Re: Renormalization



Igor Khavkine wrote: 
On 2005-04-17, jsolomon@xxxxxxxx <jsolomon@xxxxxxxx> wrote:

Eugene Stefanovich wrote:

If I understood you correctly, you are claiming that Weinberg made
a trivial mathematical mistake: he used formulas (11.2.11) and
(11.2.12) outside the region of their applicability. The most general
expression for such integrals is given in (11.A.2).
I don't have a table of integrals with me. Could somebody please
check at which values of l amd m this formula is valid?


Weinberg did not make a trivial mistake.  He explains himself in the
Appendix on page 497.  He takes the integral 11.2.12 which is divergent
for values of d->4. You don't need an integral table to convince
yourself of this.  Just substitute d = 4.01 or d=3.99 into the integral
on the left hand side of 11.2.12.  This is obviously divergent.
However the expression on the right hand side goes as 1/(d-4).  For
d=3.99 it will go as 1/.01=100.  For d=4.01 it will go as 1/(-.01)=
-100.  The equality obviously does not hold for d->4.  The equality
does hold for d<2.  In which case the integral on the left is no longer
divergent.  It can be evaluated and equals the expression on the right
hand side (for d<2).  Then Weinberg substitutes the expression on the
right hand side for the integral in the equation for the vacuum
polarization tensor.  He does this even though the two expressions are
not equal for the region of interest which is d->4.



All functions involved here are analytic in d. But they have poles at
some integer values of d, like d=4. If two analytic functions agree on a
finite volume (open) subset of the complex plain, then they agree
everywhere. Even if you only know that they agree on some finite (open)
interval of the real line, they still agree everywhere. Hence if you've
convinced yourself that the equality holds for d<2, then it must hold
everywhere, including as d->4.

If nothing else convinces you, consider simply plotting the LHS and the
RHS of the equality for a range of d. Simply fix all quantities other
than d to some reasonable values and evaluate the integral as well as
the Gamma function numerically. Numerical integrals are easily
performed, especially with the help of a computer algebra system.


I think Dan is right. Take (11.2.12) with d=3.99. The integrand on the
LHS goes as k^{2.99} when k -> oo. Thus the LHS is clearly divergent.
On the other hand, the RHS expression is finite, because arguments of
the Gamma functions are not at the positions of poles.

Probably there is another way to justify what Weinberg is doing,
but eq. (11.2.12) is clearly wrong for d values around 4.

Eugene Stefanovich.


.

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