WEBVTT

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PROFESSOR: This is the answer.

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Tan of 2 k a plus delta.

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It's a little messy, no wonder.

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Just plotted this functions.

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[INAUDIBLE] Sin k plus a
cosh kappa a [INAUDIBLE]

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plus a prime over kappa
cosh a prime a sinh kappa a.

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You have sin a prime a sinh.

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The sinh causes outer
[INAUDIBLE] the same

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but this one changes.

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Centered here, k prime
over kappa cos k prime

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a cosh kappa a.

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So it's a well-defined
expression,

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if you know the
number of kappa a,

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you can calculate everything.

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Let me just make sure
you can see that.

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If you know-- if you
said, for example, ka,

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this is a natural variable.

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ka is just-- the k is related
to the energy direction.

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So ka has no limits, call it u.

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Then you have two parameters
of the square well [INAUDIBLE],,

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a depth we don't and
a height we want.

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So there's natural to
define just [INAUDIBLE]

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would have z 0 squared
which is 2m v 0

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a squared over h squared.

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You also define z 1 squared
[INAUDIBLE] as 2m v1

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a squared over h squared.

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And then the energy--

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we can do another
thing, we're going

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to find the energy divided
say, by v 1 to be little e.

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And that's reasonable
because these energies

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is compared always with
v 1, and we're solving--

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we've solved this
problem in the domain

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when the energy
is less than v 1,

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and that's why we'll
have kappas here,

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and if energy was
bigger than v 1,

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you would have trigonometric
functions everywhere.

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Now you can switch from
trigonometric to hyperbolic

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by analytic continuation,
letting and angle

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become imaginary, a
trigonometric function becomes

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a hyperbolic function.

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Most of us are
rather comfortable

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doing this at the beginning,
because of the matter of sine

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if you mess up a sine,
it's a big problem.

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But at the end of the day, it
actually saves a lot of work.

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So a little of that in
the homework you will see.

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But at any rate, this is
valued for e in this range.

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And therefore this ratio
is h squared k squared.

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You can put an a squared
over 2m v 1 a squared.

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And you can see
a u squared here.

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And-- I'm sorry, not there.

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A u squared here, and
the h squared divided

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here gives you a z 1.

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So this is u squared
over z 1 squared.

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A couple of extra things--

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I'm just putting it
here because if you ever

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have the curiosity of doing
this plots, this may help.

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k prime squared is
related to the energy,

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therefore u squared
and v 0 z 1 squared.

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This is-- c 0 squared,
[INAUDIBLE] 0.

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And for the other
one, kappa a squared

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is equal to z 1 squared
minus u squared.

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So everything becomes a
function of u Wherever

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you have a kappa prime a
you have a square root of u.

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Now, somebody has to give you
the values of z 0 and z 1.

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Those are the data of
the potential-- z 0

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and z 1 are numbers
you know then.

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Therefore you know kappa a
is a function of u, kappa--

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k prime a is a function
of u, and ka, which is u.

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Therefore this equation becomes
a function of u [INAUDIBLE]

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this arc tangent here,
and so for delta.

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It's messy.

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There's no-- remember I
mentioned the other day

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the fact that you could do
trigonometric identities,

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and so for tan delta here.

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But then this I don't think
simplifies when you solve that.

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It becomes just a bigger mess.

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You did solve for tan
delta, it's very messy.

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So anyway, let's leave it there.

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Are there any questions
in solving this problem?

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So in principle, we solved it.

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I didn't plot
anything, so you still

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don't have any insight
as to what's happening.

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But you've learned in
principle how to solve it.

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Any questions?

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OK, so let's plot this then.

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Maybe I should start here.

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OK.

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I'll start here.

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Oh, if I'm going to plot,
I have to choose things.

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So we'll choose z 0
squared equals to 1,

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or z 0 equals to 1.

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And z 1 squared equals to 5.

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So we put a big barrier there.

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And now let's go delta
as a function of ka or u.

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Now, from this equation, we
see that u and the little e

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must be at most 1 for our
formulas to be correct.

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So u must be up to square--

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up to z 1, because you must
keep this ratio less than 1.

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So we can plot u up
to square root of 5.

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5, here's 2.

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And that's it.

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And here we're going
to plot delta of u.

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And here is minus
pi over 2 minus pi.

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And so what does
it do, the phase?

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There no way you can guess,
I think, from this formula.

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I could not guess
from this formula.

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So we could try to imagine--

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it's possible to guess,
actually, after you've

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solved this week's homework.

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You probably will
have a good guess

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that the phase shift begins
with minus 2ka, as if

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with very little energy
you reflect back here.

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So there will be a shift that
is calculable without doing

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any work.

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So it begins linearly.

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And it represents
a time advance.

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So it goes linearly
and negative.

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That's how it begins.

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Because for very little energy,
it's going to bounce back.

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And you know that the
delay is proportional

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to this derivative.

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So must be negative like that.

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So then what does it do?

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It crosses this point,
which is almost pi,

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and here is what something
quite remarkable happens.

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That some value you star,
which is about 1.8523, I think.

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That's what I calculated.

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Let's see [INAUDIBLE].

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The face that is almost minus pi
suddenly jumps very, very fast,

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crosses pi over 2, and then
about [INAUDIBLE] something

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like this.

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I don't know what else
it does, because we

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haven't calculated it.

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But it jumps very fast.

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Almost a value of pi.

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Now, let me-- this
is quite interesting.

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If you think of what we
used to call the scattering

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amplitude as squared
was sin squared delta.

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The amplitude of the
scattered wave, sin

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squared delta, the amplitude
of the scattered wave

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is going to be quite
large here, it's

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going to be 1, which is the
maximum, as squared is here.

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So I want to keep these
two blackboards aligned.

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So here it goes like this.

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It's going to go up like this,
and do this, and broadly go

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down, and then
very sharply go up.

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More sharply, at least.

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Go up and then
something like that.

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So here it is.

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This point over here has a
strong scattering amplitude,

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but there's nothing too
dramatic happening here.

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This sine corresponds to
time advanced, the derivative

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is negative.

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And time advanced cannot
be too large, as you know.

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On the other hand,
here is time delayed,

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and apparently they
can't be very large.

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So we think this must
be the resonance,

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and this is not a resonance,
even though the scattering

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amplitude is bigger.

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So we continue here,
and plot the time--

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interestingly, the
amplitude inside the well.

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So here it is.

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How does the wave function
become, through this constant,

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inside the well?

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And indeed, the this confirms
that nothing very special

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is happening here.

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What happens now is some
sort of behavior like this,

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and a big jump here, in which
the amplitude apparently--

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I have not quite
confirmed this number,

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it's at least a value 3.

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[INAUDIBLE] very
large and short.

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I should have room for one more
plot [INAUDIBLE] the star plot

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of this thing is--

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I'll do it compressed here.

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The total delay, 1
over a d delta dk.

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Well, it begins
negative, and remember

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that when we did the
1 over a, d delta dk,

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this is a pure number.

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It expresses the delay
in terms of the time

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that it would take to
travel the inside region.

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So how many-- if you would
get a 1, or a minus 1,

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it's just a delay of
the size of the time

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needed to travel back and forth.

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So actually this goes
a little negative

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at the beginning-- we
know the derivative is

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like that-- and
when you plot this,

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you see that it's
very sharp, and it's

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a value of about fourteen.

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Fourteen times gets delayed from
what you would expect naturally

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of the time that it should have
spent travelling back and forth

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A gigantic time delay.

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A peak in the time delay.

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Peak in time delay.

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Peak in the amplitude
inside the well.