fish diagram calculation in Minkowski space

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Barrow - 20 Oct 2007 21:26 GMT

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

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Raymond Manzoni - 21 Oct 2007 21:22 GMT

Barrow a écrit :

> Dear all,
> I am studying QFT and wanna complete the fish diagram calculation in
[quoted text clipped - 17 lines]
>
> 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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Raymond Manzoni - 23 Oct 2007 04:46 GMT

Raymond Manzoni a écrit :
(snip long derivation)

> so that f(s)-log(s) will be -2 and finally
> for m2/s > 1/4 :
[quoted text clipped - 4 lines]
> I(m2,s)= log(m2) +
> sqrt(1-4*m2/s)*log((sqrt(1-4*m2/s)+1)/(sqrt(1-4*m2/s)-1)) - 2

Sorry for the laborious derivation... :-(
The last equation may be found quickly by rewriting
I(m2)= int_0^1 log(m2 - s*x*(1-x)) dx
as
I(m2)= int_{-1/2}^{1/2} log(t^2-a^2) dt + log(s)
with a^2= 1/4-m2/s and t= x-1/2
I(m2)= int_{-1/2}^{1/2} log(t-a) + log(t+a) dt +log(s)
and use int log(x) dx= x*log(x)-x to get

I(m2)= (t-a)*log(t-a)+(t+a)*log(t+a)-2*t |_{-1/2}^{1/2} +log(s)
= t*log(t^2-a^2) + a*log((t+a)/(t-a)) |_{-1/2}^{1/2} -2 + log(s)
= log(1/4-a^2)+log(s)+ a*log((1/2+a)*(-1/2-a)/((-1/2+a)*(1/2-a)))-2
= log(m2)+ 2*a*log((1/2+a)/(1/2-a))- 2
up to a minus sign or too ;-)

Raymond

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Barrow - 23 Oct 2007 07:38 GMT

> Raymond Manzoni a ?crit :
> (snip long derivation)
[quoted text clipped - 24 lines]
>
> Raymond

Many thanks for your clear and elegant derivation! You are so cool~!
I'm really appreciated.

However, I still have some problems to consult you (I'm bad at complex
analysis :-( )
Actually, first time I met the tough integral, i.e. I(m2), I tried to
rewrite the integrand as log[(x-a)(x-b)], then to integrate it by
part. But I don't know how to deal with each situation, I mean, when s

> 4m^2 or s < 4m^2, the singular points x = a and x = b are at

different positions, I don't know if the positions of the
singularities affect the solution.(actually, the solution in the case
of s > 4m^2 should have imaginary part, but for the case of s < 4m^2,
the solution is real)
I can find that, when s > 4m^2, the singularities a and b are near
the real axis between the interval [0,1]. Otherwise I can do nothing.
(:-P) From your previous article, the case of s > 4m^2, there is an
imaginary part if we note log(-1) = i*pi. (Here is part of your answer
log((sqrt(1-4*m2/s)+1)/(sqrt(1-4*m2/s)-1)) <-- the denominator is
negative)

The sign of the imaginary part if quite important for physics since
it affects the scattering process. But I can't work it
out...Nevertheless, your instruction really helped me a lot. Again,
you are so cool!

Sincerely

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Raymond Manzoni - 28 Oct 2007 18:59 GMT

Barrow a écrit :

> Many thanks for your clear and elegant derivation! You are so cool~!
> I'm really appreciated.
[quoted text clipped - 22 lines]
>
> Sincerely

Thank you very much!

I sent you a formal answer concerning int log(t^2-a^2) dt to your
gmail GRseminar account (to resume my suggestion: use the principal
branch of log for any 'a' not real).
Concerning the more physical point of view your problem is handled
with care in Pierre Ramond's "Field Theory : A Modern Primer" in the
euclidean domain (page 121-122) and continued in the Minkowski space
(page 149-150). You should be able to read these here :
http://www.amazon.com/gp/reader/0201304503/ by searching
[Minkowski space] and looking at the given pages

Hoping it helped more!
Raymond

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Natural Science Forum / Physics / Research / October 2007

fish diagram calculation in Minkowski space