Why does $k^z/E$ have dimensions of inverse velocity rather than velocity?

Question

I'm studying quantum field theory and I want to prove the cross section. In Peskin's book, equation 4.77 says that:

$$ \frac{1}{\left | \frac{k_{A}^{z}}{E_{A}}-\frac{k_{B}^{z}}{E_{B}}\right |}=\frac{1}{\left | v_{A}-v_{B} \right |}. \tag{4.77} $$

This means that: $$ \frac{k_{A}^{z}}{E_{A}}=v_{A} $$ Where $k_{A}$ is the momentum of particle $A$, and $v_{A}$ is the velocity.

My problem is $E_{A}$ has the dimension: $\left [ M \right ]\left [ L^{2} \right ]\left [ T^{-2} \right ]$, and $k_{A}$ has the dimension $\left [ M \right ]\left [ L\right ]\left [ T^{-1} \right ]$. Doing $$\frac{k_{A}^{z}}{E_{A}}$$ gives the dimension of $$ \frac{\left [ M \right ]\left [ L \right ]\left [ T^{-1} \right ]}{\left [ M \right ]\left [ L^{2}\right ]\left [ T^{-2} \right ]}=\left [ L^{-1}\right ]\left [ T\right ] $$ which is the dimension of the reciprocal of velocity. But the text shows that it has the dimension of velocity. Why is that?

Answer 1

Peskin and Schroeder use natural units, where $c = 1$, or alternatively where factors of $c$ (and $\hbar$) are omitted. If you restore the necessary constants to make the units work out in an SI-like system, you would have $$\frac{k^z}{E} \to \frac{k^z c^2}{E} = v$$