displaystyle E'_K \eqsp \frac{1}{2}Mv_0^2
\eqsp \frac{1}{2}(2m)\left(\frac{1}{2m}\right)^2
\eqsp \frac{1}{4m}
$

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

$\displaystyle E_K \eqsp E'_K + E'_R
\eqsp \frac{1}{4m} + \frac{1}{4m}
\eqsp \frac{1}{2m}
$

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 ( $ p_0=Mv_0=(2m)(v_r/2) =
mv_r=m\cdot(1/(2m))=1/2$), and the angular momentum in the body-fixed frame ( $ L'=I\omega= (2mr^2)(v'_r/r)=(2mr^2)(1/(2mr))=r/2$) 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.


Angular Velocity Vector

When working with rotations, it is convenient to define the angular-velocity vector as a vector $ \underline{\omega}\in{\bf R}^3$ 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 $ \vert\vert\,\underline{\omega}\,\vert\vert $ should obviously equal the angular velocity $ \omega $. It is convenient also to work with a unit-length variant $ \underline{\tilde{\omega}}\isdeftext \underline{\omega}/ \vert\vert\,\underline{\omega}\,\vert\vert $.

As introduced in Eq.$ \,$(B.8) above, the mass moment of inertia is given by $ I=mR^2$ where $ R$ is the distance from the (instantaneous) axis of rotation to the mass $ m$ located at $ \underline{x}\in{\bf R}^3$ . In terms of the angular-velocity vector $ \underline{\omega}$, we can write this as (see Fig.B.6)

$\displaystyle I$ $\displaystyle =$ $\displaystyle mR^2
\eqsp m\cdot \left\Vert\,\underline{x}-{\cal P}_{\underline{\omega}}(\underline{x})\,\right\Vert^2$  
  $\displaystyle =$ $\displaystyle m\cdot \left\Vert\,\underline{x}-(\underline{\tilde{\omega}}^T\underline{x})\underline{\tilde{\omega}}\,\right\Vert^2
\protect$ (B.14)

where

$\displaystyle {\cal P}_{\underline{\omega}}(\underline{x}) \isdefs \frac{\under...
...ga}\eqsp (\underline{\tilde{\omega}}^T\underline{x})\underline{\tilde{\omega}}
$

denotes the orthogonal projection of $ \underline{x}$ onto $ \underline{\omega}$ (or $ \underline{\tilde{\omega}}$) [451]. Thus, we can project the mass position $ \underline{x}$ onto the angular-velocity vector $ \underline{\omega}$ and subtract to get the component of $ \underline{x}$ that is orthogonal to $ \underline{\omega}$, and the length of that difference vector is the distance to the rotation axis $ R$, as shown in Fig.B.6.

Figure: Mass position vector $ \underline{x}$ and its orthogonal projection $ {\cal P}_{\protect\underline{\omega}}(\underline{x})$ onto the angular velocity vector $ \underline{\omega}$ for purposes of finding the distance $ R$ of the mass $ m$ from the axis of rotation $ \underline{\tilde{\omega}}$.
\includegraphics[width=1.5in]{eps/pxov}

Using the vector cross product (defined in the next section), we will show (in §B.4.17) that $ R$ can be written more succinctly as

$\displaystyle R \eqsp \left\Vert\,\underline{x}-{\cal P}_{\underline{\omega}}(\...
...\eqsp \left\Vert\,\underline{\tilde{\omega}}\times \underline{x}\,\right\Vert.
$


Vector Cross Product

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

$\displaystyle \underline{x}\times \underline{y}$ $\displaystyle =$ $\displaystyle \left\vert \begin{array}{ccc}
\underline{e}_1 & \underline{e}_2 &...
...ine{e}_3\\ [2pt]
x_1 & x_2 & x_3\\ [2pt]
y_1 & y_2 & y_3
\end{array}\right\vert$  
  $\displaystyle =$ $\displaystyle (x_2 y_3 - y_2 x_3)\underline{e}_1 + (x_3y_1 - y_3x_1) \underline{e}_2 + (x_1y_2- y_1x_2) \underline{e}_3$  
  $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc}
0 & -x_3 & x_2\\ [2pt]
x_3 & 0 & -x_1\\ ...
...}\right]
\left[\begin{array}{c} x_1 \\ [2pt] x_2 \\ [2pt] x_3\end{array}\right]$  
    $\displaystyle \protect$ (B.15)

where $ \underline{e}_i$ denote the unit vectors in $ {\bf R}^3$. The cross-product is a vector in 3D that is orthogonal to the plane spanned by $ \underline{x}$ and $ \underline{y}$, 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 $ y\times x
= -\; x\times y$. This is one example of a host of identities that one learns in vector calculus and its applications.

Cross-Product Magnitude

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

$\displaystyle \left\Vert\,\underline{x}\times \underline{y}\,\right\Vert \eqsp ...
...dot\left\Vert\,\underline{y}\,\right\Vert \cdot \vert\sin(\theta)\vert \protect$ (B.16)

where

$\displaystyle \sin(\theta)\eqsp \sqrt{1-\cos^2(\theta)}
$

with

$\displaystyle \cos(\theta)
\isdefs \frac{\left<\underline{x},\underline{y}\rig...
...c{x_1y_1+x_2y_2+x_3y_3}{\sqrt{x_1^2+x_2^2+x_3^2}\cdot\sqrt{y_1^2+y_2^2+y_3^2}}
$

(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 $ \underline{x}\times\underline{y}= \mathbf{X}\underline{y}$ using the first matrix form in the third line of the cross-product definition in Eq.$ \,$(B.15) above. Then

\begin{eqnarray*}
(\underline{x}\times\underline{y})^2
&=&
\underline{y}^T\mat...
...rline{x}^{\tiny\perp}}\underline{y}_{\underline{x}^{\tiny\perp}}
\end{eqnarray*}

where $ \mathbf{E}=[\underline{e}_1,\underline{e}_2,\underline{e}_3]$ denotes the identity matrix in $ {\bf R}^3$, $ {\cal P}_{\underline{x}}$ denotes the orthogonal-projection matrix onto $ \underline{x}$ [451], $ {\cal P}_{\underline{x}}$ denotes the projection matrix onto the orthogonal complement of $ \underline{x}$, $ \underline{y}_{\underline{x}^{\tiny\perp}}$ denotes the component of $ \underline{y}$ orthogonal to $ \underline{x}$, and we used the fact that orthogonal projection matrices $ {\cal P}$ are idempotent (i.e., $ {\cal P}^2 ={\cal P}$) and symmetric (when real, as we have here) when we replaced $ {\cal P}_{\underline{x}^{\tiny\perp}}$ by $ {\cal P}^T_{\underline{x}^{\tiny\perp}}{\cal P}_{\underline{x}^{\tiny\perp}}$ above. Finally, note that the length of $ \underline{y}_{\underline{x}^{\tiny\perp}}$ is $ \vert\vert\,\underline{y}\,\vert\vert \cdot\vert\sin(\theta)\vert$, where $ \theta$ is the angle between the 1D subspaces spanned by $ \underline{x}$ and $ \underline{y}$ in the plane including both vectors. Thus,

$\displaystyle (\underline{x}\times\underline{y})^2
\eqsp \vert\vert\,\underline...
...line{x}\,\vert\vert ^2 \vert\vert\,\underline{y}\,\vert\vert ^2\sin^2(\theta),
$

which establishes the desired result:

$\displaystyle \left\Vert\,\underline{x}\times\underline{y}\,\right\Vert = \vert...
...t\vert \cdot \vert\vert\,\underline{y}\,\vert\vert \cdot\vert\sin(\theta)\vert
$

Moreover, this proof gives an appealing geometric interpretation of the vector cross-product $ \underline{x}\times\underline{y}$ as having magnitude given by the product of $ \vert\vert\,\underline{x}\,\vert\vert $ times the norm of the difference between $ \underline{x}$ and the orthogonal projection of $ \underline{x}$ onto $ \underline{y}$ ( $ \vert\vert\,\underline{x}\times\underline{y}\,\vert\vert = \vert\vert\,\underl...
... \vert\vert\,\underline{x}-{\cal P}_{\underline{y}}(\underline{x})\,\vert\vert $) or vice versa ( $ \vert\vert\,\underline{x}\times\underline{y}\,\vert\vert = \vert\vert\,\underl...
... \vert\vert\,\underline{y}-{\cal P}_{\underline{x}}(\underline{y})\,\vert\vert $). 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 $ \underline{x}$ and $ \underline{y}$ according to the right-hand rule. This orthogonality can be checked by verifying that $ \mathbf{X}\underline{x}=\mathbf{Y}\underline{y}=\underline{0}$. 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$ and $ \underline{y}'=[ \vert\vert\,\underline{y}\,\vert\vert \cos(\theta), \vert\vert\,\underline{y}\,\vert\vert \sin(\theta),0]^T$ in which case $ \underline{x}'\times\underline{y}' =
\vert\vert\,\underline{x}\,\vert\vert [0,...
...}\,\vert\vert \vert\vert\,\underline{y}\,\vert\vert \sin(\theta)\underline{e}_3$. Thus, the cross product points ``up'' relative to the $ (\underline{x},\underline{y})$ plane for $ \theta
\in(0,\pi)$ and ``down'' for $ \theta\in(0,-\pi)$.


Mass Moment of Inertia as a Cross Product

In Eq.$ \,$(B.14) above, the mass moment of inertia was expressed in terms of orthogonal projection as $ I = mR^2 = m\cdot
\vert\vert\,\underline{x}-{\cal P}_{\underline{\omega}}(\und...
...erline{\tilde{\omega}}^T\underline{x})\underline{\tilde{\omega}}\,\vert\vert ^2$, where $ \underline{\tilde{\omega}}\isdeftext \underline{\omega}/ \vert\vert\,\underline{\omega}\,\vert\vert $. In terms of the vector cross product, we can now express it as

$\displaystyle I \eqsp m\cdot(\underline{\tilde{\omega}}\times \underline{x})^2 ...
...cdot\sin(\theta_{\underline{\tilde{\omega}}\underline{x}})\right]^2
\eqsp mR^2
$

where $ R= \vert\vert\,\underline{x}\,\vert\vert \sin(\theta_{\underline{\tilde{\omega}}\underline{x}})$ is the distance from the rotation axis out to the point $ \underline{x}$ (which equals the length of the vector $ \underline{x}-{\cal P}_{\underline{\omega}}(\underline{x})$).


Tangential Velocity as a Cross Product

Referring again to Fig.B.4, we can write the tangential velocity vector $ \underline{v}$ as a vector cross product of the angular-velocity vector $ \underline{\omega}$B.4.11) and the position vector $ \underline{x}$:

$\displaystyle \underline{v}\eqsp \underline{\omega}\times \underline{x} \protect$ (B.17)

To see this, let's first check its direction and then its magnitude. By the right-hand rule, $ \underline{\omega}$ points up out of the page in Fig.B.4. Crossing that with $ \underline{x}$, again by the right-hand rule, produces a tangential velocity vector $ \underline{v}$ pointing as shown in the figure. So, the direction is correct. Now, the magnitude: Since $ \underline{\omega}$ and $ \underline{x}$ are mutually orthogonal, the angle between them is $ \pi /2$, so that, by Eq.$ \,$(B.16),

$\displaystyle \left\Vert\,\underline{\omega}\times \underline{x}\,\right\Vert \...
...,\right\Vert\cdot\left\Vert\,\underline{x}\,\right\Vert \eqsp \omega R \eqsp v
$

as desired.


Angular Momentum

The angular momentum of a mass $ m$ rotating in a circle of radius $ R$ with angular velocity $ \omega $ (rad/s), is defined by

$\displaystyle L \eqsp I\omega
$

where $ I=mR^2$ denotes the mass moment of inertia of the rigid body (§B.4.4).

Relation of Angular to Linear Momentum

Recall (§B.3) that the momentum of a mass $ m$ traveling with velocity $ v$ in a straight line is given by

$\displaystyle p = m v,
$

while the angular momentum of a point-mass $ m$ rotating along a circle of radius $ R$ at $ \omega $ rad/s is given by

$\displaystyle L \eqsp I\omega,
$

where $ I=mR^2$. The tangential speed of the mass along the circle of radius $ R$ is given by

$\displaystyle v \eqsp R\omega.
$

Expressing the angular momentum $ I$ in terms of $ v$ gives

$\displaystyle L \isdefs I\omega \eqsp I\frac{v}{R} \isdefs mR^2\frac{v}{R} \eqsp Rmv \eqsp Rp. \protect$ (B.18)

Thus, the angular momentum $ L$ is $ R$ times the linear momentum $ p=mv$.

Linear momentum can be viewed as a renormalized special case of angular momentum in which the radius of rotation goes to infinity.


Angular Momentum Vector

Like linear momentum, angular momentum is fundamentally a vector in $ {\bf R}^3$. The definition of the previous section suffices when the direction does not change, in which case we can focus only on its magnitude $ L=I\omega$.

More generally, let $ \underline{x}_m\in{\bf R}^3$ denote the 3-space coordinates of a point-mass $ m$, and let $ \underline{v}_m=\dot{\underline{x}}_m$ denote its velocity in $ {\bf R}^3$. Then the instantaneous angular momentum vector of the mass relative to the origin (not necessarily rotating about a fixed axis) is given by

$\displaystyle \underline{L}\isdefs m\, \underline{x}_m \times \underline{v}_m \eqsp m\, \underline{x}_m \times (\underline{\omega}\times\underline{x}_m) \protect$ (B.19)

where $ \times$ denotes the vector cross product, discussed in §B.4.12 above. The identity $ \underline{v}_m=\underline{\omega}\times\underline{x}_m$ was discussed at Eq.$ \,$(B.17).

For the special case in which $ \underline{v}_m$ is orthogonal to $ \underline{x}_m$, as in Fig.B.4, we have that $ x_m\times\underline{v}_m$ points, by the right-hand rule, in the direction of the angular velocity vector $ \underline{\omega}$ (up out of the page), which is satisfying. Furthermore, its magnitude is given by

$\displaystyle \left\Vert\,\underline{L}\,\right\Vert \eqsp m\left\Vert\,\underl...
...ft\Vert\,\underline{v}_m\,\right\Vert
\eqsp mRv
\eqsp mR^2\omega
\eqsp I\omega
$

which agrees with the scalar case.

In the more general case of an arbitrary mass velocity vector $ \underline{v}_m$, we know from §B.4.12 that the magnitude of $ \underline{x}_m\times\underline{v}_m$ equals the product of the distance from the axis of rotation to the mass, i.e., $ \vert\vert\,\underline{x}_m\,\vert\vert $, times the length of the component of $ \underline{v}_m$ that is orthogonal to $ \underline{x}$, i.e., $ \vert\vert\,\underline{v}_m\,\vert\vert \sin(\theta)$, 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].

Angular Momentum Vector in Matrix Form

The two cross-products in Eq.$ \,$(B.19) can be written out with the help of the vector analysis identityB.23

$\displaystyle \underline{x}\times (\underline{y}\times\underline{z}) \eqsp \und...
...underline{z}^T\underline{x})-\underline{z}\cdot(\underline{x}^T\underline{y}).
$

This (or a direct calculation) yields, starting with Eq.$ \,$(B.19),
$\displaystyle \underline{L}$ $\displaystyle =$ $\displaystyle m\, \underline{x}_m \times (\underline{\omega}\times\underline{x}...
...line{x}^T\underline{x}) - \underline{x}\cdot(\underline{x}^T\underline{\omega})$  
  $\displaystyle =$ $\displaystyle m\,\left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}- \underline{x}\underline{x}^T\right)\underline{\omega}$  
  $\displaystyle \isdef$ $\displaystyle \mathbf{I}\,\underline{\omega}
\protect$ (B.20)

where

$\displaystyle \mathbf{I}\underline{\omega}\eqsp
\left[\begin{array}{ccc}
I_{1...
...begin{array}{c} \omega_1 \\ [2pt] \omega_2 \\ [2pt] \omega_3\end{array}\right]
$

with $ I_{ii}=m\left(\sum_{j=1}^3x_j^2 - x_i^2\right)$, and $ I_{ij}=-mx_ix_j$, for $ i\ne j$. That is,
$\displaystyle \mathbf{I}\eqsp m\left[\begin{array}{ccc}
x_2^2+x_3^2 & -x_1x_2 &...
...line{x}\,\right\Vert^2\mathbf{E}- \underline{x}\underline{x}^T\right).
\protect$     (B.21)

The matrix $ \mathbf{I}$ 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,

$\displaystyle \underline{L}\eqsp \sum_i m_i \left(\left\Vert\,\underline{x}_i\,...
...e{x}_i^T\right)\underline{\omega}
\,\isdefs \, \mathbf{I}\,\underline{\omega}.
$

Since $ \underline{\omega}$ 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 $ \underline{L}$ is given by the mass moment of inertia tensor $ \mathbf{I}$ times the angular-velocity vector $ \underline{\omega}$ representing the axis of rotation.

Note that the angular momentum vector $ \underline{L}$ does not in general point in the same direction as the angular-velocity vector $ \underline{\omega}$. We saw above that it does in the special case of a point mass traveling orthogonal to its position vector. In general, $ \underline{L}$ and $ \underline{\omega}$ point in the same direction whenever $ \underline{\omega}$ is an eigenvector of $ \mathbf{I}$, as will be discussed further below (§B.4.16). In this case, the rigid body is said to be dynamically balanced.B.24


Mass Moment of Inertia Tensor

As derived in the previous section, the moment of inertia tensor, in 3D Cartesian coordinates, is a three-by-three matrix $ \mathbf{I}$ 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 $ \underline{\omega}$ is always fixed at $ \underline{0}$ in the space (typically located at the center of mass). Therefore, the moment of inertia tensor $ \mathbf{I}$ 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 $ \underline{\tilde{\omega}}=\underline{\omega}/ \vert\vert\,\underline{\omega}\,\vert\vert $ as

$\displaystyle I \eqsp \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde{\omega}}. \protect$ (B.22)

Since rotational energy is defined as $ (1/2)I\omega^2$ (see Eq.$ \,$(B.7)), multiplying Eq.$ \,$(B.22) by $ \omega^2$ gives the following expression for the rotational kinetic energy in terms of the moment of inertia tensor:

$\displaystyle E_R \eqsp \frac{1}{2}\, \underline{\omega}^T\mathbf{I}\,\underline{\omega} \protect$ (B.23)

We can show Eq.$ \,$(B.22) starting from Eq.$ \,$(B.14). For a point-mass $ m$ located at $ \underline{x}$, we have

\begin{eqnarray*}
I &=& m \left\Vert\,\underline{x}-(\underline{\tilde{\omega}}^...
...nderline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde{\omega}}
\end{eqnarray*}

where again $ \mathbf{E}$ denotes the three-by-three identity matrix, and

$\displaystyle \mathbf{I}\eqsp m \left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}-\underline{x}\underline{x}^T\right), \protect$ (B.24)

which agrees with Eq.$ \,$(B.20). Thus we have derived the moment of inertia $ I$ in terms of the moment of inertia tensor $ \mathbf{I}$ and the normalized angular velocity $ \underline{\tilde{\omega}}$ for a point-mass $ m$ at $ \underline{x}$.

For a collection of $ N$ masses $ m_i$ located at $ \underline{x}_i\in{\bf R}^3$, we simply sum over their masses to add up the moments of inertia:

$\displaystyle \mathbf{I}\eqsp \sum_{i=1}^N m_i \left(\left\Vert\,\underline{x}_i\,\right\Vert^2\mathbf{E}
-\underline{x}_i\underline{x}_i^T\right)
$

Finally, for a continuous mass distribution, we integrate as usual:

$\displaystyle \mathbf{I}\eqsp \frac{1}{M}\int_V \rho(\underline{x}) \left(\left...
...underline{x}\,\right\Vert^2\mathbf{E}
-\underline{x}\underline{x}^T\right)\,dV
$

where $ M=\int_V\rho(\underline{x})dV$ is the total mass.

Simple Example

Consider a mass $ m$ at $ \underline{x}=[x,0,0]^T$. Then the mass moment of inertia tensor is

$\displaystyle \mathbf{I}\eqsp m \left(\left\Vert\,\underline{x}\,\right\Vert^2\...
...ay}{ccc}
0 & 0 & 0\\ [2pt]
0 & 1 & 0\\ [2pt]
0 & 0 & 1
\end{array}\right].
$

For the angular-velocity vector $ \underline{\omega}=[\omega,0,0]^T$, we obtain the moment of inertia

\begin{displaymath}
I \eqsp \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\...
...{array}{c} 1 \\ [2pt] 0 \\ [2pt] 0\end{array}\right]m \eqsp 0.
\end{displaymath}

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$, we get

$\displaystyle I \eqsp \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde...
...]
\left[\begin{array}{c} 0 \\ [2pt] 1 \\ [2pt] 0\end{array}\right] \eqsp m x^2
$

which is what we expected.


Example with Coupled Rotations

Now let the mass $ m$ be located at $ \underline{x}=[1,1,0]^T$ so that

\begin{eqnarray*}
\mathbf{I}&=& m \left(\left\Vert\,\underline{x}\,\right\Vert^2...
... & 0\\ [2pt]
-1 & 1 & 0\\ [2pt]
0 & 0 & 2
\end{array}\right].
\end{eqnarray*}

We expect $ \underline{\omega}=[1,1,0]$ to yield zero for the moment of inertia, and sure enough $ \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tilde{\omega}}=0$. Similarly, the vector angular momentum is zero, since $ \mathbf{I}\,\underline{\omega}=\underline{0}$.

For $ \underline{\omega}=[1,0,0]^T$, the result is

\begin{displaymath}
\mathbf{I}\eqsp
\begin{array}{r}\left[\begin{array}{ccc} 1 ...
...egin{array}{c} 1 \\ [2pt] 0 \\ [2pt] 0\end{array}\right]m = m,
\end{displaymath}

which makes sense because the distance from the axis $ \underline{e}_1$ to $ \underline{x}$ is one. The same result is obtained for rotation about $ \underline{\omega}=\underline{e}_2$. For $ \underline{\omega}=\underline{e}_3$, however, the result is $ I=2m = m \vert\vert\,\underline{x}\,\vert\vert ^2$, as expect.


Off-Diagonal Terms in Moment of Inertia Tensor

This all makes sense, but what about those $ -1$ off-diagonal terms in $ \mathbf{I}$? Consider the vector angular momentumB.4.14):

$\displaystyle \underline{L}\eqsp \mathbf{I}\,\underline{\omega}\eqsp
m\left[\b...
...begin{array}{c} \omega_1 \\ [2pt] \omega_2 \\ [2pt] \omega_3\end{array}\right]
$

We see that the off-diagonal terms $ I_{ij}$ correspond to a coupling of rotation about $ \underline{e}_i$ with rotation about $ \underline{e}_j$. That is, there is a component of moment-of-inertia $ I_{ij}$ that is contributed (or subtracted, as we saw above for $ \underline{\omega}=[1,1,0]^T$) when both $ \omega_i$ and $ \omega_j$ 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 $ \underline{x}=\underline{0}$ 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:

$\displaystyle \underline{L}\eqsp \mathbf{I}\,\underline{\omega}
$

If $ \underline{\omega}$ is an eigenvector of $ \mathbf{I}$, then we have

$\displaystyle \underline{L}\eqsp \mathbf{I}\,\underline{\omega}\eqsp \lambda\underline{\omega}
$

where the (scalar) eigenvalue $ \lambda$ is called a principal moment of inertia. If we set the rigid body assocated with $ \mathbf{I}$ rotating about the axis $ \underline{\omega}$, then $ \lambda$ 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).

Positive Definiteness of the Moment of Inertia Tensor

From the form of the moment of inertia tensor introduced in Eq.$ \,$(B.24)

$\displaystyle \mathbf{I}\eqsp m \left(\left\Vert\,\underline{x}\,\right\Vert^2\mathbf{E}-\underline{x}\underline{x}^T\right), \protect$

it is clear that $ \mathbf{I}$ is symmetric. Moreover, for any normalized angular-velocity vector $ \underline{\tilde{\omega}}$ we have

\begin{eqnarray*}
I &=& \underline{\tilde{\omega}}^T\mathbf{I}\,\underline{\tild...
...\cal P}_{\underline{x}}(\underline{\tilde{\omega}})\right] \ge 0
\end{eqnarray*}

since $ \underline{\tilde{\omega}}$ is unit length, and projecting it onto any other vector can only shorten it or leave it unchanged. That is, $ \vert\underline{\tilde{\omega}}^T{\cal P}_{\underline{x}}(\underline{\tilde{\omega}})\vert \le 1$, with equality occurring for $ \underline{x}=\alpha\underline{\tilde{\omega}}$ for any nonzero $ \alpha\in{\bf R}$. Zooming out, of course we expect any moment of inertia $ I$ for a positive mass $ m$ to be nonnegative. Thus, $ \mathbf{I}$ is symmetric nonnegative definite. If furthermore $ \underline{\tilde{\omega}}$ and $ \underline{x}$ are not collinear, i.e., if there is any nonzero angle between them, then $ \mathbf{I}$ is positive definite (and $ I>0$). 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 $ \underline{\omega}$ 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 $ \mathbf{I}$ (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 $ MR^2/2$, was $ \underline{e}_3$ (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 $ MR^2/4$).

Symmetry of the rigid body about any axis $ \underline{\omega}_s$ (passing through the origin) means that $ \underline{\omega}_s$ 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 $ \underline{\omega}_s$ axis, intersecting at the $ \underline{\omega}_s$ axis. Because of the circular symmetry about $ \underline{\omega}_s$, the two principal moments of inertia in that plane are equal. Thus the moment of inertia tensor can be diagonalized to look like

$\displaystyle \mathbf{I}= \left[\begin{array}{ccc}
I_1 & 0 & 0\\ [2pt]
0 & I_2 & 0\\ [2pt]
0 & 0 & I_2
\end{array}\right],
$

where $ I_1$ is the principal moment of inertia about $ \underline{\omega}_s$, and $ I_2$ 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 $ I_1=2I_2$.



Rotational Kinetic Energy Revisited

If a point-mass is located at $ \underline{x}$ and is rotating about an axis-of-rotation $ \underline{\tilde{\omega}}$ with angular velocity $ \omega $, then the distance from the rotation axis to the mass is $ r= \vert\vert\,\underline{x}-{\cal P}_{\underline{\tilde{\omega}}}(\underline{...
...nderline{\tilde{\omega}}\underline{\tilde{\omega}}^T)\underline{x}\,\vert\vert $, or, in terms of the vector cross product, $ r= \vert\vert\,\underline{\tilde{\omega}}\times\underline{x}\,\vert\vert $. The tangential velocity of the mass is then $ r\omega$, so that the kinetic energy can be expressed as (cf. Eq.$ \,$(B.23))

$\displaystyle E_R \eqsp \frac{1}{2}m\left\Vert\,\underline{\tilde{\omega}}\times\underline{x}\,\right\Vert^2\omega^2 \eqsp \frac{1}{2}I\omega^2 \protect$ (B.25)

where

$\displaystyle I \eqsp m\left\Vert\,\underline{\tilde{\omega}}\times\underline{x}\,\right\Vert^2.
$

In a collection of $ N$ masses $ m_i$ having velocities $ \underline{v}_i$, 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

$\displaystyle E_R \eqsp \frac{1}{2}\, \underline{\omega}\cdot \underline{L}\eqsp \frac{1}{2} I \omega^2,
$

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 $ \underline{\omega}\cdot\underline{L}=\underline{\omega}\cdot \mathbf{I}\underl...
...^2 \underline{\tilde{\omega}}^T\mathbf{I}\underline{\tilde{\omega}}= I \omega^2$.


Torque

Figure B.7: Application of torque $ \tau $ about the origin given by a tangential force $ f_t$ on a lever arm of length $ R$.
\includegraphics[width=1.1in]{eps/torque}

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 $ f_t$ times the moment arm (length of the lever arm) $ R$

$\displaystyle \tau \isdefs f_tR \protect$ (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, $ f_t$ is the force applied at the end of the wrench by one's hand, orthogonal to the wrench, while the moment arm $ R$ is the length of the wrench. Doubling the length of the wrench doubles the torque. This is an example of leverage. When $ R$ is increased, a given twisting angle $ \theta$ is spread out over a larger arc length $ R\theta$, thereby reducing the tangential force $ f_t$ required to assert a given torque $ \tau $.

For more general applied forces $ \underline{f}\in{\bf R}^3$, we may compute the tangential component $ \underline{f}_t$ by projecting $ \underline{f}$ onto the tangent direction. More precisely, the torque $ \tau $ about the origin $ \underline{0}$ applied at a point $ \underline{x}\in{\bf R}^3$ may be defined by

$\displaystyle \underline{\tau}\isdefs \underline{x}\times \underline{f} \protect$ (B.27)

where $ \underline{f}$ is the applied force (at $ \underline{x}$) and $ \times$ denotes the cross product, introduced above in §B.4.12.

Note that the torque vector $ \underline{\tau}$ 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

$\displaystyle \tau \isdefs \vert\vert\,\tau\,\vert\vert \eqsp \vert\vert\,\unde...
...{x}\,\vert\vert \cdot \vert\vert\,\underline{f}\,\vert\vert \cdot\sin(\theta),
$

where $ \theta$ denotes the angle from $ \underline{x}$ to $ \underline{f}$. We can interpret $ \vert\vert\,\underline{f}\,\vert\vert \sin(\theta)$ as the length of the projection of $ \underline{f}$ onto the tangent direction (the line orthogonal to $ \underline{x}$ in the direction of the force), so that we can write

$\displaystyle \tau \eqsp \vert\vert\,\underline{x}\,\vert\vert \cdot \vert\vert\,\underline{f}_t\,\vert\vert
$

where $ \vert\vert\,\underline{f}_t\,\vert\vert = \vert\vert\,\underline{f}\,\vert\vert \sin(\theta)$, thus getting back to Eq.$ \,$(B.26).


Newton's Second Law for Rotations

The rotational version of Newton's law $ f=ma$ is

$\displaystyle \tau \eqsp I\alpha, \protect$ (B.28)

where $ \alpha\isdeftext \dot{\omega}$ denotes the angular acceleration. As in the previous section, $ \tau $ is torque (tangential force $ f_t$ times a moment arm $ R$), and $ I$ is the mass moment of inertia. Thus, the net applied torque $ \tau $ equals the time derivative of angular momentum $ L=I\omega$, just as force $ f$ equals the time-derivative of linear momentum $ p$:

\begin{eqnarray*}
\tau &=& \dot{L} \,\eqss \, I\dot{\omega}\,\isdefss \, I\alpha\\ [5pt]
f &=& \dot{p} \,\eqss \, m\dot{v}\,\isdefss \, ma
\end{eqnarray*}

To show that Eq.$ \,$(B.28) results from Newton's second law $ f=ma$, consider again a mass $ m$ rotating at a distance $ R$ from an axis of rotation, as in §B.4.3 above, and let $ f_t$ denote a tangential force on the mass, and $ a_t$ the corresponding tangential acceleration. Then we have, by Newton's second law,

$\displaystyle f_t \eqsp ma_t
$

Multiplying both sides by $ R$ gives

$\displaystyle f_tR \eqsp ma_tR \isdefs m\dot{v}_tR \isdefs m\dot{\omega}R^2 \eqsp
I\dot{\omega} \eqsp I\alpha.
$

where we used the definitions $ \omega=v_tR$ and $ I=mR^2$. Furthermore, the left-hand side is the definition of torque $ \tau=f_tR$. Thus, we have derived

$\displaystyle \tau\eqsp I\alpha
$

from Newton's second law $ f_t=ma_t$ applied to the tangential force $ f_t$ and acceleration $ a_t$ of the mass $ m$.

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:

$\displaystyle \dot{p_t}=ma_t\;\;\;\Leftrightarrow\;\;\; \dot{L}=I\alpha
$


Equations of Motion for Rigid Bodies

We are now ready to write down the general equations of motion for rigid bodies in terms of $ f=ma$ for the center of mass and $ \tau=I\alpha$ 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 $ \underline{v}$ of its center of mass, and
(2)
its angular velocity $ \underline{\omega}$ about its center of mass.

The linear motion is governed by Newton's second law $ \underline{f}=M\dot{\underline{v}}$, where $ M$ is the total mass, $ \underline{v}$ is the velocity of the center-of-mass, and $ \underline{f}$ is the sum of all external forces on the rigid body. (Equivalently, $ \underline{f}$ 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:

$\displaystyle \underline{\tau}\eqsp \dot{\underline{L}} \eqsp \mathbf{I}\,\dot{\underline{\omega}} \eqsp \mathbf{I}\,\dot{\underline{\omega}} \protect$ (B.29)

where $ \tau $ is the vector torque defined in Eq.$ \,$(B.27), $ \underline{L}$ is the angular momentum, $ \mathbf{I}$ is the mass moment of inertia tensor, and $ \underline{\omega}$ 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 $ \underline{L}$ 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 $ \underline{\tau}$ 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 ( $ \underline{0}$) 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 ( $ \mathbf{I}$ 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 frameB.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]

$\displaystyle \underline{L}\eqsp \sum \underline{x}\times \underline{p}
$

where the sum is over all mass particles in the rigid body, and $ \underline{p}$ 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

$\displaystyle \underline{\tau}\eqsp \dot{\underline{L}} \eqsp \frac{d}{dt}\sum ...
...m (\underline{v}\times\underline{p}+ \underline{x}\times \dot{\underline{p}}).
$

However, $ \underline{v}\times \underline{p}=\underline{v}\times m\underline{v}=\underline{0}$, so that

$\displaystyle \underline{\tau}\eqsp \dot{\underline{L}} \eqsp \sum \underline{x}\times \dot{\underline{p}}
\eqsp \sum \underline{x}\times \underline{f}
$

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 $ \underline{\omega}$. Then the total torque on the rigid body becomes [270]

$\displaystyle \underline{\tau}\eqsp \dot{\underline{L}} + \underline{\omega}\times \underline{L}. \protect$ (B.30)

Similarly, the total external forces on the center of mass become

$\displaystyle \underline{f}\eqsp \dot{\underline{p}} + \underline{\omega}\times\underline{p}.
$

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 $ \mathbf{I}=$diag$ (I_1,I_2,I_3)$. In this frame, the angular momentum is simply

$\displaystyle \underline{L}\eqsp \left[\begin{array}{c} I_1\omega_1 \\ [2pt] I_2\omega_2 \\ [2pt] I_3\omega_3\end{array}\right]
$

so that the term $ \underline{\omega}\times\underline{L}$ becomes (cf. Eq.$ \,$(B.15))

\begin{eqnarray*}
\underline{\omega}\times\underline{L}&=&
\left\vert \begin{arr...
...1\,\underline{e}_2 +
(I_2-I_1)\omega_1\omega_2\,\underline{e}_3.
\end{eqnarray*}

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:

\begin{eqnarray*}
\tau_1 &=& I_1 \dot{\omega}_1 + (I_3-I_2)\omega_2\omega_3\\
\...
...a_1\\
\tau_3 &=& I_3 \dot{\omega}_3 + (I_2-I_1)\omega_1\omega_2 \end{eqnarray*}

These are call Euler's equations:B.29Since these equations are in the body-fixed frame, $ I_i$ is the mass moment of inertia about principal axis $ i$, and $ \omega_i$ is the angular velocity about principal axis $ i$.


Examples

For a uniform sphere, the cross-terms disappear and the moments of inertia are all the same, leaving $ \tau_i=I\omega_i$, for $ i=1,2,3$. 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 $ \tau_i=I\omega_i$, for $ i=1,2$, given $ \omega_3=\tau_3=0$ (no spin). Note, however, that if we replace the circular cross-section of the cylinder by an ellipse, then $ I_1\ne I_2$ and there is a coupling term that drives $ \dot{\omega}_3$ (unless $ \tau_3$ happens to cancel it).


Properties of Elastic Solids

Young's Modulus

Young's modulus can be thought of as the spring constant for solids. Consider an ideal rod (or bar) of length $ L$ and cross-sectional area $ S$. Suppose we apply a force $ F$ to the face of area $ S$, causing a displacement $ \Delta L$ along the axis of the rod. Then Young's modulus $ Y$ is given by

$\displaystyle Y \isdefs \frac{\mbox{Stress}}{\mbox{Strain}} \isdefs \frac{F/S}{\Delta L/L}
$

where

\begin{eqnarray*}
F &=& \mbox{total applied force}\\
S &=& \mbox{area over whic...
...a L/L &=& \mbox{\emph{strain} = displacement per unit length}\\
\end{eqnarray*}

For wood, Young's modulus $ Y$ is on the order of $ 10$ N/m$ \null^2$. For aluminum, it is around $ 70$ (a bit higher than glass which is near $ 65$), and structural steel has $ Y\approx 200$ [180].

Young's Modulus as a Spring Constant

Recall (§B.1.3) that Hooke's Law defines a spring constant $ k$ as the applied force $ F$ divided by the spring displacement $ x$, or $ F = k x$. 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 $ L$), and consider compression by $ \Delta L$ along the length dimension. The length of each spring in the bundle is the length of the bar, so that each spring constant $ k$ must be inversely proportional to $ L$; in particular, each doubling of length $ L$ doubles the length of each ``spring'' in the bundle, and therefore halves its stiffness. As a result, it is useful to normalize displacement $ \Delta L$ by length $ L$ and use relative displacement $ \Delta L/L$. 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 $ S$ of the bar. Therefore, the force applied to each spring is proportional to the total applied force $ F$ divided by the cross-sectional area $ S$. Thus, Hooke's law for each spring in the bundle can be written

$\displaystyle \frac{F}{S} = Y \frac{\Delta L}{L}
$

where $ Y$ 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 $ F$ used to stretch it. It is therefore directed along the axis of the string. A force $ F$ must be applied at the endpoint on the right, and a force $ -F$ 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 $ F + (-F) = 0$. (A nonzero force on a massless point would produce an infinite acceleration.)

If the cross-sectional area of the string is $ S$, then the tension is given by the stress on the string times $ S$.


Wave Equation for the Vibrating String

Consider an elastic string under tension which is at rest along the $ x$ dimension. Let $ \mathbf{i}$, $ \mathbf{j}$, and $ \mathbf{k}$ denote the unit vectors in the $ x$, $ y$, and $ z$ directions, respectively. When a wave is present, a point $ \mathbf{p}=(x,0,0)$ originally at $ x$ along the string is displaced to some point $ \mathbf{a}=\mathbf{p}+d\mathbf{p}$ specified by the displacement vector

$\displaystyle d\mathbf{p}= \mathbf{i}\xi + \mathbf{j}\eta + \mathbf{k}\zeta.
$

Note that typical derivations of the wave equation consider only the displacement $ \eta$ in the $ y$ 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 $ \mathbf{q}=(x+d
x,0,0)$ along the string can be specified as

$\displaystyle d\mathbf{q}= \mathbf{i}(\xi+d\xi) + \mathbf{j}(\eta+d\eta) + \mathbf{k}(\zeta+d\zeta).
$

Let $ K$ denote string tension along $ x$ when the string is at rest, and $ \mathbf{K}$ denote the vector tension at the point $ \mathbf{p}$ in the present displaced scenario under analysis. The net vector force acting on the infinitesimal string element between points $ \mathbf{p}$ and $ \mathbf{q}$ is given by the vector sum of the force $ -\mathbf{K}$ at $ \mathbf{p}$ and the force $ \mathbf{K}+ (\partial \mathbf{K}/\partial x)d
x$ at $ \mathbf{q}$, that is, $ (\partial \mathbf{K}/\partial x)dx$. 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 $ \epsilon \,dx$, where $ \epsilon $ denotes the mass per unit length of the string at rest. Applying Newton's second law gives

$\displaystyle \frac{\partial \mathbf{K}}{\partial x} = \epsilon \frac{\partial^2 \mathbf{p}}{\partial t^2}$ (B.31)

where $ d x$ 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 $ \mathbf{K}$ in terms of the tension $ K$ of the string at rest, the elastic constant of the string, and geometrical factors. The displaced string element $ \mathbf{p}\mathbf{q}$ is the vector

$\displaystyle d{\bf s}$ $\displaystyle =$ $\displaystyle \mathbf{i}(dx + d\xi) + \mathbf{j}d\eta + \mathbf{k}d\zeta$ (B.32)
  $\displaystyle =$ $\displaystyle \left[\mathbf{i}\left(1+\frac{\partial \xi}{\partial x}\right)+
\...
...artial \eta}{\partial x} +
\mathbf{k}\frac{\partial \zeta}{\partial x}\right]dx$ (B.33)

having magnitude

$\displaystyle ds = \sqrt{\left(1+\frac{\partial \xi}{\partial x}\right)^2 + \le...
...\eta}{\partial x}\right)^2 + \left(\frac{\partial \zeta}{\partial x}\right)^2}.$ (B.34)

Non-Stiff String

Let's now assume the string is perfectly flexible (zero stiffness) so that the direction of the force vector $ \mathbf{K}$ is given by the unit vector $ d{\bf s}/ds$ 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 $ \mathbf{K}$ at any position is the rest tension $ K$ plus the incremental tension needed to stretch it the fractional amount

$\displaystyle \frac{ds - dx}{dx} = \frac{ds}{dx} - 1.
$

If $ S$ is the constant cross-sectional area of the string, and $ Y$ is the Young's modulus (stress/strain--the ``spring constant'' for solids--see §B.5.1), then

$\displaystyle \left\vert\mathbf{K}\right\vert = K+ SY\left(\frac{ds}{dx} - 1\right),
$

so that

$\displaystyle \mathbf{K}= \left[K+ SY\left(\frac{ds}{dx} - 1\right)\right]\frac{d{\bf s}}{ds}$ (B.35)

where no geometrical limitations have yet been placed on the magnitude of $ \mathbf{p}$ and $ \partial\mathbf{p}/\partial x$, 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 $ \partial\xi/\partial x$, $ \partial\eta/\partial x$, and $ \partial\zeta/\partial x$ be small compared with unity. This is the same assumption ( $ \vert\partial\eta/\partial x\vert\ll 1$) necessary to derive the usual wave equation for transverse vibrations only in the $ y$-$ x$ plane.

When (B.35) is expanded into a Taylor series in the strains, and when only the first-order terms are retained, we obtain

$\displaystyle \mathbf{K}= \mathbf{i}\left(K+ SY \frac{\partial \xi}{\partial x}...
...frac{\partial \eta}{\partial x} + \mathbf{k}K\frac{\partial \zeta}{\partial x}.$ (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 $ \mathbf{K}$, it is found that (B.31) separates into the three wave equations
$\displaystyle \frac{\partial^2 \xi}{\partial x^2}$ $\displaystyle =$ $\displaystyle \frac{1}{c_b^2} \frac{\partial^2 \xi}{\partial t^2}$ (B.37)
$\displaystyle \frac{\partial^2 \eta}{\partial x^2}$ $\displaystyle =$ $\displaystyle \frac{1}{c_t^2} \frac{\partial^2 \eta}{\partial t^2}$ (B.38)
$\displaystyle \frac{\partial^2 \zeta}{\partial x^2}$ $\displaystyle =$ $\displaystyle \frac{1}{c_t^2} \frac{\partial^2 \zeta}{\partial t^2}$ (B.39)

where $ c_b=\sqrt{SY/\epsilon }$ is the longitudinal wave velocity, and $ c_t=\sqrt{K/\epsilon }$ 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.,

$\displaystyle \left\vert\frac{\partial \xi}{\partial x}\right\vert \ll 1, \quad
\left\vert\frac{\partial \eta}{\partial x}\right\vert \ll 1,$   and$\displaystyle \left\vert\frac{\partial \zeta}{\partial x}\right\vert \ll 1.
$


Wave Momentum

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]

$\displaystyle \epsilon {\ddot \xi}$ $\displaystyle =$ $\displaystyle \left(SY+K\right) \xi^{\prime\prime} +
SY\eta^\prime\eta^{\prime\prime}$ (B.40)
$\displaystyle \epsilon {\ddot \eta}$ $\displaystyle =$ $\displaystyle K \eta^{\prime\prime} +
SY\left(\frac{3}{2}(\eta^\prime)^2\eta^{\...
...rime}
+\xi^{\prime}\eta^{\prime\prime} + \eta^{\prime}\xi^{\prime\prime}\right)$ (B.41)
  $\displaystyle \approx$ $\displaystyle K\eta^{\prime\prime},$ (B.42)

where $ \xi$ and $ \eta$ denote longitudinal and transverse displacement, respectively, and the commonly used ``dot'' and ``prime'' notation for partial derivatives has been introduced, e.g.,
$\displaystyle {\dot \xi}$ $\displaystyle \isdef$ $\displaystyle \frac{\partial \xi}{\partial t}$ (B.43)
$\displaystyle {\xi^{\prime}}$ $\displaystyle \isdef$ $\displaystyle \frac{\partial \xi}{\partial x}.$ (B.44)

(See also Eq.$ \,$(C.1).) We see that the term $ SY\eta^\prime\eta^{\prime\prime}$ 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 $ u$.


Volume Velocity of a Gas

The volume velocity $ U$ of a gas flow is defined as particle velocity $ u$ times the cross-sectional area $ A$ of the flow, or

$\displaystyle U(t,x) = u(t,x) A(x)
$

where $ x$ denotes position along the flow, and $ t$ denotes time in seconds. Volume velocity is thus in physical units of volume per second (m$ \null^3$/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 $ \rho$ remains constant. This follows directly from conservation of mass in a flow: The total mass passing a given point $ x$ along the flow is given by the mass density $ \rho$ times the integral of the volume volume velocity at that point, or

$\displaystyle M(t_0:t_f,x) = \rho\int_{t_0}^{t_f} U(t,x) dt.
$

As a simple example, consider a constant flow through two cylindrical acoustic tube sections having cross-sectional areas $ A_1$ and $ A_2$, respectively. If the particle velocity in cylinder 1 is $ u_1$, then the particle velocity in cylinder 2 may be found by solving

$\displaystyle u_1 A_1 = u_2 A_2
$

for $ u_2$.

It is common in the field of acoustics to denote volume velocity by an upper-case $ U$. Thus, for the two-cylinder acoustic tube example above, we would define $ U_1\isdeftext u_1A_1$ and $ U_2\isdeftext u_2A_2$, so that

$\displaystyle U_1 = U_2
$

would express the conservation of volume velocity from one tube segment to the next.


Pressure is Confined Kinetic Energy

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.

$\displaystyle p = \frac{1}{3}\rho \left<u^2\right>
$

Here, $ \langle u^2\rangle $ denotes the average squared particle velocity in the gas. (The constant $ 1/3$ 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 $ M$ be the total mass of a gas confined to a rectangular volume $ V = Aw$, where $ A$ is the area of one side and $ w$ the distance to the opposite side. Let $ \overline{u}_x$ denote the average molecule velocity in the $ x$ direction. Then the total net molecular momentum in the $ x$ direction is given by $ M\vert\overline{u}_x\vert$. Suppose the momentum $ \overline{u}_x$ is directed against a face of area $ A$. A rigid-wall elastic collision by a mass $ M$ traveling into the wall at velocity $ \overline{u}_x$ imparts a momentum of magnitude $ 2M\overline{u}_x$ to the wall (because the momentum of the mass is changed from $ +M\overline{u}_x$ to $ -M\overline{u}_x$, and momentum is conserved). The average momentum-transfer per unit area is therefore $ 2M\overline{u}_x/A$ at any instant in time. To obtain the definition of pressure, we need only multiply by the average collision rate, which is given by $ \overline{u}_x/(2w)$. That is, the average $ x$-velocity divided by the round-trip distance along the $ x$ dimension gives the collision rate at either wall bounding the $ x$ dimension. Thus, we obtain

$\displaystyle p \isdefs \frac{2M\overline{u}_x}{A}\cdot \frac{\overline{u}_x}{2w} \eqsp \rho \overline{u}_x^2
$

where $ \rho=M/V$ is the density of the gas in mass per unit volume. The quantity $ \rho\overline{u}_x^2/2$ is the average kinetic energy density of molecules in the gas along the $ x$ dimension. The total kinetic energy density is $ \rho\overline{u}^2/2$, where $ \overline{u}=\sqrt{\overline{u}_x^2+\overline{u}_y^2+\overline{u}_z^2}$ 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 $ \overline{u}_x^2=\overline{u}_y^2=\overline{u}_z^2 = \overline{u}^2/3$, so that

$\displaystyle p = \frac{1}{3}\rho \overline{u}^2.
$


Bernoulli Equation

In an ideal inviscid, incompressible flow, we have, by conservation of energy,

$\displaystyle p + \frac{1}{2}\rho u^2 + \rho g h =$   constant

where

\begin{eqnarray*}
p &=& \mbox{pressure (newtons/m$^2$\ = kg /(m s$^2$))}\\
u &=...
...\
\mbox{\lq\lq Inviscid''} &=& \mbox{\lq\lq Frictionless'', \lq\lq Lossless''}
\end{eqnarray*}

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 $ h$, 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.
\includegraphics{eps/bernoulli-effect}

Figure B.8 illustrates the Bernoulli effect for the case of a reservoir at constant pressure $ p_m$ (``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 $ u$. To conserve energy, the pressure within the acoustic channel must drop down to $ p_m - \rho u^2/2$. 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].


Air Jets

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 $ p_m$) and the outside air (pressure $ p_m - \rho u^2/2$). Therefore, any pressure recovery would erode the pressure drop and hence the flow velocity $ u$.

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

Acoustic intensity may be defined by

$\displaystyle \zbox {\underline{I} \isdefs p \underline{v}}
\quad \left(\frac{\...
...mall Time}}
\eqsp
\frac{\mbox{\small Power Flux}}{\mbox{\small Area}}\right)
$

where

\begin{eqnarray*}
p &=& \mbox{acoustic pressure} \quad \left(\frac{\mbox{\small ...
...ad \left(\frac{\mbox{\small Length}}{\mbox{\small Time}}\right).
\end{eqnarray*}

For a plane traveling wave, we have

$\displaystyle \zbox {p = R v}
$

where

$\displaystyle \zbox {R \isdefs \rho c}
$

is called the wave impedance of air, and

\begin{eqnarray*}
c &=& \mbox{sound speed},\\
\rho &=& \mbox{mass density of ai...
...ume}}\right),\\
v &\isdef & \left\vert\underline{v}\right\vert.
\end{eqnarray*}

Therefore, in a plane wave,

$\displaystyle \zbox {I = p v = Rv^2 = \frac{p^2}{R}.}
$


Acoustic Energy Density

The two forms of energy in a wave are kinetic and potential. Denoting them at a particular time $ t$ and position $ \underline{x}$ by $ w_v(t,\underline{x})$ and $ w_p(t,\underline{x})$, respectively, we can write them in terms of velocity $ v$ and wave impedance $ R=\rho c$ as follows:

\begin{eqnarray*}
w_v &=& \frac{1}{2} \rho v^2 \eqsp \frac{1}{2c} R v^2 \quad\le...
...ad\left(\frac{\mbox{\small Energy}}{\mbox{\small Volume}}\right)
\end{eqnarray*}

More specifically, $ w_v$ and $ w_p$ may be called the acoustic kinetic energy density and the acoustic potential energy density, respectively.

At each point in a plane wave, we have $ p(t,\underline{x})=R\,v(t,\underline{x})$ (pressure equals wave-impedance times velocity), and so

\begin{eqnarray*}
w_v &=& \frac{1}{2c} R v^2 = \frac{1}{2}\cdot \frac{I}{c}\\
w_p &=& \frac{1}{2c} \frac{p^2}{R} = \frac{1}{2} \cdot \frac{I}{c},
\end{eqnarray*}

where $ I(t,\underline{x})\isdef p(t,\underline{x})\,v(t,\underline{x})$ denotes the acoustic intensity (pressure times velocity) at time $ t$ and position $ \underline{x}$. Thus, half of the acoustic intensity $ I$ in a plane wave is kinetic, and the other half is potential:B.30

$\displaystyle \frac{I}{c} = w = w_v+w_p = 2w_v = 2w_p
$

Note that acoustic intensity $ I$ has units of energy per unit area per unit time while the acoustic energy density $ w=I/c$ has units of energy per unit volume.


Energy Decay through Lossy Boundaries

Since the acoustic energy density $ w=w_v+w_p$ 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:

$\displaystyle E(t) = \iiint\limits_V w(t,\underline{x}) \,dV
$

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 $ V$ in time interval $ \Delta t$ must equal the acoustic intensity $ \underline{I}(t,\underline{x})$ exiting the volume, times $ \Delta t$ (approximating $ I$ as constant between times $ t$ and $ t+\Delta t$):

$\displaystyle E(t+\Delta t) - E(t) = -\Delta t \iint\limits_A \underline{I}\cdot \underline{n}\, dA
$

The term $ \underline{I}(t,\underline{x})\cdot\underline{n}(\underline{x})$ is the dot-product of the (vector) intensity $ \underline{I}$ with a unit-vector $ \underline{n}$ chosen to be normal to the surface at position $ \underline{x}$ along the surface. Thus, $ \underline{I}\cdot \underline{n}$ is the component of the acoustic intensity $ \underline{I}$ exiting the volume normal to its surface. (The tangential component does not exit.) Dividing through by $ \Delta t$ and taking a limit as $ \Delta t\to 0$ yields the following conservation law, originally published by Kirchoff in 1867:

$\displaystyle \frac{d}{dt}E = \frac{d}{dt}\iiint\limits_V w(t,\underline{x}) \,...
...imits_A \underline{I}(t,\underline{x})\cdot \underline{n}(\underline{x})\, dA.
$

Thus, the rate of change of energy in an ideal acoustic volume $ V$ 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.


Ideal Gas Law

The ideal gas law can be written as

$\displaystyle PV \eqsp nRT \eqsp NkT \protect$ (B.45)

where

\begin{eqnarray*}
P &=& \mbox{total pressure (Pascals)}\\
V &=& \mbox{volume (c...
...ture}\index{absolute temperature\vert textbf} (degrees Kelvin).}
\end{eqnarray*}

The alternate form $ PV=NkT$ comes from the statistical mechanics derivation in which $ N$ is the number of gas molecules in the volume, and $ k$ is Boltzmann's constant. In this formulation (the kinetic theory of ideal gases), the average kinetic energy of the gas molecules is given by $ (3/2) kT$. Thus, temperature is proportional to average kinetic energy of the gas molecules, where the kinetic energy of a molecule $ m$ with translational speed $ v$ is given by $ (1/2)mv^2$.

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 $ m$ of the gas in volume $ V$ is given by $ m=nM$, where $ M$ is the molar mass of the gass (about 29 g per mole for air). The air density is thus $ \rho=m/V$ so that we can write

$\displaystyle P \eqsp \frac{R}{M} \rho T.
$

That is, pressure $ P$ is proportional to density $ \rho$ at constant temperature $ T$ (with $ R/M$ 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 $ P_0$. The physical pressure is then $ P=P_0+p$, where $ p$ is the usual acoustic pressure-wave variable. That is, we are only concerned with small pressure perturbations $ p$ in typical audio acoustics situations, so that, for example, variations in volume $ V$ and density $ \rho$ 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].


Isothermal versus Isentropic

If air compression/expansion were isothermal (constant temperature $ T$), then, according to the ideal gas law $ PV=nRT$, the pressure $ P$ would simply be proportional to density $ \rho$. 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 $ S$) 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 $ V$, for example, not only does the pressure $ P$ increase (§B.7.3), but the temperature $ T$ 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.


Adiabatic Gas Constant

The relative amount of compression/expansion energy that goes into temperature $ T$ versus pressure $ P$ can be characterized by the heat capacity ratio

$\displaystyle \gamma \isdefs \frac{C_P}{C_V}
$

where $ C_P$ is the specific heat (also called heat capacity) at constant pressure, while $ C_V$ 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 $ C_p=C_v+R$, where $ R$ is the ideal gas constant (introduced in Eq.$ \,$(B.45)). Thus, $ \gamma>1$ for any ideal gas. The extra heat absorption that occurs when heating a gas at constant pressure is associated with the workB.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 $ \gamma$ 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 $ \gamma$, we have

$\displaystyle P_1V_1^\gamma \eqsp P_2V_2^\gamma, \protect$ (B.46)

where $ \gamma\approx 1.4$ for dry air at normal temperatures. Thus, if a volume of ideal gas is changed from $ V_1$ to $ V_2$, the pressure change is given by

$\displaystyle P_2 \eqsp P_1 \left(\frac{V_1}{V_2}\right)^{1/\gamma}
$

and the temperature change is

$\displaystyle T_2 \eqsp T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1}.
$

These equations both follow from Eq.$ \,$(B.46) and the ideal gas law Eq.$ \,$(B.45).

The value $ \gamma=1.4$ is typical for any diatomic gas.B.31 Monatomic inert gases, on the other hand, such as Helium, Neon, and Argon, have $ \gamma\approx 1.6$. Carbon dioxide, which is triatomic, has a heat capacity ratio $ \gamma=1.28$. We see that more complex molecules have lower $ \gamma$ values because they can store heat in more degrees of freedom.


Heat Capacity of Ideal Gases

In statistical thermodynamics [175,138], it is derived that each molecular degree of freedom contributes $ R/2$ to the molar heat capacity of an ideal gas, where again $ R$ 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, $ C_v=(3/2)R$. This means we expect

$\displaystyle \gamma\isdefs \frac{C_p}{C_v}\eqsp \frac{C_v+R}{C_v} \eqsp
\frac{3/2+1}{3/2} \eqsp 5/3,
$

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

$\displaystyle \gamma\isdefs \frac{C_p}{C_v}\eqsp \frac{C_v+R}{C_v} \eqsp
\frac{5/2+1}{5/2} \eqsp 7/5\eqsp 1.4,
$

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


Speed of Sound in Air

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

$\displaystyle c = \sqrt{\gamma R_m T},
$

where, as discussed in the previous section, the adiabatic gas constant is $ \gamma=1.4$ for dry air, $ R_m=286$ is the ideal gas constant for air in meters-squared per second-squared per degrees-Kelvin-squared, and $ T$ 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 $ x$ is given by

$\displaystyle I(x) = I_1 e^{-x/\xi},
$

where

\begin{eqnarray*}
I(x) &=& \hbox{intensity $x$\ meters from the source}\\
& & ...
...ox{(depends on frequency, temperature, humidity, and pressure).}
\end{eqnarray*}

Tables B.1 and B.2 (adapted from [314]) give some typical values for air.


Table B.1: Attenuation constant $ m = 1/\xi $ (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 ($ H_2O$) and/or carbon dioxide ($ CO_2$), the loss and dispersion due to $ N_2$ and $ O_2$ 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 $ 2\pi$ 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.


Wave Equation in Higher Dimensions

The wave equation in 1D, 2D, or 3D may be written as

$\displaystyle \left(\nabla ^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} \right) z(\underline{x},t) \eqsp 0, \protect$ (B.47)

where, in 3D, $ z(\underline{x},t)$ denotes the amplitude of the wave at time $ t$ and position $ \underline{x}\in{\bf R}^3$, and

$\displaystyle \nabla ^2 \isdefs \nabla \cdot \nabla \isdefs \nabla ^T\nabla \eq...
...tial x^2}
+ \frac{\partial^2}{\partial y^2}
+ \frac{\partial^2}{\partial z^2}
$

denotes the Laplacian operator in Euclidean coordinates. (In general coordinates, it is often denoted by $ \Delta$.) 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.

Plane Waves in Air

Figure B.9 shows a 2D $ xy$ cross-section of a snapshot (in time) of the sinusoidal plane wave

$\displaystyle p(x,y,z) = p_0 + \cos(k_x x + k_y y)
$

for $ k_x = 2\pi 5$ and $ k_y=2\pi 5/2$, with $ x$ and $ y$ in the range $ [0,1)$.

Figure: Gray-scale density plot of the $ xy$ cross-section of a sinusoidal plane wave $ p(t,\underline{x}) = \cos\left(\omega t -
\underline{k}^T\underline{x}\right)$, at $ t=0$ with vector wavenumber $ \underline{k}^T=[10\pi, 5\pi, 0]$.
\includegraphics[width=\twidth]{eps/planewave}

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 $ p(t,\underline{x}) = \cos\left(\omega t -
\underline{k}^T\underline{x}\right)$, for some fixed time $ t$ and $ \underline{x}$ in the $ (x,y,0)$ plane.
\includegraphics{eps/planewaveangle}

The direction of travel and spatial frequency are indicated by the vector wavenumber $ \underline{k}$, 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

$\displaystyle p(t,\underline{x}) \eqsp p_0 + A\cos\left(\omega t - \underline{k}^T\underline{x}+ \phi\right), \quad \underline{x}\in{\bf R}^3 \protect$ (B.48)

where p(t,x) is the pressure at time $ t$ (seconds) and position $ \underline{x}\in{\bf R}^3$ (3D Euclidean space). The amplitude $ A$, phase $ \phi$, and radian frequency $ \omega $ are ordinary sinusoid parameters [451], and $ \underline{k}$ is the vector wavenumber:

$\displaystyle \underline{k}\eqsp \left[\begin{array}{c} k_x \\ [2pt] k_y \\ [2p...
... \cos{\beta} \\ [2pt] \cos{\gamma}\end{array}\right] \isdefs k\,\underline{u},
$

where
  • $ \underline{u}= $ (unit) vector of direction cosines
  • $ k = 2\pi/\lambda = $ (scalar) wavenumber along travel direction
Thus, the vector wavenumber $ \underline{k}= k\,\underline{u}$ contains
  • wavenumber along the travel direction in its magnitude $ k=\left\Vert\,\underline{k}\,\right\Vert$
  • travel direction in its orientation $ \underline{u}= \underline{k}/k$
Note that wavenumber units are radians per meter (spatial radian frequency).

To see that the vector wavenumber $ \underline{k}= k\,\underline{u}$ has the claimed properties, consider that the orthogonal projection of any vector $ \underline{x}$ onto a vector collinear with $ \underline{u}$ is given by $ (\underline{u}^T\underline{x})\underline{u}$ [451].B.35Thus, $ (\underline{u}^T\underline{x})\underline{u}$ is the component of $ \underline{x}$ lying along the direction of wave propagation indicated by $ \underline{u}$. The norm of this component is $ \vert\vert\,(\underline{u}^T\underline{x})\underline{u}\,\vert\vert =\vert\underline{u}^T\underline{x}\vert$, since $ \underline{u}$ is unit-norm by construction. More generally, $ \underline{u}^T\underline{x}$ is the signed length (in meters) of the component of $ \underline{x}$ along $ \underline{u}$. This length times wavenumber $ k$ gives the spatial phase advance along the wave, or, $ \theta(\underline{x})=k\cdot(\underline{u}^T\underline{x}) \isdeftext \underline{k}^T\underline{x}$.

For another point of view, consider the plane wave $ \cos(\underline{k}^T\underline{x})$, which is the varying portion of the general plane-wave of Eq.$ \,$(B.48) at time $ t=0$, with unit amplitude $ A=1$ and zero phase $ \phi=0$. The spatial phase of this plane wave is given by

$\displaystyle \theta(\underline{x}) \isdefs \underline{k}^T\underline{x}\eqsp k_x x + k_y y + k_z z.
$

Recall that the general equation for a plane in 3D space is

$\displaystyle \alpha x + \beta y + \gamma z =$   constant

where $ \alpha$, $ \beta$, and $ \gamma$ are real constants, and $ x$, $ y$, and $ z$ are 3D spatial coordinates. Thus, the set of all points $ \underline{x}^T=(x,y,z)$ yielding the same value $ \theta(\underline{x})=\theta_0$ define a plane of constant phase $ \theta_0$ in $ {\bf R}^3$.

As we know from elementary vector calculus, the direction of maximum phase advance is given by the gradient of the phase $ \theta(\underline{x})$:

$\displaystyle \underline{\nabla }\theta(\underline{x}) \isdefs
\left[\begin{ar...
...rray}{c} k_x \\ [2pt] k_y \\ [2pt] k_z\end{array}\right] \isdefs \underline{k}
$

This shows that the vector wavenumber $ \underline{k}$ is equal to the gradient of the phase $ \theta(\underline{x})$, so that $ \underline{k}$ points in the direction of maximum spatial-phase advance.

Since the wavenumber $ k$ 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 $ \theta(\underline{x})$ along $ \underline{k}$, i.e.,

$\displaystyle k \isdefs d_{\underline{\nabla \theta}}\theta(\underline{x}) \isd...
...ta(\underline{x})}{\delta \left\Vert\,\underline{\nabla \theta}\,\right\Vert}.
$

An explicit calculation yields

$\displaystyle k \eqsp \left\Vert\,\underline{\nabla \theta}\,\right\Vert \eqsp \sqrt{k_x^2+k_y^2+k_z^2} \isdefs \left\Vert\,\underline{k}\,\right\Vert
$

as needed.

Scattering of plane waves is discussed in §C.8.1.


Solving the 2D Wave Equation

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

$\displaystyle \left(\nabla ^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2} \right)
z(\underline{x},t) \eqsp 0.
$

where

$\displaystyle \nabla ^2 \isdefs \nabla \cdot \nabla \isdefs \nabla ^T\nabla \eqsp
\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}.
$

The 2D wave equation is obeyed by traveling sinusoidal plane waves having any amplitude $ A$, radian frequency $ \omega $, phase $ \phi$, and direction $ \underline{u}$:

$\displaystyle z(\underline{x},t) = A\,e^{j\phi}\,e^{j(\omega t - \underline{k}^T\underline{x})}
$

where $ \underline{k}=k\underline{u}$ denotes the vector-wavenumber, $ k=\omega/c$ denotes the wavenumber (spatial radian frequency) of the wave along its direction of travel, and $ \underline{u}$ 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:

$\displaystyle k\eqsp \vert\underline{k}\vert\eqsp \frac{\omega}{c}
$

In particular, the wave can travel in any direction, with any amplitude, frequency, and phase. The only constraint is that its spatial frequency $ k$ is tied to its temporal frequency $ \omega $ 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 $ x$ axis, we have

$\displaystyle z(\underline{x},t) \eqsp A\,e^{j\phi}\,e^{j\omega t - kx} + A\,e^{j\phi}\,e^{j\omega t + kx} \eqsp 2A\,e^{j(\omega t + \phi)}\,\cos(kx) \protect$ (B.49)

which is a standing wave along $ x$.


2D Boundary Conditions

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 $ X\times Y$ meters.

Similar to the derivation of Eq.$ \,$(B.49), we can subtract the second sinusoidal traveling wave from the first to yield

$\displaystyle z(\underline{x},t) \eqsp A\,e^{j\phi}\,e^{j\omega t - kx} - A\,e^{j\phi}\,e^{j\omega t + kx}
\eqsp 2A\,e^{j(\omega t + \phi+\pi/2)}\,\sin(kx)
$

which satisfies the zero-displacement boundary condition along the $ y$ axis. If we restrict the wavenumber $ k$ to the set $ m\pi/X$, where $ m$ is any positive integer, then we also satisfy the boundary condition along the line parallel to the $ y$ axis at $ x=X$. Similar standing waves along $ y$ will satisfy both boundary conditions along $ (x,0)$ and $ (x,Y)$.

Note that we can also use products of horizontal and vertical standing waves

$\displaystyle z(\underline{x},t) \eqsp A\,e^{j\omega t + \phi}\,\sin(kx)\sin(ky)
$

because, when taking the partial derivative with respect to $ x$, the term $ \sin(ky)$ 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:

$\displaystyle z(\underline{x},t) \eqsp e^{j\omega t} \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} A_{mn}e^{j\phi_{mn}}\,W_{mn}(x,y) \protect$ (B.50)

where

$\displaystyle W_{mn}(x,y) \isdefs
\sin\left(m\frac{\pi}{X}x\right)\sin\left(n\frac{\pi}{Y}y\right).
$

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 $ (0$-$ X,0$-$ Y)$. Since sinusoids at different frequencies are orthogonal, the solution building-blocks $ W_{mn}(x,y)$ are orthogonal under the inner product

$\displaystyle \left<f,g\right> \isdefs \int_0^X\int_0^Y f(x,y)\overline{g}(x,y)\,dx\,dy.
$

It remains to be shown that the set of functions $ W_{mn}(x,y)$ 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 $ z(x,y)$ over the membrane (subject to the boundary conditions, of course), we can find the amplitude of each excited mode by simple projection:

$\displaystyle Z_{mn} \isdef \frac{\left<z,W\right>}{\left<W,W\right>}
$

Showing completeness of the basis $ W_{mn}(x,y)$ 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 $ \underline{x}_s\in{\bf R}^3$. Then the acoustic complex amplitude at position $ \underline{x}_l\in{\bf R}^3$ is given by

$\displaystyle p(\underline{x}_l;\underline{x}_s) = p_1(\underline{x}_s) \frac{e...
...derline{x}_l-\underline{x}_s\vert}}{\vert\underline{x}_l-\underline{x}_s\vert}
$

where $ p_1(\underline{x}_s)$ denotes the complex amplitude one meter from the point source in any direction, and $ k=2\pi/\lambda$ 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 $ M$ loudspeakers, which are often modeled as a point source for each speaker.


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