Theoretical Mechanics IPSP

The Lagrange formalism provides an effective approach to derive the EOM for generalized coordinates. We first provide a derivation in a Cartesian coordinate frame. Then we discuss how the EOM for generalized coordinates are determined.

In Section 5.5 we saw that a mobile will be at rest in a position characterized by the coordinate vector $\mathbf x$ when the leading order correction $\delta\mathbf x \, \cdot \nabla \Phi(\mathbf x)$ to its potential energy $\Phi(\mathbf x)$ vanishes for every perturbation $\delta\mathbf x$ of the position. In the following we denote the leading order corrections term of the Taylor expansion as variation.

In Section 5.5 we showed that $\delta\Phi( \mathbf x_0 ) = 0$ for every critical point $\mathbf x_0$ where the system is (and remains) at rest. We now also account fort explicitly time-dependent potentials $\Phi( \mathbf x, t )$ and consider the variations $\delta\mathbf x(t)$ of time dependent trajectories $\mathbf x(t)$ with $t \in [t_I, t_F]$. Here $\delta\mathbf x(t)$ describes the deviation of the perturbed trajectory from the reference trajectory $\mathbf x(t)$ at time $t$, and it is understood that $\delta\mathbf x(t_I) = \delta\mathbf x(t_F) = \mathbf 0$ Now we have \begin{align*} \delta \Phi( \mathbf x, t ) &= \delta\mathbf x \cdot \nabla_{\mathbf x} \Phi( \mathbf x, t ) = - \delta\mathbf x \cdot \mathbf F( \mathbf x, t ) = - \delta\mathbf x \cdot m \, \ddot{\boldsymbol x} \end{align*} The velocity and acceleration for the perturbed trajectory $\mathbf x + \delta\mathbf x$ are $\dot{\boldsymbol x} + \delta\dot{\boldsymbol x}$ and $\ddot{\boldsymbol x} + \delta\ddot{\boldsymbol x}$ such that \begin{align*} \frac{\mathrm{d}}{\mathrm{d} t} \left( m\, \dot{\boldsymbol x} \cdot \delta\mathbf x \right) = m\, \ddot{\boldsymbol x} \cdot \delta\mathbf x + m\, \dot{\boldsymbol x} \cdot \delta\dot{\boldsymbol x} = m\, \ddot{\boldsymbol x} \cdot \delta\mathbf x + \delta \frac{ m\, \dot{\boldsymbol x}^2 }{2} \end{align*} where $T = m\, \dot{\boldsymbol x}^2 / 2$ is the kinetic energy. Hence, we can express the variation of the potential as \begin{align*} \delta \Phi( \mathbf x, t ) &= - \frac{\mathrm{d}}{\mathrm{d} t} \left( \delta\mathbf x \cdot m \, \dot{\boldsymbol x} \right) + \delta T( \dot{\boldsymbol x} ) \\ \quad\Rightarrow\quad \delta\bigl( T( \dot{\boldsymbol x} ) - \Phi( \mathbf x, t ) \bigr) &= - \frac{\mathrm{d}}{\mathrm{d} t} \left( \delta\mathbf x \cdot m \, \dot{\boldsymbol x} \right) \end{align*} The difference between the kinetic and potential energy is a total time derivative. Integrating the expression over time from $t_I$ to $t_F$ therefore provides

The integral vanishes because $\mathbf x$ is fixed a the start and the end point. Up to mathematical identities that are always true we only used Newton's law $\mathbf F( \mathbf x, t ) = m \ddot{\boldsymbol x}$ to arrive at this conclusion. This observation is denoted as the principle of least action. Rather than on Newton axioms we may therefore base mechanics on the principle of least action.

Remark 6.4. The principle is called the principle of least action. However, it only requires that the action has a critical point. There are many examples in physics where the action takes a saddle point, rather than a minimum.

The principle provides an alternative way to determine the EOM that proceeds as follows. \begin{align*} 0 &= \delta S[\mathbf x(t), \dot{\boldsymbol x}(t)] = \int_{t_I}^{t_F} \mathrm{d} t \: \delta\mathcal L( \mathbf x(t), \dot{\boldsymbol x}(t), t ) \\ &= \int_{t_I}^{t_F} \mathrm{d} t \: \left[ \delta\dot{\boldsymbol x} \: \nabla_{\dot{\boldsymbol x}} \mathcal L(\mathbf x, \dot{\boldsymbol x}, t) + \delta\mathbf x \: \nabla_{\mathbf x} \mathcal L(\mathbf x, \dot{\boldsymbol x}, t) \right] \\ &= \int_{t_I}^{t_F} \mathrm{d} t \; \delta\mathbf x \: \left[ \left( -\frac{ \mathrm{d} }{\mathrm{d} t} \nabla_{\dot{\boldsymbol x}} \mathcal L(\mathbf x, \dot{\boldsymbol x}, t) \right) + \nabla_{\mathbf x} \mathcal L(\mathbf x, \dot{\boldsymbol x}, t) \right] \end{align*} In the last step we performed a partial integration¹⁾. The integral must vanish for every choice of the variation $\delta\mathbf x$. In particular we may choose a function $\delta\mathbf x$ that takes the same sign as the square bracket whenever it does not vanish. However, in that case the integral is strictly positive unless the square bracket vanishes. This provides the EOM of the dynamics in terms of the Euler-Lagrange equation.

The principle of least action is an application of variational calculus to the action integral, Equation 6.2.3. In order to provide a better intuition of the mathematical concept of the variation, we demonstrate now how one can derive a differential equation of the shortest path on a plane.

We describe a curve from the origin, $(0,0)$ to the position $(x_e,y_e)$ in the plane by a function $f(x)$ with $f(0)=0$ and $f(x_e)=y_e$. Hence, the curve follows the coordinates $\mathbf q = (x,f(x))$, and according to Remark 3.7 the length of a curve is determined by the line integral \begin{align*} L[f(x)] &= \int_0^{x_e}\mathrm{d} x \: \mathcal{D}_{\text{plane}}\bigl( f(x), f'(x) \bigr) \\ \text{with}\quad \mathcal{D}_{\text{plane}}\bigl( f(x), f'(x) \bigr) &= \left\lvert \frac{\mathrm{d} \mathbf q}{\mathrm{d} x} \right\rvert = \sqrt{ 1 + \bigl(f'(x) \bigr)^2 } \end{align*} where $f'(x) = \mathrm{d} f(x) / \mathrm{d} x$. Paths of minimal length must therefore be solutions of the Euler-Lagrange equation \begin{align*} 0 = \frac{\partial \mathcal D}{\partial f} = \frac{\mathrm{d}}{\mathrm{d} x} \frac{\partial \mathcal D}{\partial f'} = \frac{\mathrm{d}}{\mathrm{d} x} \frac{f'(x)}{\sqrt{ 1 + \bigl(f'(x) \bigr)^2 }} \end{align*} which implies that there is a constant $K$ with \begin{align*} K = \frac{f'(x)}{\sqrt{ 1 + \bigl(f'(x) \bigr)^2 }} \quad &\Rightarrow \quad K^2 \; \left( 1 + \bigl(f'(x) \bigr)^2 \right) = \left( f'(x) \right)^2 \\ \quad &\Rightarrow \quad f'(x) = \frac{K}{\sqrt{1-K^2}} = \text{const} \end{align*} Consequently, the shortest connection between two points in the plane is a straight line, where the slope is constant. We urge the reader to go through the steps of the derivation of the Euler-Lagrange equation for this problem, and to take not of the important requirement that the variation of the path $\delta \mathbf q$ must vanish at both endpoints of the trajectory. An example of a variational problem where this requirement is relaxed is provided in Problem 6.4.

We describe a curve on a cylinder with radius $R$ by adopting cylinder coordinates and specifying $z(\theta)$ for the range $\mathcal I = [\theta_I, \theta_E]$. The values at the boundary of $\mathcal I$ will be denoted as $z(\theta_I) = z_I$ and $z(\theta_E) = z_E$. Hence, the curve follows the coordinates $\mathbf q = z(\theta) \:\hat{\boldsymbol z} + R \: \hat{\boldsymbol r}(\theta)$ along a path with direction \begin{align*} \mathbf q'(\theta) = \frac{\mathrm{d} \mathbf q}{\mathrm{d}\theta} = z'(\theta) \:\hat{\boldsymbol z} + R \: \hat{\boldsymbol \theta}(\theta) \end{align*} The length of this curve is determined by the line integral \begin{align*} L[z(\theta)] &= \int_{\theta_I}^{\theta_E}\mathrm{d} \theta \: \mathcal{D}_{\text{cyl}}\bigl( z(\theta), z'(\theta) \bigr) \\ \text{with}\quad \mathcal{D}_{\text{cyl}}\bigl( z(\theta), z'(\theta) \bigr) &= \left\lvert \frac{\mathrm{d} \mathbf q}{\mathrm{d} \theta} \right\rvert = R \; \sqrt{ 1 + \bigl(z'(\theta)/R \bigr)^2 } \end{align*} The distance function $\mathcal{D}_{\text{cyl}}$ of this problem is identical to the one for the plane, up to replacing $f(x)$ by $z(\theta)$ and $x$ by $R\,\theta$. Hence, the solutions will be paths of the form

\begin{align*} z(\theta) = z_I + \frac{z_E - z_I}{ \theta_E - \theta_I + 2\pi \, n } \; (\theta-\theta_I) \quad \text{with} \quad n \in \mathbb Z \end{align*} When one makes sure that $\theta_E \in ( \theta_I-\pi , \theta_I+\pi )$ then the solution for $n=0$ represents the shortest path from $z_I$ to $z_E$. For $\theta_E - \theta_I = \pi$ the path for $n=0$ and $n=-1$ have the same length. All other paths represent local minima of $L$. Small perturbations of the path will increase the length. However, trajectories that reach the final point with a smaller number of loops around the cylinder will in general be shorter. An example is shown in Figure 6.4.

A catenoid is the surface of revolution of the hyperbolic cosine function. We describe a curve on a catenoid with radius $\cosh z$ at height $z$ by adopting cylinder coordinates and specifying $\theta(z)$ for the range $\mathcal I = [z_I, z_E]$ ²⁾. The values at the boundary of $\mathcal I$ are now denoted as $z(\theta_I) = z_I$ and $z(\theta_E) = z_E$. Hence, the curve follows the coordinates $\mathbf q = z \:\hat{\boldsymbol z} + \cosh z \: \hat{\boldsymbol r}(\theta(z))$ along a path with direction \begin{align*} \mathbf q'(z) = \frac{\mathrm{d} \mathbf q}{\mathrm{d} z} = \hat{\boldsymbol z} + \sinh z(\theta) \:\hat{\boldsymbol r}(\theta(z)) + \cosh z(\theta) \: z'(\theta) \: \hat{\boldsymbol \theta}(\theta(z)) \end{align*} The length of this curve is determined by the line integral \begin{align*} L[z(\theta)] &= \int_{z_I}^{z_E}\mathrm{d} z \: \mathcal{D}_{\text{cat}}\bigl( \theta(z), \theta'(z) \bigr) \\ \text{with}\quad \mathcal{D}_{\text{cat}}\bigl( \theta(z), \theta'(z) \bigr) &= \left\lvert \frac{\mathrm{d} \mathbf q}{\mathrm{d} \theta} \right\rvert = \sqrt{ \bigl( 1 + \theta'(z)^2 \bigr) \; \cosh^2 z + \sinh^2 z } \end{align*} Consequently, the Euler-Lagrange function takes the form \begin{align*} 0 &= \frac{\partial \mathcal{D}_{\text{cat}}}{\partial \theta} = \frac{\mathrm{d}}{\mathrm{d} z} \frac{\partial \mathcal{D}_{\text{cat}}}{\partial \theta'} &= \frac{\mathrm{d}}{\mathrm{d} z} \frac{\theta'(z) \: \cosh^2 z}{ \sqrt{ \bigl( 1 + \theta'(z)^2 \bigr) \; \cosh^2 z + \sinh^2 z } } \end{align*} The variable $\theta$ is cyclic and we denote the entailed conservation law as $K$. Rearranging terms provides \begin{align*} \theta'(z) = K \; \sqrt{ \frac{1+\tanh^2 z}{\cosh^2 z - K^2} } \end{align*} such that \begin{align*} \theta(z) = \theta_I + K \; \int_{z_I}^{z} \mathrm{d} z \; \sqrt{ \frac{1+\tanh^2 z}{\cosh^2 z - K^2} } \end{align*} The shortest paths on the catenoid must be determined by numerical evaluation of this integral. Problem 6.5 extends the present discussion to situations where one minimizes the surface area of a soap film, rather than a feature of a one-dimensional object. Problem 6.14 addresses extremal paths on a sphere. Unless two points lie exactly on opposite sides of the sphere (like North and South pole) there are exactly two trajectories of extremal length. One of them is the shortest trajectory. The other one is a saddle point.

The Euler-Lagrange equations derive from a variational principle stating that the gradient of the Lagrange function with respect to the phase-space coordinate $\mathbf\Gamma = (\mathbf x, \dot{\boldsymbol x})$ must vanish for physically admissible trajectories. This holds for all directions in phase space. However, generalized coordinates do not qualify as a vector such that some care is needed to derive their EOM.

Let $\mathbf q$ be the generalized coordinates of a system and $\mathbf x(\mathbf q)$ the associated configuration vector of the system. It will be provided in Cartesian coordinates from the point of view of an observer who is at rest. Hence, $\mathbf x$ is a vector with all properties discussed in Chapter 2. In contrast, $\mathbf q$ will in general only be a tuple of functions that provide a convenient parameterization of valid configurations. We address the situation where the forces in the system are conservative, arising from a potential energy $\Phi\bigl( \mathbf x(\mathbf q), t \bigr)$. Moreover, we assume that the potential energy can be represented as a sum of $\Phi_c\bigl( \mathbf x(\mathbf q), t \bigr)$ and $U\bigl( \mathbf x( \mathbf q), t \bigr)$. The contribution $\Phi_c\bigl( \mathbf x(\mathbf q), t \bigr)$ accounts for forces that constraint the coordinates of the system such that they comply with positions $\mathbf x(\mathbf q)$. The part $U\bigl( \mathbf x( \mathbf q), t \bigr)$ accounts for all other forces. We will now explore the implications of the principle of least action for variations of the path that refer only to accessible coordinates. For the $k$th coordinate of the variation we write \begin{align*} \delta x_k &= x_k( \mathbf q + \delta\mathbf q, t ) - x_k( \mathbf q, t ) = \sum_{ \nu = 1 }^d \frac{\partial x_k}{ \partial q_\nu } \: \delta q_\nu \end{align*} and for the associated time derivative we have \begin{align*} \delta\dot x_k = \frac{\mathrm{d}}{\mathrm{d} t} \delta x_k &= \sum_{ \nu = 1 }^d \frac{\partial \dot x_k}{ \partial q_\nu } \: \delta q_\nu + \sum_{ \nu = 1 }^d \frac{\partial x_k}{ \partial q_\nu } \: \delta\dot q_\nu \end{align*} As a consequence the variation of the Lagrangian takes the form \begin{align*} \delta \mathcal L &= \delta\mathbf x \cdot \nabla_{\mathbf x} \mathcal L + \delta\dot{\boldsymbol x} \cdot \nabla_{\dot{\boldsymbol x}} \mathcal L = \delta\mathbf x \cdot ( \mathbf F_c + \mathbf F_e ) + \delta\dot{\boldsymbol x} \cdot m \dot{\boldsymbol x} \end{align*} where $ \mathbf F_c $ represent the constraint forces. We consider variations $\delta\mathbf x$ that relate trajectories complying with the constraints such that $ \delta\mathbf x \cdot \mathbf F_c = 0$. Therefore, in the setting of generalized coordinates one need not account for constraint forces³⁾. We will now express the variation of the Lagrangian in terms of the variations of the generalized coordinates, \begin{align*} \delta \mathcal L &= \sum_{k=1}^D \left[ \delta x_k \; \frac{ \partial \mathcal L }{ \partial x_k } + \delta\dot x_k \; \frac{ \partial \mathcal L }{ \partial\dot x_k } \right] \\ &= \sum_{k=1}^D \Biggl[ \left( \sum_{ \nu = 1 }^d \frac{\partial x_k}{ \partial q_\nu } \: \delta q_\nu \right) \; \frac{ \partial \mathcal L }{ \partial x_k } + \sum_{ \nu = 1 }^d \left( \frac{\partial \dot x_k}{ \partial q_\nu } \: \delta q_\nu + \frac{\partial x_k}{ \partial q_\nu } \: \delta\dot q_\nu \right) \frac{ \partial \mathcal L }{ \partial\dot x_k } \Biggr] \\ &= \sum_{ \nu = 1 }^d \delta q_\nu \sum_{k=1}^D \left( \frac{\partial x_k}{ \partial q_\nu } \: \frac{ \partial \mathcal L }{ \partial x_k } + \frac{\partial\dot x_k}{ \partial q_\nu } \: \frac{ \partial \mathcal L }{ \partial\dot x_k } \right) + \sum_{ \nu = 1 }^d \delta\dot q_\nu \; \sum_{k=1}^D \frac{\partial x_k}{ \partial q_\nu } \; \frac{ \partial \mathcal L }{ \partial\dot x_k } \end{align*} On the other hand \begin{align*} \frac{ \partial \mathcal L\bigl( \mathbf x(\mathbf q, t), \dot{\boldsymbol x}(\mathbf q, \dot{\boldsymbol q}, t), t \bigr) }{ \partial q_\nu } &= \sum_{k=1}^D \left( \frac{\partial x_k}{ \partial q_\nu } \: \frac{ \partial \mathcal L }{ \partial x_k } + \frac{\partial\dot x_k}{ \partial q_\nu } \: \frac{ \partial \mathcal L }{ \partial\dot x_k } \right) \\ \frac{ \partial \mathcal L\bigl( \mathbf x(\mathbf q, t), \dot{\boldsymbol x}(\mathbf q, \dot{\boldsymbol q}, t), t \bigr) }{ \partial\dot q_\nu } &= \sum_{k=1}^D \frac{\partial \mathcal L\bigl( \mathbf x(\mathbf q, t), \dot{\boldsymbol x}(\mathbf q, \dot{\boldsymbol q}, t), t \bigr) }{ \partial\dot x_k } \frac{ \partial\dot x_k }{ \partial\dot q_\nu } \\ &=\sum_{k=1}^D \frac{\partial \mathcal L }{ \partial\dot x_k } \;\; \frac{ \partial }{ \partial\dot q_\nu }\!\! \left( \frac{\partial x_k}{\partial t} + \sum_{\mu=1}^d \frac{\partial x_k}{\partial q_\mu} \: \dot q_\mu \right) \\ &=\sum_{k=1}^D \frac{\partial \mathcal L }{ \partial\dot x_k } \; \frac{\partial x_k}{\partial q_\nu} \end{align*} Therefore, \begin{align*} \delta S &= \int \mathrm{d} t \delta \mathcal L = \int \mathrm{d} t \sum_{ \nu = 1 }^d \left( \delta q_\nu \: \frac{ \partial \mathcal L }{ \partial q_\nu } + \delta\dot q_\nu \; \frac{ \partial \mathcal L }{ \partial\dot q_\nu } \right) \\ &= \int \mathrm{d} t \sum_{ \nu = 1 }^d \delta q_\nu \: \left( \frac{ \partial \mathcal L }{ \partial q_\nu } - \frac{\mathrm{d}}{\mathrm{d} t} \frac{ \partial \mathcal L }{ \partial\dot q_\nu } \right) \end{align*} The equations of motion are derived from the Lagrangian, Definition 6.5, by Algorithm 6.1.

In the next section we will apply the formalism to models with a single degree of freedom.

Problem 6.3: Shortest path on a catenoid
In footnote 2 we pointed out that it is a good idea to parameterize the paths on a body of revolution in terms of $z(\theta)$ rather than $\theta(z)$. Adopt the latter parameterization and work out for yourself that this leads to a second order ODE while the former one provides a conserved quantity, and subsequently immediately a first order ODE.

Problem 6.4: Fermat's principle

Fermat's principle states that a light beam propagates along a path minimizing the flight time. When passing from air into glass it changes direction according to Sellius' refraction law. Here, we consider a setting where the beam starts in air at the position, $(x,y)=(0,0)$, to the top left in the figure, with coordinates where $\hat x$ points downwards and $\hat y$ to the right. The path of the light is described by a function $y(x)$. We require that beam passes from air into the glass at the position $(a,u)$ such that it will eventually proceed through the prescribed position $(b,w)$ in the glass. The speed of light in air and in glass will be denoted as $c_A$ and $c_G$, respectively.

a) Show that the time of flight $T$ for a (hypothetical) trajectory $y(x)$ with derivative $y'(x)$ can be determined as follows \begin{align*} T = c_A^{-1} \; \int_0^a \mathrm{d} x \: \sqrt{ 1 + (y'(x))^2 } + c_G^{-1} \; \int_a^b \mathrm{d} x \: \sqrt{ 1 + (y'(x))^2 } \, . \end{align*} b) In the following we consider a glas body with a planar surface, and align the coordinates such that glass surface is aligned parallel to the $y$-axis. Hence, we know that the light passes from air to glass at the fixed position $a$, but we still have to determine $u$. Determine $\delta T$ for a variation $y(x) + \delta y(x)$ of the trajectory. What does this imply for $\left. \delta y(x)\right|_{x=0}$, $\left. \delta y(x)\right|_{x=a}$ and $\left. \delta y(x)\right|_{x=b}$? What does it imply for the boundary terms that arise from the integration by parts, when determining $\delta T$?

c) Show that the beam must go in a straight line in air and in glass. Show that this implies that \begin{align*} T(u) = \frac{1}{c_A} \; \sqrt{ u^2 + a^2 } + \frac{1}{c_G} \; \sqrt{ (w-u)^2 + (b-a)^2 } \, . \end{align*} Derive Snellius' law from the condition that $0 = \mathrm{d} T(u) / \mathrm{d} u$.

d) Snellius' Law can also be directly obtained from Fermat's principle. How?

Problem 6.5: Stability of soap films
When a soap film is suspended between two rings, it takes a cylinder-symmetric shape of minimal surface area. We discuss here the form of the film for rings of radius $R_0$ and $R_1$ positioned at the height $x_0$ and $x_1$, respectively. At the Mathematikum in Gießen there is a nice demonstration experiment: $x_0$ is the surface height of soap solution in a vessel around the platform where the children are standing, and $x_1$ is the height of the ring pulled upwards by the children.

a) Let $w(x)$ be the radius of the cylinder-symmetric soap films at the vertical position $x$. Sketch the setup and mark the relevant notations for the problem.

b) Show that the surface area $A$ of the soap film takes the form \begin{align*} A = \int_{x_0}^{x_1} \mathrm{d} x \: w(x) \: f( w'(x) ) \, , \end{align*} Here, the factor $f( w'(x) )$ takes into account that the area is larger when the derivative $w'(x) = \mathrm{d} x / \mathrm{d} x$ increases. Determine the function $f( w'(x) )$ in this expression.

c) Show that $A$ is extremal for shapes $w(x)$ that obey the differential equation \begin{align*} w''(x) = \frac{ 1 + (w'(x))^2 }{w(x)} \, . \end{align*}

d) Determine the solutions of the differential equation. Hint: Rewrite the equation into the form \begin{align*} \frac{ w'(x) \; w''(x)}{ 1 + (w'(x))^2 } = \frac{w'(x)}{w(x)} \, . \end{align*}

e)Consider now solutions with $-x_0 = x_1 = a$ and $R_0 = R_1 = R$, and denote the radius at the thinnest point of the soap film as $w_0$. Show that $w_0$ is the solution of \begin{align*} \frac{ R }{ a } = \frac{w_0}{a} \: \cosh\frac{a}{w_0} \, . \end{align*}

f)Sketch $R/a$ as function of $a/w_0$. For given $R$ and $a$ you can then find $w_0$. For small separation of the rings you should find two solutions. What happens when one slowly rises the ring? Will an adult ever manage to pull up the ring to head height before the film ruptures?