What Is Deep Inelastic Scattering and How Does It Reveal Quark Structure?

[komigen]

Hey guys,
I have some questions about deep inelastic scattering. I have read a bit about it, but don't really understand the theroy behind it. I know that it is a method to be able to study quarks inside, example with an electron hitting a proton target one can be able to study the quarks of the proton. but my questions are:
1) how can we briefly describe what is meant by inelastic scattering compared to elastic scattering?

2) draw a rough (feynmann) diagram to show a deep inelastic scattering process, whereby an electron scatters of a fixed target proton. ( Here i think of this diagram to be the electron and proton will interact and as a result an electorn, with different energy and momentum) is introduced with hadrons). is it right? and in this case how can we be able to see inside the protons since the proton now produces new particles?

3) using the feynmann diagram or, otherwise, prove the relation:
x=-q²/2P_quark.q
where x is Bjorken x( the fraction of momentum the scattered quark has from the initial proton). P_quark is the momentum of the quark and q is the momenum transffered by the scattering force midiator

d) can we directly measure x, if not how can we infer it from experiment?

I really need to understand these questions for my exam, will we great if someone can help me

 

[tom.stoer]

Regarding 1)
Inelastic scattering means that the particles in the initial and final states differ, i.e.that the proton "does not survive"
elastic: e p => e' p'
inelastic: e p => e' X

Regarding 2)
Data extracted from the DIS process combine two things:
i) non-perturbative information regarding the momentum fraction x carried by the quark described via struture functions F(x)
ii) perturbative information regarding the scattering of the electron with the free (quasi-free with correction) quark with fraction x
So we can zoom into the proton via extracting F(x) from the inclusive cross section; F(x) is process independent

Regarding 4)

$$x = \frac{Q^2}{2M\nu}$$

In the proton rest frame Q² = -q² and nu = (E-E') are the momentum transfer and the energy transfer between proton and electron. Therefore x can be extracted via detecting the scattered electron and measuring its energy and momentum.

 

[komigen]

Regarding 1)
Inelastic scattering means that the particles in the initial and final states differ, i.e.that the proton "does not survive"
elastic: e p => e' p'
inelastic: e p => e' X

Regarding 2)
Data extracted from the DIS process combine two things:
i) non-perturbative information regarding the momentum fraction x carried by the quark described via struture functions F(x)
ii) perturbative information regarding the scattering of the electron with the free (quasi-free with correction) quark with fraction x
So we can zoom into the proton via extracting F(x) from the inclusive cross section; F(x) is process independent

Regarding 4)

$$x = \frac{Q^2}{2M\nu}$$

In the proton rest frame Q² = -q² and nu = (E-E') are the momentum transfer and the energy transfer between proton and electron. Therefore x can be extracted via detecting the scattered electron and measuring its energy and momentum.

Thanx, But I don't really understand 2), I can draw the digram, but don't understand the 2 points you wrote.
In 4 you wrote it's allowed to measure x directly, but we don't know P_quark right? how can we measure the momentum of the quark in this case, and which quark do they mean, u or d? or is it the mass of the proton now since it is at rest?

 

[tom.stoer]

We do neither know nor measure the quark momentum directly. Instead we prepare / measure the initial / final state energy and momentum of the electron, therefore we know Q² and nu; now we can calculate x. As we know that the elementary process is between electron and quark (not the proton) we can extract the x as fraction of the momentum carried by the quark.

For one scatterer (= the proton itself) the distribution of x must be peaked at x=1; for three scatterers (three quarks inside the proton) we expect a peak at x=1/3; but as there are additonal sea-quarks visible in a certain domain this peak is smeared which is the effect of F(x).

To get this idea (just the kinematics) forget about bound quarks and think about three collinear, free quarks moving exactly with P/3.

 

[komigen]

now I got it, But in this case M is the mass on the quark, but we have quarks u,u,d for the proton, how can we know which mass we are going to use?

 

[tom.stoer]

no, M is the mass of the proton

http://www.phy.uct.ac.za/courses/phy400w/particle/dis.pdf

 

[daqpan]

On a related note, I have a probably stupid question:

We can calculate the scattering angle from the inelasticity according to :

\begin{equation}
y=\frac{1-\text{cos}(\theta)}{2}
\end{equation}

so that:

\begin{equation}
\theta=\text{cos}^{-1}(1-2y)
\end{equation}

From simulation I know y. My question is this: is $\theta$ in radians or degrees?

 

[jtbell]

is $\theta$ in radians or degrees?

It depends on the degree/radian setting of your calculator.