where did the i come from

[Mike James]

I'm reading through:

Title: Regularization and renormalization of gauge fields
Author: Veltman, M.J.G.
Hooft, G. 't
Year: 1972
which can be obtained from:
http://igitur-archive.library.uu.nl/phys/2005-0622-155148/UUindex.html

As an exercise I have been trying to do the integrals in appendix A -
just to make sure I can.
For the first one A5 that is I get the same answer but real i.e. no
factor of i in front of the pi.
Am I missing something or is there a mistake?
mikej

[Mike James]

Mike James wrote:
> I'm reading through:
>
> Title: Regularization and renormalization of gauge fields
> Author: Veltman, M.J.G.
> Hooft, G. 't
> Year: 1972
> which can be obtained from:
> http://igitur-archive.library.uu.nl/phys/2005-0622-155148/UUindex.html
>
> As an exercise I have been trying to do the integrals in appendix A -
> just to make sure I can.
> For the first one A5 that is I get the same answer but real i.e. no
> factor of i in front of the pi.
> Am I missing something or is there a mistake?
> mikej
>
I'm still stuck.
Any one?
mikej

[CarlB]

On Jan 7, 1:31Â*pm, Mike James <mike.ja...@infomaxgroup.co.uk> wrote:
> Am I missing something or is there a mistake?
> mikej

I'm no expert, but it sure looks like both sides need to be real. So
I'd
say that the paper has an error. This sort of thing is not uncommon.

Carl

[Igor Khavkine]

On Jan 8, 4:49 am, CarlB <c...@brannenworks.com> wrote:
> On Jan 7, 1:31 pm, Mike James <mike.ja...@infomaxgroup.co.uk> wrote:
>
> > Am I missing something or is there a mistake?
> > mikej
>
> I'm no expert, but it sure looks like both sides need to be real. So
> I'd say that the paper has an error. This sort of thing is not
> uncommon.

That's not necessarily true. The integral of 1/(z-i0^+) over the real
line is imaginary, i*pi to be precise. The imaginary part appears due to
the prescription that avoids the poles of the integrand.

Glancing at the appendix or 't Hooft and Veltman's paper, he seems to
refer to section 3 for the precise prescription of how the integrals
should be taken. Perhaps therein lies the key.

Hope this helps.

Igor

[Mike James]

Igor Khavkine wrote:
>
> That's not necessarily true. The integral of 1/(z-i0^+) over the real
> line is imaginary, i*pi to be precise. The imaginary part appears due to
> the prescription that avoids the poles of the integrand.


I have read section 3 - doesn't seem to help.

> Glancing at the appendix or 't Hooft and Veltman's paper, he seems to
> refer to section 3 for the precise prescription of how the integrals
> should be taken. Perhaps therein lies the key.
>
> Hope this helps.
>
> Igor
>

Its is exactly the reason you quote that means I can't rely on the
"integral of real equals real result" approach... but I still can't get
a factor of i out of the integral. Thanks
mikej

[Igor Khavkine]

On Jan 10, 5:59 am, Mike James <mike.ja...@infomaxgroup.co.uk> wrote:

> > Glancing at the appendix or 't Hooft and Veltman's paper, he seems to
> > refer to section 3 for the precise prescription of how the integrals
> > should be taken. Perhaps therein lies the key.
>
> Its is exactly the reason you quote that means I can't rely on the
> "integral of real equals real result" approach... but I still can't get
> a factor of i out of the integral. Thanks

OK, I've looked at the integral (A5) and what is said about it in
section 3. There 't Hooft and Veltman refer only to the usual +-
i*epsilon prescription. This is the prescription that allows Wick
rotations to be used to evaluate loop integrals in Feynman diagrams.
It is explained in any introductory book on field theory.

Here's how to do the integral (A5). It is

I(k) = int dp/(p^2+2kp+m^2)^a ,

where the integration is done over R^n, with n an arbitrary dimension.

1. Wick rotate the integration over the time component of the
momentum: p_0 = i p_4. Now the integral is over the n-dimensional
Euclidean space and there is that extra factor if i in front that
confused you.

2. Shift the integration variable from p to p+k, the denominator now
becomes (p^2+m^2-k^2)^a.

3. Since the integral is now rotationally invariant, switch to radial
coordinates:

I(k) = i A(n) int dp p^(n-1)/(p^2+m^2-k^2)^a ,

with p now denoting the radial coordinate and A(n) is the area of the
unit (n-1)-sphere.

4. A(n) is given in equation (A3), while the radial integral can be
evaluated with the help of equation (A4), with beta=n-1 and M^2=m^2-
k^2:

I(k) = i [2 pi^(n/2)/G(n/2)] [1/2 G(n/2)G(a-n/2)/G(a)/(m^-k^2)^(a-n/
2)]
= [i pi^(n/2)/(m^2-k^2)^(a-n/2)] [G(a-n/2)/G(a)] .

The last line is precisely equation (A5) of 't Hooft and Veltman's
paper. The other integrals (A6)-(A10) can be obtained from derivatives
of (A5) with respect to k.

Hope this helps.

Igor

[Mike James]

Igor Khavkine wrote:
>
> That's not necessarily true. The integral of 1/(z-i0^+) over the real
> line is imaginary, i*pi to be precise. The imaginary part appears due to
> the prescription that avoids the poles of the integrand.


I have read section 3 - doesn't seem to help.

> Glancing at the appendix or 't Hooft and Veltman's paper, he seems to
> refer to section 3 for the precise prescription of how the integrals
> should be taken. Perhaps therein lies the key.
>
> Hope this helps.
>
> Igor
>

Its is exactly the reason you quote that means I can't rely on the
"integral of real equals real result" approach... but I still can't get
a factor of i out of the integral. Thanks
mikej

[Mike James]

Igor Khavkine wrote:
> On Jan 10, 5:59 am, Mike James <mike.ja...@infomaxgroup.co.uk> wrote:
>
>>> Glancing at the appendix or 't Hooft and Veltman's paper, he seems to
>>> refer to section 3 for the precise prescription of how the integrals
>>> should be taken. Perhaps therein lies the key.
>> Its is exactly the reason you quote that means I can't rely on the
>> "integral of real equals real result" approach... but I still can't get
>> a factor of i out of the integral. Thanks
>
> OK, I've looked at the integral (A5) and what is said about it in
> section 3. There 't Hooft and Veltman refer only to the usual +-
> i*epsilon prescription. This is the prescription that allows Wick
> rotations to be used to evaluate loop integrals in Feynman diagrams.
> It is explained in any introductory book on field theory.
>
>

I though it might be something to do with Wick rotation but I've always
been a bit vague as to what the process of actually doing the rotation
was - even though I've looked it up and looked at examples.

Thanks for working it though its really helped me follow what is going
on. My mistake was reading the appendix as an abstract list of integrals
that had little to do with physics apart from the symbols use and I
didn't follow the idea that they were working them out using Wick rotation.
Thanks again.