LOG#040. Relativity: Examples(IV).

Example 1. Compton effect.

Let us define as “a” a photon of frequency $\nu$ . Then, it hits an electron “b” at rest, changing its frequency into $\nu'$ , we denote “c” this new photon, and the electron then moves after the collision in certain direction with respect to the line of observation. We define that direction with $\theta$ .

We use momenergy conservation:

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

We multiply this equation by $P_{\mu c}$ to deduce that

$P^\mu_a P_{\mu c}+P^\mu_{b}P_{\mu c}=P^\mu_c P_{\mu c}+P^\mu_d P_{\mu c}$

Using that the photon momenergy squared is zero, we obtain:

$P^\mu_a P_{\mu c}+P^\mu_bP_{\mu c}=P^\mu_dP_{\mu c}$

$P^\mu _a=\left(\dfrac{h\nu}{c},\dfrac{h\nu}{c},0,0\right)$

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

$P^\mu_c=\left(\dfrac{h\nu'}{c},\dfrac{h\nu'}{c}\cos\theta,\dfrac{h\nu'}{c}\sin\theta,0\right)$

Remembering the definitions $\dfrac{c}{\lambda}=\nu$ and $\dfrac{c}{\lambda'}=\nu'$ and inserting the values of the momenta into the respective equations, we get

$\dfrac{h^2}{c^2}\nu\nu'\left(1-\cos\theta\right)+mh\nu'=mh\nu$

$\dfrac{h^2}{\lambda\lambda'}\left(1-\cos\theta\right)+\dfrac{mhc}{\lambda'}=mhc\dfrac{\lambda'-\lambda}{\lambda\lambda'}$

$\boxed{\Delta \lambda\equiv \lambda'-\lambda=\dfrac{h}{mc}\left(1-\cos\theta\right)}$

$\boxed{\dfrac{1}{\nu'}-\dfrac{1}{\nu}=\dfrac{h}{mc^2}\left(1-\cos\theta\right)}$

$\boxed{\dfrac{1}{\omega'}-\dfrac{1}{\omega}=\dfrac{\hbar}{mc^2}\left(1-\cos\theta\right)}$

$\boxed{\dfrac{\omega'}{\omega}=\mbox{Energy transfer}=\left[1+\dfrac{\hbar \omega}{mc^2}\right]^{-1}}$

It is generally defined the so-callen electron Compon’s wavelength as:

$l\lambda_C=\dfrac{h}{mc}$

$l \bar{\lambda_C}=\dfrac{\hbar}{mc}\approx 2.42\cdot 10^{-12}m$

Remark: There are some current discussions and speculative ideas trying to use the Compton effect as a tool to define the kilogram in an invariant and precise way.

Example 2. Inverse Compton effect.

Imagine an electron moving “to the left” denoted by “a”, it hits a photon “b” chaging its frequency into another photon “c” and the electron changes its direction of motion, being the velocity $-u_b$ and the angle with respect to the direction of motion $\theta$ .

The momenergy reads

$P^\mu_a=\left(\dfrac{h\nu}{c},\dfrac{h\nu}{c},0,0\right)$

$P^\mu_b=\left(\gamma_b mc,-\gamma_b m u_b,0,0\right)$

$P^\mu_c=\left(\dfrac{h\nu'}{c},-\dfrac{h\nu'}{c},0,0\right)$

Using the same conservation of momenergy than above

$\dfrac{2EE'}{c^2}+\gamma_b mE'-\gamma_b m\dfrac{u_b}{c}E'=\gamma_b m E+\gamma_b \dfrac{mu_b E}{c}$

Supposing that $u_b\approx c$ , and then $1-u_b/c\approx \dfrac{1}{2}\left(1+\dfrac{u_b}{c}\right)\left(1-\dfrac{u_b}{c}\right)=\dfrac{1}{2}\left(1-\dfrac{u_b^2}{c^2}\right)=\dfrac{1}{2}\dfrac{1}{\gamma_b^2}$

Thus,

$\dfrac{2EE'}{c^2}+\dfrac{mE'}{2\gamma_b}=2\gamma_b mE$

$\dfrac{E'}{E}=\dfrac{2\gamma_b m}{\dfrac{2E}{c^2}+\dfrac{m}{2\gamma_b}}=\dfrac{4\gamma_b^2}{1+\dfrac{4\gamma_b E}{mc^2}}$

This inverse Compton effect is important of importance in Astronomy. PHotons of the microwave background radiation (CMB), with a very low energy of the order of $E\approx 10^{-3}eV$ , are struck by very energetic electrons (rest energy mc²=511 keV). For typical values of $\gamma_b >>10^8$ , the second term in the denominator dominates, giving

$E'\approx \gamma_b\times 511keV$

Therefore, the inverse Compton effect can increase the energy of a photon in a spectacular way. If we don’t plut $u_b\approx c$ we would get from the equation:

$\dfrac{2EE'}{c^2}+\gamma_b mE'-\gamma_b m\dfrac{u_b}{c}E'=\gamma_b m E+\gamma m \dfrac{mu_b E}{c}$

$\gamma_b m E'\left(1-\dfrac{u_b}{c}+\dfrac{2E}{\gamma_b mc^2}\right)=\gamma_b m E\left(1+\dfrac{u_b}{c}\right)$

$\boxed{\dfrac{E'}{E}=\dfrac{1+\dfrac{u_b}{c}}{1-\dfrac{u_b}{c}+\dfrac{2E}{\gamma_b mc^2}}}$

If we suppose that the incident electron arrives with certain angle $l \alpha_i$ and it is scattered an angle $\alpha_f$ . Then, we would obtain the general inverse Compton formula:

$\boxed{\dfrac{E'_f}{E'_i}=\dfrac{1-\beta_i\cos\alpha_i}{1-\beta_i\cos\alpha_f+\dfrac{E'_i}{\gamma_i mc^2}\left(1-\cos\theta\right)}}$

$\boxed{\dfrac{E'_\gamma}{E_\gamma}=\dfrac{1-\beta_i\cos\alpha_i}{1-\beta_i\cos\alpha_f+\dfrac{E_\gamma}{E_e}\left(1-\cos\theta\right)}}$

In the case of $\alpha_f \approx 1/\gamma<<1$ , i.e., $\cos\alpha_f\approx 1$ , and then

$\dfrac{E'}{E}\approx \dfrac{1-\beta_i\cos\alpha_i}{1-\beta_i}\approx \left(1-\beta_i\cos\alpha_i\right)2\gamma_i^2$

In conclusion, there is an energy transfer proportional to $\gamma_i^2$ . There are some interesting “maximal boosts”, depending on the final energy (frequency). For instance, if $\gamma_i\approx 10^3-10^5$ , then $E_f\approx \gamma_i^2\times 511 keV$ provides:

a) In the radio branch: $1GHz=10^9Hz$ , a maximal boost $10^{15}Hz$ . It corresponds to a wavelength about 300nm (in the UV band).

b) In the optical branch: $4\times 10^{14}Hz$ , a maximal boost $10^{20}Hz\approx 1.6MeV$ . It corresponds to photons in the Gamma ray band of the electromagnetic spectrum.

Example 3. Bremsstrahlung.

An electron (a) with rest mass $m_a$ arrives from the left with velocity $u_a$ and it hits a nucleus (b) at rest with mass $m_b$ . After the collision, the cluster “c” moves with speed $u_c$ , and a photon is emitted (d) to the left. That photon is considered “a radiation” due to the recoil of the nucleus.

The equations of momenergy are now:

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

$2P\mu_{a}P_{d\mu}+2P^\mu_bP_{d\mu}=2P\mu_cP_{d\mu}+2P^\mu_dP_{d\mu}$

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

$P^\mu_a=\left(\gamma_am_ac,\gamma_am_au_a,0,0\right)$

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

$P^\mu_d=\left(\dfrac{E}{c},-\dfrac{E}{c},0,0\right)$

$(m_ac)^2+2\gamma_am_am_bc^2+(m_bc)^2-2\gamma_am_a(1+\beta_b)E-2m_bE=(m_a+m_b)^2c^2+0$

$2E\left(\gamma_am_a(1+\beta_a)+m_b\right)=2(\gamma_a-1)m_am_bc^2$

$\boxed{E=\dfrac{(\gamma_a-1)m_am_bc^2}{\gamma_a m_a(1+\beta_a)+m_b}}$

In clusters of galaxies, typical temperatures of $T\sim 10^7-10^8K$ provide a kinetic energy of proton and electron at clusters about $1.3-13keV$ . Relativistic kinetic energy is $E_k=(\gamma_a-1)m_ac^2$ and it yields $\gamma_a\sim 1.0025-1.025$ for hydrogen nuclei (i.e., protons $p^+$ ). If $\gamma_am_a(1+\beta_a)<<1$ , then we have $E\approx (\gamma_a-1)m_ac^2=(\gamma_a-1)\times 511keV$ . Then, the electron kinetic energy is almost completely turned into radiation (bremsstrahlung). In particular, bremsstrahlung is a X-ray radiation with $E\sim 1.3-13keV$ .