Class
Notes
These notes are NOT intended to be a complete record
of what has been covered in class. Students are responsible for
all material discussed in class and in the assigned readings, not
just what appears here. I will try (but not always succeed) to be brief
here. I will focus on material not adequately discussed in the text.
Mon,
Tues, Wed, Thurs, 12/9-12
- Sorry I have no time to
write this up in any detail. We covered (from the course plan page):
two
spatially separated souce interference, sound interferometer
|
Michelson interferometer,
combined double/single
slit interference patterns
|
circular aperture, N slits, diffraction gratings, resolving power
|
diffraction grating, spectra, electron
(matter wave) diffraction |
This material is all covered
fairly nicely in the textbook, except for the electron diffraction.
Thurs,
12/5 no class: snow
Wed,
12/4
- interference
from two spatially separated sources: the phase difference is then determined
by the path length difference. If the sources lie symmetrically on the y-axis,
we get a simple formula for the amplutude and intensity on an x=const line
far from the sources. For the intensity, I = 4 cos^2(D phi/2) I_0, where the
phase difference D phi = 2 pi d sin(theta)/lambda, theta is the angle from
the x-axis, lambda is the wavelength, and I_0 is the intensity at theta=0.
Tues, 12/3
- more on polarization: We had an extensive discussion about
how the quarter wave plate works to produce circular polarization from linear
polarization. One key idea is that we can choose to write, say, a vertical
polarization as a superposition of two linear polarizations at 45 degrees
tilted to the left and right, for which the index of refraction of the birefringent
material is different. Then, because the material acts linearly on the electromagnetic
input, we can analyze separately what happens to the two 45 degree poalrizations,
and add the results. A good question was raised about the time delay: if
the incoming wave has finite extent, i.e. is a wavepacket rather than an infinite
wave train, then if it passes through enough birefringent material the two
polarization wavepackets will separate spatially so they no longer overlap.
That's true. But for light coming from atomic transitions the wavepackets
are very long: since it takes more than a nanosecond for the atomic transition,
the emitted wavepacket is more than one light-nanosecond long, i.e. 30 cm.
(This is huge compared to the wavelength which is, say, 0.5 microns, so it
contains 600,000 wavelengths.) Thus it would take quite a large thickness
of birefringent material to separate the two component wavepackets.
- interference: we considered thin film interference, in three
examples: Newton's
rings formed with a convex lens touching a flat glass plate, an oil
slick on water, and a soap
film. In each case two reflected waves combine, coming from reflection
off the top or bottom interface, with some relative phase. If they differ
by an integer times 2pi they add constructively and give a maximum. If they
differ by an integer plus a half times 2pi they cancel (interfere destructively)
and produce a minimum. The relative phase is governed by two factors:
1) the phase
change of each upon reflection, and
2) the extra path length of the wave that reflects off the bottom interface.
1) It turns out that
there is a pi phase change upon reflection if the second medium has a higher
index of refraction than the first. If both reflections involve a phase
change or both do not, then the reflection produces no relative phase shift.
If only one reflection changes phase there is a relative phase shift of pi.
In the example of oil (n=1.2) on water (1.33) both reflections produce a
phase change, while in the soap film the top reflection changes phase while
the bottom doesn't. For the lens and plate it is the reverse.
2) Aside from the possible phase change upon reflection, the total phase
delay acquired by the wave that reflects off the bottom surface is the number
of wavelengths times 2pi. If the reflection is perpendicular to the film
or gap of width d, the numer of wavelengths is 2d/lambda_n, where lambda_n
is the wavelength in the medium with index of refraction n (if any), lambda_n
= lambda/n.
Thus, the conditions for a maximum are, with m=0,1,2,3,...:
2d = m lambda_n
if
both or neither change phase upon reflection
2d = (m+1/2) lambda_n
if one changes phase upon reflection
For a minimum, just interchange
these.
Mon, 12/2
- Circular polarization: (See the supplement and/or textbook.)
Superposition of two in-phase linear polarizations yields a new linear polarization,
the vector sum of the two. However superposition of two equal amplitude orthogonal
linear polarizations 90 degrees out of phase yields right or left circular
polarization. If the amplitudes are not equal one gets elliptical
polarization. Conversely, linear polarization can be expressed as a superposition
of right and left circular polarization. According to the book's definition,
the electric field vector at a fixed position rotates clockwise when viewed
head on for a right circular polarization. Most treatments, especially involving
quantum processes, use the opposite definition, since the angular momentum
of right circular polarized light is then positive using the right hand rule
in the direction of propagation.
- Optical activity can be understood as a result of the fact that there
is a different index of refraction for right and left circular polarization.
As explained in class, and in this week's homework, that results in a rotation
of the linear polarization direction.
- Circular polarization can be produced by a circular current acting
as a source for the wave, either macroscopically with an antenna, or microscopically
with atomic transitions. Circular polarization can also be produced from
linear polarization using a "quarter wave plate". This is
a material with a different index of refraction for, say, vertical and horizontal
polarizations. (This is called a birefringent material.) If
light with 45 degrees linear polarization enters the material, the passage
results in a phase shift of the vertical polarization component relative
to the horizontal one. If the thickness of the plate is adjusted to produce
a 90 degree relative phase shift (for a particular wavelength), the light
emerges circularly polarized.
Wed, 11/27
- Liquid crystal display: works on the principle of crossed polarizers,
with a mirror below if not back-lit. Between the polarizers is a liquid
crystal that has chain molecules arranged in a corkscrew that rotates the
polarization direction so as to pass the light through. When a voltage is
placed across the plates in one spot, the molecules (being electrically
polarized) line up in the direction of the elctric field and so eliminate
the corkscrew effect that rotates the polarization of the light, thus producing
a dark spot.
- Polarization by reflection: Light reflecting off an interface
tends to be transmitted if it's polarization direction is not parallel
to the interface. In fact, at a critical angle, called Brewster's
angle or the polarizing angle, the reflected wave is 100 % polarized
in the plane of the interface. This angle of incidence theta_B is such
that the refracted and reflected waves are at right angles to each other,
which means that
tan(theta_B)
= n_2/n_1. (Brewster's angle)
- Polarization by
scattering: light scattering at a right angle is polarized in the direction
that is perpendicular to the plane formed by the incident and scattered waves.
- Optical activity: rotation of the polarization plane by passing
though a "chiral" ("handed") medium, such as sugar syrup. The biologically
produced sugar molecules have a preferred handedness (dextrose rather than
levulose), which somehow produces this rotation of the polarization direction.
The amount of rotation for a given sample depends on the wavelength of
the light, hence if the sugar syrup is placed between two crossed polarizers,
white light incident on the first polarizer emerges from the second polarizer
colored, due to the unequal mix of intensities for the different wavelengths
passed by the second polarizer. For demonstrations, see
http://www.physics.umd.edu/deptinfo/facilities/lecdem/services/demos/demosm8/m8-01.htm
http://www.physics.umd.edu/deptinfo/facilities/lecdem/services/demos/demosm8/m8-03.htm
Tues, 11/26
- Polarization: definition, demonstration with radio wave generator.
Amplitude of wave received by diploe antenna making an angle theta with
respect to the tranmitting antenna is E_0 cos(theta), where E_0 is the incident
amplitude. The intensity received goes as teh square of the amplitude, hence
I = I_0 cos^2(theta).
(Malus' law)
A polarizing filter
absorbs (or reflects) one linear polarization and transmits the orthogonal
one. We demonstrated this with microwaves. The transmitted intensity follows
the same relation just discussed, which is known as Malus' Law.
Optical polarizing filters absorb one polarization and transmit the other.
They effectively pass current in one direction, along chain molecules,
but with resistance that absorbs that part of the wave. Optical sources
of light are usually unpolarized to begin with, since they contain many
accelerating charges with random orientations.
Mon, 11/25:
- Mach cone: when source moves faster thant he wave speed in a
medium, it emits a shock wave in the form of a cone with vertex at the instantaneous
position, with an opening angle from the axis given by sin(theta) = v_w/v_s.
The ratio v_s/v_w is called the Mach number. In the case
of charged particles in a medium this is called Cerenkov radiation.
The cone orientation and angle indicate the direction and speed of the
source, so the effect is used in high energy charged particle detectors,
usually with the medium being water.
- Doppler effect for EM radiation: the frequency shift can only
depend on the relative motion, since there is no meaning to "motion relative
to the medium". How can this be right? Something must be missing in the
derivation of the previous results (see Mon, 11/18). The case of moving
source and moving observer differ by terms of order (v_s/v_w)^2 and (v_o/v_w)^2.
In the case of EM waves, we should take into account another effect of this
same order: the time dilation effect. The period of the source as measured
in the rest frame of the source T_s,s is not the same as the period of the
source as measured in the rest frame of the observer T_s,o. Rather, T_s,o
= T_s,s/Sqrt[1 - (v/c)^2], hence f_s,o = f_s,s Sqrt[1 - (v/c)^2]. Taking
this into account, we derived the relativistic Doppler shift formula:
f_o,o = f_s,s Sqrt[(c+v)/(c-v)] = f_s,s Sqrt[(1+v/c)/(1-v/c)] =~ f_s,s
(1 + v/c) for v/c << 1.
For small v/c this agrees with both the moving source and moving obserber
results derived before. By the way, rather than using the time dilation
formula to derive the Doppler formula, one can turn it around and use the
fact that the Doppler formula must depend only on the relative motion of
the source and observer to derive the time dilation formula.
- examples of EM Doppler: police and baseball Doppler radar; mass of
Saturn by measuring the orbital speed of Saturn's rings using the difference
between the blueshift of spectral lines from the approaching side and the
redshift from the receding side; mass of giant black hole in the center
of a galaxy by a similar method using absorption lines of orbiting gas.
Thurs, 11/21: went
over exam 2.
Wed, 11/20: exam 2
Tues, 11/19: review for exam 2
Mon, 11/18
- Doppler effect: The effect of motion
on frequency and wavelength. Motion relative to WHAT? One
can consider motion of the source or observer relative to the medium
in which the wave propagates, as well as motion of the source and
observer relative to each other. For this class we restrict to motion
along the wave propagation direction. We consider two settings: I. (non-relativistic)
waves in a medium, and II. waves at the speed of light in vacuum. Under
setting I we consider three cases. Let f_s be the source frequency and
f_o be the observed frequency, and let v_w,s,o be the wave speed, the source
velocity, and the observer velocity respectively. Then by examining the
time between the observer's reception of successive wavefronts we deduced
(for cases A & B) the following relations:
A. source at rest, observer moving: f_o = f_s (1 + v_o/v_w)
[v_o>0 toward source]
B. source moving, observer at rest: f_o = f_s /(1
- v_s/v_w) [v_s>0 toward observer]
C. both moving:
f_o = f_s (1
+ v_o/v_w) /(1 - v_s/v_w)
= f_s (v_w + v_o)/(v_w - v_s)
Note that A and B
differ, but for velocities small compared to the wave speed they are approximately
equal for equal relative velocities. Example: Consider sound wave
in air at speed 340 m/s. If v_o or v_s are 10 m/s, then since 10/340 is
around 0.03 we have in either case f_0 =~ 1.03 f_s. Compare this to a musical
interval of a half step, which is 2^(1/12) = 1.06. So the note change is
a quarter step. For speeds of 20 m/s the shift would be about a half step.
Another example: Police siren and speeder. This is on the homework
this week. We discussed at length the different cases, whether and how
the frequency and wavelength are changed.
Thur, 11/14
- refraction in inhomogeneous media: examples: atmosphere
of the earth, mirage over hot sand, Einstein's original calculation of
bending of starlight by the Sun (he thought at that time that the speed
of light was position dependent and larger farther from the earth. These
can all be understood using wavefront propagation and Snell's law, but
another point of view is Fermat's principle of least time.
A nice example is the lifeguard and swimmer: to miniize the time from
his bench to the swimmer the lifeguard runs farther on the beach where
he runs faster, so that he can swim a shorter distance in the water where
he moves more slowly. Another example: seismic waves in the earth, generated
by an earthquake or (in the past) nuclear detonation tests. Both compressional
and shear waves are generated. The former have the higher wave speed. The
waves refract and bend back up to the surface of the earth, where they reflect
and go back down again. Shear waves cannot penetrate the liquid core. From
arrays of seismic measurements the properties of the interior of the earth
can be reconstructed, as in tomography. The earth also has free oscillations,
with normal modes, which depend on the structure of the earth and hence
provide information about that structure.
- spectrum of white light from a prism: Due to dispersion.
refractive index slightly different for different colors (i.e. different
frequencies). Blue refracted more than red since it is closer to the
resonant frequencies (UV) of the dipoles in the matter. Explained
how a rainbow works: when parallel light impinges on a spherical water
droplet, the light emerges preferentially at the maximum angle of refraction,
called the rainbow angle. This is demonstrated and explained nicely in
this rainbow
applet.
Wed, 11/13
- Huygen's principle (1678) is what we implicitly used when
propagating the wavefronts to derive Snell's law. Huygen's principle
can be mathematically derived from the wave equation, including the
direction dependence of the amplitude of the secondary wavelets. The
theory of this was completed in the 19th century, by Helmholtz (1859)
and Kirchoff (1882). Note Huygen's principle does not apply for the wave
equation in two dimensions, or any even number of dimensions. This is related
to the fact, mentioned earlier, that the shape of a circular wave pulse is
not preserved in 2d.
- Total internal relflection. A nice example not in the
text: a light source under water in a lake sends out rays in all directions,
but only those within a certain cone refract out into the air. Beyond
a critical angle they reflect back underwater. Conversely, if you are
looking around underwater, you see above you in a cone the full hemisphereabove
the water, and outside the cone angle you see reflections of things such
as fish under the water.
Tues, 11/12
- Speed of light: Maxwell's equations imply EM waves
travelling at speed c = 1/Sqrt[mu_0 epsilon_0] = 3 x 10^8 m/s = 3 x
10^5 km/s...relative to whom?? How could this be the correct speed independent
of the motion of the observer? If I chase a sound wave, by travelling
at 340 m/s, it will appear as a stationary pattern of compression and
rarefaction, not propagating at all. In the sound case, my motion
can be reckoned relative to the air in which the sound propagates. Therefore
one might think---and Maxwell did think---that EM waves propagate in a
medium, called the ether, and the equations can only be applied
in the rest frame of that medium. Either Maxwell's equations only apply
in the rest frame of the ether, or there is something wrong with the apparently
obvious conclusion that by chasing a light wave you could make it stand still.
Ever since Galileo it was clear that the laws of physics do not
distinguish different states of relative unaccelerated motion, so that
all inertial reference frames are equivalent. It would therefore be surprising
if Maxwell's equations applied only in a preferred frame. Moreover,
Einstein pointed out that the phenomena of electrodynamics only depend
on the relative motion of things, supporting the idea that Maxwell's
equations apply in any inertial frame. For example he pointed
to Faraday's law: whether the magnetic flux changes because a magnet
is moving relative to a fixed loop of wire, or because the loop of wire
moves in the opposite way relative to a fixed magnet, one observes the
same emf in the loop. That is, the emf depends only on the relative
motion. In the first case the emf is due to the induced electric
field produced by the changing magnetic field, while in the second case
it is due to the magnetic Lorentz force on the charges that are made to
move along with the moving wire. If Maxwell's equations do apply
in any inertial frame however, the only way to explain the fact that the
speed of light is the same for all observers is if there is something
profoundly wrong with how we relate the time and space measurements of one
observer to those of another. Thinking along these lines led Einstein to
the theory of relativity.
- How fast is c? Light travel time to the sun: 150 Mkm/(3 x 10^5
km/s) = 500 s ~ 8 minutes. To moon: (3.8 x 10^5 km)/(3 X 10^5 km/s) ~
1.3 s. Orbital speed of earth 29.8 km/s ~ 10^-4 c (not so slow, eh?). Electron
in H atom: ~ alpha c = (e^2/hbar c) c ~ (1/137) c (zipping right along,
but not quite so fast that relativistic corrections are very important for
electrons in H atoms).
- Speed of
light in matter: use dielectric constant and magnetic relative permeability
(~ 1 in non ferromagnetic matter) in Maxwell's eqns, find wave speed
(kappa_e kappa_m)^-1/2 c. What's going on? Electric field stretches
dipoles in matter, which "slows down" the wave. Really, the wave keeps
going, but the stretched dipoles generate a secondary wave that is superposed
with the incoming wave to produce a net wave that has a lower phase velocity.
See Ch. 31 of vol. 1 of The Feynman Lectures, "The Origin of the
Refractive Index", for a great explanation of this. The response
of the medium depends on the relation between the natural frequency of
the dipoles in the medium and the wave frequency, so the dielectric constant
is frequency dependent and hence so is the wave speed. In a simple model
of the diples as harmonic oscillators with natural frequency w0 Feynman
derives a formula for the index of refraction: n = c/v: n = 1+ Ne^2/(2eps0m(w0^2-w^2)).
So you see that the closer to the resonance the greater the index of refraction
differs from 1 by a greater amount. Below resonance the wave is slowed
down. Above resonance it is speeded up! Examples of refractive indices...(see
textbook here and for much of what follows).
- Reflection and transmission at a nonzero angle of incidence:
law of reflection and Snell's law. Example of sound and light
at an air/water interface.
- Wavefronts and rays, derived Snell's law from a picture
of propagating wavefronts.
Mon, 11/11
- Radiation pressure: EM waves have momentum equal to 1/c
times their energy. One way to understand this is to think about a particle
first. Nonrelativistically, p=mv. The correct relation between momentum
and velocity in relativity is obtained by replacing m by E/c^2, thus p
= (E/c^2)v. This E is the total energy, rest energy mc^2 plus the
kinetic energy. For v<<c, E is approximately just mc^2, so the relativistic
momentum agrees with the nonrelativistic one. More generally, E = mc^2/Sqrt[1-(v^2/c^2)]
= mc^2 + 1/2 mv^2 + 3/8 m v^4/c^2 + ... . Setting v =c is only possible
if m = 0, i.e. something can go at the speed of light only if it has zero
rest mass. In this case, the momentum is given by p = E/c. This suggests
that EM waves have a momentum density equal to 1/c times their energy density.
An EM wave of intensity I thus has a momentum flux per unit time
per unit area of I/c. If the radiation is absorbed, this rate of change
of momentum produces a force per unit area, i.e. a radiation pressure,
equal to I/c. If the radiation is reflected, the pressure is 2I/c. We
looked at the example of a 60W light bulb viewed from a distance of one
meter: I = 60W/(4pi m^2) =~ 5 W/m^2, and pressure = I/c =~ 1.7 10^-8 N/m^2.
Small, but not zero. The same effect holds the sun up! And can be used
to produce fusion using high powered lasers.
How does the EM wave push on charges? Answer: the magnetic field
must do it, since the electric field is transverse to the wave direction.
This can be used to understand in another way the relation p = E/c.
Consider an EM wave propagating in the z direction with electric field
E_x in the x direction and magnetic field B_y=E_x/c in the y direction.
The rate of change of momentum of a charge q in the z-direciton is then
dp_z/dt = F_z
= q(vxB)_z = q v_x B_y = q v_x E_x /c = F.v/c
= (dE/dt)/c = d(E/c)/dt.
(The step where
F.v appears is justified by the fact that only the electric
force contributes, since the magnetic force is perpendicular to v.)
Thus, the charge absorbs z component of momentum at 1/c times the rate
at which it absorbs energy from the wave. This makes it plausible that
the wave contains momentum in that proportion to energy.
- Electromagnetic spectrum: there is no length scale in
Maxwell's equations, so waves of all wavelengths are "equivalent".
Moreover, what is a low frequency to one observer is a short wavelength
to another observer running towards the direction the wave is coming
from. Since the wave speed is c in any inertial reference frame, this
means that the wavelength is changed inversely to the frequency by such
relative motion. As far as we know so far, there is no violation of relativity,
and all wavelengths travel at exactly the same speed. I discussed the
different parts of the EM spectrum, much as the book does. Gave examples
of gamma ray bursts and cosmic microwave background radiation, and lengthening
of wavelengths due to the expansion of the universe.
Thurs,
11/7
- Displacement
current: recall we introduced the displacement current density
j_d so that
div(j+j_d)
= 0
(1)
would hold identically,
as required by consistency of the Ampere-Maxwell law: curl B
= mu_0 (j+j_d). This gave us j_d = epsilon_0
E,t. Going back to the integral form, this corresponds to \oint
B.dl = mu_0 (i + i_d), where i = \int j.dA
is the current
through a surface spanning the loop and i_d = \int j_d.dA
= epsilon_0 d/dt
(\int E.dA) is the displacement current though
a surface spanning the loop. Thus i_d is epsilon_0 times the rate of change
of electric flux through the loop. Integrating (1) over any volume,
and converting into a surface flux integral using the divergence theorem,
we get that the sum of i + i_d into the closed surface bounding
the volume is zero. That is, any charge current flowing into a closed surface
is balanced by a displacement current flowing out of the surface. We applied
these ideas to understand the magnetic field at a distance r from a wire
with a circular capacitor (see figure 38-24 in problem 1 on page 879
of HRK). No matter what surface spans the Amperian loop, we get the same
value for the right hand side of the Maxwell-Ampere law. We can even take
the surface to miss the wire althogether by going between the capacitor
plates! In that case the whole contribution comes from the displacement
current. The magnetic field at a point next to the capacitor at distance
R from the wire is thus the same as at a point far from the capacitor, as
long as R is outside the capacitor (neglecting the fringing fields).
- Generation
of electromagnetic waves: See section 38-4 of HRK. A shaking charge
will "shake the electric field lines" and generate waves, sort of like
transverse waves on a string. The changing electric field induces a
magnetic field via the Ampere-Maxwell law, which then induces a further
electric field via Faraday's law, and the whole thing propagates away
at the speed of light. Two examples: 1) X-ray generation: high speed
electrons slam into a target, and as they decelerate shake off X-rays,
2) radio waves generated by a radio transmitting anetenna with oscillating
currents. A nice visualization of the electric field lines from a wiggling
charge is in this radiating
charge applet . I demonstrated dipole radiation in class
with the demo
K8-42: RADIOWAVES - ENERGY AND DIPOLE PATTERN. We observed the fact
that the wave amplitude is zero along the axis of the antenna, and the
electric field vector is in the plane spanned by the antenna and the
line joining the observation point to the antenna. That is, the wave is
polarized in that direction. We also saw that the intensity
falls off as the distance from the transmitter grows. This pattern, and
the time-development of the waves, is nicely represented in a visualization
of dipole radiation made at MIT. One thing about this visualization
that really bugs me is they don't tell us how the electric field lines
diplayed at each instant of time are selected. Is the point on the axis
just poapagated at the speed of light, or what?
- Radiation
pressure: it turns out, as we'll discuss more on Monday, that light
with energy density u has momentum density u/c, so if it is absorbed or
refelcted by a surface it transfers momentum and therefore exerts a force,
called radiation pressure. This is fairly important to us, since it is
what holds the sun up!
Wed,
11/6
- Electromagnetic
waves: With the displacement current term the Maxwell equations
are more symmetric..., but
how could it have been missed before Maxwell? The displacement
current density is (epsilon_0 mu_0) E,t. That dimensionful coefficient
is very small: (epsilon_0 mu_0)= (3 10^8 m/s)^-2, so it will not
be important compared to any charge current unless the electric field
is changing rapidly enough. It does have a profound effect though: not
only can a changing magnetic field produce an electric field (Faraday's
law of induction), but a changing electric field can produce a magnetic
field. This raises the possibility of a self-sustaining electromagnetic field
in vacuum (i.e. with no charges or currents). We showed in fact that the
vacuum Maxwell equations imply that E and B satisfy the wave equation with
a wave speed c = (epsilon_0 mu_0)^(-1/2) = 3 10^8 m/s = the speed of light!
For plane symmetric waves depending only on z and t, and propagating only
in the k (+z) direction, the electric and magnetic fields are tightly
linked to each other and to k:
E.k = 0
= B.k, E.B = 0, E
= cB, E x B parallel to k
The energy
density in an electromagnetic field is u = (1/2) epsilon_0 E^2 + (1/2
mu_0) B^2. These two terms are equal in a plane, unidirectional
wave, which is reminiscent of the equality of kinetic and potential energy
in string or sounds waves. The intensity is I = uc = EB/mu_0 = c epsilon_0
E^2 = c B^2/mu_0. The energy flux vector is called the Poynting
vector: S = (E x B)/mu_0. Since the electric
and magnetic fields are perpendicular, the magnitude of the Poynting
vector is equal to the intensity, while the direction is the direction
of propagation of the plane wave. In fact, the Poynting vector describes
the field energy flux in all situtations, not just for plane waves.
Demos:
K8-01: ELECTROMAGNETIC WAVE - MODEL
K8-05:
ELECTROMAGNETIC PLANE WAVE MODEL
K8-03:
LIGHT NANOSECOND (we also discussed picoseconds (10^-12
s) and femtoseconds (10^-15 s)---one light-femtosecond is about 3/4
the wavelength of blue light.)
Tues,
11/5
- Discussed
aspects of the homework problems:
1) the integral for the time translated wavepacket is not just
a complex Gaussian (due to the finite range of integration); 2)
dispersion of light in vacuo: the fractional difference
of the speeds in part 4(ii) will be proportional to a; 3) the equation
of motion F = ma for the nth mass on the linear chain is obtained by adding
the forces from the two adjacent springs, -k(x_n - x_(n-1)) + k(x_(n+1)
- x_n).
- Maxwell's equations: we recalled various facts about
vector calculus, in particular Stoke's theorem and the divergence theorem,
and used these to infer the differential form of Maxwell's equations
from the integral form. The supplement Differential form
of Maxwell's equations and electromagnetic waves covers this topic.
We saw that the equations are inconsistent with charge conservation unless
Maxwell's displacement current term is added to Ampere's law. Charge
conservation is expressed locally by the continuity equation, rho_t +
div j = 0.
Mon,
11/4
- Showed
how to get the dispersion relation from the differential equation
for waves: assume a solution of the form exp(ikx -w(k)t) and insert
into the equation, which becomes an algebraic equation for w(k).
- Schrodinger's equation
for a particle in one dimension: i hbar psi_t = - (hbar^2/2m) psi_xx. A brief
(one page) synopsis
of quantum mechanics is given in the supplements. We looked at
the dispersion relation and the interpretation in terms of energy and
momentum, and discussed the probability interpretation and the spreading
of the wavepacket, as well as theHeisenberg uncertainty relation.
- Wavepackets: looked in more detail at wavepackets constructed
with square and Gaussian window functions A(k) (see notes from 10/30).
Explicitly, consider the two normalized amplitude functions:
A1(k) = (1/2Dk) for k_0-Dk<k<k_0+Dk and 0 otherwise; and the
Gaussian A2(k) = (1/Sqrt[pi] Dk)exp[-((k-k0)/Dk)^2]. These give for
Y(x) = \int dk A(k) exp(ikx) respectively Y1(x) = exp(ik_0 x) sin(x
Dk)/(x Dk) and Y2(x) = exp(ik_0 x) exp[-(x Dk/2)^2]. You can see
graphs of these here. It is remarkable how much difference
the smoothing of A(k) makes in localizing Y(x).
Discussed how the latter is more localized than the former,
and explained the limitation Dx Dk >~ 1 in terms of the requirement
of phase cancellation to localize the waves: consider exp(i k Dx)+exp(i(k+Dk)Dx).
These will first cancel completely when Dx Dk = pi. So you see that
there is going to be some inverse realtion between the localization in
x and the localization in k. Multiplying by hbar, this becomes
Dx Dp >~ hbar, Heisenberg's uncertainty relation.
We also discussed
propagation
and spreading of wavepackets, and interpreted the graphs in terms
of the group velocity and its variation over different k's in the wavepackets.
Thurs,
10/31
- group velocity: showed transparencies
that illustrate group velocity: the two transparencies have black
stripes, made with slightly different magnifications on a copy machine,
so one set of stripes is slightly farther apart than the other. When
they are placed one atop the other a pattern of beats is seen. When
the shorter "wavelength" one is moved the beat pattern moves with it,
much more quickly than the sheet itself. When the longer wavelength pattern
is moves the beat pattern moves opposite. The beat pattern moves at what
is called the group velocity. We analyzed this for a pair of harmonic
waves:
exp(ik_1
x - iw_1 t) + exp(ik_1 x - iw_1 t) =
exp(ik_0 x - iw_0 t) 2 cos(Dk
x - Dw t)
where k_0
= (k_1 + k_2)/2, and Dk = (k_2 - k_1)/2, and similarly for w. The
first factor is the carrier wave. It is rapidly varying.
The cosine factor is a slowly varying modulation if k_1 and k_2 are
close to each other. It defines an envelope, which moves at the speed
Dw/Dk, which is the group velocity. In the limit Dk -> dk, this
becomes dw/dk.
- Surface
waves on water: We can get a lot of information from dimensional
analysis. On shallow water of depth h the wave speed might depend on
h, g, and rho, the mass density of the water. However the only combination
with dimensions of speed is Sqrt[gh]. It turns out this is in fact the
wave speed, with a numerical coefficient of unity. On very deep
water the depth h can't matter, so there is no way to make a speed independently
of the wavenumber k. Waves on deep water are thus dispersive, even if
they have linear wavefronts (i.e. not just for circular waves). Using
k, one can form Sqrt[g/k]. It turns out that with a coefficient of unity
this is the phase velocity. That is, w(k) = Sqrt[gk]. The group velocity
dw/dk is thus 1/2 the phase velocity. (Check this.) The shallow water
dispersion relation applies when the wavelength is much longer than the
depth, and the deep water relation holds in the opposite case. The general
case turns out to be governed by w = Sqrt[gk tanh(kh).] When kh <<
1, tanh(kh) ~= kh, so this reduces to the shallow water case. When kh >>
1, tanh(kh) ~= 1, so it reduces to the deep water case.
- Watched a film loop that illustrated the meaning of phase
and group velocity, and how to make a localized wavepacket by superposing
component waves of different wavelengths.
Wed,
10/30
- Dispersion:
w(k) not proportional to k, so phase velocity w(k)/k
not same for different k. Thus wavepackets spread, or disperse. The relation
w = w(k) is called the dispersion relation for the waves.
The group velocity , i.e. the speed of the center of a
wavepacket centered on the wavenumber k, is given by dw/dk. An example
I gave was waves on a chain of masses connected by springs, with equilibrium
spacing a, for which w(k) = (2v_0/a) sin(ka/2). When ka << 1, this
is approximately v_0 k. But when ka = 2 pi it is zero! As you'll see in
the homework, this is actually the same as k =0 in disguise. When ka
= pi one gets the maximum possible w, which turns out to correspond to
a standing wave with vanishing group velocity, corresponding to a normal
mode with alternate masses vibrating in opposite directions.
- One can make a localized wavepacket by superposing an infinite
number of waves of different wavenumbers, by integrating:
Y(x,t) = \int
dk A(k) exp(ikx - iw(k)t).
If
A(k) is peaked around a central value k_0 with width Dk, then Y(x,0)
will be peaked around x=0 with width Dx ~ 1/ Dk. That is, the more sharply
localized in k, the more weakly localized in x, and vice versa.
The form of the wavepacket
at time t will depend on the form of the dispersion relation w(k).
Tues,
10/29
- Matching conditions for waves at an interface
between two media: the for both string waves and sound waves, the
displacements must match, and the forces must be equal and opposite
(Newton's third law). In the sound case, the two bulk moduli generally
differ, as well as the mass density differing. Treated the case of string in
class, allowing for different
tensions, which could be achieved e.g. with the help of a frictionless
rod and massless ring. This is all treated carefully and clearly in
the textbook, Chapter 8, pp. 256-264. In the first part the problem is
treated assuming the tensions in the two strings are equal. The second
part allows for different tensions, and shows that the relevant quantity
for each medium is the impedance, Z = T/v. If the impedances
match, then there is no reflection at all, even if the density and tension
both differ. I talked about the idea of avoiding reflection by putting
in a gradual interpolation between the impedances of two media. We have
a nice demo of impedance
interpolation. (One can also make an impedance match without even
having another wave medium, but rather just a kind of damper. We also
have a demo of impedance-matched
wave absorption.
Mon,
10/28 No class (fire in Physics building)
Thur,
10/24
- Discussed the difference between 2d and
3d circular/spherical waves: the latter preserve pulse shape, the
former not. We could see this by a trick: make a 2d wave using a 3d
wave with cylindrical symmetry. The same argument cannot be used to
show that 1d waves are not shape preserving, since the integral over
the plane diverges.
- Discussed CO_2 molecule model, and the reason for the
disagreement between the model and the experimental results. The
key thing (I think) is that the position of one oxygen affects the
spring between the carbon and the other oxygen. See discussion in
the homework 5.2d solution.
- mid-semester course evaluations
- Partial reflection and transmission
at an interface between two media: examples of string, sound from air
to water, and light from air to glass. The boundary conditions are 1)
equal displacements and 2) equal slopes in the case of equal string tensions,
equal pressures in the case of sound waves. We had some trouble understanding
the reason for the equal slopes and pressures. I explained it interms
of an infinitesimal mass element at the interface: this has finite acceleration,
but infinitesimal mass, hence Newton's second law says that it must
have infinitesimal net force on it. It was clear that this line of argument
was difficult for the students to follow, I'm not sure why. Next Tuesday
I gave a simpler reason for this boundary condition, using Newton's
third law.
Wed,
10/23
- Reflection
from fixed and free ends. Treated with the method of the virtual
pulse. See Ch. 8, pp. 253-256.
Tues,
10/22
- Energy density in a 3d plane wave is better measured
per unit volume (= per unit area per unit length) rather than
per unit length as for waves on strings. The energy of sound waves
in a slab of thickness dx and area A is dK = 1/2 rho A (s_t)^2 dx. For
a purely unidirectional wave s_t = +/- v s_x, so dK = 1/2 rho v^2 A (s_x)^2.
As you'll show in this week's homework, the potential energy is dU =
1/2 K A (s_x)^2
= 1/2 rho v^2 A (s_x)^2.
THis is equal to the kinetic energy for directional waves.
- Intensity of a unidirectional wave: let rho_E be the
energy density. The energy in a slab is rho_E A dx, which passes a
fixed plane in a time dx/v, hence the intensity is (rho_E A dx)/( dx/v) = rho_E v, i.e.
(energy density) x (wave speed).
- Plane waves in other directions: introduce the wave
vector k = k n, where n
is a unit vector. The magnitude of the wave vector k is the wave number.
A sinusoidal plane wave in the n direction can then be written A sin(k.x
- wt), where x is the position vector.
- Plane waves are an idealization: far from
the source, in a small enough region, all waves will look like plane
waves. On a larger scale, another useful idealization is spherical
waves, which have constant amplitude on spheres of constant
radius about some origin. Far from the origin in a small enough solid
angle these look like plane waves as well, except that their amplitude
decreases as 1/r. Why? If the energy is conserved, and all flows at
rate v, then the rate of energy flow of an outgoing wave through a sphere
of radius r must be independent of r. Denoting the intensity at radius
r by I(r), this says that I(r) 4pi r^2 = constant, so I(r) ~ 1/r^2.
On the other hand, energy density is proportional to the square of the
amplitude A(r), hence A(r) ~ 1/r. In fact, the general spherical
wave solution to the wave eqn has the form f(r-vt)/r + g(r+vt)/r, where
the two terms correspond to outgoing and ingoing spherical waves respectively.
We illustrated this with the example of using intensity to measure the
distance to a star.The intrinsic luminosity L, i.e the energy per unit
time emitted, must be equal to the flux of energy through a large sphere
at the location of the earth, a distance d from the star: L = I(d)(4pi
d^2). This can be solved for d = Sqrt[4pi L/I(d)]. If we know L, we can
measure the distance to the star just by measuring the intensity of its
light as viewed at the location of earth.
- Wave
eqn in 3d: s_tt = v^2 (s_xx + s_yy + s_zz)
= v^2 div.grad s = v^2
Laplacian s. Although the first form refers explicitly
to the x,y,z axes, the laplacian differential operator div.grad is
independent of the orientation of the axes. This should be reasonably
evident from the fact that it can be written as the dot product
of the gradient, a vector operator, with itself. It can be shown---in fact you show in
the homework---that the spherical solutions to this equation really have the
form indicated in the previous paragraph. You might think that the 2d case
is just as straightforward, but this turns out not to be true! It is more
complicated...
Mon,
10/21
-
discussed the curving of the exam: what and why; please see
me if you have concerns about the course, your grade, etc.
- Superposition of waves, contined: Similarly electromagnetic
waves can be superposed, due to the linearity of Maxwell's equations.
The electric field vector at each point in sapce and time is the vector
sum of the ones produced from all different sources. A radio receiver
works by tuning to a resonant curcuit at a certain frequency, to pluck
out the signal from one transmitter...which thanks to linearity is completely
undisturbed by the presence of the other signals.
- Energy and superposition: two facts derived
from waves on strings and applicable also to planar sound waves:
1) purely left
or right moving waves have equal kinetic and potential energy densities;
2) the total energy density of a general wave is the
sum of the right moving and left moving energy densities.
We showed
this by working out the kinetic and potential energy densities.
I think the same thing is shown in the textbook.
- Sound waves: these propagate in 3d, so the displacement
field s is a function of four variables: s(x,y,z,t), but for a
plane wave it is just a function of x and t: s(x,t).
The energy density is then properly given as energy per unit area
per unit length, i.e. energy per unit volume. We discussed the form
this takes for kinetic and potential energy. The intensity
is the energy per unit time per unit area carried by the wave. For a
purely unidirectional wave this is given by Intensity = (energy density)
x (wave speed).
- Bulk
modulus vs. Young's modulus: the book says that the former is always
larger than the latter, and gives a reason, however this seems not
to be borne out in the data. I think the reason given is specious.
Also the some of the numerical values given in the book disagree with
other sources. In particular, for aluminum Y is 7 10^10 N/m^2, not 6.
Thurs,
10/17
speed of sound in gases:
air(79% N2, 21% O2)
0 C
|
331 m/s
|
air
20 C
|
343 m/s
|
He
0 C
|
965 m/s
|
H2
0 C
|
1284 m/s
|
Why the difference with type? Why
the dependence on temperature? For sound in air, Newton
figured he could assume the air is at constant temperature and found
K_isothermal = p (see notes for 10/9/02). Thus he found v = Sqrt[p/rho],
which yields 289 m/s at standard temperature (0 C) and pressure (1 atm).
Not a bad start, but something is obviously wrong. On the other hand,
something's right: using the ideal gas law pV = NkT, we get p/rho = kT/m,
where m is the mass of the molecule, so v = Sqrt[kT/m]. The square root
dependence on temperature explains the temperature dependence of the speed
of sound in air: Sqrt[(273+20)/273] x 331 = 343! Also the square root
dependence on the mass explains the ratio of the speed in air to the speed
in hydrogen: atomic
weight is 28 for N2 and 32 for O2, compared with 2 for hydrogen, so
the ratio should be slightly higher than Sqrt[28/2] x 331=1238, and
lower than Sqrt[32/2]
x 331=1324. Not bad. However the ratio to the speed in helium doesn't
work out: the atomic weight is 4, and Sqrt[28/4] x 331 = 876, whereas
the speed in helium is 965 m/s.
What's wrong is that the gas is not at
a constant temperature in the wave. Instead, it is at constant energy,
i.e. adiabatic: no heat flow in or out. The adiabatic
bulk compressibility differs from p by a factor called the adiabatic
index gamma, which depends on the type of gas (and also
on the temperature in a stepwise fashion---see below): K_adiabatic
= gamma
p, hence v = Sqrt[gamma kT/m], which agrees with
observation when the correct adiabatic index is used. The fact that
the wave is not at constant temperature is plausible when one looks
at the rms velocity of the molecules: kinetic theory of gases shows
that kT/m = <v_x^2>, the average of the squared x-component of
velocity, so the isothermal wave speed formula of Newton would yield
a wave speed equal to the rms molecule speed in the direction of propagation.
But if the wave is going as fast as the average molecule, then it seems
plausible that there isn't enough time for thermal equilibrium to be
maintained with the surroundings.
Physics of the adiabatic index: If the volume
of gas is compressed a given amount, the pressure goes up. The amount
by which the pressure rises depends on how many degrees of freedom
there are other than the translational ones that give rise to pressure.
The adiabatic index is determined the the number of accessible degrees
of freedom (d.o.f.) D of a molecule, hence is different for monatomic
(3 d.o.f.) , diatomic (5 d.o.f.), and polyatomic (6 d.o.f.) molecules.
It is equal to the ratio c_P/c_V of the specific heat at constant pressure
to the specific heat at constant volume. It is also given by the formula
gamma = 1 + 2/D. Thus gamma_air = 1 + 2/5 = 7/5, while gamma_He = 1 + 2/3 = 5/3.
For air then, the sound speed should be Sqrt[7/5] times Newton's value
of 289, i.e. 342 m/s. Pretty good, but not dead on. Not sure why.
For the ratio of the speeds in helium to air, we have the square root
of the ratio of the adiabatic indices: Sqrt[(5/3)/(7/5)]=Sqrt[25/21]=1.09.
Multiplying this by the 876 found above from the mass ratio alone gives
955 m/s, which is pretty darn close. The extra mass of the O2 in air
probably explains the difference.
The number of accessible degrees
of freedom D actually depends on temperature in a stepwise fashion, because
of quantum mechanics! For temperatures around 0 C
it turns out that D is indeed given by 5 for N2, O2, and H2, while it
is 3 for He. However, the general quantum story here is that, according
to the equipartition of energy, each accessible d.o.f.
has on average 1/2 kT of energy, and this holds in the sound wave,
at the local temperature which varies with the local pressure. However,
quantum mechanics asserts that the angular momentum of anything,
and molecules in particular, is quantized in units of Planck's constant
hbar. The energy of rotation can be written as L^2/2I, where I is the
moment of inertia. The squared angular momentum of the molecule is quantized
as N(N+1)hbar^2, N=0,1,2,3,...hence the lowest energy of rotation
is E_0 = hbar^2/I. If this smallest energy is much larger than
1/2 kT, then the rotation is not accessible, and doesn't carry any
energy. It turns out that the rotational degrees of freedom of N2 and
O2 are accessible way below 0 C, while those of H2 are fully
accessible just around 0 C, since the moment of inertia of H2 is much
smaller due to the low mass of hydrogen compared to nitrogen and oxygen
(the size of the molecules is comparable). (Molecules also have vibrational
degrees of freedom, but these are unaccessible except for much higher
temperatures, due to the quantization of the energy of the vibrational
motion.)
superposition of traveling waves: Because
of linearity of the wave equation: the sum of two solut
ions is a solution.
This means that right and left moving wave pulses just travel through
each other. Were it not for this we could not make out the sounds
from different instruments in an orchestra, for example. An example
shows that left and right moving pusles ona string can cancel exactly
at a moment, so there is no potential energy. At that point, however,
the string has transverse velocity, and all the energy is kinetic.
Wed,
10/16 traveling waves: So far we discussed
the equation of motion of string, and the normal modes that result.
A string can also carry travelling waves, which must also be described
by the same equation of motion. To see this, we began with a normal
mode, in the form
y(x,t)
= A sin kx cos wt.
Terminology:
time
|
w: angular frequency
|
f = w/2pi: frequency
|
T = 1/f = 2pi/w: period
|
space
|
k: wave number
|
k_book = k/2pi
|
lambda = 1/k_book = 2pi/k: wavelength
|
space/time
|
v = w/k = lambda.f = lambda/T: wave
speed
|
|
|
Using a trig identity, we can rewrite the above
normal mode as a superposition of right and left moving traveling
waves:
y(x,t) = (A/2){sin[k(x-vt)] + sin[k(x+vt)]}
The first term is right-moving at speed v, the second
is left-moving. Why? Consider any function of the form y(x,t) =
f(x-vt). If t is increased by Dt and x is increased by Dx =vDt then
the argument of f is unchanged, so y(x+vDt, t+Dt) = y(x,t). That is,
the shape of the function y(x,t) at time t+Dt is exactly the same as
it was at time t, but just shifted over to the right by a distance vDt.
Similarly for y(x,t) = g(x+vt), moving to the left. Hence v really
is the wave speed.
Fact (to be proved in the homework): the general
solution to the wave equation is of the form
y(x,t) = f(x-vt) + g(x+vt)
Wave speeds: string: Sqrt[T/mu], rod: Sqrt[Y/rho],
bulk(e.g. sound): Sqrt[K/rho], where T= tension, Y= Young's (stretch)
modulus,
K=bulk modulus, mu= mass/length, and rho=mass/volume.
I discussed sound in air, but let me put all of this in Wednesday's
notes, since we finished it then...
Tues,
10/15 discussed solutions to exam 1
Mon,
10/14 Exam 1
Thur,
10/10 review
for exam 1, went over last year's exam1 (see supplements for a copy)
Wed,
10/9
- microscopic energy of atomic springs
underlying Young's modulus: see notes for Tuesday
- Bulk modulus: (see notes for Tuesday).
Bulk modulus of gas: Newton was the first to compute this, as
far as I know. He used it to compute the speed of sound. He knew
Boyle's law: pV = constant for a gas at constant temperature. Thus
0 = d(pV) = dp V + p dV, so dp = - p (dV/V). Compare this to
the definition of Bulk modulus: F/A = -K DV/V. The force here is
due to the overpressure, that is the pressure over (or under)
the ambient pressure. Hence it is the same as dp, so Boyle's law
gives us K=p. It turns out that in a sound wave the temperature is
NOT constant, so Boyle's law does not apply. Instead we need the "adiabatic"
Bulk modulus, which is a constant times p. More on this later.
- Discussed some hw problems. For the piano wire, you
don't know the cross sectional area, but both the tension and mass
per unit length are proportional to it, so it cancels out in v. For
the CO2 molecule, clarified the assumption about the nature of the
other normal mode, and how to impose the condition that the center
of mass remains at rest. For the energy in the vibrating string, the kinetic
energy is an integral over the string: K = \int dK = \int 1/2 dm v^2
= \int 1/2 M (dx/L) (y_t)^2. When there are two (or more) modes simultaneously
excited, you need to show that the cross terms between the two modes vanish
upon integration from 0 to L.
Tues,
10/8
- normal modes of aluminum
rod demo: first three normal modes excited, by stroking the rod
while holding it at the different node locations.
- Young's modulus:
intrinsic property of a material that measures its stiffness, independent
of the particular size and shape of the chunk of material being
considered. Stress = F/A is proportional to strain = DL/L (fractional
change in length) . The proportionality constant is Y, the Young's
modulus: F/A = - Y DL/L. (The minus sign is because the book likes
to let F be the reaction force of the stretched stuff on whatever agent
is stretching it.) The relation to the spring constant k of the rod
is seen by writing this as F = - (AY/L) DL, so k = AY/L. The Young's
modulus of aluminum is 6 x 10^10 N/m^2, and the propotionality holds
for strains less than 0.1 % or so.
- microscopic picture of Young's modulus: material
is like a bunch of microscopic springs Hooked together (pun intended).
Spring constants add in parallel and their inverses add in series,
so the total spring constant of the rod is k = (N_parallel/N_series)
k_0, where k_0 is the microscopic
sping constant for each atomic spring. Say the
longitudingal distance between the atoms is a_0, the transverse
distance between them is b_0, and the cross sectional area and length
of the rod are A and L. Then N_parallel = A/b_0^2 and N_series = L/a_0,
so k = (A/L)(a_0 k_0/b_0^2). Thus Y = a_0 k_0/b_0^2
in this crude model. This is a formula for Y in terms of microscopic
properties of the material. [It is fun to put in the numbers for
aluminum just to see what microscopic spring constant k_0 = Y b_0^2/a_0
results. Aluminum is a cubic crystal, with a_0=b_0 = 4 x 10^-11 m.
Hence k_0 = Y a_0 = (6 x 10^10 N/m^2)(4 x 10^-11
m)=2.4 N/m.
Is this reasonable? Work out the potential energy when the spring
is stretched by an amount a_0:
U = 1/2 k_0 a_0^2 = 1/2 Y a_0^3 = (0.5) (6 x 10^10
N/m^2)(4 x 10^-11
m)^3 = 1.9 x 10^-19 J = 1.2 eV. This says that
to pull apart one atom pair spring to a distance equal to the interatomic
spacing would cost an energy equal to 1.2 eV, which is quite reasonable
in order of magnitude, since a typical covalent bond energy is
of the order of an electron volt...]
- wave equation for compressional waves:
we derived this
by applying F = ma to the longitudinal motion of each little bit
Dm of the rod. See textbook for details. The key step was to note that
the local strain is equal to s_x, the partial derivative of the displacement
wrt x, and to use this to write the force of the material on the right
hand side of Dm as AYs_x. The net force is the difference between this and
the force on the left, which gives approximately AYs_xx Dx. Equating this
to (rho A Dx) s_tt yields s_tt = (Y/rho) s_xx, where rho is the mass
per unit volume. This is the equation of motion for the rod and it is mathematically
identical to the equation for transverse vibrations of a stretched string.
The same story goes for sound waves except there we speak of not the
Young's modulus, but the bulk modulus K, defined by F/A = - K DV/V, where
V is the volume. If the change of volume is only along one direction then
DV/V=DL/L so the bulk and Young's moduli are the same. (For the rod this
is not the case, since when the ron stretches it gets a bit skinnier in
the transverse direction, and when it compresses it gets a bit fatter.)
Mon, 10/7:
- Wavelength: distance for sin
function to complete one cycle. If mode function is f(x) = A sin((w/v)
x + phi), then wavelength satisfies (w/v) lambda = 2pi, or w = 2pi
v/lambda. Since ordinary frequency is f = w/2pi, this is the same
as
(wavelength)*(frequency)=(wavespeed),
or
(wavelength)/(period)=(wavespeed)
-
Dependence of normal mode frequencies and shapes on boundary conditions:
one fixed and one free end gives the spectrum of frequencies w_n =
(n - 1/2) pi v/L. The mode functions in this case have an odd number
of quarter wavelengths. If both ends are free, the frequencies are
the same as if both are fixed, but the mode functions are shifted by
pi/2 in phase, i.e. nodes and anti-nodes are interchanged. Sound waves
in a tube is an example: an open end of a tube is a node for pressure
(since the pressure must match the ambient air pressure in the room)
but an anti-node for displacement. At a closed end of the tube it is
the reverse. We did a demo with two identical length tubes, one with both
ends open and the other with one open and one closed end. The fundamental
of the closed/open tube has one quarter wavelength, while that of the
open/open tube has one half wavelength, hence the latter has a frequency
double the former, i.e. one octave above.
Thur,
10/3:
- normal modes of string:
The general solution to the equation for the
normal mode amplitude function (see the previous lecture) is f(x)
= A sin((w/v) x + phi). There must be something
missing, since this holds for ANY w,
whereas we know that only certain special frequencies are
allowed for normal modes. What is missing is the boundary conditions,
which enforce the fact that the ends of the string are not moving:
y(0,t) = 0 = y(L,t). The condition at x = 0 implies that phi = 0 (or
pi, but that can be absorbed in a change of the sign of A), and the
condition at x = L then implies that the frequency must be such that
wL/v is an integer multiple of pi. Therefore the frequencies of the
normal modes are multiples of a lowest one w_n = n w_1, where w_1 = pi
v/L. The lowest mode is called the fundamental, and the rest
are called harmonics.
- The amplitude function in the n^th mode is thus
f_n(x) = A_n sin(n pi x/L). The points where the amplitude vanishes
are called nodes. The number of nodes is equal to n+1, including
the two nodes at the endpoints. The maxima of the amplitude are
called anti-nodes. The integer n is the number of half-cycles
of the sin function that fit in the length L. The distance for one
cycle is called the wavelength, and it is given in the n^th
mode by lambda_n = 2L/n.
- The total number of normal modes is infinite, since
the system has an infinite number of masses, continuously strung
together. However, there are really only a finite number of atoms
in the string, and we know that a system with N degrees of freedom
has N normal modes, so there can only be a finite number. When the
wavelength is shorter than the inter-atomic spacing the continuous treatment
of the string fails.
- In the n^th mode, we have y_n(x,t) = A_n sin(n pi x/L)
cos(n pi v t/L). From
this we can find the speed of a point on the string at position
x and the slope of the string using partial derivatives with respect
to t and x respectively. The maximum speed is thus A_n n pi v/L and the maximum slope is A_n n pi/L.
- Different boundary conditions: if
an end of the string is free to move up and down (say if it is tied to a
massless ring that slides up and down a frictionless post) then the slope
of the string must vanish at that end, i.e. it must lie at an anti-node.
Why? Think of the force on the ring: it is pulled vertically only on one
side, so the vertical force would be finite while the mass is zero. This
would cause an infinite acceleration that would flatten out the string. Put
differently, since the acceleration of the ring is finite, and the mass is
zero, the force must be zero, which is only the case if the slope is zero.
Wed, 10/2:
- Equation of motion of string:
A string under tension
is like an infinite number of coupled oscillators. The restoring
force is due to the tension T. The inertia is from the mass per unit
length (linear mass density) mu. We consider only small transverse
vibrations. Then each bit of string only moves perpendicular to the
equilibrium line. The string configuration is described by a function
y(x,t), the displacement of the bit of string at coordinate x and time
t. We assume the slope always remains much less than unity, i.e. y_x <<1,
where y_x denotes the partial derivative of y wrt x. Under these assumptions
also the tension is constant to a good approximation. To find the equation
of motion we consider a little section of string of mass Dm = mu Dx, and
apply Newton's second law to it. The acceleration of this bit in the y direction
is y_tt, the second partial derivative of y wrt t. Hence we have mu Dx y_tt
= F_y. The force is the y-direction is due to the y-component of the tension,
T_y. Similar triangles give T_y/T_x = y_x, but T_x is approximately
T, hence T_y = T y_x + terms of higher order in the small quantity y_x.
Now the net force in the y-direction comes from the mis-match in the slopes
at x and x+Dx:
F_y = T_y(x+Dx)
- T_y(x) = T [ y_x(x+Dx) - y_x(x) ] = T y_xx(x) Dx +
O(Dx^2).
Thuse we have Dx y_tt = F_y
= T y_xx(x) Dx + O(Dx^2). Dividing through by Dx and taking
the limit Dx->0 then gives
y_tt = v^2 y_xx,
(string
wave equation)
where v^2 = T/mu is what turns
out to be the square of the wave speed. (Check that it has the dimensions
of speed squared.)
-
Normal modes of string: Assume each point undergoes SHM with
same frequency but different amplitude: y(x,t) = f(x) cos wt.
Then y_tt = -w^2 y and y_xx = f_xx cos wt = (f_xx/f) y, so the
amplitude must satisfy
f_xx = -(v/w)^2
f
(string normal mode equation)
Tues,
10/1:
- Examples of coupled oscillators
and normal modes: 2,3,4 coupled pendula, Wilberforce
pendulum, pair
of hanging masses.
- superposition of normal modes.
- energy transfer
in coupled oscillators viewed as one oscillator forcing the other;
if they have the same or nearby frequencies there is resonance, and
most or all of the energy can be transferred.
-molecules have resonant frequencies, and they
absorb electromagnetic (infrared) radiation of the resonant
frequencies. This produces absorption lines, by which chemical
species can be detected from far away. In optical fibers, one tries
to use frequencies that do not resonate with any atomic structures
in the fiber material.
- if you know linear algebra: the general problem
of finding the normal mode frequencies and amplitudes amounts to
finding the eigenvalues and eigenvectors of a matrix.
Mon,
9/30:
- coupled oscillators:
demo
of two pendula connected by a spring. Motion can start in
one mass, and then transfer to the other and back again. So the
motion of one mass exhibits beats. On the other hand there are two
very special motions which are simple harmonic oscillations, in which
all parts of the system oscillate with the same angular frequency and
fixed amplitudes. These are called normal modes. We found the
normal mode frequencies first by just writing down F=ma for these two
motions.
(See textbook for the details). Then we went back
and solved it more generally, not assuming any special relation
between the motion of the two pendula. Adding and subtracting we
found that the normal mode coordinates q_1 = x_A
+ x_B and q_2 = x_B - x_A each satisfy simple harmonic oscillator
equations, with different natural frequencies. Using the general
solution for these normal mode coordinates we can go back and
find the solution for the coordinates of the masses x_A = (q_1 - q_2)/2
and x_B = (q_1 + q_2)/2. For example,
if the initial conditions are x_A(0) = A, x_B(0)
= 0, and the time derivatives both vanish, then the corresponding
initial conditions on q_1 and q_2 are q_1(0) = A, q_2(0) =
-A, and the time derivates vanish. Thus q_1 = A cos(w_1 t) and q_2 = - A cos(w_2 t), which
corresponds to x_A = (A/2)[ cos(w_1 t) + cos(w_2 t) ] and x_B = (A/2)[cos(w_1 t) - cos(w_2 t) ].
Each of these exhibits beats
at the beat frequency equal to the difference of the normal mode
frequencies (w1 - w2).
Thur, 9/19, Mon-Thur 9/22-26
(I may try to fill this in later...haven't
had the time)
power in AC circuits
series RC circuit
complex impedance, parallel RC circuit
RLC circuit, complex impedance
power & resonance
transients, damped driven
oscillator & resonance
Wed, 9/18:
- example of expressing sum of
oscillations with same frequency in the form Re[A e^i(wt+a)]:
2 cos(wt) + 3 sin (wt) = Re[2 e^iwt] + Re[-3i
e^iwt] = Re[(2-3i)e^iwt] = Re[Sqrt[13]e^i(wt - tan^-1(3/2))]. Another method is to just set the left
hand side equal to Acos(wt + a) = A cos wt cos a - A sin wt
sin a, and equate the coefficients of the cos and sin terms: A
cos a = 2 and A sin a = -3, which has the solution A = Sqrt[13] and
a = tan^-1(3/2).
- Transients: when we
demonstrated the driven oscillator on Tuesday using the torsional
oscillator demo, it didn't behave quite like the steady
state solution would indicate. In fact, the amplitude seemed
to slowly oscillate from being large to being small and back again.
The reason is that I did not turn on the damping force, so the transients
were not going away. What is going on here is that in addition to the
steady state solution there is another component to the motion. The
general solution has the form x(t) = x_ss(t) + x_free(t), where x_free(t)
is any solution to the free oscillator equation. (The fact that we
can just add any solution x_free(t) is due to the fact that
x appears linearly in the equation of motion.) More explicitly,
x(t) = A_ss cos(wt -d) + A_free cos(wt + a), so x(t) is the sum of two
harmonic oscillations at the same frequency with different amplitude
and different phase. This produces beats, as we have previously heard
with a guitar string and now see with the torsional oscillator. What
is different in the damped case is only that x_free(t) is a solution
to the damped free oscillator equation. Those solutions all have the property
that they die away exponentially in time, due to the damping. Hence
at late times only the steady state solution remains! The greater the
damping, the faster the total solution approaches the steady state solution.
- steady state solution in the driven case:
we found this using the complex eponential method. The result:
x(t) =
Re[C e^(iwt)], where C = (F_0/m)/(w0^2 - w^2 + i gamma w). Put
differently, C = A e^(-id), where A = (F_0/m)/Sqrt[(w0^2 - w^2)^2
+ (gamma w)^2] and tan d = gamma w/(w0^2 - w^2). Note that
the denominator of C has a positive imaginary part, hence its phase
is between 0 and pi, so d runs from 0 to pi. Features:
(1) When w =
w_0, the amplitude is finite: A = F_0/(m gamma w), and
the phase is -pi/2. The position thus lags the force by 90 degrees.
The smaller the damping, the larger the amplitude.
(2) For w < w_0 the position lags the force
by less than 90 degrees, while for w > w_0 it lags by more
than 90 degrees, approaching 180 degrees for w >> w_0.
(3) The amplitude is not maximized at exactly
w = w_0 unless gamma = 0. Rather it is maximized below w_0.
Tues,
9/17:
- we found the combination of the
two solutions in the overdamped case for the intial condition
v(0)=0. both solutions are involved. for large damping, the
more slowly damped solution dominates.
- driven, undamped oscillator: we found
the steady state solution. it has the following features:
(1)
the amplitude is determined, not free to be specified;
(2) the amplitude diverges as w approaches
w_0, which is called resonance;
(3) x is in phase with F below resonance
(w < w_0), and x is 180 degrees out of phase with F above resonance
(w>w_0).
-
if a damping term is added, then the larger the oscillation
the larger the rate of energy loss by damping. Thus, instead of
diverging, the amplitude will reach a maximum at (or near---see
Wednesday) w = w_0.
Mon, 9/16:
- Please don't use calculators
to do the complex number manipulations on the homework (like
converting between polar and cartesian form). You need to really
understand how to do this yourself! (You can of course use the caculator
to evaluate trig functions.)
- more on the damped oscillator (see textbook):
- damping force power drain: P = Fv = -bv^2, leads to
decrease of the energy of the oscillator.
- meaning of gamma: dimensions are 1/T; 1/(time
for energy to drop to 1/e of initial value); 2/(time for amplitude
to drop to 1/e of initial value;gamma = |dE/dt|/E,
the fractional rate of change of energy loss. Note however
that this only holds on the average. The instantaneous rate of
energy loss is zero when the oscillator is at rest at the maxima.
- Quality factor Q = w_0/gamma, which is approximately
(2pi)(# of cycles for energy to drop by 1/e)
- overdamped case, gamma/2 > w_0: There is no
oscillation. The general solution is a combination of two different
exponential decay rates!
- critically damped case: gamma/2 = w_0: the
exponential method yields only one solution, x(t) = Cexp(-(gamma/2)
t). But there must be two independent parameters in the general
solution, corresponding to the freedom to choose the initial position
and velocity! What is the other solution?? To find it, one can take
the limit of the overdamped case as gamma/2 approaches w_0. Alternatively,
look at the very special case where there is no damping and no restoring
force: d^2x/dt^2 = 0. This is just a free particle, and the general solution
is x(t) = C + Dt. This might lead you to guess that the solutoin for
the critically damped oscillator is simply x(t) = (C + Dt)exp(-(gamma/2)
t). This guess is correct, as you can easily verify by plugging it into
the equation of motion.
- I ended class today with a question: if you start out
an over-damped harmonic oscillator at rest and displaced some distance
from equilibrium, what will its subsequent motion be? Will it
involve both of the damping rates, and if so with what relative
weight?
Thu,
9/12:
- example of
beats with equal amplitudes and zero phase shift: A1=A2=A and
d1=d2=0 (see textbook).
- damped oscillator (see textbook).
- to be emphasized: the reason the complex method works
is that (1) the equation of motion is linear, and (2) the coefficients
are real. In detail: the equation is of the form x'' + r
x' + s x = 0, where the prime denotes derivative wrt time and r
= gamma = b/m and s = w0^2 = k/m. We replace this by the same equation
for a complex function z(t) = x(t) + i y(t), z'' + r z'
+ s z = 0. The real and imaginary parts of z both separately
satify the orginal equation:
0 = (x + iy)''
+ r (x + iy)' + s (x + iy) = (x'' + r x' + s x) + i(y''
+ r y' + s y).
If a complex number is zero,
then both its real and imaginary parts must be zero, hence it follows
that (x'' + r x' + s x) = 0 and (y'' + r y' + s y) = 0.
- To better
understand why this worked, consider a case where it would not
work: suppose the original equation were x'' + q x^2 = 0...
- We inserted
z(t) = C e^pt and saw that the differential equation became
an algebraic equation, since the operation of differentiation
became the operation of multiplication by p! Solving the quadratic
equation we obtained the solution for the under-damped case, i.e.
when gamma/2 < w0.
Wed,
9/11:
- z = x + iy
is the cartesian representation and z = r exp(i q) is
the polar representation of the complex number z. r = |z|
is the modulus of z and q is the argument of z, Arg[z]
(defined up to an integer multiple of 2pi). It is also called
the phase of z.
- cos q = [exp(iq) + exp(-iq)]/2, and sin q = [exp(iq)
- exp(-iq)]/2i.
- did several examples of computing modulus, phase, inverse,
polar and cartesian forms of complex numbers, both geometrically
and algebraically.
- circular or rotating vector representation of harmonic
oscillation: x(t) = A cos(w t + d) = Re[z (t)], where z(t) = A
exp[i(w t + d)]. The complex number z(t) rotates counterclockwise
with angular frequency w on a circle of radius A in the complex
plane. In this way a harmonic oscillation in x is represented as the
projection on the x-axis of a uniform circular motion in the plane.
- Beats: when two oscillating signals with different
frequencies are combined the total signal has an intensity that
oscillates with a beat frequency that is the difference of the two
individual frequencies. This can be understood nicely using the
rotating vector representation: Say x1(t) = A1 cos(w1 t + d1)
and x2(t) = A2 cos(w2 t + d2) are the two signals to be added.
Then x_tot(t) = Re[z1(t) + z2(t)], where z1(t) = A1 exp[i(w1 t + d1)]
and similarly for z2(t). The resultant z1(t) + z2(t) has maximum
modulus when the two angles are lined up, i.e. when w1 t + d1 = w2
t + d2 + 2pi n, i.e. when t = [2pi/(w1-w2)]n + (d2-d1)/(w1-w2). That
means the interval T_beat between successive maxima of the modulus
is T_beat = 2pi/(w1-w2). Hence the beat frequency is w_beat = w1-w2.
(The exact line-up generally occurs when z1(t) + z2(t) is not along
the real axis, however if w1 and w2 are close to each other, there will
be many trips around the circle while the two are almost lined
up, which will produce the observed beats.
(You may like to look at this visual
applet illustrating beats, and this audio one.)
Tues,
9/10:
- direction angle
for a complex number is q = tan^-1(y/x) if x>0, i.e. in the right half
of the complex plane. For x<0 it is pi + tan^-1(y/x), while on the imaginary
axis it is + pi/2 for y>0 and - pi/2 for y<0.
- geometrical representation of complex numbers: a
complex number is a vector on the complex plane. Addition of complex
numbers corresponds to vector addition. Mutliplication by a
real number is scalar multiplication of the vector. Multiplication
by i is counter-clockwise rotation through 90 degrees, i.e. pi/2
radians. Thus i^2 = ii = counter-clockwise rotation through 180 degrees,
which is the same as multiplication by -1. So you can understand
the "number" i as an operation of pi/2 rotation
in the complex plane.
- complex exponential function, Euler's identity: exp(i
q) = cos q + i sin q.
- multiplication by exp(iq) corresponds to counterclockwise
rotation through the angle q.
- z^w = exp(w ln z). So i^i = exp(i ln i). But i = exp(i
pi/2), so ln i = i pi/2, so i^i = exp(-pi/2). Note however that
the ln function is multi-valued: you can add any inter multiple
of i 2pi to the exponent and change nothing, since exp(i 2pi n)
= 1 for all integers n. Thus to be more general, i^i = exp(-(pi/2 +
2pi n).
- complex harmonic oscillator solution to the equation
d^2z/dt^2 = -w^2 z
z(t) = (A +
iB) exp(iwt) = (A cos wt - B sin wt) + i(A sin wt + B cos wt)
the real and imaginary parts
are separately solutions, and in fact by choosing A and B one gets
the most general solution this way.
Mon,
9/9:
- molecular potential
example
- took photos of the class
- Complex numbers: oscillator equation d^2x/dt^2
= - w^2 x. Exponential solution: x(t) = A e^pt works if p^2
= - w^2, i.e. p = w Sqrt[-1]. So invent a new number, i = Sqrt[-1],
called the "imaginary unit". Then i^2 = -1. A real oscillator position
should be described by a real number, not an imaginary one. We shall
use imaginary (more generally, complex) numbers to construct REAL
solutions.
- Define general complex number as z = x + iy. x
= Re[z] = real part of z; y = Im[z] = imaginary part of z (sometimes
iy called "imaginary part of z"). Addition defined by z1 + z2
= (x1 + x2) + i (y1 + y2), multiplication defined to satisfy the
usual rules: communtative, associative, distributive. Hence z1 z2
= (x1 x2 - y1 y2) + i (x1 y2 + y1 x2).
- Do we have to invent more numbers? No! All equations can
be solved with complex numbers. E.g. sin z = 2, and e^z = -1
have solutions (see later).
- fundamental theorem of algebra: any polynomial
z^n + c_(n-1) z^(n-1) + ... + c2 z^2 + c_1 z + c_0 can be factorized
to the form (z - w1)(z - w2)...(z - wn), so any nth order polynomial
equation has exactly n solution ("roots").
- complex plane, representation of complex numbers
as two-dimensional vector with magnitude or "modulus" |z|
= Sqrt[x2 + y2] and direction angle q = tan^-1(y/x).
Thurs,
9/5:
- theme: everything
is (approximately) a harmonic oscillator near equilibrium.
- Simple pendulum (point mass at the end of a massless
string): NOT a harmonic oscillator...except for small enough
amplitude. If q is the angle, we showed that d^2q/dt^2 = -(g/l)
sin q. Since sinq = q - q^3/3! + q^5/5! - ...this is approximately
the h.o. equation when q is not too large, and the angular
frequency is Sqrt[g/l]. Another way to look at it: the potential
energy of the pendulum is U(q) = mgl(1-cos q) = mgl (q^2/2 - q^4/4!
+ ...), so the potential energy function is approximately a parabola
for small enough q.
- Galileo noticed that the period of a pendulum is (nearly)
independent of the amplitude, and used that to make a good clock
for timing physics experiments.
- Dimensional analysis would tell us T(m,l,g,q_max) = f(q_max)
Sqrt[g/l], without any other analysis! But it tells us nothing
about the function f(q_max) of the dimensionless maximum angle q_max.
As we showed above, for small q_max, f is approximately 1. The next
correction is -q_max^2/16. (The honors students will show this
in the first homework.) (In class I mistakenly said it was -q_max/16,
without the square.) The exact result is not an elementary function,
but can be expressed as a definite integral.
- Physical pendulum: extended rigid body suspended
from a point. Also approximately a h.o.
- completely general situation: make Taylor expansion of
potential about the point x_e:
U(x) = U(x_e)
+ U'(x_e) (x - x_e) + 1/2! U''(x_e) (x - x_e)^2 + 1/3! U'''(x_e)
(x - x_e)^3 + ...
where the primes denote derivates.
If x_e is a point of stable equilibrium, then U'(x_e) = 0 and
(assuming it is nonzero) U''(x_e) > 0. Thus sufficiently near
the bottom of the bowl we have the approximation U(x) = U(x_e) + 1/2 U''(x_e)
(x - x_e)^2 which has the parabolic form of a harmonic oscillator potential
with effective spring constant
k_eff = U''(x_e)
The frequency of small oscillations
about x_e is thus Sqrt[U''(x_e)/m].
Wed,
9/4:
- more syllabus
discussion
- dimensional analsis: how to do it and what it's
good for:
(1) Catch errors, (2) infer answers without solving the
whole problem, (3) check answers, (4) imbue formulae with meaning.
For example, using just dimensional analysis we can find that
the angular frequency of the harmonic oscillator must be of the
form # Sqrt[k/m], where # is a dimensionless number the same
for all oscillators. So we find how the frequency depends on k and m
without solving the differential equation of motion!
- important observation: argument of trig functions
and exponential must be dimensionless!
- energy conservation for the harmonic oscillator
- derivation of equation of motion from energy conservation:
0= dE/dt = d/dt(1/2
m v^2 + U) = mv dv/dt + dU/dx dx/dt.
If v is not zero, this implies
m dv/dt = - dU/dx, i.e. ma = F, Newton's 2nd law.
- example
of the LC oscillator: Total energy: E = 1/2 L I^2 + 1/2 (1/C)
Q^2. Perfect analogy to a mechanical oscillator:
| x |
v = dx/dt |
m |
k |
k/m |
| Q |
I = dQ/dt |
L |
1/C |
1/LC |
The total energy in the
LC circuit is conserved, and oscillates between electric field energy
stored in the capacitor and magnetic field energy stored in the
inductor. Q oscillates like a sin (or cosine) function of time with
angular frequency
1/Sqrt[LC].
You can also find the equation of motion not from energy conservation
but from Kirchoff's law that the sum of the voltage drops around
a closed loop is zero. In this case that gives: L dI/dt + Q/C = 0,
or dI/dt = -(1/LC) Q. Since I = dQ/dt this is just the harmonic oscillator
equation.
Tue,
9/3:
- Syllabus discussion
-
Course intoduction, everything is waves
-
mathematical techniques used in course: complex number, ordinary
& partial differential equations, Taylor expansions, vector
calculus.
-
Harmonic oscillator: prototype of all oscillations and waves:
- restoring force and oscillatory motion
- (second order, linear, ordinary) differential equation and general
solution
- two initial conditions (e.g. position and velocity at time t=0)
required to determine solution, hence general solution has two
adjustable parameters. The way we wrote it, these are the amplitude
and phase angle.
- amplitude, angular frequency, frequency, period, phase angle