I More rigorous Euler-Lagrange derivation

[romsofia]

TL;DR Summary

What really is the “variation”?

Sorry if there are other threads on this, but after a discussion with a friend on this (im in the mountains, so no books, and my googlefu isn't helping), I realize that my understanding of the variational principles arent exactly... great! So, maybe some one can help.

Start with a functional defined by: $S= \int L(q(t), \dot{q}(t), t) dt$ where $\dot{q} = \frac{dq}{dt}$ we “vary” the functional, in the following manner: $\delta S = \int (\frac{\partial L}{\partial q} \delta q + \frac{\partial L}{\partial \dot{q}} \delta \dot{q})dt$

And so, from there i know how to get the EL to pop out from integration and some arguments about boundary conditions. The issue he had, and where I am also lacking is, what REALLY is that $\delta$?

I've always treated it similar to a derivative, and essentially all we are doing is taking a chain rule when doing $\delta S$, but then i can't really justify the $\delta q$ since the chain rule in that spot would be $\frac{dq}{dt}$.

Basic question that I should know(it is just calculus of variations), but better to finally learn it properly, than go off by handwaving because my muscle memory can write it down properly!

 

[weirdoguy]

My thesis supervisor has a nice paper that summarizes geometrical (fibre bundle) viewpoint of mechanics: https://arxiv.org/abs/1306.2744
Starts at page 4. She had another paper that also had nice summary, but I can't find it on arXiv. Weird.
Also papers of Paweł Urbański: https://arxiv.org/search/math?searchtype=author&query=Urbanski,+P
may be a good read.

 

[vanhees71]

$\delta q$ is simply a deviation of the path in configuration space from the trajectory of the particle, which is defined as the stationary path of the action.

In Hamilton's principle the variation is over all paths with fixed endpoints $q(t_1)$ and $q(t_2)$, and time is not varied. From this you get
$$\delta \dot{q}=\mathrm{d}_t \delta q.$$
Then you have
\begin{equation*}
\begin{split}
\delta S &= \delta \int_{t_1}^{t_2} \mathrm{d} t L(q,\dot{q},t)\\
&=\int_{t_1}^{t_2} \mathrm{d} t \left (\delta q \cdot \frac{\partial L}{\partial q} + \delta \dot{q} \cdot \frac{\partial L}{\partial \dot{q}} \right) \\
&=\int_{t_1}^{t_2} \mathrm{d} t \delta q \cdot \left ( \cdot \frac{\partial L}{\partial q} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}} \right) + \delta q(t_2) \cdot \left . \frac{\partial L}{\partial \dot{q}}\right|_{t=t_2} - \delta q(t_1) \cdot \left . \frac{\partial L}{\partial \dot{q}} \right |_{t=t_1}.
\end{split}
\end{equation*}
Since by definition $\delta q(t_1)=\delta q(t_2)=0$ the boundary terms vanish and you finally get
$$\delta S=\int_{t_1}^{t_2} \mathrm{d} t \delta q \cdot \left ( \frac{\partial L}{\partial q} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}} \right).$$
Since this must hold for all functions $\delta q(t)$ the term in the parentheses must vanish, which leads to the Euler-Lagrange Equations,
$$\frac{\partial L}{\partial q} = \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}}.$$
For a mathematically rigorous treatment of analytical mechanics, see

V. I. Arnold, Mathematical Methods of Classical Mechanics, Springer (1989)

 

[andresB]

And so, from there i know how to get the EL to pop out from integration and some arguments about boundary conditions. The issue he had, and where I am also lacking is, what REALLY is that $\delta$?

So, as you pointed out, the action is a functional of paths in the $(q,\dot{q})$ space, the tangent bundle of configuration space.

A path is a function from the real numbers to this tangent bundle $\mathbb{R}\ni t\rightarrow(q(t),\dot{q}(t))$.
We can define a slightly different path by $q^{'}(t)=q(t)+\varepsilon\eta(t)$ so that $\dot{q}^{'}(t)=\dot{q}(t)+\varepsilon\dot{\eta}(t)$. Notice that $\delta q=\varepsilon\eta(t)$ and $\delta\dot{q}=\varepsilon\dot{\eta}(t)$ represents how far away is the new path from the old one i.e., how much we have deformed the original path.

We can now ask, what is the value of the action for this new path? We can find it by a Taylor expansion of the Lagrangian
$$
S' =\int_{t_{1}}^{t_{2}}L(q^{'}(t),\dot{q}^{'}(t))dt=\int_{t_{1}}^{t_{2}}L(q(t)+\varepsilon\eta(t),\dot{q}(t)+\varepsilon\dot{\eta}(t))dt
\approx\int_{t_{1}}^{t_{2}}dt\left[L(q(t),\dot{q}(t))+\varepsilon\eta(t)\frac{\partial L}{\partial q}+\varepsilon\dot{\eta}(t))\frac{\partial L}{\partial\dot{q}}\right]dt$$

So, we can conclude that the first order variation of the action is
$$\delta S=\int_{t_{1}}^{t_{2}}\left(\varepsilon\eta(t)\frac{\partial L}{\partial q}+\varepsilon\dot{\eta}(t))\frac{\partial L}{\partial\dot{q}}\right)dt=\int_{t_{1}}^{t_{2}}\left(\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial\dot{q}}\delta\dot{q}\right)dt$$

 

[romsofia]

Thanks for your help, I'll read them in more detail, and see if I have any questions! The discussion of configuration spaces is bringing back memories of non-holonomic constraints for some reason!

 

[vanhees71]

If it comes to non-holonomic constraints, don't use the distorted 3rd edition of Goldstein's textbook. The 2nd edition is fine though.