MARKUS KLUTE: Welcome back to 8.20, Special Relativity.
In this last example of relativistic kinematics,
we want to investigate scattering--
in this specific case, a scattering
of a photon on an electron at rest.
So we have as an initial state a photon, an electron at rest,
and then the photon is scattered and we also
observe a scattered electron.
There's one important piece of physics
here, which we add without further explanation, which
is the Planck-Einstein relation, which
relates the energy of the photon to the frequency of the photon
or the wavelength.
This is fundamentally important in quantum physics,
and can be explained or tested with the photoelectric effect
for which Einstein received the Nobel Prize.
So what we want to do here is find the wavelength shift,
so delta lambda, which is the wavelength
of the incoming photon minus the wavelength
of the outgoing photon, as a function of the scattering
angle zeta, as shown in this picture here.
OK, so again, this is an activity
I want you to work on and try to find out this.
The algebra here is not trivial, but knowing
how to set up a problem like this is important.
So let's try.
So the way to set this up is to write this four vector
relation, or you could just simply write down
energy conservation and momentum conservation.
So you have an initial state, the before,
and the final state, the after, where
you simply add the four vectors of the initial electron
and photon and set this equal to the scattered electron
and scattered photon.
Now, we are interested in a quantity delta
lambda, which is related to the change in energy of the photon.
So therefore, it brings the four vector of the four
scattered photon over here to this side,
and builds a square, which allows us then
to use our invariant information in the scattering process.
When we explore the squared here,
we find the photon four vectors squared
for the scattered and the unscattered photon,
minus 2 times the product of the two four vectors.
Now, the mass of the photon is 0,
and then hence the invariant mass
is 0, too, so this invariant four vector is 0.
So this cancels and this cancels.
And then we know that the mass of the electron
is the mass of the electron, the initial momentum
of the electron is 0, and we just for the further,
not to get confused, we said C equal to 1.
So then we just go through a sequence of algebra
here, making use of information that those guys here are simply
the mass of the electron.
And we move things around a little bit
and then find this equation here,
which relates the energies of the two photons
[INAUDIBLE] the scattering angle, which
we get from the scattered product of the [? three ?]
momentum of the photon to the change in electron
energy, which is the energy of the electron minus the mass.
OK.
And then we start using the Einstein relation here.
And again, a little bit of algebra
then brings us to delta lambda equal h over me
times 1 minus cosine theta.
So this relates the shift in wavelengths
to the scattering angle of the photon.
Important.
If you want to recall this, the most important part
through this problem is setting up this first equation here,
which relates the energy and the momentum,
or the four vector of those particles,
before and after the collision.
And again, then it takes a little bit of practice.
But the way to approach most of this problem
is to make use of the invariant four vector squared,
or the invariant mass of the objects involved,
if we know the masses of the object involved.
OK.