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MARKUS KLUTE: Welcome back
to 8.20 Special Relativity.

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In this section, we're going
to talk about the emission

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and the absorption of photons.

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So you can think
about a scenario

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where you have an atom
which emits a photon,

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like in this picture here.

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And to create a new atom,
which we call the atom prime,

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in the absorption process,
you have a collision

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between a photon and the atom.

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And you make a new atom,
again, indicated by a prime.

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So I want you to actually work
this out in two activities.

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The first one is on absorption.

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So here we have a stationary
particle, the atom,

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with a mass M0, and
it's struck by a photon.

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We have this scenario here.

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And the photon has an
energy, Q, and becomes--

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and there's a new particle
coming out with the rest mass

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M0 prime and
recoiling velocity, it

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has to have some
sort of momentum,

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v. And so the question
is, how can we

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now find the mass of this new
particle and the velocity?

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So please stop the video,
and try to work this out.

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So the way to work this out is
by writing energy and momentum

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equation.

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You can do this just
with a four-vector.

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even though we are not
performing any Lorentz

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transformation in this
part, you can just

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write on the four-vector.

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And this needs to
be conserved energy.

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And momentum need to be
conserved in this correlation.

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All right, so we have the
mass of this initial particle

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at rest.

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So the energy is M 0 C square.

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The energy of the photon is
Q. The momentum of the photon

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is Q over c.

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And so then the final particle--
the outcoming new particle

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has its own mass,
but it has a boost.

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So the total energy is M 0
prime gamma using the velocity

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of this particle, C square.

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And the momentum is M 0
prime gamma times v. So

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that's after the correlation.

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And so then you can
use the energy relation

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in order to get to the mass.

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This is a function
of the velocity here.

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And you can get the velocity
by looking at the momentum.

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So you basically solve for this
v or v over c in this case.

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And then if you want, you
can add this back in here

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in order to get the
mass as a function

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of the mass of the
initial particle

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and the energy of the photon.

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The second example
now is the emission.

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So here we have a stationary
atom with a rest mass M0.

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And it emits the
photon, this energy Q.

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And it becomes the new
atom with rest mass

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M0 prime and recoiling
with the velocity.

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Now given the two masses, what
is the energy of the photon?

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Q.

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So we set this up in
the same way as before.

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Now we start with
this particle at rest.

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And the energy is M 0 c squared.

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Momentum is 0.

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And that's equal to
the energy and momentum

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of the new particle and the
photon, in the very same way

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as we set this up before.

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So now what you
want to do here is

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use our invariants of the
four-vector using the energy

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and momentum relation.

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So the mass of this new particle
times c square squared is equal

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to the energy squared .

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The energy squared-- you
just bring this over here--

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is M0 c squared minus Q squared.

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And then the momentum is
simply of this new particle,

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is simply Q square.

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This you cannot solve for Q.
And you find this relation here.

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So that's the energy
of the outgoing photon.

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If you now define Q0 as
the difference in masses

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or the difference of the
masses times c square,

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then you can write the
energy of the photon

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as Q0 times 1 minus
Q0 over 2M c square.

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And that's always smaller
than the difference in masses.

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So if you now analyze this
and look at this some more

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and try to understand
what it means,

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what we find is that the
emission and the absorption,

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they only occur at
a very precise value

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of the energy of the photon.

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It also means that an
emitted photon can only

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be reabsorbed if the
particles are moving

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with just the right velocity.

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So if you're an atom,
and the photon goes out,

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and your neighboring atom is
just sitting there next to you

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at rest as you were at
rest before, it is not

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able to reabsorb the photon.

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So we learned just by using
special relativity quite a bit

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about the physics
involved in absorption

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and emission of photons.