Question 3.5: The Cross Section as a Function of x and y Derive (3.45) and......
The Cross Section as a Function of x and y
Derive (3.45) and (3.46).
{\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}E^{\prime}\mathrm{d}\Omega}}={\frac{E^{\prime}}{2\pi M_{\mathrm{N}}E_{y}}}{\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}x\mathrm{d}y}}\ . (3.45)
\left.\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}x\,\mathrm{d}y}\right|_{\mathrm{eN}}=\frac{8\pi M_{\mathrm{N}}E\alpha^{2}}{Q^{4}}\times\Biggl[{x y}^{2}{F}_{1}^{\mathrm{eN}}(Q^{2},x)+\Biggl(1-y-{\frac{M_{\mathrm{N}}x y}{2E}}\Biggr){F}_{2}^{\mathrm{eN}}({Q}^{2},x)\Biggr]. (3.46)
Equations (3.44), (3.39), and (3.30) yield
y,=\,\frac{\nu}{M_{\mathrm{N}}E}=\frac{E-E^{\prime}}{E}\ . (3.44)
x=\frac{Q^{2}}{2\nu}=\frac{-q^{2}}{2P\cdot q}=\frac{-q^{2}}{-q^{2}}=1\ . (3.39)
Q^{2}=2\;p\cdot p^{\prime}=2(E E^{\prime}-p\cdot p^{\prime})=4E E^{\prime}\sin^{2}\left(\frac{\theta}{2}\right)\;. (3.30)
y={\frac{E-E^{\prime}}{E}}=1-{\frac{E^{\prime}}{E}}\ ,x=\frac{Q^{2}}{2(E-E^{\prime})M_{\mathrm{N}}}=\frac{2E E^{\prime}}{(E-E^{\prime})M_{\mathrm{N}}}\sin^{2}\left(\frac{\theta}{2}\right)\ . (1)
Owing to the cylindrical symmetry of the problem we have in addition
{\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}E^{\prime}\mathrm{d}\Omega}}={\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}E^{\prime}2\pi\sin(\theta)\mathrm{d}\theta}}\ . (2)
Therefore we only have to evaluate the Jacobi determinant
{\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}E^{\prime}\mathrm{d}\Omega}}={\frac{1}{2\pi\sin(\theta)}}\;\left|{\frac{\partial(x,y)}{\partial(E^{\prime},\theta)}}\right|{\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}x\mathrm{d}y}} =\left|\begin{matrix} \frac{2E\sin^{2}\!\left(\frac{\theta}{2}\right)}{M_{N}}\left(\frac{E^{\prime}+(E-E^{\prime})}{(E-E^{\prime})^{2}}\right) & \frac{2E E^{\prime}}{(E-E^{\prime})M_{N}}\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \\ -\frac{1}{E} & 0 \end{matrix} \right|\frac{1}{2\pi\;\sin(\theta)}\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}x\mathrm{d}y} ={\frac{1}{2\pi\sin(\theta)}}{\frac{2E^{\prime}}{(E-E^{\prime})M_{\mathrm{N}}}}{\frac{\sin(\theta)}{2}}{\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}x\,\mathrm{d}y}}={\frac{E^{\prime}}{2\pi M_{\mathrm{N}}E y}}{\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}x\,\mathrm{d}y}}\ . (3)
This is identical to (3.45). With the help of the definitions introduced in (3.39)– (3.44) the double differential cross section for electron–nucleon scattering consequently becomes
{\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}x\mathrm{d}y}}={\frac{8\pi M_{\mathrm{N}}y E E^{\prime}\alpha^{2}}{Q^{4}}}\biggl[{\frac{2}{M_{\mathrm{N}}}}\sin^{2}\left({\frac{\theta}{2}}\right)F_{1}^{\mathrm{eN}}(Q^{2},x)+{\frac{\cos^{2}\left({\frac{\theta}{2}}\right)}{y E}}F_{2}^{\mathrm{eN}}(Q^{2},x)\biggr]. (4)
Here we have employed (3.32) and (3.43). Now we replace θ by x and y, using (1):
\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}E^{\prime}\mathrm{d}\Omega}\bigg|_{\mathrm{eN}}=4E^{\prime2}\frac{\alpha^{2}}{Q}\bigg[2\,\sin^{2}\left(\frac{\theta}{2}\right)W_{1}\left(Q^{2},\nu\right)+\cos^{2}\left(\frac{\theta}{2}\right)W_{2}\left(Q^{2},\nu\right)\bigg]\,. (3.32)
M_{\mathrm{N}}W_{1}^{\mathrm{eN}}(Q^{2},\,\nu)=F_{1}^{\mathrm{eN}}(Q^{2},\,x)\ , \\\frac{\nu}{M_{\mathrm{N}}}W_{2}^{\mathrm{eN}}(Q^{2},\nu)=F_{2}^{\mathrm{eN}}(Q^{2},x)\ ,\\\frac{\nu}{M_{\mathrm{N}}}W_{3}^{\nu\mathrm{N}}(Q^{2},\nu)=F_{3}^{\mathrm{eN}}(Q^{2},x)\ . (3.43)
\sin^{2}\left(\frac{\theta}{2}\right)=\frac{(E-E^{\prime})M_{\mathrm N}}{2E E^{\prime}}x=\frac{M_{\mathrm N}}{2E^{\prime}}x y\;\;, (5)
\frac{\mathrm{d}^{2}\sigma}{\mathrm{d}x\mathrm{d}y}=\frac{8\pi M_{\mathrm{N}}y E E^{\prime}\alpha^{2}}{Q^{4}}\biggl[\frac{x y}{E^{\prime}}F_{1}^{\mathrm{eN}}(Q^{2},x)+\frac{1-\frac{M_{\mathrm{N}}}{2E^{\prime}}x y}{y E}F_{2}^{\mathrm{eN}}(Q^{2},x)\biggr]=\frac{8\pi M_{\mathrm{N}}E\alpha^{2}}{Q^{4}}\biggl[x y^{2}F_{1}^{\mathrm{eN}}(Q^{2},x)+\biggl(1-y-\frac{M_{\mathrm{N}}}{2E}x y\biggr)F_{2}^{\mathrm{eN}}(Q^{2},x)\biggr]\ , (6)
which yields (3.46).