# Lagrangian Mechanics

Lagrangian mechanics is a reformulation of classical mechanics introduced by Joseph-Louis Lagrange in 1788. Here we discuss the use of generalized co-ordinates and the connection to Newtonian mechanics and Hamilton's principle.

# Statement

A system described by generalized coordinates $\ q_1, q_2, ..., q_n$ obeys Lagrange's equation

${\partial{L}\over \partial q_i} - {\mathrm{d} \over \mathrm{d}t}\left({\partial{L}\over \partial{\dot{q_i}}}\right) = 0.$ for all $\ i = 1, 2, ..., n$

where $\ L = L (t, q_1, q_2, ..., q_n, \dot{q_1}, \dot{q_2}, ..., \dot{q_n})$ is the Lagrangian of the system.

We may write $\ (t, q, \dot q)$ in place of $\ (t, q_1, q_2, ..., q_n, \dot{q_1}, \dot{q_2}, ..., \dot{q_n})$.

# Generalized coordinates

The state of a system of $\ N$ particles in 3D space can be uniquely defined using no more than $\ 3N$ independent quantities. The number of such quantities required is called the number of degrees of freedom. Any $\ n$ such quantities $\ q_1, q_2, ..., q_n$ are called generalized co-ordinates, and $\dot{q_1}, \dot{q_2}, ..., \dot{q_n}$ generalized velocities. These quantities are not necessarily Cartesian.

# Conservation laws

Consider a system of $n$ particles, and let $\ \vec q_i$ be the $i^{th}$ particle’s position in space, and let $\ \dot{\vec q_i} = \frac{d \vec q_i}{dt}$ be its velocity. We define the generalized momentum of the $i^{th}$ particle as

$\ \vec p_i = {\partial{L}\over \partial{\dot{\vec q_i}}} = \left( \frac{\partial L}{\partial \dot{q_{ix}}}, \frac{\partial L}{\partial \dot{q_{iy}}}, \frac{\partial L}{\partial \dot{q_{iz}}} \right)$

and the generalized force acting on the $i^{th}$ particle as

$\ \vec F_i = {\partial L\over \partial \vec q_i} = \left( \frac{\partial L}{\partial q_{ix}}, \frac{\partial L}{\partial q_{iy}}, \frac{\partial L}{\partial q_{iz}} \right)$

In this notation, Lagrange's equations are written

$\dot{\vec p_i} = \vec F_i, \quad i = 1,...,n$

We now discuss the symmetries and conservation laws implied by Lagrange's equation. A symmetry refers to a family of transformations $\ t \longmapsto t ^\prime, \vec q_i \longmapsto {\vec{q_i}}^{\prime}$ that leads to zero first-order changes in the Lagrangian. Three standard symmetries in classical mechanics are

1. Invariance under translations of time $\Rightarrow$ the Lagrangian has no explicit time dependence

2. Invariance under translations of space $\Rightarrow$ the Lagrangian does not depend on $\ \vec q_i$ for all $i=1,...,n$.

3. Invariance under rotations about an axis

Each one implies a conservation law: a conserved quantity or an integral of Lagrange’s equations can be identified.

Proposition 1 (Conservation of energy): Suppose the Lagrangian $\ L(t,\vec q_1,...,\vec q_n,\dot{\vec q_1},...,\dot{\vec q_n})$ is invariant under the transformations $\ t \longmapsto t + \epsilon$. Then the Hamiltonian of the system is conserved.

Proof: Since the Lagrangian is invariant under translations of time, it has no explicit dependence on time. So $\ \frac{\partial L}{\partial t} = 0$

Since the Hamiltonian, or the energy of the system, is defined by

$H = \sum_{i=1}^n \dot{\vec q_i} \cdot \frac{\partial L}{\partial \dot{\vec q_i}} - L$

it follows that

\begin{align} \frac{\partial H}{\partial t} & = \sum_{i=1}^n \ddot{\vec{q_i}} \cdot \frac{\partial L}{\partial \dot{\vec{q_i}}} + \sum_{i=1}^n \dot{\vec{q_i}} \cdot \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\vec{q_i}}} \right) - \frac{\partial L}{\partial t} \\ & = \sum_{i=1}^n \ddot{\vec{q_i}} \cdot \frac{\partial L}{\partial \dot{\vec{q_i}}} + \sum_{i=1}^n \dot{\vec{q_i}} \cdot \frac{\partial L}{\partial \vec{q_i}} - \frac{\partial L}{\partial t} \\ & = \frac{\partial L}{\partial t} - \frac{\partial L}{\partial t} \\ & = 0-0 \\ & = 0 \\ \end{align}

Therefore, $\ H$ is conserved. When $\ L = T - V$, where

$\ T = \sum_{i=1}^n \frac{1}{2}m_i \dot{\vec q_i}^2$ is the kinetic energy and
$\ V = V(\vec q_1,...,\vec q_n)$ is the potential energy,
( $\dot{\vec q_i}^2 = \dot{\vec q_i} \cdot \dot{\vec q_i}$ )

then $\ \vec p_i = m_i \dot{\vec q_i}$ and

$\ H = \sum_{i=1}^n m_i \dot{\vec q_i}^2 - L = 2T - (T - V) = T + V$.

In this case, the total energy is conserved.

Proposition 2 (Conservation of Momentum): Suppose the Lagrangian is invariant under the transformations $\ \vec{q_i} \longmapsto \vec{q_i} + \vec a$, where $i \in {1,...,n}$. Then the generalized momentum $\ \vec p_i$ is conserved.

Proof: Since the Lagrangian does not depend on $\ \vec q_i$, then

$\ \vec 0 = \frac{\partial L}{\partial \vec{q_i}} = \dot{\vec{p_i}}$

Thus, $\ p_i$ is conserved.

Corollary 1: If the Lagrangian is invariant under the transformation $\ \vec{q_i} \longmapsto \vec{q_i} + \vec a$, for all $i \in {1,...,n}$, then the total momentum $\ \vec{P} = \textstyle \sum_{i=1}^n \vec{p_i}$ is conserved.

This follows immediately from Proposition 2.

Corollary 2: If the Lagrangian is invariant under the transformation $\ q_{iz} \longmapsto q_{iz} + a$, then the z-component of the generalized momentum, i.e. $\ p_{iz}$, is conserved.

The proof is the same as that of Proposition 2, with the subscript $i$ replaced by $iz$, so we obtain the result from Lagrange's equation for the $z$-coordinate. The statement in Corollary 2 also holds if we replace $z$ with $x$ or $y$ instead. A coordinate on which the Lagrangian has no explicit dependence is called cyclic. In general, if a coordinate is cyclic, then the generalized momentum corresponding to that coordinate is conserved.

Proposition 3 (Conservation of Angular Momentum): Suppose that the Lagrangian is invariant under an infinitesimal rotation of the system. Then the component of angular momentum along the axis of rotation is conserved.

Proof: Let the vector $\ \delta \vec{\phi}$ represent the infinitesimal rotation, whose magnitude is the angle of rotation $\ \delta \phi$, and whose direction is that of the axis of rotation, so the direction of rotation is that of a right-handed screw driven along $\ \delta \vec{\phi}$. Let $\ \delta \vec{q_i}$ denote the resulting change in the position $\ \vec q_i$ of the $i^{th}$ particle in the system undergoing rotation. Let $\ \theta$ be the angle that $\ \vec{q_i}$ makes with $\ \delta \vec{\phi}$. Then

$\ | \delta \vec{r} | = \delta \phi | \vec{q_i} | \sin \theta$.

The direction of $\ \delta \vec{q_i}$ is perpendicular to the plane spanned by $\ \vec{q_i}$ and $\ \delta \vec{\phi}$. Hence

$\ \delta \vec{q_i} = \delta \vec{\phi} \times \vec{q_i}$

When the system rotates, the velocities of the particles also change direction. The velocity increment relative to a fixed system of coordinates is

$\ \delta \dot{\vec{q_i}} = \delta \vec{\phi} \times \dot{\vec{q_i}}$

Since the Lagrangian is unchanged by the rotation,

\ \begin{align} 0 & = \delta L \\ & = \sum_{i=1}^n \left( \frac{\partial L}{\partial \vec{q_i}} \cdot \delta \vec{q_i} + \frac{\partial L}{\partial \dot{\vec{q_i} } } \cdot \delta \dot{\vec{q_i}} \right) \\ & = \sum_{i=1}^n \left( \dot{\vec{p_i}} \cdot \left( \delta \vec{\phi} \times \vec{q_i} \right) + \vec{p_i} \cdot \left( \delta \vec{\phi} \times \dot{\vec{q_i}} \right) \right) \\ \end{align}

Permuting factors and taking $\ \delta \vec{\phi}$ outside the sum,

\begin{align} 0 & = \delta \vec{\phi} \cdot \sum_{i=1}^n \left( \vec{q_i} \times \dot{\vec{p_i}} + \dot{\vec{q_i}} \times \vec{p_i} \right) \\ & = \delta \vec{\phi} \cdot \frac{d}{dt} \left( \sum_{i=1}^n \vec{q_i} \times \vec{p_i} \right) \end{align}

$\ \delta \vec{\phi}$ does not change with time, so

$\ \frac{d}{dt} \left( \delta \vec{\phi} \cdot \sum_{i=1}^n \vec{q_i} \times \vec{p_i} \right) = 0$

Since the angular momentum of the system is the vector

$\ \vec M = \sum_{i=1}^n \vec{q_i} \times \vec{p_i}$

this implies that the component of angular momentum along $\ \delta \vec{\phi}$, i.e. the axis of rotation, is conserved.

Corollary 3: If the Lagrangian is invariant under any rotation of the system, then the angular momentum $\ \vec M$ is conserved.

Proof: As shown in the proof of Proposition 3, we have

$\ \delta \vec{\phi} \cdot \frac{d \vec M}{dt} = 0$

for arbitrary $\ \delta \vec{\phi}$. Therefore,

$\ \frac{d \vec M}{dt} = \vec 0$

Remark: Propositions 2 and 3 are actually consequences of a more general result called Noether's Theorem. Thus, we can describe conservation of momentum and of angular momentum as being consequences of the invariance of the action functional under a family of spatial translations and a family of rotations, respectively. The idea of Noether's Theorem is to find a conserved quantity given any symmetry. Thus, once we know the symmetries of a given problem, we can easily find conserved quantities.

References:

• Calculus of Variations by I.M. Gelfand & S.V. Fomin (translated and edited by Richard A. Silverman), 1991.
• Introductory Classical Mechanics with Problems and Solutions, by David Morin, 2004.
• Course of Theoretical Physics Vol. 1: Mechanics, 3rd ed., by Landau and Lifshitz; first published in 1960.

To be completed by Alex.Baiyun 23:19, 21 May 2010 (UTC)

# The Lagrangian for a test particle in a Newtonian system

Consider a test particle of mass $\ m$ placed inside a Newtonian system. Then, the Lagrangian of the particle in Cartesian co-ordinates $\vec x = (x_1, x_2, x_3)$ is given by

$\ L = T - V$

where

$\ T = \frac 12 m \dot{\vec{x}} \cdot \dot{\vec{x}}$ is the kinetic energy of the particle, and
$\ V = V(\vec x)$ is the potential energy.

Hence,

${\partial{L}\over \partial x_i} = - {\partial V\over \partial x_i}$ for all $\ i=1,2,3$

and

${\partial{L}\over \partial{\dot{x_i}}} = m \dot x_i$ for all $\ i=1,2,3$

Substituting into Lagrange's equation, we obtain

$\ m \ddot x_i + {\partial V \over \partial x_i} = 0$ for all $\ i=1,2,3$

or

$\ m \ddot{\vec x} + \vec{\nabla} V = 0$

This qualifies Lagrangian mechanics as a reformulation of Newtonian mechanics.

# Hamilton's principle

For a given system, we define the action from $\ t_1$ to $\ t_2$ as

$\ S = \int_{t_1}^{t_2} L (t, q, \dot q) dt$

with conditions $\ q(t_1)=q^{(1)}$ and $\ q(t_2)=q^{(2)}$.

Hamilton's principle states that the evolution $\ q$ of the system from $\ t_1$ to $\ t_2$ is an extremum of $\ S$. In other words, δS = 0.

Consequently, from the Euler-Lagrange equation, we obtain

${\partial{L}\over \partial q_i} - {\mathrm{d} \over \mathrm{d}t}{\partial{L}\over \partial{\dot{q_i}}} = 0.$ for all $\ i = 1, 2, ..., n$

as expected.

This principle is also called the principle of least action.