Why do you perform transformation from rest frame to infinite momentum frame?

Answer 1

In the infinite momentum frame, the final and initial wave functions of the particles have the same momentum, which is important in some studies. It is a subset of the Breit Frames.

An example can be found at refernce: DOI: 10.1103/PhysRevC.78.032201

IN part this reads:

In the nonrelativistic case, the form factors are simply

the Fourier transforms of the rest frame spatial distributions,

and the charge and magnetization mean square radii are

derived from the slope of the form factors at Q2

= 0. In the

relativistic case, this interpretation is not correct because the

wave functions of the initial and final nucleons have different

momenta and therefore differ, invalidating a probability or

density interpretation. This is addressed by working in the Breit

frame, where the magnitude of the initial and final nucleon

momenta are identical. However, one needs boost corrections

of order Q2 / m2 , where m2 is the mass of the constituent

particles to which the boost is applied, to relate the rest frame

and moving nucleon wave functions. These corrections are

model dependent [5], so the use of the Breit frame does not

provide a precise, model-independent measure of the spatial

distribution of the nucleon.

A recent work showed that it is possible to obtain a model-

independent nucleon charge density [6].

Answer 2

Transforming from the rest frame to the infinite momentum frame simplifies calculations in high-energy physics by allowing for a simpler description of particle interactions. In the infinite momentum frame, the speed of light becomes infinite, making the calculations more manageable. This transformation can help reveal the underlying symmetries and dynamics of the system under consideration.