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MARKUS KLUTE: Welcome back
to 8.20, Special Relativity.

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In this last example of
relativistic kinematics,

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we want to investigate
scattering--

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in this specific
case, a scattering

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of a photon on an
electron at rest.

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So we have as an initial state
a photon, an electron at rest,

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and then the photon is
scattered and we also

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observe a scattered electron.

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There's one important
piece of physics

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here, which we add without
further explanation, which

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is the Planck-Einstein
relation, which

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relates the energy of the photon
to the frequency of the photon

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or the wavelength.

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This is fundamentally
important in quantum physics,

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and can be explained or tested
with the photoelectric effect

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for which Einstein
received the Nobel Prize.

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So what we want to do here
is find the wavelength shift,

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so delta lambda, which
is the wavelength

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of the incoming photon
minus the wavelength

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of the outgoing photon, as
a function of the scattering

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angle zeta, as shown
in this picture here.

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OK, so again, this
is an activity

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I want you to work on
and try to find out this.

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The algebra here is not
trivial, but knowing

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how to set up a problem
like this is important.

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So let's try.

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So the way to set this up
is to write this four vector

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relation, or you could
just simply write down

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energy conservation and
momentum conservation.

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So you have an initial
state, the before,

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and the final state,
the after, where

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you simply add the four
vectors of the initial electron

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and photon and set this equal
to the scattered electron

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and scattered photon.

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Now, we are interested
in a quantity delta

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lambda, which is related to the
change in energy of the photon.

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So therefore, it brings
the four vector of the four

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scattered photon over
here to this side,

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and builds a square,
which allows us then

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to use our invariant information
in the scattering process.

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When we explore
the squared here,

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we find the photon
four vectors squared

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for the scattered and
the unscattered photon,

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minus 2 times the product
of the two four vectors.

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Now, the mass of
the photon is 0,

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and then hence
the invariant mass

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is 0, too, so this
invariant four vector is 0.

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So this cancels
and this cancels.

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And then we know that
the mass of the electron

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is the mass of the electron,
the initial momentum

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of the electron is 0, and
we just for the further,

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not to get confused,
we said C equal to 1.

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So then we just go through
a sequence of algebra

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here, making use of information
that those guys here are simply

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the mass of the electron.

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And we move things
around a little bit

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and then find this
equation here,

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which relates the energies
of the two photons

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[INAUDIBLE] the
scattering angle, which

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we get from the scattered
product of the [? three ?]

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momentum of the photon
to the change in electron

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energy, which is the energy of
the electron minus the mass.

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OK.

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And then we start using
the Einstein relation here.

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And again, a little
bit of algebra

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then brings us to delta
lambda equal h over me

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times 1 minus cosine theta.

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So this relates the
shift in wavelengths

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to the scattering
angle of the photon.

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Important.

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If you want to recall this,
the most important part

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through this problem is setting
up this first equation here,

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which relates the
energy and the momentum,

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or the four vector
of those particles,

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before and after the collision.

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And again, then it takes
a little bit of practice.

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But the way to approach
most of this problem

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is to make use of the
invariant four vector squared,

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or the invariant mass
of the objects involved,

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if we know the masses
of the object involved.

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OK.