# Trajectories piecewise smooth?

In my studies of calculus and real analysis I have found the proofs of several theorems, commonly used in physics, such as those concerning the conservativity of fields, for example like

If $\mathbf{F}:A\subset\mathbb{R}^3\to\mathbb{R}^3$ is a vector field of class $C^1(A)$, the following propositions are equivalent:

1. $\mathbf{F}$ possesses a potential function;
2. give two paths $\gamma_1,\gamma_2\subset A$ with the same ordered endpoints, the equality $\int_{\gamma_1}\mathbf{F}\cdot d\mathbf{r}=\int_{\gamma_2}\mathbf{F}\cdot d\mathbf{r}$ holds;
3. give a closed path $\gamma\subset A$, the circuitation of the field along it is $\oint_{\gamma}\mathbf{F}\cdot d\mathbf{r}=0$.

under the assumption that the paths that are piecewise smooth curves according to the following definition: they admit a continuous parametrisation $\mathbf{r}:[a,b]\to\mathbb{R}^3$, $\mathbf{r}\in C[a,b]$, such taht $\mathbf{r}$ is of class $C^1$ and $\|\mathbf{r}'(t)\|\ne 0$ except at most a finite number of points $t_i$ where $\lim_{t\to t_i^+}\mathbf{r}'(t)$ and $\lim_{t\to t_i^-}\mathbf{r}'(t)$ exist and are finite. All the texts of mathematics that I have seen do not even define $\int_\gamma \mathbf{F}\cdot d\mathbf{r}$, which is $\int_a^b \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)dt$, if $\gamma$ is not piecewise smooth.

In elementary mechanics textbooks I have never found an explicit wording of the assumptions under which such theorems hold, although they are ubiquitously used.

Is it usually implicitly assumed, in physics, that trajectories and paths are piecewise smooth curves according to the definition that I have quoted? I $\infty$-ly thank you for any answer!

"The answer is that since we are proud physicists and not nitpicking mathematicians we will just wing it when the need arises"
This quote is taken from A. Zee's Quantum Field Theory in a Nutshell, and it summarizes the attitude of physicists to mathematics. (At least in an undergraduate level)

Since we are physicists, most of our mathematics isn't rigorous. As such, we tend to cut corners, until it blows up in our faces. And when it does, we go over it a bit more carefully, or wing it.

You can see this, for example, in the Dirac delta "function". Ask any mathematician, and they will tell you (Justifiably) that this is not a function, and physicists don't use it properly. It can be seem as a functional, or a distribution. But it works. (Examples of lack of rigor blowing up in our faces include this)

Mostly, physicists ignore rigor because results still work, and because rigor would take a lot more time. If you want to properly do Analytical Mechanics (A standard undergraduate 2nd year course), you would have to finish an undergraduate degree in mathematics, as the subjects to formalize this include Lie Algebra, and manifolds, and symplectic geometry. (See this for another question on rigor, which presents a bit why it is so problematic in physics)

But, if you do want rigor in physics, there's the area of mathematical physics, which formalizes these things, and there is plenty of literature on the subject. (See for example this Math SE post on the subject)

This got a bit longer than I originally intended, so here is the bottom line:
A physicist would not care about it, a mathematician would probably formalize it and say all physical curves are piecewise smooth. (They're mostly composed of straight lines and circles actually)

• Very nice post! I can only add one other component to this: usually the physical limitations of a model or theory like classical mechanics appear way before we reach the problems with mathematical rigor. Some mathematicians/mathematical physicists are now (righteously so!) analyzing the details of integrability of Hamiltonian systems (and I love the results). Is that of real importance to physicists who are now on the search for dark matter, though? Hardly. – CuriousOne Jun 27 '15 at 17:57
• ...quite a problem "when it blows in the faces"... ;-) I thank you very much!!! – Self-teaching worker Jun 27 '15 at 19:10