Transformation of the Lorentz factor when a relativistic particle partially absorbs energy from a photon? I came across this paper (on arXiv) that says when a particle with high Lorentz factor $\Gamma$ meets a radiation beam and absorbs some energy $\epsilon$, it's Lorentz factor transforms like so (Eq. (1) of the text):
$$\Gamma_1 \approx \frac{\Gamma}{\sqrt{1 + 4\Gamma^2\epsilon/mc^2}}\tag{1}$$
Even though the author states it's "immediate" from energy-momentum conservation and proceeds to give a physical description of what this means for the interaction, it's not at all obvious to me how to derive such a result, nor what kind of approximations went into deriving it.
Could someone help me derive it please? Answers with detailed steps are appreciated. TIA.
EDIT: Thanks for the detailed answers! I'll award the bounty to @AlexGhorbal for the initial take that inspired the detailed answer by @robphy, and accept the response from @robphy for the relevant physical intuition.
 A: I could be wrong but I am getting a slightly different answer. The initial particle with mass $m$ and high Lorentz factor $\Gamma\gg 1$ has four-momentum $p^{\mu}\approx\Gamma m c (1,0,0,1)$ while the photon it absorbs has $p_{\gamma}^{\mu}=\frac{\epsilon}{c}(1,0,0,-1)$ assuming a head on collision in the $z$ direction. The particle after it has absorbed the photon has mass $m_1$ and four-momentum $p_1^{\mu}=\Gamma_1m_1(c,0,0,v_1)$ with $v_1$ its unknown velocity. Four-momentum conservation now gives
$$p^{\mu}+p_{\gamma}^{\mu}=p_1^{\mu}$$
and squaring both sides we find
$$p^{\mu}p_{\mu}+2p^{\mu}{p_{\gamma}}_{\mu}+{p_{\gamma}}^{\mu}{p_{\gamma}}_{\mu}={p_1}^{\mu}{p_1}_{\mu}$$
$$\Rightarrow -m^2c^2+2\left[-\Gamma m c\left(\frac{\epsilon}{c}\right)+\Gamma m c\left(\frac{-\epsilon}{c}\right)\right]\approx-m_1^2c^2$$
$$\Rightarrow m_1\approx m\sqrt{1+\frac{4\Gamma\epsilon}{mc^2}}$$
On the other hand, by simple energy conservation we have
$$\Gamma_1 m_1 c^2=\Gamma m c^2+\epsilon\approx\Gamma m c^2$$
assuming $\epsilon\ll\Gamma m c^2$, which is just the same as assuming that $\Gamma_1\gg1$ because if $\epsilon\sim\Gamma m c^2$ then the particle would be significantly slowed and no longer be ultra-relativistic.
Thus we finally get
$$\Gamma_1\approx\Gamma \frac{m}{m_1}\approx\frac{\Gamma}{\sqrt{1+\frac{4\Gamma\epsilon}{mc^2}}}$$
which is nearly what you have given. If you or anyone finds a mistake please let me know. Otherwise the paper might be wrong.
A: (See UPDATE below)

Inspired by @AlexGhorbal 's attempt,
I present a trigonometric derivation of Alex's result with an approximation used only at the last step.
(I was playing around with the equations... and the result fell out.)

The methods shown here are useful for other problems in special relativity,
not just the question in the OP.

I'm going to use rapidity $\theta$, where $v=\tanh\theta$ and thus
 the Lorentz factor $\Gamma=\cosh\theta$ and $v\Gamma =\sinh\theta$ and the Doppler factor $k=\exp\theta$.
Using components in the lab frame, energy-momentum conservation yields
\begin{align}
N\cosh\phi &=M\cosh\theta+\epsilon\\
N\sinh\phi &=M\sinh\theta-\epsilon
\end{align}
By addition, we obtain
$$\boxed{
\begin{align}
N\exp\phi &=M\exp\theta,
\end{align}
}
$$
which is essentially the Doppler Effect, where $K=\exp(\theta-\phi)=(N/M)$ is relative-Bondi-Doppler factor.
By subtraction, we obtain
$$
\boxed{\begin{align}
N\exp(-\phi) &=M\exp(-\theta)+2\epsilon,
\end{align}}
$$
[The above boxed equations represent
"energy-momentum conservation in light-cone coordinates" (in the eigenspace)]
We wish to eliminate $N$ (the mass of the final particle resulting from the absorption of the photon)
By multiplication,
$$\boxed{N^2=M^2+2\epsilon M\exp\theta}$$
which is the related to Alex's $\Rightarrow$ equation, but without using any approximation.
(Of course, this is the square-norm of the 4-momentum equation
$\tilde N=\tilde M+\tilde \gamma$.)
Re-write this as
\begin{align}
\left(\frac{N}{M}\right)^2 &=1+\frac{2\epsilon}{M}\exp\theta \\
K^2 &=1+\frac{2\epsilon}{M}\exp\theta \\
(\exp(\theta-\phi))^2 &=1+\frac{2\epsilon}{M}\exp\theta \\
\end{align}
(This is the a-ha moment!) Taking square-roots,
\begin{align}
\exp(\theta-\phi) &=\sqrt{1+\frac{2\epsilon}{M}\exp\theta} \\
\end{align}
and solving for $\exp\phi$, we have
\begin{align}
\exp\phi &=\frac{\exp{\theta}}{\sqrt{1+\frac{2\epsilon}{M}\exp\theta}} \\
\end{align}
This is a physics relationship between the lab frame's Doppler factors of the initial and final particle, the mass $M$ of initial particle and the photon energy $\epsilon$ in the lab frame. That is,
$$
\boxed{\boxed{
\begin{align}
k_N &=\frac{k_M}{\sqrt{1+\frac{2\epsilon}{M}k_M}} \\
\end{align}
}}
$$
which is an exact result.

UPDATE
The double-boxed result is obtained more directly by division of the first pair of boxed equations, rather than multiplication then some additional algebra to eliminate $N$.
\begin{align}
\boxed{\exp(2\phi) 
= \frac{M\exp\theta}{M\exp(-\theta)+2\epsilon}}
&= \frac{\exp(2\theta)}{1+\frac{2\epsilon}{M}\exp\theta}\\
\exp\phi 
&= \frac{\exp\theta}{\sqrt{1+\frac{2\epsilon}{M}\exp\theta}},
\end{align}
which is the double-boxed equation (when the Doppler-$k$ is introduced)!
In summary,

*

*express energy-momentum conservation in light-cone coordinates
(by adding and subtracting energy-momentum conservation in rectangular coordinates).

*divide those equations (which relate the square-of-$N$'s-Doppler-factor with $M$'s-Doppler-factor, mass $M$, and the lab-frame energy $\epsilon$ of the beam 
-- this is Mermin's aspect-ratio of the "causal diamond of N" (Mermin's "light rectangle")
to obtain the square of the double-boxed-equation

*In the large $\Gamma$ limit, the Doppler-factor $k \approx 2\Gamma$ (as shown below).

*If desired, one could carry out the expansion for $k$ further to get the next order in the approximation in terms of the Lorentz factors $\Gamma$.


Note
\begin{align}
k=\exp\theta &=\cosh\theta+\sinh\theta\\
&= \cosh\theta+\sqrt{\cosh^2\theta-1}\\
&=\Gamma+\sqrt{\Gamma^2-1}
\end{align}

Now, because the particles are assumed to have large Lorentz factors,
we can make the approximation that:
for large $\Gamma$, we have $k\approx 2\Gamma$,
and thus,
\begin{align}
2\Gamma_N &\approx \frac{2\Gamma_M}{\sqrt{1+\frac{2\epsilon}{M}(2\Gamma_M)}} \\
\Gamma_N &\approx \frac{\Gamma_M}{\sqrt{1+\frac{4\epsilon}{M}\Gamma_M}} \\
\end{align}
...in agreement with Alex's final result...
 but
using only one approximation, once [for $k_N$ and $k_M$]
at this last step here at the very end.
We see the approximate relation between Lorentz-factors
is based on a similar but exact relation between Doppler factors.
