Spacelike and timelike photons I know the definition of space-like and time-like intervals. How do you show that in some reaction the virtual photon(s) is spacelike/timelike?
On page 191 Peskin and Schroeder, it says "since $q^2<0$ for a scattering process..." Why is this statement true? I don't see it straight away
For example in Bahba scattering discussed on this page.
 A: The same way you determine if the interval is space-like or time-like. In fact you do it by computing the square of the four-momentum and examining the sign.
Which sign is space-like and which time-like is a matter of convention, and varies from source to source. I like to compute the squared-interval as
$$ (\Delta s)^2 = (\Delta t)^2 - (\Delta \vec{x})^2 \,,$$
and therefore the square of the four momentum as
$$ q^2 = E^2 - \vec{p}^2 \,,$$
(where $E$ and $p$ are those of the exchange particle) so positive is time-like and negative is space-like.
Of course I often actual use $Q^2 = -q^2$ as the figure of merit, so the signs change again, but that just reflect the kind of reaction I was working on at the start of my career.
A: Suppose a scattering process with a 3- particle  vertex : $A \to  B + \gamma$. Here we suppose that particles $A$ and $B$ are massive, with the same mass $m$, and  $\gamma$ is the "virtual" photon. Let $a,b, q$ be the momenta of the particles $A,B,\gamma$.
You have : 
$q^2 
=(a-b)^2 \\= (a_0 - b_0)^2 - ( \vec a - \vec b)^2
\\=(a_0^2 - \vec a^2) + (b_0^2 - \vec b^2) - 2(a_0b_0 - \vec a . \vec b )
\\=2(m^2-a_0b_0 + \vec a . \vec b)$
We may always choose a frame where $\vec a = - \vec b, a_0=b_0$, so we get :
$q^2 = 2(m^2-a_0^2-\vec a^2)= 4(m^2-a_0^2) < 0$
