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Jan 27, 2015 at 23:43 answer added ritter timeline score: 1
Jan 27, 2015 at 21:35 comment added ritter Digging more into this I now believe that referring to "spacelike" ($q^2<0$) just means that they are talking about scattering. Later in the article they refer to the timelike ($q^2>0$) range and this is no longer a scattering process. The $q^2>0$ region generally means annihilation processes. Although here more questions arise. How do you measure the pion FF in the timelike region, i.e. when there's by definition no scattering..?
Jan 27, 2015 at 20:58 comment added ritter The vector $q$ is defined as the momentum transfer originating from the scattered pion $q_\mu=k_\mu-k'_\mu$ with $k$ and $k'$ the pion 4-momentum before and after the scattering. Not sure if that helps..
Jan 27, 2015 at 20:52 comment added ritter I believe the vector $q$ connects the first scattering vertex (incoming pion, photon, outgoing pion) with the second vertex (incoming e, photon, outgoig e). Thus, the question is, why is this energy range considered as being space-like?
Jan 27, 2015 at 20:50 comment added ritter Of course, one can also talk about a vector being space/time/light-like; same as two points in a vector space always define a vector, namely the one pointing from the first to the second point.
Jan 27, 2015 at 0:56 comment added Ryan Unger @dmckee: I'm not sure how it pertains to this question though. If the four-momentum is not null, then the photon has to be virtual. I'm not sure what the physical meaning of space-like energy is.
Jan 27, 2015 at 0:17 comment added dmckee --- ex-moderator kitten @0celo7 That's an answer. Why not post it as one.
Jan 26, 2015 at 19:49 comment added Ryan Unger You can of course talk about a spacelike distance, but more generally it refers to a spacelike vector. In -+++, a timelike vector has negative Minkowski norm. A spacelike one has positive norm.
Jan 26, 2015 at 17:01 history asked ritter CC BY-SA 3.0