We outline the various stages of the calculation. First,...

We outline the various stages of the calculation. First, show that for $\gamma^{*}(q) q(p) \rightarrow q\left(p^{\prime}\right)$, $$ \overline{|\mathscr{M}|^{2}}=2 e_{i}^{2} e^{2} p \cdot q $$ where we have averaged over transverse polarization states of the incoming $\gamma^{*}$. From Section $4.3$, we have $$ F d \dot{\sigma}_{T}=\overline{\left.|\mathscr{}|\right|^{2}}(2 \pi)^{4} \delta^{(4)}\left(p^{\prime}-p-q\right) \frac{d^{3} p^{\prime}}{2 p_{0}^{\prime}(2 \pi)^{3}} $$ where $F$ is the $\gamma^{*} \mathrm{q}$ flux factor. Calculate $F \hat{\sigma}_{T}$ by making use of $(6.47) .$ Use $F \hat{\sigma}_{0}=8 \pi^{2} \alpha$, see (10.5). To determine the parton model prediction for $F_{2} / x$ of (10.4), we input in (10.8) the following cross section ratio for $\gamma^{*} \mathrm{q} \rightarrow \mathrm{q}$ : $$ \frac{1}{\partial_{0}}\left(\hat{\sigma}_{T}+\hat{\sigma}_{L}\right)=e_{i}^{2} \delta(1-z), $$ see $(10.10)$ and (10.11). After substitution, we obtain $$ \frac{F_{2}\left(x, Q^{2}\right)}{x}=\sum_{i} e_{i}^{2} \int_{x}^{1} \frac{d y}{y} f_{i}(y) \delta\left(1-\frac{x}{y}\right)=\sum_{i} e_{i}^{2} f_{i}(x) $$ An identical expression is found for $2 F_{1}$. The parton model results of (9.13) and (9.14) are indeed reproduced.Show that the parton model diagram, Fig. 10.1, gives $$ \begin{aligned} &\frac{\hat{\sigma}_{T}\left(z, Q^{2}\right)}{\hat{\sigma}_{0}}=e_{i}^{2} \delta(1-z) \\ &\hat{\sigma}_{L}\left(z, Q^{2}\right)=0 \end{aligned} $$

We outline the various stages of the calculation. First, show that for $\gamma^{*}(q) q(p) \rightarrow q\left(p^{\prime}\right)$,
$$
\overline{|\mathscr{M}|^{2}}=2 e_{i}^{2} e^{2} p \cdot q
$$
where we have averaged over transverse polarization states of the incoming $\gamma^{*}$. From Section $4.3$, we have
$$
F d \dot{\sigma}_{T}=\overline{\left.|\mathscr{}|\right|^{2}}(2 \pi)^{4} \delta^{(4)}\left(p^{\prime}-p-q\right) \frac{d^{3} p^{\prime}}{2 p_{0}^{\prime}(2 \pi)^{3}}
$$
where $F$ is the $\gamma^{*} \mathrm{q}$ flux factor. Calculate $F \hat{\sigma}_{T}$ by making use of $(6.47) .$ Use $F \hat{\sigma}_{0}=8 \pi^{2} \alpha$, see (10.5).

To determine the parton model prediction for $F_{2} / x$ of (10.4), we input in (10.8) the following cross section ratio for $\gamma^{*} \mathrm{q} \rightarrow \mathrm{q}$ :
$$
\frac{1}{\partial_{0}}\left(\hat{\sigma}_{T}+\hat{\sigma}_{L}\right)=e_{i}^{2} \delta(1-z),
$$
see $(10.10)$ and (10.11). After substitution, we obtain
$$
\frac{F_{2}\left(x, Q^{2}\right)}{x}=\sum_{i} e_{i}^{2} \int_{x}^{1} \frac{d y}{y} f_{i}(y) \delta\left(1-\frac{x}{y}\right)=\sum_{i} e_{i}^{2} f_{i}(x)
$$
An identical expression is found for $2 F_{1}$. The parton model results of (9.13) and (9.14) are indeed reproduced.Show that the parton model diagram, Fig. 10.1, gives
$$
\begin{aligned}
&\frac{\hat{\sigma}_{T}\left(z, Q^{2}\right)}{\hat{\sigma}_{0}}=e_{i}^{2} \delta(1-z) \\
&\hat{\sigma}_{L}\left(z, Q^{2}\right)=0
\end{aligned}
$$

Key Concepts

Deep Inelastic Scattering

Deep inelastic scattering (DIS) is a process in which a high-energy lepton (or photon) scatters off a hadron, probing its internal structure at very short distance scales. The virtual photon from the lepton interacts with an individual parton, allowing researchers to explore the momentum distribution of the partons inside the hadron and extract structure functions that characterize the hadron's internal dynamics.

Delta Function and Momentum Conservation

The Dirac delta function plays a crucial role in scattering theory by enforcing energy-momentum conservation in the interaction. In the calculation of differential cross sections, the delta function ensures that only those events that satisfy the conservation laws contribute to the measured cross section, and it often simplifies the integration over final state momenta by localizing the integrals to well-defined kinematical conditions.

Flux Factors and Differential Cross Sections

The flux factor in scattering processes quantifies the relative motion of the incoming particles and is essential for normalizing the differential cross section of the interaction. It appears in the relation between the squared invariant amplitude and the measurable differential cross section, ensuring that both energy-momentum conservation and particle density effects are correctly incorporated in theoretical predictions.

Parton Model

The parton model is a framework used in high-energy physics to describe the internal structure of hadrons as being composed of quasi-free constituents called partons (quarks and gluons). It provides the basis for understanding deep inelastic scattering by treating the interaction between an incoming virtual photon and a single parton, simplifying the complex dynamics into factorized contributions from parton distributions and elementary cross sections.

Structure Functions

Structure functions, such as F2 and F1, encapsulate the distribution of momentum and spin among the partons inside a hadron. These functions are determined experimentally through deep inelastic scattering processes and are directly related to the parton distribution functions. They provide essential tests of quantum chromodynamics (QCD) and the validity of the parton model by linking the measured cross sections to the internal structure of the hadron.

Invariant Amplitude and Averaging

In scattering processes, the invariant amplitude squared (|M|^2) is a key quantity that determines the probability of a given interaction. In theoretical calculations, averaging over the initial state polarizations (for instance, the transverse polarizations of the virtual photon) and summing over final states is necessary to compare with experimental observables. This averaging ensures that the calculated cross sections are representative of the physical process.

Transcript

00:01 Hi friends the average kinetic energy for the motion of frequency omega is given by half n into m average velocity at a given frequency is square say equation one average kinetic energy will be half of total energy here a not is h cross omega y2 but a not is h cross omega y 2 but a not is h cross omega so average kinetic energy you will get h cross omega by 2 using 1 we can write h cross omega by 2 is called to half of n into m square of average velocity so square of average velocity you will get h cross omega upon 2 and m we have v square omega is is called to omega square or omega square and m is called to m is called to roe so from two we can write v square omega just a moment omega square or omega square average is called to h cross omega upon two roe v so r square average you will get h cross upon 2 row v to average value of 1 by omega average value of 1 by omega having the value 0 omega d d omega d omega d omega d omega it is to be h cross upon 2 ro b g omega d x to be h cross omega say equation number three the density state is given by d omega into t omega is equal to three times v four pi k square d k upon two pi whole cube using omega is called to b k we will get d as a function of omega into d omega is equal to three times v 4 pi omega square d omega divided by two pi whole cube v that is 3b omega square t omega 2 pi square v in 3 if we substitute we will get average value of r square to be h cross upon 2 row b 3 v upon 2 pi square bq zero integration zero to omega t omega t omega so average value of r square you will get three h cross omega square t upon eight row pi square b cube now second part we will start in one day density state will be 3 into 2 l d k upon 2 pi that is 3 l d omega by pi using third equation average value of r square will be h cross upon 2 row l 3l upon pi into v integration 0 to omega d d omega y omega which is tending to infinity so r square average having the value infinite the average kinetic energy of motion in 3d average kinetic energy in 1d you will write get 1 by 2 nc or an average value of square of its velocity it is equal to e0 by 2 r h cross omega by 4 so average value of half nc omega square average value of r square omega will be h cross omega by 4 so it is to be r square omega average value to be h cross omega by 2 c n so we can write r square average value to be h cross upon 2c n integration from 0 to omega d omega omega in omega into d function of omega d omega d for one bowl density straight will be 2 l d k by 2 pi that is l d omega d omega d omega into pi into b substitute the value.

08:21 Average value of r square, you will get h cross upon 2c n 0 to omega d...

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