Analytical mechanics
Newtonian mechanics
A conserved force is the gradient of a scalar function: $\mathbf{F}=-\nabla V(\mathbf{x})$.
Energy of a system only with conserved forces is conserved:
\[\begin{equation} E = \frac{m}{2}\left(\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}\right)^2 + V(\mathbf{x}),\ \frac{\mathrm{d}E}{\mathrm{d}t} = 0. \tag{eq:NE} \end{equation}\]Any one-dimensional force F(x) only explicitly dependent on x is conserved:
\[\begin{equation} V(x) = -\int^x F(\xi) \mathrm{d}\xi. \tag{eq:NP} \end{equation}\]Lagrangian formalism
State of the system is described by generalized coordinates ${q_i} (i = 1,\cdots,N)$ in the configuration space M (a manifold). Generalized velocity is $\dot{q_i}$.
Principle of the least action
Integral form
The action is a functional from the trajectory $q(t)$ to a real number, defined by the integral of the Lagrangian from the initial to the final state of the system:
\[\begin{equation} S[q(t), \dot{q}(t)] = \int_{t_i}^{t_f} L(q,\dot{q}) \mathrm{d}t. \tag{eq:HA} \end{equation}\]Hamilton’s principle/principle of the least action: The physically realized trajectory corresponds to an extremum of the action.
Variation form
Suppose q(t) is a path realizing an extremum of S. Consider a variation δq(t) of the trajectory such that $\delta q(t_i) = \delta q(t_f) = 0$. Under this variation,
\[\begin{equation} \begin{split} \delta S &= \int_{t_i}^{t_f} L(q+\delta q, \dot{q}+\delta \dot{q}) - L(q, \dot{q}) \mathrm{d}t\\ &= \int_{t_i}^{t_f}\frac{\partial L}{\partial q}\delta q\mathrm{d}t+\int_{t_i}^{t_f}\frac{\partial L}{\partial \dot{q}}\delta \dot{q}\mathrm{d}t + \mathcal{O}(\delta q^2) + \mathcal{O}(\delta \dot{q}^2)\\ &= \int_{t_i}^{t_f}\frac{\partial L}{\partial q}\delta q\mathrm{d}t+\int_{t_i}^{t_f}\frac{\partial L}{\partial \dot{q}}\mathrm{d}\delta q + \mathcal{O}(\delta q^2)\\ &= \int_{t_i}^{t_f}\frac{\partial L}{\partial q}\delta q\mathrm{d}t-\int_{t_i}^{t_f}\delta q\mathrm{d}\frac{\partial L}{\partial \dot{q}} + \mathcal{O}(\delta q^2)\\ &= \int_{t_i}^{t_f}\delta q\ \mathrm{d}t\ \left(\frac{\partial L}{\partial q}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}}\right) + \mathcal{O}(\delta q^2)\\ &\approx \int_{t_i}^{t_f}\delta q\ \mathrm{d}t\ \left(\frac{\partial L}{\partial q}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}}\right). \end{split} \tag{eq:PLA} \end{equation}\]Euler-Lagrange equation
q(t) is an extremum, so $\delta S=0\ \forall\ \delta q$, which implies
\[\begin{equation} \frac{\partial L}{\partial q_k}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}_k}=0\quad \text{for }k=1,\cdots,N, \tag{eq:ELE} \end{equation}\]which is the Euler-Lagrange equation.
Using generalized momentum $p=\dfrac{\partial L}{\partial \dot{q}}$,
\[\begin{equation} \frac{\partial L}{\partial q_k}=\frac{\mathrm{d}p_k}{\mathrm{d}t}. \tag{eq:ELEk} \end{equation}\]Let $L=\frac{1}{2}m\dot{q}^2-V(q)$, then (eq:ELEk) reduces to $p = m\dot{q},\ m\ddot{q}+\frac{\partial V}{\partial q}=0$.
Functional derivative
Using a single degree of freedom s or t, the parameter of the configuration trajectory, a functional derivative, considering the variation from the stationary path q(s) to another path q(t), can be defined by
\[\begin{equation} \begin{split} \frac{\delta S[q,\dot{q}]}{\delta q(s)}: &=\lim_{\varepsilon\to 0}\frac{1}{\varepsilon} \left\{S[q(t),\dot{q}(t)]-S[q(s),\dot{q}(s)]\right\}\\ &=\lim_{\varepsilon\to 0}\frac{1}{\varepsilon} \left\{ S[q(s)+\varepsilon\delta(t-s), \dot{q}(s)+\varepsilon\frac{\mathrm{d}} {\mathrm{d}t}\delta(t-s)] -S[q(s),\dot{q}(s)] \right\}, \end{split} \tag{eq:FD} \end{equation}\]where $\varepsilon\delta(t-s)$ is a parametric expression of $\delta q$. $\varepsilon$ can be thought of a distance in configuration space between the two trajectories q(s), q(t), where they have parameter values of s and t, and $\delta(t-s)$ the unit vector pointing from point q(s) to point q(t). Note that the s and t here represent not a single value of the parameter (time), but the whole process of them ranging from the initial to the final state.
Therefore, substituting $\delta q=\varepsilon\delta(t-s)$ into (eq:PLA) yields
\[\begin{equation} \begin{split} \delta S &= \int_{t_i}^{t_f}\mathrm{d}t\ \varepsilon\ \left( \frac{\partial L}{\partial q}(s) -\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{q}}(s) \right) \delta(t-s) + \mathcal{O}(\varepsilon^2). \end{split} \tag{eq:PPLA} \end{equation}\]In this flavor the Euler-Lagrangian Equation may be written as
\[\begin{equation} \frac{\partial L}{\partial q}(s) = \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial\dot{q}}(s) \tag{eq:PELE} \end{equation}\]Symmetry of Lagrangian
Cyclic coordinate and momentum conservation
If Lagrangian L is independent of $q_k$ (but not $\dot{q}_k$), then $q_k$ is called cyclic. The conjugating momentum is conserved, according to (eq:PELE).
Symmetry
Infinitesimal symmetry operation: $q_k(t)\to q_k(t)+\delta q_k(t)$ on the path $q_k(t)$ from $t_i$ till $t_f$. Since L is independent of $q_k$, this Lagrangian should remain constant under the operation, at any point of the path.
For a particle in a spherically symmetric potential, the longitude angle $\phi$ is a cyclic coordinate. The corresponding conservative momentum is the angular momentum around the z axis.
Remarks:
- $L$ is equivalent to $L+\frac{\mathrm{d}Q}{\mathrm{d}t}\ \forall\ Q=Q(q)$.
- Newtonian mechanics is realized as an extremum of the action, but the action itself is defined on any path in the configuration space.
Hamiltonian formalism
Hamiltonian is the Legendre transformation of the Lagrangian with respect to $\dot{q}$, into the phase space $(q_k,p_k)$.
\[\tag{eq:Hd} H(q,p):=\sum_k p_k\dot{q}_k - L(q,\dot{q}),\]This transformation requires the Jacobian determinant to be non-zero.
\[\color{red}\det{\mathcal{J}} = \det{\frac{\partial p_i}{\partial \dot{q}_j}} = \det{\frac{\partial^2 L}{\partial \dot{q}_j\partial\dot{q}_i}} \neq 0. \tag{eq:detJ}\]For Lagrangians without cross terms between the velocities like $L = \frac{1}{2}\sum_i{m_i\dot{q}_i^2}-V$, (eq:detJ) is simply $m_i\neq 0$.
An intuitive example is:
$L=\sum_k\frac{1}{2}m\dot{q}_k^2-V(q) = T-V$
$L+H = \sum_k p_k\dot{q_k}=\sum_k m\dot{q}_k^2 = 2T$
$H = \sum_k\frac{1}{2}m\dot{q}_k^2+V(q) = T+V$
Under an infinitesimal variation of $q_k$ and $p_k$,
\[\begin{equation} \begin{split} \delta H &= \sum_k\left[ \delta p_k\dot{q}_k + p_k\delta\dot{q}_k - \frac{\partial L}{\partial q_k}\delta q_k - \frac{\partial L}{\partial \dot{q}_k}\delta \dot{q}_k \right]\\ &= \sum_k\left[ \delta p_k\dot{q}_k - \frac{\partial L}{\partial q_k}\delta q_k \right], \end{split} \tag{eq:delH} \end{equation}\]corresponding terms
\[\frac{\partial H}{\partial p_k} = \dot{q}_k,\quad \frac{\partial H}{\partial q_k} = -\frac{\partial L}{\partial q_k}. \tag{eq:HE1}\]Substituting the Euler-Lagrange equation, we get
\[\frac{\partial H}{\partial p_k} = \dot{q}_k,\quad \frac{\partial H}{\partial q_k} = -\dot{p}_k, \tag{eq:HE2}\]Substituting $L = \frac{1}{2}m\dot{\mathbf{q}}^2-V{\dot{\mathbf{q}}}$, then (eq:HE2) correspond to definition of momentum and Newton’s equation respectively.
Poisson bracket
\[[A,B] := \sum_k\left( \frac{\partial A}{\partial q_k}\frac{\partial B}{\partial p_k} -\frac{\partial A}{\partial p_k}\frac{\partial B}{\partial q_k}. \right) \tag{eq:PB}\]The Poisson bracket is a Lie bracket, namely it is linear, antisymmetric, and Jacobi identical.
Jacobi identity: (see for Landau’s book)
\[[[A,B],C] + [[C,A],B] + [[B,C],A] = 0. \tag{eq:JI}\]Fundamental Poisson brackets
\[[p_i,p_j]=[q_i,q_j]=0,\quad [q_i,p_j]=\delta_{ij}. \tag{eq:FPB}\]Using Poisson brackets, the time development of $A(q,p)$ can be written as
\[\frac{\mathrm{d}A}{\mathrm{d}t}=[A,H]. \tag{eq:AP}\]From (eq:AP) :
- $[A,H]=0$ implies that A is conserved.
- Hamilton’s equations can be written as
Noether’s theorem
Suppose the Hamiltonian of a system is invariant under an infinitesimal coordinate transformation $q_k\to q’_k=q_k+\varepsilon f_k(q)$, then
\[Q=\sum_k p_kf_k(q) \tag{eq:NT}\]is conserved.
Proof
The coordinate transform yields
\[\begin{equation} \begin{split} \frac{\partial q_i'}{\partial q_j} &= \frac{\partial}{\partial q_j}\left[q_i+\varepsilon f_i(q)\right] = \delta_{ij} + \varepsilon\frac{\partial f_i(q)}{\partial q_j} + \mathcal{O}(\varepsilon ^2),\\ \frac{\partial q_i}{\partial q_j'} &= \frac{\partial}{\partial q_j'}\left[q_i'-\varepsilon f_i(q)\right] = \delta_{ij} - \varepsilon\frac{\partial f_i(q)}{\partial q_j'} + \mathcal{O}(\varepsilon ^2), \end{split} \tag{eq:Lij} \end{equation}\]where
\[\begin{equation} \begin{split} \frac{\partial f_i(q)}{\partial q_j'} &=\sum_k \frac{\partial q_k}{\partial q_j'} \frac{\partial f_i(q)}{\partial q_k}\\ &=\sum_k\left[ \delta_{kj} - \varepsilon\frac{\partial f_k(q)}{\partial q_j'} + \mathcal{O}(\varepsilon ^2) \right] \frac{\partial f_i(q)}{\partial q_k}\\ &=\frac{\partial f_i(q)}{\partial q_j} + \mathcal{O}(\varepsilon), \end{split} \tag{eq:Lji0} \end{equation}\]so
\[\frac{\partial q_i}{\partial q_j'} = \frac{\partial}{\partial q_j'}\left[q_i'-\varepsilon f_i(q)\right] = \delta_{ij} - \varepsilon\frac{\partial f_i(q)}{\partial q_j} + \mathcal{O}(\varepsilon ^2). \tag{eq:Lji1}\]Using (eq:Lji1), since $H$ is invariant, the momentum transforms as
\[\begin{equation} \begin{split} p_k\to p_k' &= -\frac{\partial H}{\partial q_k'} = -\sum_l \frac{\partial q_l}{\partial q'_k} \frac{\partial H}{\partial q_l}\\ &= -\sum_l \left[ \delta_{kl} - \varepsilon\frac{\partial f_l(q)}{\partial q_k} + \mathcal{O}(\varepsilon ^2) \right] \frac{\partial H}{\partial q_l}\\ &= -\frac{\partial H}{\partial q_k} + \varepsilon\frac{\partial f_l(q)}{\partial q_k} \frac{\partial H}{\partial q_l} + \mathcal{O}(\varepsilon ^2)\\ &= p_k - \varepsilon\frac{\partial f_l(q)}{\partial q_k} q_l + \mathcal{O}(\varepsilon ^2). \end{split} \tag{eq:MT} \end{equation}\]The Hamiltonian is invariant implies that:
\[\begin{equation} \begin{split} 0 &=H(q_k',p_k')-H(q_k,p_k)\\ &=\sum_k\left[ \frac{\partial H}{\partial q_k} \varepsilon f_k(q) - \frac{\partial H}{\partial p_k} \varepsilon \sum_j p_j\frac{\partial f_j(q)}{\partial q_k} \right]\\ &=\varepsilon \sum_k\left[ \frac{\partial H}{\partial q_k} \frac{\partial Q}{\partial p_k} - \frac{\partial H}{\partial p_k} \frac{\partial Q}{\partial q_k} \right]\\ &= [H, Q] = \frac{\mathrm{d}Q}{\mathrm{d}t}. \end{split} \tag{eq:HQ} \end{equation}\]which shows that Q is conserved.$\qquad\square$
Noether’s theorem shows that to find a conserved quantity is equivalent to finding a transformation which leaves the Hamiltonian invariant.
Generator of the transform
Since $q_i$ and $p_k$ between each other, $\frac{\partial q_i}{\partial p_k}=0$, the Poisson bracket
\[[q_i, Q] = \sum_k\left[ \frac{\partial q_i}{\partial q_k} \frac{\partial Q}{\partial p_k} - \frac{\partial q_i}{\partial p_k} \frac{\partial Q}{\partial q_k} \right] = \sum_k\delta_{ik} f_k = f_i(q), \tag{eq:qiQ}\]which shows that $\delta q_i = \varepsilon f_i(q)=\varepsilon[q_i, Q]$. In this sense $Q$ is called the generator of the transformation $\delta q_i$.
| System | 1D particle | 2D particle |
|---|---|---|
| Lagrangian | $L = \frac{m\dot{r}^2}{2}$ | $L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-V(r)$ |
| Hamiltonian | $H = \frac{p^2}{2m}$ | $H=\frac{p_r^2}{2m}+\frac{p_{\theta}^2}{2mr^2}+V(r)$ |
| Symmetry | Translation | Rotation around $z$ |
| Transform of q | $r\to r + \varepsilon$ | $\theta\to \theta+\varepsilon$ |
| Transform of p | $p\to p$ | $p_{\theta}\to p_{\theta}$ |
| Generator Q | $p$ | $p_{\theta}=mr^2\dot{\theta}$ |
| Conservation law | Momentum | Angular momentum around $z$ |
References
This article is based on:
- Nakahara, Mikio. Geometry, Topology, and Physics. 2nd ed. Graduate Student Series in Physics. Bristol ; Philadelphia: Institute of Physics Publishing, 2003.
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