A What is the energy dependence of the Equivalent photon approximation?

[ribella]

Hi,
What is the energy dependence of the Equivalent photon approximation? For this approach to be valid, what is the maximum center of mass-energy. As know, this approach is an energy-dependent approach. Can this approach be used to calculate, for example, at a center of mass energy of 100 TeV? Is there any problem with the approach in terms of physics?

 

[RGevo]

For what type of process do you have in mind?

A probe of the elastic form factor of a nucleon/nucleus? Or something like the inelastic photon pdf of a nucleus?

Typically this is governed by the probe energy / exchange energy Q^2 through the photon its self. If this energy scale is sufficiently small compared to other scales in the scattering process, then it is a decent approximation. If there are other energy scales such as a fermion mass that is mf~Q, then the EPA (assuming the photon is essentially on shell) neglects power corrections of the form Q/mf which can be important for a precision theory description.

 

[ribella]

Thank you for your answer. For example, let's assume that the center of mass energy for the gg->e-e+ process (Here, g is EPA photon) is 100 TeV. In this case, is the EPA approximation valid for the incoming photons at these energies?

 

[RGevo]

I would not be confident that the EPA is a good approximation in this case.

(Do you mean the photon photon CoM or the hadronic one? That wasnt clear to me)

Even if the CoM is high, I think it is possible that the cross-section may still be dominated by low-Q (corresponding the virtuality probe of the nucleus/nucleon that is giving you these EPA photons).

Depending if you are considering a nucleus or a nucleon form factor (im not sure what you have in mind exactly) these distributions-the photon flux-may peak at small Q (eg at values below the electron mass). If that were the case, terms like Q/m_electron power corrections are absent in the EPA approach.

Note that a form factor like woods-saxon peaks as Q tends to zero (ie when you resolve the whole photon field of the nucleus).