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naima
Sep8-09, 03:53 PM
Hi all

I found this (http://en.wikipedia.org/wiki/Cross_section_(physics)#Relation_to_the_S_matrix) on wikipedia:

{d\sigma \over d\Omega} = (2\pi)^4 m_i m_f {p_f \over p_i} |T_{fi}|^2

Has it the dimension of a surface? (I only see M*M)
It is the first time I read this formula about differential cross section.

Astronuc
Sep8-09, 04:08 PM
Hi all

I found this (http://en.wikipedia.org/wiki/Cross_section_(physics)#Relation_to_the_S_matrix) on wikipedia:

{d\sigma \over d\Omega} = (2\pi)^4 m_i m_f {p_f \over p_i} |T_{fi}|^2

Has it the dimension of a surface? (I only see M*M)
It is the first time I read this formula about differential cross section. Yes, the microscopic cross-section σ has units of area.

The differential cross-section is a measure of how the microscopic cross-section, σ, changes with respect to the solid angle, Ω.

naima
Sep8-09, 11:31 PM
Of course, but what are the dimensions of mi mf pi pf and T?
and how do you get L*L?

naima
Sep11-09, 04:12 AM
I am sure that there is a c = 1 or a hbar = 1 in this formula which disables one to do dimesional calculus.
Do you know how this equality is deduced?