Meaning of eq.(4.64) in Peskin & Schroeder? In an introduction to the scattering matrix in the relativistic QFT setting, the book tries to generalize the non-relativistic Breit-Wigner formula (4.63):
$$
f(E)\propto\frac1{E-E_0+i\Gamma/2}.
$$
The reasoning goes as follows:

If we call the 4-momentum of the unstable particle $p$ and its mass $m$, we can make a relativistically invariant generalization of (4.63):
  $$
\frac1{p^2-m^2+im\Gamma}\approx\frac1{2E_{\mathbf p}(p^0-E_{\mathbf p}+i(m/E_{\mathbf p})\Gamma/2)}.
$$
  The decay rate of the unstable particle in a general frame is $(m/E_{\mathbf p})\Gamma$, in accord with relativistic time dilation. Although the two expressions in (4.64), i.e. the above equation, are equal in the vicinity of the resonance, the left-hand side, which is manifestly Lorentz invariant, is much more convenient.

My question: whose momentum and mass do we mean by $p$ and $m$? What are their relations with the $E$ and $E_0$ in eq.(4.63)?
I mean, usually the convention is $p^0=E_{\mathbf p}=\sqrt{m^2+|\mathbf p|^2}$, but this is clearly not the case here. Should we interpret the mass $m$ as the resonant energy $E_0$? Should we interpret the 0th-component of the 4-momentum $p^0$ as the energy of the incident particle $E$? If so, where does $E_{\mathbf p}$ come from? And how is $m/E_{\mathbf p}$ related to the time dilation factor?
So what happened between eq.(4.63) and eq.(4.64)? Many thanks!
 A: It is $p^0$ that replaces Breit-Wigners $E$, and $E_{\bf p}= \sqrt{m^2+{\bf p}^2}$ that replaces their  resonance energy $E_0$. P&S's expression  then reaches a maximum as  $p^0$ approaches $E_{\bf p}$.
A: I decide to write a detailed answer in case someone else comes across! Special thanks to @mike stone!
The whole system has a non-zero momentum, and this plays a part in the definition of the mass $m$ of the unstable particle!
The whole system has a 3-momentum $\mathbf p$ (from the incident particle) and a resonant energy $E_0$. When treated as an unstable particle, the resonant energy is associated to the energy of the unstable particle $E_{\mathbf p}:=E_0$. The rest mass of the unstable particle satisfies the formula $E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}$.
During the scattering process, however, the energy of the system may not be precisely the same as the resonant energy. The energy $E$ in the non-relativistic case is promoted to $p^0$ in the relativistic case.
The approximation in eq.(4.64) works as follows:
In the vicinity of resonance, 
$$p^0=E_{\mathbf p}+\epsilon.$$
By definition,
$$p^2=(p^0)^2-\mathbf p^2=(p^0)^2-E_{\mathbf p}^2+m^2.$$
Therefore, to first order of $\epsilon$,
$$p^2-m^2=2\epsilon E_{\mathbf p}=2E_{\mathbf p}(p^0-E_{\mathbf p}).$$
