Legendre transform and momentum/velocity

[redtree]

I apologize for the simplicity of the question. I have been reading a paper on the Legendre transform (https://arxiv.org/pdf/0806.1147.pdf), and I am not understanding a particular step in the discussion.

In the paper, Equation 16, where $\mathcal{H} = \sqrt{\vec{p}^2 + m^2}$:
\begin{equation}
\begin{split}
\vec{v} &\doteq \frac{d\mathcal{H}}{d \vec{p}}
\\
&= \frac{\vec{p}}{\sqrt{\vec{p}^2 + m^2}}
\end{split}
\end{equation}

Such that, Equation 17
\begin{equation}
\begin{split}
\vec{p} &= m \vec{v} \left(\frac{1}{\sqrt{1-\vec{v}^2}} \right)
\end{split}
\end{equation}

However, I don't understand the derivation of Equation 17 from Equation 16. I assume I am missing something very simple, but I am not seeing it.

 

[Orodruin]

Square (16) and solve for $\vec p^2$ in terms of $\vec v$. Reinsert in the denominator and solve for $\vec p$.

 

[redtree]

Nice; simple; I figured; thanks!