Relating momentum fraction to rapidity in a high-energy collision It's a well-known result in particle physics that in an underlying interaction like this:

assuming $p_0^-,p_{0\perp},m_0\ll p_0^+$, $m_1$ and $m_2$ are small, but $p_{1\perp}$ and $p_{2\perp}$ are large, the rapidity ($y = \frac{1}{2}\ln\frac{p^+}{p^-}$) differences between the final particles and the initial particle are
$$\begin{align}
y_1 - y_0 &= \ln\frac{1}{\xi} \\
y_2 - y_0 &= \ln\frac{1}{1 - \xi}
\end{align}$$
perhaps neglecting some terms subleading in $E_0$.
How would I show this? Pretty much all the references I've looked at (Kovchegov and Levin, Halzen and Martin, lots of papers) seem to take it as prerequisite knowledge. I did find this paper which gives a similar equation for the rapidity gap in diffractive scattering, but the derivation there relies on a nontrivial assumption about the distribution of produced particles which I don't think should be necessary.
Naturally, I've tried playing around with various kinematic equations involving rapidity, but I can't put together anything resembling the desired result.
 A: I'll take the simple, one-dimensional case. Since $\sqrt{2} p^+ = E + p$ and $\sqrt{2} p^- = E - p$, it works out that
$$
p^- = \frac{m^2}{2p^+}
$$
So, the rapidity of a particle is
$$
y = \frac{1}{2} \ln \frac{p^+}{p^-} = C + \ln \frac{p^+}{m}
$$
and therefore
$$
y_1 - y_0 = - \ln \frac{1}{\xi} + \ln \frac{m_0}{m_1} \\
y_2 - y_0 = - \ln \frac{1}{1-\xi} + \ln \frac{m_0}{m_2}
$$
So, I need to re-express the log of the ratios of the masses in terms of $\xi$.
Conservation of energy-momentum requires that
$$
m_0^2 = \frac{m_1^2}{\xi} + \frac{m_2^2}{1-\xi}
$$
I haven't yet used the condition that the masses are small, and I don't see how to use it naturally. In the limit that $m_1 = m_2 = m'$, the conservation law states that
$$
\frac{m_0^2}{m'^2} = \frac{1}{\xi(1-\xi)}
$$
and thus that
$$
\ln \frac{m_0}{m_1} = \ln \frac{m_0}{m_2} = \frac{1}{2} \ln \frac{1}{\xi} + \frac{1}{2} \ln \frac{1}{1-\xi}
$$
This clearly isn't working - it makes
$$
y_1 - y_0 = \frac{1}{2} \ln \frac{\xi}{1-\xi} \\
y_2 - y_0 = \frac{1}{2} \ln \frac{1-\xi}{\xi}
$$
