# Explicit calculation, non-relativistic limit of $p^2-m^2$ in L. Ryder p.189

This could be related to this question or Eq. (18) p.12 but at the moment I am just very pragamatically in need of an basic explicit calculation to understand the paragraph between (6.36) and (6.37) p.189 of QFT by Lewis H.Ryder.

In the notes by Padmanabhan, he takes the limit $$c\to \infty$$ but here we obviously are considering $$c=1$$... In fact I'm not sure at all about what Ryder means by $$p^2$$ here: if it's on-shell then it's $$0$$. We may try to give explicit expression, e.g. $$p^0:=\frac{E}{c}= \gamma m c\ ,\quad \mathbf{p}=\gamma m \mathbf{v}\ ,\quad \gamma = \frac{1}{\sqrt{1- \mathbf{v}^2/c^2}}$$ and then make a Taylor expansion of $$\gamma$$ (very stupid) because of course the following will give 0 (expansion of 0 is 0...): $$p^2-m^2 c^2 = \gamma^2 (m^2 c^2 - m^2 \mathbf{v}^2) - m^2 c^2 = \cdots =0$$

So obviously we must understand $$p^2$$ as off-shell. And if we look at the result, we must introduce what he calls $$T$$. How is that related to $$E$$?, what does he mean by $$\mathbf{p}$$? probably the non-relativistic one? but isn't that the kinetic energy $$T$$?

• Can you tell us what eq 6.36 and 6.37 say? Not all of us have copies of Ryder. – mike stone Sep 14 '20 at 18:16

## 1 Answer

I found Ryder on amazon "look inside" and the equations are there. It is indeed off shell. If we define the off-shell energy as $$E=m+T$$, with $$|{\bf p}|$$, $$T$$ small I think he means that the denominator is approximately
$$p^2-m^2= E^2-{\bf p}^2-m^2 =(E-\sqrt{{\bf p}^2+m^2})(E+\sqrt{{\bf p}^2+m^2)}\\ \approx (E-m -{\bf p}^2/2m)(E+m) \approx (T-{\bf p}^2/2m) 2m$$ The last equation is the unnumbered equation in the relevent paragraph before 6.37.