# LOG#039. Relativity: Examples(III).

**Posted:**2012/10/07

**Filed under:**Physmatics, Relativity |

**Tags:**atom, boost, decay, effective mass, exercises, frame, pair creation, photon absorption, photon emission, Relativity, rest mass Leave a comment

**Example 1. Absorption of a photon by an atom.**

In this process, we have from momenergy conservation:

If the atom is the rest frame, before absorption we get

Description: an atom “a” at rest with mass absorbs a photon “b” propagating in the x-direction turning itself into an excited atom “c”, moving in the x-axis (suffering a “recoil” after the photon hits it).

The atom after aborption has

Therefore, since the photon verifies:

and it is true in every inertial frame. Thus,

Then,

Expanding the square root

In this case,

Atom’s rest mass increases by an amount up to first order in the Planck’s constant, and it decreases up to second order in h a quanity due to motion ( “recoil”). Therefore,

In this way, identifying terms:

In the laboratory frame, the excited atom velocity is calculated by momenergy conservation. It is simple:

Then, we obtain that:

where we have used in the last step .

**Example 2. Emission of a photon by an atom.**

An atom c at rest, with the rest mass, emits a photon b with frequency in the x-direction, turning itself into a non-excited atom “a”, with . What is the energy shift ?

where

is in the rest frame of “a”.

in the rest from “c” reads

in the rest frame of “c” is

in the rest frame of “c” is

In this way, we have

From this equations we deduce that

And from the definition we get

**Note:** the photon’s energy IS NOT equal to the difference of the atomic rest energies but it is less than that due to the emission process. This fact implies that the atom experieces “recoil”, and it gains kinetic energy at the expense of the photon. There is a good chance for the photon not to be absorbed by an atom of some kind. However, “resonance absorption” becomes problematic. The condition for recoilless resonant absorption to occur nonetheless, e.g., the reabsorption of gamma ray photons by nuclei of the some kind were investigated by Mössbauer. The so-called Mössbauer effect has been important not only to atomic physics but also to verify the theory of general relativity. Furthermore, it is used in materials reseach in present time as well. In 1958, Rudolf L. Mössbauer reported the 1st reoilless gamma emission. It provided him the Nobel Prize in 1961.

**Example 3. Decay of two particles at rest.**

The process we are going to study is the reaction

The particle C decays into A and B. It is the inverse process of the completely inelastic collision we studied in a previous example.

From the conservation of the tetramomentum

Choose the frame in which the following equation holds

Let be the laboratory frame velocities in the rest frame of “C”. Then, we deduce

From these equations

Therefore, the total kinetic energy of the two particles A,B is equal to the mass defect in the decay of the particle.

**Example 4. Pair production by a photon. **

Suppose the reaction , in which a single photon ()decays into a positron-electron pair.

That is, .

Squaring the momenergy in both sides:

In the case of the photon:

In the case of the electron and the positron:

We calculate the components of momenergy in the center of mass frame:

with . Therefore,

so

This equation has no solutions for any positive solution of the photon energy! It’s logical. In vacuum, it requires the pressence of other particle. For instance is the typical process in a “photon collider”. Other alternative is that the photon were “virtual” (e.g., like QED reactions ). Suppose, alternatively, the reaction . Solving this process is hard and tedious, but we can restrict our attention to the special case of three particles staying together in a cluster C. In this way, the real process would be instead . In the laboratory frame:

we get

and in the cluster reference frame

Squaring the momenergy:

In the absence of an extra mass M, i.e., when , the energy would be undefined, and it would become unphysical. The larger M is, the smaller is the additional energy requiere for pair production. If M is the electron mass, and , the photon’s energy must be twice the size of the rest energy of the pair, four times the rest energy of the photon. It means that we would obtain

and thus

In general, if we would deduce:

**Example 5. Pair annihilation of an electron-positron couple.**

Now, the reaction is the annihilation of a positron-electron pair into two photons, turning mass completely into (field) energy of light quanta. implies the momenergy conservation

where “a” is the moving electron and “b” is a postron at rest. Squaring the identity, it yields

Then, we deduce

while we do know that

for the electron/positron and

since they are photons. The left hand side is equal to , and for the momenergy in the right hand side

Combining both sides, we deduce

The only solution to the right hand side to be not zero is when we select and and we plug values with DIFFERENT signs. In that case,

From previous examples:

and we evaluate it in the laboratory frame to give

The last two boxed equations allow us to solve for

If we insert this equation into the first boxed equation of this example, we deduce that

or

Solving for this last equation