When you hold a loose string by its two ends, and just let it dangle in space, it looks an awful lot like a parabola. I mean, on first sight, who wouldn't have thought that? It was only upon closer inspection that we realized in fact the shape formed is not a parabola, but a catenary (see the graph of the hyperbolic cosine function, for instance). This demonstrates that sometimes immediate intuition can be wrong. Now then, my question is how do we know for sure that vectors are the correct description of reality when one discusses things like force or momentum? Intuitively, again, they seem highly reasonable, but do we have absolutely certain evidence that they are the correct model of the real world?

  • $\begingroup$ The vector description breaks down on the quantum scale. We use vectors in classical mechanics for no other reason than they predict the correct numbers (up to a level of uncertainty). $\endgroup$
    – Charlie
    Aug 13, 2020 at 19:10
  • $\begingroup$ "in fact the shape formed is not a parabola" To a second order approximation, it actually is, and for many purposes, that's good enough to be a correct description of reality. $\endgroup$
    – Sandejo
    Aug 13, 2020 at 19:13
  • $\begingroup$ So vectors are only convenient for approximating reality, then? $\endgroup$ Aug 13, 2020 at 19:14
  • $\begingroup$ How do you distinguish between a "correct description" and an approximation? $\endgroup$
    – Sandejo
    Aug 13, 2020 at 19:15
  • $\begingroup$ Well, I guess that means having a theoretically sound basis versus just a numerical approximation. $\endgroup$ Aug 13, 2020 at 19:16

1 Answer 1


Physics is an empirical science. The only things that have physical meaning are experimental measurements. One of our goals as physicists is to write down mathematical models, the outputs of which correspond to these measurements. The only way to check the validity of a model is to compare the outputs of the model to the measurements. There is no way to ever guarantee or prove that a given model is correct, in the sense of a mathematical proof. The best we can ever do is acquire evidence that our model agrees with experiments.

Since the only things with physical meaning are measurements, the mathematical objects and techniques we use within our models have no meaning in the physical world. They are only tools for calculating outputs that we can compare with experiments.

If we use a model like newtonian mechanics that uses vectors to produce predictions that agree with experiments, we accept the validity of that model, including the mathematical tools it applies. But this is absolutely not a proof that vectors provide a description of reality, nor can we ever hope to have such a proof — because experiments cannot prove anything, and because experimental measurements are the only description of reality.

Having said all that, vectors are very natural and generic mathematical objects that show up any time we use a model that is linear. Linear models are typically the simplest models to understand and compute with. (Perhaps this is because of the way our brains evolved.) Much of what we do in physics is to reduce complicated problems to something linear, in which case vectors will always show up.

For example, in general relativity, spacetime is not a linear space. But if we zoom in to an infinitesimally small region of spacetime, we can model this small region by a linear space, which we can describe using vectors. Then we can describe the complicated spacetime essentially by gluing together these linear spaces.

Another example is that in physics we are interested in the action of symmetries on our model. To study how these symmetries act, we can look in particular at how they act on linear spaces — that is, on vectors. This is called representation theory. Representation theory is essential to our current understanding of quantum mechanics.

In spite of the fact that vectors do not and cannot have any physical meaning, it is likely that vectors will always show up in our physical models, because we will always want to be able to reduce complicated problems to more easily understood linear problems.

  • $\begingroup$ Thanks for the very thorough response! This answers my question. $\endgroup$ Aug 13, 2020 at 19:55
  • 1
    $\begingroup$ +1. I would add that we don't have a complete understanding of physics yet. That is, there are some areas where we don't know which mathematical tools to use to model the experiments. $\endgroup$
    – mmesser314
    Aug 13, 2020 at 23:05
  • $\begingroup$ Good point. Indeed, it is possible — perhaps even likely — that those tools have not yet been developed. $\endgroup$
    – d_b
    Aug 13, 2020 at 23:10

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.