In order to calculate the cross-section of an interaction process the following formula is often used for first approximations: sigma=(2 pi)/(h v_i) |M_fi|^2 q(E_f)V

powerinojSs

powerinojSs

Answered question

2022-11-22

In order to calculate the cross-section of an interaction process the following formula is often used for first approximations:
σ = 2 π v i | M f i | 2 ϱ ( E f ) V
M f i = ψ f | H i n t | ψ i
M f i = ψ f | H i n t | ψ i
ϱ ( E f ) = d n ( E f ) d E f = 4 π p f 2 ( 2 π ) 3 V v f
The derivation of this equation in the context of the non relativistic Schrödinger equation. Use this formula in the relativistic limit: v i , v f c , p f E f / c
Very often books simply use this equation with matrix element derived from some relativistic theory, e.g. coupling factors and propagators from the Dirac equation or Electroweak interaction. How is this justified?

Answer & Explanation

Gunnar Molina

Gunnar Molina

Beginner2022-11-23Added 8 answers

Fermi's golden rule still applies in the relativistic limit, and can be rewritten in a Lorentz invariant fashion. Starting with the transition probability
W i f = 2 π | m i f | 2 ρ ( E ) ,
to have W Lorentz invariant we'd like both the matrix element | m i f | 2 and the density of final states ρ ( E ) to be invariant.
This can be done by shifting a few terms around. A little bit of handwaving to motivate it: The wave function ψ (which is in the matrix element) has to be normalized by | ψ | 2 d V = 1, which gives us a density (of probability to encounter a particle) of 1 / V. Now, a boosted observer experiences length contraction of 1 / γ, which changes the density to γ / V. To obtain the correct probability again, we should re-normalize the wave function to ψ = γ ψ by pulling the Lorentz factor out.
So we intoduce a new matrix element
| M i f | 2 = | m i f | 2 i = 1 n ( 2 γ i m i c 2 ) = | m i f | 2 i = 1 n ( 2 E i ) 2
(this is for an n-body process). Now the transition probability (here in differential form) becomes:
d W = 2 π | M i f | 2 ( 2 E 1 ) 2 ( 2 E 2 ) 2 1 ( 2 π ) 3 n d 3 p 1 d 3 p 2 δ ( p 1 μ + p 2 μ + p μ )
The delta function is there to ensure conservation of momentum and energy. Now we can regroup the terms:
d W = 2 π | M i f | 2 2 E 1 2 E 2 d L I P S
The density of states/"phase space" d ρ is replaced by a relativistic version, sometimes called the Lorentz invariant phase space d L I P S , which is given by
d L I P S = 1 ( 2 π ) 3 n i = 1 n d 3 p i 2 E i δ ( i = 1 n p i μ p μ ) .
The nice thing about the relativistic formula for d W is that, in the case you are scattering particles off one another, it immediately shows us three important contributions: not only the matrix element and phase space, but also the flux factor 1 / s (where s = ( p 1 μ + p 2 μ ) 2 is the Mandelstam variable, and in case the masses are negligible, s 2 E). This flux factor is responsible for the general 1 / Q 2 falling slope when you plot cross section over momentum transfer Q = s , which comes entirely from relativistic kinematics.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Relativity

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?