A Calculation of a Form Factor at large ##q^2##

  • #1
MathematicalPhysicist
Gold Member
4,699
369
TL;DR Summary
A question on a calculation in the book of Mueller, Perturbative QCD, from page 180.
On page 180 they write the following passage:
The form factors for the heavy hadrons are normalized by the constraint that the Coulomb contribution to the form factor equals the total hadronic charge at ##q^2=0##. Further, by the correspondence principle, the form factor should agree with the standard non-relativistic calculation at small momentum transfer. All of these constraints are satisfied by the form:
[tex]F^M_{(0,0)}(q^2)=e_1\frac{16\gamma^4}{(q^2+\gamma^2)^2}(\frac{M_H^2}{m_2^2})^2 \bigg( 1-\frac{q^2}{4M_H^2}\frac{2m_2}{m_1}\bigg)+1\leftrightarrow 2 .[/tex]
At large ##q^2## the form factor can also be written as:
[tex]F^M_{(0,0)}=e_1 \frac{16\pi\alpha_s f_M^2}{9q^2}(\frac{M_H^2}{m_2^2})+(1\leftrightarrow 2), f_M/(2\sqrt{3})=\int_0^1dx \phi(x,Q)[/tex]
where ##f_M=(6\gamma^3/\pi M_H)^{1/2}## is the meson decay constant.

My question is how do I get ##F^M_{(0,0)}## from ##F^M_{(0,0)}(q^2)## for large ##q^2##.
I tried using Mathematica's free input to get the series ##1/(1+x^2)^2=1-2x^2+3x^4+O(x^6)##, in our case ##x=\gamma/q##.
But it doesn't seem to fit the result from the book, perhaps there's something I am missing, any help?
A remark, ##\gamma## is Lorentz factor.

Thanks!
 
Last edited:

Answers and Replies

  • #2
MathematicalPhysicist
Gold Member
4,699
369
No one knows?
 

Suggested for: A Calculation of a Form Factor at large ##q^2##

Replies
0
Views
500
Replies
28
Views
605
Replies
1
Views
512
Replies
28
Views
507
Replies
1
Views
663
Replies
3
Views
102
  • Last Post
Replies
2
Views
1K
Replies
2
Views
359
Replies
49
Views
2K
  • Last Post
Replies
7
Views
1K
Top