Equivalent definitions of vectors.

In maths a vector is an object that obeys some axioms of a vector space. But in physics a vector can be thought as an object which is invariant under rotations of the system's coordinates.

Is there a way to prove this equivalence?

  • $\begingroup$ More on the notion of a vector: physics.stackexchange.com/q/65084/2451 and physics.stackexchange.com/q/41211/2451 $\endgroup$ – Qmechanic Sep 16 '13 at 18:50
  • 3
    $\begingroup$ But in physics a vector can be thought as an object which is invariant under rotations of the system's coordinates. I would consider a scalar to be an object that is invariant under rotations. A vector transforms under rotations, e.g., under a 180-degree rotation it flips its direction. Is there a way to prove this equivalence? They're not equivalent. The mathematical definition includes all kinds of objects that have nothing to do with space or rotations. E.g., you could have a vector whose 19 components represented the production of 19 factories. $\endgroup$ – Ben Crowell Sep 16 '13 at 19:07
  • $\begingroup$ @Ben that would probably make a good answer. $\endgroup$ – David Z Sep 16 '13 at 20:40
  • $\begingroup$ @BenCrowell I think "invariant under change of coordinates" means, that if you have a vector point at some specific point in space and then change your coordinate system, the vector will still point to this same point. In the new coordinate system it has of course different components, but it still represents the same thing, namely this point. Therefore I think your answer is wrong, since it is indeed invariant under a change of coordinate system, because it continues to represent the same thing. $\endgroup$ – philmcole Feb 12 '18 at 19:46

What's happened is that you've been victimized by a standard physicist abuse of a non-distinction (I myself was a victim of this once) between a vector space, and a vector space that has been equipped with a certain notion of how elements of that space should transform (which is an additional mathematical input).

Both physicists and mathematicians use the term "vector" to mean an element of some vector space. In classical mechanics, for example, the velocity of a particle moving in three dimensions is modeled as an element of $\mathbb R^3$. In quantum mechanics, the state of a system is modeled as an element of a Hilbert space, a special kind of complex vector space that can either be finite-, or infinite-dimensional. There are lots of other examples as well.

So where does this idea of transformation come in?

Well in physics, it's relevant ask what happens to certain vectors when we do some physical transformation on the system we are studying. If we take the single particle and its velocity vector as an example, we could ask ourselves what happens to that mathematical vector $\mathbf v$ when we physically rotate the axes with respect to which we've determined its components. If we go out and do this in the real world, we find that the components of the position of the vector $\mathbf x$ get rotated, \begin{align} \mathbf x \to R\mathbf x \end{align} and nothing happens to our time measurements, so our determination of the velocity vector, which is just the time-derivative of the position vector, also get's rotated; \begin{align} \mathbf v \to R\mathbf v \end{align} Notice that there was no such notion of transformation built into the original mathematical description of velocity vectors as elements of $\mathbb R^3$, it's an extra consideration we make as physicists. In other words, what we're doing here is saying:

"Let's consider acting on elements of the vector space $\mathbb R^3$ by rotations in a particular way that is physically meaningful.

Once you have determined such a physically meaningful notion of transformation, you then ask

"If I build some vector out of other vectors that are defined to transform in this way, then will the resulting vector transform in the same way?"

This is precisely what we did with the position and velocity. We defined the position to change in a particular (physically motivated and meaningful) way under rotations, and then we noticed that if position transforms like this, then velocity does too.

If you want to mathematically formalize the additional notion of transforming vectors in vector spaces, then you would appeal to a mathematical field called representation theory in which one studies different mathematical ways that one can act on elements of vector spaces by elements of (among other things) groups. In particular, one considers, in addition to a vector space $V$ mappings $\rho:G\to \mathrm{GL}(V)$ from a group (or algebra, etc.) to the set of invertible, linear operators on a vector space $V$. One then chooses to study certain of these representations according to the physical context, namely according to which representation gives a type of transformation that is physically meaningful.

In the example above with position an velocity, we are implicitly considering the representation $\rho:\mathrm{SO}(3)\to \mathrm{GL}(\mathbb R^3)$ defined by $\rho(R) = R$, where here $\mathrm{SO}(3)$ denotes the group of rotations of three-dimensional Euclidean space, and we are then saying

"Define position vectors to transform as $\mathbf x \to \rho(R) \mathbf x = R\mathbf x$ under rotations, and now notice that velocity can be shown to transform in the same way."

  • $\begingroup$ +1 I'd add that most physicists see a vector as something in the real world that can be measured as having a magnitude and a direction, despite it being taught in schools to kids. The mathematician then abstracts this definition, stripping it of "direction" and "magnitude", just as a topologist strips space of "distance", leaving the rest of us dazed and confused. $\endgroup$ – Larry Harson Sep 17 '13 at 19:43
  • $\begingroup$ Not much related: $\mathrm{O(3)}$ is defined as the 3x3 orthogonal matrices. Could we define it instead, by one of its representations? $\endgroup$ – jinawee Dec 18 '13 at 20:05
  • $\begingroup$ @jinawee If the representation were faithful, then yes. $\endgroup$ – joshphysics Dec 20 '13 at 1:28
  • 1
    $\begingroup$ @goblin If that were true, then how would one account for vector spaces in physics for which there may not be a physically natural inner product? Take, for example, the tangent space to the phase space (a symplectic manifold) of a system in classical mechanics. $\endgroup$ – joshphysics Feb 13 '18 at 3:02
  • 1
    $\begingroup$ @DannyHansen Let me know if this clarifies: physical positions are mathematically modeled as elements $\mathbf x$ of $\mathbb R^3$. One can show empirically that physically rotating the positions of objects in the real world corresponds in the mathematical model to acting on $\mathbf x$ by a suitable element of $\mathbb R^3$. So the action of $\mathrm{SO}(3)$ on $\mathbb R^3$ defined by $\mathbf x\mapsto R\mathbf x$ is the physically relevant one. This group action can then induce actions of $\mathrm{SO}(3)$ on other mathematical objects we attempt to build out of mathematical positions. $\endgroup$ – joshphysics Jul 30 at 19:06

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.