Is the total cross section a Lorentz Invariant? In Peskin and Schroeder's book (P&S), on the botton of page 106, the authors say that the total cross section transforms as its only non-invariant factor, namely:
$$
{1 \over E_{A} E_{B} |v_A - v_B|} 
$$
Where $E_i$ and $v_i$ are energies and velocities of the incoming particles ($i=A,B$). The authors then conclude that the cross section itself is not invariant. They actually go as far as rationalizing that it transforms as an area should (invariant to boosts in one direction but not on the other two).
This is at odds with many other sources (here is an example, look for equation 3.18 and 3.19), where an invariant cross section is obtained and that very factor is found out to be:
$$
{1 \over F} = {1 \over \sqrt{(p_A.p_B)^2-m_A^2 m_B^2}} = {1 \over \sqrt{|E_A\vec{p}_B-E_B\vec{p}_A|^2-|\vec{p}_A\times\vec{p}_B|^2}}
$$
Where $F$ is the so called Møller's invariant ﬂux factor ($p_i$ are the four momenta, and $m_i$ the masses). And the conclusion here is that the cross section is Lorentz invariant.
Of course, the second expression reduces to the first in any frame that $\vec{p}_A\times\vec{p}_B = 0$ (in particular the center of mass frame or the usual colinear bean "laboratory frame").
I am under the impression that P&S assume such a frame in more than one step of the calculation and that's why their result is frame dependent, but that means that their conclusion is wrong. Am I missing something?
 A: Peskin and Schroeder assume the laboratory frame, as is evident from the top of page 106: 

The difference $|v_A-v_B|$ is the relative velocity of the beams as viewed from the laboratory frame.

As a result, as you (and they) have also pointed out, the cross section is not Lorentz invariant. They furthermore explain that it is only invariant with respect to boosts along the $z$-direction. There is no contradicton in their derivation an conclusion. 
A: I present a deviation of the differential cross section, that (at least i hope so) makes things clearer. We rely on the following definition:
$$
 \mathrm d P =  \mathrm d \sigma \times F.
$$
Here $\mathrm d P$ is the transition probability per time and volume, with resulting particles $\{1,\dots,n\}$ carrying momenta $\{p_{f1},\dots, p_{fn}\} = p_f$. The incoming particles are labeled (as in P&S) with $A$ and $B$. F is the (relativistic) incoming flux density. We should express the flux density in terms of Lorentz covariants, in order to get a Lorentz covariant form of $\mathrm d \sigma$. We will come back to this later.
Lets start with the Transition probability. The incoming particles in the "infinite past" are considered as free particles (aswell as the final states in the "infinite future"). We define an initial state as the following wavepacket:
$$
\vert i \rangle = \vert \phi_A \phi_B \rangle_\text{in} = \int \dfrac{\mathrm d^3 k_A}{(2\pi)^3 2 k_A^0} \int \dfrac{\mathrm d^3 k_B}{(2\pi)^3 2 k_B^0} \tilde \phi_A(\vec k_A) \tilde \phi_B(\vec k_B) \vert \vec k_A \vec k_B \rangle_{\text{in}}.
$$
The typical restriction $k_i^0 = \sqrt{\vec k_i^2 + m_i^2}$ is made implicitly. Further, we use the abbreviation
$$
\mathrm{d} \tilde p = \dfrac{\mathrm{d}^3 p}{(2\pi)^3 2 p^0}
$$
from now on. The final state, can be a definite momentum eigenstate, as long as the detectors mainly measure momentum, and do not resolve positions at the level of de Broglie wavelenghts. E.g.
$$
\vert f \rangle = \vert \vec p_{f1},\dots,\vec p_{fn} \rangle_\text{out}.
$$
Before we continue, we make a few general statements on the initial wavepackets. The normalization of the momentum eigenstates is given as
$\langle \vec p \vert \vec p^\prime\rangle = (2\pi)^3 2 p^0 \delta(\vec p - \vec p^\prime)
$ and thus for a wavepacket
$$
\vert \phi \rangle = \int \mathrm d \tilde p \tilde \phi(\vec p) \vert \vec p \rangle,
$$
we obtain
$$
1 \stackrel{!}{=}\langle \phi \vert \phi \rangle = \int \mathrm d \tilde p \vert \tilde \phi(\vec p) \vert^2.
$$
We can define $\tilde \phi(\vec p)$ as Dirac sequence, e.g.
$$
\tilde \phi_\epsilon(\vec p) = \dfrac{1}{\sqrt{2\pi \epsilon}} \exp\left(-\dfrac{(\vec p - \vec p_A)^2}{2 \epsilon} \right) \stackrel{\epsilon \to 0}{\to} \delta(\vec p - \vec p_A).
$$
I.e. we can construct a wavepacket, with momenta tightly concentrated around a definite momentum $\vec p_A$:
$$
\vert \phi_\epsilon \rangle \stackrel{\epsilon \to 0}{\to} \vert \vec p_A \rangle.
$$
Let now $\vert \phi \rangle$ a wavepacket with concentrated momenta around $\vec p_A$. In QFT the position representation of such a wavepacket is defined as
$$
\phi(x) = \langle 0 \vert \Phi(x) \vert \phi \rangle = \int \mathrm d \tilde p e^{-ipx} \tilde \phi(\vec p),
$$
with the scalar field operator
$$
\Phi(x) = \int \mathrm d \tilde p \left(e^{-ipx} a(\vec p) + e^{ipx} b^\dagger(\vec p) \right).
$$
One could also use spinor or vector fields. For simplicity the further deviation is presented, using a scalar field operator. The final results are, however, valid for general fields. Now we define the following current (similar to the probability current in QM):
\begin{align}
j^\mu(x) &= i \left( \phi(x) \partial^\mu \phi^*(x) - \phi^*(x) \partial^\mu \phi(x) \right) \\
&=  \int \mathrm d \tilde p \int \mathrm d \tilde p^\prime \tilde \phi^*(\vec p) \tilde \phi(\vec p^\prime) (p^\mu + p^{\prime \mu}) e^{i(p-p^\prime)x} \\
&\simeq 2 p_A^\mu \int \mathrm d \tilde p \int \mathrm d \tilde p^\prime \tilde \phi^*(\vec p) \tilde \phi(\vec p^\prime) e^{i(p-p^\prime)x}\\
&= 2 p^\mu_A \vert \phi(x) \vert^2.
\end{align}
We interpret $j^0(x) = \rho(x)$ as particle density and $\vec j(x) = (j^1(x),j^2(x),j^2(x))$ as particle flux density. The transition probability $P(A,B \to 1,\dots,n \text{ with momenta }p_f)$ is given by
\begin{align}
P(A,B \to 1,\dots,n \text{ with momenta }p_f) = \left(\prod_f \mathrm d \tilde p_f\right) \vert {}_{\text{out}}\langle \vec p_f \vert \phi_A \phi_B \rangle_{\text{in}}\vert^2.
\end{align}
It holds
\begin{align}
{}_{\text{out}}\langle \vec p_f \vert \vec k_A \vec k_B \rangle_\text{in} &=  \langle \vec p_f \vert S \vert \vec k_A \vec k_B \rangle \\
&= \underbrace{\langle \vec p_f \vert \vec k_A \vec k_B \rangle}_{\text{unscattered part}} + \langle \vec p_f \vert \underbrace{S- \mathbb 1}_{T} \vert \vec k_A \vec k_B \rangle
\end{align}
(For a detailed definition of S, and the "in" and "out" states see P&S). The transition amplitude $\mathcal M$ is now defined as
\begin{align}
\langle \vec p_f \vert T \vert \vec k_A \vec k_B \rangle = i (2\pi)^4 \delta(k_A +k_B - {\textstyle\sum} p_f) \mathcal M(\{k_i\} \to p_f).
\end{align}
The physical interesting case is that the unscattered part vanishes (i.e. $\langle p_f \vert \phi_A \phi_B\rangle \simeq 0$). We furthermore assume, that the wavepackets $\vert \phi_A \phi_B \rangle$ are concentrated around $\vec p_A$ and $\vec p_B$. We obtain
\begin{align}
P(A,B \to 1,\dots,n \text{ with momenta }p_f) &=\left(\prod_f \mathrm d \tilde p_f\right) \left( \prod_{i=A,B} \int \mathrm d \tilde k_i \int \mathrm d \tilde k_i^\prime \tilde \phi_i^*(\vec k_i^\prime) \tilde \phi_i(\vec k_i) \right) \\
&\quad \times (2\pi)^4 \delta(k_A + k_B - {\textstyle\sum} p_f) (2\pi)^4 \delta(k_A^\prime + k_B^\prime - {\textstyle\sum} p_f) \\
&\quad \times \mathcal M^*(\{k_i^\prime\} \to p_f) \mathcal M(\{k_i\} \to p_f).
\end{align}
Since the momenta of the wavepackets are concentrated around $\vec p_A$ and $\vec p_B$ it holds
\begin{align}
\mathcal M^*(\{k_i^\prime\} \to p_f) \mathcal M(\{k_i\} \to p_f) \simeq \vert \mathcal M(\{\vec p_A ,\vec p_B\} \to p_f ) \vert^2,
\end{align}
and
\begin{align}
(2\pi)^4 \delta(k_A + k_B - {\textstyle\sum} p_f) (2\pi)^4 \delta(k_A^\prime + k_B^\prime - {\textstyle\sum} p_f) &= (2\pi)^4 \delta(k_A + k_B -(k_A^\prime + k_B^\prime)) (2\pi)^4 \delta(k_A^\prime + k_B^\prime-{\textstyle\sum} p_f) \\
&\simeq \int \mathrm d^4 x e^{ix(k_A^\prime + k_B^\prime - k_A -k_B)} (2\pi)^4 \delta(p_A + p_B - {\textstyle\sum} p_f) .
\end{align}
Back inserting yields
\begin{align}
P(A,B \to 1,\dots,n \text{ with momenta }p_f) &=\left(\prod_f \mathrm d \tilde p_f\right) (2\pi)^4 \delta(p_A + p_B -{\textstyle\sum} p_f)  \\ &\quad \times \int \mathrm d^4 x \vert \phi_A(x) \vert^2 \vert \phi_B(x) \vert^2 \vert \mathcal M(\{\vec p_A ,\vec p_B\} \to p_f ) \vert^2.
\end{align}
Comparison with the definition of $\mathrm d P$ yields
\begin{align}
\mathrm d P = \mathrm d \Phi_f \vert \phi_A(x) \vert^2 \vert \phi_B(x) \vert^2 \vert \mathcal M(\{\vec p_A ,\vec p_B\} \to p_f ) \vert^2,
\end{align}
with
\begin{align}
\mathrm d \Phi_f = \left(\prod_f \mathrm d \tilde p_f\right) (2\pi)^4 \delta(p_A + p_B -{\textstyle\sum}p_f),
\end{align}
the LIPS (Lorentz invariant phase space) element. We define the incoming flux density
as
\begin{align}
F = \sqrt{(j_A \cdot j_B)^2 - j_A^2 j_B^2}.
\end{align}
This is manifest Lorentz invariant, since $j_i^\mu$ are Lorentz vectors by construction. With the result from above we obtain
\begin{align}
F &= \sqrt{(4 \vert \phi_A(x)\vert^2 \vert \phi_B(x)\vert^2 p_A \cdot p_B)^2 - 16 \vert \phi_A(x)\vert^4 \vert \phi_B(x)\vert^4 p_A^2 p_B^2 \vert} \\
&= 4  \vert \phi_A(x)\vert^2 \vert \phi_B(x)\vert^2 \sqrt{(p_A \cdot p_B)^2 - m_A^2 m_B^2}.
\end{align}
The final result for the differential cross section is
\begin{align}
\mathrm d \sigma = \mathrm d \Phi_f \dfrac{\vert \mathcal M(\{p_A, p_B\} \to p_f)\vert^2}{4 \sqrt{(p_A \cdot p_B)^2 - m_A^2 m_B^2}}.
\end{align}
This expression is manifest Lorentz invariant. Lets see, if we can restore the result of P&S, where the incoming particles have collinear velocities $\vec v_A, \vec v_B$, e.g. in the center of masse frame of the incoming particles. In this case $p_A$ and $p_B$ are given as
\begin{align}
p_A = (E_A,\vec p), \quad p_B = (E_B,-\vec p).
\end{align}
Inserting yields
\begin{align}
4 \sqrt{(p_A\cdot p_B)^2 - m_A^2 m_B^2} &= 4 \sqrt{(E_A E_B + \vec p^2)^2 - (E_A^2 - \vec p^2)(E_B^2 - \vec p^2)} \\
&= 4 E_A E_B \sqrt{\vec v_A^2 + \vec v_B^2 - 2 \vec v_A \vec v_B} \\
&= 2 E_A 2E_B \vert \vec v_A - \vec v_B \vert. \\
\Rightarrow \mathrm d \sigma &= \mathrm d \Phi_f\dfrac{\vert \mathcal M(\{p_A, p_B\} \to p_f)\vert^2}{2 E_A 2E_B \vert \vec v_A - \vec v_B \vert}.
\end{align}
This agrees with the formula for the differential cross section, given in (4.79) in P&S. This specific expression for the differential cross section, is only form invariant under boosts in the direction of $\vec p$. But the general result above is valid in all frames. The only problematic statement P&S made, in my opinion, is to call $\vert \vec v_A - \vec v_B \vert$ the relative velocity. In special relativity, $\vert \vec v_A - \vec v_B \vert$ cannot be interpreted as "relative velocity". E.g. with $\vec v_A = c \hat e$ and $\vec v_B = -c \hat e$, we obtain
\begin{align}
\vert \vec v_A - \vec v_B \vert = 2c.
\end{align}
A proper definition of the relativistic relative velocity is
\begin{align}
v_\text{rel} = \dfrac{\sqrt{(p_A \cdot p_B)^2 - m_A^2 m_B^2}}{p_A \cdot p_B} = \dfrac{\sqrt{(\vec v_A -\vec v_B)^2 - (\vec v_A \times \vec v_B)^2}}{1-\vec v_A \cdot \vec v_B}.
\end{align}
A: I think I understand what Qmechanic means. The first expression for 1/F is written in a manifestly Lorentz invariant way, as it is written in terms of a dot product and masses only. This however does not mean that the cross section is a Lorentz invariant quantity. The Lorentz invariance of the expression only means that sigma can be computed in any frame using the momenta measured in that specific frame.
Let me give you another example to make this clearer: the energy of a scattering particle in the CM frame (or in the lab frame) can be expressed in terms of the Mandelstam variables and masses  (see exercise 3.25 of Griffiths). These "expressions" are thus also Lorentz invariant in the sense that they will give the same answer when computed in any inertial frame. Nevertheless, the answer will just give the energy in that specific frame, which is clearly not a Lorentz invariant quantity!
