In Section 6.3 we discussed the EOM of systems with one degree of freedom. In the present section this analysis is extended to systems with two and more degrees of freedom. Again the discussion will be based on examples:
1. We first discuss uniform motion where there are no fores acting on the particle. Here, we learn about the concept of cyclic observables, and how momentum conservation is related to translation invariance.
2. Subsequently, we address the case of uniform acceleration. In this case there only is translation invariance in the directions normal to the acceleration. The according components of the momentum are still preserved.
3. For systems with harmonic, Hookian forces we explore the relation between rotation invariance and the conservation of angular momentum. Moreover, we introduce the concepts of an effective potential and of a rotation barrier.
4. Finally, we discuss the spherical pendulum. It constitutes an example with a generic, non-trivial coupling of the evolution of its different DOF. We will learn how to exploit conservation laws to decompose the dynamics into distinct dynamics with a single degree of freedom.

When there are no forces acting on the particle we have \begin{align*} T &= \frac{m}{2} \: \dot{\mathbf x}^2 = \frac{m}{2} \; \sum_i \dot x_i^2\\ V &= 0 \end{align*} Note that the Lagrange function $\mathcal L = T-V$ only depends on the velocity $\dot{\mathbf q}$ and not on the position $\mathbf q$. As a consequence, the Euler-Lagrange equation for the coordinate $x_i$ requires that the acceleration of the particle vanishes, \begin{align*} m \: \ddot x_i = \frac{\mathrm{d}}{\mathrm{d} t} \frac{ \partial\mathcal L }{ \partial \dot x_i } = \frac{ \partial\mathcal L }{ \partial x_i } = 0 \end{align*} Alternatively, one may also express this result in the form of momentum conservation (cf. Section 3.4.3), \begin{align*} \dot{\mathbf P} = \frac{\mathrm{d}}{\mathrm{d} t} \Bigl( m \: \dot{\mathbf x} \Bigr) = \frac{\mathrm{d}}{\mathrm{d} t} \nabla_{ \dot{\mathbf x} } \mathcal L = \nabla_{\mathbf x} \mathcal L = \mathbf 0 \end{align*} Momentum conservation therefore is an immediate consequence of the Euler-Lagrange equation for systems where the Lagrange function does not depend on the coordinates. The general case is summarized in the following definition:

Remark 6.9. The constant value of $C$ is determined by the initial conditions on $\dot q_i$ and on the other coordinates.

Based on the concepts introduced in problem 4.35, this finding generalizes as follows to a translation-invariant dynamics of $N$ particles

Proof. The kinetic energy of the dynamics is \begin{align*} T = \sum_{i=1}^N \frac{m_i}{2} \: \dot{\mathbf q}_i^2 \end{align*} and by assumption $V(\mathbf q, t)$ does not depend on $\dot{\mathbf q}$. Hence, \begin{align*} \sum_k \nabla_{\dot{\mathbf q}_k} \mathcal L = \sum_k \nabla_{\dot{\mathbf q}_k} T = \sum_i \frac{m_i}{2} \: \left( \sum_k \nabla_{\dot{\mathbf q}_k} \dot{\mathbf q}_i^2 \right) = \sum_{i=1}^N m_i \: \dot{\mathbf q}_i = \mathbf P \end{align*} Consequently, the time derivative of $\mathbf P$ amounts to \begin{align*} \frac{\mathrm{d} \mathbf P}{\mathrm{d} t} = \frac{\mathrm{d}}{\mathrm{d} t} \sum_k \nabla_{\dot{\mathbf q}_k} \mathcal L = \sum_k \nabla_{\mathbf q_k} \mathcal L = - \sum_{i<j} \left( \sum_k \nabla_{\mathbf q_k} \Phi_{ij}\bigl( \lvert \mathbf q_i - \mathbf q_j \rvert, t \bigr) \right) \end{align*} since $T$ does not involve the coordinates $\mathbf q_k$. Now, we observe that \begin{align*} \nabla_{\mathbf q_k} = \lvert \mathbf q_i - \mathbf q_j \rvert = \left\{ \begin{aligned} \frac{ \mathbf q_i - \mathbf q_j }{ \lvert \mathbf q_i - \mathbf q_j \rvert } \quad& \text{for}\quad k=i \\ \frac{ \mathbf q_j - \mathbf q_i }{ \lvert \mathbf q_i - \mathbf q_j \rvert } \quad& \text{for}\quad k=j \\ 0 \qquad & \text{else} \end{aligned} \right . \end{align*} We denote the derivative of $\Phi(d,t)$ with respect to its $d$ as $\Phi'(d,t)$ and observe that the sum over $k$ will have two non-vanishing contributions, \begin{align*} \frac{\mathrm{d} \mathbf P}{\mathrm{d} t} &= - \sum_{i<j} \left[ \frac{ \mathbf q_i - \mathbf q_j }{ \lvert \mathbf q_i - \mathbf q_j \rvert } \; \Phi_{ij}'\bigl( \lvert \mathbf q_i - \mathbf q_j \rvert, t \bigr) + \frac{ \mathbf q_j - \mathbf q_i }{ \lvert \mathbf q_i - \mathbf q_j \rvert } \; \Phi_{ij}'\bigl( \lvert \mathbf q_i - \mathbf q_j \rvert, t \bigr) \right] \\ &= \mathbf 0 \\ \end{align*} qed

A particle experiencing free flight is subject to a constant force, i.e., a force that is uniform in space and that does not depend on time. A common example is a particle of mass $m$ close to the Earth surface. The particle is subject to a constant acceleration $\mathbf g$ (Section 4.2) that gives rise to a potential \begin{align*} V(\mathbf q) &= -m\: \mathbf g \cdot \mathbf q \end{align*} For Cartesian coordinates $\mathbf q = (q_1,q_2,q_3)$ where the first coordinate axis is aligned antiparallel to $\mathbf g$ the Lagrange function takes the from \begin{align*} T &= \frac{m}{2} \: \Bigl( \dot q_1^2 + \dot q_2^2 + \dot q_3^2 \Bigr) \\ V &= m\,g\: q_1 \\ \mathcal L &= T-V = \frac{m}{2} \: \Bigl( \dot q_1^2 + \dot q_2^2 + \dot q_3^2 \Bigr) - m\,g\: q_1 \, . \end{align*} In this setting $q_2$ and $q_3$ are cyclic variables such that the corresponding components of the momentum, $\mathbf P = m \: \mathbf q$, are conserved, \begin{align*} \dot P_2 = \dot P_3 = 0 \, . \end{align*} Moreover, for the first component we find \begin{align*} m \: \ddot q_1 = \frac{\mathrm{d}}{\mathrm{d} t} \frac{ \partial\mathcal L }{ \partial \dot q_1 } = \frac{ \partial\mathcal L }{ \partial q_1 } = m\, g \end{align*} such that we recover the EOM Equation 4.2.1 with the solution, Equation 4.2.2 The impact of constant external forces on a set of $N$ particles is discussed in Problem 6.10.

A force $\mathbf F$ that is proportional to the distance $\mathbf q$ from some reference position is denoted as a harmonic, or Hookian spring force, $\mathbf F = -k \: \mathbf q$ (cf Remark 3.4). One readily checks that it derives from a potential \begin{align*} V = \frac{k}{2} \: \mathbf q^2 \end{align*} such that the Lagrange function for a particle of mass $m$ takes the form \begin{align*} \mathcal L &= \frac{m}{2} \: \dot{\mathbf q}^2 - \frac{k}{2} \: \mathbf q^2 \end{align*} Accordingly, the Euler-Lagrange equations provide the EOM \begin{align} \ddot{\mathbf q} = -\frac{k}{m} \: \mathbf q \,. \tag{6.4.1} \end{align} The equation for each component can be solved independently, providing \begin{align*} q_i(t) = A_i \: \sin\bigl( \alpha_i + \omega \, t \bigr) \qquad\text{with}\quad \omega = \sqrt{ \frac{k}{m} } \, . \end{align*} Here the amplitudes $A_i$ and the phase factors $\alpha_i$ are integration constants that must be chosen in accordance with the initial condition $\bigl(\mathbf q (t_0), \dot{\mathbf q}(t_0) \bigr)$. Even though we found an explicit solution is non-trivial to sort out how the trajectories will look like. One can gain additional insight into their shape by describing the motion in terms of spherical coordinates, Theorem 6.3, \begin{align*} \mathbf q &= R(t) \, \hat{\boldsymbol R}\bigl(\theta(t), \varphi(t)\bigr) \\ \dot{\mathbf q} &= \dot R(t) \: \hat{\boldsymbol R}\bigl(\theta(t), \varphi(t)\bigr) + R(t) \: \dot\theta(t) \: \hat{\boldsymbol \theta}\bigl(\theta(t), \varphi(t)\bigr) + R(t) \: \sin\theta(t) \: \dot\varphi(t) \: \hat{\boldsymbol \varphi}\bigl(\theta(t), \varphi(t)\bigr) \end{align*} such that \begin{align*} T &= \frac{m}{2} \; \left( \dot R^2(t) + R^2(t) \: \dot\theta^2(t) + R^2(t) \: \dot\varphi^2(t)\: \sin^2\theta(t) \right) \\ V &= \frac{k}{2} \: R^2(t) \end{align*} The potential is rotationally symmetric. It depends only on $R$, and neither on $\theta$ nor on $\varphi$. Moreover, the kinetic energy also does not involve $\varphi$. Hence, $\varphi$ is cyclic. It provides the conservation law \begin{align} C_\varphi = \frac{ \partial\mathcal L}{ \partial\dot\phi } = m \: R^2 \: \sin^2\theta(t) \: \dot\varphi \tag{6.4.2} \end{align} Moreover, the Euler-Lagrange equation for $\theta$ provides \begin{align} \nonumber m \: R^2(t) \: \ddot\theta + 2\,m \: R \: \dot R \: \dot\theta &= \frac{\mathrm{d}}{\mathrm{d} t} \left( m \: R^2 \: \dot\theta \right) = \frac{\mathrm{d}}{\mathrm{d} t} \frac{ \partial\mathcal L }{ \partial \dot\theta } \\ &= \frac{ \partial\mathcal L }{ \partial\theta } = m \: R^2(t) \: \dot\varphi^2(t)\: \sin\theta \: \cos\theta \tag{6.4.3} \end{align} This equation has a fixed point for $\theta = \pi/2$. We therefore choose the polar coordinates such that the initial position $\mathbf q$ and the initial velocity $\dot{\mathbf q}$ lie in the equatorial plane of the coordinate system, i.e., we require that $\theta(t_0) = \pi/2$ and that \begin{align*} 0 &= \hat{\boldsymbol \theta}\left( \frac{\pi}{2}, \varphi_0 \right) \cdot \dot{\mathbf q}(t_0) \\ &= \hat{\boldsymbol \theta}\left( \frac{\pi}{2}, \varphi_0 \right) \cdot \left[ \dot R(t_0) \: \hat{\boldsymbol R}\left( \frac{\pi}{2}, \varphi_0 \right) + R_0 \: \dot\theta(t_0) \: \hat{\boldsymbol \theta}\left( \frac{\pi}{2}, \varphi_0\right) + R_0 \: \dot\phi( t_0) \: \hat{\boldsymbol \varphi}\left(\frac{\pi}{2}, \varphi_0\right) \right] \\ &= R_0 \: \dot\theta(t_0) \\ \Rightarrow\quad& \dot\theta(t_0) = 0 \qquad\text{unless}\quad R_0 = 0 \, , \end{align*} and for for $R_0 =0$ one can still choose the coordinates such that $\dot\theta(t_0) = 0$. For this choice the angular momentum will take the constant value \begin{align*} \mathbf L = R \: \hat{\boldsymbol R} \times m\, \left( \dot R \: \hat{\boldsymbol R} + R\,\dot\varphi \:\hat{\boldsymbol \varphi} \right) = m\,R^2\: \dot\varphi \; \hat{\boldsymbol R} \times \hat{\boldsymbol \varphi} = C_\varphi \: \hat{\boldsymbol \theta} \end{align*}

Hence, Equation 6.4.3 expresses the conservation of the modulus of angular momentum, and the motion is confined to a plane due to the conservation of the direction of the angular momentum. This is in line with the analogous observation in the Kepler problem, Section 4.7.1. For $\theta=\pi/2$ and $\dot\theta=0$ the EOM for $R(t)$ takes the form \begin{align*} m \: \ddot R = \frac{\mathrm{d}}{\mathrm{d} t} \frac{ \partial\mathcal L }{ \partial \dot R } = \frac{ \partial\mathcal L }{ \partial R } = m \, R \: \dot\varphi^2 - k \, R \end{align*} We employ angular-momentum conservation, $C_\varphi = m\,R^2\:\dot\varphi$, to eliminate $\dot\varphi$ in this equation. Subsequently, we introduce the dimensionless time, $\tau = \omega \, (t-t_0)$, and distance, $r = R/R_c$, based on the frequency $\omega = \sqrt{k/m}$ and the length scale $R_c = \sqrt{C_\varphi/m\,\omega}$: \begin{align} \frac{\mathrm{d}^2 r}{\mathrm{d}\tau^2} = \frac{1}{r^3} - r \tag{6.4.4} \end{align} Multiplication by $2\, \dot r$ and integration provides \begin{align*} \dot r^2(\tau) - \dot r^2(0) &= \int_0^\tau \mathrm{d}\tau \: \frac{\mathrm{d} r^2}{\mathrm{d}\tau} = -\int_0^\tau \mathrm{d}\tau \: \frac{\mathrm{d} r}{\mathrm{d}\tau} \; \frac{\mathrm{d}}{\mathrm{d} r} \left( \frac{r^2}{2} + \frac{1}{2\,r^2} \right) \\ &= -\int_{r_0}^{r(\tau)} \mathrm{d} r \; \frac{\mathrm{d}}{\mathrm{d} r} \left( \frac{r^2}{2} + \frac{1}{2\,r^2} \right) = - \left( \frac{r^2}{2} + \frac{1}{2\,r^2} \right)_{\!\! r_0}^{\!\! r(\tau)} \end{align*} This entails conservation of the dimensionless energy, \begin{align} e = \frac{\dot r^2 }{2} + \frac{r^2}{2} + \frac{1}{2\,r^2} \tag{6.4.5} \end{align} This amounts to the EOM, Equation 6.4.4, of a particle at position $r(\tau)$ that follows a dynamics with energy conservation, Equation 6.4.5, where it has a dimensionless kinetic energy $\dot r^2/2$ and a potential energy $r^2/2 + r^{-2}/2$. Hence, we refer to it as effective potential for the $r$ dynamics.

Remark 6.10. If the system conserves energy, then $e$ will often amount to the non-dimensionalized energy.

Remark 6.11. In Section 6.3.2 and Section 6.3.4 we encountered settings where the angle $\varphi$ was controlled by an external motor such that the energy is not conserved (cf. Problem 6.7). In this case the constraint on $\varphi$ allows us to math the non-inertial dynamics of the rotating system into an autonomous dynamics with the effective potentials, Equation 6.3.5, Equation 6.3.7, respectively.

In the coordinate plane the trajectories of the harmonic oscillator that take the form of closed ellipses. This is most transparently confirmed by choosing the Cartesian coordinates $\mathbf q = (q_1,q_2,q_3)$ in Equation 6.4.1 such that $q_3$ vanishes and adopting a complex variable $z=q_1 + \mathrm{i}\, q_2$. The EOM will then take the form $\ddot z = -\omega^2 \, z$, and it has the general solution \begin{align*} z(t) = z_1 \: \mathrm{e}^{\mathrm{i}\,\omega (t-t_0)} + z_2 \: \mathrm{e}^{-\mathrm{i}\,\omega (t-t_0)} \end{align*} We express the complex integration constants $z_1$ and $z_2$ in terms of four real numbers $A$, $B$, $\alpha$, and $\beta$, \begin{align*} z_1 &= \frac{A+B}{2} \: \mathrm{e}^{\mathrm{i}\, (\beta+\alpha)} & z_2 &= \frac{A-B}{2} \: \mathrm{e}^{\mathrm{i}\, (\beta-\alpha)} \end{align*} and find

\begin{align} z(t) = \mathrm{e}^{\mathrm{i}\, \beta} \; \left[ A \: \cos\bigl(\alpha+\omega\, (t-t_0)\bigr) + \mathrm{i} \: B \: \sin\bigl(\alpha+\omega\, (t-t_0)\bigr) \right] \tag{6.4.6} \end{align}

The amplitudes $A$ and $B$ describe the length of the semi-axes of the ellipse. The angle $\beta$ rotates the ellipse, i.e., it describes the orientation of the axes. The angle $\alpha$ is a phase factor that establishes the position on the ellipse where the particle started at time $t_0$. A sketch of the resulting trajectory is provided in Figure 6.16. We observe that the trajectory keeps a minimum distance $B$ to the origin. For the harmonic potential this distance can be calculated explicitly by observing that $\dot R$ vanishes at the minimum distance $R_{\text{min}}$ such that energy conservation entails

\begin{align} \nonumber E &= \frac{1}{2\,m} \: \frac{ C_\varphi^2 }{ R_{\text{min}}^2 } + \frac{k}{2} \: R_{\text{min}}^2 \\ \Rightarrow\quad R_{\text{min}}^2 &= \frac{E}{k} \; \left( 1 - \sqrt{ 1-\left( \frac{C_\varphi}{k\,E} \right)^2 } \right) \tag{6.4.7} \end{align} Hence, the trajectory can go through the origin iff $C_\varphi=0$. This observation holds for every system with energy conservation and a potential that obeys cylinder symmetry¹⁾.

Proof. We adopt cylinder coordinates $(R,\varphi,z)$ with a symmetry axis $\hat{\boldsymbol z}$ aligned along $\hat{\boldsymbol n}$. Then the kinetic energy and $L_n$ amount to \begin{align*} T &= \frac{m}{2} \: \left( \dot z^2 + R^2 \: \dot\varphi^2 \right) \\ C_\varphi &= m \: R^2 \: \dot\varphi \end{align*} As a consequence, we have \begin{align*} E - \varphi_{\text{min}} &\geq E - U(R, z) = T \geq \frac{m}{2} \:R^2 \: \dot\varphi^2 = \frac{C_\varphi^2}{2m\:R^2} \end{align*} The bound is obtained by solving this inequality for $R$. qed

Remark 6.12. Equation 6.4.8 implies that also in general the rotation axis can only be approached for $C_\varphi=0$.

Remark 6.13. It is not necessary that the potential $U(R, z)$ has a minimum for $R=0$. The argument works in exactly the same way when it has a maximum.²⁾. An example is provided in Problem 6.12.

Remark 6.14. Energy conservation always provides an implicit equation for $R_{\text{min}}$. In general this equation can not be solved for $R_{\te