Relativistic collision and then quesion about cerenkov radiation

In summary, the problem at hand involves an incident photon beam being directed into a fog chamber, resulting in the observation of a Compton electron with a momentum of 1.5 Mev/c emitted at a 10º angle to the incident beam. The goal is to determine the energy of the incident and scattered photons, using conservation laws of energy and momentum and the relativistic relation E^2 = p^2c^2 + m^2c^4. To solve for the minimum energy required for the Compton electron to be detected using Cerenkov radiation (n=1.33), the 4-momenta of the photon and the electron must be applied, along with defining two angles \theta and \alpha.
  • #1
jonjacson
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Homework Statement


a)

We have an incident photon beam into a fog chamber, and we observe a compton electron with a moment of 1,5 Mev/c emitted in a 10º angle to the incident beam. ¿Which is the energy of the incident and scattered photons?

b)

Find the minimum energy that should have the photon for the compton electron could be detected using cerenkov radiation (n=1,33).

Homework Equations



4-moment of the photon and the electron

Conservation laws of energy and momentum

And the relativistic relation:

[tex]
E^2=p^2 c^2+ m^2c^4.
[/tex]

The Attempt at a Solution



I must apply conservation of moment and energy of the reaction [tex]\gamma[/tex] [tex]\epsilon[/tex] --->[tex]\gamma'[/tex] [tex]\epsilon'[/tex] :

[tex] p'_\gamma = p_e + p_\gamma -p'_e [/tex]

and then my problem is to find every component of the 4vector, i am not sure if this is right:

I have some general doubts:

I have seen the expression of the 4vector energy momentum in three different ways:

-In the minkowsky space: [tex]\textbf{P}=(m\textbf{v}, icm)[/tex]

-In another book i found: [tex]P=(mc\gamma,m\gamma\textbf{v})[/tex] then he uses this relationships: [tex]\textbf{p}=m\gamma\textbf{v}[/tex] and [tex]E=mc^2\gamma[/tex] to arrive at:

[tex]\textbf{P}=(E/c,\textbf{p})[/tex]

And i don't know which should i apply, in one of them the energy is the P_0 component, in the other you have the imaginary number i, i will use the equation without the i component.

before the collision:

-I will supose that the electron is at rest so p=0 and :

[tex]P_e=(m_ec,0,0,0) [/tex]

-For the photon, the mass is zero so :

[tex] E^2=p^2c^2[/tex] so the 4vector is:

[tex]P_\gamma=(E/c,E/c,0,0)[/tex]

¿Is anything wrong until now?.

And after the collision, I must define two angles[tex]\theta[/tex] and [tex]\alpha[/tex], which are the angles between the electron and the photon to the incident beam.

So now we have:

-The 4vector of the electron is:

[tex]P'_e=(E'_e/c,p_e cos(\theta),p_e sin(\theta),o), p_e [/tex] is the moduli of the thridimensional momentum

-for the photon:

[tex]P'_\gamma=(E'_\gamma/c,p_\gamma cos(\alpha),p_\gamma sin(\alpha),0)[/tex]Well , before applying conservation equations I would like to check if this is wrong ¿is it?

--------------------------------------------------------------------

b)

I think that i need understand part a) to solve part b).
 
Last edited:
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  • #2
Perhaps ¿Did i made anything wrong explaining the problem?:confused:
 

1. What is relativistic collision?

Relativistic collision is a type of collision that occurs between particles at speeds close to the speed of light. This type of collision follows the principles of Einstein's theory of relativity, which states that the laws of physics are the same for all observers moving at constant velocities.

2. How does relativistic collision differ from non-relativistic collision?

In non-relativistic collision, the particles involved have speeds much lower than the speed of light, so the effects of relativity are not significant. In relativistic collision, the particles have speeds close to the speed of light, causing an increase in energy and mass, and a change in the laws of physics.

3. What is the significance of relativistic collision in particle physics?

Relativistic collision is crucial in particle physics as it allows scientists to study the properties and behavior of particles at high energies. This helps in understanding the fundamental building blocks of the universe and the forces that govern them.

4. What is Cerenkov radiation?

Cerenkov radiation is a type of electromagnetic radiation that is emitted when a charged particle, such as an electron, moves through a medium at a speed greater than the speed of light in that medium. It appears as a faint blue light and is commonly seen in nuclear reactors and particle accelerators.

5. How is Cerenkov radiation related to relativistic collision?

Cerenkov radiation is a byproduct of relativistic collision as high-speed particles collide with a medium, causing them to exceed the speed of light in that medium and emit this characteristic blue light. Therefore, the presence of Cerenkov radiation is an indication of relativistic collision occurring.

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