35.1 Lagrangian mechanics

35.1 Lagrangian mechanics

In the Lagrangian formalism, instead of a phase space ${\mathbf{R}}^{2d}$ of positions $q_j$ and momenta $p_j$ , one considers just the position (or configuration) space ${\mathbf{R}}^d$ . Instead of a Hamiltonian function $h(\mathbf{q},\mathbf{p})$ , one has:

Definition (Lagrangian)

The Lagrangian L for a classical mechanical system with configuration space ${\mathbf{R}}^d$ is a function

$$L:(\mathbf{q},\mathbf{v})\in {\mathbf{R}}^d\times {\mathbf{R}}^d\to L(\mathbf{q},\mathbf{v})\in \mathbf{R}$$

Given diferentiable paths in the configuration space defined by functions

$$\gamma :t\in [t_1,t_2]\to {\mathbf{R}}^d$$

which we will write in terms of their position and velocity vectors as

$$\gamma (t)=(\mathbf{q}(t),\dot{\mathbf{q}}(t))$$

one can define a functional on the space of such paths:

Definition (Action)

The action $S$ for a path $\gamma$ is

$$S[\gamma ]={\int }_{t_1}^{t_2}L(\mathbf{q}(t),\dot{\mathbf{q}}(t))dt$$

The fundamental principle of classical mechanics in the Lagrangian formalism is that classical trajectories are given by critical points of the action functional. These may correspond to minima of the action (so this is sometimes called the “principle of least action”), but one gets classical trajectories also for critical points that are not minima of the action. One can define the appropriate notion of critical point as follows:

Definition (Critical point for S)

A path $\gamma$ is a critical point of the functional $S[\gamma ]if$

$$\delta S(\gamma )\equiv \frac{d}{ds}S({\gamma }_s{)}_{|s=0}=0$$

where

$${\gamma }_s:[t_1,t_2]\to {\mathbf{R}}^d$$

is a smooth family of paths parametrized by an interval $s\in (-ϵ,ϵ)$ , with ${\gamma }_0=\gamma$

We’ll now ignore analytical details and adopt the physicist’s interpretation of $\delta$S as the first-order change in $S$ due to an infinitesimal change $\delta \gamma =$ $(\delta \mathbf{q}(t),\delta \dot{\mathbf{q}}(t))$ in the path.

When $(\mathbf{q}(t),\dot{\mathbf{q}}(t))$ satisfy a certain diferential equation, the path $\gamma$ will be a critical point and thus a classical trajectory:

Theorem (Euler-Lagrange equations)

One has

$$\delta S[\gamma ]=0$$

for all variations of $\gamma$ with endpoints $\gamma (t_1)$ and $\gamma (t_2)$ fixed $if$

$$\frac{\partial L}{\partial q_j}(\mathbf{q}(t),\dot{\mathbf{q}}(t))-\frac{d}{dt}\left(\frac{\partial L}{\partial {\dot{q}}_j}(\mathbf{q}(t),\dot{\mathbf{q}}(t))\right)=0$$

for $j=1,\cdots ,d.$ . These are called the Euler-Lagrange equations.

Proof. Ignoring analytical details, the Euler-Lagrange equations follow from the following calculations, which we’ll just do for $d=1$ , with the generalization to higher d straightforward. We are calculating the first-order change in $S$ due to an infinitesimal change $\delta \gamma =(\delta q(t),\delta \dot{q}(t))$

$$\begin{array}{rl}\delta S[\gamma ]& ={\int }_{t_1}^{t_2}\delta L(q(t),\dot{q}(t))dt\\ & ={\int }_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}(q(t),\dot{q}(t))\delta q(t)+\frac{\partial L}{\partial \dot{q}}(q(t),\dot{q}(t))\delta \dot{q}(t)\right)dt\end{array}$$

But

$$\delta \dot{q}(t)=\frac{d}{dt}\delta q(t)$$

and, using integration by parts

$$\frac{\partial L}{\partial \dot{q}}\delta \dot{q}(t)=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\delta q\right)-\left(\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right)\delta q$$

so

$$\begin{array}{rl}\delta S[\gamma ]& ={\int }_{t_1}^{t_2}\left(\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right)\delta q-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\delta q\right)\right)dt\\ & ={\int }_{t_1}^{t_2}\left(\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}\right)\delta qdt-\left(\frac{\partial L}{\partial \dot{q}}\delta q\right)(t_2)+\left(\frac{\partial L}{\partial \dot{q}}\delta q\right)(t_1)\end{array}\text{\hspace{1em}(35.1)}$$

If we keep the endpoints fixed so $\delta q(t_1)=\delta q(t_2)=0$ , then for solutions to

$$\frac{\partial L}{\partial q}(q(t),\dot{q}(t))-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}(q(t),\dot{q}(t))\right)=0$$

the integral will be zero for arbitrary variations $\delta \mathbf{q}$

As an example, a particle moving in a potential $V(\mathbf{q})$ will be described by a Lagrangian

$$L(\mathbf{q},\dot{\mathbf{q}})=\frac{1}{2}m|\dot{\mathbf{q}}{|}^2-V(\mathbf{q})$$

for which the Euler-Lagrange equations will be

$$-\frac{\partial V}{\partial q_j}=\frac{d}{dt}(m{\dot{q}}_j)=m{\ddot{q}}_j$$

This is just Newton’s second law, which says that the force due to a potential is equal to the mass times the acceleration of the particle.

Given a Lagrangian classical mechanical system, one would like to be able to find a corresponding Hamiltonian system that will give the same equations of motion. To do this, we proceed by defining (for each configuration coordinate $q_j)$ a corresponding momentum coordinate $p_j$ by

$$p_j=\frac{\partial L}{\partial {\dot{q}}_j}$$

Then, instead of working with trajectories characterized at time t by

$$(\mathbf{q}(t),\dot{\mathbf{q}}(t))\in {\mathbf{R}}^{2d}$$

we would like to instead use

$$(\mathbf{q}(t),\mathbf{p}(t))\in {\mathbf{R}}^{2d}$$

where $\begin{array}{r}{p}_j=\frac{\partial L}{\partial {\dot{q}}_j}\end{array}$ and identify this ${\mathbf{R}}^{2d}$ (for example at $t=0)$ as the phase space of the conventional Hamiltonian formalism.

The transformation

$$\left(q_j,{\dot{q}}_k\right)\to \left(q_j,p_k=\frac{\partial L}{\partial {\dot{q}}_k}\right)$$

between position-velocity and phase space is known as the Legendre transform, and in good cases (for instance when $L$ is quadratic in all the velocities) it is an isomorphism. In general though, this is not an isomorphism, with the Legendre transform often taking position-velocity space to a lower dimensional subspace of phase space. Such cases are not unusual and require a much more complicated formalism, even as classical mechanical systems (this subject is known as “constrained Hamiltonian dynamics”). One important example we will study in chapter 46 is that of the free electromagnetic field, with equations of motion the Maxwell equations. In that case the configuration space coordinates are the components $\left(A_0,A_1,A_2,A_3\right)$ of the vector potential, with the problem arising because the Lagrangian does not depend on ${\dot{A}}_0$

Besides a phase space, for a Hamiltonian system one needs a Hamiltonian function. Choosing

$$h=\sum _{j=1}^d{p}_j{\dot{q}}_j-L(\mathbf{q},\dot{\mathbf{q}})$$

will work, provided the relation

$$p_j=\frac{\partial L}{\partial {\dot{q}}_j}$$

can be used to solve for the velocities ${\dot{q}}_j$ and express them in terms of the momentum variables. In that case, computing the diferential of h one finds (for $d$ = 1, the generalization to higher $d$ is straightforward)

$$\begin{array}{rl}dh& =pd\dot{q}+\dot{q}dp-\frac{\partial L}{\partial q}dq-\frac{\partial L}{\partial \dot{q}}d\dot{q}\\ & =\dot{q}dp-\frac{\partial L}{\partial q}dq\end{array}$$

So one has

$$\frac{\partial h}{\partial p}=\dot{q},\frac{\partial h}{\partial q}=-\frac{\partial L}{\partial q}$$

but these are precisely Hamilton’s equations since the Euler-Lagrange equations imply

$$\frac{\partial L}{\partial q}=\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}=\dot{p}$$

While the Legendre transform method given above works in some situations, more generally and more abstractly, one can pass from the Lagrangian to the Hamiltonian formalism by taking as phase space the space of solutions of the Euler-Lagrange equations. This is sometimes called the “covariant phase space”, and it can often concretely be realized by fixing a time $t=0$ and parametrizing solutions by their initial conditions at such a $t=0$ . One can also go directly from the action to a sort of Poisson bracket on this covariant phase space (this is called the “Peierls bracket”). For a general Lagrangian, one can pass to a version of the Hamiltonian formalism either by this method or by the method of Hamiltonian mechanics with constraints. Only for a special class of Lagrangians though will one get a non-degenerate Poisson bracket on a linear phase space and recover the usual properties of the standard Hamiltonian formalism.

Chapter contents · Next section: 35.2 Noether’s theorem and symmetries in the Lagrangian formalism · 中文版本

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