# LOG#040. Relativity: Examples(IV).

**Posted:**2012/10/07

**Filed under:**Physmatics, Relativity |

**Tags:**Astrophysics, bremsstrahlung, compton effect, exercises, inverse compton effect, Relativity Leave a comment

**Example 1. Compton effect. **

Let us define as “a” a photon of frequency . Then, it hits an electron “b” at rest, changing its frequency into , 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 .

We use momenergy conservation:

We multiply this equation by to deduce that

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

Remembering the definitions and and inserting the values of the momenta into the respective equations, we get

or

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

**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 and the angle with respect to the direction of motion .

The momenergy reads

Using the same conservation of momenergy than above

Supposing that , and then

Thus,

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 , are struck by very energetic electrons (rest energy mc²=511 keV). For typical values of , the second term in the denominator dominates, giving

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

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

In the case of , i.e., , and then

In conclusion, there is an energy transfer proportional to . There are some interesting “maximal boosts”, depending on the final energy (frequency). For instance, if , then provides:

a) In the radio branch: , a maximal boost . It corresponds to a wavelength about 300nm (in the UV band).

b) In the optical branch: , a maximal boost . It corresponds to photons in the Gamma ray band of the electromagnetic spectrum.

**Example 3. Bremsstrahlung.**

An electron (a) with rest mass arrives from the left with velocity and it hits a nucleus (b) at rest with mass . After the collision, the cluster “c” moves with speed , 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:

In clusters of galaxies, typical temperatures of provide a kinetic energy of proton and electron at clusters about . Relativistic kinetic energy is and it yields for hydrogen nuclei (i.e., protons ). If , then we have . Then, the electron kinetic energy is almost completely turned into radiation (bremsstrahlung). In particular, bremsstrahlung is a X-ray radiation with .