- Circular polarization stick model.

- Michelson interferometer:
discussion of principle; use by Michelson & Morley to test for the motion of the earth relative to a hypothetical medium---the "ether"---in which electromagnetic waves were thought to travel; use in LIGO
to detect gravitational waves; use in measuring the index of refraction of gas; demonstration with laser light and with white light (which shows beautiful colors due to the differing interference of different wavelengths for a given path length difference).

- Interference of sound waves in Quincke's tubes

- Intro to Young's double slit interference experiment; demo with two speakers (microphone pickup didn't show the effect very well).

-DEMOS shown today:

     M3-01 MICHELSON INTERFEROMETER - LASER LIGHT
     M3-02 MICHELSON INTERFEROMETER - WHITE LIGHT
     H2-23 INTERFERENCE - KLINGER TRANSPARENT SLIDES
     H2-24 AUDIBLE YOUNG'S EXPERIMENT - MIKE AND SCOPE
     H2-25 QUINCKE'S INTERFERENCE TUBES
     M9-03 CIRCULAR POLARIZATION - STICK MODEL

- intro to spacetime diagram of Doppler effect (supplement)

- Mach cone, Cerenkov effect, sonic boom

- circular polarization (supplement) (NOTE: definition of right and left circular polarization in the supplement
is opposite to that in the textbook. Use the textbook version in the homework problem.)

- optical activity: chiral molecules in solution produce different indices of refraction for right and left circular polarization. This produces a phase shift between these components of light passing through Karo syrup which leads to a rotation of the polarization direction. When placed between polarizing filters, the syrup thus produces beautiful colors that chage as the angle between the polarizing filters changes.

- how a liquid crystal display (LCD) works.

- INTERFERENCE: demo of reflection from a soap film, explanation in terms of interference.

- DEMO's shown today:

     M4-02 NEWTON'S RINGS - PROJECTION
     M4-21 SOAP FILM INTERFERENCE - SIMPLE LARGE VERSION
     M8-03 OPTICAL ACTIVITY IN KARO SYRUP CYLINDERS
     M8-01 POLAROIDS AND KARO SYRUP

- total internal reflection of microwaves on wax prism. Frustrated total internal reflection.*

- polarizing filters, Malus' Law for transmitted intensity.

- superposition of linear polarizations to make new linear polarization; unpolarized light

- polarization by reflection: Brewster's angle

- polarization by scattering: example, polarization of scattered sunlight

* Today in class we did the demo of total internal reflection of microwaves from a wax prism.
Damon asked why, if there is an oscillating electric field on the far side of the prism surface
at which the reflection occurs, does not a wave propagate off in that direction. His intuition
was coming from the self-consistent process whereby a changing electric field produces a
changing magnetic field and vice versa. I think the missing intuition is that it is just
difficult to see what this implies about the spatial dependence of the fields in general.

There is an easy way to see that there can be a non-propagating oscillating field. This is
called an "evanescent wave". The trick: USE THE COMPLEX EXPONENTIAL METHOD!

Consider the prism surface to be the z=0 plane, with the prism at z<0, and look at a field that
depends just on x, z, and t, and has the complex form E(x,z,t) = E_0 exp(-iwt + ikx - mz)
outside the prism in vacuum for z>0. We know that this should satisfy the wave equation in
vacuum, which implies that w^2 = c^2(k^2 - m^2). As long as k>m this gives a real w, so an
oscillating solution in time. As z-> infinity E -> 0, so the field amplitude decays away
exponentially. As z-> - infinity the field amplitude blows up, however if there is a wax prism
at z<0 so the equation does not apply there.

We should really check if all the Maxwell equations can be satisfied, not just the second order
wave equation for each field component. To do so we'd have to specify the electric and magnetic
field vectors. For example, suppose only the y-component of the electric field is nonzero and
has the above form. Try the same form for the magnetic field, except in an arbitrary fixed
direction, plug into the Maxwell equations, and see if there is a solution for some direction of
B... Let me know if you find a solution like this!

- introduced polarization

- Doppler effect: relation between wave frequency at the source, observed wave frequency,  and motion of source, observer, and medium.  Extremely useful since frequency measurements give velocity information. See RHK v. 1 readings, and supplement page on case of electromagnetic waves. Also I didn;t discuss it in class but I put up a supplement page showing a derivation of the Doppler effect in the general case of moving source and observer, using a spacetime diagram. It's neat. Check it out.

- Using Huygen's principle to derive Snell's law

- Snell's law for media with inhomogeneous wave speed: e.g. mirage, seismic waves inside the earth

- Fermat's principle of least time: deriving Snell's law, understanding inhomogeneous media

- total internal reflection

- the electromagnetic wave spectrum

- laws of reflection and refraction (Snell's law) in terms of rays...

- concept of wave front : surface of constant wave amplitude, and ray : line perpendicular to wave fronts.  Reflection and refraction in terms of wave fronts.

- index of refraction, and its frequency dependence.

- Huygen's principle ("poor man's wave equation")

- Discussed momentum of em waves. To understand this we need the relativistic relation between energy and momentum. Instead of p = mv we have p = (E/c^2)v, where E is the total energy, including rest energy if any. For a particle with mass m,  E = mc^2/Sqrt[1 - (v/c)^2], so for low velocities these two expressions for momentum agree. A v approaches c, we have p = E/c. This holds for both massice particles and masseless "particles", and applies to electromagnetic radiation. Thus radiation with energy density u has momentum density u/c, and radiation with intensity I has a momentum flux of I/c. If this radiation is absorbed by a surface, it transfers a momentum I/c per unit time per unit area to the surface, which is to say that it exerts a
pressure, called radiation pressure, of this magnitide on the surface. If the radiation is reflected rather than absorbed, the momentum transfer is doubled, so the pressure is 2I/c. We computed the pressure 10 cm from a 100 W light bulb, and found 2.65  microNewtons/m^2. This is small but not THAT small. On the other hand without radiation pressure you would not be here: it holds the sun up!

- Showed a demo of refraction of light. Explained how refraction can be understood as being due to the slowing down of the part of the wavefront in the lucite. More on refraction Wednesday.

- "cutoff" when driving frequency is above frequency at which waves can propagate, as in chain of masses connected by springs

- the surprising amt of energy in gravitational waves from far away

- Bose-Einstein condensates: matter waves consisting af many particles in the same quantum state

- Electromagnetic waves: smallness of m₀e₀. Explains why noone noticed before Maxwell. Derivation of wave equation from Maxwell's
equations. Electromagnetic plane wave properties. (See supplement for details.) Generation of EM waves from oscillating charges and dipoles.
Energy density, intensity, and energy flux vector (Poynting vector).

-Demos: see lecture demo link.
K8-01 ELECTROMAGNETIC WAVE - MODEL
     K8-05 ELECTROMAGNETIC PLANE WAVE MODEL
     K8-42 RADIOWAVES - ENERGY AND DIPOLE PATTERN
    K8. L RADIATING DIPOLE ANTENNA
     L OSCILLATING CHARGES

- Differential form of Maxwell's equations.

- Maxwell's displacement current: The key point: without the displacement current, the equations are inconsistent with the change continuity equation. Why? Consider Ampere's law, curl B = m₀ j. For any vector field B, div curl B = 0, so Ampere's law requires div j = 0. But this is only true if the charge density is not changing! To correct this inconsistency Maxwell replaced j by j + e₀ d_tE. The second term is called the displacement current density. Since div E = r/e₀, the sum of the true current density and the displacement current density is divergenceless, restoring consistency.

- Integral form of Maxwell's equations

- Charge density r and current density j. Charge conservation expressed by the continuity equation,
r_t + div j = 0.

- Explained group velocity by reference to a wavepacket, as explained in the notes from the book by Blandford and Thorne. Briefly, consider a wave packet psi(x,,t) = \int A(k) exp(ikx -iw(k)t), where the amplitude A(k) is peaked at some value k_0 and has a width small compared to k_0. Imagine doing the integral. As k sweeps through the values near k_0, the integrand is generally oscillating and will tend to average out to zero. The integral will be largest for those values of x and t such that the phase (kx - w(k)t) is stationary with respect to changes of k. The condition that the derivative of the phase with respect to k be zero is 0 = d/dk(kx - w(k) t)  = x - (dw/dk) t. Therefore the integral is largest when x = (dw/dk) t, which means that the wave packet moves with the group velocity dw/dk, the derivative being evaluated at k_0 where the amplitude A(k) peaks.

- Example of the waves on the one-dimensional crystal (chain of masses connected by springs) from the last homework: the group velocity VANISHES at k = +/- Pi/a. This corresponds to the normal mode of the crystal in which alternating masses move in opposite ways. Nothing is propagating.

-Example of plane waves in 3 dimensions, also from the book of Blandford and Thorne. A plane wave exp(ik.x - iwt) is constant on the planes perpendicular to the wave vector k. The phase velocity has magnitude w/|k| and direction k/|k|. The group velocity for a wave packet composed of such plane waves is the gradient of w with respect to k, grad_k w. The group velocity need not be parallel to k. Plane waves may seem artificial , but any expanding wave in three dimensions will look locally like a piece of a plane wave when it gets sufficiently far from the source. For example,  a perfectly spherical  explosion will generate sound waves where the pressure variation is constant on spheres. Far form the source, these spheres look locally flat, and the wave locally appears to be a plane wave propagating in the radial direction.

- QUANTUM  MECHANICS

Wave function and Schrodinger equation for a free particle
Schrodinger's equation for a particle in one dimension:

i hbar psi_t =  - (hbar^2/2m) psi_xx

where hbar is Planck's constant h divided by 2Pi, m is the mass of the particle, and psi is the wave function. Because of the factor of i on the left hand side, all solutions to the Schrodinger equation must be complex. Numerically,  hbar ~= 2/3 eV-fs = (6.63/2Pi )  x 10^(-34) J-s. For macroscopic systems hbar is a TINY number, but for atomic systems it is of order unity. More specifically, the binding energy of an electron in an atom is of the order of eV (electron volt)  and the time to complete one orbit around the nucleus is of the order of one fs (1 fs = femtosecond = 10^(-15) s).

Physical meaning
The physical meaning of psi  is a "probability density amplitude", that is, the probability of finding the particle between x=a and x=b at time t is the integral from a to b of  |psi(x,t)|^2, the squared modulus of psi(x,t). Since the particle must be SOMEWHERE the integral of |psi(x,t)|^2 over all space must be 1. This condition is automatically preserved in time for solutions of the Schrodinger equation. A particle doesn't have a definite position, it has only a probability of being found in any given region.

Energy and momentum
The above equation describes a force-free particle, so it is to quantum mechanics what Newton's first law is to particle mechanics. For a wave function of the form  Aexp(-iwt)exp(ikx), Schrodinger's equation implies the dispersion relation hbar w = (hbar  k)^2/2m. This is the same as the relation E = p^2/2m between energy and momentum for a free particle, if one adopts the correspondence E = hbar w and p = hbar k. Equivalently, E = h nu and p = h/lambda, where nu and lambda are the frequency and wavelength. Lambda is called the de Broglie wavelength of the particle. In terms of E and p the above free particle Schrodinger wave is A exp(- iEt/hbar) exp(ipx/hbar). Note that although the Schrodinger equation does not at first look like a wave equation, since the time derivative is only first order instead of second order, its solutions are nevertheless like those of the wave equation. This is due to the magic of i. Replace the i by 1 on the left hand side and you get a completely different beast: the diffusion equation.

Wave packets, the uncertainty relation, and velocity
A particle with a definite momentum is evenly spread out in space, equally likely to be found anywhere, since |Aexp(-iwt)exp(ikx)|=|A|. (Actually this is a bit of an embarassment, since the total probability of being anywhere must be unity. No value of the amplitude A will produce this normalization. Therefore a particle with a precisely defined momentum is not physically meaningful. There must always be at least a very small spread in momentum.) A more localized particle is described by a wave packet  containing a spread of momenta. A particle localized perfectly at one point is described by the Dirac delta function, which consists of ALL momenta. (This extreme case is also not physically meaningful.) In general , the spread in position is at least inversely proportional to the spread in momentum: (Delta x) (Delta p) >= hbar/2 (it turns out the exact minimum is hbar /2 if Delta x and Delta p are defined as the standard deviations from the mean).  This is called the Heisenberg uncertainty relation.  As a wave packet  peaked at the wave number k evolves, it spreads out, but its center moves with the group velocity dw/dk evaluated at k. This velocity is hbar k/m , swhich corresponds to p/m , the particle velocity. So the velocity of a classical particle corresponds in fact to the group velocity of the quantum wave packet for the particle! (The phase velocity is half of this.)

External forces
If a force is acting on the particle, the Schrodinger equation must be modified, If the force is -dV/dx for some potential energy function V(x), the term V psi must be added to the right hand side of the Schrodinger equation,  i hbar psi_t =  - (hbar^2/2m) psi_xx + V(x) psi. This equation is to quantum mechanics what Newton' secondlaw (F = ma) is to particle mechanics.

Energy levels
Although a quantum particle can not have a definite position, it can have a definite energy. For example in an atom, the allowed definite energies form a discrete set and are called the quantizedenergy levels of the atom. Since energy corresponds to frequency, the energy levels correspond to solutions of the Schrodinger equation with a definite frequency, together with some spatial dependence: psi(x,t) = exp(-iEt/hbar) f(x). The spatial part f(x) must satisfy the time-independent Schrodinger equation  - (hbar^2/2m) f_xx + V(x) f = E f, written here in one dimension.Only for particular values of the energy E are there solutions to the spatial equation for f satisfying the boundary conditions. For a quantum particle, the "boundary condition" on f is that the total probability is one, i.e. \int |f|^2 dx = 1. These definite energy level solutions are in perfect analogy with the normal modes of a continuous vibrating system, like a string for example. So the energy levels of a quantum particle are the normal modes of the Schrodinger wave.

- Continued with dispersion and group velocity. Showed a film loop with a nice demonstration of features of group velocity of water waves: 1) phase velocity greater than group velocity, 2) forming a localized wave packet by adding several component waves.

- Explained how to form a localized wavepacket by adding an infinite number if harmonic components. If all k's are added with equal amplitudes one gets an infinite spike at the origin, and zero everywhere else. This is called the Dirac delta-dunction. If one adds with a "window function" A(k) one gets something more tame: phi(x) = \int  dk A(k) exp(ikx). I showed phi(x) in two cases, a square window and a Gaussian.
For the square window phi(x) falls like 1/x. For the Gaussian phi(x) falls exponentially.

- If the window function A(k) is peaked areound some k_0, and narrow compared to k_0, then the wavepacket that results will move with a group velocity equal to dw/dk evaluated at k_0. You can see this in the Mathematica notebook I posted on the homework assignment for this week. I'll try to copy this material to the supplemental link of the course web page. Evolving in time, a wavepacket in the case of dispersion spreads in time, but the center of it still moves with a characteristic speed, the groups velocity.

- In the case of waves on water w = Sqrt[gk], so that v_ph = Sqrt[g/k] and v_g = 1/2 v_ph: for every k the phase velocity is twice the group velocity.

- Dispersion and group velocity. Explained difference between phase velocity and group velocity. The textbook covers this in chapter 7. Phase velocity is the velocity of a single harmonic wave, v_ph = w/k. If w(k)/k is the same for all k, the waves are non-dispersive , otherwise, they are dispersive, since the wave packets, made from superposing harmonic components, disperse. Group velocity is the velocity of an interference pattern. We looked at just a pair of harmonic waves, and showed that the spatial pattern of beats moves in an envelope that moves at speed (Delta w)/(Delta k).  This was illustrated nicely with two overlayed transparencies with parallel black stripes of slightly different spacing. This is a Moire pattern.

-Examples of dispersive waves are waves on the surface of water, and waves on a chain of masses, as in hw8.

- Energy flux and intensity. I sent out an email discussion of this. Let me know if you need another copy. In class, we clarified (hopefully) some confusion between (energy per unit time) and (energy per unit time per unit area). For transverse waves on a string it makes no sense to talk about the area. Rather we can fix a point and talk about the energy per unit time flowing past that point, which is equal to the energy per unit length times the wave speed. The SI unit for this is the watt = W = J/s. For a wave that is spread out in the transverse direction, such as a sound wave or a compressional wave in a material, we can talk about the intensity,  which is the energy per unit time flowing through a unit area. The SI unit for intensity is W/m^2. The intensity is equal to the energy per unit volume times the wave speed. If a particular surface is fixed, the energy per unit time flowing though it is called the energy flux...although I can't guarantee that nobody will ever use the word "energy flux" synonymously with "intensity".

- Discussed  the 3-d wave equation Psi_tt = v^2 (del-squared) Psi, where del-squared = div grad = (divergence of the gradient) is also called the Laplacian operator. I emphasized that this operator is the dot product of two "vector operators", so it is a scalar, and in particular it is independent of the orientation of the coordinate axes. For a plane wave, i.e. a wave that depends on only one spatial coordiate, such as Psi(x,y,z,t) = Psi(x,t), this wave equation reduces the the 1d wave equation we have been using. We can also consider spherical waves, for which Psi depends only on r and t where r is the radius from some origin. The book quotes the fact that for functions of
r alone, the Laplacian becomes d^2/dr^2 + (2/r) d/dr. A convenient trick is to simplify this further by factoring a 1/r out of Psi, since [del-squared] (u/r) = (1/r) u_rr . This shows that Psi(r,t) = (1/r) u(r,t) satisfies the 3d wave equation if and only if u(r,t) satisfies the 1-d wave equation u_tt = v^2 u_rr, i.e. if and only if u(r,t) = f(r-vt) + g(r+vt) for any two functions f and g. The functions f and g describe outgoing and ingoing spherical waves respectively.

- Intensity of spherical waves: As an outgoing spherical wave pulse expands, the energy in it is conserved, but spread out over a greater area. The energy that flows through a surface of radius r per unit time is I(r,t) (4 Pi r^2). In order for this rate to be the same when the pulse reaches another radius, the intensity must be inversly proportional to 1/r^2.  That this is so can also be seen from the formula (Intensity) = (energy density) x (wave speed). The energy density is proportional to the square of the wave amplitude Psi(r,t) = (1/r) f(r-vt), which is inversly proportional to r.

- Dispersive and non-dispersive waves: (See French, pp. 230-234.) The waves we have dealt with so far keep their shape while propagating. The whole pulse simply translates by an amount v dt in a time dt. These are called non-dispersive waves. In general however waves do disperse, because different wavelength components propagate with different speeds! (For example, think about light propagating through water. The different speed for different frequencies produces the rainbow.) We started to look at the consequences of this for the motion of wave packets. More on Monday.

- Discussed the midterm for 20 minutes.

- Discussed reflection & transmission, in particular what is going on in the details of Fig 8-3. I said the students should read the math carefully there to understand how and why the pulse shape changes (squeezed on the right hand side). Physically, the reason is that the pulse is moving at a different speed on the right hand side. I also emphasized the limits of small and large mu_2 being like free and fixed ends, and discussed energy conservation here. In the case of mu_2-> 0, there is a transmitted wave, but it carries zero energy in the limit mu_2->0.

- Mentioned example of light partially reflecting and partially transmitting at a window. That' s why you can often see yourself reflected in a window but you can also see though the window.

- Started waves in 3d. See pg. 244.

- exam1

- Derived form of reflected pulses at fixed and free ends. Also covered case of a transition from one medium to another, where the wave speed changes. See pp. 253-259.

-Reviewed material for the exam: transients; physical nature of the two solutions for an overdamped oscillator; expressing z = a+ib as Aexp(i*alpha), and sin wt + 2cos(wt+pi/4) +cos(wt) as Re(Aexp(i(wt+alpha)); relation between initial conditions and normal mode amplitudes and phases. (Did I forget anything?)

- Regarding energy density, if the solution is a general one, y(x,t) = f(x-vt) + g(x+vt), then the kinetic and potential energy densities are
dK/dx = 1/2 T (-f' + g')^2
dU/dx = 1/2 T (f' + g')^2.
Three interesting remarks can be made:

1. If f=0 or g=0, dK/dx = dU/dx. For a purely rightmoving or leftmoving wave, the kinetic and potential energy densities are equal.

2. dE/dx = d(K+U)/dx = Tf'^2 + Tg'^2, that is, the TOTAL energy density is always just the sum of the total energy densities from the f and g solutions themselves.  The cross terms cancel.

3. dK/dx=0 if and only is f' = g', which is the same as f = g + C. If we assume that there is some point where both f and g are zero (as must be true if they have finite extent), then C=0, so f = g. Putting back in the arguments, this means f(x-vt) = g(x+vt). At t=0, it means just f(x) = g(x). Similarly, dU/dx=0 if and only if f = -g.

- We also watched two film loops of waves on springs. One showed superposition of pulses very nicely, the other showed reflection of pulses at fixed and free ends.

-Energy stored in a displaced string consists of transverse kinetic energy and potential energy of stretching. We worked out the amount of each form of energy per unit length, i.e, the inetic energy density and potential energy density. This is in the textbook (pp. 237-240.) For a purely left or rightmoving wave these two densities are equal.

- Measured the speed v of sound in aluminum, by measuring the length L of the aluminum
rod and the frequency f of the fundamental mode. Then  computed v = 2fL. This compared well with Sqrt[Y/r].

- Next we looked at sound. I'm going to go into this in some detail here, since I think the subject involves a beautiful application of statistical, thermodynamic, atomic, and Newtonian physics all together, and is not difficult to understand in elementary terms. Most of this material is in the textbook, but it is spread out and I think not as transparently explained as below.

First we listed speeds in different gases at different temperatures:
Air, 0 C             331 m/s
Air, 20 C           343 m/s
Helium, 0 C      965 m/s
Hydrogen, 0 C 1284 m/s
We'd like to understand how the sound speeds depend on temperature and on the atomic properties of these gases.

The analysis of the longitudinal waves in the rod can be applied to sound waves in
a gas, replacing the Young's modulus Y by the bulk modulus K. The question is then what is K, e.g., for air? The physical picture is that a gas is "elastic" not because of interatomic forces, but because of the pressure caused by collisions with the moving molecules. Newton was the first to consider this problem. He measured the speed of sound by timing echoes, and computed it using the above formula. I'm not sure on the history, but I think he computed K as follows.

The definition of bulk modulus is through the relation F/A = -K dV/V. In applying this to sound in air, the relevant force is due to the overpressure, that is the pressure responsible for the change in volume, which is over and above (or below) the ambient average pressure. Calling this overpressure dp, we have dp = -K dV/V, so K = -V dp/dV.

Newton assumed the temperature was constant, and used Boyle's law, pV=const. at constant temperature. Thus 0 = d(pV) = Vdp + pdV, so -Vdp/dV = p. This says that, quite simply, K=p.  So Netwon calculates v =  Sqrt[K/r] =  Sqrt[p/r]. Putting in pressure and density of air at standard temperature yields v = 289 m/s, whereas the sound speed is really v = 343 m/s. Newton thought this was pretty good, and it is indeed rather impressive that he got anything close, but clearly there is something wrong.

What's wrong is the assumption that the temperature is constant. The question is, is there enough time, as the wave goes by, for a little region that is momentarily compressed, and therefore heated, to conduct its heat away and come to equilibrium with the surrounding mass of air?  To assess this, let's compare the wave speed to the average speed of the molecules. The latter is determined by kinetic theory. The average energy of each (accessible) degree of freedom is 1/2 kT. (The "accessible" is stuck in there because according to quantum mechanics, if the minimum energy that a degree of freedom can store is quantized and much greater than kT, then it cannot be appreciably excited at temperature T, and therefore stores no energy.) Therefore the average of the x-component of the kinetic energy is 1/2 m <v_x^2> = 1/2 kT, so the root-mean-square of each component of the velocity is Sqrt[kT/m]. Now according to the ideal gas law, pV=NkT, so kT/m = pV/Nm = p/r, so according to Newton's calculation the rms speed of a molecule is EQUAL to the wave speed. Therefore, on average, the molecules are going just fast enough to transfer the wave action. There is no time to dilly dally around colliding with their neighbors in order to establish equilibrium. One student objected that the rms velocity is only an average, and there are molecules with larger velocity. That's true, however the relative number with higher velocities is not very large. In fact it decreases exponentially like exp(-1/2mv^2/kT), according to the Maxwell distribution.

To do better than Newton, therefore, we must give up the idea that the air in the sound wave maintains a uniform, constant temperature. Rather when a little region is compressed, it heats up adiabatically. So how do we compute K = -V dp/dV under the conditions of adiabatic compression? We need to know what is constant. The answer: energy! The work done on a volume of gas is dW = -pdV, and this must be equal to the change of internal energy of this gas, dU, since no heat has time to flow out of the volume. So to go further we need to know something about the internal energy.

For an ideal gas, U = N u, where N is the number of molecules and u is the average energy of one molecule. This is "ideal" in the sense that it neglects repulsive or attractive interaction energy between the molecules. This is a good approximation for air, since the molecules are electrically neutral and pretty far apart, of order  ten times their diameter. The average energy of one molecule is u = D 1/2kT, where D is the number of accessible degrees of freedom of the molecule. For helium this is just 3: the three translational degrees of freedom of position. For nitrogen N_2, or oxygen O_2, it is 3+2=5, three for the cm position and 2 for the angle of the axis. For sulfer hexafluoride SF_6 it is 3+3=6, since there is no symmetry axis and it therefore takes three angles to specify the orientation. In principle the molecules with more than one atom also have vibrational degrees fo freedom, but they are fairly inaccessible at room temperature.

Now we can compute the adiabatic bulk compressibility:

dW = -pdV = dU = d(Nu) = d(D/2 NkT) = D/2 d(pV).

Notice the difference from the isothermal (dT=0) case: instead of d(pV)=0, we have

d(pV) = -2/D pdV,

so

K = -V dp/dV = (1+2/D) p.

After all that analysis, the difference is just a factor of 1+2/D! So does that fix the disagreement? YES! Air is almost totally N_2 and O_2. Both of these have D=5, so we can just use the above result. The correction factor to the speed is Sqrt[7/5] = 1.18. Multiplying 289 m/s by 1.18 we get 342 m/s!! WOW! The factor 1 +2/D is usually called g, the adiabatic index. It is also given by the ratio of the specific heat at constant pressure to the specific heat at constant volume. I'm not sure but I think the result K = g p holds even for non-ideal gases, if g is defined in terms of the specific heats in this way. Thus the speed of sound in a gas is given by v = Sqrt[g p/r].

Finally, let's work out the temperature and mass dependence of the speed of sound.
We saw above using the ideal gas law that  p/r = kT/m. This shows that for a gas consisting of a single type of molecule, the sound speed is determined just by the temperature and the mass, independent of the pressure or density! This yields for the speed of sound v = Sqrt[g kT/m]. We checked whether this gives the right ratio of the speed of sound for Helium and air at room temperature. For helium g = 5/3 and for N_2 and O_2  g = 7/5. N_2 has 2x14=28 amu and O_2 has 2x16=32 amu. In class we used the average, m=30 amu, and got a result that was about 3% too large. But there is  roughly 79% N_2 and 21% O_2, so it would have been a better approximation to take it all N_2. Sqrt[30/28]=1.035, so that probably explains the difference. (It would be interesting to work out how to account for the mixture properly. I invite you to try!) As for the temperature dependence, the ratio of the speeds at T= 20 C = 293.15 K  and T= 0 C = 273.15 K  is Sqrt[293/273]=1.036, so the speed should be 3.6% faster at 20 C than at 0 C. This agrees nicely with the numbers given above.

- Next, we demonstrated standing waves (i.e. normal modes) on a long spring, and then on the same spring demonstrated travelling wave pulses, and superpositions of these.

- Talked about the wave speed for longitudinal compressional waves on an aluminum bar, and in air (sound waves). The speed is v = Sqrt[Y/r] for the bar and v = Sqrt[K/r] for sound,
where Y is the Young's modulus and K is the bulk modulus. These speeds can be inferred
from dimensional analysis, up to an overall numerical coefficient (which turns out to be 1!).

- Recall definition of Young's elastic modulus:
(force on rod per unit area) = Y (fractional change in length of rod), i.e. F/A = Y DL/L.
This can also be written as (stress) = Y (strain). The Young's modulus applies when the
stress is entirely along one direction.

-  The bulk modulus measures the elasticity when the stress is in all directions,
as a uniform pressure, and is defined by F/A = K DV/V.

- For sound in air, it turns out that K = (1 - 2/D)p, where p is the pressure, and D is the
effective number of degrees of freedom. More on this Friday.

- Derived the wave equation for longitudinal compressional waves on an aluminum
bar. I followed pretty much the way the book does it.

- Went over angular frequency, frequency, wave number, wavelength, period, wave speed,
and the relations between all of these.

- Analyzed the relation between the solutions to the mode equation that satisfy given boundary conditions, and the simple picture of fitting parts of wavelengths into a given length in a way that satsifies the conditions for location of the nodes and/or antinodes determined by the boundary conditions.

 - Derived normal modes of a guitar string by assuming y(x,t) = cos(wt)f(x). Boundary conditions  f(x=0)=0 and f(x=L)=0 selected f(x) = sin(n Pi x/L), and corresponding frequencies w_n = n w_1, with the fundamental frequency w_1 = Pi v/L.
- The zeros of f(x) are nodes, and the maxima are anti-nodes.
- Since every motion can be expanded in normal modes, every string shape function f(x)
can be expressed as a combination of the form Sum_n  A_n sin(n Pi x/L). This is called a
 Fourier series.
- Aluminum rod and longitudinal vibrations. The deformation of the rod is described by a function s(x,t), giving the longitudinal displacement of the part of the bar originally at x. Different boundary conditions: ends are free.
- If string end is free the bc is different too: suppose a ring of mass m is tied to the string end and passed over a frictionless post, so it can slide up and down. The vertical force on the mass comes then entirely from the string on one side, so we have m a_y = T y_x if the ring is on the left side. As the mass goes to zero, we get infinite acceleration unless the slope y_x goes to zero. Thus the slope at such a free end must vanish.
- Air also has compressional vibrations: sound. The restoring force is due to the pressure, which can be traced back to atomic collisions. The configuration of the air in a pipe can be described by the pressure p(x,t) (or by the average displacement of the molecules). At an open end, the pressure is fixed to be "room pressure", approximately 1 atmosphere, so it is a pressure node. At a closed end the pressure is maximal: an anti-node. We saw that a pipe open at both ends has different harmonics than one open at only one end. The fundamental of the latter is an octave lower than that of the former.
- Travelling wave solutions: TRICK: Let d_t be partial derivative wrt t, and d_x similarly for x. Then the string equation can be written

(d_t d_t - v^2 d_x d_x) y = 0.

Since partial derivatives commute, this can be rewritten as

(d_t - v d_x)(d_t + v d_x) y = 0.

Therefore y is a solution if it satisfies

(d_t + v d_x) y = 0    OR (d_t - v d_x) y = 0.

The first equation says simply that y(x,t) = f(x - vt), where f is ANY function! Similarly the second equation says y(x,t) = g(x + vt), where g is ANY function. The general solution is just a linear combination of the two. (We haven't proved this, but it's true.)

- If t is increased by T and x is increased by vT then the combination x - vt stays the same. Therefore if y(x,t) = f(x - vt), then y(x + vT, t + T) = y(x,t). This means that the shape of y just shifts without distortion though a distance vT in a time T: the solution describes a wave pulse moving to the right at speed v!

- Normal modes of four coupled pendula in a line.
- Normal modes of a guitar string; nodes, harmonics.
- Derived the equation of motion for a string! The configuration of the string is described by a function y(x,t) giving the displacement of the bit of string at position x at time t in the direction perpendicular to the string. We restrict to the case of small oscillations, so the slope of the string is always much less than 1. For example, if the guitar string moves 1mm and the length is 50 cm the slope is-oops! in class I said 1/50, but it should be 1/500! Also the length of the string is almost constant, so the string tension T is almost constant. We simplify the analysis using these facts. I'll reproduce the derivation here, since it might be helpful to see a slightly different discussion than the one in the textbook.

The string is characterized by the tension T and mass per unit length m.We look at a small segment of string stretching from x to x + Dx, with mass Dm = m Dx, and apply Newton's second law to the y-component of its motion,

  F_y = Dm a_y.

In the end we take the limit as Dx approaches zero, so we neglect many terms along the way that will not affect the result in this limit. The acceleration is just the second partial derivative y_tt of y w.r.t. t, so the right hand side is

  Dm a_y  =   m  y_tt Dx.

The force is the sum of the force pulling on the left and the force pulling on the right. The y-component of the force on the left is  -T y_x(x), where y_x is the first partial derivative of
y w.r.t. x. (The time dependence of the function y is suppressed here.) The y-component of the force on the right is +T y_x(x+Dx).  The net force in the y-direction is therefore
F_y = T [y_x(x+Dx) - y_x(x)]. Since we are going to take the limit where Dx goes to zero, we can expand y_x(x+Dx) = y_x(x) + y_xx(x) Dx + ... using Taylor's theorem and neglect the terms involving the square and higher powers of Dx. (y_xx here is the second partial derivative of y w.r.t. x.)  This yields

F_y = T y_xx Dx

The force F_y goes to zero as Dx goes to zero, because the pull on the left exactly cancels the pull on the right. The mass also goes to zero as Dx goes to zero, so both sides of Newton's second law go to zero. However before letting  Dx go to zero we can divide both sides by Dx.
This yields a finite equation in the limit, which applies at the point x on the string:

y_tt = (T/m) y_xx.

This is called a partial differential equation, since it involves the partial derivatives of the function y(x,t). It holds at every point along the string.

- "Superposition" of normal modes
- beats from the viewpoint of the complex plane
- normal modes of other systems, e.g. the Wilberforce pendulum and more than two coupled
  oscillators
- first look at the continuous string as a system of coupled oscillators

- Coupled oscillators, basically the material in pp. 119-126 of French.

- Discussed example of parallel RC circuit in detail.
- Discussed power in AC circuits.

- Discussed the analogy between series RLC circuits and damped harmonic oscillators.
- Showed oscilloscope demonstration of resonance in an RLC circuit.
- Introduced the method of complex impedance for AC circuits.

- First we discussed several homework problems.
1)We explained that in solving the driven pendulum problem, we assume the small amplitude approximation so we can use the sho equation. Once we're done, we check whether the oscillation amplitude is in fact consistent with the small amplitude approximation. Since the problem didn't specify explicitly the length of the pendulum, we need to work that out from the given period.
2) We discussed the nature of the third order differential equation describing the radiation reaction force on a radiating charged particle.
3) We discussed the comparison of damping rates for overdamped and critically damped ( or underdamped) cases. The main observation is that at late enough times, the term with the slowest exponential decay will eventually be larger than any other term.
-  Physics of power input: rate of work done by external force is Fv, which is positive if F and v have the same sign and negative otherwise. Thus for example for a driven undamped oscillator the net work done in a cycle is ZERO. Below resonance, the force and the position are exactly in phase, so the force and the velocity are 90 degrees out of phase. This means that the sign of the work done by the external force switches every quarter cycle. The same holds above resonance.
Of course it must work like this, since the undamped oscillator is not dissipating any energy, so there must be no average energy input. For a damped oscillator, on the other hand, the phase shift is intermediate between 0 and 180 degrees. In fact, at resonance,    the phase shift is exactly -90 degrees, and therefore the force and velocity are in phase, so the external force is always doing positive work.
-  Transients: The steady state solution of the ddo equation of motion is  found by assuming a harmonic time dependence. The amplitude and phase are uniquely determined. But the general solution must have two free paramters, corrensponding for example to the initial values x(0) and v(0). So how do we find the general solution? We exploit the linearity of the equation. If x1 and x2 are both solutions to the "inhomogeneous equation" (i.e. the equation with the external forcing term), then x2 - x1 is a solution to the homogeneous equation (i.e. the equation for the free, damped oscillator).Thus x2 = x1 + x_homog. But x2 is arbitrary, so we see that the general solution can be written in the form x = x_particular + x_homog, where x_particular is any solution to the inhomogeneous equation, and x_homog is the general solution to the homogeneous equation. The steady state solution x_ss is one very special particular solution to the inhomogeneous equation, so it is natural to use that one. Thus we have for the general solution x = x_ss + x_homog. The two free parameters in h_homog are determined for example by imposing the initial conditions. In class we worked out the example in detail where w = w_0,
x(0)=0=v(0). The homogeneous solution is exponentially damped at large times, leaving only the steady state solution.

- Reviewed ddo material from Monday.
- Energy and power in ddo. Basic idea is that the time average of energy is constant in steady state, so average energy drained by damping must equal average energy input. Time average power input P_avg(w)  is maximum at w = w₀. The full width at half maximum of P_avg(w) (FWHM) of
is approximately (exactly?) g. One can thus read off  w₀ and g from the resonance curve.

- Damped, driven harmonic oscillator (ddho). We first analyzed the undamped case, then turned to the damped case. This is all discussed in the textbook. One thing not explained in the textbook is  how the ambiguity of the arctan should be resolved to find the phase lag d. By looking at the complex amplitude this ambiguity is resolved. The result is that d goes from 0 to p rather than from -p/2 to p/2.

- Damped harmonic oscillator, d²x/dt² + g dx/dt + w₀² x = 0.  A complex function of the form z(t) = C e^pt is a solution if p satisfies the quadratic equation p² + g p + w₀² =0, i.e. p = - g/2 +/- Sqrt[(g/2)²- w₀²].  Identify three cases: overdamped, underdamped, and critically damped, according as g/2  is greater than, less than, or equal to w₀. We looked at the behavior in these cases in detail. In the overdamped case, there are two decaying solutions, with diffferent decay rates, the general solution being a linear combination of these two. In the limit of very large damping, one of the two solutions becomes constant in time. In the underdamped case, there is oscillation with a definite frequency, lower than w₀, and the amplitude decreases within an exponentially decaying envelope. In the critically damped case,  we get only one decaying solution for p, but we know there must be two independent solutions because the equation of motion is a second order equation. This means that our luck ran out: the exponential assumtpion for the form of the general solution has failed. The other solution can be found by taking a limit of the overdamped case as the damping approaches the critical value. The result is x(t) = (A +Bt) exp(-gt/2). Note that when there is no damping or restoring force, the critical solution becomes x(t) = A + Bt, which is just the solution for a free particle.  We determined the relation between A and B and the initial position and velocity of the system. The critically damped case is exploited to engineer devices that damp out motion as smoothly and quickly as possible. You might think the overdamped case would be better, but it isn't, since it has two solutions, one that damps more quickly and one that damps less quickly than the critical case, and the less damped solution would generally be present.
- Quality factor Q = w₀/g . Higher Q means less damping relative to oscillating.  We expressed the frequency shift in terms of Q, as well as the energy loss rate: the amplitude drops by a factor e  in a time 2/g, which corresponds to about Q/p oscillations. The energy is proportional to the square of the amplitude, so it drops by a factor of e in half this time.
- Driven harmonic oscillator: demonstrated this with a torsional oscillator. Energy is fed into the system with a harmonic time dependence, while energy is drained out by the damping force. A steady state is approached when the damping has had time to damp the transients. In the steady state, the amplitude is largest when the driving frequency is near the natural frequency. This is called resonance. (If there were no damping, this steady state amplitude would be infinite, since the driver would keep feeding in energy.) When the driving frequency is below resonance, the response in the steady state is in phase with the driving force. When the driving frequency is is above resonance, the response is out of phase.

-  discussed the effective spring constant problems from hw1
-  discussed series addition of springs
-  Noted exp(i Pi/2) = i, exp(i Pi) + 1 = 0
-  Noted Re(z) = (z + z*)/2, cos q = (exp iq + exp -iq)/2,  sin q = (exp iq - exp -iq)/2i.
-  showed how to find the polar form of complex numbers, and noted care is needed due
   to  the ambiguity of the arctan. We take Sqrt and arctan of positive numbers to be positive.
   The information about which quadrant a complex number is in then must be put in "by hand".
   For example,
        2 - 3i = Sqrt[13] exp(-i arctan(3/2)),        whereas
       -2+3i = Sqrt[13] exp(-i arctan(3/2)+iPi).
-  Showed how to put, e.g., 2 cos wt + 3 sin wt in the form Re(Aexp(is)), in two different ways:
   Method 1: Use cos wt = Re( exp(iwt)) and sin wt = Re(-i exp(iwt)); so 2 cos wt + 3 sin wt =
   Re((2-3i)exp(iwt)). Using the above we get Re(Sqrt[13]exp(iwt -i arctan(3/2)).
   Method 2: Write cos and sin in terms of complex exponentials and collect terms. This gives
  Re((2-3i)exp(iwt)).
-  Reviewed solution of h.o. by complex method: guess solution of form xhat(t) = C e^pt,
   allowing both C and p to be complex. The real solution we are after is then the real
   part of xhat, x(t) = Re(xhat(t)). The general complex solution has two complex constants
   of integration: xhat(t) = C₊exp(iw₀t) + C_- exp(-iw₀t) . The real part of this is
   Re(C₊exp(iw₀t) + C- exp(-iw₀t) ) = Re((C₊+ C_- *)exp(iw₀t)), which is obtained by using
   Re(z) = (z + z*)/2. So only the combination C₊+ C_- * affects the real part of xhat(t).

- Complex exponential: Euler's formula exp(iq) = cos q + i sin q. Derived this in two ways:
  (1) by comparing the series expansions of the two sides, and (2) by showing that the two
  sides are equal at q = 0 and they satisfy the same first order differential equation, df/dq = iq.
- Cartesian and polar forms of complex numbers.
- rotating vector description of SHM
- Solved the harmonic oscillator equation d²x/dt² + w₀² x = 0 by assuming x(t) = C e^pt. Found  an  algebraic equation for p, whose solutions are +/- i w₀. The real solution is the real part. See  p. 43 of French. To be emphasized: Since the equation itself is linear, and the coefficients are real, it follows that if z(t) = x(t) + i y(t) is a complex solution, then x(t) and y(t) are both real solutions to the equation. Also, the constant C is freely specified.
- Damped harmonic oscillator: consider a damping force F = -b v, where b is a constant and v = dx/dt. The damped harmonic oscillator eqn can be written as d²x/dt² + g dx/dt + w₀² x = 0, where g = b/m. We solved this by again assuming an x(t) of the above exponential form. See p. 62 of French.

Explained problem 7: The "relaxed length" l₀  is the length of the spring when it exerts
no force. When it has a length l the force is -k(l - l₀), and the potential energy is 1/2(l - l₀)².

Complex numbers:
- imaginary unit i = Sqrt[-1],  i² = -1.
- z = x+ iy, x = Re[z], y = Im[z]
- addition, multiplication, division of complex numbers
- fundamental theorem of algebra: any polynomial can be factorized to the form
  (z - w₁)(z - w₂)...(z - w_n), so any nth order polynomial equation has exactly n roots.
- complex plane, representation of complex numbers as vector with magnitude
  or "modulus"  |z| = Sqrt[x² + y²] and direction q = tan^-1(y/x)
        z₁ + z₂  as  vector addition
        iz is z rotated 90 degrees counterclockwise
       i² = ii = rotation by 180 degrees = -1: geometrical picture of Sqrt[-1]!
        z₁z₂ has magnitude  |z₁| |z₂| and direction q₁ + q₂
- complex conjugate: z* = x - iy,  (or z "bar", which I can't type)
       z is real iff z* = z,    z is imaginary iff z* = -z
      (z₁z₂)* = z₁*z₂*
       zz* = |z|²

Energy conservation:
- conservative forces & potential energy
- energy conservation as a consequence of F=ma
- F=ma as a consequence of energy conservation for one dimensional systems
- total mechanical energy of an oscillator
- electrical oscillator: RL circuit
- pendulum: harmonic oscillator in the small angle approximation
- Taylor expansions in general, and expansion of cosine
- physical pendulum: Kinetic energy in terms of moment of inertia and
  angular velocity; potential energy in terms of height of center of mass
- Young's elastic modulus

Went over syllabus.

Harmonic oscillator:
- differential equation and general solution
- amplitude, angular frequency, frequency, period, phase shift