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It probably seems ridiculously naive of me to ask such a basic question, but I have a need to use accurate language. Typically, I think of a displacement as a directed line segment whose end-points are the locations of a particle at an initial time, and at a subsequent time.

Or, in the absence of such a particle, a displacement is a directed line segment between two points.

A displacement is not a vector. A vector is defined either as a unique mathematical object in an abstract vector space, or as an equivalence class of parallel directed line segments of equal magnitude and sense, or some variant, thereof.

So, by my definition of displacement: Consider a particle initially at point $\mathscr{A}_0$, which moves along an arbitrary path to point $\mathscr{A}_1$, which is one meter distance from $\mathscr{A}_1$. A second particle is initially located at $\mathscr{B}_0$, which is not colocated with $\mathscr{A}_0$. The second object moves along any other arbitrary path to point $\mathscr{B}_1$, one meter distance from $\mathscr{B}_0$, along a line parallel to the line determined by $\mathscr{A}_0$ and $\mathscr{A}_1$. The displacements of these particles are equal in direction and magnitude, but are not identical displacements.

So, I ask, is my definition of displacement commonly accepted?

For some of my motivation for asking this question, see the introductory chapters in Gravitation by Charles W. Misner, Kip S. Thorne & John Archibald Wheeler or the introductory chapters in Modern Classical Physics Optics, Fluids, Plasmas, Elasticity, Relativity, and Statistical Physics, by Kip S. Thorne & Roger D. Blandford.

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So, I ask, is my definition of displacement commonly accepted?

No. You're defining an alternative notion, which is not standard in the context of Newtonian mechanics. By the standard definition, your two displacements are the same displacement.

If you look at old physics textbooks, ca. 1930, you can find variations in these definitions, e.g., I believe in Millikan and Gale they define a vector as having a location. But nowadays the definitions are standardized so that displacements are vectors, and vectors are "portable."

For some of my motivation for asking this question, see the introductory chapters in Gravitation by Charles W. Misner, Kip S. Thorne & John Archibald Wheeler or the introductory chapters in Modern Classical Physics Optics, Fluids, Plasmas, Elasticity, Relativity, and Statistical Physics, by Kip S. Thorne & Roger D. Blandford.

In the context of general relativity, neither positions nor finite displacements are vectors. Only infinitesimal displacements are vectors, and vectors that live at different points cannot be compared except by parallel transport. Also, "vector" in this context refers to a four-vector with the correct transformation properties for a four-vector.

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  • $\begingroup$ In general relativity, as on any curved manifold, there is a vector space tangent to every point. My motivation is derived from the distinction between events as space-time locations (where a thing really happens,) and coordinates. Coordinates being the addresses we assign to events. A space-time interval is "the separation between two events", Taylor, Wheeler, Spacetime Physics. But they equivocate between definitions of interval. Sometimes it means space-time "distance". Other times it is the segment of the un-accelerated world line between two events, etc. $\endgroup$ – Steven Thomas Hatton Jun 16 '18 at 22:11
  • $\begingroup$ Bah! In The Meaning of Relativity, Einstein uses the term interval, applied to Euclidean 3-space to mean essentially the same thing I mean by displacement. He uses displacement to mean the "motion" of a point along a curve, as in parallel displacement. $\endgroup$ – Steven Thomas Hatton Jun 16 '18 at 22:51

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