LOG#039. Relativity: Examples(III).

Example 1. Absorption of a photon by an atom.

In this process, we have from momenergy conservation:

P^\mu_a P_{a\mu}+2P^\mu_a P_{b\mu}+P^\mu_b P_{b\mu}=P^\mu_c P_{c\mu}

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

P^\mu_a =(m_a c,0,0,0)


Description: an atom “a” at rest with mass m_a 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 absorption has

P^\mu_c=(m_c c,0,0,0)

Therefore, since the photon verifies:

P^\mu_b P_{b\mu}=\left(\dfrac{h\nu}{c}\right)^2-\left(\dfrac{h\nu}{c}\right)^2=0

and it is true in every inertial frame. Thus,

l (m_a c)^2+2m_a c\dfrac{h\nu}{c}+0=(m_a c)^2


\boxed{m_c=\sqrt{m_a^2+2m_a\dfrac{h\nu}{c}}=m_a\sqrt{1+2\dfrac{h\nu}{m_a c^2}}}

Expanding the square root

m_c\approx m_a\left[ 1+\dfrac{h\nu}{m_a c^2}-\dfrac{1}{2}m_a\left(\dfrac{h\nu}{m_a c}\right)^2+\mathcal{O}(h^3)\right]

In this case,

m_c\approx m_a+\dfrac{h\nu}{ c^2}-\dfrac{1}{2}\left(\dfrac{h\nu}{m_a c}\right)^2

Atom’s rest mass increases by an amount \dfrac{h\nu}{c^2} up to first order in the Planck’s constant, and it decreases up to second order in h a quanity -\dfrac{1}{2}m_a\left(\dfrac{h\nu}{m_a c^2}\right)^2 due to motion ( “recoil”). Therefore,

\dfrac{1}{2}m_a\left(\dfrac{h\nu}{m_a c^2}\right)^2=\dfrac{1}{2}m_a\left( \dfrac{u_a}{c}\right)^2

In this way, identifying terms: u_c=\dfrac{h\nu}{m_a c}

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

\dfrac{h\nu}{c}=\gamma_c m_c u_c

m_a c^2+h\nu=\gamma_c m_c c^2

\dfrac{h\nu}{u_c c}=m_a+\dfrac{h\nu}{c^2}

Then, we obtain that:

u_c=\dfrac{h\nu c}{m_a c^2+h\nu}=\dfrac{c}{\left(\dfrac{m_a c^2}{h\nu}\right)+1}\approx \dfrac{h\nu}{m_a}

where we have used in the last step m_ac^2>>h\nu.

Example 2. Emission of a photon by an atom.

An atom c at rest, with m_c the rest mass, emits a photon b with frequency \dfrac{E_b}{h}=\nu in the x-direction, turning itself into a non-excited atom “a”, with m_a. What is the energy shift \Delta E=E_c-E_a?

P^\mu_a P_{a\mu}+P^\mu_b P_{b\mu}=P^\mu_c P_{c\mu}


P^\mu_a is P^\mu_a =(m_a,0,0,0) in the rest frame of “a”.

P^\mu_a in the rest from “c” reads P^\mu_a =\left(\gamma_a m_a c, -\dfrac{E_b}{c},0,0\right)

P^\mu_b in the rest frame of “c” is P^\mu_b =\left(\dfrac{E_b}{c},\dfrac{E_b}{c},0,0\right)

P^\mu_a in the rest frame of “c” is P^\mu_c=(m_c c,0,0,0)

In this way, we have

(m_a c)^2+\gamma _a m_a E_b +\left( \dfrac{E_b}{c}\right)^2=\gamma _am_am_c c^2

\gamma_a m_a c^2+E_b=m_c c^2

m_a^2c^2+\dfrac{E_b}{c^2}\left(m_c c^2-E_b\right)^2=m_c^2c^2-m_c E_b

m_a^2c^2+m_cE_b=m^2_c c^2-m_c E_b


From this equations we deduce that

E_b=\dfrac{m_c c^2-m_a^2c^2}{2m_c}=\dfrac{E_c^2-E_a^2}{2E_c}=(E_c-E_a)\left(1-\dfrac{E_c-E_a}{2E_c}\right)

And from the definition E_c-E_a we get E_b=\Delta E\left(1-\dfrac{\Delta E}{2E_c}\right)

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 C\rightarrow AB

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

(m_A c)^2+2\gamma_A\gamma_Bm_Am_B(c^2-u_Au_B)+(m_Bc)^2=(m_C c)^2

Choose the frame in which the following equation holds


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



P^\mu_C=(m_C c,0,0,0)

From these equations (m_Ac)^2+\gamma_A\gamma_Bm_Am_B(c^2-u_Au_B)=\gamma_Am_Am_C c^2







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 \gamma \rightarrow e^+e^-, in which a single photon (\gamma)decays into a positron-electron  pair.

That is, h\nu\rightarrow e^+e^-.


Squaring the momenergy in both sides:

P^\mu_a P_{\mu a}=P^\mu_c P_{\mu c}+2P^\mu_c P_{\mu d}+P^\mu_d P_{\mu d}

In the case of the photon: P^\mu_a P_{a\mu}=0

In the case of the electron and the positron: P^\mu_c P_{\mu c}=P^\mu_d P_{\mu d}=-(m_e c)^2=-(mc)^2

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

P^\mu_c=\left( \dfrac{E_c}{c},p_{cx},p_{cy},p_{cz}\right)

P^\mu_d=\left( \dfrac{E_d}{c},-p_{dx},-p_{dy},-p_{dz}\right)

with E_c=E_d=mc^2. Therefore,

2 P^\mu_c P_{\mu d}=-2\left( \dfrac{E_c^2}{c^2}+p_x^2+p_y^2+p_z^2\right)


-2(mc)^2-2\left( \dfrac{E_c^2}{c^2}+\mathbf{p}^2\right)=0

2(mc)^2+2\left( \dfrac{E_c^2}{c^2}+\mathbf{p}^2\right)=0

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 \gamma \gamma \rightarrow e^+e^- is the typical process in a “photon collider”. Other alternative is that the photon were “virtual” (e.g., like QED reactions e^+e^-\rightarrow \gamma^\star\rightarrow e^+e^-). Suppose, alternatively, the reaction AB\rightarrow CDE. Solving this process is hard and tedious, but we can restrict our attention to the special case of three particles C,D,E staying together in a cluster C. In this way, the real process would be instead AB\rightarrow C. In the laboratory frame:


we get

P^\mu_A=\left( \dfrac{E_A}{c},\dfrac{E_A}{c},0,0\right)

P^\mu_B=\left( Mc,0,0,0\right)

and in the cluster reference frame


Squaring the momenergy:





In the absence of an extra mass M, i.e., when M\rightarrow 0, the energy E_A 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 M=m, 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 \gamma =4\rightarrow


\beta=\sqrt{\dfrac{15}{16}}=\dfrac{\sqrt{15}}{4}\approx 0.97

v=\dfrac{\sqrt{15}}{4}c\approx 0.97c

In general, if m\neq M 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. l e^+e^-\rightarrow \gamma \gamma implies the momenergy conservation


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

l P^\mu_aP_{\mu a}+2P^\mu_aP_{\mu b}+P^\mu_bP_{\mu b}=P^\mu_cP_{\mu c}+2P^\mu_cP_{\mu c}+P^\mu_dP_{\mu d}

Then, we deduce


P^\mu_b=\left( m_e c,0,0,0\right)

while we do know that

P^\mu_aP_{\mu a}=P^\mu_bP_{\mu b}=-(m_e c)^2 for the electron/positron and

P^\mu_cP_{\mu c}=P^\mu dP_{\mu d}=0 since they are photons. The left hand side is equal to -2(m_e c)^2+2E_am_e, and for the momenergy in the right hand side


P^\mu_d=\left( \dfrac{E_d}{c},p_{dx},0,0\right)

Combining both sides, we deduce

(m_a c)^2+E_a m_e=\dfrac{E_cE_d}{c^2}-p_{cx}p_{dx}

The only solution to the right hand side to be not zero is when we select p_{cx}=\pm \dfrac{E_c}{c} and p_{dx}=\pm\dfrac{E_d}{c} and we plug values with DIFFERENT signs. In that case,

\boxed{(mc)^2+E_a m=\dfrac{2E_cE_d}{c^2}}

From previous examples:

P^\mu_aP_{\mu_b}+P^\mu_bP_{\mu b}=P^\mu_c P_{b\mu}+P^\mu_aP_{\mu b}

and we evaluate it in the laboratory frame to give

\boxed{E_a m+(mc)^2=E_c m+E_d m}

The last two boxed equations allow us to solve for E_d


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

(mc)^2+mE_a=\dfrac{2E_c}{c^2}\left( E_a-E_c+mc^2\right)


l \dfrac{1}{2}mc^2\left(mc^2+E_a\right)=-E_c^2+E_c(E_a+mc^2)

Solving for E_c this last equation


\boxed{E_c^{1,2}=E_d^{1,2}=\dfrac{\left(E_a+mc^2\right)\pm \sqrt{\left(E_a+mc^2\right)\left(E_a-mc^2\right)}}{2}}

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