- #1
MathematicalPhysicist
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Hi. on page 95 , I am not sure how did he derive the second term on the RHS of equation (13.16).
http://books.google.co.il/books?id=...V4QSnhYFA&ved=0CBsQ6AEwAA#v=onepage&q&f=false
I mean if I plug back I should get:
[tex] i\theta(x^0-y^0)\int_{4m^2}^\infty \rho(s) \int \tilde{dk} e^{ik(x-y)} + i\theta(y^0-x^0)\int_{4m^2}^\infty \int \tilde{dk} e^{-ik(y-x)} [/tex]
And since [tex] \int \tilde{dk} e^{\pm ik(x-y)} = \int \frac{d^3 k}{(2\pi)^3 2k^0} e^{\pm ik(x-y)} = \frac{\delta(x-y)}{2k^0}[/tex]
But I don't see how do we arrive from all of the above to the epression: [tex] \frac{1}{k^2+s-i\epsilon}[/tex] in the integral, anyone cares to elaborate?
Thanks.
P.S
I forgot to mention that: [tex] k^0 =\sqrt{\vec{k}^2 +s} [/tex]
where s is one of Mandelstam parameters, I believe it's standard in the literature.
http://books.google.co.il/books?id=...V4QSnhYFA&ved=0CBsQ6AEwAA#v=onepage&q&f=false
I mean if I plug back I should get:
[tex] i\theta(x^0-y^0)\int_{4m^2}^\infty \rho(s) \int \tilde{dk} e^{ik(x-y)} + i\theta(y^0-x^0)\int_{4m^2}^\infty \int \tilde{dk} e^{-ik(y-x)} [/tex]
And since [tex] \int \tilde{dk} e^{\pm ik(x-y)} = \int \frac{d^3 k}{(2\pi)^3 2k^0} e^{\pm ik(x-y)} = \frac{\delta(x-y)}{2k^0}[/tex]
But I don't see how do we arrive from all of the above to the epression: [tex] \frac{1}{k^2+s-i\epsilon}[/tex] in the integral, anyone cares to elaborate?
Thanks.
P.S
I forgot to mention that: [tex] k^0 =\sqrt{\vec{k}^2 +s} [/tex]
where s is one of Mandelstam parameters, I believe it's standard in the literature.
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