- #1

- 43

- 4

Here (page 6) https://people.phys.ethz.ch/~pheno/PPP/PPP2.pdf it is claimed

Where ##Q^2## is not actually "momentum" but its the square of the four momentum transferred by the photon.The key factor for investigating the proton substructure is the wavelength of the probing

photon, which is related to the transferred momentum by

$$\lambda\sim \frac{1}{\sqrt{Q^2}}$$

I can't understand why this is used, since the square root of square of four momentum is

$$\sqrt{p\cdot p}=\sqrt{|\textbf{p}|^2-E^2/c^2}$$

While De Broglie relation usually involves the three momentum ##\textbf{p}##

$$\lambda\sim \frac{1}{|\textbf{p}|}$$

So is in this case ##\sqrt{Q^2}## approximately the three momentum? Or is the De Broglie relation in the relativistic case to be written using ##\sqrt{Q^2}## instead of three momentum?