64

I had this same problem, too. The trick with it is realizing that there's an important difference between Taylor series and Taylor approximations or polynomials, whose behavior is described by Taylor's theorem. Yes, very often I suspect a common mistake is that you first see Taylor polynomials and theorem, and then you get Taylor series and that becomes the ...


49

Something I posted on reddit answers this question quite well, I think: "Rational" and "irrational" are properties of numbers. Quantities with units aren't numbers, so they're neither rational nor irrational. A quantity with units is the product of a number and something else (the unit) that isn't a number. By choosing the unit you use to express a ...


38

The key reason is that we want to understand the behavior of the system in the neighborhood of the state rather than at the state itself. Take the equation of motion for a simple pendulum, for example: $$\ddot{\theta} = -\frac{g}{\ell}\sin(\theta)$$ If we take the limit where $\theta \rightarrow 0$, we find $\ddot{\theta}= 0$, and we would conclude that ...


37

The complex exponentials are eigenfunctions of the derivative and integral operators. So if you're analyzing linear differential equations, and using Fourier series, then you can consider each term on its own. If you use Taylor series you have to consider interactions between one term and other terms in the series. (This is also why we often write our ...


34

There are some problems with using quaternions to describe spacetime. Quaternions have two important properties: (1) they form a four-dimensional vector space; (2) you can multiply quaternions together.[1] The first property is obviously very suggestive, but it's no different from the usual four-vectors that we already use in special relativity. To ...


32

The first thing that must be said is that the question is not really specific enough: Applications to what exactly are you looking for? To me, a book on algebraic geometry and mirror symmetry, and how it relates to mirror symmetry as physicists know it, is very relevant and interesting. However, I have the feeling that this is not exactly what you're looking ...


32

No, physics is not rigorous in the sense of mathematics. There are standards of rigor for experiments, but that is a different kind of thing entirely. That is not to say that physicists just wave their hands in their arguments [only sometimes ;) ], but rather that it does not come even close to a formal axiomatized foundation like in mathematics. Here's an ...


32

No, there is not nor can there be a similar statement in physics. That is because we can know all there is to know about the mathematical systems we construct; after all, we have set them up ourselves (but then Gödel's incompleteness theorem tells us that there can be features of certain systems that remain unknowable; sorry for butchering what the theorem ...


30

I depends on the book you've chosen to read. But usually some basics in Calculus, Linear Algebra, Differential equations and Probability theory is enough. For example, if you start with Griffiths' Introduction to Quantum Mechanics, the author kindly provides you with the review of Linear Algebra in the Appendix as well as with some basic tips on probability ...


27

Arthur Suvorov gives a nice comment, I am just going to give a list of a few specific physical problems I can think of from the top of my head. Yang Mills existence and mass gap (Millenium Prize) and generally the problematic of rigorous definitions and constructions of quantum field theories Navier Stokes equations and smoothness (also Millenium) - it's ...


27

As ACuriousMind has already noted, you can geometrically interpret the length of the cross product of two vectors as the area of the parallelogram (or as twice the area of the triangle) spanned by them, and (the absolute values of) its components as the areas of the projections of that parallelogram onto the coordinate planes. As for the dot product of two ...


27

This is definitely not a dumb question. If we work in a (linear) Hilbert space, then our inner product $\langle \cdot,\cdot \rangle$ induces the usual natural flat metric (given by $d(\psi,\phi) = || \psi - \phi ||$). However, often we take the viewpoint that our states are elements of projective Hilbert space $\mathbb CP^n$. Then it is more natural to ...


24

People tend to take Gödel's theorem and bend it, stretch it, misstate it, misapply it, and generally do things to it that, if you did them to a cockroach in Texas, would get you arrested for animal cruelty. But there is a book, Franzén (2005), that should be enough to inoculate any responsible adult against such naughty behavior. Some points made by Franzén: ...


24

The Stone-Weierstrass theorem says that any continuous function on a compact interval is arbitrarily well approximated by polynomials. Thus, as long as we're only interested in explaining experimental results (and not in the exact solutions of theoretical models), series expansions are plenty good enough. That is, for whatever we'd like to describe, there is ...


23

Here's a mathematician's stab at an answer. The tl;dr version: "symmetric" and "self-adjoint" are the same thing for bounded operators, whilst they are the same thing for unbounded operators only insofar as the Aharonov--Bohm effect doesn't exist! A densely defined operator $A$ on your Hilbert space $H$ is symmetric if for any vectors $\xi$ and $\eta$ in ...


21

No, nothing in physics depends on the validity of the axiom of choice because physics deals with the explanation of observable phenomena. Infinite collections of sets – and they're the issue of the axiom of choice – are obviously not observable (we only observe a finite number of objects), so experimental physics may say nothing about the validity of the ...


20

There's no escaping Lie theory if you want to understand what is going on mathematically. I'll try to provide some intuitive pictures for what is going on in the footnotes, though I'm not sure if it will be what you are looking for. On any (finite-dimensional, for simplicity) vector space, the group of unitary operators is the Lie group $\mathrm{U}(N)$, ...


20

Sure it does. One-dimensional creatures can take an object of mass $m$, attach it to a spring $k$ and they will find out the period of oscillations if this system is proportional to $\sqrt{m/k}$. The coefficient would be some strange number approximately equal to $6.28$, but not an integer or a rational (actually it's $2*\pi$). Then one day some advanced ...


19

There are many attempts for a physical proof of the Riemann hypothesis. The major work in this direction was summarized in a recent review by: Schumayer and Hutchinson. One of these attempts was proposed by: Berry and Keating. Their suggestion is within the framework of the Hilbert–Pólya conjecture, according to which, the Hilbert–Pólya Hamiltonian, whose ...


18

To give an example where convergence of Cauchy sequences is important: time-evolution is typically calculated as $$ |\psi(t)\rangle = e^{i\hbar^{-1} \hat{H}\cdot t}|\psi_0\rangle $$ now, the exponential of an operator is defined1 by $$ e^{\hat{A}} = \sum_{i=0}^\infty\frac{\hat{A}^i}{i!} $$ where the sum in turn is defined by $$ \Bigl(\sum_{i=0}^b\frac{\...


18

No. There is no proof of this. What we do know is that the models that result from assuming it allow us to correctly predict the outcome of all experiments to date. Therefore, it is accepted for the same reason that most assumptions in physics are accepted: it works.


18

Great question. One reason that complex exponential expansions (which end up turning on sines and cosines for real-valued problems) are more natural that Taylor series expansions is that they don't require picking a special point to expand around. In many situations the differential equation is translationally invariant, and there's no natural point to ...


17

The set of irrational numbers densely fills the number line. Even assuming that quantum mechanics doesn't disable the preimse of your question, the probability that you will randomly pick an irrational number out of a hat of all numbers is roughly $1 - \frac{1}{\infty} \approx 1$. So the question should be "is it possible to have an object with rational ...


17

This is notation from Distribution Theory in Functional Analysis. The theory of distributions is meant to make things like the Dirac Delta rigorous. In this context, just to give you one overview, a distribution is a functional on the space of test functions. We define the space of test functions over $\mathbb{R}$ as $\mathcal{D}(\mathbb{R})$ being the ...


16

You can divide vectors with clifford ("geometric") algebra. The geometric product of vectors is associative: $$abc = (ab)c = a(bc)$$ And the geometric product of a vector with itself is a scalar. $$aa = |a|^2$$ These are all the properties required to define a unique product of vectors. All other properties can be derived. I'll sum them up, however: ...


16

If you want to learn topology wholesale, I would recommend Munkres' book, "Topology", which goes quite far in terms of introductory material. However, in terms of what might be useful for physics I would recommend either: Nakahara's "Geometry, Topology and Physics" Naber's "Topology, Geometry and Gauge Fields: Foundations" Personally, I haven't read much ...


16

One of the big reasons not discussed above is Fourier theory -- any function $f(x)$ can be expressed in the form $f(x) = \int dk\, A(k)e^{ikx}$, which basically means that any function can be decomposed into an infinite sum of sines and cosines. Since this is the case and dealing with sine and cosine is mathematically simpler than the general case of ...


16

The length of the cross product of two vectors is the area of the parallelogram spanned by them, so the square-meters are the correct unit as well as geometrically meaningful - it's really an area. The $x$-component is the area of the projection of the parallelogram onto the $y$-$z$-plane, the $y$-component the area of the projection onto the $z$-$x$-plane ...


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