In Chapter 8, we evaluated the ep →eX cross section for the electron to be scattered into the d

Step 1: Start by understanding the variables and their relationships. We have: - ( E ) and ( E'

Question

In Chapter 8, we evaluated the ep $\rightarrow$ eX cross section for the electron to be scattered into the $d E^{\prime} d \Omega$ element in the target proton rest frame (the laboratory frame). Show that $$ d E^{\prime} d \Omega=\frac{\pi}{E E^{\prime}} d Q^2 d \nu=\frac{2 M E}{E^{\prime}} \pi y d x d y, $$ where $x$ and $y$ are the dimensionless variables $$ x=\frac{Q^2}{2 M \nu}, \quad y=\frac{p \cdot q}{p \cdot k_{(\mathrm{lab} .)}}=\frac{\nu}{E}, $$ see (8.30). The allowed kinematic region $(0 \leq x, y \leq 1)$ is shown in Fig. 9.3.

   In Chapter 8, we evaluated the ep $\rightarrow$ eX cross section for the electron to be scattered into the $d E^{\prime} d \Omega$ element in the target proton rest frame (the laboratory frame). Show that
$$
d E^{\prime} d \Omega=\frac{\pi}{E E^{\prime}} d Q^2 d \nu=\frac{2 M E}{E^{\prime}} \pi y d x d y,
$$
where $x$ and $y$ are the dimensionless variables
$$
x=\frac{Q^2}{2 M \nu}, \quad y=\frac{p \cdot q}{p \cdot k_{(\mathrm{lab} .)}}=\frac{\nu}{E},
$$
see (8.30). The allowed kinematic region $(0 \leq x, y \leq 1)$ is shown in Fig. 9.3.

Quarks and leptons: introductory course in modern particle physics

Francis Halzen, Alan… 1st Edition

Chapter 9, Problem 4

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Step 1

We have: - $ E $ and $ E' $ as the initial and final energies of the electron, - $ \Omega $ as the solid angle into which the electron is scattered, - $ Q^2 $ as the negative of the square of the four-momentum transfer $ q $, - $ \nu $ as the energy Show more…

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In Chapter 8, we evaluated the ep $\rightarrow$ eX cross section for the electron to be scattered into the $d E^{\prime} d \Omega$ element in the target proton rest frame (the laboratory frame). Show that $$ d E^{\prime} d \Omega=\frac{\pi}{E E^{\prime}} d Q^2 d \nu=\frac{2 M E}{E^{\prime}} \pi y d x d y, $$ where $x$ and $y$ are the dimensionless variables $$ x=\frac{Q^2}{2 M \nu}, \quad y=\frac{p \cdot q}{p \cdot k_{(\mathrm{lab} .)}}=\frac{\nu}{E}, $$ see (8.30). The allowed kinematic region $(0 \leq x, y \leq 1)$ is shown in Fig. 9.3.

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