(a) The squared magnitude of a four-vector could be negative.

Question

Quizplus · Accepted Answer

The squared magnitude of a four-vector in special relativity, defined as $s^2 = -(c\tau)^2 + \mathbf{x}^2$, where $c$ is the speed of light, $\tau$ is the proper time, and $\mathbf{x}$ is the spatial part of the four-vector, can be negative for spacelike intervals where the spatial separation is greater than the temporal separation when measured in the same units. 
(a) For example, the displacement four-vector $\Delta \boldsymbol{s}$ between space like events has a squared magnitude of $\Delta s^{2}<0$ by definition.

(A) the Squared Magnitude of a Four-Vector Could Be Negative