Download This chapter is the second on electromagnetic waves. We begin with

This chapter is the second on electromagnetic waves. We begin with a discussion of electromagnetic
waves traversing a slab containing a transparent material (eg glass, plastic) with a constant mu and
epsilon – such as a transparent Class-A dielectric. Essentially what happens is that the light
propagation velocity gets lowered from c = 1/sqrt(epsilon_0 mu_0) to v = 1/sqrt(epsilon mu) which is
written as v= c/n where n > 1 is the index of refraction. We discuss the boundary conditions and as a
first exercise we apply them to a light wave at normal incidence on a transparent slab to calculate the
fraction of the light reflected from the boundary. We get the same reflection coefficient for light
normally incident on a dielectric as for a mechanical string wave hitting the junction between a light
and heavy string. In both cases the reflection amplitude ratio is r = (v2 – v1)/(v2 + v1) where for v1
and v2 are the speed of propagation for EM waves in the two media, or the wave velocities for the
stretched string. Although the physics of E&M waves and string waves is very different, it is amazing
that the reflection and transmission properties are the same. We turn next to a discussion of light at
non-normal incident and derive Snell’s law for the refraction of a light wave. We also obtain
“Brewster’s” angle which is particular incident angle where the reflected wave is 100% polarized in a
plane parallel to the dielectric boundary. We discuss total internal reflection and evanescent waves.
We discuss electromagnetic waves propagating conducting materials and derive the skin depth (d) in
which the electrical field in an E&M wave falls to 1/e of its initial value. Higher frequency waves have
shorter skin depths. We next describe a model for the index of refraction of in gases which is useful
for the identification of subatomic particles by Cerenkov detectors. We conclude with a discussion of
wave guides where microwaves travel down conducting channels. There is some interesting and
very useful physics here – much of which involves applying boundary conditions on E and B fields.
We also get some QM previews.
1
This slide begins by reviewing many of the changes of Maxwell’s Equations for
materials with magnetization and polarization. Recall our basic model is to write the
induced dipole moment per unit volume (P and M) as proportional to E and H times
electric or magnetic susceptibilities. In order to get consistent dimensions, we write
the P proportionality constant as epsilon_0 chi_e where chi_e is dimensionless. This
is because P has dimensions of charge/m^2 and E has dimensions of
charge/(epsilon_0 m^2). For the case of linear media where the polarization and
magnetization are induced by external electric and magnetic fields and are
proportional to the product of susceptibilities and fields, one can take into account
the material by writing Maxwell’s equations in terms of auxiliary fields such as D and
H, which are related to E and B through modified Ampere (mu) and Faraday
constants (epsilon). If these constants are uniform (eg not dependent on position
inside the slab) one gets the same wave equations for E and B as in vacuum except
epsilon_0 mu_0 gets replaced by epsilon mu – which is equivalent to replacing c by
c/n where n is the index of refraction. In most transparent materials mu is
approximately mu_0 but epsilon is significantly larger than epsilon_0, and hence
usually n = sqrt(1+chi_e).
2
We next summarize the difference in the E&M wave fields, energy expressions, and
boundary conditions for materials where mu and epsilon can be defined. E and B
are still transverse waves which are perpendicular to each other and transverse to
vec-k. But the associated B in media is larger than B in vacuum for a given E-field
since B=E/c changes to B = nE/c. The energy density and Poynting vector
expressions are the same as in vacuum except epsilon_0 , mu_0 goes to epsilon,
mu for the material through which the waves travel. I next review the boundary
conditions for continuity across the dielectric boundary. The component of the Efield parallel to the dielectric surface (E||) is continuous across the boundary. The
component of B transverse to the boundary (B_perp) is continuous as is the
transverse component of D or (D_perp). Finally H|| is continuous across the
boundary. The cartoons illustrate the reasons for these four boundary conditions by
showing Gaussian pill boxes and loops. The D_perp and B_perp follow from zero
divergence of the D and B field owing to the assumption of no free charge density
on the boundary and no magnetic charge ever. The continuity of H|| follows from
applying the integral form of the Ampere-Maxwell equations to the indicated loop.
Under the assumption of no free surface current and a zero displacement current in
the limit of an infinitesimal loop and finite E derivatives the total line integral must be
zero and hence H|| just inside the material equals H|| just outside. Similarly E||
continuity follows the integral form of Faraday’s law applied with the assumption of a
vanishing magnetic flux as the transverse dimension of the EMF loop becomes
infinitesimal. If mu and epsilon can be defined, we can convert the H and D BC to E
and B with appropriate mu and epsilon constants.
3
Consider normal incidence to a boundary with different transparent material slab that join on the 1-2
boundary at z=0. We write the complex electrical fields for the incident, reflected, and transmitted
waves with amplitudes E_I,E_R, and E_T. At normal incident we only need E|| and H|| boundary
conditions. Our E BC at z=0 is E_I + E_R = E_T which relates the three amplitudes. We have
drawn the E-fields assuming that the reflected wave is parallel to the incident light, but we need to
find the real sign from solving the BC. The B-field just inside region 1 is B_I – B_R. We need the (-)
sign between B_I and B_R since we need E cross B to point in the respective propagation direction
for both waves. We convert the H|| BC to a second E|| BC by using the fact that B_I = E_I/v1 and
B_R = E_R/v1 and B_T = E_T/v2. We use algebra to get the E_R/E_I and E_T/E_I amplitude ratios
in terms of beta. The more useful expressions describing reflection and transmission involve the
ratios of intensity or Poynting vectors. The time-averaged Poynting vector is just E* cross B/2mu and
B is E/v which means S = E^2/(2 mu v). We define T and R which are the ratios of the transmitted
intensity to incident intensity and the ratios of the reflected intensity to incident intensity. T and R are
easy to obtain from our field ratios and can be easily cast in terms of beta. If we assume that
insulating dielectrics are insulators then J=0 and E dot J=0 which means that: (1) there is no energy
dissipated in heat, (2) the divergence of S is 0 and (3) the Poynting vector in region 1 equals the
Poynting vector in region 2 (which you can show with Gauss’s Law pill box straddling the boundary).
This means R+T = 1 if we can ignore interference between the incident and reflected wave. In
homework you will show that this interference is zero and indeed our R + T expressions add to 1. If
we assume that the dielectrics are not magnetic materials (usually true for transparent materials) we
can write beta = v1/v2 = n2/n1 which is the ratio of the indices of refraction of the two slabs. When
written in terms of propagation velocities the ratio of the two amplitudes is exactly the same
expression as for transverse waves on a string – a remarkable coincidence! Lets try this formula out.
For a light wave traveling from air (n=1) to a pair of glasses (n = 1.5) , the R formula tells us that
R=[(1.0-1.5)/(1.5+1.0)]^2 = 4% meaning 4% of the light is reflected from the dielectric boundary –
noticeable but not huge. The E-field of reflected wave is 180 degrees out of phase with the incident
wave.
4
We now consider light approaching the boundary (at z = 0) between two dielectric slabs at non-normal
incidence. The result depends on the polarization of the incident light, which in this example we pick to be in
the x-z plane which is also the reflection-refraction plane. You will work out the case where vec E_I is
transverse to the plane (|| to y-axis) in homework. The figure shows the k-vectors and E-field directions for the
incident and reflected waves in slab 1 and the transmitted wave in slab 2. Three angles theta-1, theta_R, and
theta_T = theta_2 are defined for the incident, reflected, and refracted (or transmitted) wave which give the
angle of propagation with respect to the normal (z-axis). There are more unknowns and field-components for
the non-normal incident case since we don’t know theta_R or theta_T or differently put, the directions of the
three k vectors. Lets begin by finding the k-vectors. The critical observation is that any boundary condition on
the fields must apply at all points on the z=0 plane. Each boundary condition will be of the form ()exp(i k_I dot r
– i omega t) + ()exp(i k_R dot r – i omega t) = ()exp(i k_T dot r – i omega t) where the () corresponds to certain
components of E or B and k_I, k_R, and k_T correspond to the wave vectors of the incident, reflected, and
transmitted wave. We assume all three waves have the same omega. Since () are constants once the incident
wave is specified, the only way our BC can be satisfied at all points on the boundary is if the three exp(i k dot r)
are the same for all x,y on the boundary or vec r = (x,y,0). This translates to all three x components being equal
( k_I_x = k_R_x = k_T_x ). All three y components are also equal. ( k_I_y = k_R_y = k_T_y ) but they are all
zero. Once we know how the magnitude of the k-vectors are related, we can use these relations to relate the
three directions specified by theta_1 , theta_R and theta_T. We can relate each of the k-vec magnitudes to the
common frequency omega, since we know each wave must satisfy the wave equation or del^2 E = (partial^2
E/partial t^2)/v^2. Hence |k| = (omega/v)= n omega/c for each of the three waves. We get our angle relation by
equating the x components of the k-vector for all three waves where we extract the x-component by applying
trig three triangles in the figures. Since the reflected and incident wave are both in slab 1 and have the same n,
it follows that theta_1 = theta_R. The sin(theta) for the transmitted and incident waves depends on the ratio of
n1/n2 and the relationship is known as the (hopefully) familiar Snell’s law. We thus rather easily obtain the two
laws of geometrical optics. Now we calculate field strengths and intensity ratios by using two boundary
conditions that apply to the E-fields: continuity of E|| and D_perp = epsilon E_perp. The E|| components depend
on the cosine of the normal angles and the E_perp components depend on the sine of the normal angles. The
cosines ratios are called alpha and the sine ratios are called beta. We can easily calculate alpha and beta
using Snell’s law when we need to. We get fairly simple expressions for the reflected to incident E-field ratio (or
r) and the transmitted to incident field ratio or (t) in terms of alpha and beta. It is interesting that we don’t need to
use the magnetic boundary conditions. This derivation is known as the Fresnel equations.
5
We next wish to compute the reflection and transmission intensity ratios or T and R.
Again these are basically the propagation velocity ratios times the ratio of the
square of the electric field. Inserting the field ratios we found on the previous slide,
we have simple R and T expressions as a function of alpha and beta. Interestingly
enough we find that R+T does not sum to 1 suggesting a violation of the
conservation of energy. What happened? The trick is we need to be careful to
interpret the Poynting vector when applying the conservation of energy. Lets
consider the time-average energy flowing in and out of an area on the dielectric
boundary that we parameterize as d area-vec which we direct normal to the
boundary. The power flowing into the area is the integral of the Poynting vector
dotted into the area vector. For the left (n1) side we include the Poynting vector of
the incident and reflected wave. The reflected wave has a negative contribution
since energy is flowing against the “inward” da vector. For the (n2) side we use the
Poynting vector of the transmitted (eg refracted) ray. To conserve energy we want
these two powers to be the same, and we find our field ratios do indeed conserve
power. We show the field ratio and intensity ratios as a function of theta_1 for the
“glass” case where n2/n1 = 1.5.
6
We note that the R coefficient disappears when alpha = beta, this means that none
of the light polarized in the refractive plane is reflected and thus the reflected light is
100% polarized perpendicular to the refractive plane. The magic incident angle
which insures this is called Brewster’s angle or theta_1 = theta_B. We set alpha to
beta which gives us conditions relating the sines and cosines of theta_1 and
theta_2. We then eliminate theta_2 using the fact that sin^2 + cos^2 = 1. We can
solve the resultant equation and for sin(theta_1) as a function of beta. We can find
cos(theta_1) from sin(theta_1). The tangent of the angle of incidence is particularly
simple : tan (theta_1)=beta = n2/n1=tan(theta_B). We can then use Snell’s law to
find sin(theta_2) and we find sin(theta_2) = cos(theta_1) which means that theta_1
and theta_2 are complementary angles which implies that the transmitted ray is at
right angles to the reflected ray when we are at Brewster’s angle. In homework you
will compute R for the case where the polarization is transverse to the scattering
plane. We plot R versus theta_1 for the case we worked out (red) and the case you
will work out (blue). As you can see when theta_1 is close to zero we recover the 1d result and the reflection is very small for either polarization. You can see that the
red curve kisses the axis when theta_1 is at Brewster’s angle. Finally note that the
blue reflection curve is much higher than the red once theta_1 exceeds roughly 60
degrees. This is the case of glancing reflections and over much of this range the
reflected polarization is parallel to the road. This is why Polaroid sunglasses are
aligned to block the road parallel polarization are so effective at reducing road glare.
7
Let us reconsider total internal reflection which you first learned about in Physics 212. We have an
incident and reflected wave in a slow medium with n > 1. The dielectric boundary is at z=0. By
Snell’s law the sin (theta_R) will be larger than sin_theta by a factor of n. But what happens if n
sin(theta) >1. Since -1<= sin (theta) <= 1, evidently there can be no refracted light wave and we say
100% of the light is reflected from the boundary. But this does not mean that there are no electric or
magnetic fields past z > 0. These z > 0 fields form what is known as a “evanescent wave”. We will
learn that the evanescent wave transmits no energy past the z=0 plane. The simplest case is when
the incident E is polarized in the y direction. In order to satisfy the BC we know that k_x for the
reflected, and transmitted wave must equal k_x of the incident wave which is just k sin(theta) or (n
omega/c)sin(theta). We know that the electric and magnetic fields for the transmitted (or evanescent
wave) satisfy the wave equation which means |k_T| = omega/c since the transmitted wave
propagates with v=c. This allows us to compute the k_z component of the transmitted wave since we
know k_y = 0 and we know |k_T|. Assuming that we have a large enough theta for total internal
reflection, n sin(theta)>1 which means that the k for transmitted wave has an imaginary z component.
Since the wave propagates as exp(i vec k dot vec r) its z dependence is exp (i [+/-i] gamma z) =
exp(+/- gamma z). We throw out the exp(+ gamma z) possibility since it leads to an infinite E-field at
z = infinity. Hence the E field falls off exponentially with z in the z > 0 region. So there is an electric
field but is there energy flow? To answer this , we need to construct the Poynting vector which
requires we know the B-field. For a harmonic wave, Faraday’s law allows us to compute the B field
and we get the standard result that B is proportional to the cross product of k_T and E but now k_T is
complex. We thus get the time averaged S in terms of double cross product which we can simplify
using the bac –cab rule. Power flow in the z > 0 region is proportional to z-hat dotted into S. Only the
z component of k_T will contribute which is imaginary and our time average depends on the real part
of z-component times i gamma E* dot E which is imaginary. Hence there is no energy flow into the
z>0 region but there are electric and magnetic fields. In fact you can demonstrate the evanescent
field by placing a matching prism close to the total internal reflection prism. This demonstration is
usually done with microwaves with a wavelength on the order of cm and a diagonally cut block of
rock salt forms a good prism. When the two prisms are placed at a separation on the order or less
than the wavelength, the microwaves can blast through and be detected – indicating that fields really
exist in the gap region. A very similar thing happens for electrons tunneling through a energetically
forbidden barrier in quantum mechanics. The wave function in the barrier also falls exponentially.
Not only is quantum tunneling real but it is technologically important since it leads to scanning,
tunneling microscopes.
8
We change topics and discuss the propagation of E&M waves in a conductor. This is remarkably simple to
explain and calculate. We start with Maxwell’s equations for the curl of the E and B fields. The only substantive
change is to add the Ampere’s Law term which is mu J and for an Ohmic material this is J is due to the electrical
field and thus J = sigma E. We then set up the magnetic wave equation in the standard way by taking the curl
of the curl. We get the usual wave equation along with a mu sigma (partial B/ partial t ) term. Amazingly
enough we get the same wave equation for the electric field. We begin with the case of an EM wave normally
incident on a conducting slab. We start with the familar exp(i kz – i omega t) form for plane waves propagating
in the z direction into the conductor. The electrical field is transverse to the z-direction– lets say the x-hat
direction. Inserting the exponential form into the modified wave equation, we get a complex k with a real and
imaginary part. The real part of k describes how the phase of the E&M wave varies with position and is
essentially the same expression as the k for waves traveling in a dielectric with epsilon and mu. The imaginary
part is sigma omega mu. This imaginary part will cause the fields to fall off exponentially in z/d where d is the
“skin depth”. The skin depth is easy to calculate in the limit where the imaginary part of k^2 is much larger than
the real part which is the good conductor limit. We can also write our good conductor limit condition in terms of
frequency. Our limit corresponds to the wave frequency being much smaller than the reciprocal of the RC time
constant of the conductor, which as you recall from Physics 435 depends on only epsilon and sigma and not the
electrode geometry. I briefly review the reasoning we used to calculate the time constant for two arbitrary
electrodes immersed in a uniform conductor. This is a nice argument where we compute the current leaving an
electrode and the charge stored on the electrode in terms of an E-field integral just outside of one electrode
using Gauss’s law and J. We can think of the RC time constant as the characteristic time for charge to reach
equilibrium (eg cancel). If the period of the EM wave is much longer than the discharge time we are in the good
conductor limit. In this limit the skin depth is given by d=sqrt{2/(omega mu sigma)} which we obtained by taking
the sqrt of i written in polar form. Here is some data on the skin depth versus frequency for various materials on
a log-log plot where power law translate to straight lines. I plotted a 1/sqrt{omega} as a dotted line and you can
see all the material curves are parallel to this. The Cu,Al, and Pb skin depth lines get larger as the conductivity
gets smaller as one expects from our formula. The Fe line has the smallest skin depth even though it is a
relatively poor conductor. But iron is magnetic which more than compensates for its relatively poor conductivity
which makes sense since our skin depth expression is proportional to 1/sqrt(mu). Good conductors have very
small skin depths and are opaque to visible light. To illustrate this we use green light with a wavelength of 510
nm. For Cu (an excellent conductor) we have a very small skin depth for green light of a few nm. For water, a
very poor conductor, we have a 10 meter skin depth which is quite transparent.
9
One other interesting feature of harmonic E&M waves traveling in a conductor is
that the magnetic phase differs from the electric phase where as they have the
same phase when traveling in vacuum. We can find the B-field from Faraday’s law
and again find vec B = vec k cross vec E/omega which is the same relation for
vacuum traveling waves. But now vec k is complex and its imaginary parts will
create an additional phase shift for the magnetic field compared to the electric field.
In the good conductor limit k is approximately (1+i)/d which means that the B-field
phase is shifted relative to the E-field by 45 degrees.
As we will show our conductor model implies that high frequency microwaves travel
on a surface layer of a good conductor with a thickness on the scale of the skin
depth. We essentially get the same behavior as for EM waves normally incident on
a conductor that we just discussed as we will show. We put the current density in
the z-hat direction of a cylindrical coordinate system which means we can just take
the usual (scalar) Laplacean of the z-component and parameterize J as a function
of s. We tabulate the skin depth in microns for 4 conductors for a 10 GHz
microwave. The layer is very thin at these frequencies. You will show in HW that as
the skin depth decreases with increasing frequency, it takes increasingly more
power to transport current over distances. At some point it is cheaper to replace
wires with wave guides.
10
We write out the equation for current density in the high conductivity-high frequency limit. We obtain
a double derivative term (T1) and a single derivative term (T2). We expect to recover an exponential
solution in s similar to the exponential fall off in z/d for EM waves impinging normally on a conductor
discussed the previous slide since in both cases E is parallel to the conductor and close to the
surface a large round wire will look like a conducting slab. If we try this rough solution we see the
ratio of the second to the first term is proportional to d/s and we can drop T2 in the limit s >> d. This
will be the case with s near the surface of a round wire with a radius ( R) much larger than the skin
depth (d). Dropping the T2 term and switching variables to xi = R –s which is zero on the surface, we
get the same solution for k as we got for EM waves impinging normally on the surface as expected.
We have the same skin depth behavior on how J falls with increasing xi and the same phase
variation as a function of xi which can cause the current deep inside the wire to actually have the
opposite phase of the current near the surface as illustrated in the cartoon.
What if the radius of the wire is comparable to the skin depth? Since we essentially have a wave
equation describing E (or J) with a complex k with cylindrical symmetry, we can borrow our old result
on the RF capacitor from Physics 435. There the solution was a Bessel function in k s. Presumably
the same thing must be true for this problem except now we have a complex k. Of course the form of
the capacitor E field (peaked towards s=0 and slowly falling with increasing s), is very different from
our situation of maximum E falling exponentially as for deeper s. Just like cosh is essentially cosine
with a complex argument, the complex Bessel function can be written as a real function called “Ber”
and an imaginary function called “Bei”. These are called Kelvin functions and their properties can be
easily googled.
11
We turn next to a simple model for dispersion or the variation of the index of refraction with
wavelength or frequency. This is responsible for separation of colors in a beam of white light by a
prism. We will concentrate on the index of refraction for gases which I found to be very useful in my
research. For non-magnetic materials (where mu=mu_0) the index of refraction is the sqrt(epsilon
/epsilon_0) = sqrt(1+chi) where chi is the electric susceptibility. The susceptibility is the induced
dipole moment per unit volume divided by the electrical field that induced it. In order to compute the
susceptibility, we model the dielectric molecules in the as mass spring systems. Such a model can
describe molecular states by natural mass-spring frequencies (omega_0= sqrt{k/m}). We can put a
realistic “width” in the spectral lines by broadening out the mass-spring resonances with velocity
damping. We begin by writing F=MA for the oscillating mass. This includes the force due to the
electrical field (qE_0 cos (omega t)), the spring force -m omega_0^2 x and the damping force –m
gamma dx/dt. We quickly switch to exponential forms. We compute the steady state (eg ignore the
homogeneous or transient solution) solution. Not surprisingly the oscillation amplitude is proportional
to the external E-field, and maximizes as omega approaches omega_0, but does not go to infinity
because of the damping contribution to the denominator. To get the susceptibility chi , we multiply by
an additional q (to get a dipole moment), multiply by the molecular density N (to get the dipole
moment per unit volume) and divide by the epsilon_0 electrical field (to convert dipole moment
density to susceptibility). The damping term makes our susceptibility a complex number. Since our
“molecule” consists of a single spring-mass system, we just have a single resonant frequency.
Actual dielectrics have a mix of frequencies. We include this feature in our model by considering a
mix of spring-mass frequencies with “strengths” f_j, natural frequencies omega_j and damping
coefficients of gamma_j. For transparent, solid dielectrics such as glass or plastic the important
resonant frequencies are in the UV and hence lie above the frequencies of visible light. This means
the susceptibility and index of refraction gets larger as the frequency increases. Hence a glass prism
will bend blue light more than red light. We now consider the index of refraction of gases. For gases,
the molecular density N is very low (down by a factor of ~1000 relative to solids) which means we
can expand sqrt(1 + chi) as 1+chi/2. We write an expansion for n in this limit and write the real and
imaginary part of n and k. The imaginary part will cause the fields to die off as they get deeper in the
dielectric (as it did in our discussion of skin depth), while the real part is the traditional index of
refraction.
12
Here is a table of the index of refraction of gases at STP. All of them are fairly close
to 1 but the difference from 1 varies considerably from 36 ppm for He to 490 ppm
for CO2. Generally speaking to get a large index of refraction one wants lots of
resonances which implies complicated gases with many atomic or molecular states.
As an illustration we note that the (n-1) value for hydrogen is over three times larger
than that for He. This is because He is monatomic and missing the molecular
vibration pieces which are present for diatomic hydrogen. One can more or less
tune the index of refraction of gases, even at STP by using gas mixtures. One use
for tuned gas mixtures is particle identification through the Cerenkov effect (circa
1930). Cerenkov radiation is emitted by subatomic particles that travel faster than
the speed of light in a medium or v > c/n. Gas Cerenkov counters are particularly
useful in high energy particles where the charged particle created in a high energy
collision travel very close to the speed of light. As we will show in the relativity
chapters, the velocity of particle in the high energy limit is given by the boxed
expression and at accelerator labs such as Fermilab is within a few parts per million
of the speed of light. One can often identify particles just by the presence or
absence of a Cerenkov light in a system consisting of say three Cerenkov counters
with different n. The reason this is possible is that there are only five reasonably
stable charged particles( e mu pi K P) that can make it through the several meter
long counters without decaying so only a few yes-no Cerenkov decisions are
necessary to uniquely identify particles.
13
Here is an example of a spectrometer (for the FOCUS experiment) which I helped to build. Three
Cerenkov counters are used, Nitrous oxide (aka laughing gas for high refractive index), He-N2
(medium index), and He (low index). These are interspersed among magnets, and particle detectors
used to precisely measure the particle momenta. Each Cerenkov counter features an array of
photomultiplier tubes (you can see their photocathode in the image) so that several particles can be
identified at the same time. The histogram (on semi-log scale) shows how the FOCUS Cerenkov
system is used to clean up backgrounds to one of the signals that FOCUS was designed to study.
The relativity lectures will describe how the mass of the parents can be reconstructed by measuring
the momentum of the daughters using the illustrated spectrometer. In this case the parent, called a
D-meson, decays in about 413 picoseconds into a kaon and three pions. If one just plots the mass for
all 4 track combinations, one sees a Gaussian bump (or peak) at the right mass but over a very large
smooth background of random combinations. If one requires one of the tracks is more likely to be a
kaon than pion, much of the background disappears and the signal becomes much easier to resolve.
If the likelihood requirement is increased, a much cleaner peak (albeit with some event loss) results.
If one requires that the remaining three tracks have a “pion-like” Cerenkov response, one gets a very
clean charm peak which is easy to study. We developed a monitoring tool based on the probability
that the two tracks that “verticize” in a vee have a Cerenkov response consistent with the decay
kshort => to two pions which is plotted as a function of run number for the three Cerenkov counters.
We monitored the Cerenkov light since the kshorts are easy to identify through their decay topology.
We noticed that the Cherenkov likelihood of all three counters fluctuated in a similar way and finally
tracked it down to the common barometric pressure. Recall that the susceptibility contribution to the
index of refraction is proportional to the molecular density which for a gas is proportional to the gas
pressure which for an atmospheric counter proportional to the ambient barometric pressure. Our
Cerenkov system applied to kshort decays has the side benefit of acting as an accurate (albeit
expensive) barometer – but fortunately has other uses as well!
14
We begin our discussion of wave guides by reviewing the boundary conditions for a
perfect conductor since we assume our wave guides are constructed out of perfect
conductors. Basically the electric and magnetic fields vanish inside of the
conductors but if a wave is transported by the wave guide in the regions outside of
the conductor, there will be fields outside of the conductor. The standard BC of E||
and B_perp continuity apply which means that there can be a E_perp and B|| just
outside of the conductor. The way this can happen is by surface charges and
surface currents. A surface charge can cancel the E_perp inside of the conductor
and allow an E_perp just outside of the conductor. Similarly a surface current can
cancel the B|| inside of the conductor while allowing a B|| just outside of the
conductor. We can find the required surface charge using a Gauss’s law pill box
and the surface current using an Amperean loop. The surface charge is related to
the surface current via the continuity equation which is the divergence of the surface
current vec K cancels the rate of change of the surface charge. The continuity
equation provides an important check of the fairly complicated fields present in a
wave guide. To reiterate – E|| and Bperp must disappear just outside the walls of
wave guides.
15
We begin with a very simple wave guide that illustrates most of the important points that apply to the general rectangular
wave guides which we discuss on the next slide. Here the E&M wave bounces between two infinite conducting sheets
separated by a distance a. We consider the case where the wave is polarized along the y axis and travels along the z-axis.
We write the electrical field as a superposition of an “incident” harmonic wave (in exponential form) traveling upwards with an
x component of K_1 and a “reflected” wave traveling downwards with a x-component of –k_1. The “incident” and “reflected”
wave vectors have the same z-component of k_3 and are 180 degrees out of phase (since the reflected wave has a (-)
amplitude). We put in this (-) sign so that E_y vanishes at x=z=0 since E|| must vanish at the surface. In homework, you will
further justify this form for the reflected wave but it makes sense since the k_1 must change sign so it travels in the opposite
direction as the incident wave and k_3 must be the same so the angle of reflection is the same as the angle of incidence like
a pool shot off of an ideal cushion. We have drawn E and B fields for the incident and reflected waves. You can see E cross
B points in the k-direction as it should for both the incident and reflected wave. This solution is also consistent with the
surface continuity equation. The E-field is zero at the boundaries , it has no E_perp component so the surface charge
density is always sigma=0. This means the surface current vec-K should have no divergence. As you can see from the
figure, the B_x components cancel but there is a net B-field parallel to the z-axis at the conductor surfaces which means
there will be an oscillating current density in the x-hat cross z-hat direction which is the y-direction. But since we have an
infinite plane in the y direction the B field and current density cannot depend on y thus divergence of K = partial K_y/partial y
=0 which is consistent with a a constant, zero charge surface density. As usual, our convention is the physical field is the real
part of the complex field hence E is proportional to y-hat sin(k1 x) cos(k3 z – omega t). This automatically vanishes on the
lower plate at z=0 but the E|| BC says it must also vanish on the upper plot at x=a. We thus must choose k1 so we have a
node of the sine function at x=a which means k1 = m pi /a where m = 1,2,3, …. Essentially k1 is “quantized”. Since E
satisfies the wave equation, we know |k| = omega/c = sqrt(k1^2 +k3^2). Hence k_3 which we will call k = sqrt(omega^2 –
omega_m^2)/c where omega_m = c m pi/a. Unless omega > omega_m , k will be imaginary and the E-field will die off rather
than propagate. We call omega_m the “cutoff frequency”. We will find that all wave guides (with one interesting exception)
have cutoff frequencies. This one is very simple to understand since we know the E-field must disappear at the two
conductors. These “nodes” must be separated by at least one half-wavelength so lambda/2 <= a. We know we know for
waves the speed of light is the product of lambda and frequency and thus lambda (omega/2pi) =c or lambda = 2 pi c/omega.
Inserting this in our inequality on lambda we have omega > pi c/a which corresponds to our m=1 cutoff frequency. The E&M
waves that propagate down a waveguide are characterized by whether or not their fields are transverse to the direction of
propagation. In this case the waves (initial plus reflected) are propagating along the z-axis. Since E is along the y direction,
it is transverse to the z-axis so it represents a TE of “transverse” electric wave. The B-field points along the z-axis and is
parallel to propagation direction hence we do not have a TEM wave where both the electric and magnetic fields are
transverse to the propagation direction. Rectangular wave guides, like this one, support either TE or TM but not TEM waves.
Finally lets look at the velocity of propagation. One measure is the phase velocity or the velocity that corresponds to kz –
omega t = 0 => v_phase = omega/k. If we compute this we find v_phase > c but nothing can travel faster than the speed of
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light! Evidently the phase velocity is bogus. A more complicated analysis (involving wave packets) indicates that energy and
information flows at the group velocity which is partial omega/partial k rather than the simple ratio of omega/k. The group
velocity is less than the speed of light and in fact is c cos(alpha) where alpha is the angle the incident or reflected wave makes
with respect to the z-axis as shown in the figure. Having the energy flow with this speed makes a great deal of intuitive sense
since as alpha -> pi/2 the waves bounce between the two conductors without moving in the z-direction at all.
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We next consider the case where the wave guide has finite extent of “b” in y as well
as “a” in x. We use a traveling wave in z and t and where the “transverse” factors
are functions of x and y. We use Maxwell’s equations and boundary conditions to
find the transverse functions and only the sourceless, curl equations are relevant.
These relate the curl of one field to the time derivative of the other. The problem
with these curl equations is that the electric and magnetic fields are badly coupled.
We adopt a rather unusual strategy to uncouple them. We endeavor to write the
transverse field components (Ex,Ey,Bx ,By) in terms of derivatives of the
longitudinal components Ez and Bz. This allows us to write all of the
electromagnetic fields in terms of either Ez(x,y) or Bz(x,y) since we show towards
the end of this chapter that for a hollow wave guide we can either have Ez =0
(transverse electric or TE ) or Bz =0 (TM) but not both (no TEM waves). We
illustrate the strategy for Ex. To solve for Ex algebraically, we take advantage of the
fact that our traveling wave form allows us to write the z-derivative and the tderivative of Ex in terms as ik Ex and –i omega Ex. The time derivative of Ex
involves the x component of Faraday’s law and the z derivative of Ex involves the y
component of the Ampere Maxwell law. The bad news is that these AM law and
Faraday’s law components also involve the time derivative of By and the z
derivative of By. But the very good news is that these By derivatives can be written
as simple multipliers of By and thus getting rid of the By term is simple. With some
simple “complex” algebra, we can write Ex and all the other field components in
terms of derivatives of either Bz or Ez. All we need are the Ez or Bz functions.
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We start with TM solutions by finding the equation for Ez. We start with the wave
eqn applied to our Ez form and try the standard separation of variable approach.
Our X and Y factors are linear combination of sines and cosines. The vanishing of
Ez on each x=0 or y=0 plane means we can only use the sines and not the cosines.
The vanishing on the x=a or y=b planes we need an integral number of ½
wavelengths which implies k1 = m pi/a and k2=n pi/b. We thus have a cut off
frequency that depends on m and n where m and n can be m,n = 1,2,3 ... We
excluded m = 0 for two reasons: (1) if m or n is zero we have E_z =0 as well as
Bz=0 which implies a forbidden TEM wave. (2) If m=0, k1=0 and the y derivative of
sin(k1 x) sin(k2 y) vanishes since the factor sin(k1 x) vanishes, and the x derivative
vanishes since it is proportional to k1cos(k1 x) =0 and thus all three components of
E vanish according to our derivative rules. The same two flaws occur for n=0. Thus
the lowest cutoff TM mode is TM11 where we use the notation TMmn. In the next
slide, we will see that it is possible to have TE01 or TE10 waves, however.
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Lets illustrate the general wave guide technique for the a by b wave guide that we discussed earlier.
We have the choice of solving for TE or TM waves. Lets choose TE waves as we did before which
means Ez=0 and Bz is non-zero. We start by solving the modified wave equation for Bz. Since this
is a partial differential equation in x and y we will use a separation of variable solution of the form
X(x)Y(y) – just like we did for Laplace’s equation in electrostatics. Since our Bz equation involves
double derivatives for X or Y we will have two solutions for X and two solutions for Y or 4 possible
solutions. We will pick the solution that satisfies the BC which are conditions on the transverse fields
so we will use the four Bx,By,Ex, and Ey equations. But lets start with Bz. We insert the Bz =
X(x)Y(y) form, perform the usual separation of variable manipulation (i.e. divide by XY) and obtain a
simple linear DE of the form partial X/partial x = - k1^2 X and partial Y/partial x = - k2^2 Y that have
trig solutions of the form X = sin k1 x or X= cos k1 x and Y=sin k2 y or Y= cos k2 y. Inserting these
solution into the Bz equation, we get an expression for k as a function of the constants of separation
k1 and k2 and the frequency omega. But which of the 4 possible solutions will match BC? We can
try all four or think about the solutions. We want the Bx to vanish at x =0 and By to vanish y =0
because of the B_perp BC. The Bx and By involve derivatives of Bz with respect to x and y. If we
choose X = sin(k1 x), Bx will be the derivative of sin(k1 x) which is cos(k1 x) and won’t vanish at x=0.
Evidently we want the cosine solution so Bx will be sin(k1 x) and match our BC. The same reasoning
leads us to select Y= cos(k2 y) so that By vanishes at y=0. We thus have a unique choice for our
solutions which gives us (sin k1 x)( cos k2 y) solutions for Bx and (cos k1 x)( sin k2 y) for By. We
also want Bx to vanish at x=a and By to vanish at y= b which means that k1 and k2 must be chosen
to be integral multiples of pi/a and pi/b. We can insert these expressions into our k expression to get
the cutoff frequency omega_mn. All integer choices are possible except for m=n=0 which results in
Bz =constant, Bx=By=0 and our derivative rules means there are no E fields either. It is possible for
either m or n to be zero however since (1) Bz won’t vanish meaning we don’t have a TEM mode and
(2) if m=0 , there can still be non-zero y derivatives of Bz which can create non-zero Ex. Thus with
the usual convention that a > b, the lowest frequency mode that can be transmitted by a rectangular
wave guide is TE10.
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We can insert our Bz = cos(mpi x/a) cos(n pi y/a) choice into our Ex, and Ey expressions. Ez is of
course 0 since we are solving for TE waves. We find that Ex of the form cos(m pi x/a) sin(n pi y/b)
and Ey of the form sin(m pi x/a) cos(n pi y/b) . Lets establish that these solutions satisfy the E|| BC.
For example Ex should vanish at y=0,b since Ex is parallel to the left and right side of the wave guide
and the sin(n pi y/b) factor insures this will happen. Similarly Ey should vanish at x=0,a and it does
because of the sin(m pi x/a) factor. Although E|| must vanish on the conducting walls of the wave
guide, Eperp can be non-zero just outside of the surface since it can be canceled inside the
conductor by a surface charge density. We illustrate the locations of the surface charge for a TE_11
wave. For example on the x=0 plane, the maximum Ex occurs at y=b/2 and this will be where the
maximum surface charge density. There will be an opposite charge density on the x=a surface
because of the cos(pi x/a) factor. We also show the E-fields at these locations which must be
perpendicular to the surfaces. Now, of course, these surface charges vary harmonically because of
the exp(i k z -i omega t) factor common to all electric and magnetic fields. Because the surface
charge at a particular spot on the wave guide has a sinusoidal time variation, it must be fed by
surface currents according to the surface continuity equation. We can find these surface currents
using the magnetic fields as vec K = (vec eta cross vec B)/mu0 where vec eta is the surface normal.
You perform a check of the surface continuity equation using the E and B fields for a TM mode in
homework.
It is remarkably simple to prove surface current conservation on the x=0 plane for the TE or TM wave
of general wave guide just from the form of the boundary conditions. Since E|| must disappear just
outside of a conductor only the Ex component can exist on the x=0 plane. Similarly there can be only
a By and Bz component since Bperp=0 just outside of a conductor. We can then find the surface
current and charge from the usual BC expressions sigma=epsilon_0 xhat dot E and K= xhat cross
B/mu_0. We next construct the divergence of K and the time derivative of sigma in terms of Ex , By,
and Bz which are functions of y,z, and t. We see that if the fields satisfy the Ampere-Maxwell law, we
automatically satisfy the continuity equation. After all the Maxwell term was added to Ampere’s law
in order to conserve current.
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We next show that our 1-d example electric field matches the general TE form in the limit b
approaches infinity. Ex is proportional to sin(n pi y/b) which vanishes as b approaches infinity so we
only have an Ey component which matches our intuitive example where vec E propto y-hat. The
cos(n pi y/b) factor goes to 1 in the infinite b limit and hence in our example Ex has no y factor but
has the same sin(m pi x/a) factor as the general solution. We also review the B-field for this harmonic
wave. The B-field is easily computed using Faraday’s law and has a non vanishing z-component .
The E-field is tangential to and disappears on the conductor surfaces so there is no surface charge
density. There is a surface current since there is a B|| at the surface in the z-hat direction with both
Bx and Bz components but only the Bz component is present on the x=0 surface which implies the
surface current is in the x-hat cross vec y-z or y-hat direction. But the Ky surface current, only varies
in the z direction and has no divergence. Hence the surface current will not result in an oscillating
surface charge. Using wave guide terminology these are essentially TE_m0 waves (or perhaps 0m
given b > a) with a Bz-field of the form Bz propto cos(m pi x/a) cos(n pi y/b) => cos(m pi x/a) as b
becomes infinite. This is, of course, consistent with our general TE field results where the x and y
components are given as x and y derivatives of Bz. For example you can take our Ey form, compute
Bz and Bx using Faraday’s law and show they are consistent with our transverse field expressions
such as Bx = i k/[(w/c)^2 –k^2] partial Bz/partial x or Ey= i k/[(w/c)^2 –k^2] (-omega partial Bz/partial
x).
Here are some pictures of TE and TM waves which I found on the internet. We indicate the position
of a two slots that can be used to inject fields or perturb the waves. The TE wave (top) shows surface
currents (in blue) converging or diverging from the boundaries of the top plate cutaway which creates
the oscillating surface charge density shown on the previous slide. You can also see there are E
fields in the y direction which emanate from this oscillating surface charge density. There is evidently
a node in the surface currents in the center of the cutaway plate. We finally note that all of the electric
field lines (shown in black) are perpendicular to the z axis. The TM picture shows that all magnetic
fields (green) are transverse to the z-axis. Interestingly enough the surface currents (blue) are
parallel to the z axis and disappear in the corners. You will see why in homework. You can see that
the surface currents also disappear in center of the top cutaway plate and are strongest in the center
of the two x edges. This suggests a divergence and an oscillating surface charge. Indeed there are
electric fields perpendicular to the surface at these current maxima.
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We know that coax cables can transport currents and energy. They can be thought of as a special case of a wave guide. We
will explore the TEM mode for the coaxial “wave guide”. Recall that there is no TEM mode for a rectangular wave guide.
What makes a coaxial wave guide so special? We begin by writing the E and B fields in “wave guide” notation as transverse
field functions times exp (i k z – i omega t). We begin with the x and y components of the Faraday and Ampere-Maxwell laws.
These are particularly simple and useful since they involve the x and y derivatives of Ez or Bz which vanish, and the z
derivatives of the transverse fields which just bring down factors of i k. We can find the relationship between omega and k
(e.g. the dispersion relation) by demanding that the relationship between Ex and By from the Ampere -Maxwell law is
consistent with the relationship from Faraday’s law. We find consistency as long as k = omega /c. We would get the same
omega-k relationship for Ey and Bx consistency. This is the same k-omega relationship for plane waves traveling through
vacuum.
This is an interesting result since it implies that there is no cutoff frequency for the TEM coaxial wave– just like plane waves
traveling in vacuum. By contrast, all of the TE_mn or TM_mn waves in the rectangular wave guides have cutoff frequencies
that depend on m and n which represent the minimum frequencies that can be transported down the wave guide. Evidently
we can transmit even zero frequency waves down a coax cable. But of course this had to happen since we have discussed
static (i.e. DC) magnetic energy flow down a coaxial cable. The TEM fields are also remarkably simple. Since there are no
Ez or Bz fields, both the (two-dimensional) divergence and curl vanish. In fact we get exactly the same equations as we
would have in electrostatics or magnetostatics for the case of an infinite coax with no z dependence that we considered in
Physics 435. The vanishing of the two-D curl means we can define a scalar electric potential (i.e. voltage). The vanishing of
the 2-D divergence means the voltage satisfies Laplace’s equations. We can simply recycle the old static solutions and
apply them to this electrodynamic case where there are harmonic waves traveling down the wave guide. The foregoing
made no specific reference to the coax cable and could equally apply to the case of a rectangular wave guide or any hollow
wave guide for that matter. But for a hollow wave guide we have an empty region, where Laplace’s equation holds which is
enclosed by a constant potential (on the metal that forms the guide). The uniqueness theorem says that only one solution
exists – the solution with constant potential or no fields. Hence no TEM modes with non-zero fields can exist in a hollow
wave guide. We reached the same conclusion using the wave guide equations. The transverse fields (such as Ex) depend
on transverse derivatives of Ez and Bz but for a TEM mode Ez = Bz =0. But the coax case we are considering here is not
hollow and can support a non-constant potential in the empty region and to solve it we can recycle the Physics 212 (or 435)
solutions for a charged cylinder with linear charge density lambda, or an infinite wire, carrying current I. These must be the
solutions for our coaxial “wave guide”!
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But the difference between the TEM wave and the static solutions is that the current and linear
charge densities are linked by Maxwell’s Equations. We write E in terms of a more convenient
variable A, which is the strength of the electric field at s=a. We compute B from E using Faraday’s
law since this is often the easiest way for a harmonic wave. Interestingly enough, you can see that Bexpression is the same as B = z-hat cross E/c which is exactly the same relation between E and B as
for a wave in a vacuum propagating along the z-axis. Lets check the surface continuity equation with
these forms on the inner conductor surface. We compute sigma from the Eperp at s=a and vec-K
from B|| at s=a. As expected vec K points in z-hat direction which is the direction of current flow for
our infinite wire. Now lets check the continuity equation. The time derivative of sigma brings in a
factor of –i omega and the divergence of K brings in a factor of i k. We can get agreement with the
continuity equation as long as mu_0 epsilon_0 = k/(c omega) which given that we showed k/omega =
1/c means we satisfy the continuity equation as long as mu_0 epsilon_0 =1/c^2 which is always true.
We can use our sigma and K-vec expressions to get the linear charge density lambda, and the
current I in terms of A and exp(i k z – i omega t). We thus see that the current and linear charge
densities form surface waves but remarkably their fields are of the old static form. It is also interesting
to compute the total transmitted power by constructing the Poynting vector. We use the complex
time-average form of S. Indeed the power is transported in the z-direction which is the propagation
direction of the wave. We can integrate the Poynting vector over the area between a and b. In the
conservation chapter we worked out the power flowing through a static coaxial cable with charge
density lambda and current I. If we insert the I and lambda “amplitudes” in terms of A into the static
expression, we find the static power is just double the time-average power as one would expect.
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We conclude this long chapter by discussing box-like, resonant cavities. We write
an oscillating electric field where each component is built of a separation of variable
type solution consisting of products of trig functions in x, y, and z. To satisfy the E||
BC we need the x component to disappear on the y-z walls, the y component to
disappear on the x-z walls and the z component to disappear on the x-y walls. One
way to do this is to construct Ex out of a cosine in x and sines in y and z and choose
the arguments to have an integral number of ½ wavelengths in each dimension. We
also need zero divergence for the E-field which translates to E1 k1 + E2 k2 +E3
k3=0. Hence there are only two independent amplitudes and the 3rd amplitude
solves the zero divergence condition. It seems like one could also satisfy the E|| BC
by making a product of three sines in each dimension rather than two sines and a
cosine. But this alternative form doesn’t work since each term in the divergence will
have one cosine and two sines and thus one cannot have the divergence vanish
everywhere no matter how one adjusts the relative amplitudes apart from setting all
fields to zero. The electric fields still must satisfy a wave equation which means k^2
= k1^2 +k2^2 +k3^2=(omega/c)^2 and k1, k2, k3 are integral multiples of pi/lengths
to satisfy the ½ wavelength condition. As a result only certain frequencies will
resonate, i.e. oscillate forever for perfect conductors. The two independent
amplitudes can be organized into TE and TM amplitudes. This is very reminiscent of
the particle in a box as a model for quantum dots.
You will see essentially the same physics for a particle in a 3-d box in quantum
mechanics having specific energy levels.
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