Jorriss said:
You need to look at better sources, Lagrangian mechanics has a rigorous framework.
micromass said:
Blatant to you maybe. I certainly know that it's rigorous enough, and I don't think I'm alone in this.
I'll show you how I see rigorous Lagrangian mechanics. Let's see what happens.
We fix a dimension N=1,2,3,\ldots. We fix some function L:\mathbb{R}^{1+2N}\to\mathbb{R}, and assume that all its second order partial derivatives exist, and are continuous. We agree on a custom, that the parameter of L is often denoted as (t,x,\dot{x}), where t\in\mathbb{R}, x\in\mathbb{R}^N and \dot{x}\in\mathbb{R}^N. We assume that spacetime points (t_A,x_A) and (t_B,x_B)\in\mathbb{R}^{1+N} are fixed. We define sets \mathcal{X} and \mathcal{X}_0 in the following way:
<br />
\mathcal{X}= \big\{x\in C^2([t_A,t_B],\mathbb{R}^N)\;\big|\; x(t_A)=x_A,\; x(t_B)=x_B\big\}<br />
<br />
\mathcal{X}_0 = \big\{x\in C^2([t_A,t_B],\mathbb{R}^N)\;\big|\; x(t_A)=0,\; x(t_B)=0\big\}<br />
Now \mathcal{X}_0 is a vector space, while in most cases \mathcal{X} will not be, at least not with the ordinary addition and scaling. By using the ordinary Euclidean norm in \mathbb{R}^N we make \mathcal{X} into a metric space by setting
<br />
d(x,y) = \sqrt{\int\limits_{t_A}^{t_B} \|x(t)-y(t)\|^2dt} + \underset{t_A\leq t\leq t_B}{\textrm{max}} \|\dot{x}(t) - \dot{y}(t)\|<br />
for all x,y\in\mathcal{X}. We make \mathcal{X}_0 into a norm space by setting
<br />
\|x\| = \sqrt{\int\limits_{t_A}^{t_B} \|x(t)\|^2dt} + \underset{t_A\leq t\leq t_B}{\textrm{max}} \|\dot{x}(t)\|<br />
for all x\in\mathcal{X}_0. Now these metric and norm spaces are related in an obvious way. For example d(x,x+h)=\|h\| holds for all x\in\mathcal{X} and h\in\mathcal{X}_0. Then we define a mapping S:\mathcal{X}\to\mathbb{R} by setting
<br />
S(x) = \int\limits_{t_A}^{t_B} L\big(t,x(t),\dot{x}(t)\big)dt<br />
Now S is a continuous mapping. (A non-trivial claim! The continuity is considered with respect to the defined metric, of course.) For all x\in\mathcal{X} there exists a continuous linear mapping DS(x):\mathcal{X}_0\to\mathbb{R} and a real coefficient M(x)\in\mathbb{R} such that
<br />
\big|S(x+\eta) - S(x) - DS(x)\eta\big|\leq M(x)\|\eta\|^2<br />
holds for all \eta\in\mathcal{X}_0 with the additional condition \|\eta\|\leq 1. With a fixed x this linear mapping is unique. (This means that the mentioned inequality does not hold with any other linear mapping even if the coefficient M(x) was replaced with a new coefficient too.) If x is a local minimum or a local maximum of S, then DS(x)=0. (A non-trivial claim! The locality is considered with respect to the mentioned metric.) Also the relation
<br />
DS(x)=0\quad\Longleftrightarrow\quad D_t\nabla_{\dot{x}} L\big(t,x(t),\dot{x}(t)\big) = \nabla_x L\big(t,x(t),\dot{x}(t)\big)\quad\quad\forall\; t_A\leq t\leq t_B<br />
holds.
How do you like this theorem? Not very elegant, but I like this. I could mention this as one of my answers to the opening post's question.
I have few questions to those who claim that mainstream Lagrangian mechanics is rigorous. Firstly, have you seen anything like this anywhere ever? I know that this kind of mathematics can be found from some mathematics books, but they are not books on physics. I have never seen a book on physics or mechanics that would give results like this. Have you? Another thing is that do you think that I'm making things too complicated here? Do you think that Lagrangian mechanics can be made rigorous more easily, without these kind of function spaces? I would be interested to hear more.
