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

(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 would appear in front of the current. To solve the equation, we introduce the Green function, , define by

(7.134) |

Its Fourier representation is

(7.135) |

where , for .

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

(7.136) |

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

The Feynman propagator for electromagnetic radiation is

(7.137) |

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

(7.138) |

since

(7.139) |

In a more general gauge

(7.140) |

It carries Lorentz indices because the photon is a spin-1 particle. In our case, the tensor is just proportional to so that the tensor indices can be discarded for convenience by defining .

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

The -matrix is

(7.141) |

is the current of the electron. The problem now is to decide what to choose for the proton current ? Its reasonable to try

(7.142) |

where is the proton charge. and represent the initial and final plane wave solutions for a free proton. Using plane-wave solutions for a proton gives

(7.143) |

where and are the initial and final proton four-momentum and is its mass.

We now get for the -matrix

(7.144) |

Notice the symmetry in electron and proton variables.

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

For each line and intersection of the graph there corresponds a unique factor in the -matrix. In addition, always contains a four-dimensional -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 between the electron and proton. The amplitude for the virtual photon to propagate between the two currents is . At the points (or vertices) on which the photon lands there operate factors sandwiched between spinors 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 and follow the line through until the end, inserting vertices and propagators in the right order, until you reach the outgoing state . There is a uniform rule for the factors of : for each vertex and for each internal line in the graph.

We now form a transition rate per unit volume by dividing by the time interval of observation and by the spatial volume of the interaction region. This gives

(7.145) |

where

(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 -function

(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 and by the number of target particles per unit volume, which is just . Since the normalization of the wave function was performed in such a way that there is just one particle in the normalized volume . 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 is

(7.148) |

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

(7.149) |

The physics lies in , the square of the invariant amplitude. There is a factor 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 . Each final particle gives rise to the factor . The following combination forms a Lorentz invariant volume element in momentum space as can be seen from

(7.150) |

which is invariant provided is time-like, as is the case here.

In the factor , is the flux and for collinear beams, which is the number of particles per unit area which run by each other per unit time,

(7.151) |

which is the particle density times the relative velocity.

We have required that the velocity vectors are collinear. When is combined with the normalization factor for two incident particles, it forms a Lorentz invariant expression

(7.152) |

The last expression can be seen by

(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

(7.154) |

The two-particle Lorentz-invariant phase space is

(7.155) |

where is the Mandelstram variable.

Averaging over initial and summing over final spins

(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

(7.157) |

were is the lepton (i.e. electron) tensor and the hardonic (i.e. proton) tensor:

(7.158) |

and similarily

(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

(7.160) |

The second trace is of the same form

(7.161) |

The square of the invariant amplitude now becomes

(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 in the laboratory frame in which the initial proton is at rest:

(7.163) | |||

(7.164) | |||

(7.165) |

We also write

(7.166) |

and use

(7.167) |

to get

(7.168) |

Using , we calculate

(7.169) |

and require

(7.170) |

We obtain

(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,

(7.172) |

where

(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

(7.174) |

To evaluate the matrix element we re-express in terms of the electron recoil using energy-momentum conservation and

(7.175) |

Using , we have

(7.176) |

In the limit , conservation of energy gives

(7.177) |

Thus

(7.178) |

The -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

(7.179) |

This derivation has treated the proton as a heavy electron of mass . 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.

2004-03-18