# De Broglie relation using four momentum in particle physics

• I
In particle phyisics four-momentum is used and De Broglie relation is used to understand what lenghts can be "seen" in an experiment.
Here (page 6) https://people.phys.ethz.ch/~pheno/PPP/PPP2.pdf it is claimed

The key factor for investigating the proton substructure is the wavelength of the probing
photon, which is related to the transferred momentum by
$$\lambda\sim \frac{1}{\sqrt{Q^2}}$$
Where ##Q^2## is not actually "momentum" but its the square of the four momentum transferred by the photon.

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?

Related High Energy, Nuclear, Particle Physics News on Phys.org
jedishrfu
Mentor
Not sure if this will help but wikipedia has an article on matter-waves:

https://en.wikipedia.org/wiki/Matter_wave

and toward the end they describe how they use four-vectors to reduce the deBroglie relations to a single equation:

https://en.wikipedia.org/wiki/Matter_wave

Four-vectors
Main article: Four-vector
Using four-vectors, the De Broglie relations form a single equation:

which is frame-independent.

Likewise, the relation between group/particle velocity and phase velocity is given in frame-independent form by:

where

Four-momentum

Four-wavevector

Four-velocity

crick