Physics I Honors
Monday 25 August
Fall 2008
First class! Name and course on board, read roll, introduce TA’s.
Distribute outline, lecture, and lab schedules. Discuss them.
New material for today: Start “thinking like a physicist.” Two new ideas, namely
(1) Solving problems with dimensional analysis, and
(2) F = ma is a “differential equation”
Today’s Topic #1: Dimensional Analysis
Given a problem, ask “What is the important physics?”. Then list the quantities that have
something to do with that physics. In many cases, you can combine these things to get the
answer, at least to within a factor of 2 or π or something like that.
Convenient notation:
[distance] = L
[time] = T
[mass] = M
[temperature] = K
Example: What is the radius of a black hole of mass M ? This is a combination of gravity
and relativity, so use M , G, and c. Newton’s constant G has units m3 /kg-sec2 , so [G] =
L3 M −1 T −2 . (Easy to see this from F = GmM/r2 = ma.)
Let the radius be R = M x Gy cz where we need to find x, y, and z. This gives
[R] = M x L3y M −y T −2y Lz T −z = M x−y L3y+z T −2y−z = L1
(1)
so x = y, 3y + z = 1, and 2y + z = 0. The second and third of these say that y = 1, and so
x = 1 and z = −2. Therefore we get
R = GM/c2
The “correct answer” is 21 GM/c2 , called the Schwarzchild Radius.
(Hint: Useful website for units and constants is
http://pdg.lbl.gov/2008/reviews/consrpp.pdf)
1
(2)
Practice Exercise
You’ve heard of “String Theory.” The length of strings is supposedly the “Planck Length,”
since gravity is unified with quantum mechanics and relativity at this distance. Gravity is
governed by Newton’s gravitational constant G, quantum mechanics by Planck’s constant h̄,
and relativity by the speed of light c. Combine these to find the Planck Length. (Note that
the dimensions of h̄ are distance×momentum, or L2 M/T .) Look up values of G, h̄, and c
and compare the Planck Length to the size of a proton (10−15 m).
2
Physics I Honors
Monday 25 August
Fall 2008
Today’s Topic #2: Newton’s Second Law as an “Equation of Motion”
The most important of Newton’s Laws is the second. For motion in one dimension, we write
F = ma. (We’ll get to F~ = m~a later this week.) Physicists refer to this as the “equation of
motion” for a particle moving through time in one dimension of space.
Let x(t) be the position x of a particle at time t. Then the velocity is v(t) = dx/dt ≡ ẋ.
(Have you all seen this notation before??) Acceleration is a(t) = dv/dt = v̇ = ẍ. So
Newton’s Second law can be written as
ẍ =
1
F (x, t)
m
(3)
where we make it clear that the force F may itself depend on position and/or time.
Equation 3 is called a “differential equation.” Unlike an algebraic equation, whose solution
is a value for a quantity like, say, x, the solution of a differential equation is a function, say
x(t). Since x(t) describes the motion of a particle with mass m acted upon by a force F (x, t),
we call Eq. 3 the “equation of motion.”
Simple example. Let F (x, t) = F0 , a constant independent of both x and t. Then
F0
dv
=
dt
m
F0
dx
=
t + v0
v(t) =
dt
m
1 F0 2
x(t) =
t + v0 t + x0
2m
ẍ = v̇ =
(4)
(5)
(6)
where the last line is the solution to this equation of motion. This might look more familiar
to you from your high school physics course, if we write a = F0 /m. You probably called
this “motion under constant acceleration”, but of course a constant acceleration implies a
constant force, so it’s the same thing.
Each step in the integration above involves a “constant of integration.” I called these constants v0 for the first step, and x0 for the second step. These are standard notations, and
in fact good ones. The constant v0 is the velocity at time t = 0, also called the “initial
velocity.” Similarly, x0 is the “initial position.” In order to solve any equation of motion, we
will need to specify these “initial conditions.”
3
Practice Exercise
Let y(t) describe the one dimensional motion of a particle with mass m that moves vertically.
Assume that the particle is acted on by a (constant) force F = −mg. (Isn’t that peculiar, a
force that is exactly proportional to the quantity m that appears in the equation of motion?)
Assume that the particle starts out at y = 0 with an initial (upward) velocity +V . Find the
maximum height to which the particle climbs, in terms of m, g, and V . Make a sketch of
y(t) as a function of t. What is the shape of the curve? (Mathematicians have a name for
it, and I’m sure you’ve heard it before.)
4
Physics I Honors
Thursday 26 August
Fall 2008
This class will concentrate mainly on mathematics. First we will review vectors, and and
apply them to three dimensional motion. Then we’ll talk about Taylor series, how they can
be used in approximations, and then apply them to complex numbers.
Today’s Topic #1: Vectors and Motion in Three Dimensions
Introduction: The vector equation F = ma is a shorthand for three equations, namely
Fx = max
and
Fy = may
and
Fz = maz
~ on board. Unit vectors are n̂ in book, n̂ on board.
Notation: A in book, A
The length of a vector is |A| = A, so |n̂| = 1.
If you add two vectors, the result is still a vector: C = A + B
A number times a vector is still a vector:
C = bA
This means that A = AÂ.
The “dot product” between two vectors is a scalar: A · B = AB cos θ where θ is the angle
between the two vectors. This is the projection of A onto B or vice versa.
The “cross product” between two vectors is a vector: A × B = C (Watch out! Some day
you’ll learn about polar and axial vectors!) The magnitude of C is AB sin θ. The direction of
C is given, by convention, by the “right hand rule”. This means that A×B = −B×A.
Vectors can be written in terms of coordinates in any given coordinate system or reference
frame. For an (x, y, z) coordinate system, define base vectors î, ĵ, and k̂, all orthogonal (so
their mutual dot products are zero) and with î × ĵ = k̂, etc. Then
A = Ax î + Ay ĵ + Az k̂
A · B = Ax Bx + Ay By + Az Bz
A × B = (Ay Bz − Az By )î + (Az Bx − Ax Bz )ĵ + (Ax By − Ay Bx )k̂
(7)
(8)
(9)
(1) General Motion in Three Dimensions. A vector r(t) = x(t)î + y(t)ĵ + z(t)k̂ gives
the motion of a particle in three dimensions. The velocity vector is v(t) = ṙ(t) and the
acceleration vector is a(t) = r̈(t). The (vector) equation of motion is r̈(t) = F(x, t)/m.
(2) Examples. Work W = F·d is a dot product. Torque τ = r×F is a cross product.
(3) Uniform circular motion. The motion r(t) traces out a circle at a constant rate.
Let ω be the number of radians per second traced out,
i.e. ω/2π revolutions per second. (Review radians?)
r(t) = r(î cos ωt + ĵ sin ωt)
(10a)
v(t) = ωr(−î sin ωt + ĵ cos ωt)
(10b)
N ote : v · r = 0
a(t) = −ω 2 r(î cos ωt + ĵ sin ωt) = −ω 2 r(t) (10c)
N ote : a = ω 2 r = v 2 /r
(10d)
The force that is responsible for this acceleration must
point back to the center of the circle!
5
Practice Exercise Find the period P of the “conical pendulum” pictured below, in terms
of the hanging angle θ, the string length L, and the acceleration g due to gravity.
The mass m executes uniform circular
orbits with radius R. If the speed is v,
then the period is just the time it takes
to execute one orbit, namely P = 2πR/v.
So, you need to figure out v in terms of R,
or perhaps in terms of the string length L
and the hanging angle θ, since R = L sin θ.
Since we know that F = ma, we just need
to figure out what the force F is. Clearly,
this force is provided by the tension T , but
only by part of it.
6
Physics I Honors
Thursday 26 August
Fall 2008
Topic #2: Taylor Series, Approximation Techniques, and Complex Numbers
It makes sense that we can approximate any function by some high order polynomial:
10
8
6
4
Zeroth order
First order
Second order
Full function
2
0
0
1
2
3
4
The function is expanded about the point
x0 = 1. Shown are the “zeroth order” (i.e.
a constant) approximation, the first order,
second order, and the full function. The
zeroth order is just the value of the function at x0 = 1, and the first order is the
tangent line at that point. The second order approximation comes close to the full
function in the neighborhood of x = x0 .
f (x) = a0 + a1 (x − x0 ) + a2 (x − x0 )2 + a3 (x − x0 )3 + · · ·
f 00 (x0 ) = 2 · a2
f 000 (x0 ) = 3 · 2 · a3
1
f (x) = f (x0 ) + f 0 (x0 )(x − x0 ) + f 00 (x0 )(x − x0 )2 + · · ·
2!
f (x0 ) = a0
f 0 (x0 ) = a1
(11)
(12)
(13)
1
α(α − 1)x2 + · · ·
(14)
2!
1
1
1
1
e x = 1 + x + x2 + x 3 + x4 + x5 + · · ·
(15)
2!
3!
4!
5!
1
1
sin(x) = x − x3 + x5 + · · ·
(16)
3!
5!
1
1
cos(x) = 1 − x2 + x4 + · · ·
(17)
2!
4!
Obviously, if |x| 1, i.e. “x is small”, then only a few terms (maybe just one) is needed in
order to get an accurate approximation.
Examples : (1 + x)α = 1 + αx +
Complex numbers. There√is a number i such that i2 = −1. Numbers formed by a “real”
number times i, like 2i, −i 3, or xi where x is a “real” number, are called “imaginary”
numbers. Don’t let the words fool you, though. All of these are perfectly valid numbers. The
imaginary numbers are a group outside integers, rational, and irrational (real) numbers.
Numbers like 3 + 2i and z = x + iy are called “complex” numbers. They have a “real part”
and an “imaginary part”. We write, forp
z = x + iy where x and y are real, x = Re(z) and
y = Im(z). The “modulus” of z is |z| = x2 + y 2 .
Now, here’s a neat thing:
1
1
1
1
eix = 1 + ix + (ix)2 + (ix)3 + (ix)4 + (ix)5 + · · ·
2!
3! 5!
4!
1 2 1 4
1 3 1 5
= 1 − x + x + ··· + i x − x + x + ···
2!
4!
3!
5!
= cos(x) + i sin(x)
This is called Euler’s Formula. It is very useful in physics and engineering!
7
Practice Exercise
First, a “math rule.” The derivative of an exponential is an exponential. That is
f (x) = eax
=⇒
f 0 (x) = aeax
Use this to find the solution to the equation of motion for a particle of mass m subject to
a force F (x) = −kx where k is a positive constant. Write down the equation of motion as
ẍ(t) = F/m. Then show that x(t) = Ceiωt is a solution to the equation of motion, for any
value of C, so long as ω has one of two possible values. What are those values?
Next, write the “general solution” x(t) as the sum of two functions Ceiωt , one for each of
the two possible values of ω and with two different values of C. (Call them C1 and C2 .) Is it
obvious to you that this sum of solutions is also a solution to the equation of motion?
Finally, determine the two constants C1 and C2 from the initial conditions x(0) = A and
ẋ(0) = 0. Simplify the result as much as possible. (Euler’s formula will be handy.) What
kind of motion does this describe.
8
Physics I Honors
Thursday 4 September
Fall 2008
Let’s try to stick with the “20-20-20” plan, that is new material for 20 minutes, 20 minutes
to work on a practice exercise, and then 20 minutes to talk about the solution.
Today’s Topic #1: Applications of Newton’s Laws
Discuss “Free Body Diagrams.” Statics first. Use example of one block A sitting atop another
block B (page 68 in K&K) and take care to note the weight of each block, the normal force
on B from the lower surface, and the force F1 (F2 ) which block B (A) exerts on block A
(B). Hence F1 = MA g and N = F2 + MB g. Third law says F1 = F2 so N = (MA + MB )g,
which is just what you expect, of course.
Pretty much the same thing for two masses hanging, connected by a wire under tension. The
top wire hangs both masses! Tell the story of the Hyatt Regency disaster, July 17, 1981 in
Kansas City, Missouri: http://en.wikipedia.org/wiki/Hyatt Regency walkway collapse
Dynamics is very similar, just don’t set a = 0. See practice problem and also homework.
Let’s do the conical pendulum. (See Example
2.8 in your textbook.) Find the period P in
terms of the angle θ, the string length L, and
the acceleration g due to gravity. The speed
of m is v = Rω, so P = 2πR/v. To find v, use
F = ma with a = ω 2 R = v 2 /R. (Recall Equations 10.) Find F = T sin θ with T cos θ = mg.
Putting it all together, we have
R2
2R
=
(2π)
v2
a
Rm
Rm cos θ
= (2π)2
= (2π)2
T sin θ
mg sin θ
s
s
R cos θ
L cos θ
P = 2π
= 2π
g sin θ
g
P 2 = (2π)2
Practice Exercise
Find the acceleration a of the coupled masses on the left, in
terms of m1 , m2 , and g. The mass m1 slides on a horizontal,
frictionless surface, m2 hangs freely, and the string that
connects them is massless. The pulley is also massless, and
gives no frictional resistance to motion.
!"
!#
Start by drawing free body diagrams of m1 and m2 . Identify
all forces on each of the two masses, and then work with the
components that are in the direction of the acceleration.
9
Topic #2: The Everyday Forces of Physics
Start by making a list. Use vector notation!
• Tension, the pull of a “massless” string: T
• Weight, i.e. gravity near the Earth’s surface: W = −mg k̂
• Springs, i.e. “Hooke’s Law”: F = −kx
• Frictional forces
– Kinetic friction: fk = µk N “opposing the direction of motion”
– Static friction: fs ≤ µs N “as big as it needs to be to keep the object from moving”
Note that µs > µk for the same two surfaces. Demo with book on desk!
• Viscous forces: Fv = −Cv
• Inverse square “central” forces (Draw some pictures here.)
– Newtonian gravity: Fa→b = −(GMa Mb /r2 )r̂a→b
– Electrostatic attraction: Fa→b = +(kQa Qb /r2 )r̂a→b
Note: General central force is just Fa→b = f (r)r̂a→b
Let’s study the motion of an object under the influence of a drag force. (See Example 2.16.)
Assume a mass m is moving through a viscous medium. The equation of motion is
−Cv = ma = m
dv
dt
or
dv
C
=− v
dt
m
Now integrate. The “math trick” you need is d(ln x)/dx = 1/x. We have
Z v
Z
v
dv
C t
C
dt
or
ln
=−
=− t
or
v(t) = v0 e−Ct/m
v
m
v
m
0
v0
0
(18)
(19)
Note use of initial condition notation. The behavior is clear. The particle moves with
velocity v0 at t = 0, but then slows down exponentially.
Note: Study the “shell theorem” in Note 2.1. Basically, a spherical object attracts an external
mass gravitationally, where the equivalent object mass is all that is contained within a sphere
or radius intersecting the external mass.
Practice Exercise
Express the gravitational acceleration near the Earth’s surface, g, in terms of G, and the
mass ME and radius RE of the Earth. Start by putting an “external mass” m on the surface
of the Earth. Then calculate the gravitational attraction of the Earth on this mass, using
the formula for Newtonian gravity.
Cavendish devised an sensitive experiment that measured G in 1798. People say he “weighed
the Earth.” What do they mean? (Note that Eratosthenes measured the Earth’s radius
somewhere around 200 B.C.)
10
Physics I Honors
Monday 8 September
Fall 2008
Today’s Topic #1: Momentum and Systems of Many Particles
P
Newton actually wrote the SecondPLaw as
F = dp/dt where momentum p = mv. If mass
m doesn’t change with time then
F = mdv/dt = ma, so we recover the old version.
Now imagine an object colliding with a brick wall. The collision will change the momentum
of the object. The force of the collision varies over the collision time. (Draw a figure; See
Example 3.9 in K&K.) We can use Newton’s second law to write an expression involving the
integral of the force and the change in momentum:
Z t2
F(t)dt ≡ J
∆p =
t1
We call J the “impulse.” The “average force” Favg is J divided by ∆t = t2 − t1 .
Now imagine two objects colliding with each other. Each exerts an “equal and opposite”
momentum on each other. That is J1 = −J2 , or ∆p1 + ∆p2 = ∆(p1 + p2 ) = 0. In other
words, the total momentum P ≡ p1 + p2 is conserved in collisions. Note that this doesn’t
work if there is some “external” force acting on the masses.
Now look at two particles again, but not just collisions. Allow some external force Fext to
act as well as internal forces. Then
X
dp1 dp2
dP
=
+
= m1 a1 + m2 a2 =
Fext
dt
dt
dt
The last equality follows because the internal forces cancel in the sum. It is convenient to
write P = M vcm where M = m1 + m2 and vcm = drcm /dt with
m1 r1 + m2 r2
1
=
(m1 r1 + m2 r2 )
(20)
m1 + m2
M
P
in which case Newton’s Second Law reads
Fext = Mr̈cm = M acm . We call rcm the center
of mass. (Is “ center of momentum” better?) For more than two particles
rcm =
rcm =
N
m1 r1 + m2 r2 + · · · + mN rN
1 X
=
mn rn
m1 + m2 + · · · + mN
M n=1
(21)
Don’t forget that this is really two (or three) equations for xcm , ycm , and (if three) zcm . The
figures below show how this all works out:
11
Practice Exercise
An “elastic” collision is one in which kinetic energy is conserved. (The kinetic energy of an
object with mass m and speed v is K = 12 mv 2 . The total kinetic energy is the sum of the
individual object’s kinetic energies. We will learn more about this later.) Let two objects,
one with mass m and the other with mass 2m collide elastically in one dimension. They are
initially heading towards each other with speed v. Find their speeds and directions after the
collision, in terms of m and v.
Topic #2: Calculating the Center of Mass of Solid Objects
If we are working on a solid object, we could sum over all the atoms that make it up, but
that would be nuts. Instead, we use calculus, and assume that the object is continuous.
Then the sum becomes an integral. That is
Z
Z
1
M = dm
and
rcm =
dm r
M
We usually write the infinitesimal mass dm as a density (usually called ρ or σ) times a
volume element dV (or some other measure), i.e. dm = ρ(r)dV . This allows the density to
vary as a function of position inside the solid object. The rest is in the details.
R
Here we can slice up an eggplant, making the point that dV is the sum of small pieces.
Here is a simple example. What is the volume of a right circular cone of height h and base
radius R? Let z measure the height, and slice the cone up into thin disks, each of thickness
dz. The radius of each disk is R(h − z)/h = R(1 − z/h), so the volume of each disk is
πR2 (1 − z/h)2 dz. Therefore, the volume of the cone is
Z
Z h
Z h
h
2z z 2
2
2
2
+ 2 = πR2
dV =
πR (1 − z/h) dz = πR
1−
h
h
3
0
0
or one-third of the base area times the height. Of course this is volume, not mass, but that’s
basically the same thing for constant density. If the density is not a constant, you’d need to
include the functional form of ρ(r) and that would be a harder problem
Practice Exercise
Find the position of the center of mass of a long thin rod of mass M and length L. The
density of the material making up the rod changes linearly from zero at one end. Start by
realizing that the mass of a short segment of length dx is dm = σ(x)dx where σ(x) is the
linear mass density. Write a formula for σ(x), and determine any parameters by fixing the
mass of the rod to be M . Then, find the position of the center of mass.
12
Physics I Honors
Thursday 11 September
Fall 2008
Today’s Topic #1: Work and Kinetic Energy in One Dimension
Let’s use some calculus to manipulate F = mdv/dt, in one dimension:
Z
b
Z
F (x)dx = m
a
a
b
dv
dx = m
dt
Z
a
b
dv dx
dt = m
dt dt
Z
a
b
1
1
vdv = mvb2 − mva2
2
2
We write F = F (x) but it could be a constant, for example F = −mg:
Z yb
p
1
1
dy = −mg(−h) = mv 2 − m02 =⇒ v = 2gh
−mg
2
2
ya
(22)
(23)
for an object falling through a distance h = ya − yb starting from rest. We already knew this
from solving the equation of motion (Eq. 6) but here we’ve done this a different way.
A harder, but still easy, example is a spring force F = −kx. Equation 22 becomes
1
1
1
1
− kx2b + kx2a = mvb2 − mva2
2
2
2
2
(24)
You will play around with this on the practice exercise.
These are both examples of something called the Work-Energy Theorem. Define the
work (in one dimension) done by a force F (x) on a mass m through a path xa → xb as
Rb
Wba = a F (x)dx. Define the kinetic energy of that object as K = 12 mv 2 , for velocity v.
Then Eq. 22 says that
Wba = Kb − Ka ≡ ∆K
(25)
That is, the work done on the object equals the change in kinetic energy. Of course, the
work done for a constant force is just the product of the force times the displacement.
Let’s use this to calculate “escape velocity.” A rocket
starts at ra = Re with (escape) velocity v. It ends up
at very large distance, rb → ∞, with v = 0. (If the
initial velocity were larger, it would have a finite speed at
infinite distance. If smaller, it would fall back to Earth.)
Z
∞
Re
r
∞
GMe m GMe m
1 2
2GMe
GMe m
dr =
−
=−
= 0 − mv =⇒ v =
2
r
r
Re
2
Re
Re
(26)
We will consider all these ideas from a purely “energy” point of view on Monday.
Practice Exercise An object with mass m is acted on by a spring force F = −kx. It has
zero velocity when x = A. Use Eq. 24 to find the velocity v = V when x = 0, in terms of k,
m, and A. Explain why this is consistent with x(t) = A cos ωt, which solves the equation of
motion so long as ω 2 = k/m.
13
Topic #2: Work and Kinetic Energy in Several Dimensions
How do we generalize Eq. 22 to recognize that force is a vector, i.e. F = mdv/dt? Assuming
we want to end up at the work energy theorem, we can try working backwards:
d 2
d
dv dv
dv
dr dv
(v ) =
(v · v) = v ·
+
· v = 2v ·
=2 ·
(27)
dt dt
dt
dt
dt dt dt
Z b
Z b Z b
Z b
1 d 2
1 2
dr dv
dv
d
=
∆K =
mv
m (v )dt =
m ·
dt =
m
· dr (28)
2
dt
dt dt
dt
a 2
a
a
a
Z b
F(r) · dr
(29)
or
∆K =
a
That’s all well and good formally, but what does the last integral mean? The motion r(t)
traces out some path in space, and the integral is “along the path.” (See the figures below.)
Indeed, we call it a “path integral” or “line integral.” (The latter terminology is more
common. The former has a larger connotation when studying quantum mechanics.)
A couple of simple things to notice:
• For a constant force F acting along a straight line path with displacement d = ∆r, the
work done by the force is W = F · d
• For an object moving in a circular path, the “centripetal” force keeping it moving this
way does no work on the object, because F(r) · dr = 0 everywhere along the path.
• If the path is “closed”, like a circle orH some other closed loop, then the line integral
done all the way around is written as F · dr
Practice Exercise
H
Find the work W = F · dr done around the
closed rectangular loop shown, for two forces (a)
F(x, y) = y î, and then for (b) F(x, y) = xî. Treat
the closed loop as four, straight line paths I, II, III,
and IV, and calculate the work done on each path.
The total work done is just the sum of the work
done over each of the four straight line segments.
The force in (b) is called “conservative” and the
force in (a) is called “non-conservative.” We will
be discussing all this a lot more.
14
Physics I Honors
Monday 15 September
Fall 2008
Today’s Topic #1: Potential Energy; Conservation of Energy
Consider a mass m on a spring, i.e. F (x) = −kx, and take the time derivative of
1 2 1 2
mv + kx
2
2
E =
(30)
dE
= mv v̇ + kxẋ = v(mv̇ + kx) = v(F + kx) = 0
dt
(31)
We say that E is a “constant of the motion.” It is “conserved quantity.” Note that
d(kx2 /2)/dx = kx. In general, let F (x) = −dU/dx for some function U (x). So we have
dE
dU
dU dx
dU
1 2
= mv v̇ +
= mv v̇ +
=v F+
=0
(32)
E ≡ mv + U (x) =⇒
2
dt
dt
dx dt
dx
We call U (x) the “potential energy” and E the “total energy.” Relating this to work,
b
Z
Z
F (x)dx = −
Wba =
a
a
b
dU
dx = −
dx
Z
b
dU = − [Ub − Ua ]
(33)
a
That is, the work done from a to b is the negative of the change in potential energy.
Now generalize to three dimensions. This is simple, in principle. We just write
Z b
F(r) · dr
Ub − Ua = U (rb ) − U (ra ) = −Wba = −
(34)
a
but the line integral can (in principle) depends on the path! If it does, then Eq. 34 makes
no sense. We can only define a potential energy for forces with path independent work, also
known as “conservative” forces. These let us write down an energy that is conserved.
Non-conservative forces lead to a loss of (mechanical) energy. For F = Fc +Fnc , we have
Kb − Ka =
total
Wba
Z
b
Z
F · dr = − [Ub − Ua ] +
=
a
b
nc
Fnc · dr =⇒ Eb − Ea = Wba
(35)
a
The energy lost in going from a to b is just the work done by the non-conservative forces.
Finally, let’s define power. Power is just the rate at which work is done. That is
P ≡
dW
F · dr
dr
=
=F·
=F·v
dt
dt
dt
(36)
Practice Exercise
An object with mass m falls at a constant (that is, “terminal”) velocity v through the
atmosphere. If the drag force is f = −bv, find b in terms of m, g, and v. Then, calculate (a)
the change in total mechanical energy ∆E when the object falls through a distance h, and
(b) the work W nc done by the non-conservative forces. Finally, show that ∆E = W nc .
15
Topic #2: Energy Diagrams
We gain insight about motion by looking more literally at energy conservation, i.e.
r
1
2
2
E = mẋ + U (x)
so
ẋ = v = ±
[E − U (x)]
2
m
(37)
Velocity can be to the left or right, but “speed”
(≡ |v|) is determined. The square root leads
to turning points to the motion, which limit
the position based on the energy. See the figure, and discuss motions for different values of
E. Relate force plot to the energy plot. Discuss behaviors of particles with various Ei , and
mention oscillations. Draw the analogy with
“up and down a hill.” Note that, in principle, one can get the motion x(t) by integrating
Eq. 37. We’ll do this in a practice problem.
Example: The Pendulum. A simple pendulum is a mass m attached to a massless string
of length ` which swings in a plane through an angle φ. (See page 255 in your textbook.)
Analyzing this in terms of “F = ma” is a little tricky, but we can do it. It is two dimensional,
but the two “axes” change with time. One is radial (in which there is no motion) and the
other is “angular” along the path, a variable we’ll call s = `φ. We have
g
(38)
Fs = −mg sin φ = ms̈
so
φ̈ = − sin φ
`
Nobody knows how to solve this problem analytically. However, if the pendulum never swings
through large angles, then sin φ ≈ φ and we recover the “harmonic oscillator” equation of
motion, with ω 2 = g/`.
We can also look at this in terms of energy. The potential energy in terms of φ is
U (φ) = mgh = mg(` − ` cos φ) = mg`(1 − cos φ)
(39)
Plot this as a function of φ and oscillate about the minimum at φ = 0. (Also point out the
equilibrium point at φ = π.) This leads into a discussion of “small oscillations”, i.e.
φ2
g s2
φ2 φ4
+
+ ···
≈ mg` = m
(40)
U (φ) = mg` 1 − 1 −
2!
4!
2
` 2
which is a “harmonic oscillator” potential with k replaced by mg/`, i.e. k/m → g/`.
Practice Exercise
Use Eq. 37 with U (x) = mgx and E = mgh to find x(t). Assume that x = h when t = 0.
(This describes Rthe motion of something
falling from rest at a height h, right?) Split dx/dt
R
so that you get dx on the left and dt on the right. It should be obvious after integrating
that the constant of integration is zero for this initial condition. You’ll need the following
obscure calculus rule:
√
Z
dx
2 a + bx
√
=
(41)
b
a + bx
where a and b are constants.
16
Physics I Honors
Thursday 18 September
Fall 2008
Today’s Topic: Conservation Laws and Particle Collisions
Textbook (Sec.14.4): “No new physical principles are involved; the discussion is intended to
illustrate ideas we have already discussed.”
Pfinal = m1 v10 + m2 v20
Pinitial = m1 v1 + m2 v2
Pinitial = Pfinal
m1 v1 + m2 v2 = m1 v10 + m2 v20
(42)
(43)
(44)
Elastic and inelastic collisions. If the (internal) forces that cause the collision are conservative, then mechanical energy wil be conserved. Far away from the colliding point, there are
no forces, so no (internal) potential energies, and conservation of energy means conservation
of kinetic energy. If non-conservative forces act, then energy can be lost. We write
Ki = Kf + Q
i.e.
Q = Ki − Kf ≥ 0
If Q = 0 then the collision was elastic. Using the same notation as above
1
1
1
1
m1 v12 + m2 v22 = m1 v102 + m2 v202 + Q
2
2
2
2
(45)
(46)
Example 4.18: Elastic Collision of Two Balls. (Recall Sept.8 practice exercise.) Put
m1 = m and m2 = 3m, v1 = v and v2 = −v. Conserving momentum and energy,
mv − 3mv = mv10 + 3mv20
1 2 1
1 02 1
mv + 3mv 2 =
mv1 + 3mv202
2
2
2
2
(47)
(48)
Equation 47 gives
v10 = −2v − 3v20
which we can then plug into Eq. 48 to find
1
1
2v 2 =
(2v + 3v20 )2 + 3mv202
2
2
= 2v 2 + 6vv20 + 6v202
or
0 = vv20 + v202
17
(49)
(50)
(51)
(52)
This equation has two solutions for v20 , each of which gives v10 by Eq. 49, namely
or
v20 = −v
v20 = 0
v10 = +v
v10 = −2v
=⇒
=⇒
(53)
(54)
The first of these is just the initial conditions. If nothing happens when the two balls pass
near each other, then nothing happens. If they do in fact collide, however, then the second
ball stops, and the first one fires backward at twice the velocity that it started with.
Collisions and the Center of Mass. The velocity of the center of mass is
V=
m1 v1 + m2 v2
m1 + m2
(55)
Let’s call this the “laboratory frame.” We can change our frame of reference to the “center
of mass frame” by imagining that we are observing the collision by moving along with the
center of mass. In this frame, the velocities of the two particles are observed to be
m2
(v1 − v2 )
m1 + m2
m1
= v2 − V =
(v2 − v1 )
m1 + m2
v1C = v1 − V =
(56)
v2C
(57)
so that the momenta in the center of mass frame are given by
m1 m2
(v1 − v2 ) = µv
m1 + m2
m1 m2
=
(v2 − v1 ) = −µv
m1 + m2
p1C = m1 v1C =
(58)
p2C = m2 v2C
(59)
where we have made the definitions
v ≡ v1 − v 2
and
µ≡
m1 m2
m1 + m2
(60)
The total momentum p1C + p2C = 0 in the center of mass frame, as we expect. It is easy
0
0
to show that for elastic collisions in the center of mass, v1C
= v1C and v2C
= v2C . Your
textbook has the details.
We call v the “relative velocity” and µ the “reduced mass.” You will come back to these
quantities over and over again as you study the physics of collisions, both classically and
quantum mechanically.
Practice Exercise
Suppose that the two balls in Example 4.18 stick together after the collision, i.e. v10 = v20 .
Calculate the Q value for the collision.
18
Physics I Honors
Monday 22 September
Fall 2008
First exam is this Thursday, in class, at 10am. Today, we will do some math.
A Review of Some Calculus Rules
R
f (x) f 0 (x)
f (x)dx
1
xa
axa−1
xa+1
a+1
ex
ex
ex
cos x − sin x
sin x
Chain rule :
d
df du
f (u) =
dx
du dx
1
x
f 0 (x)
—
ln x
sin x
cos x
f (x)
1
x
Product rule :
R
f (x)dx
ln x
—
− cos x
d
du
dv
uv = v(x) + u(x)
dx
dx
dx
(61)
Partial Derivatives, Gradient, and Force & Potential Energy in 3D
The derivative (with respect to x) of a function f (x) is just the rate at which f (x) is changing,
that is ∆f /∆x when the changes are very small. We write f 0 (x) = df /dx, where the right
side of the equation is a much more descriptive notation.
Now suppose that f is a function of two (or more) variables x and y, that is f (x, y). We
can ask how f changes with respect to x if y is held constant, or vice versa. We write these
derivatives as ∂f /∂x and ∂f /∂y, respectively, and call them “partial derivatives.” There are
other notations, but they are all bad. In many cases df /dx and ∂f /∂x can be quite different,
but we will not get into this kind of confusing situation in this course!
Suppose x changes by an amount dx, and y changes by an amount dy. How much does
f (x, y) change? The answer is pretty obvious, namely
df =
∂f
∂f
dx +
dy
∂x
∂y
(62)
So now, consider the work dW done by a force F(r) over a distance dr. The change in
potential energy U (r) = U (x, y, x) (assuming that one can be defined) is
dU = −dW = −F(r) · dr
(63)
Write the left side using Eq. 62 and write out the right side in terms of components, so
∂U
∂U
∂U
dx +
dy +
dz = −Fx dx − Fy dy − Fz dz
∂x
∂y
∂z
(64)
Clearly the x component of F(r) is just −∂U/∂x and so forth. It is nice to write this in a
simple notation using vectors, by defining the gradient operator ∇ as follows:
∇ ≡ î
∂
∂
∂
+ ĵ
+ k̂
∂x
∂y
∂z
(65)
In this case, Eq. 64 is really just a statement that
F(r) = −∇U (r)
We frequently refer to F(r) as a “vector field” and U (r) as a “scalar field.”
19
(66)
A Useful Theorem about the Curl of a Vector Field
&"
&$
!"#$%
Imagine a tiny, closed rectangular path in the (x, y) plane, with side
lengths dx and dy, and with the lower left corner at the point (x, y).
It is very simple to calculate the work done by a force field F(x, y)
while going around this path, since the force is approximately constant along each of the straight legs. We have
I
F · dr = F (x, y)dx + F (x + dx, y)dy
− F (x + dx, y + dy)dx − F (x, y + dy)dy
(67)
Collecting some terms, and ignoring a change in y, say, when moving in the x direction,
I
F · dr = [F (x + dx, y) − F (x, y + dy)] dy − [F (x + dx, y + dy) − F (x, y)] dx
F (x + dx, y) − F (x, y + dy) F (x + dx, y + dy) − F (x, y)
=
−
dxdy
dx
dy
∂F
∂F
−
=
dxdy = [∇ × F]z dxdy = [∇ × F] · dA
(68)
∂x
∂y
We call ∇ × F the “curl” of the field F, and have dA ≡ dxdy and the vector dA pointed
perpendicular to the plane bounded by the tiny loop.
Now imagine an arbitrary surface A bounded by some curve C. You can “tile” A into a
bunch of tiny rectangles, and the loop integrals around any two adjacent rectangles cancels
on the border. In other words, the tiny loop integrals (68) end up giving the loop integral
around C. Mathematically, this says that
Z
I
[∇ × F] · dA
(69)
F(r) · dr =
C
A
This is called Stokes’ Theorem. Obviously, a force field F(r) is conservative if ∇ × F = 0.
This is automatically satisfied if F(r) = −∇U (r).
A Different Useful Theorem about the Divergence of a Vector Field
Now consider a vector field E(r) and a tiny box in three dimensional space, with sides
dx, dy, and dz. The “flux” out of the box near the point (x + dx, y, z) is pretty clearly
Ex (x + dx, y, z)dydz. Play a game similar to what we did with Stoke’s theorem, building an
arbitrary volume V out of a bunch of tiny boxes. For two adjacent boxes, the flux out of one
cancels the flux into the other, so you end up measuring the flux through the entire (closed)
surface A that contains the volume. It isn’t hard to go from here to show that
I
Z
E(r) · dA =
[∇ · E] dV
(70)
A
V
Physicists call this Gauss’ Theorem; mathematicians call it the Divergence Theorem. (The
quantity ∇ · E is called the “divergence” of the field E.) We will find some important uses
of Gauss’ theorem this semester when we study fluid behavior. Next semester, you will use
Gauss’ theorem in your study of electromagnetism.
20
Physics I Honors
Monday 29 September
Fall 2008
Hand back Exam #1. Average was 74.1. Questions? Tell me. Grades ≤ 60 see me.
Angular Momentum of a Particle
Consider an object moving in two dimensions under the action of a central force
F = Fx î + Fy ĵ. The figure shows that
Fy /Fx = y/x
or
Fy − yFx = 0
Now consider a quantity ` defined as
` ≡ m (xẏ − y ẋ)
(71)
This quantity is a constant of the motion:
d`
= m (ẋẏ + xÿ − ẏ ẋ − yẍ)
dt
= m (xÿ − yẍ) = xFy − yFx = 0
If we define L ≡ r × (mv) = r × p, then we see that ` = Lz . We call L the angular
momentum of a particle, and it is conserved for motion in a central force field.
It is convenient to think of L in terms of parallel and perpendicular components of r. If we write r = r⊥ + rk then
L = r × p = r⊥ + rk × p
= r⊥ × p + rk × p = r⊥ × p (72)
so that
` = Lz = |r⊥ | |p|
(= |r| |p⊥ |)
(73)
We call |r⊥ | the “moment arm” of the angular momentum.
In 1609 Kepler told the world that planets “sweep out equal
areas in equal times.” Newtonian gravity is a central force,
and Kepler’s statement is a consequence of angular momentum
conservation. Following the figure at the right
1
dA (r + dr)(rdθ)
2
so
dA
1 dθ
1 ds
= r2
= r
dt
2 dt
2 dt
(74)
But ds/dt = v⊥ = p⊥ /m so dA/dt = rp⊥ /2m = Lz /2m.
Practice Exercise
Angular momentum depends on the choice of coordinate system. Show this explicitly for
something falling straight down, starting from rest at y = 0. What is L for something falling
along the y-axis? Do it again for a line parallel to the y-axis, along the line x = A?
21
Torque
If angular momentum is constant for an object acted on by a central force, then how do we
get the angular momentum to change? By brute force, we see that
dL
d
dr
dp
=
(r × p) =
×p+r×
=r×F
dt
dt
dt
dt
(75)
where we used F = dp/dt and v × p = 0 with v = dr/dt and p = v. We write τ ≡ r × F,
where τ is the torque. It is convenient to think of torque as force times a moment arm.
Imagine a bunch of masses hanging from a string. The torque from gravity is (y is “up”)
!
X
X
X
rn × Fn =
rn × (−mn g ĵ) =
mn rn × −g ĵ = rcm × −g ĵ
(76)
n
n
n
If we hang from the center of mass, then all moment arms are measured from here and
rcm = 0. So, there is no torque and the object is balanced. Sometimes we call the center of
mass the “center of gravity.”
Practice Exercise
Calculate the torque for each of the two cases in the previous practice exercise, due to a
mass m falling under the influence of gravity, and show that in each case, τ = d`/dt.
Angular Momentum and Fixed Axis Rotation
So what is the motion resulting from a torque? Consider one single particle, inside the solid
object, located at r and acted on by F. It can only move in a circle of radius r and the force
in that direction is F sin θ. If the path length it travels is called s, then Newton’s Second
Law says F sin θ = ms̈ = mrφ̈. Multiply both sides by r, and we have τ = (mr2 )α, where
α = φ̈ is called the angular acceleration. The quantity mr2 has replaced mass when we
describe the motion in angular variables.
Since α is the same for all parts of the object, and since τ is either from one force at one
point, or due to a sum of such things, Newton’s Second Law for a solid object becomes
X
X
mn rn2
(77)
τ = Iα
where
I=
n
is called the rotational inertia or moment of inertia. Notice that I is a property of the solid
object and the specified axis. (Do you get the feeling that I is not really a “scalar” but not
a vector either? If so, you are right. It is actually something called a “tensor.”)
You can prove something called the Parallel Axis Theorem which states that the rotational
inertia of any body about an arbitrary axis equals the rotational inertia about a parallel axis
through the center of mass plus the total mass times the squared distance between the two
axes. Mathematically, we write I = Icm + M h2 . See Example 6.9 in your textbook.
Rotational Inertia of Solid Objects
Of course, for solid objects, we don’t do the sum over
R 2 all the masses in Eq. 77 to get the
rotational inertia. Instead,
we do the integral I = r dm. This is just like calculating the
R
center of mass, that is r dm, but with another factor of r. See Example 6.8 in your textbook.
We will be doing more of these sorts of things, include homework problem 6.8.
22
Physics I Honors
Thursday 2 October
Fall 2008
Solid objects both rotate and translate. There is a lot of physics associated with these motions. See Chap.6. Rather than derive and review it all, here are the rules (Table 6.1):
Pure rotation (about a fixed axis)
L = Iω
τ = Iα
K = 12 Iω 2
Rotation plus translation (“0” means “CM”)
Lz = I0 ω + (R × M V)z
τz = τ0 + (R × F)z with τ0 = I0 α
K = 12 I0 ω 2 + 21 M V 2
Good examples to review include the Atwood’s Machine (Example 6.10), Kater’s Pendulum
(6.12), and the “Disk on Ice” (6.15). We’ll do two applications in class today.
Objects Rolling Down a Plane. (Refer to Example 6.16)
Consider some object with radius b, mass M , and I0 = βM b2 , rolling down a ramp. How
does the shape (i.e. β) affect the time it takes for the object to get to the bottom of the
ramp?
P
First analyze it using just
F = ma. This says that
W sin θ − f = M g sin θ − f = M a
Friction makes it roll, with torque bf , so
bf = I0 α = (βM b2 )(a/b) = βM ab
Solving,
a = g sin θ/(1 + β)
(78)
A different approach uses torque and angular momentum.
τz = τ0 + (R × F)z = −bf + b(f − W sin θ) = −bM g sin θ
(The normal force cancels W cos θ so no net torque.)
Lz = I0 ω +(R×M V)z = −βM b2 ω −bM V = −(1+β)M b2 ω
Then put τ = L̇z = −(1 + β)M b2 α = −(1 + β)M ba and get the same answer as before.
So, a larger of β gives a smaller acceleration and takes longer to get down the ramp.
A third way to look at the problem is in terms of energy. The potential energy at the start
is Ui = W h = M gh and there is no kinetic energy. At the bottom of the ramp, the CM has
speed V so Kf = 12 I0 ω 2 + 21 M V 2 = 12 (βM b2 )(V /b)2 + 21 M V 2 = 12 (1 + β)M V 2 . Then,
s
1
2gh
U i = Kf
=⇒
M gh = (1 + β)M V 2
=⇒
V =
(79)
2
1+β
and, again, large β travels more slowly at the bottom. (Friction is irrelevant here. Why?)
Practice Exercise
Use one dimensional kinematic relations, for something moving through a distance d with
acceleration a, to show that Eq. 79 is the same as Eq. 78.
23
Kinematics of an Ultraelastic Rough Ball
See Richard L. Garwin, American Journal of Physics 37(1969)88. It is posted on our website.
(Discuss how to look up papers in “e-Journals” using http://library.rpi.edu/)
Play with the ball a little, as in Figure One and then as in Figure Three.
The analysis is based on two important assumptions about the ball and its collisions:
(a) The collision is elastic, conserving the sum rotational+translational kinetic energy
(b) There is no slip in contact between the ball and a surface, i.e. the ball is “rough”
The figure shows the coordinate system and nomenclature
used in the paper. We use “b” for “before” and “a” for
“after” the collision. The abbreviation C = ωR is used for
brevity, where R is the radius of the ball. The moment of
inertia (about the CM) of the ball is I0 = βM R2 with β =
2/5. (Garwin uses α instead of β, but I switched so that there
is no confusion with our notation for angular acceleration.)
The kinetic energy of the ball is just
1
1
1
1
1
K = I0 ω 2 + M V 2 = βM R2 ω 2 + M V 2 = M (V 2 + βC 2 )
2
2
2
2
2
so Kb = Ka . The angular momentum about the contact point (see the figure) is just
L = I0 ω − R(M V ) = βM R2 ω − M V R = M R(βC − V )
(80)
(81)
and since all of the forces in the collision pass through the contact point, Lb = La . This immediately gives a simple result, since Eqs. 80 and 81 imply, with conservation applied,
β(Cb2 − Ca2 ) = Va2 − Vb2
and
β(Cb − Ca ) = −(Va − Vb )
(82)
Ca + Va = −(Cb + Vb )
(83)
These equations are divided to yield
Cb + Ca = −(Va + Vb )
or
Garwin calls V + C the “parallel velocity,” the velocity tangent to the ball’s surface at the
contact point. We have shown that this velocity is conserved in magnitude, but reverses in
direction after a collision with the wall or the floor. Subracting Eq. 83 from 82 yields
Ca =
2
β−1
Cb −
Vb
β+1
β+1
and then
Va = −
2β
β−1
Cb −
Vb
β+1
β+1
(84)
Let’s apply this to a super ball bouncing off the floor. Pause for a moment to realize that
the velocity V⊥ perpendicular to the floor is conserved in the collision. (Why?) If a ball falls
vertically downward, then Vb = 0 and Va = −2β/(β + 1)Cb = −(4/7)Cb for β = 2/5, i.e. a
solid sphere. That is, it bounces to the side with a transverse velocity that depends on how
fast it is spinning.
If on the other hand, a ball that is not spinning (Cb = 0) bounces at a 45◦ angle so that
Vb = V⊥ , then Va = (1 − β)/(1 + β)Vb = (3/7)V⊥ . That is, it bounces up at an angle (with
respect to the vertical) θa = tan−1 (Va /V⊥ ) = tan−1 (3/7) = 23◦ < 45◦ . Try it and see!
See Garwin’s paper for other calculations, including the trajectory for bouncing off the
underside of a table.
24
Physics I Honors
Monday 6 October
Fall 2008
Today we recognize angular velocity ω, angular momentum L, and torque τ as full-fledged
vector objects. The physical implications are interesting.
What exactly is a vector? Not angular position!
I have told you that vectors are “just bookkeeping” but that’s not really true. To be useful
in physics, they need to have some obvious (?) mathematical properties. In fact, if I try to
?
specify the angular position of an object with the construct θ = θx î + θy ĵ + θz k̂ where θx ,
θy , and θz are angles with respect to fixed axes, then I will get into trouble quickly.
Show the trick with the textbook. Rotations do not commute but vector addition must!
However, infinitesimal rotations do commute. (See Note 7.1 in K&K.) The trick is that the
part that that does not commute enters in second order in the angles. So I can write
ω=
dθx
dθy
dθz
î +
ĵ +
k̂ = ωx î + ωy ĵ + ωz k̂
dt
dt
dt
(85)
for angular velocity. For its direction, see the following figures of a rotating rigid body:
Pick some particle in the object. Its position is r and its velocity is v = dr/dt. The
magnitude of v is (r sin φ)dθ/dt. In other
words, we write
dθ
dr
= v = n̂ × r = ω × r
dt
dt
(86)
where we define ω ≡ (dθ/dt)n̂. In particular, ω points in the direciton of the rotation
axis, specified by the unit vector n̂.
Example: Angular Momentum of a Rotating Skew Rod
Find the angular momentum L of P
the two masses around the
rod midpoint. By definition, L = (ri × pi ). For each mass
r ⊥ p and r × p is perpendicular to the rod! Since r ⊥ p we
have |r × p| = lp = lmv = lmωl cos α so L = 2mωl2 cos α.
P 2
For this axis I =
mr = 2m(l cos α)2 so L 6= Iω! The
equality is violated in both magnitude and direction, although
Lz = L cos α = Iω. This is because the “object” is asymmetric
and our non-tensor formulation of rotational inertia is naive.
Practice Exercise
Suppose a particle rotates counter clockwise in the xy plane. Let θ be the polar angle,
relative to the x-axis. Then, its angular velocity would be ω = ω k̂ where ω = θ̇. Show that
equation 86 gives the correct vector velocity v = ω(xĵ − y î) = ωrθ̂. Draw a picture with the
particle at a couple of points, and sketch r and v.
25
Example: Torque on the Rotating Skew Rod
The (direction of the) angular momentum of the skew rod changes with time. What torque
causes this? To see this, let’s find dL/dt and interpret the result. We can do
L(t) = L sin α cos ωtî + L sin α sin ωtĵ + L cos αk̂
dL
τ =
= ωL sin α −î sin ωt + ĵ cos ωt
so
τ = ωL sin α
dt
(87)
(88)
Alternatively, we can use a geometric approach. Looking down
the axis from the top, on the horizontal (“h”) plane, we see that
|∆Lh | = Lh ∆θ. This is the only component of L that changes.
Therefore
dLh
dθ
dL
=
= Lh
= (L sin α)(ω)
(89)
dt
dt
dt
so we get the same result as before. The torque is zero if α = 0, and also if α = 90◦ (L = 0).
The torque is from forces which hold the vertical bearing onto the rotation axis. Can you see
why the torque zero in both cases? (Imagine you’re holding the bearing in your hand.)
τ=
The Gyroscope
A simple gyroscope is a spinning wheel on a horizontal axis, supported on one end only:
Contrary to our intuition, it does not fall down. Instead, gravity exerts a torque which causes
it to precess about a vertical axis which passes through the support point.
Formally, this is easy to see. The angular momentum vector L is along the spinning axis. The
torque about the support point is only the weight W, since the normal force N has no moment
arm about this point. The torque τ has magnitude τ = lW and direction perpendicular to
both W and L. So L changes in direction, not magnitude, tracing the circle in the horizontal
plane, shown on the left. Calculating the precession angular frequency Ω
lW
lW
dL
= LΩ
so
Ω=
=
(90)
τ = lW =
dt
L
I0 ωs
A simple model makes it easier to see, physically, why the gyroscope
precesses. Consider two masses on an arm, rotating in the horizontal
plane. Instantaneously apply a torque with opposing forces F up and
down on the two masses, so there is no net force and the center of mass
doesn’t move. If the force acts for time ∆t, then ∆p = m∆v = F∆t
for each mass. One mass moves up and the other moves down. Thus
∆φ ≈
∆v
F ∆t
2lF ∆t
τ
=
=
= ∆t
v
mv
2lmv
L
so
which is the same result that we got in Equation 90.
26
dφ
τ
=
dt
L
(91)
Physics I Honors
Thursday 9 October
Fall 2008
Different format today. We will spend most of the period doing “worksheets” which you will
turn in. These will be counted together as one homework assignment. The point is to help
you get up to speed on the material we’ve been covering, especially the use of vectors to
describe motion, both translation and rotation, in two and three spatial dimensions.
Worksheet #1: Angular velocity for motion in a plane.
Collect the worksheets, and work the problem out on the board.
Rigid Bodies and Conservation of Angular Momentum
For a symmetric rigid body, we now know that the angular momentum vector is parallel to the angular velocity vector. We write
X
L = Iω
where
I=
mi ri2
is the rotational moment of inertia about the rotation axis. We say
that L is conserved if there are no external torques but nobody can
prove this is so, starting from Newton’s laws! See the diagrams on
the right. The torques τ ij = ri ×fij cancel only if fij works along the
line ri − rj . That is, fij = −fji (Newton’s Third Law) is not enough.
There is some higher reason for angular momentum conservation.
Worksheet #2: Conservation of Angular Momentum
Collect the worksheets, and work the problem out on the board.
General Angular Momentum of Rigid Bodies: The Rotational Inertia Tensor
If all position vectors ri , for the particles of mass mi within a rigid body, are measured with
respect to the center of mass, then the angular momentum vector is
X
X
X
L=
r×p=
r × mi ṙi =
mi ri × (ω × ri )
i
i
i
You can reduce this expression with some tedium. (See K&K.) You’ll find, for example,
"
#
"
#
"
#
X
X
X
Lz =
mi (x2i + yi2 ) ωz −
mi xi zi ωx −
mi yi zi ωy
or
L = I · ω (92)
i
i
i
where I is the rotational inertia tensor, a 3 × 3 matrix of numbers. This is a general formulation
angular momentum of a rigid body. Note that if ω = ω k̂ then Lz = Iωz where
Pof the
2
I=
mr just as we had before. Also, as we have seen, if the “off diagonal elements” of I
are nonzero, then there are other terms in L. (How much do you guys know about matrices?
Can I tell you that the axes which diagonalize I are called the principle axes?)
Worksheet #3: The Rotational Inertia Tensor for the Skew Rod
Collect the worksheets, and work the problem out on the board, if there is time.
27
Physics I Honors
Name:
Thursday 9 October
Fall 2008
Worksheet #1: Angular velocity for motion in a plane
Consider a particle which moves in the (x, y) plane. Answer the following questions:
A. In one sentence, why does r(t) = x(t)î + y(t)ĵ tell us the position of the particle?
B. The particle moves in a circle with angular velocity ω. It starts at (x, y) = (r, 0). Find
the functions x(t) and y(t) in terms of r, ω, and t.
C. Find the velocity vector v(t) = ṙ(t) in terms of r, ω, and t, and unit vectors î and ĵ.
D. The angular velocity is ω = ω k̂. Take the cross product ω × r(t) and show that it is the
same as v(t). Refer to Sec. 1.4 in your book for the cross product of the unit vectors.
28
Physics I Honors
Name:
Thursday 9 October
Fall 2008
Worksheet #2: Conservation of Angular Momentum for Rigid Bodies
A student sits in a swiveling chair that allows her to rotate about a vertical axis. Her
rotational inertia is I0 . She is holding a wheel on an axle. The wheel has rotational inertia
IW . The axle is oriented vertically. Initially, the student is not moving, but the wheel is
spinning with an angular velocity ω, counter clockwise as viewed from the top.
A. What is the magnitude and direction of the initial angular momentum vector Li ?
B. The student turns the wheel and axle over by 180◦ so that it again is oriented vertically,
but now the wheel rotates clockwise as seen from the top. If the student’s angular velocity
is now Ω, write the final angular momentum Lf in terms of I0 , IW , ω, and Ω. Be careful of
signs. You can ignore the mass of the wheel as compared to the mass of the student.
C. Find Ω in terms of I0 , IW , and ω.
29
Name:
Thursday 9 October
Physics I Honors
Fall 2008
Worksheet #3: The Rotational Inertia Tensor for the Skew Rod
Use Eq. 92 to calculate the angular momentum of the skew rod
which we studied last class. The angular velocity is ω = ω k̂.
A. Using ρ as defined at the left, derive Lz in terms of m, ρ, and ω.
B. By rearranging x, y, and z in Eq. 92, show that
"
#
"
#
X
X
Lx = −
and
Ly = −
mi xi zi ω
mi yi zi ω
i
i
C. Complete the following table for the coordinates of the two masses in terms of ρ, ω, and h:
Particle 1
x1 = ρ cos ωt
y1 =
z1 =
Particle 2
x2 =
y2 = −ρ sin ωt
z2 = h
D. Derive Lx and Ly in terms of m, ρ, h, and ω. Compare to our result from last class.
30
Physics I Honors
Tuesday 14 October
Fall 2008
Today we finally (!?) leave rotations of rigid bodies, but we will use those topics later.
Accelerating Reference Frames
Newton’s first law of motion says something like “An object at rest stays at rest unless some
force acts on it. Also, an object in motion doesn’t change its direction or speed unless some
force acts on it.” These statements are fundamental assumptions and cannot be proven. In
fact, they aren’t always true. We say that an “inertial reference frame” is one in which
Newton’s first law, sometimes called the “law of intertia”, holds.
A reference frame in which Newton’s law doesn’t hold is an “accelerating reference frame.”
The second law really embodies the first law. The second law says that
X
F = mA
(93)
P
so that if there are no forces ( F = 0) then A = 0, i.e. the first law. If you are “inside” the
frame that is accelerating, then you feel something that’s like a force, but isn’t really there.
So, invent a “fictitious force” Ffict = −mA and everything is fine. Here’s a simple example:
Analyze this two ways, for an observer standing on the
street, and for someone inside the car.
PFfict
P Fx
Fy
Outside (“inertial”) Inside (“accelerating”)
0
−mAî
T sin θ
T sin θ + Ffict
T cos θ − mg
T cos θ − mg
So, the person outside the car says that T sin θ = mA, while the person inside the car says
that T sin θ + Ffict = T sin θ − mA = 0. These are the same equations. This is really pretty
trivial, but more complex situations are more complicated.
Practice Exercise
Here’s a slightly more complex situation, from Example 8.2 in your textbook.
The plank accelerates to the right with acceleration A. The
cylinder rolls without slipping on the plank. To an observer
in an inertial frame, the cylinder accelerates to the right due
to the frictional force f . To an observer on the plank, a
fictional force Ffict = M A acts on the center of mass and
accelerates the cylinder to the left. If the acceleration of the
cylinder is a0 to the observer on the plank, then the torque
f R accelerates the cylinder by α0 via f R = I0 α0 = I0 a0 /R
where I0 = M R2 /2.
So, first, write x = x0 + X to locate the cylinder in the inertial frame relative to the accelerating frame, where X locates the plank. Use this to explain why the acceleration in the
inertial frame is a = a0 + A. Then, use the relationships between rolling and translational
motion above, to find the accelerations a and a0 in terms of A.
31
Galilean Relativity
No let’s be a little more formal about this idea of “reference frames.” In particular, we will
show that Newton’s laws of motion are the same in different “inertial” reference frames, that
is, two frames that are moving with constant velocity with respect to each other.
For this discussion, I am using the notation and figures from Sec. 11.4 in your textbook, as
opposed to Sec. 8.2. Both are called “The Galiliean Transformations”, but the latter is a
more common formulation.
Two observers follow motion of the same particle. It is located at r in the coordinate system of one of them, and at
r0 in the coordinate system of the other. The origin of the
“primed” system is located at R in the unprimed system.
We therefore have
r0 = r − R
where
R = vt
(94)
That is, the primed coordinate system moves with a (constant) velocity v relative to the
unprimed system. We let v be along the x (and x0 ) axis.
Obviously ṙ0 = ṙ − v and r̈0 = r̈. Therefore, even though the velocity of the particle is
shifted by v in one
P coordinate system relative to the other, the accelerations are the same.
This means that
F = ma in both systems, so Newton’s second law is intact. In fact, you
cannot tell which frame you are in by studying the particle’s motion. Notice also that, for
two particles a and b, r0a −r0b = ra −rb . This means that the distances and directions between
the two particles are the same in both frames, and so are two-body forces like gravity.
(This all makes some plausible, albeit ultimately wrong, assumptions. For example, it assumes that lengths are measured the same by both observers, but this ends up being incorrect.
Leave this for now, but remember it when you come to study special relativity.)
Now assume that some sort of pulse is emitted at t = 0 from
the origin in the (x, y) system. Its position along the x axis
in the (x, y) system is
x = ct
where I’ve assumed that its velocity is c. Its position along
the x0 axis in the (x0 , y 0 ) system is therefore, from Eq. 94,
x0 = x − vt = ct − vt = (c − v)t
In other words, if you follow the pulse in the (x0 , y 0 ) system, then it appears to move in the x0
direction with a velocity dx0 /dt = c − v. Of course, this is just what you’d expect. However,
as you will learn next semester, the laws of electromagnetism (aka Maxwell’s Equations)
say that the speed of light is c in any inertial frame. This inconsistency is what led to the
development of special relativity, and the eventual recognition that Newton’s laws are only
an approximation in a world where all velocities are small compared to c.
Practice Exercise
Show that the force of gravity is the same between two objects a and b, regardless of whether
they are viewed in a primed or unprimed system, related by the transformation Eq. 94.
32
Physics I Honors
Thursday 16 October
Fall 2008
Thanks to Brian for covering class today (and lab yesterday!)
Remember: Exam #2 is two weeks from today.
Note: Section 8.4 in book on Equivalence Principle is interesting, but we’re skipping it.
Rotating Coordinate Systems
Recall from last class: Frame accelerating with acceleration A doesn’t obey Newton’s laws,
but you can fix things by inventing a fictitious force Ffict = −mA.
Familiar example: For a turntable rotating at constant angular velocity Ω, there is an inward
acceleration Ω2 R where R is the distance from the rotation axis. So, if you are on the
turntable you feel a fictitious force mΩ2 R outward. This is called “centrifugal force.”
Our goal today is to be more formal with this concept of “rotating coordinate systems.”
See the figure. One set or coordinates rotates
about the other with angular velocity Ω. The
origins coincide, that is, we are only concerned
with rotation, not translation. The (x, y, z)
system is inertial. At time t all three axes
coincide. Rotation is about the z (z 0 ) axis.
Our job is to relate the derivatives of the vector r(t) in the inertial (unprimed) and rotating
(primed) frames. Examine the change in this vector between times t and t + ∆t, below:
In short, the rotating observer is “fooled.” After the time ∆t has elapsed, he measures a
change ∆r0 based on the correct location of the final position, but a “wrong” starting position
because his axes have rotated. The figure makes it clear that
∆r(t) = [r0 (t) + ∆r0 (t)] − r(t)
(95)
We know how to relate r0 (t) and r(t). They are connected through the angular velocity
vector Ω = Ωk̂ according to Equation 86, that is
r0 (t) − r(t) = (Ω × r)∆t
(96)
If we combine equations 95 and 96 we can write
∆r(t)
∆r0 (t)
=
+Ω×r
∆t
∆t
33
(97)
This is essential. If we just let ∆t → 0, we can relate the velocities in the two frames as
vinertial = vrotating + Ω × r
(98)
This proof relies only on the geometric properties of r, and really relates the derivative of this
vector when viewed by two different coordinate systems. So, we could just as well write
dr
dr
dB
dB
=
+Ω×r
or, generally,
=
+ Ω × B (99)
dt in
dt rot
dt in
dt rot
for any vector B. (See also Note 8.2.) For example, the acceleration vector becomes
dvin
dvin
ain =
=
+ Ω × vin
dt in
dt rot
d
=
(vrot + Ω × r)
+ Ω × (vrot + Ω × r)
dt
rot
= arot + 2Ω × vrot + Ω × (Ω × r)
(100)
where we have assumed that the rotation vector Ω does not change with time. Thus, the
fictitious force in the rotating coordinate system
Ffict = −2mΩ × vrot − mΩ × (Ω × r)
(101)
has two terms. The second term has magnitude mΩ2 r for a position r in the rotation plane,
so is just our old friend the centrifugal force. The first term is new. It depends on the
velocity vrot in the rotating frame, and is called the coriolis force.
Example 8.7. Bead sliding without friction on a radial rod,
rotating with angular velocity ω. Use the rotating coordinate
system. The vector ω × r is in the horizontal plane, upward
in the lower figure. So, Fcent = −mω × (ω × r) is outward,
and Fcor = −2mω×vrot is as shown for vrot radially outward.
X
Radial
F = ma
so
mω 2 r = mr̈
X
Transverse
F = ma
so
N − 2mω ṙ = 0
The radial equation gives r(t) = Aeωt + Be−ωt where A and B are determined from initial
conditions. Thus N = 2mω 2 (Aeωt −Be−ωt ) for the normal force of the rod on the bead.
Practice Exercise
Get some practice with the directions of coriolis and centrifugal forces on the Earth’s surface.
What is the direction of Ω? What is the magnitude and direction of Ω × r for a particle
at the equator? What if the particle is at the North Pole? Do the same for Ω × (Ω × r).
Now calculate, in terms of Ω and the radius R of the Earth, the magnitude of the centrifugal
force on a particle at the equator and at the pole.
Next, instead of being stationary, let the particle fall from some height near the Earth’s
surface. It therefore has a velocity vrot downward in the rotating frame. Go through a
similar process, and determine the magnitude and direction of the coriolis force on the
falling particle, in terms of vrot , Ω, and R.
34
Physics I Honors
Monday 20 October
Fall 2008
Motion under Central Forces
Recall our discussion of two-body motion and center of mass. We found that we can treat a
two body problem as a one-body problem with a “reduced mass.” Let’s review that.
First, write Newton’s Second Law for each of the two bodies:
m1 r̈1 = f (r)r̂
and
m2 r̈2 = −f (r)r̂
Add these two equations to see what happens to the center of mass:
m1 r̈1 + m2 r̈2 = M R̈ = 0
where M = m1 + m2 and R = (m1 r1 + m2 r2 )/M is the position of the center of mass. On the
other hand, the two equations of motion can also be combined by subtracting, hence
m1 m2
1
1
f (r)r̂
so
µr̈ = f (r)r̂
where
µ=
+
r̈ = r̈1 − r̈2 =
m1 m2
M
is called the reduced mass. We now can deal with the one-body “central force” problem.
Note that if m2 m1 (like Sun & planet) then M ≈ m2 and µ ≈ m1 , so we ignore it.
Recall that angular momentum L = µr × ṙ is conserved, in both magnitude and direction,
for central forces. One obvious conclusion is that the motion must be in a plane.
We can go further by using polar coordinates. The magnitude of the angular momentum is ` = µr2 θ̇. The total
mechanical energy is
1
1
E = µv 2 + U (r) = µ(ṙ2 + r2 θ̇2 ) + U (r)
(102)
2
2
where U (r) is the potential energy with f (r) = −dU/dr. So,
1
`2
`2
1
2
E = µ ṙ + 2 2 + U (r) = µṙ2 + Ueff (r)
where
Ueff (r) =
+ U (r) (103)
2
µr
2
2µr2
is called the effective potential energy. It is useful for analyzing central force motion in terms
of the radial coordinate. The coordinate θ obeys dθ/dt = `/µr2 , where ` is constant.
Effective potential for the potential energy of two gravitating
bodies. The solid curve is
Ueff (r) =
`2
Gm1 m2
−
2
2µr
r
(104)
For E = Emin , the radial coordinate doesn’t change. For
E < 0, the radial coordinate oscillates between min and
max values. For E ≥ 0, there is no maximum. Later!
Practice Exercise
Put m1 = m M = m2 in Eq. 104. Show that ` = mvR is the correct circular orbit.
35
Dark Matter in Galaxies
We can learn a lot about one of the most important problems in physics today, just by
combining what we know now about circular orbits, Newtonian gravity, and observations of
the Milky Way and some distant galaxies.
A circular orbit of some small mass m about some large mass M obeys Newton’s second law
in the form
r
GM
mM
v2
G 2 =m
which gives
v=
r
r
r
for the “orbital velocity”. Note that the period of revolution is T = 2πr/v so we have
T2 =
4π 2 3
r
GM
which is called Kepler’s Law of Periods. (We’ll look at this more on Thursday.) Note that this
is a way to determine the mass of the sun from the revolution periods of the planets.
One very
√ basic result from this is that the tangential rotation speed v should be proportional
to 1/ r as soon as the mass is concentrated at distances smaller than r. Galaxies should
look like this, since there is generally a big “bulge” of bright stars near the center. However,
the “galactic rotation curves” look very different! (See slides.) This is one of the strong
pieces of evidence for “dark matter” in the universe.
For some interesting reading, see Alternatives to dark matter and dark energy. Philip D.
Mannheim (UConn). Published in Prog.Part.Nucl.Phys.56:340-445,2006. You can also visit
http://arxiv.org/ and get e-Print: astro-ph/0505266.
Practice Exercise
Use the Milky Way galactic rotation curve to estimate the mass of the galaxy contained
within a distance of 16 kpc. (Note that 1 kpc=103 pc=3.1 × 1019 m.) The mass of the sun is
2 × 1030 kg. How many sun-like stars do you need to get this mass? Note that the brightness
of a typical spiral galaxy (like the Milky Way?) seems to be about 1010 times that of the
sun. (Table 21-1 in Introductory Astronomy and Astrophysics by Zeilik and Gregory.)
36
The Milky Way Galaxy
Visible Light
Infrared Light
37
38
Note: The vertical axis has a suppressed zero!
“Keplerian” prediction
for 1010 Solar Masses
1985ApJ...295..422C
39
Physics I Honors
Thursday 23 October
Fall 2008
Remember: Exam #2 is one weeks from today. Same ground rules as before.
Motion under Newtonian Gravity
We want to find the motion under Newtonian gravity in three dimensions. Assume an object
(say, a planet) with mass m orbiting a (massive) body M . (Don’t worry about reduced mass
µ distinction now.) The potential energy is (See Example 5.3 in your textbook)
U (r) = −G
C
mM
=−
r
r
where
C ≡ GmM
(105)
We use C because that makes it easy to generalize to other central force laws.
First let’s review from last class. Motion is in a plane since L = mr × ṙ is conserved for
central forces. This is easy to show using Newton’s second law:
X dL
= mṙ × ṙ + mr × r̈ = 0 + r ×
(106)
F = r × f (r)r̂ = 0
dt
where the sum of forces takes the form that defines a “central” force.
Central forces are conservative. (See Example 4.8 in your textbook.) So write f (r) = −dU/dr
for some U (r). We know that this means that the energy
1 2
1 2
2 2
E = mṙ + U (r) = m ṙ + r θ̇ + U (r)
(107)
2
2
doesn’t change with time. It is a “constant of the motion.” Note that we’ve written this in
terms of polar coordinates r and θ. See Section 1.9 of your textbook for a review.
The magnitude ` = |L| of the angular momentum is also a constant of the motion. The
component of velocity perpendicular to the moment arm r is vθ = rθ̇ by definition. So
` = mr2 θ̇
(108)
Once again, for as long as there is motion, the value of ` does not change, even though r
and θ̇ are of course functions of time. They just always combine so that ` is constant.
Now we can find the shape of the orbits. We can combine Equations 107 and 108 to
form two differential equations describing the motion as r(t) and θ(t), namely
r
dθ
`
dr
2
`2
=
and
=
[E
−
U
(r)]
where
U
(r)
≡
U
(r)
+
(109)
eff
eff
dt
mr2
dt
m
2mr2
However, the shape of the orbit is some function r(θ). So, divide the two to find
1
dθ
`
`
p
=
=
dr
mr2 (2/m)[E − Ueff (r)]
r(2mEr2 + 2mCr − `2 )1/2
(110)
The integral looks messy, but in fact it can be done analytically. (You can find it in tables.)
Integrating both sides, and putting the constant of integration, −θ0 , on the left, we have
mCr − `2
−1
√
θ − θ0 = sin
(111)
r m2 C 2 + 2mE`2
40
Divide through my mC in the parentheses, take the sine of both sides, and solve for r:
r = r(θ) =
`2 /mC
p
1 − 1 + (2E`2 /mC 2 ) sin(θ − θ0 )
Let’s clean this up. Pick θ0 = −π/2 (it’s arbitrary) and define
r
2E`2
r0
r0 ≡ `2 /mC
and
≡ 1+
>0
so
r = r(θ) =
2
mC
1 − cos θ
(112)
(113)
The parameter is called the eccentricity. The curves r(θ) are called conic sections. The
forms become familiar if we work in cartesian coordinates x = r cos θ and y = r sin θ, so
p
(114)
x2 + y 2 − x = r0
=⇒
(1 − 2 )x2 − 2r0 x + y 2 = r02
Case = 0. The curve is r = r0 , a circle. Setting mv 2 /r = p2 /mr = `2 /mr3 = C/r2 gives
r = `2 /mC = r0 . This is precisely what we know we should get for a circular orbit.
Case = 1, i.e. E = 0. In cartesian form, we see that the “orbit” is a parabola,
x=
y2
r0
−
2r0
2
(115)
Case < 1, i.e. 1 − 2 > 0. Your book works this out in polar coordinates, but let’s try it
in cartesian. First find the “ends” of the orbit by setting y = 0. This leads to
xmin = −
r0
1+
and
xmax =
r0
1−
=⇒
a≡
1
r0
(xmax − xmin ) =
(116)
2
1 − 2
Now translate to x0 = x − x0 where x0 = a + xmin = r0 /(1 − 2 ) is the halfway point between
the two ends. Substituting for x in Equation 114, and simplifying, we find
r02
(1 − )x + y =
≡ b2
2
1−
2
02
2
(117)
Clearly y = ±b are the “top” and “bottom” of the orbit. Since a2 = b2 /(1−2 ) we have
2
x0
y2
+
=1
a2
b2
(118)
which is the familiar equation of an ellipse. The parameters a and b are called the semimajor
and semiminor axes. The position x = 0 is called the focus; it is not the “center”!
Case > 1, i.e. 2 − 1 > 0. Multiply Equation 117 through by −1, but keep the sign in
front of b2 so that it is still positive. Equation 118 becomes
2
x0
y2
−
=1
a2
b2
(119)
This is the equation of a hyperbola. Note that x0 is now a negative number.
This is enough for today. You should read through your book on Kepler’s laws, but we won’t
specifically cover them in this course. They will be important, though, if you should take a
course on astrophysics.
41
Physics I Honors
Name:
Thursday 23 October
Fall 2008
Worksheet: The following shows a circular orbit of a planet around the Sun.
a. Draw on this graph a circular orbit with larger total energy.
b. Draw on this graph an elliptical orbit with the same total energy as the original circular
orbit. Make the eccentricity e of the orbit equal to 4/5. (Hint: Use the definitions of r0 , ,
and a to show that the energy E only depends on C and a.) Calculations:
c. At a point where the elliptical and (original) circular orbit cross, draw the velocity vectors
for each of the two orbits at that point. Indicate which is which.
d. Explain briefly why the magnitudes of the two velocity vectors must be the same.
e. Of the original circular orbit and your elliptical orbit, which has the largest angular
momentum? (Assuming the same orbiting mass in each case.) Explain your answer.
42
Physics I Honors
Monday 27 October
Fall 2008
Second exam is on Thursday. Everything the same as for Exam #1.
The Simple Harmonic Oscillator (Again)
First, start with a review of old stuff. Pay attention to the notation,
e.g. ω0 .
P
Newton’s P
Second Law says
F = ma = mẍ. For this
example,
F = Fspring = −kx. So we need to solve
r
k
2
−kx = mẍ =⇒ ẍ(t) = −ω0 x(t) where ω0 ≡
(120)
m
Any function x(t) proportional to cos ω0 t or sin ω0 t solves this. So, general solution is
x(t) = B sin ω0 t + C cos ω0 t
(121)
We need more information to determine B and C. Typically, we use “initial conditions.”
Everybody knows that x and x(t) mean exactly the same thing, right?
Solution using complex numbers. We solved Eq. 120 by “recognition.” A better way is
to use an ansatz, namely x(t) = Aeiωt . Substitute into Eq. 120 and solve for ω:
−ω 2 Aeiωt = −ω02 Aeiωt
ω = ±ω0
=⇒
(122)
Once again, there are two possible solutions, and so once again the general solutions is:
x(t) = A1 eiω0 t + A2 e−iω0 t
(123)
Initial conditions. Work with solution 121. Obviously x(0) = C. We also have
v(t) = ẋ(t) = ω0 B cos ω0 t − ω0 C sin ω0 t
=⇒
v(0) = ω0 B or B = v(0)/ω0
(124)
A trig identity gives us a different, more useful way, to write the solution. We write
x(t) = A cos(ω0 t + φ) = A cos φ cos ω0 t − A sin φ sin ω0 t
(125)
which we compare to 121 to see that
s
x(0)2 +
B = −A sin φ and C = A cos φ =⇒ A =
v(0)2
v(0)
and tan φ = −
2
ω0
ω0 x(0)
(126)
Energy. Have U (x) = kx2 /2 since −dU/dx = Fspring . Also K = mv 2 /2 = mẋ2 /2, so
1
1
1
E = K + U = mω02 A2 sin2 (ω0 t + φ) + kA2 cos2 (ω0 t + φ) = kA2
(127)
2
2
2
So, the total energy is the potential energy when the position x = A, the “amplitude.”
Practice Exercise
Apply the initial conditions to solution 123. Solve for A1 and A2 and then show that the
complete solution is the same as (121) with B and C in terms of x(0) and v(0). Recall
Euler’s formula, namely
eiθ = cos θ + i sin θ
43
The Damped Harmonic Oscillator
P
Now
F = Fspring + Fdamp = −kx − bv = −kx − bẋ. The equation of motion becomes
r
b
k
and ω0 ≡
(128)
mẍ = −kx − bẋ
=⇒
ẍ + γ ẋ + ω02 x = 0 where γ ≡
m
m
(The definition of ω0 is just as before.) Solving this by “recognition” is too hard, so we
immediately move on to our ansatz x(t) = Aeiωt . Inserting this into Eq. 128 gives
r
γ
γ2
=⇒
ω = i ± ω02 −
−ω 2 + iγω + ω02 = 0
(129)
2
4
p
Case 1: γ < 2ω0 . Define ω1 ≡ ω02 − γ 2 /4 and the solution to Eq. 128 becomes
x(t) = Ae−(γ/2)t cos(ω1 t + φ)
(130)
Some comments are in order. First, the oscillation frequency is ω1 < ω0 . Secondly, the
oscillation maximum is not a fixed amplitude, but decays away with a 1/e “time constant”
(γ/2)−1 = 2m/b. Finally, the exponential factor in front makes the association of A and
φ with x(0) and v(0) kind of messy. This isn’t so interesting, so we won’t worry about it.
The red plot shows a damped oscillator with γ = ω0 /4, with
the dashed line showing the exponential factor alone. It is
clear that the energy decreases with time. In fact, in the
case where γ ω0 (a situation called “lightly damped”) it
is not hard to show that
10
8
6
x(t)
4
2
0
−2
E(t) = E0 e−γt
−4
where
−6
−8
0
2
4
6
8
10
1
E0 = kA2
2
(131)
So, 1/γ ≡ τ is the “energy decay time constant.”
Time t
Case 2: γ > 2ω0 . In this case, we write the solutions to Eq. 129 as ω = iα where
s
γ γ
4ω 2
α= ±
=⇒
x(t) = C1 e−α1 t + C2 e−α2 t
1 − 20 = {α1 , α2 }
2 2
γ
(132)
where both α1 and α2 are positive real numbers, so both terms of the solution are decaying
exponential functions. The blue curve in the figure is for γ = 3ω0 , and the same initial
conditions as for the underdamped case. This situation is called “overdamped.”
Case 3: γ = 2ω0 . This case is called “critical damping.” Something is weird, though. It
looks like our ansatz has broken down, because we only have one solution, α = γ/2, when
we know there should be two. But don’t worry, math is good. See your textbook, or a book
on differential equations. The second solution in this case is proportional to te−αt , so
x(t) = D1 e−αt + D2 te−αt
(133)
is the general solution. This is the green line plotted in the figure.
Practice Exercise
Derive Eq. 131 for the decay time of a lightly damped harmonic oscillator. Follow the discussion in your textbook, pages 416 and 417. Our Eq. 131 is the book’s equation 10.16.
44
Physics I Honors
Monday 3 November
Fall 2008
Hand back exam? If so, give grade statistics. Some Notes:
• Wednesday is “make-up lab” day; need email to me or Scott by the end of today
• Preliminary lab notebooks due in class next Monday!
• Stay tuned: Testing something new, perhaps, in lab on coupled oscillations (Nov.19)
Review: Simple and Damped Harmonic Oscillations
P
The simple harmonic oscillator is based on F = −kx and has the equation of motion
ẍ(t) = −ω02 x(t)
where ω0 ≡
(134)
p
k/m. We solve this with the “ansatz” x(t) = Aeiωt . Substituting,
−ω 2 = −ω02
=⇒
ω = ±ω0
(135)
So, both x(t) = A1 eiω0 t and x(t) = A2 e−iω0 t are valid solutions. In fact, any linear combination of the two is a valid solution. We find A1 and A2 by applying the initial conditions.
P
The damped harmonic oscillator is based on
F = −kx − bẋ and has the equation of
motion
ẍ(t) = −ω02 x(t) − γ ẋ(t)
(136)
where γ ≡ b/m. We again solve this with x(t) = Aeiωt . Substituting, we find a quadratic
“characteristic equation” for ω. The solutions to this equation are
r
γ
γ2
ω = i ± ω1
where
ω1 ≡ ω02 −
(137)
2
4
(This is the “underdamped” solution, where γ < 2ω0 , which is the only case which yields
harmonic oscillations.) So, both x(t) = A1 e−γt/2 eiω1 t and x(t) = A2 e−γt/2 e−iω1 t are valid
solutions. Again, any linear combination of the two is a valid solution, and we find A1 and
A2 by applying the initial conditions. For future reference,
x(t) = A1 e−γt/2 eiω1 t + A2 e−γt/2 e−iω1 t
(138)
is the general solution to the equation of motion for a mass m attached to a spring with force
constant k and being damped out with a force proportional to velocity, with coefficient b; the
parameters ω1 and γ are defined in terms of m, k, and b by the relations given above.
Note that we could rewrite Eq. 138 simply by using e±iω1 t = cos ω1 t ± sin ω1 t and then
defining coefficients B1 and B2 appropriately in terms of A1 and A2 so that
x(t) = B1 e−γt/2 cos ω1 t + B2 e−γt/2 sin ω1 t
= Ae−γt/2 cos(ω1 t + φ)
This is exactly the same result which we got in Eq. 130 from last class.
45
(139)
(140)
The Forced Harmonic Oscillator
Now imagine that we are “forcing” the damped
Pharmonic oscillator by applying a sinusoidally
varying force as a function of time. That is
F = −kx − bẋ + F0 cos ωt where ω is now a
frequency that we can control independently. The equation of motion becomes
F0
cos ωt
(141)
m
We can solve this equation using something of a trick. Pick another variable y(t) so that
ẍ + γ ẋ + ω02 x =
F0
sin ωt
(142)
m
and define z(t) = x(t) + iy(t). The differential equation that governs z(t) is therefore
ÿ + γ ẏ + ω02 y =
F0 iωt
e
m
where we just need to recall that x(t) = <z(t) to get the motion of our oscillator.
z̈ + γ ż + ω02 z =
(143)
Applying our ansatz z(t) = Aeiωt yields something a little different than before. Instead of
finding a characteristic equation to solve for ω, we instead find an equation for the (complex)
amplitude A as a function of ω. That is
F0
F0
1
A −ω 2 + iγω + ω02 =
=⇒
A=
(144)
2
m
m ω0 − ω 2 + iγω
(Note: We can always add in a solution like Eq. 139 since it just yields zero in the differential
equation. This is where we get the constants we need to apply the initial conditions. This
part of the solution, though, dies out with time, so we usually just ignore it; it is called a
“transient.”) This is easy to interpret if we write
F0
1
γω
iφ
−1
A = Ae
where
A=
and φ = tan
(145)
m [(ω02 − ω 2 )2 + ω 2 γ 2 ]1/2
ω 2 − ω02
In other words, the motion of the forced damped oscillator
is given by x(t) = A cos(ωt + φ) where A and φ are strong
functions of the driving frequency ω. These functions
are shown on the left, for the lightly camped case where
γ ω0 . The motion has an amplitude A that becomes very
large when the driving frequency gets close to the natural
frequency ω0 . This phenomenon is called resonance.
Note the phase. The motion is in phase φ = 0 for frequencies
well below ω0 , and precisely out of phase (φ = 180◦ ) far above
ω0 . Also, it passes through φ = (−)90◦ as ω moves through
the resonance.
The resonance has a width that depends directly on γ. In fact, if we define the width ∆ω
as the distance between the two ω values for which the curve drops to half its value, then
∆ω = γ. The value Q ≡ ω0 /γ is a dimensionless number which characterizes the width,
and can be interpreted as the amount of energy lost in one oscillation. Large Q means low
damping, and is an important figure of merit for resonating systems.
46
Physics I Honors
Thursday 6 November
Fall 2008
Hand back exam, discuss grade statistics. (Average on test was 76.73.)
Waves on a Stretched String: The Wave Equation
These notes follow Chap.18 of “Physics”, by Resnick, Halliday, and Krane, 5th Ed.
What is the “equation of motion” for a string that is stretched under some tension F?
The motion is up and down, so analyze the forces
in the vertical direction on a small piece of string
at fixed x. We assume small amplitude waves so
δx can represent the length of the small piece. See
the figure. If F is the tension in the string, then
X
Fy = F sin θ2 − F sin θ1 = (δm)ay
(146)
Put δm = µδx for linear mass density µ. Obviously ay = ÿ = ∂ 2 y/∂t2 since we are working
at fixed x. Also, for small deflections, θ 1 so sin θ ≈ θ ≈ tan θ = slope = ∂y/∂x. So,
∂y ∂y ∂ 2y
1 ∂y ∂y µ ∂ 2y
F (sin θ2 − sin θ1 ) ≈ F
−
=
µδx
=⇒
−
=
(147)
∂x 2 ∂x 1
∂t2
δx ∂x 2 ∂x 1
F ∂t2
The left side Eq. 147 is just ∂ 2 y/∂x2 as δx → 0. This gives
∂ 2y
1 ∂ 2y
=
∂x2
v 2 ∂t2
(148)
where 1/v 2 ≡ µ/F . Eq. 148 is called the wave equation. It
describes a shape of the string, function y of x, which changes
in time, hence y(x, t). The solution is any function like
y(x, t) = f (x0 ) = f (x ± vt)
i.e.
x0 ≡ x ± vt
(149)
This means that the change in time is rather peculiar. The
shape f (x) moves to the right or left with speed v. This is
the definition of a (non-dispersive) wave.
Practice Exercise
Which of the following solves the wave equation, Eq. 148, for v = 1? Take derivatives, or
show which are of the form Eq. 149. Note that the sum of two waves is also a wave, i.e. the
“principle of superposition;” This is because Eq. 148 is a linear differential equation.
(A) y(x, t) = x2 − t2 ; then try y(x, t) = x2 − t2
(B) y(x, t) = sin x2 sin t
(C) y(x, t) = log(x2 − t2 ) − log(x − t)
(D) y(x, t) = ex sin t
47
Sine Waves
We almost always describe waves as if the shape f (x ± vt) is a sine function.
It is handy to describe a sine wave as follows:
2π
(x − vt)
(150)
y(x, t) = ym sin
λ
We call λ the wavelength, for obvious reasons. Note that for fixed x (say x = 0) we have
y(0, t) = ym sin [−2πvt/λ] which means that the transverse oscillations are sinusoidal with
period T = λ/v = 1/f where f (or, maybe, ν) is the frequency.
The more common way to write the equation of a sine wave is
y(x, t) = ym sin(kx − ωt − φ)
(151)
where k = 2π/λ, ω = 2πf and φ is called the phase constant.
Standing Waves
Add two sine waves of the same speed and frequency, but moving in opposite directions:
y1 (x, t) = ym sin(kx − ωt)
and
y2 (x, t) = ym sin(kx + ωt)
⇒ y(x, t) = y1 (x, t) + y2 (x, t) = [2ym cos(ωt)] sin(kx)
(152)
where the result follows from a trig identity. This wave has zero velocity (i.e. kx = k(x−0t))
but an amplitude that varies like cos ωt. It is called a “standing wave.” Watch:
Something important happens when we set up standing
waves on a stretched string that is fixed at both ends, like
a guitar string. As shown in the figure, we can only have
standing waves that “fit” into the distance L between the
two fixed ends. That is
L=n
λ
2L
v
⇒ λn =
⇒ fn = n
2
n
2L
(153)
That is, frequencies have to be integer multiples of a “fundamental” frequency v/2L. This is the basis of the harmonic sound of nearly all string and wind instruments.
48
Physics I Honors
Monday 10 November
Fall 2008
The topic we call “coupled oscillations” has far reaching implications. The formalism ends
up being appropriate for many different applications, some of which bear only a passing
resemblance to classical oscillation phenomena. This includes the mathematics of eigenvalues
and eigenvectors, for example.
These notes describe the elementary features of coupled mechanical oscillations. We use one
specific example, and describe the method of solution and the physical implications implied
by that solution. More complicated examples, and their solutions, are easy to come by, for
example on the web.1
The following figure describes the problem we will solve:
k
x1
m
kc
x2
k
m
Two equal masses m slide horizontally on a frictionless surface. Each is attached to a fixed
point by a spring of spring constant k. They are “coupled” to each other by a spring with
spring constant kc . The positions of the two masses, relative to their equilibrium position,
are given by x1 and x2 respectively.
Now realize an important point. We have two masses, each described by their own position
coordinate. That means we will have two equations of motion, one in terms of ẍ1 and the
other in terms ẍ2 . Furthermore, since the motion of one of the masses determines the extent
to which the spring kc is stretched, and therefore affects the motion of the other mass, these
two equations will be “coupled” as well. We will have to develop some new mathematics in
order to solve these coupled differential equations.
We get the equations of motion from “F = ma”, so let’s do that first for the mass on the
left, i.e., the one whose position is specified by x1 . It is acted on by two forces, the spring k
on its left and the spring kc on its right. The force from the spring on the left is easy. It is
just −kx1 .
The spring on the right is a little trickier. It will be proportional to (x2 − x1 ) since that is
the extent to which the spring is stretched. (In other words, if x1 = x2 , then the length of
spring kc is not changed from its equilibrium value.) It will also be multiplied by kc , but we
need to get the sign right. Note that if x2 > x1 , then the string is stretched, and the force on
the mass will be to the right, i.e. positive. On the other hand, if x2 < x1 , then the spring is
compressed and the force on the mass will be negative. This makes it clear that we should
write the force on the mass as +kc (x2 − x1 ).
1
See http://math.fullerton.edu/mathews/n2003/SpringMassMod.html.
49
The equation of motion for the first mass is therefore
−kx1 + kc (x2 − x1 ) = mẍ1
(154a)
We get some extra reassurance that we got the sign right on the second force, because if
k = kc , then the term proportional to x1 does not cancel out.
The equation of motion for the second mass is now easy to get. Once again, from the spring
k on the right, if x2 is positive, then the spring pushes back so the force is −kx2 . The force
from the “coupling” spring is the same magnitude as for the first mass, but in the opposite
direction, so −kc (x2 − x1 ). This equation of motion is therefore
−kx2 − kc (x2 − x1 ) = mẍ2
(154b)
So now, we can write these two equations together, with a little bit of rearrangement:
(k + kc )x1 − kc x2 = −mẍ1
−kc x1 + (k + kc )x2 = −mẍ2
(155a)
(155b)
To use some jargon, these are coupled, linear, differential equations. To “solve” these equations is to find functions x1 (t) and x2 (t) which simultaneous satisfy both of them. We can
do that pretty easily using the exponential form of sines and cosines, which we discussed
earlier when we did oscillations with one mass and one spring.
Following our discussions about single mass oscillating systems, we assume the same ansatz
and write
x1 (t) = a1 eiωt
x2 (t) = a2 eiωt
(156)
where the task is now to see if we can find expressions for a1 , a2 , and ω which satisfy the differential equations, including the initial conditions. You’ll recall that taking the derivative
twice of functions like this, brings down a factor of iω twice, so a factor of −ω 2 . Therefore, plugging these functions into our coupled differential equations gives us the algebraic
equations2
(k + kc )a1 − kc a2 = mω 2 a1
−kc a1 + (k + kc )a2 = mω 2 a2
(158a)
(158b)
This is good. Algebraic equations are a lot easier to solve than differential equations. To
make things even simpler, let’s divide through by m, do a little more rearranging, and define
two new quantities ω02 ≡ k/m and ωc2 ≡ kc /m. Our equations now become
(ω02 + ωc2 − ω 2 )a1 − ωc2 a2 = 0
−ωc2 a1 + (ω02 + ωc2 − ω 2 )a2 = 0
2
(159a)
(159b)
A mathematician would call collectively call these two equations an eigenvalue equation. The reason
becomes clearer when you write this using matrices. In this case, Eqs. 158 become
k + kc
−kc
a1
a1
= mω 2
(157)
−kc
k + kc
a2
a2
50
So, what have we accomplished? We think that some kind of conditions on a1 , a2 , and ω
will solve these equations. Actually, we can see a solution right away. In mathematician’s
language, these are two coupled homogeneous (i.e. = 0) equations in the two unknowns a1
and a2 . The solution has to be a1 = a2 = 0. Yes, that is a solution, but it is a very boring
one. All it means as that the two masses don’t ever move.
The way out of this dilemma is to turn these two equations into one equation. In other
words, if the left hand side of one equation was a multiple of the other, then both equations
would be saying the same thing, and we could solve for a relationship between a1 and a2 ,
but not a1 and a2 separately. Mathematically, this condition is just that the ratio of the
coefficients of a2 and a1 for one of the equations is the same as for the other, namely3
−ωc2
ω02 + ωc2 − ω 2
ωc4
±ωc2
ω2
ω02 + ωc2 − ω 2
−ωc2
= (ω02 + ωc2 − ω 2 )2
= ω02 + ωc2 − ω 2
= ω02 + ωc2 ∓ ωc2
=
(160)
(161)
In other words, these two equations are really one equation if
ω 2 = ω02 ≡ ωA2
(162)
ω 2 = ω02 + 2ωc2 ≡ ωB2
(163)
or, instead, if
Borrowing some language from the mathematicians, the physicist refers to ωA2 and ωB2 as
“eigenvalues.” We will also refer to A and B as “eigenmodes.”
What is the physical interpretation of the eigenmodes? To answer this, we go back to
Eq. 159a or, equivalently, Eq. 159b. (Remember, they are the same equation now.) If we
substitute ω 2 = ωA2 into Eq. 159a we find that
A
aA
1 = a2
(164)
where the superscript just marks the eigenmode that we’re talking about. In other words,
the two masses move together p
in “lock step”, with the same motion. This happens when
the frequency is ω = ±ω0 = ± k/m, and that is just what you expect. It is as if the two
masses are actually one, with mass 2m. The effective spring constant is 2k, as you learned
in your laboratory exercise, where you had one cart (with variable mass) and a spring on
each side. Therefore, we expect this double-mass to oscillate with ω 2 = (2k)/(2m) = k/m.
Good. This all makes sense.
Now consider eigemode B. Substituting ω 2 = ω02 + 2ωc2 into Eq. 159a, we find that
B
aB
1 = −a2
(165)
In other words, the two masses oscillate “against” each other, and with a somewhat higher
frequency. In this case, it is interesting to see the difference between ωc ω0 (in other
words, a very weak coupling spring) and ωc ω0 (strong coupling spring). For the weak
3
A mathematician would say that we are setting the determinant equal to zero.
51
coupling spring, the frequency is just the same as ω0 , and that makes sense. If the two
masses aren’t coupled to each other very strongly, then they act as two independent masses
m each with their own spring constant
√ k. On the other hand, for a strong coupling between
the masses, the frequency is ω = ωc 2, which gets arbitrarily large. Sometimes, this “high
frequency mode” can be hard to observe.
A simple experimental test of this result is to set up two identical masses with three identical
springs. In other words,
ωc = ω0 . In that case ωB2 = 3ωA2 and therefore the frequency for
√
mode B should be 3 times larger than the frequency for mode A. We should be able to
make this test as a demonstration in class.
We have been talking about general properties of the two eigenmodes. Of course, this
doesn’t tell you how the system behaves given certain starting values, that is, specific initial
conditions.
For this, we need to understand that the two eigenmodes can be combined. In fact, the general motion of each of the masses really needs to be written as a sum of the two eigenmodes.
Each of these comes with a positive and negative frequency, just as it did when we applied
all this to the motion of a single mass and spring. (Recall that the sum and difference of
a positive and negative frequency exponential, are equivalent to a cosine and sine of that
frequency.) Incorporating what we’ve learned about the relative motions of modes A and B,
we can write the motions as follows:
x1 (t) = aeiωA t + be−iωA t + ceiωB t + de−iωB t
x2 (t) = aeiωA t + be−iωA t − ceiωB t − de−iωB t
(166a)
(166b)
where a, b, c, and d are constants which are determined from the initial conditions. Note
that these four constants appear as they do in Eqs. 166 because each frequency term must
match up between the equation for x1 (t) and x2 (t), with the correct relative sign, in order
to solve the coupled differential equations.
An obvious set of initial conditions is starting mass #1 from rest at x = x0 , and with mass
x2 from rest at its equilibrium position. This leads to the solution
x1 (t) = (x0 /2) [cos(ωA t) + cos(ωB t)]
x2 (t) = (x0 /2) [cos(ωA t) − cos(ωB t)]
(167)
(168)
The last laboratory experiment for this course will investigate this motion. Some number
of setups will be using new electronics that allows each mass to be followed separately and
simultaneously.
On to the plots. Two cases are shown, one with ωc = omega0 , and the other with ωc = ω0 /4.
For each case, we plot x1 (t) with x2 (t), and also x1 (t) + x2 (t) with x1 (t) − x2 (t). The first
case shows clearly that the motion is somewhat complicated for either mass, but clearly
splits into two eigenmodes with rather different frequencies for the sum and difference. The
second case shows something much closer to a beat pattern, with the energy shifting from
one mass to the other and then back again.
52
ω0=1
ωc=1
20
x1
x2
15
10
5
0
−5
−10
−15
−20
0
5
10
15
ω0=1
20
25
30
ωc=1
20
x1+x2
x1−x2
15
10
5
0
−5
−10
−15
−20
0
5
10
15
53
20
25
30
ω0=1
ωc=0.25
20
x1
x2
15
10
5
0
−5
−10
−15
−20
0
20
40
60
ω0=1
80
100
120
ωc=0.25
20
x1+x2
x1−x2
15
10
5
0
−5
−10
−15
−20
0
20
40
60
54
80
100
120
Physics I Honors
Thursday 13 November
Fall 2008
Today we start a new general topic, namely Thermodynamics. I will introduce it to you
by discussing the related subject of Statistical Mechanics. This is a way of simplifying
complicated systems by averaging over their properties, an approach that should work if
they are complicated enough.
The Ideal Gas
Let’s imagine a box filled with a very large number of particles. The particles are very small
and don’t interact with each other. There are so many particles, and they are moving so
quickly, that all you feel is a steady pressure on the walls.
See the figure. Put N gas molecules each of mass m in a
cubic box of side length L. Molecule has x velocity vx . It
bounces off the wall and incurs momentum change 2mvx .
The time between bounces is 2L/vx so Fx (2L/vx ) = 2mvx .
The pressure on the wall at x = L is the sum of the forces
of all the molecules, divided by the area of the wall. That is
1 X mvx2
Nm 1 X 2
1 X
Fx = 2
=
vx
(169)
p= 2
L
L
L
V
N
where we recognize that the volume of the cube is V = L3 . Now the quantity in parentheses
is the average value of vx2 , written hvx2 i. There is nothing special about the x direction, and
the number N is very large, so we expect that hvx2 i = hvy2 i = hvz2 i = 31 hv 2 i where we call hv 2 i
the “mean square velocity.” Since all molecules have the same mass, hKi = 12 mhv 2 i is the
average kinetic energy of the molecules. So, we can now write
2
pV = N hKi
(170)
3
Now we invent something called temperature which measures the amount of energy per
particle in the gas. For historical reasons, the temperature T is measured in different units
called Kelvin (K). The conversion constant from temperature to energy is called k, known
as Boltzmann’s constant, and has the value k = 1.38 × 10−23 J/K. We assign half a unit of
kT to each of the three directions in which the particle can move. In other words
3
=⇒
pV = N kT
(171)
hKi = kT
2
This is known as the ideal gas law. You may have already seen this in a chemistry course,
but chemists count particles in a different way. They prefer to count moles where a mole is
Avogadro’s number NA = 6.02 × 1023 of particles. That is, N = nNA where n is the number
of moles. Chemist’s also write R ≡ NA k, that is pV = nRT .
Practice Exercise
A one liter cubical box is 10 cm on a side. Assume it is filled with molecules of oxygen (atomic
mass 32 amu) at room temperature (T = 300K) and under a pressure of one atmosphere
(105 Pa, where a pascal is one newton per square meter). Calculate the number N of oxygen
molecules in the box, the average speed, i.e hv 2 i1/2 , and the average number of collisions
with the box walls in one second.
55
Practical Aspects of Temperature
The kind of temperature we’ve just defined is called absolute temperature. Sometimes it is
called the Kelvin temperature scale. As far as physics is concerned, it is the only kind of
temperature that matters.
There are other common temperature scales used in everyday life, although neither is used
much by physicists. The two ordinary scales are Fahrenheit (TF ) and Celsius (TC ), i.e.
9
(172)
TF = TC + 32
5
This is no historical introduction to temperature. Initially people wanted to measure how
“hot” or “cold” something was. They found some material with a property that was sensitive
this quantity. Then they defined temperature scales by pinning two values to the behavior
of the particular material property, that is, boiling and freezing of pure water in the case of
Celsius, and body temperature and freezing point of brine in the case of Fahrenheit.
TC = T − 273.15
and
Then we came across the more physical meaning of temperature, based on internal energy,
and the absolute temperature scale was born. It was chosen to have the same size of “degree”
as Celsius, but put the zero of the scale at the place where there would be no internal kinetic
energy. Sometimes we call that point “absolute zero.”
Thermal Expansion: A Material Property that Used to Define Temperature
Most solids expand when they get hot. The reason is subtle.
Higher temp means faster
molecular motion. Solids
are like atoms connected by
“anharmonic” springs, and
average position shifts to
larger separations as the energy increases. So, solids expand.
The effect is rather linear and not large for typical temperatures. For solids, we write
∆L = αL ∆T
(173)
where α is called the coefficient of linear expansion and expresses the fractional change in
length (∆L/L) per degree of temperature change, and varies from solid to solid. Since the
fractional change is small, we expect ∆A = 2αA ∆T for expansion of surface area, and
∆V = 3αV ∆T for volume. (Remember your Taylor expansions.) Note that fluids (like
water!) are more complicated than solids.
Practice Exercises
(1) Determine the temperature at which the Fahrenheit and Celsius scales coincide, that is,
give the same value for the temperature. Have you ever been outside on a day or night when
this was the air temperature?
(2) Explain why our definition hKi ≡ 32 kT for the temperature of an ideal gas, would not
make sense if we wrote it in terms of TF or TC , instead of absolute temperature T .
56
Physics I Honors
Monday 17 November
Fall 2008
Energy, Work, and Heat
We are going to start talking about systems like the one
shown on the right. It is a cylinder of gas with a piston
that moves without friction. Weight on the top of the
piston keeps the gas in an “equilibrium” state. That is,
all changes are slow ones. The gas has some internal energy that we will call U . (For an ideal gas, U = 3N kT /2,
the average kinetic energy of a molecule times the number of molecules.) Work can be done on the piston by
the gas, and work can be done on the gas by the piston.
This figure plots pressure versus volume for the gas in the
cylinder. (It is called a “pV diagram.”) Both quantities
can be changed in different ways, like adding or subtracting gas molecules or changing the mass M sitting on top
of the piston. (There is also something called a “thermal
reservoir” but we’ll get to that in a minute.) We write
that work W is done on the gas by the piston. We have
dW = F dh = −pAdh = −pdV
(174)
where A is the area of the piston and F = −pA is the force acting on the gas, in the direction
opposite to that which increases h.
Now consider taking the gas from “state” B (p = pf and V = Vi ) to state D (p = pf and
V = Vf ) along path #1. The pressure p = pf , a constant, along this path. Therefore the
work done on the gas is
Z
Wpath #1 = −
pdV = −pf (Vf − Vi ) = −pf ∆V
(175)
path #1
In contrast, along path #2, the volume does not change, so the work done along this path,
connecting state C to state D, is zero. From there, it is simple to see that if we go from B
to D instead of along path #1, but rather down to A, then across to C and then back up to
D, the work done on the gas is
Wpath
BACD
= −pi (Vf − Vi ) = −pi ∆V 6= Wpath
#1
(176)
So, the work done in going from one state to another will depend on the “path” taken on
the pV diagram. We refer to U , p, V , and also the temperature T , as “state variables”
because they only depend on the properties of the gas at any one time. Work W is not a
state variable; it depends on the path taken to get between different states.
Next let’s think about what happened to the gas in going from B to D along path #1. For
simplicity, assume it is an ideal gas. Also assume that the gas supply valve is turned off, so
57
no gas molecules are added or taken away. Then since pV = N kT , the temperature must
have increased, since p = pf has not changed but the volume has increased from Vi to Vf .
Since the temperature has gone up, so has the internal energy U . The work done on the
gas, Eq. 175, is negative, so that would have decreased the internal energy. Where did the
extra internal energy come from? It came from the thermal reservoir, by transferring energy
in the form of heat. Heat (Q) is a new concept. It is a different way of transferring energy,
other than work (W ), in a thermodynamic system.
Energy is conserved. Always. In thermodynamic systems as well as mechanical ones. Our
expression of energy conservation for the gas in the cylinder can be written as
Q + W = ∆U
(177)
This is called the First Law of Thermodynamics. Note the signs. If heat is added to a
system, then Q is positive. If the gas expands against the piston, then W is negative.
Specific Heat Capacity
We pause for a moment and talk about heat in a general sense. This will be useful later.
Add heat and the internal energy goes up, so the temperature increases. For solids and
liquids, the volume doesn’t change, so work is not an issue. For all materials, the amount
by which the temperature increases depends on the mass and the specific properties of the
material. This is all collected into the definition of the specific heat capacity c, namely
c=
1 Q
m ∆T
(178)
where m is the mass of the object, Q is the heat added or lost, and ∆T is the change in temperature. The specific heat capacity is tabulated in many places for lots of substances.
We write C = mc for “heat capacity.” Clearly, C depends on how much matter is present.
Specific heat capacity depends on temperature, but it is not a strong function of temperature
for most substances for common temperatures. Under these conditions
Z Tf
cdT = mc(Tf − Ti )
(179)
Q=m
Ti
For an ideal gas under constant volume, Q = ∆U = (3/2)N k∆T . Therefore C = (3/2)N k =
(3/2)(N/NA )NA k = (3/2)nR. We write CV = 3R/2 for the “molar heat capacity” at
constant volume. Real, monatomic gases show molar specific heat capacities very close to
this value. Diatomic or polyatomic gases show higher values, reflecting the additional degrees
of freedom in those molecules.
For an ideal gas that expands under constant pressure, W = −p∆V = −N k∆T . Therefore
Q = ∆U − W = (5/2)N k∆T and the molar heat capacity at constant pressure is Cp =
5R/2. Again, real monatomic gases agree with this very well. In addition, one expects that
Cp − CV = R = 8.31 J/mol-K. We commonly write
γ≡
Cp
5
=
CV
3
for a monatomic gas. A table is included which demonstrates these concepts.
58
(180)
Cyclic Processes: An Introduction
The world makes a lot of use of so-called cyclic processes in thermodynamics. These are
processes that operate in a cycle, ending up at the same place from which they start. We’ll
do more with these next class, but let’s start with a simple one and see what we can learn
from it.
The figure shows a cyclic process for an ideal gas. It goes
first from A to B by reducing the pressure at constant volume, taking out heat Q1 and lowering the temperature;
then to C by increasing the volume at constant pressure,
in which case heat Q2 is added to raise the temperature
and work W2 < 0 is done on the gas; and then back to
A by reducing the volume and increasing the pressure in
such a way that the temperature remains constant.
Since the process is cyclic, the change in internal energy around the cycle must be zero.
Therefore, by the first law of thermodynamics, Q + W = 0 where Q and W represent the
net work around the cycle. As drawn in the figure, W2 < 0. Recall Eq. 175. On the other
hand, W3 > 0 since the volume decreases. Furthermore, W3 > −W2 since the area under the
curve defined by path #3 is larger than that for path #2. The work done along path #1 is
zero.
Therefore, the net work is W = W2 + W3 > 0, and so the net heat Q < 0. That is, net work
is done on the gas and heat leaves it. We call such a thing a refrigerator. If instead the cycle
operated clockwise, then heat is added and the gas does work on the piston. We call this
an engine. These conclusions will always be the case for cyclic processes, regardless of the
curves that bound them.
The constant volume path #3 is called an isotherm. It follows the curve p = N kT3 /V , where
T3 is the temperature at both C and A. Therefore, the work done along this path is
Z VA
Z VC
VC
dV
= N kT3 log
>0
(181)
pdV = N kT3
W3 = −
V
VA
VC
VA
as advertised. Recall that we know that Q3 = −W3 since the path is an isotherm.
Adiabatic Processes in an Ideal Gas
Left to Next Class
A process carried out in “thermal isolation” is called adiabatic. It is characterized by Q = 0,
so W = ∆U = (3/2)N k∆T = (3/2)nR∆T = nCV ∆T . If the temperature change is
infinitesimal, then we can write dW = −pdV = nCV dT . The ideal gas law gives d(pV ) =
pdV + V dp = nRdT or V dp = nCV dT + nRdT = nCp dT . Dividing these gives us
V dp
nCp dT
Cp
=−
=−
= −γ
pdV
nCV dT
CV
=⇒
dp
dV
= −γ
p
V
Integrating this equation is simple. For limits called A and B, it just says that
γ
VB
VA
pB
= −γ log
= log
or
pV γ = constant
log
pA
VA
VB
This is the equation that describes an adiabatic path on a pV diagram.
59
(182)
(183)
Physics I Honors
Fall 2008
Useful Tables for the Study of Thermodynamics
Taken from Physics, by Halliday, Resnick, and Krane, Fifth Edition
Heat Capacities of Some Substances
Substance Specific Heat Capacity (J/kg-K)
Lead
129
Tungsten
135
Silver
236
Copper
387
Carbon
502
Aluminum
900
Brass
380
Granite
790
Glass
840
Mercury
139
Water
4190
Coefficients of Linear Expansion
Substance
α (10−6 /K)
Ice
51
Lead
29
Aluminum
23
Brass
19
Copper
17
Steel
11
Ordinary glass
9
Pyrex glass
3.2
Invar alloy
0.7
Fused quartz
0.5
Molar Heat Capacities of Ideal and Real Gases
Cp
CV
Cp − CV
Gas
(J/mol-K) (J/mol-K) (J/mol-K)
γ
Monatomic
Ideal
20.8
12.5
8.3
1.67
Helium
20.8
12.5
8.3
1.67
Argon
20.8
12.5
8.3
1.67
Diatomic
Ideal
H2
N2
O2
29.1
28.8
29.1
29.4
20.8
20.4
20.8
21.1
8.3
8.4
8.3
8.3
1.40
1.41
1.40
1.40
Polyatomic
Ideal
CO2
NH3
33.3
37.0
36.8
24.9
28.5
27.8
8.3
8.5
9.0
1.33
1.30
1.31
60
Physics I Honors
Fall 2008
Homework Assignment Due Thursday 20 November
(1) The following diagram shows the problem in coupled oscillators that we solved in class:
k
x1
m
kc
x2
k
m
Using the parameters ω02 ≡ k/m and ωc2 ≡ kc /m, find the motions x1 (t) and x2 (t) for the
following sets of initial conditions:
a) x1 (0) = x2 (0) = A and ẋ1 (0) = ẋ2 (0) = 0
b) x1 (0) = x2 (0) = 0 and ẋ1 (0) = −ẋ2 (0) = V
c) x1 (0) = x2 (0) = ẋ2 (0) = 0 and ẋ1 (0) = V
(2) As a result of a temperature rise of 32 C◦ , a bar
with a crack at its center buckles upward, as shown
in the figure on the right. The distance L0 = 3.77 m
between the bar supports is fixed, and the coefficient
of linear expansion of the bar is 25 × 10−6 /K. Find x,
the distance to which the center rises.
(3) In an experiment, 1.35 moles of oxygen (O2 , a diatomic molecule) are heated at constant pressure, starting at 11◦ C. How much heat must be added to the gas to double its
volume?
(4) Gas within a chamber undergoes the process shown in the pV diagram below:
Find the net heat (in Joules) added to the system during one complete cycle.
61
Physics I Honors
Thursday 20 November
Fall 2008
Review: Statistical Mechanics and Thermodynamics
Concentrate for now on gases. For a monatomic ideal gas, the “equation of state” is
2
pV = hKiN kT = nRT
3
(184)
which relates the state variables p, V , and T to the number of molecules N through a
fundamental constant k. The internal energy U = N (3/2)kT = N hKi where hKi is the
average molecular kinetic energy. The “3” counts the number of degrees of freedom,i.e.
three because all a monatomic gas molecule can do is move in three directions. If, for
example, the molecule is diatomic and can rotate, there are five degrees of freedom.
If the gas container has movable walls, then work W can be done on the gas. If it is in
contact with some kind of thermal bath, then heat Q can be transferred into the gas. These
change the state of the gas by an amount the depends on the “path” taken between the
states. The first law of thermodynamics says that
∆U = W + Q
(185)
where by convention, W is the work done on the gas. In effect, this equation defines Q.
Entropy: A New State Variable
Now we define a new state variable S called entropy. Actually, rather than defining entropy
itself, we’ll define the change in entropy ∆S between two states. We have
Z
dQ (186)
∆S ≡
T reversible
The notation implies that the path for the integral is “reversible”, that is, connected by a
series of equilibrium states. For example, ∆S = ∆Q/T for a reversible isothermal path.
Here’s a trickier example, a class problem of “free expansion.” The walls are perfectly insulated, so no heat comes
in to the gas. Opening the stopcock lets gas molecules move
into the evacuated chamber, until there is equal density
everywhere. What is the change in entropy?
You are tempted to say ∆S = 0 since ∆Q = 0. However, the
process
shown is not reversible. We need to compute ∆S =
R
dQ/T along some reversible path. Since no work is done
and no heat is transferred, ∆U = 0. So, choose an isothermal
expansion from Vi to Vf . Therefore dQ = −dW = pdV and
Z
Vf
∆S =
Vi
pdV
= Nk
T
Z
Vf
Vi
dV
= N k ln
V
Vf
Vi
62
> 0 (187)
The Second Law of Thermodynamics
Note that free expansion of a gas (Eq. 187) always yields an increase in entropy. This is a
system completely isolated from its surroundings, and the expansion process is irreversible.
In a reversible process, like isothermal expansion or compression, heat is transferred through
a common thermal reservoir, so we don’t expect any change in net entropy if we include both
the gas and the thermal bath with which it is in contact.
So, now we state the Second Law of Thermodynamics, namely “When changes occur within
a closed system, its entropy either increases (for irreversible processes) or remains constant
(for reversible processes). It never decreases.” In other words ∆S ≥ 0.
Adiabatic Processes in an Ideal Gas
A process carried out in “thermal isolation” is called adiabatic. It is characterized by Q = 0,
so W = ∆U = (3/2)N k∆T = (3/2)nR∆T = nCV ∆T . If the temperature change is
infinitesimal, then we can write dW = −pdV = nCV dT . The ideal gas law gives d(pV ) =
pdV + V dp = nRdT or V dp = nCV dT + nRdT = nCp dT . Dividing these gives us
nCp dT
Cp
dp
dV
V dp
=−
=−
= −γ
=⇒
= −γ
pdV
nCV dT
CV
p
V
Integrating this equation is simple. For limits called A and B, it just says that
γ
pB
VB
VA
log
= −γ log
= log
or
pV γ = constant
pA
VA
VB
(188)
(189)
This is the equation that describes an adiabatic path on a pV diagram.
The Carnot Engine
The figure on the right shows a cyclic process
which is bounded by two isotherms (at two
temperatures TL < TH ) and two adiabats which
connect those two temperatures. This is called a
Carnot cycle. The cycle proceeds clockwise, so
negative work is done on the gas from A to B and
heat QH enters the gas. During the isothermal
contraction C to D, the work is positive and so
heat QL leaves the gas. (Remember that ∆U = 0
along an isotherm, and Q = 0 along an adiabat.)
Since ∆U = 0 around the cycle, we must have
WDA + WH + QH + WBC + WL + QL = 0 (190)
Now WH < 0 and WL > 0. Futhermore, since paths H and L are isotherms, WH + QH = 0 =
WL + QL . Therefore WDA + WBC = 0, and also QH = −WH > 0 (heat enters along H) and
QL = −WL < 0 (heat exits along L). Since (by areas) |WL | < |WH |, we have |QL | < |QH |,
so net heat enters the gas. Furthermore, the work done around the cycle is
W = WDA + WH + WBC + WL = WH + WL < 0
and the gas does work on the piston. Heat in, work out. This is an engine.
63
(191)
How efficient is this engine? In other words, how much work do you get out, for heat that
you need to put in? Let us define an engine efficiency just that way, namely
≡
|W |
|WH + WL |
|QH | − |QL |
|QL |
=
=
=1−
|QH |
|QH |
|QH |
|QH |
(192)
We can go further by realizing that entropy change is zero around the cycle. That is
∆S =
QH QL
|QH | |QL |
+
=
−
=0
TH
TL
TH
TL
(193)
Therefore |QL |/|QH | = TL /TH and the efficiency of a Carnot engine is
=1−
TL
TH
(Carnot engine efficiency)
(194)
Of course, operated counter-clockwise around the cycle, a Carnot engine becomes a Carnot
refrigerator. Instead of efficiency we speak of coefficient of performance. This is the ratio
of what you “want”, i.e. the heat QL taken out of the low temperature reservoir, and what
you pay for, namely the work W you have to do on the gas. We write
K=
|QL |
TL
|QL |
=
=
|W |
|QH | − |QL |
TH − TL
(Carnot refrig coeff of performance)
(195)
A very large coefficient of performance is needed for a refrigerator that operates between two
temperatures that are close to each other.
There is no such thing as a perfect engine, i.e. one with = 1. This requires TL = 0
for a cold reservoir, that is one with no internal energy. There is also no such thing as a
perfect refrigerator, i.e. one with an infinite coefficient of performance. This would mean
that |QH | = |QL | ≡ Q and ∆S = Q/TH − Q/TL < 0, which violates the second law.
These statements can be put together to show that “No real engine can have an efficiency
greater than that of a Carnot engine working between the same two temperatures.” That
is, Eq. 194 is the most efficient an engine can ever be. In other words, by Eq. 192, any real
engine must always discharge some heat QL to its surroundings.
Don’t Forget: Short homework assignment due in class on Monday.
64
Physics I Honors
Fall 2008
Homework Assignment Due Monday 24 November
1) Two identical blocks, each of mass M and made from a material with specific heat capacity
c, are kept thermally isolated from each other and from their surroundings. One block is at
temperature T1 and the other is at temperature T2 > T1 . The blocks are then put in thermal
contact with each other, but still thermally isolated from their surroundings.
a. Use the relationship between heat transfer and specific heat capacity to prove that the
final equilibrium temperature of the two blocks together is T = (T1 + T2 )/2.
b. Find the change in entropy for the system consisting of the two blocks. (You can
imagine a reversible process by which one block is taken slowly to its final temperature,
and then the other block is taken to this same temperature, but contact with a slowly
changing thermal bath.)
2) The figure on the right shows a cyclic
process, plotted as a T S diagram, rather
than a pV diagram.
a. Explain why this is a Carnot cycle.
b. Calculate the heat that enters.
c. Calculate the work done on the gas.
65
Ludwig Boltzmann is one of the few people who “understood” entropy. He came up with a way to understand it in terms of disorder, which he wrote as W . His result, in the form of the equation
S = k log W , is etched on his grave stone.
66
Physics I Honors
Monday 24 November
Fall 2008
Remember that homework and final lab notebooks are due on Monday 1 Dec.
How to Build a Star: Hydrostatics of really big ball of gas
Start with the balance of pressure against gravitation. Then use ideal gas law to relate to
temperature. (Will assume that stars like the sun have uniform density.) Finish with the
behavior of white dwarf stars, aka “degenerate Fermi gas.”
More reading (in order of increasing sophistication):
• Introductory Astronomy and Astrophysics, 4th Edition, Zeilik and Gregory, esp. Chap.10
• Order-of-magnitude “theory” of stellar structure, George Greenstein, Am. J. Phys. 55,
804 (1987); Erratum Am. J. Phys. 56, 94 (1988)
• Stars and statistical physics: A teaching experience, Roger Balian and Jean-Paul
Blaizot Am. J. Phys. 67, 1189 (1999)
Begin. First, set up the notation. Build our star with shells of radius r. Total mass of star
is M and radius is R. Density is ρ = M/(4πR3 /3) assumed to be independent of r. Write
M̃ (r) be the mass enclosed at radius r, i.e. M̃ (R) = ρ(4πr3 /3) = M (r/R)3 .
Now, balance pressure p(r) against gravity. Take a small piece of area A, thickness dr and
mass dm = ρAdr, at radius r. Outward force from difference in pressure must equal the
gravitational force from mass inside radius r. (Recall “shell theorems.”) Write it out:
pA − (p + dp)A = G
M̃ (r)ρAdr
M̃ (r)dm
=G
2
r
r2
or,
M̃ (r)ρ(r)
dp
= −G
(196)
dr
r2
This equation expresses “hydrostatic equilibrium” for a large, spherical self-gravitating object
like a star (or a planet, or a moon). It is written here in a way that holds for any density
function ρ(r), but we will be taking ρ to be a constant.
Let’s use this to determine the pressure pC at the center of a star, i.e. r = 0. Make the
replacements for constant density in Eq. 196 to get
r 3 M
1
3GM 2
dp
= −GM
=
−
r
dr
R 4πR3 /3 r2
4πR6
=⇒
p(r) = pC −
3GM 2 2
r
8πR6
At the surface of the star, the pressure is zero “by definition”, i.e. p(R) = 0. Therefore,
pc =
3GM 2
8πR4
(197)
Next let us try to estimate the temperature at the center of a star. This requires us to know
the “equation of state” which relates the gas pressure and density to the temperature, for
the matter that makes up the star. Understanding the equation of state is a big deal, and
67
an area of active research interest. For now, we will use an equation of state based on an
ideal gas, that is Eq. 184, namely pV = N kT .
If we assume that the star is made of N identical particles with mass m, then ρ = N m/V and
p = ρkT /m = (3/4π)(M/m)kT /R3 . In fact, stars are hot, so hot that matter is a “plasma”
of free nuclei and electrons. The nuclei are from mostly hydrogen, about 25% helium, and
a few percent of heavier elements. For now, though, it is close enough to assume the star is
100% hydrogen atoms. Therefore, the temperature at the center of a star is
TC =
1 4πR3 m
1 GmM
pC =
k 3 M
k 2R
(198)
Can you see why we would not be able to get this answer from dimensional analysis?
The White Dwarf Star
We found some properties of stars assuming they consisted of an ideal gas, but the assumptions used to get Eq. 184 don’t always hold. Instead, we need another equation of
state.
In deriving Eq. 184 we wrote Fx = ∆p/∆t = (2px )/(2L/vx ), and that is still fine. For
particles, this gives a pressure P = N Fx /L2 = N pv/V . (We temporarily switch to P
avoid confusion with momentum. Also, we are making a rough estimate, so put px and
to p and v.) Our concern will be with electrons, so write an electron density ne = N/V
which case P = ne pv.
N
to
vx
in
At very high densities, the ideal gas law assumptions are violated. The electrons are so close
together, they start to violate the Pauli Exclusion Principle. The distance between them is
1/3
1/3
d ≈ 1/ne and the Heisenberg Uncertainty Principle says that pd ≈ h̄ so that p ≈ h̄ne .
Putting v = p/me and switching back to p for pressure, the equation of state becomes
p=
ne h̄n1/3
e
h̄n1/3
e /me
h̄2 5/3
h̄2 ρ 5/3
h̄2 5/3
n ∼
n =
=
me e
me
me m
(199)
where n is the number density of (what would have been) atoms, and we make the (somewhat
incorrect) assumption that the star is still made up of hydrogen atoms with mass m. This
is an odd sort of equation of state. Combined with Eq 197 it leads to a star whose radius
decreases with mass like 1/M 1/3 .
As the mass increases, and the radius decreases, the electrons are more and more confined
until they are moving as fast as they can, i.e. v = c. The equation of state becomes
ρ 4/3
4/3
4/3
p = ne h̄n1/3
c
=
h̄c
n
∼
h̄c
n
=
h̄c
e
e
m
(200)
Combined this with Eq 197 gives an equation where the radius drops out! One solves for a
mass that is the highest possible mass of a white dwarf star. Doing a more careful job leads
to a value of 1.44 solar masses. This is the “Chandrasekhar limit” after the physicist who
first derived it. What do you suppose would happen for a star of mass larger than this?
68
Physics I Honors
Fall 2008
Homework Assignment Due Monday 1 December
(1) This problem asks you to carry through some numerical calculations that lead to an
understanding of what it is like in the center of the Sun. The following web sites contains
useful data:
http://pdg.lbl.gov/2008/reviews/consrpp.pdf
http://pdg.lbl.gov/2008/reviews/astrorpp.pdf
a. Use Eq. 197 to estimate the central pressure of the Sun. Express it in units of atmospheres, where one atmosphere is the air pressure at the surface of the Earth.
b. Use Eq. 198 to estimate the central temperature. Express it in Kelvin (K).
c. Recall that temperature is related to the average kinetic energy of the gas particles.
What is the average kinetic energy of the hydrogen atoms in the center of the Sun?
Express the answer in electron volts (eV). What does this tell you about the kind of
matter in the center of the Sun? (A typical “binding energy” of an electron in an atom
is around 10 eV.)
(2) Show that Eq. 199 leads to a “star” with radius R and mass M where R ∝ 1/M 1/3 .
69
Physics I Honors
Monday 1 December
Fall 2008
Last class! I will miss all of you, but don’t be a stranger.
Hand in lab books. Third midterm exam is this Thursday. Final exam is Wednesday, 3 Dec
3-6pm, this room.
Conservation Laws and the Continuity Equation
Our last class will concentrate on how to write down a general conservation principle. Then,
we’ll apply this to fluid dynamics.
The basic idea is simple. Imagine some substance that is conserved. In other words, it
cannot appear or disappear unless it is somehow supplied. If we talk about the amount Q
of this substance in some region R of space, then the only way to change this amount is to
let some flow into (or out of) the region.
Make this statement mathematical. The rate of change of Q is dQ/dt. Define some quantity
J = J(r, t) to be the “flux density” of the substance. That is, it is the amount of the
substance flowing (perpendicularly) through some small area per unit time. So, through
some small surface dA = dAn̂, an amount J · dA = JdA cos φ flows per unit time. Here, n̂
is a unit vector perpendicular to the surface, and φ is the angle between n̂ and J.
If we let S denote the surface of R, we then write our conservation condition as
I
dQ
= − J · dA
dt
S
(201)
The negative sign is there because of the convention that elements of area dA point outward
on S. Therefore, if the integral is positive, there is a net flux out of R so Q decreases.
Now let ρ = ρ(r, t) be the substance density. Then Q is the integral of ρ over R, and
Z
Z
dQ
∂ρ
d
ρ(r, t)dV =
=
dV
(202)
dt
dt R
R ∂t
The derivative becomes a partial derivative inside the integral, because all we care about is
the explicit dependence on time. That’s all that would be left after we carry out the integral
over space.
For the right side of Eq. 201 we will make use of a result we first mentioned in Eq. 70.
Imagine that R is small and rectangular, with one corner at (x, y, z) and with sides of length
dx, dy, and dz. In this case
I
J · dA = Jx (x + dx, y, z)dydz − Jx (x, y, z)dydz
S
+ Jy (x, y + dy, z)dxdz − Jy (x, y, z)dxdz
+ Jz (x, y, z + dz)dxdy − Jz (x, y, z)dxdy
Jx (x + dx, y, z) − Jx (x, y, z)
+ · · · dxdydz
=
dx
∂Jx ∂Jy ∂Jz
=
+
+
dxdydz
∂x
∂y
∂z
70
(203)
Now we can divide any arbitrary region R into little rectangular boxes that butt up against
each other. The flux out of one box goes into the one next to it, so in the end, all that matters
are the sides of the boxes along the surface S. We also have dV = dxdydz, and
∂Jx ∂Jy ∂Jz
∇·J=
+
+
(204)
∂x
∂y
∂z
which is called the “divergence” of J(r, t). Therefore, for any region R,
I
Z
J · dA =
∇ · J dV
(205)
S
R
which is known at the Divergence Theorem or Gauss’ Theorem. So, combining Eq. 202 with
Eq. 205, we turn Eq. 201 into
Z ∂ρ
+ ∇ · J dV = 0
(206)
R ∂t
which holds for any region R. In particular, if we make R very tiny, then there is no
appreciable variation over the volume, and the therefore the integrand must be zero. In
other words, our statement of conservation becomes
∂ρ
+∇·J=0
(207)
∂t
This is called the Continuity Equation. It finds application in all areas of physics, including
the conservation of material in fluid mechanics; conservation of electric charge in electromagnetism; conservation of energy and momentum in general relativity; and conservation of
probability in quantum mechanics.
Let’s make this physical with the specific example of fluid flow. In this case, the conserved
substance is the fluid mass, and density ρ(r, t) is just ordinary mass density. To identify
J(r, t), consider the tiny fluid mass flowing through a small element of area dA in a time
dt. Let the velocity of this tiny mass be v(r, t), and n̂ be a unit vector perpendicular to the
surface of dA. Then the size of the volume element of mass swept out is v(r, t)dt · n̂dA, and
the mass which passes through dA is ρv(r, t) · n̂dtdA = J · dAdt. In other words, J(r, t) = ρv
is the amount of mass flowing (perpendicularly) through some small area per unit time.
The figure shows flow from P to Q in a closed
region. Specialize to ρ = ρ(r), independent of
time. Then Eq. 201 says that the integrated
flux over the surface is zero. Assume the areas
A1 and A2 on the right are small, so that the
velocities v1 and v2 do not vary much over
them, and are perpendicular to them. Then
ρ1 v1 A1 = ρ2 v2 A2
(208)
If the fluid is “incompressible” (liquids), that
is, the density ρ(r, t) =constant everywhere,
then the continuity equation becomes
v1 A1 = v2 A2
(209)
I.e., water flows faster in constricted pipes.
71
Physics I Honors
Fall 2008
Practice Exercises on the Continuity Equation
(1) A pipe of diameter 34.5 cm carries water moving at 2.62 m/sec. How long will it take to
discharge 1600 m3 of water? Answer: 1h 49 min.
(2) The figure below shows the flow of an incompressible liquid through a straight pipe with
a constriction midway between the input and output:
Let z measure the distance along the axis of the pipe, with x and y perpendicular to z.
What are Jx and Jy at values of z near points P and Q? For a value of z near the midpoint
between P and Q (i.e., in the middle of the constriction), make a sketch of Jx as a function
of x. Use your sketch, and Eq. 207, to explain why |v2 | > |v1 |.
(3) The figure shows the confluence of two streams
to form a river. One stream has a width of 8.2 m,
depth of 3.4 m, and current speed of 2.3 m/sec. The
other stream is 6.8 m wide, 3.2 m deep, and flows
at 2.6 m/sec. The width of the river is 10.7 m and
the current speed is 2.9 m/sec. What is its depth?
Answer: 3.9 m
(4) The figure shows a familiar effect, namely the “necking down”
of a stream of water when it falls under gravity. (Water adheres
to itself, so that no pipe is necessary to keep the mass continuous.)
Use the continuity equation and conservation of mechanical energy
to explain this effect. If the cross sectional area A1 is 1.2 cm2 and
that of A2 is 0.35 cm2 , and their separation h = 45 mm, at what
rate R does water flow from the tap? Answer: R = 34 cm3 /sec.
72