Intuitive Lagrangian Mechanics

[lsusr]

Link post

Lagrangian mechanics is an alternative formulation of Newtonian mechanics. Newtonian mechanics solves movements using successive approximations. Lagrangian mechanics figures out the whole path all at once. Lagrangian mechanics is useful for solving problems subject to certain constraints. It is also a prerequisite to understanding quantum field theory and the path integral formulation of quantum mechanics.

In the classical formulation of Lagrangian mechanics, the Lagrangian $L$ is the difference between a particle’s kinetic energy $T$ and potential energy $V$.

$$L=T-V$$

The action $S$ of a particle is the path integral of its Lagrangian.

$$S={\int }_{t_i}^{t_f}Ldt$$

The path of a particle extremizes $S$.

$$\delta S=0$$

We therefore derive the Euler–Lagrange equations.

$$\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{q}}_j}\right)=\frac{\partial L}{\partial q_j}$$

Does this make sense to you? Because subtracting potential energy from kinetic energy does not make conceptual sense to me. Besides, what, conceptually, is “action” and why should it be extremized? Lagrangian mechanics make better sense in special relativity.

Special Relativity

To make things more intuitive, let’s look at Lagrangian mechanics in the context of special relativity, from which classical Lagrangian mechanics is an approximation. I will set $c=1$ because space and time ought to have the same units. I will set $m_0=1$ for convenience because the rest mass of our particle does not change.

The most important number in special relativity is the Lorentz factor $\gamma$, the instantaneous ratio of coordinate time $t$ to proper time $\tau$.

$$\gamma =\frac{dt}{d\tau }=\frac{1}{\sqrt{1-v^2}}$$

We can^[1] take the action $S$ to be the length of the particle’s world line between proper times ${\tau }_1$ and ${\tau }_2$. (Note that $\gamma$ is a function of $t$.)

$$S={\int }_{{\tau }_1}^{{\tau }_2}d\tau ={\int }_{t_1}^{t_2}\frac{1}{\gamma }dt$$

Locally maximizing $S$ equals locally maximizing the proper time experienced by a particle. The Lagrangian $L$ is the expression inside of the coordinate time integral.

$$L={\gamma }^{-1}=\sqrt{1-v^2}=\alpha$$

If you think of all matter particles as moving through spacetime at a speed of 1 along the hypotenuse of a right triangle with one spatial leg and one temporal leg then the particle’s spatial velocity is $v=\dot{q}$ and its temporal velocity is $\alpha$.

$${\alpha }^2+v^2=1$$

$$S={\int }_{t_1}^{t_2}\alpha dt$$

The Laplacian $L$ is simply $\alpha$, the particle’s temporal velocity. The action $S$ is the integral of the temporal velocity. Therefore, extremizing (maximizing^[2]) the action $S$ equals maximizing proper time.

Hamilton’s principle, that the evolution of a system is a stationary point of the action functional $S$, naturally follows.

$$\delta S=0$$

$$\frac{d}{dt}\frac{\partial \alpha }{\partial v}=\frac{\partial \alpha }{\partial q}$$

Why add $V$ to $\alpha$?

Under general relativity, time slows down for a particle in a gravity well. In other words, a gravity well decreases the particle’s temporal velocity. (This generalizes to the other fundamental forces.) Flipping this around, increases to a particle’s potential increase its temporal velocity. It therefore makes intuitive sense to add potential $V$ to special relativistic temporal velocity $\alpha$.

$$L=\alpha +V$$

Thus we arrive at the Euler–Lagrange equations.

$$\frac{d}{dt}\frac{\partial L}{\partial v}=\frac{\partial L}{\partial q}$$

Classical Lagrangian mechanics follows as a second order approximation.

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1. ↩︎

   The action $S$ also includes a factor of $-m_0{c}^2$. We have already declared $m_0=c=1$. I have removed the factor of $-1$ too because mechanics in general and $\delta S=0$ in particular are symmetric with respect to time parity. It does not matter to the Euler–Lagrange equations whether $S$ is flipped by a minus sign.

2. ↩︎

   Conventionally, $S$ is minimized. Since I removed the minus sign from $S$ we maximize it instead.

[Shmi]

Your definition of “intuitive” is unlikely to match that of most participants here.

[Adele Lopez]

FWIW this did make Lagrangian mechanics feel more intuitive for me. I wish it showed the derivation of the classical version as a second order approximation though.

[simon]

FWIW, if I were asked before reading this why you substract the potential energy from the kinetic energy, I wouldn’t have had a quick answer—I think “it’s minimizing the overall time dilation from gravity and speed” is a really neat way to think about it.

As to why time dilation would be relevant, if you’ve read QED by Richard Feynman he has a visualization where you think of each (version of a) particle of having a clock with a hand that goes around and around, and you add up the hands of all the clocks for all the particles that took all the different paths.

In the end only the (versions of the) particle that took the paths very close to where the time taken is extremized add up to the final result, everything else cancels because tiny differences in path lead to opposite directions of the clock hand.

[Shmi]

I’ve read the post a couple of times over, and I still don’t have an intuitive understanding of why one would subtract potential energy from kinetic, despite having done graduate work in general relativity. Yes, extremizing action comes from the stationary phase approximation of the path integral, and yes, following a path with “low potential energy” makes you arrive to the destination younger, just like moving faster does (yet arriving at the same instant as those moving slower), but first, it’s not obvious why the former is so, and second, why it would matter in non-gravitational physics, especially in classical mechanics. I would like to see an intuitive argument where the difference between kinetic and potential energies makes sense.

[lsusr]

Well, there’s always the classical derivation. :)

This article comes from trying to understand A Simple Introduction to Particle Physics, “notes…intended for a student who has completed the standard undergraduate physics and mathematics courses”.

[DaemonicSigil]

Nice. Some people in the comments were asking about the actual Taylor expansion for kinetic energy. So here it is (assuming $v<<c=1$): $$S={\int }_{{\tau }_1}^{{\tau }_2}md\tau ={\int }_{t_1}^{t_2}m\sqrt{1-v^2}dt\approx {\int }_{t_1}^{t_2}m(1-\frac{v^2}{2}+\dots )dt$$ (Since the derivative of $\sqrt{x}$ is $\frac{1}{2\sqrt{x}}$, we have $\sqrt{1+x}\approx (1+x/2)$ for small $x$.) So, up to an additive constant, we have $L=-\frac{m{v}^2}{2}$. Also, if we want to account for non-gravitational potential energy as well, we can note that in special relativistic units, $E=m$ in free space, and $E=m+V$ in a potential. ($E$ being the energy of the particle measured in that particle’s frame.) So, assuming $V<<m$: $$L=E\tau =(m+V)\sqrt{1-v^2}\approx (m+V)(1-\frac{v^2}{2})\approx m+V-\frac{m{v}^2}{2}$$

[habryka]

Note that you can use actual LaTeX in your comments by pressing CMD/​CTRL+4 in the editor and typing LaTeX in there (or using the toolbar option for inserting math).

[DaemonicSigil]

Ah thanks, I’ve got it now. My browser seems to not like CMD-4, but putting dollar signs in markdown worked.

[AlexMennen]

I’m confused about the motivation for $L=\alpha +V$ in terms of time dilation in general relativity. I was under the impression that general relativity doesn’t even have a notion of gravitational potential, so I’m not sure what this would mean. And in Newtonian physics, potential energy is only defined up to an added constant. For $\alpha +V$ to represent any sort of ratio (including proper time/​coordinate time), V would have to be well-defined, not just up to an arbitrary added constant.

I also had trouble figuring out the relationship between the Euler-Lagrange equation and extremizing S. The Euler-Lagrange equation looks to me like just a kind of funny way of stating Newton’s second law of motion, and I don’t see why it should be equivalent to extremizing action. Perhaps this would be obvious if I knew some calculus of variations?

[Charlie Steiner]

Well, it makes sense for the effective field theory form of GR, for light at least.

The key to remembering how to derive the Euler-Lagrange equation (for me) is to remember that the variation in L vanishes at the boundary. This is what’s going to let you do an integration by parts and throw away the constant term. Actually, once you have an intuitive grasp of what’s going on, it’s kind of fun to derive generalized EL equations for Lagrangians with more complicated stuff in them.