How Is Bjorken Scale Variable x Calculated in Neutrino-Proton Collisions?

[genloz]

Hi! I'm struggling with this question:
A neutrino of energy 10 GeV collides with a proton at rest producing a 9 GeV muon deflected at an angle of 0.01 radians away from the initial neutrino direction. The collision is used to probe the momentum distribution of down quarks in the proton.

Show that the requirement that the quark four-momentum be almost zero before and after the collision gives the Bjorken scale variable x. Calculate x for this process.

I know that it follows something along the lines of
(P'+q)^2=0
P'^2+2P'*q+q^2=2xP*q-Q^2=0
x=Q^2/(2P*q)

but I'm a little uncertain about what all the letters represent and how it's all derived...
Could anyone lend a hand explaining please?

 

[AndyW]

Hi there! I'll try my best to explain the derivation of the Bjorken scale variable x in this scenario.

Firstly, let's define the variables we are working with:

P = four-momentum of the proton (before collision)
P' = four-momentum of the proton (after collision)
q = four-momentum transfer (neutrino - muon)
Q^2 = square of the four-momentum transfer

Now, let's look at the equation you mentioned:

(P'+q)^2 = 0

This equation represents the conservation of four-momentum in the collision. Since the proton is at rest before the collision, its four-momentum is simply its rest mass energy (P = m_p). After the collision, the proton's four-momentum becomes P' = (E_p', p_p'), where E_p' is the energy and p_p' is the momentum of the proton.

The four-momentum transfer, q, can be calculated by taking the difference between the four-momentum of the initial and final particles (q = P' - P). This results in the following equation:

P'^2 + 2P'q + q^2 = 0

Substituting the values for P' and q, we get:

(E_p')^2 + 2E_p'q + (p_p')^2 - (E_p)^2 = 0

Since the proton is at rest before the collision, its momentum (p_p) is zero. Therefore, the equation simplifies to:

(E_p')^2 + 2E_p'q - (E_p)^2 = 0

Now, let's look at the momentum distribution of down quarks in the proton. The Bjorken scale variable x is defined as:

x = Q^2 / (2P * q)

where Q^2 is the square of the four-momentum transfer, P is the four-momentum of the proton, and q is the four-momentum transfer.

In our scenario, we can calculate x by substituting the values we have derived:

x = (E_p')^2 / (2E_p'q)

Since we know that the quark four-momentum should be almost zero before and after the collision, we can assume that E_p' is approximately equal to the proton's rest mass energy (m_p'). Therefore, the equation simplifies to:

x = m_p'^2 / (

 

[sir-pinski]

Sure, I'd be happy to explain the concept of the Bjorken scale variable x and how it relates to the given scenario. The Bjorken scale variable x is a key concept in the field of high energy physics, specifically in the study of deep inelastic scattering (DIS). It is defined as the ratio of the energy transfer (Q^2) to the square of the center of mass energy (s) of the colliding particles. In other words, it represents the fraction of the total energy of the colliding particles that is carried by the interacting partons (quarks and gluons).

In the given scenario, a neutrino with energy 10 GeV collides with a proton at rest, producing a 9 GeV muon deflected at an angle of 0.01 radians. This process can be described by the equation P+q=P'+q', where P and q are the initial four-momenta of the proton and neutrino respectively, and P' and q' are the final four-momenta of the proton and muon respectively.

To understand how this relates to the Bjorken scale variable x, let's look at the conservation of four-momentum in this process. Before the collision, the total four-momentum of the system is P+q, and after the collision, it is P'+q'. We can rewrite this as P+q=P'+q', and rearrange it to get P'-P=q-q'. This equation represents the conservation of four-momentum, where the difference between the initial and final momenta is equal to the momentum transfer (q-q').

Now, we can use this equation to calculate the Bjorken scale variable x. Since we are interested in the momentum distribution of down quarks in the proton, we can assume that the proton is made up of three valence quarks (two up quarks and one down quark). This means that the initial four-momentum of the proton, P, is equal to the four-momentum of the down quark inside the proton. Similarly, the final four-momentum of the proton, P', is equal to the four-momentum of the down quark after the collision.

Substituting these values into the conservation of four-momentum equation, we get P'-P=q-q', which can be rewritten as P'^2-P^2=2P'*q-2P*q. Since the quark four-momentum