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:


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)



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



\boxed{\Delta \lambda\equiv \lambda'-\lambda=\dfrac{h}{mc}\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:


\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_b=\left(\gamma_b mc,-\gamma_b m u_b,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}


\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 \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)}}


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:









\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.


Leave a Reply

Please log in using one of these methods to post your comment:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s