# Size of transversal momenta in Multi-Regge kinematics?

Considering a scattering process in which $$2$$ incoming particles annihilate and produce $$n-2$$ other particles, one can consider the particle momenta $$p_i^\mu$$ (with $$i=1,2,3,...,n$$) to be in so called "Multi-Regge kinematics":

For that we introduce light-cone coordinates in the 0-th and 3-rd momentum components $$k_i^+=p_i^0+p_i^3$$ and $$k_i^-=p_i^0-p_i^3$$ (with transversal components unchanged $$\vec k_{i\perp}=(p_i^1,p_i^2)$$ ).

Then we take incoming momenta as

$$k_1=(k_1^+,k_1^-,\vec k_{1\perp})=(0,-\sqrt{s},\vec 0)~~~,~~~k_2=(k_2^+ ,k_2^-,\vec k_{2\perp})=(-\sqrt{s},0,\vec 0)$$

where $$s$$ is the squared center of mass energy. Simultaneously, for the outgoing momenta we impose the following constraints in their +-components:

$$k_3^+\gg k_4^+\gg... \gg k_n^+$$

while the transverse components are all comparable in magnitude $$\vec k_{3\perp}\sim\vec k_{4\perp}\sim...\sim \vec k_{n\perp}$$.

My question is:

In the multi-Regge kinematics, how does the magnitude of each $$\vec k_{i\perp}$$ relate to the corresponding $$k_i^+$$ and $$k_i^-$$? E.g. is it implied that components of $$\vec k_{3\perp}$$ are smaller in magnitude than $$k_3^+$$ and/or $$k_3^-$$? Also, is e.g. $$k_3^-$$ implied to be smaller in magnitude than $$k_3^+$$?

The answer to your first question is dictated by the on-shell condition: $$(k_{i \perp})^2 = k^+_i k^-_-$$. In words: the magnitude of the transversal momentum of particle $$i$$ is the geometric mean of the $$k_+$$ and $$k_-$$ of particle $$i$$.

In particular, if you are holding the former fixed while taking $$k_3^+ \gg \cdots \gg k_n^+$$, this implies that $$k_3^- \ll \cdots \ll k_n^-$$.

It is usually implied that each "$$\gg$$" is by the same order of magnitude, so that when we write $$a \gg b \gg c$$ etc, we mean for example that $$b/a$$ is the same order of magnitude as $$c/b$$, etc. Let's call this common order of magnitude $$\epsilon \ll 1$$. Similarly when I use "$$\ll$$" for the $$k^-$$ components, when I write $$z \ll y \ll x$$ etc, that means $$z/y$$, $$y/x$$ etc. are all $${\cal O}(\epsilon)$$.

So we have $$k_4^+ = {\cal O}(\epsilon k_3^+)$$, $$k_5^+ = {\cal O}(\epsilon^2 k_3^+)$$, etc, which we can summarize by saying that $$k_j^+ = {\cal O}(\epsilon^{j-3} k_3^+)$$.

It remains to decide how to scale $$k_3^+$$ with $$\epsilon$$, and this is most conveniently done by making a choice that is "symmetric but flipped" between particles 3 and $$n$$ in the sense that $${\cal O}(k_n^+) = {\cal O}(1/k_3^+)$$, and $${\cal O}(k_{n-1}^+) = {\cal O}(1/k_4^+)$$, etc.

Ultimately, this fixes $$k_j^+ = {\cal O}(\epsilon^{-\frac{n+3}{2}+j})$$ and hence $$k_j^- = {\cal O}(\epsilon^{\frac{n+3}{2}-j})$$ so that their product is $${\cal O}(1)$$ for all $$j$$, and hence so is $$\vec{k}_{j \perp}$$.

After a google search I found a paper where this is spelled out with some clarity: page 7 of arXiv:1112.6365.