The spacetime interval is invariant, represented mathematically as ##ds^2=g_{ab}dx^a dx^b##, while the spacetime metric ##g_{ab}## is a tensor. Although these statements are closely related, they are not equivalent; the interval is a scalar derived from the contraction of the metric tensor with infinitesimal vectors. The spacetime interval, ##ds^2##, is a scalar quantity that remains unchanged under Lorentz transformations, distinguishing it from four-vectors.
PREREQUISITESPhysicists, mathematicians, and students of relativity who seek a deeper understanding of the relationship between spacetime intervals and metrics in the context of tensor analysis.
Thanks! A little confused why you referred to ##ds^2## as a scalar. Isn't it a four-vector?Matterwave said:They aren't exactly equivalent, but they are closely related. The space time interval is ##ds^2=g_{ab}dx^a dx^b##. Because the metric ##g_{ab}## is a tensor, contracting it with (infinitesimal) vectors ##dx^{a}## will yield an invariant scalar ##ds^2##.
Nope. It's the scalar product of (differential) four-vectors.jmatt said:Thanks! A little confused why you referred to ##ds^2## as a scalar. Isn't it a four-vector?
jmatt said:Thanks! A little confused why you referred to ##ds^2## as a scalar. Isn't it a four-vector?