Experience of motion

Our ordinary experience suggests that physical motion can be described by configuration paths that are continuous and smooth.¹¹ We do not see the juggling pin jump from one place to another. Nor do we see the juggling pin suddenly change the way it is moving.

Our ordinary experience suggests that the motion of physical systems does not depend upon the entire history of the system. If we enter the room after the juggling pin has been thrown into the air we cannot tell when it left the juggler’s hand. The juggler could have thrown the pin from a variety of places at a variety of times with the same apparent result as we walk through the door.¹² So the motion of the pin does not depend on the details of the history.

Our ordinary experience suggests that the motion of physical systems is deterministic. In fact, a small number of parameters summarize the important aspects of the history of the system and determine its future evolution. For example, at any moment the position, velocity, orientation, and rate of change of the orientation of the juggling pin are enough to completely determine the future motion of the pin.

Realizable paths

From our experience of motion we develop certain expectations about realizable configuration paths. If a path is realizable, then any segment of the path is a realizable path segment. Conversely, a path is realizable if every segment of the path is a realizable path segment. The realizability of a path segment depends on all points of the path in the segment. The realizability of a path segment depends on every point of the path segment in the same way; no part of the path is special. The realizability of a path segment depends only on points of the path within the segment; the realizability of a path segment is a local property.

So the path-distinguishing function aggregates some local property of the system measured at each moment along the path segment. Each moment along the path must be treated in the same way. The contributions from each moment along the path segment must be combined in a way that maintains the independence of the contributions from disjoint subsegments. One method of combination that satisfies these requirements is to add up the contributions, making the path-distinguishing function an integral over the path segment of some local property of the path.¹³

So we will try to arrange that the path-distinguishing function, constructed as an integral of a local property along the path, assumes a stationary value for any realizable path. Such a path-distinguishing function is traditionally called an action for the system. We use the word “action” to be consistent with common usage. Perhaps it would be clearer to continue to call it “path-distinguishing function,” but then it would be more difficult for others to know what we were talking about.¹⁴

In order to pursue the agenda of variational mechanics, we must invent action functions that are stationary on the realizable trajectories of the systems we are studying. We will consider actions that are integrals of some local property of the configuration path at each moment. Let q = χ ∘ γ be a coordinate path in the configuration space; q(t) are the coordinates of the configuration at time t. Then the action of a segment of the path in the time interval from t₁ to t₂ is¹⁵

$$\begin{array}{ll}S[q](t_1,t_2)={\displaystyle {\int }_{t_1}^{t_2}F[q].}\hfill & (1.1)\hfill \end{array}$$

where F[q] is a function of time that measures some local property of the path. It may depend upon the value of the function q at that time and the value of any derivatives of q at that time.¹⁶

The configuration path can be locally described at a moment in terms of the coordinates, the rate of change of the coordinates, and all the higher derivatives of the coordinates at the given moment. Given this information the path can be reconstructed in some interval containing that moment.¹⁷ Local properties of paths can depend on no more than the local description of the path.

The function F measures some local property of the coordinate path q. We can decompose F [q] into two parts: a part that measures some property of a local description and a part that extracts a local description of the path from the path function. The function that measures the local property of the system depends on the particular physical system; the method of construction of a local description of a path from a path is the same for any system. We can write F[q] as a composition of these two functions:¹⁸

$$\begin{array}{ll}F[q]=L\circ \Gamma [q].\hfill & (1.2)\hfill \end{array}$$

The function Γ takes the coordinate path and produces a function of time whose value is an ordered tuple containing the time, the coordinates at that time, the rate of change of the coordinates at that time, and the values of higher derivatives of the coordinates evaluated at that time. For the path q and time t:

$$\begin{array}{ll}\Gamma [q](t)=(t,q(t),Dq(t),\dots ).\hfill & (1.3)\hfill \end{array}$$

We refer to this tuple, which includes as many derivatives as are needed, as the local tuple. The function Γ[q] depends only on the coordinate path q and its derivatives; the function Γ[q] does not depend on χ or the fact that q is made by composing χ with γ.

The function L depends on the specific details of the physical system being investigated, but does not depend on any particular configuration path. The function L computes a real-valued local property of the path. We will find that L needs only a finite number of components of the local tuple to compute this property: The path can be locally reconstructed from the full local description; that L depends on a finite number of components of the local tuple guarantees that it measures a local property.¹⁹

The advantage of this decomposition is that the local description of the path is computed by a uniform process from the configuration path, independent of the system being considered. All of the system-specific information is captured in the function L.

The function L is called a Lagrangian²⁰ for the system, and the resulting action,

$$\begin{array}{ll}S[q](t_1,t_2)={\displaystyle {\int }_{t_1}^{t_2}L\circ \Gamma [q],}\hfill & (1.4)\hfill \end{array}$$

is called the Lagrangian action. For Lagrangians that depend only on time, positions, and velocities the action can also be written

$$\begin{array}{ll}S[q](t_1,t_2)={\displaystyle {\int }_{t_1}^{t_2}L(t,q(t),Dq(t))dt.}\hfill & (1.5)\hfill \end{array}$$

Lagrangians can be found for a great variety of systems. We will see that for many systems the Lagrangian can be taken to be the difference between kinetic and potential energy. Such Lagrangians depend only on the time, the configuration, and the rate of change of the configuration. We will focus on this class of systems, but will also consider more general systems from time to time.

A realizable path of the system is to be distinguished from others by having stationary action with respect to some set of nearby unrealizable paths. Now some paths near realizable paths will also be realizable: for any motion of the juggling pin there is another that is slightly different. So when addressing the question of whether the action is stationary with respect to variations of the path we must somehow restrict the set of paths we are considering to contain only one realizable path. It will turn out that for Lagrangians that depend only on the configuration and rate of change of configuration it is enough to restrict the set of paths to those that have the same configuration at the endpoints of the path segment.

The principle of stationary action asserts that for each dynamical system we can cook up a Lagrangian such that a realizable path connecting the configurations at two times t₁ and t₂ is distinguished from all conceivable paths by the fact that the action S[q](t₁, t₂) is stationary with respect to variations of the path.²¹ For Lagrangians that depend only on the configuration and rate of change of configuration, the variations are restricted to those that preserve the configurations at t₁ and t₂.²²

Exercise 1.3: Fermat optics

Fermat observed that the laws of reflection and refraction could be accounted for by the following facts: Light travels in a straight line in any particular medium with a velocity that depends upon the medium. The path taken by a ray from a source to a destination through any sequence of media is a path of least total time, compared to neighboring paths. Show that these facts imply the laws of reflection and refraction.²³

Paths of minimum action

We already know that the actual path of a free particle is uniform motion in a straight line. According to Euler and Lagrange, the action is smaller along a straight-line test path than along nearby paths. Let q be a straight-line test path with action S[q](t₁, t₂). Let q + ϵη be a nearby path, obtained from q by adding a path variation η scaled by the real parameter ϵ.³⁷ The action on the varied path is S[q + ϵη](t₁, t₂). Euler and Lagrange found that S[q + ϵη](t₁, t₂) > S[q](t₁, t₂) for any η that is zero at the endpoints and for any small nonzero ϵ.

Let’s check this numerically by varying the test path, adding some amount of a test function that is zero at the endpoints t = t₁ and t = t₂. To make a function η that is zero at the endpoints, given a sufficiently well-behaved function ν, we can use η(t) = (t − t₁)(t − t₂)ν(t). This can be implemented:

We can use this to compute the action for a free particle over a path varied from the given path, as a function of ϵ:³⁸

The action for the varied path, with ν(t) = (sin t, cos t, t²) and ϵ = 0.001, is, as expected, larger than for the test path:

We can numerically compute the value of ϵ for which the action is minimized. We search between, say, −2 and 1:³⁹

We find exactly what is expected—that the best value for ϵ is zero,⁴⁰ and the minimum value of the action is the action along the straight path.

Finding trajectories that minimize the action

We have used the variational principle to determine if a given trajectory is realizable. We can also use the variational principle to find trajectories. Given a set of trajectories that are specified by a finite number of parameters, we can search the parameter space looking for the trajectory in the set that best approximates the real trajectory by finding one that minimizes the action. By choosing a good set of approximating functions we can get arbitrarily close to the real trajectory.⁴¹

One way to make a parametric path that has fixed endpoints is to use a polynomial that goes through the endpoints as well as a number of intermediate points. Variation of the positions of the intermediate points varies the path; the parameters of the varied path are the coordinates of the intermediate positions. The procedure make-path constructs such a path using a Lagrange interpolation polynomial. The procedure make-path is called with five arguments: (make-path t0 q0 t1 q1 qs), where q0 and q1 are the endpoints, t0 and t1 are the corresponding times, and qs is a list of intermediate points.⁴²

Having specified a parametric path, we can construct a parametric action that is just the action computed along the parametric path:

We can find approximate solution paths by finding parameters that minimize the action. We do this minimization with a canned multidimensional minimization procedure:⁴³

The procedure multidimensional-minimize takes a procedure (in this case the value of the call to parametric-path-action) that computes the function to be minimized (in this case the action) and an initial guess for the parameters. Here we choose the initial guess to be equally spaced points on a straight line between the two endpoints, computed with linear-interpolants.

To illustrate the use of this strategy, we will find trajectories of the harmonic oscillator, with Lagrangian⁴⁴

$$\begin{array}{ll}L(t,q,v)=\frac{1}{2}m{v}^2-\frac{1}{2}k{q}^2,\hfill & (1.10)\hfill \end{array}$$

for mass m and spring constant k. This Lagrangian is implemented by⁴⁵

We can find an approximate path taken by the harmonic oscillator for m = 1 and k = 1 between q(0) = 1 and q(π/2) = 0 as follows:⁴⁶

We know that the trajectories of this harmonic oscillator, for m = 1 and k = 1, are

$$\begin{array}{ll}q(t)=A\text{cos}(t+\phi )\hfill & (1.11)\hfill \end{array}$$

where the amplitude A and the phase φ are determined by the initial conditions. For the chosen endpoint conditions the solution is q(t) = cos(t). The approximate path should be an approximation to cosine over the range from 0 to π/2. Figure 1.1 shows the error in the polynomial approximation produced by this process. The maximum error in the approximation with three intermediate points is less than 1.7 × 10⁻⁴. We find, as expected, that the error in the approximation decreases as the number of intermediate points is increased. For four intermediate points it is about a factor of 15 better.

Exercise 1.5: Solution process

We can watch the progress of the minimization by modifying the procedure parametric-path-action to plot the path each time the action is computed. Try this:

Exercise 1.6: Minimizing action

Suppose we try to obtain a path by minimizing an action for an impossible problem. For example, suppose we have a free particle and we impose endpoint conditions on the velocities as well as the positions that are inconsistent with the particle being free. Does the formalism protect itself from such an unpleasant attack? You may find it illuminating to program it and see what happens.

Lagrange equations

We will find that if L is a Lagrangian for a system that depends on time, coordinates, and velocities, and if q is a coordinate path for which the action S[q](t₁, t₂) is stationary (with respect to any variation in the path that keeps the endpoints of the path fixed), then

$$\begin{array}{ll}D({\partial }_{2}L\circ \Gamma [q])-{\partial }_{1}L\circ \Gamma [q]=0.\hfill & (1.12)\hfill \end{array}$$

Here L is a real-valued function of a local tuple; ∂₁L and ∂₂L denote the partial derivatives of L with respect to its generalized position argument and generalized velocity argument respectively.⁴⁷ The function ∂₂L maps a local tuple to a structure whose components are the derivatives of L with respect to each component of the generalized velocity. The function Γ[q] maps time to the local tuple: Γ[q](t) = (t, q(t), Dq(t), …). Thus the compositions ∂₁L ∘ Γ[q] and ∂₂L ∘ Γ[q] are functions of one argument, time. The Lagrange equations assert that the derivative of ∂₂L ∘ Γ[q] is equal to ∂₁L ∘ Γ[q], at any time. Given a Lagrangian, the Lagrange equations form a system of ordinary differential equations that must be satisfied by realizable paths.

Lagrange’s equations are traditionally written as a separate equation for each component of q:

$\begin{array}{ll}\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}^i}-\frac{\partial L}{\partial q^i}=0\hfill & i=0,\dots ,n-1.\hfill \end{array}$

In this way of writing Lagrange’s equations the notation does not distinguish between L, which is a real-valued function of three variables (t, q, $$ \dot{q}$$), and L ∘ Γ[q], which is a real-valued function of one real variable t. If we do not realize this notational pun, the equations don’t make sense as written—∂L/∂$$ \dot{q}$$ is a function of three variables, so we must regard the arguments q, $$ \dot{q}$$ as functions of t before taking d/dt of the expression. Similarly, ∂L/∂q is a function of three variables, which we must view as a function of t before setting it equal to $d/dt\left(\partial L/\partial \dot{q}\right)$.

A correct use of the traditional notation is more explicit:

$\frac{d}{dt}\left({\frac{\partial L\left(t,w,\dot{w}\right)}{\partial {\dot{w}}^i}|}_{\begin{array}{l}w=q\left(t\right)\hfill \\ \dot{w}=\frac{dq\left(t\right)}{dt}\hfill \end{array}}\right)-{\frac{\partial L\left(t,w,\dot{w}\right)}{\partial w^i}|}_{\begin{array}{l}w=q\left(t\right)\hfill \\ \dot{w}=\frac{dq\left(t\right)}{dt}\hfill \end{array}}=0,$

where i = 0, …, n − 1. In these equations we see that the partial derivatives of the Lagrangian function are taken, then the path and its derivative are substituted for the position and velocity arguments of the Lagrangian, resulting in an expression in terms of the time.

A Direct Derivation

Let q be a realizable coordinate path from (t₁, q(t₁)) to (t₂, q(t₂)). Consider nearby paths q + ϵη where η(t₁) = η(t₂) = 0. Let

$$\begin{array}{lll}g(\mathit{ϵ})\hfill & =S[q+\mathit{ϵ}\eta ](t_1,t_2)\hfill & \hfill \\ \hfill & ={\displaystyle {\int }_{t_1}^{t_2}L(t,q(t)+\mathit{ϵ}\eta (t),Dq(t)+\mathit{ϵ}D\eta (t))dt.}\hfill & (1.13)\hfill \end{array}$$

Expanding as a power series in ϵ

$$\begin{array}{ll}g(\mathit{ϵ})=g(0)+\mathit{ϵ}Dg(0)+\cdots \hfill & (1.14)\hfill \end{array}$$

and using the chain rule we get

$$\begin{array}{lll}Dg(0)\hfill & ={\displaystyle {\int }_{t_1}^{t_2}({\partial }_{1}L(t,q(t),Dq(t))\eta (t))dt}\hfill & \hfill \\ \hfill & +{\displaystyle {\int }_{t_1}^{t_2}({\partial }_{2}L(t,q(t),Dq(t))D\eta (t))dt.}\hfill & (1.15)\hfill \end{array}$$

Integrating the second term by parts we obtain

$$\begin{array}{lll}Dg(0)\hfill & ={\displaystyle {\int }_{t_1}^{t_2}({\partial }_{1}L(t,q(t),Dq(t))\eta (t))dt}\hfill & \hfill \\ \hfill & +{\partial }_{2}L(t,q(t),Dq(t)D\eta (t))\eta (t){|}_{t_1}^{t_2}\hfill & \hfill \\ \hfill & -{\displaystyle {\int }_{t_1}^{t_2}\frac{d}{dt}({\partial }_{2}L(t,q(t),Dq(t)))\eta (t)dt}.\hfill & (1.16)\hfill \end{array}$$

The increment ΔS in the action due to the variation in the path is, to first order in ϵ, ϵDg(0). Because η is zero at the endpoints the integrated term is zero. Collecting together the other two terms, and reverting to functional notation, we find the increment to be

$$\begin{array}{ll}\Delta S=\mathit{ϵ}{\displaystyle {\int }_{t_1}^{t_2}\{{\partial }_{1}L\circ \Gamma [q]-D({\partial }_{2}L\circ \Gamma [q])\}}\eta .\hfill & (1.17)\hfill \end{array}$$

If ΔS is zero the action is stationary. We retain enough freedom in the choice of the variation that the factor in the integrand multiplying η is forced to be zero at each point along the path. We argue by contradiction: Suppose this factor were nonzero at some particular time. Then it would have to be nonzero in at least one of its components. But if we choose our η to be a bump that is nonzero only in that component in a neighborhood of that time, and zero everywhere else, then the integral will be nonzero. So we may conclude that the factor in curly brackets is identically zero and thus obtain Lagrange’s equations:⁴⁸

$$\begin{array}{ll}D({\partial }_{2}L\circ \Gamma [q])-{\partial }_{1}L\circ \Gamma [q]=0.\hfill & (1.18)\hfill \end{array}$$

The Variation Operator

First we will develop tools for investigating how path-dependent functions vary as the paths are varied. We will then apply these tools to the action, to derive the Lagrange equations.

Suppose that we have a function f[q] that depends on a path q. How does the function vary as the path is varied? Let q be a coordinate path and q + ϵη be a varied path, where the function η is a path-like function that can be added to the path q, and the factor ϵ is a scale factor. We define the variation δ_ηf[q] of the function f on the path q by⁴⁹

$$\begin{array}{ll}{\delta }_{\eta }f[q]=\underset{\mathit{ϵ}\to 0}{\lim }\left(\frac{f[q+\mathit{ϵ}\eta ]-f[q]}{\mathit{ϵ}}\right).\hfill & (1.19)\hfill \end{array}$$

The variation of f is a linear approximation to the change in the function f for small variations in the path. The variation of f depends on η.

A simple example is the variation of the identity path function: I[q] = q. Applying the definition, we find

$$\begin{array}{ll}{\delta }_{\eta }I[q]=\underset{\mathit{ϵ}\to 0}{\lim }\left(\frac{(q+\mathit{ϵ}\eta )-q}{\mathit{ϵ}}\right)=\eta .\hfill & (1.20)\hfill \end{array}$$

It is traditional to write δ_ηI[q] simply as δq. Another example is the variation of the path function that returns the derivative of the path. We have⁵⁰

$$\begin{array}{ll}{\delta }_{\eta }g[q]=\underset{\mathit{ϵ}\to 0}{\lim }\left(\frac{D(q+\mathit{ϵ}\eta )-Dq}{\mathit{ϵ}}\right)=D\eta \text{with}g[q]=Dq.\hfill & (1.21)\hfill \end{array}$$

It is traditional to write δ_ηg[q] as δDq.

The variation may be represented in terms of a derivative. Let g(ϵ) = f[q + ϵη]; then

$$\begin{array}{ll}{\delta }_{\eta }f[q]=\underset{\mathit{ϵ}\to 0}{\lim }\left(\frac{g(\mathit{ϵ})-g(0)}{\mathit{ϵ}}\right)=Dg(0).\hfill & (1.22)\hfill \end{array}$$

Variations have the following derivative-like properties. For path-dependent functions f and g and constant c:

$$\begin{array}{ll}{\delta }_{\eta }(fg)[q]={\delta }_{\eta }f[q]g[q]+f[q]{\delta }_{\eta }g[q]\hfill & (1.23)\hfill \end{array}$$

$$\begin{array}{ll}{\delta }_{\eta }(f+g)[q]={\delta }_{\eta }f[q]+{\delta }_{\eta }g[q]\hfill & (1.24)\hfill \end{array}$$

$$\begin{array}{ll}{\delta }_{\eta }(cf)[q]=c{\delta }_{\eta }f[q].\hfill & (1.25)\hfill \end{array}$$

Let F be a path-independent function and g be a path-dependent function; then

$$\begin{array}{ll}{\delta }_{\eta }h[q]=(DF\circ g[q]){\delta }_{\eta }g[q]\text{with}h[q]=F\circ g[q].\hfill & (1.26)\hfill \end{array}$$

The operators D (differentiation) and δ (variation) commute in the following sense:

$$D\begin{array}{ll}{\delta }_{\eta }f[q]={\delta }_{\eta }g[q]\text{with}g[q]=D(f[q]).\hfill & (1.27)\hfill \end{array}$$

Variations also commute with integration in a similar sense.

If a path-dependent function f is stationary for a particular path q with respect to small changes in that path, then it must be stationary for a subset of those variations that results from adding small multiples of a particular function η to q. So the statement δ_ηf[q] = 0 for arbitrary η implies the function f is stationary for small variations of the path around q.

Exercise 1.7: Properties of δ

Show that δ has the properties 1.23–1.27.

Exercise 1.8: Implementation of δ

a. Suppose we have a procedure f that implements a path-dependent function: for path q and time t it has the value ((f q) t). The procedure delta computes the variation (δ_ηf)[q](t) as the value of the expression ((((delta eta) f) q) t). Complete the definition of delta:

b. Use your delta procedure to verify the properties of δ listed in exercise 1.7 for simple functions such as implemented by the procedure f:⁵¹

This implements an n-degree-of-freedom path-dependent function that depends on the local tuple of the path at each moment. You can define a literal two-dimensional path by

(define q (literal-function ’q (-> Real (UP Real Real))))

You should compute both sides of the equalities and subtract the results. The answer should be zero.

A Derivation with the Variation Operator

The action is the integral of the Lagrangian along a path:

$$\begin{array}{ll}S[q](t_1,t_2)={\displaystyle {\int }_{t_1}^{t_2}L\circ \Gamma [q].}\hfill & (1.28)\hfill \end{array}$$

For a realizable path q the variation of the action with respect to any variation η that preserves the endpoints, η(t₁) = η(t₂) = 0, is zero:

$$\begin{array}{ll}{\delta }_{\eta }S[q](t_1,t_2)=0.\hfill & (1.29)\hfill \end{array}$$

Variation commutes with integration, so the variation of the action is

$$\begin{array}{ll}{\delta }_{\eta }S[q](t_1,t_2)={\displaystyle {\int }_{t_1}^{t_2}{\delta }_{\eta }h[q]\text{where}h[q]=L\circ \Gamma [q].}\hfill & (1.30)\hfill \end{array}$$

Using the fact that

$$\begin{array}{ll}{\delta }_{\eta }\Gamma [q](t)=(0,\eta (t),D\eta (t)),\hfill & (1.31)\hfill \end{array}$$

which follows from equations (1.20) and (1.21), and using the chain rule for variations (1.26), we get⁵²

$$\begin{array}{lll}{\delta }_{\eta }S[q](t_1,t_2)\hfill & ={\displaystyle {\int }_{t_1}^{t_2}(DL\circ \Gamma [q]){\delta }_{\eta }\Gamma [q]}\hfill & \hfill \\ \hfill & ={\displaystyle {\int }_{t_1}^{t_2}(({\partial }_{1}L\circ \Gamma [q])\eta +({\partial }_{2}L\circ \Gamma [q])D\eta ).}\hfill & (1.32)\hfill \end{array}$$

Integrating the last term of equation (1.32) by parts gives

$$\begin{array}{lll}{\delta }_{\eta }S[q](t_1,t_2)=\hfill & ({\partial }_{2}L\circ \Gamma [q])\eta {|}_{t_1}^{t_2}\hfill & \hfill \\ \hfill & +{\displaystyle {\int }_{t_1}^{t_2}\{({\partial }_{1}L\circ \Gamma [q])-D({\partial }_{2}L\circ \Gamma [q])\}\eta .}\hfill & (1.33)\hfill \end{array}$$

For our variation η we have η(t₁) = η(t₂) = 0, so the first term vanishes.

Thus the variation of the action is zero if and only if

$$\begin{array}{ll}0={\displaystyle {\int }_{t_1}^{t_2}\{({\partial }_{1}L\circ \Gamma [q])-D({\partial }_{2}L\circ \Gamma [q])\}\eta .}\hfill & (1.34)\hfill \end{array}$$

The variation of the action is zero because, by assumption, q is a realizable path. Thus (1.34) must be true for any function η that is zero at the endpoints. Since η is arbitrary, except for being zero at the endpoints, the bracketed factor of the integrand is zero. So

$$\begin{array}{ll}D({\partial }_{2}L\circ \Gamma [q])-({\partial }_{1}L\circ \Gamma [q])=0.\hfill & (1.35)\hfill \end{array}$$

This is just what we set out to obtain, the Lagrange equations.

A path satisfying Lagrange’s equations is one for which the action is stationary, and the fact that the action is stationary depends only on the values of L at each point of the path (and at each point on nearby paths), not on the coordinate system we use to compute these values. So if the system’s path satisfies Lagrange’s equations in some particular coordinate system, it must satisfy Lagrange’s equations in any coordinate system. Thus the equations of variational mechanics are derived the same way in any configuration space and any coordinate system.

Harmonic oscillator

For an example, consider the harmonic oscillator. A Lagrangian is

$$\begin{array}{ll}L(t,x,v)=\frac{1}{2}m{v}^2-\frac{1}{2}k{x}^2.\hfill & (1.36)\hfill \end{array}$$

Then

$$\begin{array}{ll}{\partial }_{1}L(t,x,v)=-kx\text{and}{\partial }_{2}L(t,x,v)=mv.\hfill & (1.37)\hfill \end{array}$$

The Lagrangian is applied to a tuple of the time, a coordinate, and a velocity. The symbols t, x, and v are arbitrary; they are used to specify formal parameters of the Lagrangian.

Now suppose we have a configuration path y, which gives the coordinate of the oscillator y(t) for each time t. The initial segment of the corresponding local tuple at time t is

$$\begin{array}{ll}\Gamma [y](t)=(t,y(t),Dy(t)).\hfill & (1.38)\hfill \end{array}$$

So

$$\begin{array}{ll}({\partial }_{1}L\circ \Gamma [y])(t)=-ky(t)\text{and}({\partial }_{2}L\circ \Gamma [y])(t)=mDy(t),\hfill & (1.39)\hfill \end{array}$$

and

$$\begin{array}{ll}D({\partial }_{2}L\circ \Gamma [y])(t)=m{D}^{2}y(t),\hfill & (1.40)\hfill \end{array}$$

so the Lagrange equation is

$$\begin{array}{ll}m{D}^{2}y(t)+ky(t)=0,\hfill & (1.41)\hfill \end{array}$$

which is the equation of motion of the harmonic oscillator.

Orbital motion

As another example, consider the two-dimensional motion of a particle of mass m orbiting a fixed center of attraction, with gravitational potential energy −μ/r, where r is the distance to the center of attraction. This is called the Kepler problem.

A Lagrangian for this problem is⁵³

$$\begin{array}{ll}L(t;\xi ,\eta ;v_{\xi },v_{\eta })=\frac{1}{2}m(v_{\xi }^2+v_{\eta }^2)+\frac{\mu }{\sqrt{{\xi }^2+{\eta }^2}},\hfill & (1.42)\hfill \end{array}$$

where ξ and η are formal parameters for rectangular coordinates of the particle, and v_ξ and v_η are formal parameters for corresponding rectangular velocity components. Then

$$\begin{array}{lll}{\partial }_{1}L(t;\xi ,\eta ;v_{\xi },v_{\eta })\hfill & =[{\partial }_{1,0}L(t;\xi ,\eta ;v_{\xi },v_{\eta }),{\partial }_{1,1}L(t;\xi ,\eta ;v_{\xi },v_{\eta })]\hfill & \hfill \\ \hfill & =\left[\frac{-\mu \xi }{{({\xi }^2+{\eta }^2)}^{3/2}},\frac{-\mu \eta }{{({\xi }^2+{\eta }^2)}^{3/2}}\right].\hfill & (1.43)\hfill \end{array}$$

Similarly,

$$\begin{array}{ll}{\partial }_{2}L(t;\xi ,\eta ;v_{\xi },v_{\eta })=[m{v}_{\xi },m{v}_{\eta }].\hfill & (1.44)\hfill \end{array}$$

Now suppose we have a configuration path q = (x, y), so that the coordinate tuple at time t is q(t) = (x(t), y(t)). The initial segment of the local tuple at time t is

$$\begin{array}{ll}\Gamma [q](t)=(t;x(t),y(t);Dx(t),Dy(t)).\hfill & (1.45)\hfill \end{array}$$

So

$$\begin{array}{lll}({\partial }_{1}L\circ \Gamma [q])(t)\hfill & =\left[\frac{-\mu x(t)}{{({(x(t))}^2+{(y(t))}^2)}^{3/2}},\frac{-\mu y(t)}{{({(x(t))}^2+{(y(t))}^2)}^{3/2}}\right]\hfill & \hfill \\ ({\partial }_{2}L\circ \Gamma [q])(t)\hfill & =[mDx(t),mDy(t)]\hfill & (1.46)\hfill \end{array}$$

and

$$\begin{array}{ll}D({\partial }_{2}L\circ \Gamma [q])(t)=[m{D}^{2}x(t),m{D}^{2}y(t)].\hfill & (1.47)\hfill \end{array}$$

The component Lagrange equations at time t are

$$\begin{array}{ll}m{D}^{2}x(t)+\frac{\mu x(t)}{{({(x(t))}^2+{(y(t))}^2)}^{3/2}}=0\hfill & \hfill \\ m{D}^{2}y(t)+\frac{\mu y(t)}{{({(x(t))}^2+{(y(t))}^2)}^{3/2}}=0.\hfill & (1.48)\hfill \end{array}$$

Exercise 1.9: Lagrange’s equations

Derive the Lagrange equations for the following systems, showing all of the intermediate steps as in the harmonic oscillator and orbital motion examples.

a. An ideal planar pendulum consists of a bob of mass m connected to a pivot by a massless rod of length l subject to uniform gravitational acceleration g. A Lagrangian is $L(t,\theta ,\dot{\theta })=\frac{1}{2}m{l}^2{\dot{\theta }}^2+mgl\cos \theta$. The formal parameters of L are t, θ, and $\dot{\theta }$; θ measures the angle of the pendulum rod to a plumb line and $\dot{\theta }$ is the angular velocity of the rod.⁵⁴

b. A particle of mass m moves in a two-dimensional potential V (x, y) = (x² + y²)/2 + x²y − y³/3, where x and y are rectangular coordinates of the particle. A Lagrangian is $L(t;x,y;v_x,v_y)=\frac{1}{2}m\left(v_x^2+v_y^2\right)-V(x,y)$.

c. A Lagrangian for a particle of mass m constrained to move on a sphere of radius R is $L(t;\theta ,\phi ;\alpha ,\beta )=\frac{1}{2}m{R}^2({\alpha }^2+{(\beta \sin \theta )}^2)$. The angle θ is the colatitude of the particle and φ is the longitude; the rate of change of the colatitude is α and the rate of change of the longitude is β.

Exercise 1.10: Higher-derivative Lagrangians

Derive Lagrange’s equations for Lagrangians that depend on accelerations. In particular, show that the Lagrange equations for Lagrangians of the form L(t, q, $$ \dot{q}$$, $$ \ddot{q}$$) with $$ \ddot{q}$$ terms are⁵⁵

$$\begin{array}{ll}{D}^2({\partial }_{3}L\circ \Gamma [q])-D({\partial }_{2}L\circ \Gamma [q])+{\partial }_{1}L\circ \Gamma [q])=0.\hfill & (1.49)\hfill \end{array}$$

In general, these equations, first derived by Poisson, will involve the fourth derivative of q. Note that the derivation is completely analogous to the derivation of the Lagrange equations without accelerations; it is just longer. What restrictions must we place on the variations so that the critical path satisfies a differential equation?

The free particle

Consider again the case of a free particle. The Lagrangian is implemented by the procedure L-free-particle. Rather than numerically integrating and minimizing the action, as we did in section 1.4, we can check Lagrange’s equations for an arbitrary straight-line path t ↦ (at + a₀, bt + b₀, ct + c₀):

That the residuals are zero indicates that the test path satisfies the Lagrange equations.⁵⁷

We can also apply the Lagrange-equations procedure to an arbitrary function:⁵⁸

The result is an expression containing the arbitrary time t and mass m, so it is zero precisely when D²x = 0, which is the expected equation for a free particle.

The harmonic oscillator

Consider the harmonic oscillator again, with Lagrangian (1.10). We know that the motion of a harmonic oscillator is a sinusoid with a given amplitude, frequency, and phase:

$$\begin{array}{ll}x(t)=a\cos (\omega t+\phi ).\hfill & (1.50)\hfill \end{array}$$

Suppose we have forgotten how the constants in the solution relate to the mass m and spring constant k of the oscillator. Let’s plug in the proposed solution and look at the residual:

The residual here shows that for nonzero amplitude, the only solutions allowed are ones where (k − mω²) = 0 or $\omega =\sqrt{k/m}.$

Exercise 1.11: Kepler’s third law

A Lagrangian suitable for studying the relative motion of two particles, of masses m₁ and m₂, with potential energy V, is:

The argument m is the reduced mass of the system

$$\begin{array}{ll}m=\frac{m_1{m}_2}{m_1+m_2}.\hfill & (1.51)\hfill \end{array}$$

For gravity, the potential energy function is

where r is the distance between the two particles.

Consider the simple situation of the particles in circular orbits around their common center of mass. Construct a circular orbit and plug it into the Lagrange equations. Show that the residual gives Kepler’s law:

$$\begin{array}{ll}{n}^2{a}^3=G(m_1+m_2)\hfill & (1.52)\hfill \end{array}$$

where n is the angular frequency of the orbit and a is the distance between the particles.

Exercise 1.12: Lagrange’s Equations

Compute Lagrange’s equations for the Lagrangians in exercise 1.9 using the Lagrange-equations procedure. Additionally, use the computer to perform each of the steps in the Lagrange-equations procedure and show the intermediate results. Relate these steps to the ones you showed in the hand derivation of exercise 1.9.

Exercise 1.13: Higher-derivative Lagrangians

a. Write a procedure to compute the Lagrange equations for Lagrangians that depend upon acceleration, as in exercise 1.10. Note that Gamma can take an optional argument giving the length of the initial segment of the local tuple needed. The default length is 3, giving components of the local tuple up to and including the velocities.

b. Use your procedure to compute the Lagrange equations for the Lagrangian

$L(t,x,v,a)=-\frac{1}{2}mxa-\frac{1}{2}k{x}^2.$

Do you recognize the resulting equation of motion?

c. For more fun, write the general Lagrange equation procedure that takes a Lagrangian that depends on any number of derivatives, and the number of derivatives, to produce the required equations of motion.

Hamilton’s principle

Given a system of point particles for which we can identify the force as the (negative) derivative of a potential energy V that is independent of velocity, we have shown that the system evolves along a path that satisfies Lagrange’s equations with L = T − V. Having identified a Lagrangian for this class of systems, we can restate the principle of stationary action in terms of energies. This statement is known as Hamilton’s principle: A point-particle system for which the force is derived from a velocity-independent potential energy evolves along a path q for which the action

$S[q](t_1,t_2)={\displaystyle {\int }_{t_1}^{t_2}L\circ \Gamma [q]}$

is stationary with respect to variations of the path q that leave the endpoints fixed, where L = T − V is the difference between kinetic and potential energy.⁶¹

It might seem that we have reduced Lagrange’s equations to nothing more than $\overrightarrow{F}=m\overrightarrow{a}$, and indeed, the principle is motivated by comparing the two equations for this special class of systems. However, the Lagrangian formulation of the equations of motion has an important advantage over $\overrightarrow{F}=m\overrightarrow{a}$. Our derivation used the rectangular components x_α of the positions of the constituent particles for the generalized coordinates, but if the system’s path satisfies Lagrange’s equations in some particular coordinate system, it must satisfy the equations in any coordinate system. Thus we see that L = T − V is suitable as a Lagrangian with any set of generalized coordinates. The equations of variational mechanics are derived the same way in any configuration space and any coordinate system. In contrast, the Newtonian formulation is based on elementary geometry: In order for $D^2\overrightarrow{x}(t)$ to be meaningful as an acceleration, $\overrightarrow{x}(t)$ must be a vector in physical space. Lagrange’s equations have no such restriction on the meaning of the coordinate q. The generalized coordinates can be any parameters that conveniently describe the configurations of the system.

Constant acceleration

Consider a particle of mass m in a uniform gravitational field with acceleration g. The potential energy is mgh where h is the height of the particle. The kinetic energy is just $\frac{1}{2}m{v}^2$. A Lagrangian for the system is the difference of the kinetic and potential energies. In rectangular coordinates, with y measuring the vertical position and x measuring the horizontal position, the Lagrangian is $L(t;x,y;v_x,v_y)=\frac{1}{2}m\left(v_x^2+v_y^2\right)-mgy$. We have⁶²

This equation describes unaccelerated motion in the horizontal direction (mD²x(t) = 0) and constant acceleration in the vertical direction (mD²y(t) = −gm).

Central force field

Consider planar motion of a particle of mass m in a central force field, with an arbitrary potential energy U(r) depending only upon the distance r to the center of attraction. We will derive the Lagrange equations for this system in both rectangular coordinates and polar coordinates.

In rectangular coordinates (x, y), with origin at the center of attraction, the potential energy is $V(t;x,y)=U(\sqrt{x^2+y^2})$ and the kinetic energy is $T(t;x,y;v_x,v_y)=\frac{1}{2}m\left(v_x^2+v_y^2\right)$. A Lagrangian for the system is L = T − V:

$$\begin{array}{ll}L(t;x,y;v_x,v_y)=\frac{1}{2}m\left(v_x^2+v_y^2\right)-U(\sqrt{x^2+y^2}).\hfill & (1.61)\hfill \end{array}$$

As a procedure:

The Lagrange equations are

We can rewrite these Lagrange equations as:

$$\begin{array}{ll}m{D}^{2}x(t)=-\frac{x(t)}{r(t)}DU(r(t))\hfill & (1.62)\hfill \end{array}$$

$$\begin{array}{ll}m{D}^{2}y(t)=-\frac{y(t)}{r(t)}DU(r(t)),\hfill & (1.63)\hfill \end{array}$$

where $r(t)=\sqrt{{(x(t))}^2+{(y(t))}^2}$. We can interpret these as follows. The particle is subject to a radially directed force with magnitude −DU(r). Newton’s equations equate the force with the product of the mass and the acceleration. The two Lagrange equations are just the rectangular components of Newton’s equations.

We can describe the same system in polar coordinates. The relationship between rectangular coordinates (x, y) and polar coordinates (r, φ) is

$$\begin{array}{ll}x=r\cos \phi \hfill & \hfill \\ y=r\sin \phi .\hfill & (1.64)\hfill \end{array}$$

The relationship of the generalized velocities is derived from the coordinate transformation. Consider a configuration path that is represented in both rectangular and polar coordinates. Let $\tilde{x}$ and $\tilde{y}$ be components of the rectangular coordinate path, and let $\tilde{r}$ and $\tilde{\phi }$ be components of the corresponding polar coordinate path. The rectangular components at time t are $(\tilde{x}(t),\tilde{y}(t))$ and the polar coordinates at time t are $(\tilde{r}(t),\tilde{\phi }(t))$. They are related by (1.64):

$$\begin{array}{ll}\tilde{x}(t)=\tilde{r}(t)\cos \tilde{\phi }(t)\hfill & \hfill \\ \tilde{y}(t)=\tilde{r}(t)\sin \tilde{\phi }(t).\hfill & (1.65)\hfill \end{array}$$

The rectangular velocity at time t is $(D\tilde{x}(t),D\tilde{y}(t))$. Differentiating (1.65) gives the relationship among the velocities

$$\begin{array}{lll}D\tilde{x}(t)\hfill & =D\tilde{r}(t)\cos \tilde{\phi }(t)-\tilde{r}(t)D\tilde{\phi }(t)\sin \tilde{\phi }(t)\hfill & \hfill \\ D\tilde{y}(t)\hfill & =D\tilde{r}(t)\sin \tilde{\phi }(t)+\tilde{r}(t)D\tilde{\phi }(t)\cos \tilde{\phi }(t).\hfill & (1.66)\hfill \end{array}$$

These relations are valid for any configuration path at any moment, so we can abstract them to relations among coordinate representations of an arbitrary velocity. Let v_x and v_y be the rectangular components of the velocity and $\dot{r}$ and $\dot{\phi }$ be the rate of change of r and φ. Then

$$\begin{array}{ll}{v}_x=\dot{r}\cos \phi -r\dot{\phi }\sin \phi \hfill & \hfill \\ v_y=\dot{r}\sin \phi +r\dot{\phi }\cos \phi .\hfill & (1.67)\hfill \end{array}$$

The kinetic energy is $\frac{1}{2}m(v_x^2+v_y^2)$:

$$\begin{array}{ll}T(t;r,\phi ;\dot{r},\dot{\phi })=\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\phi }}^2\right),\hfill & (1.68)\hfill \end{array}$$

and the Lagrangian is

$$\begin{array}{ll}L(t;r,\phi ;\dot{r},\dot{\phi })=\frac{1}{2}m\left({\dot{r}}^2+r^2{\dot{\phi }}^2\right)-U(r).\hfill & (1.69)\hfill \end{array}$$

We express this Lagrangian as follows:

Lagrange’s equations are

We can interpret the first equation as saying that the product of the mass and the radial acceleration is the sum of the force due to the potential and the centrifugal force. The second equation can be interpreted as saying that the derivative of the angular momentum mr²Dφ is zero, so angular momentum is conserved.

Note that we used the same Lagrange-equations procedure for the derivation in both coordinate systems. Coordinate representations of the Lagrangian are different for different coordinate systems, and the Lagrange equations in different coordinate systems look different. Yet the same method is used to derive the Lagrange equations in any coordinate system.

Exercise 1.14: Coordinate-independence of Lagrange equations

Check that the Lagrange equations for central force motion in polar coordinates and in rectangular coordinates are equivalent. Determine the relationship among the second derivatives by substituting paths into the transformation equations and computing derivatives, then substitute these relations into the equations of motion.

Coriolis and centrifugal forces

The equations of motion of a free particle in a rotating coordinate system have additional terms. Consider a free particle moving in two dimensions. A Lagrangian is:

The rotation will be easy to describe in polar coordinates, so we transform to polar coordinates:

Now we can make a simple time-dependent transformation to rotating coordinates, with rate of rotation Omega:

Now let’s transform back to rectangular coordinates:

The new Lagrangian, in the rotating rectangular coordinate system is:

Although the transformation of coordinates is time dependent the resulting Lagrangian is independent of time.

The Lagrange equations for the free particle in the rotating coordinate system have force terms involving the angular velocity Ω:

The terms that are proportional to Ω² are called centrifugal force terms, and the ones that are proportional to Ω are called Coriolis force terms. Note that the centrifugal force terms are radial, pointing away from the center of rotation. These additional force terms are derived from the corresponding terms in the Lagrangian. The terms in the Lagrangian that are proportional to Ω² can be thought of as the negation of a centrifugal potential energy.

Because this is a free particle the velocity in the original unrotated coordinates is constant. In rotating coordinates, the Coriolis terms describe an acceleration that is perpendicular to the velocity, causing the trajectory to curve.

Lagrangians for rigidly constrained systems

The system is presumed to be made of N point masses, indexed by α, in ordinary three-dimensional space. The first step is to choose a convenient set of irredundant generalized coordinates q and redescribe the system in terms of these. In terms of the generalized coordinates the rectangular coordinates of particle α are

$$\begin{array}{ll}{x}_{\alpha }=f_{\alpha }(t,q).\hfill & (1.79)\hfill \end{array}$$

For irredundant coordinates q all the coordinate constraints are built into the functions f_α. We deduce the relationship of the generalized velocities v to the velocities of the constituent particles v_α by inserting path functions into equation (1.79), differentiating, and abstracting to arbitrary velocities (see section 1.6.1). We find

$$\begin{array}{ll}{v}_{\alpha }={\partial }_0{f}_{\alpha }(t,q)+{\partial }_1{f}_{\alpha }(t,q)v.\hfill & (1.80)\hfill \end{array}$$

We use equations (1.79) and (1.80) to express the kinetic energy in terms of the generalized coordinates and velocities. Let $\tilde{T}$ be the kinetic energy as a function of the rectangular coordinates and velocities:

$$\begin{array}{ll}\tilde{T}(t;x_0,\dots ,x_{N-1};v_0,\dots ,v_{N-1})={\displaystyle \sum _{\alpha }\frac{1}{2}{m}_{\alpha }{v}_{\alpha }^2,}\hfill & (1.81)\hfill \end{array}$$

where $v_{\alpha }^2$ is the squared magnitude of v_α. As a function of the generalized coordinate tuple q and the generalized velocity tuple v, the kinetic energy is

$$\begin{array}{lll}T(t,q,v)\hfill & =\tilde{T}(t,f(t,q),{\partial }_{0}f(t,q)+{\partial }_{1}f(t,q)v)\hfill & \hfill \\ \hfill & ={\displaystyle \sum _{\alpha }\frac{1}{2}{m}_{\alpha }{({\partial }_0{f}_{\alpha }(t,q)+{\partial }_1{f}_{\alpha }(t,q)v)}^2.}\hfill & (1.82)\hfill \end{array}$$