0

I have some questions regarding the Lagrangian Formalism, specifically using it for classical mechanics.


Questions regarding the derivation of the Euler-Lagrange Equation:

Consider some functional $$I[f]=\int_{t_1}^{t_2}L(t,f,\dot{f})dt$$ and boundary conditions $f(t_1)=x_1$ and $f(t_2)=x_2$. Let $x$ be a stationary function of $I$. Then every function $f(t)$ can be written as $x(t)+\epsilon\eta(t)$ where $\eta(t_1)=\eta(t_2)=0$ for some arbitrary non-zero $\eta(t)$ and $\epsilon$.

Does every functional have a stationary point that is a global minimum or a maximum? This statement isn't true from $f: \mathbb{R} \to \mathbb{R}$ (consider $x^3$), so why should it be true for functionals? If it is false, does the Euler-Lagrange equation only hold for some functionals?

Also, are there constraints on and $\eta$ besides the boundary conditions? One seemingly obvious one is that $I[f]$ must be defined/finite, but are there other constraints? I'm wondering this because it seems like every function (that satisfies the boundary conditions) has infinitely many representations this way and because of the use of the Leibniz integral rule later on.

With this representation of $I$, the functional becomes only a function of $\epsilon$, and the derivative evaluated at this point is $$I'(\epsilon)=\frac{d}{d\epsilon}\int_{t_1}^{t_2}L(t,f,\dot{f})dt=\int_{t_1}^{t_2}\frac{\partial L}{\partial f} \frac{\partial f}{\partial \epsilon}+\frac{\partial L}{\partial \dot{f}}\frac{\partial \dot{f}}{\partial \epsilon}dt=\int_{t_1}^{t_2}\frac{\partial L}{\partial f} \eta+\frac{\partial L}{\partial \dot{f}}\dot{\eta}dt$$

Using integration by parts, $$\int_{t_1}^{t_2} \left(\frac{\partial L}{\partial f} - \frac{d}{dt}\frac{\partial L}{\partial \dot{f}}\right)\eta dt$$.

Since $I'(0)$ must be $0$ assuming $f$ is a stationary function of I, $$\int_{t_1}^{t_2} \left(\frac{\partial L}{\partial x} - \frac{d}{dt}\frac{\partial L}{\partial \dot{x}}\right)\eta dt$$ is 0 for all $\eta$, meaning $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}=\frac{\partial L}{\partial x}$$ by the fundamental lemma of calculus of variations. Which is the Euler Lagrange equation.

Final question for this section, is it ok to write the expression as $$\dot{\frac{\partial L}{\partial \dot{x}}}=\frac{\partial L}{\partial x}$$ Since the dots just represent total time derivatives? This way of writing seems more "elegant", but I've never seen it used before.


Connection to Newton's Laws.

Using $L=T-U=\frac{1}{2}m\sum{\dot{x_i}^2}-U$ the Euler-Lagrange equation simply becomes $$-\frac{\partial U}{\partial x_i}=\dot{p_i}$$ In each of the $x_i$ directions, corresponding to Newton's second law.

Firstly, I've only seen proofs that if a function is a stationary function of the functional it will satisfy the Euler Lagrange Equation, but not the other way around. So how can we ensure that a solution to the Euler Lagrange equation for $L=T-U$ is a minimum, or even a a stationary function for the functional? All we know is that if there is a function that is a minimum it will satisfy the equation.

Also, why $L=T-U$? I can't think of an intuitive idea behind why objects would follow a trajectory that minimizes this. The only intuition I get from this is that generally $U$ is large and $T$ is small to minimize $L$, but why is that the case? (Why not $U-T$ or some other combination that has the right dimensions?)

Finally, how does the proof work with non-conservative forces? Using $L=T-U$ and the Euler-Lagrange equation is only equivalent to Newton's law if $F=-\nabla U$, but non-conservative forces don't have a potential field.


Any insight into these questions would b greatly appreciated.

Qmechanic
  • 201,751
QED
  • 313

1 Answers1

1

That's a lot of questions, you might want to break them up, or at least number them.

First, clearly not every functional has a minimum or maximum. Consider, for example, $L = f$. Then, your action is just the average value of $f$. But that doesn't have a finite minimum or maximum.

I don't think there are major restrictions on $\eta(t)$ other than the ones you mention. It has to be smooth enough to take all of the derivatives you need to, it has to be a scalar, it has to be real, etc.

Regarding your dot notation, I think it is technically correct but you shouldn't write it this way because it's non-standard and it's kind of hard to tell what the dot (meaning time-differentiation) is being applied to.

As to why $L = T-U$, this is so that the principle of least action recapitulates Newton's equations and is thus consistent with empirical observations. It's a law, not a theorem, so you don't have to prove it from something more fundamental. This sometimes bothers people because it feels less intuitive than Newtonian forces, but on a philosophical level it's a somewhat more modest assumption that applies more broadly, which makes it a desirable reformulation.

Non-conservative forces are their own big thing. See, e.g.,

Modeling external forces in Lagrangian dynamics

In general, time-dependent forces are not too bad, but dissipative (frictional) forces require rather awkward formulations, like adding an exponential time-dependent rescaling of the Lagrangian:

Euler-Lagrange equations and friction forces

David
  • 703