Displacements are vectors because they add like vectors is the answer. It is also an experimental fact.

Though rotations are displacements but not vectors.

Is there any more fundamental or intuitive reason that makes displacement a vector?

  • 3
    $\begingroup$ Rotations can be defined as vectors as well. The rotational displacement vector $\vec \theta$ is simply the vector that has a magnitude equal to the angle $\theta$ and a direction along the axis rotation (and directed according to the right-hand rule). You can define anything as vectors, when useful. It is just a mathematical tool. $\endgroup$ – Steeven Mar 15 at 11:50
  • 1
    $\begingroup$ I think you will find that things only work so well in 'nice' coordinate systems. Try to write a displacement vector in spherical coordinates, and using spherical basis. For example what is the displacement vector from $r=1,\, \theta=0,\,\phi=0$ to $r=1,\, \theta=\pi/2,\,\phi=0$? $\endgroup$ – Cryo Mar 15 at 12:19
  • $\begingroup$ More to the point what makes vector, a vector, for applications in physics, is that it transforms as a tensor (rank (1,0) tensor): en.wikipedia.org/wiki/Tensor $\endgroup$ – Cryo Mar 15 at 12:21
  • 1
    $\begingroup$ @Steeven It should be noted that this is only true if you keep your rotations along a single axis $\endgroup$ – Aaron Stevens Mar 15 at 12:34

I'd say that for most purposes we can define vectors as a quantities that can be represented as displacements, added as displacements, multiplied by scalars as we do displacements and assigned magnitudes as we do displacements. In other words displacement is the archetypal vector.

  • $\begingroup$ Yes, but then it becomes quite difficult to move from the archetypal vector to other vectors, without passing from extracting the main features of the archetype, which is the same as to provide the set of axioms of a vector space. $\endgroup$ – GiorgioP Mar 17 at 9:50
  • $\begingroup$ I entirely agree that if is trying to define a vector as an object which is a member of a vector space then we need to start with the axioms, and would need to extract them. But I'd suggest that quite a large and useful class of vectors may be defined in terms of displacement, without requiring formal vector space axioms (at least in elementary work). I'm thinking of velocity, acceleration, momentum... $\endgroup$ – Philip Wood Mar 17 at 10:21

Without overcomplicating, anything that has both a direction and a magnitude is a vector.

A displacement has both a magnitude and a direction. It is therefore a vector.

Rotations are interesting and can be misleading. One might think "well, you can rotate a thing by some angle, so there's a magnitude. And you can rotate it clockwise or counterclockwise, so there's your direction."

This looks rather like a displacement in one dimension: you can say it goes either left or right, for example.

But in one dimension, we no longer use the vector formalism for displacements; we just give the number. You could deal with it as a vector, but that would simply amount to writing the unit vector of the single dimension everywhere in your problem. May as well not call it a vector,

But in higher dimensions, displacement must be a vector, because you must specify the amount of displacement along all directions.

There is no higher-dimensional analog of rotations like this. Any rotation happens about a single axis, which is a vector, but the rotation itself is simply an amount, either positive or negative. As such, we usually think of rotation as a scalar.

You can, however, define rotational displacements. Like @Steeven's comment. It requires a direction and a magnitude, though. The direction in this case is the axis of the rotation, and the magnitude is the angle of rotation. A clockwise or counterclockwise rotation would be antiparallel vectors, with the direction detetmined by the right-hand rule.

The big idea: a vector has a direction and a magnitude. If your quantity has both, it's a vector.

  • $\begingroup$ As noticed by Aaron Stevens, rotations around different axis do have direction and magnitude, still they are not vectors. $\endgroup$ – GiorgioP Mar 17 at 9:52
  • $\begingroup$ That's exactly what I said in my answer. $\endgroup$ – flevinBombastus Mar 17 at 12:08
  • $\begingroup$ Not exactly. Your conclusion is that eveything which has direction and magnitude should be a vector. It is pecisely the opposite of what Aaron Stevens and me are saying. $\endgroup$ – GiorgioP Mar 17 at 15:23
  • $\begingroup$ I specifically treated rotations. It took up most of my answer. It's not hard to read this. $\endgroup$ – flevinBombastus Mar 17 at 19:31
  • $\begingroup$ Ok, then please, check your final sentence: a vector has a direction and a magnitude. If your quantity has both, it's a vector. $\endgroup$ – GiorgioP Mar 17 at 20:53

Stated in a slightly different way, the original question is: "why we can consider a displacement in the space as well modeled by the vectors as introduced and characterized by mathematicians?

Since the problem is if a physical object behaves as a mathematical entity, one has to start with the exact definition of that entity. It would be audacious to suggest a partial and hasty comparison of the physical and mathematical quantities. It is like a witness who has to recognize the responsible of a crime: he or she has to be sure. Would you never suggest an "intuitive recognition"?

A fundamental reason is the only way to provide an answer. And the sentence "Displacements are vectors because they add like vectors" even if incomplete, is in the right direction.

It is incomplete because the mathematical definition of vector is an element of a vector space, where a vector space is a set equipped with two operations: a sum of two vectors and a product of a vector by a scalar. These operations must obey a set of additional rules (the axioms defining a vector space). In particular, sum must be associative, i.e. $$ u + (v + w) = (u + v) + w. $$ This explains why rotations are not vectors. However, in order to show that displacements are vectors, one has to show that also the axioms connected with multiplication by a scalar are verified (and they are) by displacements.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy