Re: fish diagram calculation in Minkowski space

Barrow a écrit :
Dear all,
I am studying QFT and wanna complete the fish diagram calculation in
the Minkowski space with i-epsilon prescription to see the
"imaginary part."

I know the roughly procedures, i.e., use Feynman formula to rewrite
the integand, then perform the angular and radial integration of the
original volume integral, finally integrated over the feynman
parametrisation(from 0 to 1). There is a key integral

\int_0^1 \log{m^2 - sx(1-x) - i\epsilon}dx

This is really so hard to work out. I found so many books buy only one
"Quantum Fields" written by Bogoliubov and Shirkov mentioned the key
steps to work out this integral. But I can't follow them exactly.

Could anyone show me the key points of calculation or indicate me
any books about this fish diagram calculation "with i-epsilon
prescription" in detail.

Any help will be appreciated. Sincerely Barrow


Let's note m2= (m^2 - i epsilon) then I got (for m2/s > 1/4):

I(m2,s)= log(m2)+2*(arctan(sqrt(s/(4*m2-s)))*sqrt((4*m2-s)/s)-1)

(for m<>0 I think that the epsilon contribution will disappear at the limit epsilon -> 0 so that I won't take care of the epsilon part!)

Sketch of a proof (s is considered constant) :
I(m2)= int_0^1 log(m2 - s*x*(1-x)) dx

so that after derivation under the integral sign relatively to m2 :
dI/dm2= int_0^1 1/(m2 - s*x*(1-x)) dx
dI/dm2= int_0^1 1/(m2 + s*((x-1/2)^2-1/4)) dx
dI/dm2= int_{-1/2}^{1/2} 1/(m2/s-1/4 + t^2) dt / s
dI/dm2= int_{-1/2}^{1/2} 1/(u^2 + t^2) dt / s
( with u= sqrt(m2/s-1/4) and supposing m2/s > 1/4)
dI/dm2= 2*arctan(1/(2*u))/(u*s)

Since u= sqrt(m2/s-1/4)
du/dm2 = 1/(2*u*s) and
dI/dm2= 4*arctan(1/(2*u)) du/dm2
so that I= int 4*arctan(1/(2*u)) (du/dm2) dm2 + f(s)
= 4*int arctan(1/(2*u)) du + f(s)

Let's integrate arctan(1/(2*u)) by parts :
int arctan(1/(2*u)) du = u*arctan(1/(2*u)) + int 2*u/(4*u^2+1) du
= u*arctan(1/(2*u)) + log(u^2+1/4)/4
so that
I= 4*u*arctan(1/(2*u)) + log(u^2+1/4)+ f(s)
or
I(m2,s)= 2*sqrt(4*m2/s-1)*arctan(1/sqrt(4*m2/s-1)) + log(m2)+ f(s)-log(s)
(using u= sqrt(m2/s-1/4), 1/(2*u)= 1/sqrt(4*m2/s-1), u^2+1/4= m2/s)

to find f(s) (the term not depending of m2) we may consider the behavior of I(m2)-log(m2) as m2 -> +oo
int_0^1 log{m2 - s*x*(1-x)) - log(m2) dx -> 0
sqrt(4*m2/s-1)*arctan(1/sqrt(4*m2/s-1)) -> 1
so that f(s)-log(s) will be -2 and finally
for m2/s > 1/4 :
I(m2,s)= log(m2) + 2*(sqrt(4*m2/s-1)*arctan(1/sqrt(4*m2/s-1))-1)

for m2/s < 1/4 we may replace artan(x) by log((1+i*x)/(1-i*x))/(2*i) to get the corresponding formula :
I(m2,s)= log(m2) + sqrt(1-4*m2/s)*log((sqrt(1-4*m2/s)+1)/(sqrt(1-4*m2/s)-1)) - 2


A computation using Mathematica and s= -p^2 may be seen here :
http://www.scientificarts.com/feynman/feynman.html

Hoping this helped,
Raymond

.

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