- #1

- 6

- 0

But I can't figure out how to prove it. Can angone help me? Thanks!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- A
- Thread starter phylz
- Start date

- #1

- 6

- 0

But I can't figure out how to prove it. Can angone help me? Thanks!

- #2

CAF123

Gold Member

- 2,950

- 88

$$s = 4(\mathbf p^2 + m^2) ,\,\,\,\,t=-2\mathbf p^2 (1- \cos \theta), \,\,\,\text{and} \,\,\,\,u = -2\mathbf p^2 (1+\cos \theta) $$ As the physical domain of the variables is such that ##\mathbf p^2 \geq 0## and ## -1 \leq \cos \theta \leq 1##, means that ##s \geq 4m^2, t \leq 0, u \leq 0## with the equalities holding when the relative momentum vector ##\mathbf p = 0##.

In particular, and coming to your question now, if ##s## is above threshold, meaning that ##s## is higher than the ##4m^2##, then ##u## and ##t## are strictly less than zero. This all collectively maintains the constraint ##s+t+u = 4m^2##. So, the ##t## and ##u## channel diagrams proportional to ##1/t## and ##1/u## respectively, at the amplitude level, do not admit singular structures as ##t## and ##u## may not be nullified when ##s## is above ##4m^2.##

- #3

- 6

- 0

Thank you.

$$s = 4(\mathbf p^2 + m^2) ,\,\,\,\,t=-2\mathbf p^2 (1- \cos \theta), \,\,\,\text{and} \,\,\,\,u = -2\mathbf p^2 (1+\cos \theta) $$ As the physical domain of the variables is such that ##\mathbf p^2 \geq 0## and ## -1 \leq \cos \theta \leq 1##, means that ##s \geq 4m^2, t \leq 0, u \leq 0## with the equalities holding when the relative momentum vector ##\mathbf p = 0##.

In particular, and coming to your question now, if ##s## is above threshold, meaning that ##s## is higher than the ##4m^2##, then ##u## and ##t## are strictly less than zero. This all collectively maintains the constraint ##s+t+u = 4m^2##. So, the ##t## and ##u## channel diagrams proportional to ##1/t## and ##1/u## respectively, at the amplitude level, do not admit singular structures as ##t## and ##u## may not be nullified when ##s## is above ##4m^2.##

##1.## You mean that the Mandelstam variables ##t## and ##u## correspond to the branch cut singularities of ##t##- and ##u##-channel diagrams respectively? Why?

##2.## I can't understand your last sentence.

##3.## ##t=-2\mathbf p^2 (1- \cos \theta)^2, \,\,\,\text{} \,\,\,\,u = -2\mathbf p^2 (1+\cos \theta)^2 ##. You lost ##^2## above the ##(1- \cos \theta)## and ##(1+\cos \theta)##?

- #4

CAF123

Gold Member

- 2,950

- 88

Actually I confess that those statements were made with the tree level diagrams in mind and I implicitly assumed the propagator was massless. But the argument goes through with a massive propagator because then the t and u diagrams are proportional to 1/(t-mThank you.

##1.## You mean that the Mandelstam variables ##t## and ##u## correspond to the branch cut singularities of ##t##- and ##u##-channel diagrams respectively? Why?

What happens at loop level?

One finds that the t,u diagrams may contribute an imaginary part because of their proportionality to ##\sqrt{1-4m^2/q^2}##, where q

I think there might be another argument without using the explicit form of the diagrams (indeed P&S make their statement before starting their computation of the s channel diagram) but I don't see it yet.

Does the above make it better?##2.## I can't understand your last sentence.

Hmm I don't think so. What makes you think there is a square?##3.## ##t=-2\mathbf p^2 (1- \cos \theta)^2, \,\,\,\text{} \,\,\,\,u = -2\mathbf p^2 (1+\cos \theta)^2 ##. You lost ##^2## above the ##(1- \cos \theta)## and ##(1+\cos \theta)##?

- #5

- 6

- 0

Actually I confess that those statements were made with the tree level diagrams in mind and I implicitly assumed the propagator was massless. But the argument goes through with a massive propagator because then the t and u diagrams are proportional to 1/(t-m^{2}) and 1/(u-m^{2}) respectively and for s>4m^{2}strictly t,u<0 so one never encounters a pole at tree level. This is a general feature under 'S-matrix analyticity' .

What happens at loop level?

One finds that the t,u diagrams may contribute an imaginary part because of their proportionality to ##\sqrt{1-4m^2/q^2}##, where q^{2}stands for either t or u. Given that s>4m^{2}again we must have t,u<0 by the above. So the argument of the square root may never be negative and, as such, we are away from the branch cut and in the domain of parameter space where the diagrams are real.

I think there might be another argument without using the explicit form of the diagrams (indeed P&S make their statement before starting their computation of the s channel diagram) but I don't see it yet.

Does the above make it better?

Hmm I don't think so. What makes you think there is a square?

Thank you.

##1.## Now I understand that the ##t-## and ##u-## channel diagrams(tree level diagrams) have no branch cut singularities for ##s>4m^2##. But at the loop level, why are the ##t-## and ##u-## diagrams proportional to ##\sqrt{1-4m^2/q^2}## ?

##2.## Yeah, it is better.

##3.## Yes, you are right, I have made a mistake.

Last edited:

- #6

CAF123

Gold Member

- 2,950

- 88

As I mentioned in my last post, I think there should be an argument based on general considerations of analyticity but I don't see it. So, to see the proportionality to the kinematic square root, we can simply compute the diagrams, after all phi^4 theory is not beyond us :P

Edit: See below

Edit: See below

Last edited:

- #7

CAF123

Gold Member

- 2,950

- 88

Think it makes sense?

Share: