The absolute value of the $x$ component of a particle's four-momentum vector is always either equal to or smaller than its $t$ component.

Question

Quizplus · Accepted Answer

The four-momentum of a particle is defined as $p^\mu = (E/c, \vec{p})$, where $E$ is the energy, $c$ is the speed of light, and $\vec{p}$ is the three-momentum. The magnitude of the $t$ component (energy term) is always greater than or equal to the magnitude of any spatial component due to the relationship $E \geq |\vec{p}c|$, which follows from the relativistic energy-momentum relation $E^2 = (pc)^2 + (mc^2)^2$, where $m$ is the rest mass of the particle. This ensures that the energy (or the $t$ component of the four-momentum) is always greater than or equal to the magnitude of the momentum multiplied by $c$ (the spatial components). 
Note that $m^{2}=p_{t}^{2}-p_{x}^{2}-p_{y}^{2}-p_{z}^{2} \Rightarrow p_{t}^{2}=p_{x}^{2}+\left[p_{y}^{2}+p_{z}^{2}+m^{2}\right]$.

The Absolute Value of The xxx Component of a Particle's Four-Momentum Vector Is Always Either Equal

The Absolute Value of The $x$ Component of a Particle's Four-Momentum Vector Is Always Either Equal