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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?

Qmechanic
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SurfaceIntegral
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  • 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). – Charlie Aug 13 '20 at 19:10
  • "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. – Sandejo Aug 13 '20 at 19:13
  • So vectors are only convenient for approximating reality, then? – SurfaceIntegral Aug 13 '20 at 19:14
  • How do you distinguish between a "correct description" and an approximation? – Sandejo Aug 13 '20 at 19:15
  • Well, I guess that means having a theoretically sound basis versus just a numerical approximation. – SurfaceIntegral Aug 13 '20 at 19:16
  • What do you mean by "theoretically sound basis?" – Sandejo Aug 13 '20 at 19:19
  • A simple example would be the fundamental theorem of calculus which, when you work with it in the context of pure functions like f(x)=x^2 should give the correct answer for areas under curves, etc, but once you bring calculus into the real world, even though the theory behind it has been made rigorous and very sound, real-world results might not match up 100% with the pure results obtained from the theory. This would be different from say, if we just used a 2nd order Taylor approximation of the function, or something like that. Same question but for vectors... – SurfaceIntegral Aug 13 '20 at 19:22
  • In what real-world case does the fundamental theorem not hold? – NDewolf Aug 13 '20 at 19:26
  • Oh, it certainly holds, it's just that the empirical data might not (almost certainly won't) match up perfectly with the theoretical underpinnings of whatever the real-world phenomenon is. – SurfaceIntegral Aug 13 '20 at 19:28
  • Does this answer your question? Why don't we prove that functions used in physics are continuous and differentiable? This question focuses primarily on continuity and differentiability of functions, but the arguments used in the answers could also apply to other mathematical objects, like vectors. – Sandejo Aug 13 '20 at 19:29
  • Well, not really, unfortunately, since I would say my question is more fundamental. Once you already have a vector-valued function, for example, then you can go ahead and test its continuity or differentiability, etc, but my question is more: how do we even know (that is, guarantee) in the first place that a vector-valued function (or really vectors in general) are the proper description of reality? One person has said they break down at the quantum level (which I have to accept for now as a leap of faith because I haven't studied that stuff in depth yet). – SurfaceIntegral Aug 13 '20 at 19:33
  • My answer to that question addresses your question of how we know a certain mathematical object describes reality. – Sandejo Aug 13 '20 at 19:39
  • From what I could gather then, it sounds like physicists will simply use whatever seems to agree with the data best, whether it be in the context of vectors to describe macroscopic motion, or, for instance, running some sort of a regression to fit the data, etc, but it's impossible to be certain what the true nature of the phenomenon is and we just have to make do with very good approximations that suffice for all practical purposes. – SurfaceIntegral Aug 13 '20 at 19:44

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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.

d_b
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  • Thanks for the very thorough response! This answers my question. – SurfaceIntegral Aug 13 '20 at 19:55
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    +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. – mmesser314 Aug 13 '20 at 23:05
  • Good point. Indeed, it is possible — perhaps even likely — that those tools have not yet been developed. – d_b Aug 13 '20 at 23:10