Electron Scattering from a Dirac Proton (the effect of recoil

We treat the proton as a structureless Dirac particle. If we know the current of the proton, $J^\mu(x)$, we can calculate using Maxwell's equations the field it generates. We first find the electromagnetic field produced by the proton. The potential satisfies

$$\Box A^\mu(x) = J^\mu(x) ,$$ (7.133)

where we are working in the Lorentz gauge. We are free to choose the most conveniant gauge. We are working in the Heaviside-Lorentz system of units. In the Gaussian system of units a $4\pi$ would appear in front of the current. To solve the equation, we introduce the Green function, $D_F(x-y)$, define by

$$\Box D_F(x-y) = \delta^4(x-y).$$ (7.134)

Its Fourier representation is

$$D_F(x-y) = \int\frac{d^4q}{(2\pi)^4} e^{-iq\cdot(x-y)} D_F(q^2) ,$$ (7.135)

where $D_F(q^2)=-1/q^2$, for $q^2\ne 0$.

In analogy with the Dirac propagator, we append an infinitesimally small positive imaginary part to $q^2$. This is equivalent to adding a small negative imaginary mass

$$\fbox{$\displaystyle D_F(q^2) = \frac{-1}{q^2+i\epsilon} $}\ .$$ (7.136)

This is the so-called ``photon propagator''. The interacting photon is said to be ``virtual'', or ``off-mass-shell'', and $q^2$ is termed the mass squared of the virtual photon. That the photons are virtual is clear since for real photons, the energy-momentum vector $q_\mu$ satisfies $q^2=0$, whence the propagator becomes infinite. Likewise for the Dirac equation, the propagator becomes infinite when the quanta are on their ``mass shell''; $p^2-m^2=0$.

The Feynman propagator for electromagnetic radiation is

$$D_F(x-y) = \int\frac{d^4q}{(2\pi)^4} e^{-iq\cdot(x-y)} \left( \frac{-1}{q^2+i\epsilon} \right) .$$ (7.137)

Using Green's function techniques, the solution for the potential is

$$A^\mu(x) = \int d^4y D_F(x-y) J^\mu(y) ,$$ (7.138)

since

$$\Box_x A^\mu(x) = \int d^4y[\Box_x D_F(x-y)]J^\mu(y) = \int d^4y \delta^4(x-y)J^\mu(y) = J^\mu(x) .$$ (7.139)

In a more general gauge

$$A^\mu(x) = \int d^4y D_F^{\mu\nu}(x-y) J_\nu(y) .$$ (7.140)

It carries Lorentz indices because the photon is a spin-1 particle. In our case, the tensor $D_F^{\mu\nu}$ is just proportional to $g^{\mu\nu}$ so that the tensor indices can be discarded for convenience by defining $D_F^{\mu\nu}=g^{\mu\nu}D_F$.

We could also, of course, append a solution of the homogeneous equation here. However, the solution which we wish for $A^\mu(x)$ is that which vanishes in the absence of a transition density $J^\mu(x)$ (this is true for all our Green function equations).

The $S$-matrix is

$$S_{fi} = -ie\int d^4x \overline{\psi}_f(x) \not{\!\!A}(x) \Psi_i^{(+)}(x) ,$$

$$= -ie \int d^4x \overline{\psi}_f(x) \int d^4y D_F(x-y) \not{\!J}(y) \psi_i(x) ,$$

$$= -i\int d^4x d^4y [e \overline{\psi}_f(x)\gamma_\mu \psi_i(x)] D_F(x-y) J^\mu(y) .$$ (7.141)

$e\overline{\psi}_f(x)\gamma_\mu\psi_i(x)$ is the current of the electron. The problem now is to decide what to choose for the proton current $J^\mu(y)$? Its reasonable to try

$$J^\mu(y) = e_P \overline{\psi}_f^P(y) \gamma^\mu \psi^P_i(y) ,$$ (7.142)

where $e_P=-e$ is the proton charge. $\psi_i^P(y)$ and $\overline{\psi}_f^P(y)$ represent the initial and final plane wave solutions for a free proton. Using plane-wave solutions for a proton gives

$$J^\mu(y) = -\sqrt{ \frac{M^2}{E_f^\prime E_i^\prime}} \frac{... ...P_f-P_i)\cdot y} \overline{u}(P_f,S_f) \gamma^\mu u(P_i,S_i) ,$$ (7.143)

where $P_i$ and $P_f$ are the initial and final proton four-momentum and $M$ is its mass.

We now get for the $S$-matrix

$$S_{fi} = i\int d^4xd^4y \left[e\sqrt{\frac{m}{E_iV}} \sqrt{\frac{m}{E_fV}} e^{i(p_f-p_i)\cdot x} \overline{u}(p_f.s_f) \gamma_\mu u(p_i,s_i) \right]$$

$$\cdot \int \frac{d^4q}{(2\pi)^4} e^{-iq\cdot(x-y)} \left( \frac{-... ...e V}} e^{i(P_f-P_i)\cdot y} \overline{u}(P_f,S_f) \gamma^\mu u(P_i,S_i) \right]$$

$$= -\frac{ie^2}{V^2} \int \frac{d^4xd^4yd^4q}{(2\pi)^4} \sqrt{\frac{... ...ac{M^2}{E_i^\prime E_f^\prime}} e^{i(p_f-p_i-q)\cdot x} e^{i(P_f-P_i+q)\cdot y}$$

$$\cdot \left[ \overline{u}(p_f,s_f) \gamma_\mu u(p_i,s_i) \right] \frac{1}{q^2+i\epsilon} \left[ \overline{u}(P_f,S_f) \gamma^\mu u(P_i,S_i) \right]$$

$$= -\frac{ie^2}{V^2} \int d^4q \sqrt{\frac{m^2}{E_fE_i}} \sqrt{\frac{M^2}{E_f^\prime E_i^\prime}} (2\pi)^4 \delta(p_f-p_i-q) \delta(P_f-P_i+q)$$

$$\cdot \left[ \overline{u}(p_f,s_f) \gamma_\mu u(p_i,s_i) \right] ... ...{1}{q^2 + i\epsilon} \left[ \overline{u}(P_f,S_f) \gamma^\mu u(P_i,S_i) \right]$$

$$= -\frac{ie^2}{V^2}(2\pi)^4 \delta(P_f-P_i+p_f-p_i) \sqrt{ \frac{m^2}{E_fE_i} } \sqrt{ \frac{M^2}{E_f^\prime E_i^\prime} }$$

$$\cdot \left[ \overline{u}(p_f,s_f) \gamma_\mu u(p_i,s_i) \right] ... ..._i)^2 + i\epsilon} \left[ \overline{u}(P_f,S_f) \gamma^\mu u(P_i,S_i) \right] .$$ (7.144)

Notice the symmetry in electron and proton variables.

This gives the electron-proton scattering amplitude to lowest order in $\alpha$. Higher order interaction effects which distort the plane waves that were inserted in the currents have been ignored. The expression for the $S$-matrix may be represented by a Feynman diagram as shown in figure 7.4.

Figure 7.4: Feynman diagram for electron-proton scattering.

For each line and intersection of the graph there corresponds a unique factor in the $S$-matrix. In addition, $S_{fi}$ always contains a four-dimensional $\delta$-function expressing overall energy-momentum conservation. A solid line with an arrow pointing toward positive time represents the electron and a double line the proton. The wavy line represents the influence of the electromagnetic interaction, which is expressed in the matrix element by the reciprocal of the square of the momentum transfer, or the inverse d'Alembertian in momentum space. We refer to this line as representing a ``virtual photon'' exchanging four-momentum $q=p_f-p_i=P_i-P_f$ between the electron and proton. The amplitude for the virtual photon to propagate between the two currents is $-(q^2+i\epsilon)^{-1}$. At the points (or vertices) on which the photon lands there operate factors $e\gamma^\mu$ sandwiched between spinors $\sqrt{m/E}u(p,s)$ representing the free, real incident and outgoing particles. To get the spinor factor in expressions such as these, the rule is to start at the ingoing fermion line $u$ and follow the line through until the end, inserting vertices and propagators in the right order, until you reach the outgoing state $\overline{u}$. There is a uniform rule for the factors of $i$: $-i$ for each vertex and $i$ for each internal line in the graph.

We now form a transition rate per unit volume by dividing $\vert S_{fi}\vert^2$ by the time interval of observation $T$ and by the spatial volume $V$ of the interaction region. This gives

$$w_{fi} = \frac{\vert S_{fi}\vert^2}{VT} = (2\pi)^4 \delta^4(... ...ac{M^2}{E_f^\prime E_i^\prime} \vert\mathcal{M}_{fi}\vert^2 ,$$ (7.145)

where

$$\mathcal{M}_{fi} = [\overline{u}(p_f,s_f) \gamma_\mu u(p_i,s... ...^2 + i\epsilon} [\overline{u}(P_f,S_f) \gamma^\mu u(P_i,S_i)]$$ (7.146)

is a Lorentz invariant matrix element and will be called the invariant amplitude. The choice of this name is quite natural since the matrix element consits of scalar products of four-vectors, which is Lorentz invariant. We have used, in analogy to before, the square of the $\delta$-function

$$[(2\pi)^4\delta^4(P_f+p_f-P_i-p_i)]^2 = (2\pi)^4 \delta^4(0) (2\pi)^4 \delta^4(P_f+p_f-P_i-p_i)$$

$$\rightarrow VT(2\pi)^4 \delta^4(P_f+p_f-P_i-p_i) ,$$ (7.147)

where the four-dimensional delta-function is just the product of four one-dimensional delta-functions.

We divide the transition rate per unit volume by the flux of incident particles $\vert J_{\mathrm{inc}}\vert$ and by the number of target particles per unit volume, which is just $1/V$. Since the normalization of the wave function was performed in such a way that there is just one particle in the normalized volume $V$. To get a physical cross-section, we must sum over a given group of final states of the electron and proton corresponding to the laboratory conditions for observing the process. The number of final states of a specified spin in the momentum interval $d^3p_fd^3P_f$ is

$$V\frac{d^3p_f}{(2\pi)^3} V\frac{d^3P_f}{(2\pi)^3}$$ (7.148)

and thus the six-fold differential cross-section for transitions to the final states in the interval $\tau$ is

$$d\sigma = \int_\tau V^2 \frac{d^3p_f}{(2\pi)^3} \frac{d^3P_f}{(2\pi)^3} \frac{V}{\vert J_\mathrm{inc}\vert} w_{fi}$$

$$= \int_\tau \frac{d^3p_f}{(2\pi)^3} \frac{d^3P_f}{(2\pi)^3} \frac{m... ...4(P_f+p_f-P_i-p_i)}{\vert J_\mathrm{inc}\vert V} \vert\mathcal{M}_{fi}\vert^2 .$$ (7.149)

The physics lies in $\vert\mathcal{M}_{fi}\vert^2$, the square of the invariant amplitude. There is a factor $m/E$ for each external fermion line, that is, for each Dirac particle incident upon or emerging from the interaction. The phase-space factor for each final particle is $d^3p_f/(2\pi)^3$. Each final particle gives rise to the factor $\frac{m}{E}\frac{d^3p}{(2\pi)^3}$. The following combination forms a Lorentz invariant volume element in momentum space as can be seen from

$$\frac{d^3p}{2E} = \int_0^\infty dp_0 \frac{\delta(p_0-E)}{2p_0} d^3p$$

$$= \int_0^\infty dp_0 \delta(p_0^2-E^2) d^3p$$

$$= \int_0^\infty dp_0 \delta(p^2 - m^2) d^3p$$

$$= \int_{-\infty}^\infty d^4p \delta(p^2 - m^2) \theta(p_0) ,$$ (7.150)

which is invariant provided $p^\mu$ is time-like, as is the case here.

In the factor $1/V\vert J_\mathrm{inc}\vert$, $\vert J_\mathrm{inc}\vert$ is the flux and for collinear beams, which is the number of particles per unit area which run by each other per unit time,

$$\vert J_\mathrm{inc}\vert = \frac{\vert\vec{v}_i-\vec{V}_i\vert}{V} ,$$ (7.151)

which is the particle density times the relative velocity.

We have required that the velocity vectors are collinear. When $V\vert J_\mathrm{inc}\vert$ is combined with the normalization factor for two incident particles, it forms a Lorentz invariant expression

$$\frac{mM}{E_iE_i^\prime\vert\vec{v}_i-\vec{V}_i\vert} = \fra... ...P}_i\vert E_i} = \frac{mM}{\sqrt{(p_i\cdot P_i)^2 -m^2M^2}} .$$ (7.152)

The last expression can be seen by

$$(p_i\cdot P_i)^2 - m^2M^2 = (E_iE_i^\prime + \vert\vec{p}_i\vert\vert\vec{P}_i\vert)^2 - (E_i^2 - \vert\vec{p}_i\vert^2)({E_i^\prime}^2 - \vert\vec{P}_i\vert^2)$$

$$= 2E_i{E_i^\prime}\vert\vec{p}_i\vert\vert\vec{P}_i\vert + E_i^2\vert\vec{P}_i\vert^2 + {E_i^\prime}^2\vert\vec{p}_i\vert^2$$

$$= (\vert\vec{p}_i\vert E_i^\prime + \vert\vec{P}_i\vert E_i)^2 .$$ (7.153)

Consequently the naive flux factor in the cross-section has in general to be replaced by the Lorentz-invariant flux factor. In the case of collinear collisions, both results are idential. This shows that the total cross-section is invariant under Lorentz transformations along the direction of motion of the incident beams. We can write the invariant form

$$\fbox{$\displaystyle d\sigma = \int_\tau \frac{mM}{\sqrt{(p_... ...md^3p_f}{(2\pi)^3E_f} \frac{Md^3P_f}{(2\pi)^3E_f^\prime} $}\ .$$ (7.154)

The two-particle Lorentz-invariant phase space is

$$\mathrm{dLips}(s;p_3,p_4) = (2\pi)^4 \delta^4(p_3+p_4-p_1-p_... ... \frac{d^3p_3}{2E_3} \frac{M}{(2\pi)^3} \frac{d^3p_4}{2E_4} ,$$ (7.155)

where $s=(p_1+p_2)^2$ is the $s$ Mandelstram variable.

Averaging over initial and summing over final spins

$$\vert\overline{\mathcal{M}}_{fi}\vert^2 = \frac{1}{4} \sum_{s_f,s_i,S_f,S_i} \left\vert \overline{u}(p_f,s_... ...e^2}{q^2 + i\epsilon} \overline{u}(P_f,S_f) \gamma_\mu u(P_i,S_i) \right\vert^2$$

$$= \frac{1}{4} \textrm{Tr} \left[ \frac{(\not{\;\!\!\!p}_f+m)}{2m} \... ...gamma_\mu \frac{(\not{\;\!\!\!P}_i+M)}{2M} \gamma_\nu \right] \frac{e^4}{q^4} .$$ (7.156)

Note that squaring the amplitude which contained the scalar product of two Lorentz scalars has lead to the contraction of two tensors, i.e. a double sum. One often abbriviates this as

$$\vert\overline{\mathcal{M}}_{fi}\vert^2 = \frac{e^4}{q^4} L^{\mu\nu} H_{\mu\nu} ,$$ (7.157)

were $L^{\mu\nu}$ is the lepton (i.e. electron) tensor and $H_{\mu\nu}$ the hardonic (i.e. proton) tensor:

$$L^{\mu\nu} = \frac{1}{2} \sum_{s_i,s_f} \overline{u}(p_f,s_f)\gamma^\mu u(p_i,s_i) \overline{u}(p_i,s_i)\gamma^\nu u(p_f,s_f)$$

$$= \frac{1}{2} \textrm{Tr}\left[ \frac{\not{\;\!\!\!p}_f+m}{2m} \gamma^\mu \frac{\not{\;\!\!\!p}_i+m}{2m} \gamma^\nu\right] .$$ (7.158)

and similarily

$$H_{\mu\nu} = \frac{1}{2} \textrm{Tr}\left[ \frac{\not{\;\!\!... ...\gamma_\mu \frac{\not{\;\!\!\!P}_i+M}{2M} \gamma_\nu\right] .$$ (7.159)

The factorization remains meaningful as long as a single virtual photon is exchanged in the scattering process, even if the transition currents become more complicated than those here.

Returning to evaluating the traces, the first trace is

$$\textrm{Tr} \left[ \frac{\not{\;\!\!\!p}_f+m}{2m} \gamma^\mu \frac{\not{\;\!\!\!p}_i+m}{2m} \gamma^\nu \right] = \frac{1}{4m^2} \textrm{Tr} (\not{\;\!\!\!p}_f\gamma^\mu\not{\;\!\!\!p}_i\gamma^\nu + m^2\gamma^\mu\gamma^\nu)$$

$$= \frac{1}{4m^2} [4p_f^\mu p_i^\nu + 4p_f^\nu p_i^\mu -4p_i\cdot p_fg^{\mu\nu} + 4m^2g^{\mu\nu}]$$

$$= \frac{1}{m^2} [p_f^\mu p_i^\nu + p_i^\mu p_f^\nu - g^{\mu\nu}( p_f\cdot p_i - m^2)] .$$ (7.160)

The second trace is of the same form

$$\textrm{Tr} \left[ \frac{\not{\;\!\!\!P}_f+M}{2M} \gamma_\mu... ...u + {P_i}_\mu {P_f}_\nu - g_{\mu\nu} ( P_f\cdot P_i - M^2)] .$$ (7.161)

The square of the invariant amplitude now becomes

$$\vert\overline{\mathcal{M}}_{fi}\vert^2 = \frac{e^4}{4m^2M^2q^4} \left[ p_f^\mu p_i^\nu + p_i^\mu p_f^\nu - g^{\mu\nu} (p_f\cdot p_i - m^2) \right]$$

$$\cdot \left[ {P_f}_\mu {P_i}_\nu + {P_i}_\mu {P_f}_\nu - g_{\mu\nu} (P_f\cdot P_i - M^2) \right]$$

$$= \frac{e^4}{2m^2M^2q^4} [ (p_i\cdot P_i)(p_f\cdot P_f) + (p_i\cdot P_f)(p_f\cdot P_i)$$

$$- p_i\cdot p_f(P_i\cdot P_f - M^2) - P_i\cdot P_f(p_i\cdot p_f -m^2) + 2(p_i\cdot p_f -m^2)(P_i\cdot P_f -M^2)]$$

$$= \frac{e^4}{2m^2M^2q^4} [ (P_f\cdot p_f)(P_i\cdot p_i) + (P_f\cdot p_i)(P_i\cdot p_f)$$

$$-m^2(P_f\cdot P_i) -M^2(p_f\cdot p_i) + 2m^2M^2] .$$ (7.162)

To evaluate the scattering cross-section any further, the frame of reference has to be specified. Usually calculations take their simplist form in the center-of-mass system. However, electron-proton scatter experiments are mostly performed using a fixed target in the laroratory frame (except HERA). Therefore we evaluate $d\sigma$ in the laboratory frame in which the initial proton is at rest:

$$p_f = (E^\prime,\vec{p}^\prime),$$ (7.163)

$$p_i = (E,\vec{p}),$$ (7.164)

$$P_i = (M,0).$$ (7.165)

We also write

$$d^3p^\prime = {p^\prime}^2 dp^\prime d\Omega^\prime = p^\prime E^\prime dE^\prime d\Omega^\prime$$ (7.166)

and use

$$\frac{d^3P_f}{2E_f^\prime} = \int d^4P_f \delta^4(P_f^2-M^2) \theta(P_f^0)$$ (7.167)

to get

$$d\overline{\sigma} = \int \frac{mM}{\sqrt{(EM)^2 -m^2M^2}} \vert\overline{\mathcal{M}}_{fi}\vert^2 \frac{\delta^4(P_f+p^\prime-P_i-p)}{(2\pi)^2}$$

$$\cdot \frac{mp^\prime E^\prime dE^\prime d\Omega^\prime}{E^\prime} 2M d^4P_f \delta^4(P_f^2-M^2) \theta(P_f^0)$$

$$\frac{d\overline{\sigma}}{d\Omega^\prime} = \frac{2}{p} \int \frac{m^2Mp^\prime dE^\prime}{(2\pi)^2} \vert\ov... ...i}\vert^2 d^4P_f \delta^4(P_f^2-M^2) \theta(P_f^0) \delta^4(P_f+p^\prime-P_i-p)$$

$$= \frac{2}{p} \frac{m^2M}{4\pi^2} \int p^\prime dE^\prime \vert\ove... ...e{\mathcal{M}}_{fi}\vert^2 \delta[(P_i-p^\prime+p)^2 -M^2] \theta(M-E^\prime+E)$$

$$= \frac{m^2M}{2\pi^2p} \int_M^{M+E} p^\prime dE^\prime \vert\overli... ...{fi}\vert^2 \delta[2m^2 - 2(E^\prime-E)M -2E^\prime E + 2pp^\prime\cos\theta] .$$ (7.168)

Using $\delta[f(x-x_0)] = \delta(x-x_0)/\vert df(x)/dx\vert _{x_0}$, we calculate

$$\frac{df(E^\prime)}{dE^\prime} = -2[M+E-(pE^\prime/p^\prime)\cos\theta]$$ (7.169)

and require

$$m^2 -(E^\prime -E)M -2E^\prime E +pp^\prime\cos\theta = 0 .$$ (7.170)

We obtain

$$\frac{d\overline{\sigma}}{d\Omega^\prime} = \frac{m^2M}{4\pi... ...al{M}}_{fi}\vert^2}{M + E - (pE^\prime/p^\prime)\cos\theta} ,$$ (7.171)

For electrons of energy much less than the proton rest mass, we resurrect the earlier result of scattering in a static Coulomb field,

$$\frac{d\overline{\sigma}}{d\Omega^\prime} = \frac{m^2}{4\pi^... ...cal{M}}_{fi}\vert^2 \quad\textrm{for}\quad \frac{E}{M} \ll 1 ,$$ (7.172)

where

$$\vert\overline{\mathcal{M}}_{fi}\vert^2 = \frac{16\pi^2\alpha^2}{2m^2M^2q^4} (M^2E^2 + M^2E^2 - m^2M^2 -M^2p_f\cdot p_i + 2M^2m^2)$$

$$= \frac{8\pi^2\alpha^2}{m^2q^4} (2E^2 + m^2 - p_f\cdot p_i) \quad\textrm{for}\quad \frac{E}{m} \ll 1,$$ (7.173)

which is the Mott cross-section.

When the proton recoil becomes important, the electron may be treated as extremely relativistic and the electron rest mass is negligible with respect to the electron energy

$$\frac{d\overline{\sigma}}{d\Omega^\prime} = \frac{m^2}{4\pi^2} \frac{E^\prime/E}{1 + E/M -(E/M)\cos\theta} \vert\overline{\mathcal{M}}_{fi}\vert^2$$

$$= \frac{m^2}{4\pi^2} \frac{E^\prime/E}{1 + (2E/M)\sin^2(\theta/2)} \vert\overline{\mathcal{M}}_{fi}\vert^2 \quad\textrm{for}\quad \frac{m}{E} \ll 1 .$$ (7.174)

To evaluate the matrix element we re-express $P_f$ in terms of the electron recoil using energy-momentum conservation $P_f=P_i+p_i-p_f$ and

$$\vert\overline{\mathcal{M}}_{fi}\vert^2 \approx \frac{16\pi^2\alpha^2}{2m^2M^2q^4} \left[ (P_i+p_i-p_f)\cdot p_f P_i\cdot p_i + (P_i+p_i-p_f)\cdot p_i P_i\cdot p_f -M^2p_f\cdot p_i \right]$$

$$\approx \frac{8\pi^2\alpha^2}{m^2M^2q^4} [2P_i\cdot p_f P_i\cdot p_i + p_i\cdot p_f(P_i\cdot p_i - P_i\cdot p_f - M^2)]$$

$$\approx \frac{8\pi^2\alpha^2}{m^2M^2q^4} [2ME^\prime ME + EE^\prime(1-\cos\theta)(ME - ME^\prime - M^2)] .$$ (7.175)

Using $q^2=(p_f-p_i)^2 = -2EE^\prime(1-\cos\theta) = -4EE^\prime\sin^2(\theta/2)$, we have

$$\vert\overline{\mathcal{M}}_{fi}\vert^2 \approx \frac{8\pi^2\alpha^2}{m^2M^216E^2{E^\prime}^2\sin^4(\theta/2)} [2M^2EE^\prime + EE^\prime(1-\cos\theta)(M(E - E^\prime) - M^2)]$$

$$= \frac{\pi^2\alpha^2}{2m^2EE^\prime\sin^4(\theta/2)} \left[2 + (1-\cos\theta)\left(\frac{E - E^\prime}{M} - 1\right)\right] .$$ (7.176)

In the limit $m^2\rightarrow 0$, conservation of energy gives

$$0 = 2m^2-2(E^\prime-E)M -2E^\prime E + 2pp^\prime\cos\theta$$

$$0 \approx -2(E^\prime-E)M -2E^\prime E + 2EE^\prime\cos\theta$$

$$2EE^\prime(1-\cos\theta) = 2M(E-E^\prime)$$

$$\frac{E-E^\prime}{M} = \frac{EE^\prime}{M^2} (1-\cos\theta)$$

$$= \frac{2EE^\prime}{M^2} \sin^2(\theta/2) = -\frac{q^2}{2M^2} .$$ (7.177)

Thus

$$\vert\overline{\mathcal{M}}_{fi}\vert^2 = \frac{\pi^2\alpha^2}{2m^2EE^\prime\sin^4(\theta/2)} \left[ 2 + 2\sin^2\frac{\theta}{2} \left( -\frac{1}{2} \frac{q^2}{M^2} - 1\right) \right]$$

$$= \frac{\pi^2\alpha^2}{m^2EE^\prime\sin^4(\theta/2)} \left( \cos^2\... ...}{2M^2}\sin^2\frac{\theta}{2} \right) \quad\textrm{for}\quad \frac{m}{E} \ll 1.$$ (7.178)

The $q^2$-dependent second term is found to originate from the fact that the target is a spin-1/2 particle. This term is absence when the collision of electrons with spin-0 particles is calculated.

The differential cross-section thus becomes

$$\frac{d\overline{\sigma}}{d\Omega^\prime} = \frac{\alpha^2}{... ...)\sin^2(\theta/2)]} \quad\textrm{for}\quad \frac{m}{E} \ll 1.$$ (7.179)

This derivation has treated the proton as a heavy electron of mass $M$. The description is incomplete since it fails to take into account the proton structure and anomalous magnetic moment. A complete description of the proton leads to modifications which are important at large energies exceeding several hundred MeV, since the de Broglie wavelength of the electron is so small that the substructure of the proton becomes detectable.

In a complete treatment for very high energies (several 100 MeV), the formula has to be modified by introducing electric and magnetic form factors representing the internal structure of the proton. This yields the so-called Rosenbluth formula.

Our result would apply with great accuracy, however, to the scattering of electrons and muons which both are structureless Dirac particles.