# 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

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

$P^\mu_b=(\dfrac{E_b}{c},p_{bx},0,0)=(\dfrac{E_b}{c},\dfrac{E_b}{c},0,0)=(\dfrac{h\nu}{c},\dfrac{h\nu}{c},0,0)$

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 aborption 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,

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

Then,

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

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 $\Delta E=E_c-E_a$?

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

where

$P^\mu_a$ 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

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

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

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

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

$m_a^2c^2+m_cE_b=m_c^2c^2-m_cE_b$

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

$P^\mu_AP_{A\mu}+P^\mu_BP_{A\mu}=P^\mu_{C}P_{A\mu}$

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

$P^\mu_A=(\gamma_Am_Ac,\gamma_Am_Au_A,0,0)$

$P^\mu_B=(\gamma_Bm_Bc,\gamma_Bm_Bu_B,0,0)$

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

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

$-(m_Ac)^2+(m_Bc)^2=(m_Cc)^2-2\gamma_Am_Am_Cc^2$

$2E_Am_C=(m_A^2+m_C^2-m_B^2)c^2$

$\boxed{E_A=\dfrac{(m_A^2+m_C^2-m_B^2)c^2}{2m_C}}$

$\boxed{E_A^{kin}=T_A=\dfrac{(m_A^2+m_C^2-m_B^2)c^2}{2m_C}-m_Ac^2=\dfrac{\left((m_C-m_A)^2-m_B^2\right)c^2}{2m_C}}$

$\boxed{E_B^{kin}=T_B=\dfrac{\left((m_C-m_B)^2-m_A^2)\right)c^2}{2m_C}}$

$\boxed{E_{kin}=T=T_A+T_B=(m_C-m_A-m_B)c^2}$

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, $h\nu\rightarrow e^+e^-$.

$P^\mu_a=P^\mu_c+P^\mu_d$

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,

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

so

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

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

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:

$P^\mu_A+P^\mu_B=P^\mu_C$

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

$P^\mu_C=\left((M+2m)c,0,0,0\right)$

Squaring the momenergy:

$0+2E_AM+(Mc)^2=(M+2m)^2c^2$

$2ME_A=4Mmc^2+4m^2c^2$

$E_A=2mc^2+\dfrac{2m^2c^2}{M}$

$\boxed{E_A=2mc^2\left(1+\dfrac{m}{M}\right)}$

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

$E=4mc^2=2m_{pair}c^2$

and thus $\gamma =4\rightarrow$

$\beta^2=1-\dfrac{1}{16}=\dfrac{15}{16}$

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

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

In general, if $m\neq M$ we would deduce:

$\boxed{\beta=\sqrt{1-\dfrac{1}{4\left(1+\frac{m}{M}\right)^2}}}$

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. $e^+e^-\rightarrow \gamma \gamma$ implies the momenergy conservation

$P^\mu_a+P^\mu_b=P^\mu_c+P^\mu_c$

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

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

Then, we deduce

$P^\mu_a=\left(\dfrac{E_a}{c},p_{ax},0,0\right)$

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

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

$P^\mu_c=\left(\dfrac{E_c}{c},p_{cx},0,0\right)$

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

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 $E_d$

$\boxed{E_d=E_a-E_c+mc^2}$

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

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

or

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

Solving for $E_c$ this last equation

$\boxed{E_c^{1,2}=E_d^{1,2}=\dfrac{1}{2}\left(E_a+mc^2\right)\pm\sqrt{\dfrac{1}{4}\left(E_a+mc^2\right)^2-\dfrac{1}{2}mc^2\left(mc^2+E_a\right)}}$

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