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.


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