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simple wave eqn (+quantum +relativity +dispersion)



This note brings together two threads:

a) The "quantum question" thread asked what might be the
relativistically-correct analog to the Schrödinger equation.
b) The "dispersive waves" thread asked how hard it would be to teach
about dispersion without causing brain-blisters among the students.

I agree with Ludwik that covering flexy waves and their fourth-order wave
equation would be a bit much for the average student in the first
college-level physics course.

Let's confine the discussion to the context of students whose math skills
can handle the elementary wave equation. In units where c=1 we have:
(d/dt)^2 y - (d/dx)^2 y = 0 (eq. 1)

From there it is not a huge step to add another term:
(d/dt)^2 y - (d/dx)^2 y + w^2 y = 0 (eq. 2)

If you can find sinusoidal solutions to eq. 1 you can find the
corresponding solutions to eq. 2. It just isn't very tricky. You get a
dispersion relation that is a pair of hyperbolae.

The payoff is large. It includes:

A.1) Eq. 2 describes electromagnetic waves in a waveguide. The parameter w
is the cutoff frequency. In this context eq. 2 is called "the waveguide
equation".

A.2) In the limit of large k and large omega, the dispersion relation for
eq. 2 is asymptotically identical to the dispersion relation for eq. 1 --
straight lines.

B) This affords the opportunity to tell students that group velocity is
important.
<rant>
In high school they were probably taught that given the wavenumber (k) and
the frequency (omega) they could calculate the speed of the waves as
c = omega / k :-(

Well...
-- That's not true for light waves in flint glass (e.g. a prism).
-- That's not true for electromagnetic waves in a waveguide.
-- That's not true for electromagnetic waves in the ionosphere.
-- That's not true for Schrödinger waves.
-- That's not true for flexy waves on a spring.
-- That's sure as heck not true for waves on the surface of a pond.
-- et cetera.

The group velocity is what you (almost always) care about, and is given by:
v = d(omega) / d(k)

For small k in eq. 2, omega/k goes to _infinity_ while d(omega)/d(k) goes
to zero. A rather large difference, wouldn't you say?
</rant>

---------

.... At this point let's further restrict the discussion to the context of
students who know enough relativity and QM to be able to ask the question
of whether the Schrödinger equation is relativistically correct....

C.1) Equation 2 also gives a relativistically-correct description of the
wavefunction of a massive scalar particle such as an alpha particle (or a
golf ball at low temperatures). The parameter w is the rest mass, in the
appropriate units:
hbar w = m c^2

In this context eq. 2 is called "the massive scalar Klein-Gordon
equation". On the other hand, calling it that is sometimes pedagogically
unhelpful. If you say in the course announcement that you will be covering
the massive scalar Klein-Gordon equation some of the would-be students will
run away screaming with their hair standing on end.

C.2) To restate item (A.1) in another way, in the limit as w->0 we recover
the nondispersive equation that describes a massless particle (e.g. a
photon) in free space (not in a waveguide).

C.3) In the opposite limit, where the kinetic energy (hbar k)^2/2m is much
less than the rest mass energy, the dispersion relation is essentially
parabolic. In this regime eq. 2 makes the same predictions (e.g. same
group velocity) as the Schrödinger equation, as you can easily confirm by
expanding the dispersion relation to lowest order in k.