Lagrangian Mechanics: Hamilton’s Action Principle

Newtonian formulation of classical mechanics in terms of forces, while a foundational pillar, often confronts significant analytical hurdles in systems with intricate constraints or non-Cartesian coordinates. Its vectorial foundation, though immensely successful in many domains, can lead to intractable equations when grappling with complex geometries or coupled motions.

Analytical mechanics, specifically the Lagrangian formalism, represents a profound shift in perspective. The focus transitions from localized, vectorial forces to global, scalar quantities: the kinetic and potential energies of a system. This isn’t merely a notational convenience; it’s a re-conceptualization. Instead of painstakingly calculating every constraint force and vector component at each instant, we instead consider the system’s trajectory as a whole, seeking the path that extremizes a scalar functional—the action—over time. This viewpoint proves remarkably powerful, particularly for systems with many degrees of freedom.

One of the most compelling advantages of the Lagrangian approach lies in its elegant assimilation of constraints. By judiciously selecting a set of independent generalized coordinates, the inherent geometric restrictions on the system’s motion are implicitly woven into the Lagrangian’s very definition. This obviates the need to explicitly determine often non-conservative and unwieldy constraint forces, dramatically streamlining the mathematical treatment. Furthermore, the scalar nature of the Lagrangian simplifies derivations in curvilinear coordinate systems, relying on straightforward partial derivative calculations rather than cumbersome vector transformations. Beyond its utility in classical mechanics, the Lagrangian, and its close counterpart, Hamiltonian dynamics, are indispensable across the spectrum of theoretical physics. This analytical framework provides the essential mathematical language for understanding relativistic mechanics, general relativity, quantum mechanics, and quantum field theories etc. To truly navigate the landscape of modern physics, a firm grasp of the analytical mechanics paradigm is not merely beneficial, but foundational.

Constraints, Generalized Coordinates, and D’Alembert’s Principle

Newtonian mechanics, while a foundational pillar, often confronts significant analytical hurdles in systems with intricate constraints. To fully specify the configuration of $\displaystyle N$ particles in three-dimensional space, one typically requires $\displaystyle 3N$ Cartesian coordinates. However, in physical reality, systems are rarely unconstrained. Consider a particle sliding on a curved surface, a pendulum swinging with a fixed rod length, or the intricate motion of a rigid body; all these scenarios involve geometric restrictions that reduce the number of truly independent variables. This minimum set of independent coordinates, essential for describing the system’s configuration, defines its degrees of freedom.

Lagrangian mechanics inherently accommodates this reality by employing generalized coordinates $\displaystyle q_i$ and their corresponding generalized velocities $\displaystyle \dot{q}_i$ . These coordinates possess remarkable flexibility: they are not restricted to Cartesian axes, can mix disparate types (e.g., angles for rotational motion and lengths for translational motion), and need not even share the same physical dimensions. This inherent adaptability is a cornerstone of the Lagrangian method, allowing the system’s inherent geometry and constraints to be seamlessly incorporated into the very definition of the coordinate system. This effectively obviates the need for explicit calculations of constraint forces, which often prove cumbersome and analytically intractable in the Newtonian framework. For instance, a particle confined to move on the surface of a sphere can be described by just two angular coordinates (latitude and longitude), rather than three Cartesian coordinates coupled with an explicit algebraic equation defining the spherical surface.

The elegant transition from Newtonian mechanics to the Lagrangian framework for constrained systems is achieved through D’Alembert’s Principle. Consider a single particle acted upon by a total force $\displaystyle \mathbf {F}_{\text{total}}$ . By Newton’s second law, $\displaystyle \mathbf{F}_{\text{total}} = m \ddot{\mathbf{r}}$ .

We can decompose the total force into two components: the applied force $\displaystyle \mathbf{F}$ (forces explicitly acting on the particle, such as gravity or an external push), and the constraint force $\displaystyle \mathbf{F}_c$ (the force exerted by the constraint to maintain the particle’s restricted motion, e.g., normal force or tension). Thus, $\displaystyle \mathbf{F} + \mathbf{F}_c = m \ddot{\mathbf{r}}$ . Rearranging, $\displaystyle (\mathbf{F} - m \ddot{\mathbf{r}}) + \mathbf{F}_c = 0$ . This holds for each particle. Summing over all $\displaystyle N$ particles: $\displaystyle \sum_{k=1}^{N} (\mathbf{F}k - m_k \ddot{\mathbf{r}_k}) + \sum_{k=1}^{N} \mathbf{F}_{c,k} = 0$ . D’Alembert’s Principle emerges from the observation that for holonomic constraints, the constraint force $\displaystyle \mathbf{F}_{c,k}$ is always perpendicular to any virtual displacement $\displaystyle \delta \mathbf{r}_k$ consistent with the constraint. Hence, $\displaystyle \mathbf {F}_{c,k} \cdot \delta \mathbf{r}_k = 0 \quad \Rightarrow \quad \sum_{k=1}^{N} \mathbf{F}_{c,k} \cdot \delta \mathbf{r}_k = 0$ . Multiplying the original equation by $\displaystyle \delta \mathbf{r}_k$ and summing over all particles:
$\displaystyle \sum_{k=1}^{N} (\mathbf{F}_k - m_k \ddot{\mathbf{r}}k) \cdot \delta \mathbf{r}_k + \sum_{k=1}^{N} \mathbf{F}_{c,k} \cdot \delta \mathbf{r}_k = 0$ .

Since the second sum vanishes, we obtain D’Alembert’s Principle:

$\displaystyle \sum_{k=1}^{N} (\mathbf{F}_k - m_k \ddot{\mathbf{r}}_k) \cdot \delta \mathbf{r}_k = 0$ .

Now, express the Cartesian coordinates $\displaystyle \mathbf{r}_k$ in terms of generalized coordinates $\displaystyle q_j$ . Let $\displaystyle \mathbf{r}_k = \mathbf{r}_k(q_1, \ldots, q_n, t)$ . Then the virtual displacement is:

$\displaystyle \delta \mathbf{r}_k = \sum_{j=1}^{n} \frac{\partial \mathbf{r}_k}{\partial q_j} \delta q_j$ .

Substituting into D’Alembert’s principle:

$\displaystyle \sum_{k=1}^{N} (\mathbf{F}_k - m_k \ddot{\mathbf{r}}k) \cdot \left( \sum_{j=1}^{n} \frac{\partial \mathbf{r}_k}{\partial q_j} \delta q_j \right) = 0$ .

Interchanging sums:

$\displaystyle \sum_{j=1}^{n} \left[ \sum_{k=1}^{N} \mathbf{F}_k \cdot \frac{\partial \mathbf{r}_k}{\partial q_j} - \sum_{k=1}^{N} m_k \ddot{\mathbf{r}}_k \cdot \frac{\partial \mathbf{r}_k}{\partial q_j} \right] \delta q_j = 0$ .

Define the generalized force: $\displaystyle Q_j = \sum_{k=1}^{N} \mathbf{F}_k \cdot \frac{\partial \mathbf{r}_k}{\partial q_j}$

and the kinetic energy: $\displaystyle T = \frac{1}{2} \sum_{k=1}^{N} m_k \dot{\mathbf{r}}_k^2$ .

It follows that

$\displaystyle \sum_{k=1}^{N} m_k \ddot{\mathbf{r}}_k \cdot \frac{\partial \mathbf{r}_k}{\partial q_j} = \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_j} \right) - \frac{\partial T}{\partial q_j}$ .

Substitute into the equation:

$\displaystyle \sum_{j=1}^{n} \left[ Q_j - \left( \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_j} \right) - \frac{\partial T}{\partial q_j} \right) \right] \delta q_j = 0$ .

Because the $\displaystyle \delta q_j$ are independent, we must have:

$\displaystyle \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_j} \right) - \frac{\partial T}{\partial q_j} = Q_j$ .

For systems where the applied forces are conservative, they can be derived from a scalar potential energy function: $\displaystyle \mathbf{F}_k = -\nabla_k V$ , we have: $\displaystyle Q_j = -\frac{\partial V}{\partial q_j}$

Assuming $\displaystyle V$ depends only on coordinates, not velocities, we define the Lagrangian: $\displaystyle L = T - V$ . Then the Euler-Lagrange equations follow:

$\displaystyle \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_j} \right) - \frac{\partial L}{\partial q_j} = 0$ .

The quantity $\displaystyle \left( \frac{\partial L}{\partial \dot{q}_j} \right)$ is called generalized momentum, which in this case matches with the kinetic momentum.

The Principle of Stationary Action: Hamilton’s Principle

The very soul of Lagrangian mechanics resides in Hamilton’s Principle, the principle of stationary action. This profound variational principle asserts that the actual path a physical system takes between two specified points in time $\displaystyle t_1$ and $\displaystyle t_2$ is one for which a quantity known as the “action” $\displaystyle S$ is stationary. The action is defined as the time integral of the Lagrangian function $\displaystyle L$ along the system’s trajectory:

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

Here, $\displaystyle q_i$ are the generalized coordinates, and $\displaystyle \dot{q}_i$ are their corresponding generalized velocities. While often called “least action,” the principle mathematically implies an extremum (minimum, maximum, or saddle point), typically a minimum in physical applications. This global perspective, where the system “chooses” a path, contrasts sharply with Newton’s local, force-driven description, offering deeper insights into physical laws.

The machinery for solving such optimization problems involving functions is the calculus of variations, dedicated to finding functions that optimize functionals.

To derive the equations of motion from Hamilton’s Principle, we consider a small variation $\displaystyle \delta q_i(t)$ from the actual path $\displaystyle q_i(t)$ . Crucially, these variations must vanish at the endpoints: $\displaystyle \delta q_i(t_1)=\delta q_i(t_2)=0$ . For the actual path, the variation in the action, $\displaystyle \delta S$ , must be zero:

$\displaystyle \delta S=\delta\int_{t_1}^{t_2}L(q_i,\dot{q}_i,t)dt=0$

Expand the variation of the Lagrangian, $\displaystyle \delta L$ , using the chain rule:

$\displaystyle \delta L = \sum_i \left( \frac{\partial L}{\partial q_i} \delta q_i + \frac{\partial L}{\partial \dot{q}_i} \delta \dot{q}_i \right)$

Substituting this back into the expression for $\displaystyle \delta S$ :

$\displaystyle \delta S = \int_{t_1}^{t_2} \sum_i \left( \frac{\partial L}{\partial q_i} \delta q_i + \frac{\partial L}{\partial \dot{q}_i} \delta \dot{q}_i \right) dt$

The term involving $\displaystyle \delta\dot{q}_i$ is transformed using integration by parts, recognizing $\displaystyle \delta\dot{q}_i=\frac{d}{dt}(\delta q_i)$ :

$\displaystyle \int_{t_1}^{t_2} \frac{\partial L}{\partial \dot{q}_i} \frac{d}{dt}(\delta q_i) dt = \left[ \frac{\partial L}{\partial \dot{q}_i} \delta q_i \right]_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) \delta q_i dt$

Since $\displaystyle \delta q_i$ are zero at the endpoints ( $\displaystyle t_1$ and $\displaystyle t_2$ ), the boundary term $\displaystyle \left[ \frac{\partial L}{\partial \dot{q}_i} \delta q_i \right]_{t_1}^{t_2}$ vanishes. Thus, $\displaystyle \delta S$ simplifies to:

$\displaystyle \delta S = \int_{t_1}^{t_2} \sum_i \left( \frac{\partial L}{\partial q_i} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) \right) \delta q_i dt$

For $\displaystyle \delta S$ to be zero for arbitrary variations $\displaystyle \delta q_i(t)$ (that vanish at the endpoints), the integrand must be identically zero. This leads to exact the Euler-Lagrange equations, one for each generalized coordinate $\displaystyle q_i$ :

$\displaystyle \frac{\partial L}{\partial q_i} - \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) = 0$

Finite-Dimensional Systems:

In the realm of classical mechanics, for a vast and crucial class of systems, the Lagrangian $\displaystyle L$ is elegantly defined as the difference between the total kinetic energy $\displaystyle T$ and the total potential energy $\displaystyle V$ : $\displaystyle L=T-V$ Both $\displaystyle T$ and $\displaystyle V$ are expressed in terms of the chosen generalized coordinates $\displaystyle q_i$ and their corresponding generalized velocities $\displaystyle \dot{q}_i$ . Conventionally, $\displaystyle T$ typically depends on both $\displaystyle q_i$ and $\displaystyle \dot{q}_i$ , while for conservative forces, $\displaystyle V$ is solely a function of $\displaystyle q_i$ .

It is paramount to recognize that this seemingly simple form, $\displaystyle L=T-V$ , is not a universal dictum. Its applicability is chiefly confined to non-relativistic systems governed by conservative forces derivable from a potential. Beyond this domain, for instance, in the presence of non-conservative forces or relativistic effects, the Lagrangian adopts a more generalized structure that defies a neat decomposition into kinetic and potential energy components. The true essence of the Lagrangian is its role as the fundamental function whose variation, through Hamilton’s Principle, yields the correct equations of motion. Its precise form is thus intricately dictated by the underlying interactions and symmetries inherent to the physical system.

We will look examples now:

Example 1: Free Particle

Consider a particle of mass $\displaystyle m$ moving freely in one dimension $\displaystyle x$ .

Generalized Coordinate: $\displaystyle x$
Kinetic Energy (T): $\displaystyle T=\frac{1}{2}m\dot{x}^2$ .
Potential Energy (V): For a free particle, $\displaystyle V=0$ .
Lagrangian (L): $\displaystyle L=T-V=\frac{1}{2}m\dot{x}^2$ .
Euler-Lagrange Equation:
- Partial derivatives: $\displaystyle \frac{\partial L}{\partial\dot{x}}=m\dot{x} \quad$ and $\displaystyle \frac{\partial L}{\partial x}=0$
- Substituting: $\displaystyle \frac{d}{dt}(m\dot{x})-0=0$ , which gives $\displaystyle m\ddot{x}=0\implies\ddot{x}=0$ . This equation of motion simply states that a free particle moves with constant velocity, as expected from Newton’s first law.

Example 2: Simple Harmonic Oscillator

A mass $\displaystyle m$ attached to a spring with spring constant $\displaystyle k$ , moving along a single dimension $\displaystyle x$ .

Generalized Coordinate: $\displaystyle x$
Kinetic Energy (T): $\displaystyle T=\frac{1}{2}m\dot{x}^2$ .
Potential Energy (V): $\displaystyle V=\frac{1}{2}kx^2$ .
Lagrangian (L): $\displaystyle L=T-V=\frac{1}{2}m\dot{x}^2-\frac{1}{2}kx^2$ .
Euler-Lagrange Equation:
- Partial derivatives: $\displaystyle \frac{\partial L}{\partial\dot{x}}=m\dot{x} \quad$ and $\displaystyle \frac{\partial L}{\partial x}=-kx$
- Substituting: $\displaystyle \frac{d}{dt}(m\dot{x})-(-kx)=0$ which gives $\displaystyle m\ddot{x}+kx=0$ . This is the classic equation for a simple harmonic oscillator.

Example 3: Damped Harmonic Oscillator

For systems with non-conservative forces like friction, the standard
$\displaystyle L = T - V$ doesn’t work: To fix this, we modify the Euler-Lagrange equations by adding a dissipation term via the Rayleigh dissipation function and the equations become:

$\displaystyle \frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = -\frac{\partial \mathcal{F}}{\partial \dot{q}_i}$

Let’s illustrate this with a mass $\displaystyle m$ attached to a spring with constant $\displaystyle k$ , subject to a linear damping force $\displaystyle -b\dot{x}$ .

Generalized Coordinate: $\displaystyle x$
Lagrangian (L): This remains the conservative part of the system’s energy difference: $\displaystyle L=\frac{1}{2}m\dot{x}^2-\frac{1}{2}kx^2$ .
Rayleigh Dissipation Function (F): For linear damping, the dissipative force is $\displaystyle F_d=-b\dot{x}$ . The instantaneous power dissipated is $\displaystyle P=F_d\cdot\dot{x}=-b\dot{x}^2$ . Half this rate of energy dissipation yields the Rayleigh function: $\displaystyle \mathcal{F}=\frac{1}{2}b\dot{x}^2$ .

Substituting these into the augmented Euler-Lagrange equation:

$\displaystyle \frac{d}{dt}(m\dot{x})-(-kx)=-b\dot{x}$

This is the familiar equation of motion for a damped harmonic oscillator.

For systems with non-conservative forces like damping, specialized Lagrangian constructions can still yield equations of motion via the standard Euler-Lagrange framework (instead of adding these dissipative terms separately).

A time-dependent Lagrangian $\displaystyle \tilde{L}$ can encapsulate damping. For a damped harmonic oscillator (mass $\displaystyle m$ , spring $\displaystyle k$ , damping $\displaystyle b$ ):

$\displaystyle \tilde{L}(x, \dot{x}, t) = e^{\frac{b}{m} t} \left( \frac{1}{2} m \dot{x}^2 - \frac{1}{2} k x^2 \right)$

Applying the standard Euler-Lagrange equation $\displaystyle \frac{d}{dt}\left(\frac{\partial \tilde{L}}{\partial\dot{x}}\right)-\frac{\partial \tilde{L}}{\partial x}=0$ directly yields: $\displaystyle m\ddot{x} + b\dot{x} + kx = 0$ This approach effectively transforms the dissipative system into one described by a time-dependent action principle.

Alternatively, Bateman’s Dual System introduces an auxiliary variable to create a coupled, conservative Lagrangian. For a damped oscillator, using an auxiliary variable $\displaystyle y$ , the Lagrangian becomes:

$\displaystyle L(x, \dot{x}, y, \dot{y}) = m \dot{x} \dot{y} + \frac{b}{2}(x \dot{y} - y \dot{x}) - k x y$

Applying the Euler-Lagrange equation for $\displaystyle y$ gives:

$\displaystyle \frac{d}{dt}\left(\frac{\partial L}{\partial\dot{y}}\right)-\frac{\partial L}{\partial y}=0 \implies m\ddot{x} + b\dot{x} + kx = 0$

Example 4: Simple Pendulum

A point mass $\displaystyle m$ suspended by a rigid, massless rod of length $\displaystyle l$ in a uniform gravitational field.

Generalized Coordinate: The angle $\displaystyle \theta$ from the vertical.
Kinetic Energy (T): $\displaystyle T=\frac{1}{2}m(l\dot{\theta})^2=\frac{1}{2}ml^2\dot{\theta}^2$ .
Potential Energy (V): Taking the pivot as $\displaystyle y=0$ , the height is $\displaystyle -l\cos\theta$ . So $\displaystyle V=-mgl\cos\theta$ .
Lagrangian (L): $\displaystyle L=\frac{1}{2}ml^2\dot{\theta}^2+mgl\cos\theta$ .
Euler-Lagrange Equation:
- Partial derivatives: $\displaystyle \frac{\partial L}{\partial\dot{\theta}}=ml^2\dot{\theta} \quad$ and $\displaystyle \frac{\partial L}{\partial\theta}=-mgl\sin\theta$
- Substituting: $\displaystyle \frac{d}{dt}(ml^2\dot{\theta})-(-mgl\sin\theta)=0$ , which gives $\displaystyle ml^2\ddot{\theta}+mgl\sin\theta=0$ . That is we have : $\displaystyle \ddot{\theta}+\frac{g}{l}\sin\theta=0$ . This is the standard equation of motion for a simple pendulum.

Example 5: Inverted Pendulum

Consider a massless rod of length $\displaystyle l$ with a mass $\displaystyle m$ at its end, pivoted at the origin, but now the mass is above the pivot. This system is inherently unstable.

Generalized Coordinate: The angle $\displaystyle \theta$ from the upward vertical.
Kinetic Energy (T): $\displaystyle T=\frac{1}{2}m(l\dot{\theta})^2=\frac{1}{2}ml^2\dot{\theta}^2$ .
Potential Energy (V): Taking the pivot as $\displaystyle y=0$ , the height is $\displaystyle l\cos\theta$ . So $\displaystyle V=mgl\cos\theta$ .
Lagrangian (L): $\displaystyle L=\frac{1}{2}ml^2\dot{\theta}^2-mgl\cos\theta$ .
Euler-Lagrange Equation:
- Partial derivatives: $\displaystyle \frac{\partial L}{\partial\dot{\theta}}=ml^2\dot{\theta} \quad$ and $\displaystyle \frac{\partial L}{\partial\theta}=mgl\sin\theta$
- Substituting: $\displaystyle \frac{d}{dt}(ml^2\dot{\theta})-(mgl\sin\theta)=0$ , which gives $\displaystyle ml^2\ddot{\theta}-mgl\sin\theta=0$ . That is we have $\displaystyle \ddot{\theta}-\frac{g}{l}\sin\theta=0$ . The sign difference from the simple pendulum equation reflects the instability: for small $\displaystyle \theta$ , $\displaystyle \sin\theta\approx\theta$ , leading to $\displaystyle \ddot{\theta}\approx(g/l)\theta$ , which describes exponential divergence.

Example 6: Double Pendulum

A classic example of a chaotic system, consisting of two pendulums attached end-to-end.

Generalized Coordinates: Two angles, $\displaystyle \theta_1$ and $\displaystyle \theta_2$ , measured from the vertical for the first and second pendulums, respectively. Let $\displaystyle m_1, l_1$ be the mass and length of the first pendulum, and $\displaystyle m_2, l_2$ for the second.
Kinetic Energy (T):
- Position of $\displaystyle m_1$ : $\displaystyle (x_1,y_1)=(l_1\sin\theta_1,-l_1\cos\theta_1)$ .
- $\displaystyle T_1=\frac{1}{2}m_1(l_1\dot{\theta}_1)^2=\frac{1}{2}m_1l_1^2\dot{\theta}_1^2$ .
- Position of $\displaystyle m_2$ : $\displaystyle (x_2,y_2)=(l_1\sin\theta_1+l_2\sin\theta_2,-l_1\cos\theta_1-l_2\cos\theta_2)$ .
- The velocity squared of $\displaystyle m_2$ is $\displaystyle v_2^2=\dot{x}_2^2+\dot{y}_2^2=l_1^2\dot{\theta}_1^2+l_2^2\dot{\theta}_2^2+2l_1l_2\dot{\theta}_1\dot{\theta}_2\cos(\theta_1-\theta_2)$ .
- $\displaystyle T_2=\frac{1}{2}m_2v_2^2$ .
- Total Kinetic Energy: $\displaystyle T=T_1+T_2=\frac{1}{2}m_1l_1^2\dot{\theta}_1^2+\frac{1}{2}m_2[l_1^2\dot{\theta}_1^2+l_2^2\dot{\theta}_2^2+2l_1l_2\dot{\theta}_1\dot{\theta}_2\cos(\theta_1-\theta_2)]$ .
Potential Energy (V): Taking the pivot point as zero potential: $\displaystyle V=-m_1gl_1\cos\theta_1-m_2g(l_1\cos\theta_1+l_2\cos\theta_2)$ .
Lagrangian (L): $\displaystyle L=T-V$ .
Euler-Lagrange Equations: The derivation of the equations of motion for $\displaystyle \theta_1$ and $\displaystyle \theta_2$ from this Lagrangian is straightforward. The resulting coupled non-linear differential equations describe the complex, often chaotic, motion of the double pendulum.

For $\displaystyle \theta_1$ : $\displaystyle (m_1+m_2)l_1\ddot{\theta}_1+m_2l_2\ddot{\theta}_2\cos(\theta_1-\theta_2)+m_2l_2\dot{\theta}_2^2\sin(\theta_1-\theta_2)+(m_1+m_2)g\sin\theta_1=0$

For $\displaystyle \theta_2$ : $\displaystyle m_2l_2\ddot{\theta}_2+m_2l_1\ddot{\theta}_1\cos(\theta_1-\theta_2)-m_2l_1\dot{\theta}_1^2\sin(\theta_1-\theta_2)+m_2g\sin\theta_2=0$

Example 7: Rigid Body

Describing the rotation of a rigid body is complex in Newtonian mechanics. Lagrangian mechanics, using Euler angles as generalized coordinates, simplifies this. Euler angles ( $\displaystyle \phi,\theta,\psi$ ) define the orientation of a rigid body. The kinetic energy of a rigid body with one point fixed can be expressed in terms of these angles and their time derivatives.

Generalized Coordinates: Euler angles ( $\displaystyle \phi,\theta,\psi$ ).
Kinetic Energy (T): The angular velocity components ( $\displaystyle \Omega_1,\Omega_2,\Omega_3$ ) along the body’s principal axes of inertia are given by: $\displaystyle \Omega_1=\dot{\phi}\sin\theta\sin\psi+\dot{\theta}\cos\psi$ , $\displaystyle \Omega_2=\dot{\phi}\sin\theta\cos\psi-\dot{\theta}\sin\psi \quad$ and $\displaystyle \Omega_3=\dot{\phi}\cos\theta+\dot{\psi}$ The kinetic energy is then: $\displaystyle T=\frac{1}{2}I_1\Omega_1^2+\frac{1}{2}I_2\Omega_2^2+\frac{1}{2}I_3\Omega_3^2$ where $\displaystyle I_1,I_2,I_3$ are the principal moments of inertia.
Potential Energy (V): Depends on the specific forces (e.g., gravity for a top). For a free rigid body, $\displaystyle V=0$ .
Lagrangian (L): $\displaystyle L=T-V$ .
Euler-Lagrange Equations: Applying the Euler-Lagrange equations for $\displaystyle \phi,\theta,\psi$ yields Euler’s equations of motion for rigid body rotation. These are generally complex coupled differential equations, but the Lagrangian formulation provides a systematic path to them.

Example 8: Central Force Problem

Consider a particle of mass $\displaystyle m$ moving under the influence of a central force, meaning the force is always directed towards or away from a fixed point (the center of force) and its magnitude depends only on the distance $\displaystyle r$ from that point. This is the setup for planetary orbits (Kepler problem).

Generalized Coordinates: Polar coordinates ( $\displaystyle r,\phi$ ) are ideal due to the central symmetry.
Kinetic Energy (T): In polar coordinates, $\displaystyle T=\frac{1}{2}m(\dot{r}^2+r^2\dot{\phi}^2)$ .
Potential Energy (V): For a conservative central force, $\displaystyle V=V(r)$ . For gravity, $\displaystyle V(r)=-GMm/r$ .
Lagrangian (L): $\displaystyle L=\frac{1}{2}m(\dot{r}^2+r^2\dot{\phi}^2)-V(r)$ .
Euler-Lagrange Equations:
- For :
  - Partial derivatives: $\displaystyle \frac{\partial L}{\partial\dot{\phi}}=mr^2\dot{\phi} \quad$ and $\displaystyle \frac{\partial L}{\partial\phi}=0$
  - Substituting: $\displaystyle \frac{d}{dt}(mr^2\dot{\phi})-0=0\implies mr^2\dot{\phi}=\text{constant}$ . Since $\displaystyle \phi$ is a cyclic coordinate, $\displaystyle mr^2\dot{\phi}$ is conserved. This constant is the angular momentum, which is conserved for any central force problem.
- For :
  - Partial derivatives: $\displaystyle \frac{\partial L}{\partial\dot{r}}=m\dot{r} \quad$ and $\displaystyle \frac{\partial L}{\partial r}=mr\dot{\phi}^2-\frac{\partial V}{\partial r}$
  - Substituting: $\displaystyle \frac{d}{dt}(m\dot{r})-(mr\dot{\phi}^2-\frac{\partial V}{\partial r})=0$ which gives $\displaystyle m\ddot{r}-mr\dot{\phi}^2+\frac{\partial V}{\partial r}=0$ . This is the radial equation of motion, which, combined with angular momentum conservation, allows for the solution of the orbit.

Example 9: Charged Particle in Electromagnetic Field

This is a crucial example where the Lagrangian is not $\displaystyle T-V$ . This departure from the typical kinetic minus potential energy form is crucial, underscoring the Lagrangian’s role as a more fundamental construct in theoretical physics, primarily defined by its capacity to generate the correct equations of motion. It reveals that the interaction with the electromagnetic field necessitates a more nuanced structure for the Lagrangian, one that inherently captures the velocity-dependence of magnetic forces.

Generalized Coordinates: Cartesian coordinates ( $\displaystyle x,y,z$ ).
Lagrangian (L): For a charged particle $\displaystyle e$ with mass $\displaystyle m$ in an electromagnetic field (scalar potential $\displaystyle \Phi(\mathbf{r},t)$ , vector potential $\displaystyle \mathbf{A}(\mathbf{r},t)$ ), the Lagrangian is: $\displaystyle L = \frac{1}{2} m |\mathbf{v}|^2 + e \mathbf{A}(\mathbf{r}, t) \cdot \mathbf{v} - e \Phi(\mathbf{r}, t)$ . Here, $\displaystyle \mathbf{v}=(\dot{x},\dot{y},\dot{z})$ .
Euler-Lagrange Equation (for x-component):
- Partial derivatives: $\displaystyle \frac{\partial L}{\partial\dot{x}} = m\dot{x} + eA_x \quad$ and $\displaystyle \frac{\partial L}{\partial x} = e \left( \frac{\partial A_x}{\partial x} \dot{x} + \frac{\partial A_y}{\partial x} \dot{y} + \frac{\partial A_z}{\partial x} \dot{z} \right) - e \frac{\partial \Phi}{\partial x}$
- Substituting and expanding the total time derivative $\displaystyle \frac{d}{dt}A_x=\frac{\partial A_x}{\partial t}+\nabla A_x\cdot\mathbf{v}$ , one obtains the x-component of the Lorentz force law. Similar derivations for $\displaystyle y$ and $\displaystyle z$ yield the full vector equation: $\displaystyle m\frac{d\mathbf{v}}{dt}=e(\mathbf{E}+\mathbf{v}\times\mathbf{B})$ where $\displaystyle \mathbf{E}=-\nabla\Phi-\frac{\partial\mathbf{A}}{\partial t}$ and $\displaystyle \mathbf{B}=\nabla\times\mathbf{A}$ . This example highlights that the Lagrangian is a more fundamental concept than just $\displaystyle T-V$ , and the generalized momentum $\displaystyle p_x=m\dot{x}+eA_x$ is not simply the kinematic momentum.

Example 10: Foucault Pendulum

The Foucault pendulum demonstrates the Earth’s rotation by showing the precession of its oscillation plane. Analyzing this in a rotating frame is simplified by Lagrangian mechanics.

Generalized Coordinates: For a pendulum of mass $\displaystyle m$ and length $\displaystyle l$ at latitude $\displaystyle \lambda$ , we can use Cartesian coordinates ( $\displaystyle x,y$ ) in the horizontal plane, assuming small oscillations. The Earth’s angular velocity is $\displaystyle \mathbf{\Omega}=(0,\Omega\cos\lambda,\Omega\sin\lambda)$ .
Kinetic Energy (T): In a rotating frame, the kinetic energy includes terms related to the rotation. For a particle, $\displaystyle T=\frac{1}{2}m\mathbf{v}^2$ , where $\displaystyle \mathbf{v}$ is the velocity in the rotating frame. The transformation from inertial to rotating frame velocity is $\displaystyle \mathbf{v}_{\text{inertial}}=\mathbf{v}_{\text{rotating}}+\mathbf{\Omega}\times\mathbf{r}$ . For small oscillations, the kinetic energy can be approximated as: $\displaystyle T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)+m\Omega_z(x\dot{y}-y\dot{x})$ where $\displaystyle \Omega_z=\Omega\sin\lambda$ is the vertical component of Earth’s angular velocity. The term $\displaystyle m\Omega_z(x\dot{y}-y\dot{x})$ is a velocity-dependent potential term (similar to the magnetic part of the EM Lagrangian).
Potential Energy (V): For small oscillations, $\displaystyle V=\frac{1}{2}mg(x^2+y^2)/l$ .
Lagrangian (L): $\displaystyle L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)+m\Omega_z(x\dot{y}-y\dot{x})-\frac{1}{2}mg(x^2+y^2)/l$ .
Euler-Lagrange Equations:
- For $\displaystyle x$ : $\displaystyle \frac{d}{dt}\left(\frac{\partial L}{\partial\dot{x}}\right)-\frac{\partial L}{\partial x}=0$ $\displaystyle m\ddot{x}-(m\Omega_z\dot{y}-mgx/l)=0\implies m\ddot{x}-m\Omega_z\dot{y}+mgx/l=0$
- For $\displaystyle y$ : $\displaystyle \frac{d}{dt}\left(\frac{\partial L}{\partial\dot{y}}\right)-\frac{\partial L}{\partial y}=0$ $\displaystyle m\ddot{y}-(-m\Omega_z\dot{x}-mgy/l)=0\implies m\ddot{y}+m\Omega_z\dot{x}+mgy/l=0$ . These coupled equations describe the motion of the Foucault pendulum, showing the precession due to the Coriolis force. The terms involving $\displaystyle \Omega_z$ are precisely those originating from the Coriolis force, causing the plane of oscillation to precess in the rotating frame. This elegantly demonstrates how Lagrangian mechanics systematically handles inertial forces in non-inertial frames by simply choosing the appropriate kinetic energy formulation within that frame, rather than explicitly introducing fictitious forces ad-hoc. The velocity-dependent kinetic energy term is a direct manifestation of this frame transformation within the variational principle.

Example 11: Bead on a Rotating Hoop

A small bead of mass $\displaystyle m$ slides without friction along a circular hoop of radius $\displaystyle R$ . The hoop rotates with a constant angular speed $\displaystyle \Omega$ about a vertical diameter.

Generalized Coordinate: The angle $\displaystyle \theta$ that the bead’s position vector (from the center of the hoop) makes with the downward vertical axis.
Kinetic Energy (T): $\displaystyle T=\frac{1}{2}mR^2(\dot{\theta}^2+\Omega^2\sin^2\theta)$
Potential Energy (V): Taking the center of the hoop as $\displaystyle V=0$ , the vertical position is $\displaystyle R\cos\theta$ . So $\displaystyle V=-mgR\cos\theta$ .
Lagrangian (L): $\displaystyle L=\frac{1}{2}mR^2(\dot{\theta}^2+\Omega^2\sin^2\theta)+mgR\cos\theta$ .
Euler-Lagrange Equation:
- Partial derivatives: $\displaystyle \frac{\partial L}{\partial\dot{\theta}}=mR^2\dot{\theta}$ and
  $\displaystyle \frac{\partial L}{\partial\theta}=mR^2\Omega^2\sin\theta\cos\theta-mgR\sin\theta$
- Substituting: $\displaystyle mR^2\ddot{\theta}-(mR^2\Omega^2\sin\theta\cos\theta-mgR\sin\theta)=0$ . That is $\displaystyle mR^2\ddot{\theta}-mR^2\Omega^2\sin\theta\cos\theta+mgR\sin\theta=0$ . So we have $\displaystyle \ddot{\theta}-\Omega^2\sin\theta\cos\theta+\frac{g}{R}\sin\theta=0$ . This equation describes the bead’s motion, revealing equilibrium points and conditions for stable oscillations depending on $\displaystyle \Omega$ .

Example 12: Simple Atwood Machine

Consider two masses, $\displaystyle m_1$ and $\displaystyle m_2$ , connected by a light, inextensible string passing over a massless, frictionless pulley.

Generalized Coordinate: Let $\displaystyle x$ be the vertical position of mass $\displaystyle m_1$ (measured downwards from the pulley). Due to the inextensible string, mass $\displaystyle m_2$ is then implicitly at position $\displaystyle L-x$ (where $\displaystyle L$ is the constant total length of the string). This system has one degree of freedom.
Kinetic Energy (T): $\displaystyle T=\frac{1}{2}m_1\dot{x}^2+\frac{1}{2}m_2\dot{x}^2=\frac{1}{2}(m_1+m_2)\dot{x}^2$
Potential Energy (V): Taking the pulley level as the zero potential energy reference: $\displaystyle V=-m_1gx-m_2g(L-x)$
Lagrangian (L): $\displaystyle L=T-V=\frac{1}{2}(m_1+m_2)\dot{x}^2+m_1gx+m_2g(L-x)$
Euler-Lagrange Equation: For the coordinate :
- Partial derivatives: $\displaystyle \frac{\partial L}{\partial\dot{x}}=(m_1+m_2)\dot{x} \quad$ and $\displaystyle \frac{\partial L}{\partial x}=m_1g-m_2g=(m_1-m_2)g$
- Substituting: $\displaystyle \frac{d}{dt}((m_1+m_2)\dot{x})-(m_1-m_2)g=0$ $\displaystyle (m_1+m_2)\ddot{x}-(m_1-m_2)g=0$ . So we have $\displaystyle \ddot{x}=\frac{m_1-m_2}{m_1+m_2}g$ . This is the familiar acceleration for a simple Atwood machine, readily derived without explicitly considering the string tension as a constraint force.

Example 13: Double Atwood Machine

Consider a system where one of the weights of a simple Atwood machine is replaced by a second Atwood machine. This system has two degrees of freedom.

Generalized Coordinates: Let $\displaystyle x$ be the vertical position of the first pulley (from the fixed upper pulley), and $\displaystyle y$ be the vertical position of mass $\displaystyle m_1$ relative to the second pulley.
Kinetic Energy (T):
- Mass of first pulley (if it has mass $\displaystyle M$ ): $\displaystyle \frac{1}{2}M\dot{x}^2$ . (Assuming massless for standard setup)
- Mass $\displaystyle m_1$ : Its speed in laboratory space is $\displaystyle |\dot{x}-\dot{y}|$ . So, $\displaystyle \frac{1}{2}m_1(\dot{x}-\dot{y})^2$ .
- Mass $\displaystyle m_2$ : Its speed in laboratory space is $\displaystyle |\dot{x}+\dot{y}|$ . So, $\displaystyle \frac{1}{2}m_2(\dot{x}+\dot{y})^2$ .
- Total Kinetic Energy: $\displaystyle T = \frac{1}{2} M_{pulley} \dot{x}^2 + \frac{1}{2} m_1 (\dot{x} - \dot{y})^2 + \frac{1}{2} m_2 (\dot{x} + \dot{y})^2$ (assuming $\displaystyle M_{pulley}$ is the mass of the second pulley for generality, if $\displaystyle M_{pulley}=0$ , the first term vanishes).
Potential Energy (V): Taking the initial position of the first pulley as zero potential: $\displaystyle V=-M_{pulley}gx-m_1g(x-y)-m_2g(x+y)$
Lagrangian (L): $\displaystyle L=T-V$ .
Euler-Lagrange Equations:
- For $\displaystyle x$ : $\displaystyle \frac{\partial L}{\partial\dot{x}}=M_{pulley}\dot{x}+m_1(\dot{x}-\dot{y})+m_2(\dot{x}+\dot{y})$ and $\displaystyle \frac{\partial L}{\partial x}=M_{pulley}g+m_1g+m_2g=(M_{pulley}+m_1+m_2)g$ .
  $\displaystyle \frac{d}{dt}(M_{pulley}\dot{x}+m_1(\dot{x}-\dot{y})+m_2(\dot{x}+\dot{y}))-(M_{pulley}+m_1+m_2)g=0$
  $\displaystyle M_{pulley}\ddot{x}+m_1(\ddot{x}-\ddot{y})+m_2(\ddot{x}+\ddot{y})-(M_{pulley}+m_1+m_2)g=0$ .
- For $\displaystyle y$ : $\displaystyle \frac{\partial L}{\partial\dot{y}}=-m_1(\dot{x}-\dot{y})+m_2(\dot{x}+\dot{y})$
  $\displaystyle \frac{\partial L}{\partial y}=m_1g-m_2g=(m_1-m_2)g$
  $\displaystyle \frac{d}{dt}(-m_1(\dot{x}-\dot{y})+m_2(\dot{x}+\dot{y}))-(m_1-m_2)g=0$
  $\displaystyle -m_1(\ddot{x}-\ddot{y})+m_2(\ddot{x}+\ddot{y})-(m_1-m_2)g=0$

These two coupled second-order differential equations completely describe the motion of the double Atwood machine, again demonstrating the method’s efficiency in complex constrained systems.

Relativistic Particle Mechanics

Lagrangian mechanics extends naturally to relativistic systems, though the form of the Lagrangian changes. The construction of the relativistic Lagrangian is rooted in the principle of Lorentz invariance – the idea that the laws of physics are the same for all observers in uniform motion. This mandates that the action, $\displaystyle S = \int L dt$ , must be a Lorentz scalar.

Example 14: Relativistic Free Particle

The most fundamental Lorentz invariant scalar quantity describing a particle’s motion is its proper time, the time measured by a clock moving with the particle. The action for a free particle is chosen to be proportional to the total proper time elapsed. This fundamental principle dictates the form of the Lagrangian. The action is given by:

$\displaystyle S=-mc\int_{t_1}^{t_2}d\tau=-mc\int_{t_1}^{t_2}\sqrt{dt^2-\frac{1}{c^2}(dx^2+dy^2+dz^2)}$

This can be rewritten as: $\displaystyle S=-mc^2\int_{t_1}^{t_2}\sqrt{1-\frac{v^2}{c^2}}dt$

From this, the Lagrangian (L) for a relativistic free particle is directly identified as: $\displaystyle L=-mc^2\sqrt{1-\frac{v^2}{c^2}}$ where $\displaystyle v^2 = \dot{x}^2 + \dot{y}^2 + \dot{z}^2$ . You can see the proportionality constants are chosen such that in the non-relativistic limit $v/c \to 0$ , this reduces to the usual Lagrangian.

Let’s apply the Euler-Lagrange equation for the x-component. First, compute the partial derivative with respect to $\displaystyle \dot{x}$ . We have

$\displaystyle \frac{\partial L}{\partial\dot{x}} = -mc^2 \cdot \frac{1}{2\sqrt{1-\frac{v^2}{c^2}}} \cdot \left(-\frac{2\dot{x}}{c^2}\right) = \frac{m\dot{x}}{\sqrt{1-\frac{v^2}{c^2}}} = \gamma m\dot{x}$

where $\displaystyle \gamma=(1-v^2/c^2)^{-1/2}$ is the Lorentz factor. This term, $\displaystyle \gamma m\dot{x}$ , is precisely the relativistic momentum $\displaystyle p_x$ .

Next, compute the partial derivative with respect to $\displaystyle x$ . We have $\displaystyle \frac{\partial L}{\partial x}=0$ since $\displaystyle L$ does not explicitly depend on $\displaystyle x$ .

Substituting these into the Euler-Lagrange equation we get:

$\displaystyle \frac{d}{dt}(\gamma m\dot{x})=0$

This implies that the relativistic momentum $\displaystyle \mathbf{p}=\gamma m\mathbf{v}$ is conserved for a free particle, leading to constant velocity.

Example 15: Relativistic Charged Particle

To incorporate electromagnetic interactions into relativistic mechanics, the free particle Lagrangian is extended via the principle of minimal coupling. This fundamental postulate in physics asserts that interactions are introduced by precisely altering the free Lagrangian in a way that preserves both Lorentz invariance (the laws of physics are the same for all observers in uniform motion) and local gauge invariance (a specific symmetry related to how potentials are defined).

Lorentz Covariant Interaction: To ensure that the interaction between a charged particle and the electromagnetic field respects Lorentz invariance, the interaction term in the Lagrangian must be a Lorentz scalar — a quantity unchanged under Lorentz transformations. This is naturally constructed from the scalar product of two four-vectors:

The particle’s four-velocity: $\displaystyle u^\mu = (\gamma c,\, \gamma \mathbf{v})$

The electromagnetic four-potential: $\displaystyle A^\mu = \left( \frac{\Phi}{c},\, \mathbf{A} \right)$

Their scalar product is: $\displaystyle A^\mu u_\mu = \gamma \left( \Phi - \mathbf{A} \cdot \mathbf{v} \right)$

So the Lorentz-invariant interaction term in the action is: $\displaystyle S_{\text{int}} = -q \int A^\mu u_\mu\, d\tau$ . Converting to coordinate time $\displaystyle t$ , using $\displaystyle d\tau = \frac{dt}{\gamma}$ , we get the Lagrangian interaction term:

$\displaystyle L_{\text{int}} = -q \cdot \frac{A^\mu u_\mu}{\gamma} = q\left( \mathbf{A} \cdot \mathbf{v} - \Phi \right)$
Lagrangian with Interaction Term: By adding this term to the free relativistic particle Lagrangian, we obtain the full Lagrangian for a relativistic charged particle: $\displaystyle L=-mc^2\sqrt{1-\frac{v^2}{c^2}}+q(\mathbf{A}\cdot\mathbf{v}-\Phi)$ . The $\displaystyle q\mathbf{A}\cdot\mathbf{v}$ component of this interaction term is explicitly velocity-dependent. Note that we have the same expression in non-relativistic Lagrangian, but we add it adhoc so as to get the correct equations of motion, but now the term can be thought as being forced by requiring Lorentz invariance for the Lagrangian.
Canonical Momentum and Lorentz Force: Applying the Euler-Lagrange equations to this Lagrangian directly yields the relativistic Lorentz force law.

For example, for the x-component, the partial derivative with respect to $\displaystyle \dot{x}$ defines the relativistic canonical momentum $\displaystyle p_x$ :

$\displaystyle p_x = \frac{\partial L}{\partial \dot{x}} = \gamma m\dot{x} + qA_x$

Its structure (kinetic momentum plus a field-dependent term) is fundamental, as this canonical momentum, not just the kinematic momentum, is the quantity consistently quantized in quantum field theory. Further computation ultimately reproduces the complete Lorentz force law:

$\displaystyle m\frac{d(\gamma\mathbf{v})}{dt}=q(\mathbf{E}+\mathbf{v}\times\mathbf{B})$

where $\displaystyle \mathbf{E}=-\nabla\Phi-\frac{\partial\mathbf{A}}{\partial t}$ and $\displaystyle \mathbf{B}=\nabla\times\mathbf{A}$ .

Symmetries and Conservation Laws: Noether’s Theorem

One of the most profound and beautiful connections in theoretical physics is the relationship between the symmetries of a physical system and the conservation laws it obeys. This relationship is formally encapsulated by Noether’s Theorem, a fundamental result that transcends classical mechanics to underpin modern physics. It states that for every continuous symmetry transformation under which the Lagrangian (and thus the action) of a system is invariant, there exists a corresponding conserved quantity. This establishes that conservation laws as direct consequences of fundamental symmetries inherent in the system.

Let’s illustrate this with a general derivation. Consider a Lagrangian $\displaystyle L(q_i,\dot{q}_i,t)$ . If the system possesses a continuous symmetry, it means that the Lagrangian is invariant under an infinitesimal transformation of the coordinates $\displaystyle q_i\to q_i+\delta q_i$ , where $\displaystyle \delta q_i$ depends on some infinitesimal parameter $\displaystyle \epsilon$ . That is, $\displaystyle \delta L=0$ under this transformation.

The total time derivative of the Lagrangian is given by:

$\displaystyle \frac{dL}{dt}=\sum_i\left(\frac{\partial L}{\partial q_i}\dot{q}_i+\frac{\partial L}{\partial\dot{q}_i}\ddot{q}_i\right)+\frac{\partial L}{\partial t}$

Using the Euler-Lagrange equations, $\displaystyle \frac{\partial L}{\partial q_i}=\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}_i}\right)$ , we can substitute this into the expression for $\displaystyle \frac{dL}{dt}$ :

$\displaystyle \frac{dL}{dt}=\sum_i\left(\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}_i}\right)\dot{q}_i+\frac{\partial L}{\partial\dot{q}_i}\ddot{q}_i\right)+\frac{\partial L}{\partial t}$

The first term on the right-hand side is precisely the total time derivative of $\displaystyle \sum_i\frac{\partial L}{\partial\dot{q}_i}\dot{q}_i$ . So, $\displaystyle \frac{dL}{dt} = \frac{d}{dt} \left( \sum_i \frac{\partial L}{\partial \dot{q}_i} \dot{q}_i \right) + \frac{\partial L}{\partial t}$
Rearranging, we get: $\displaystyle \frac{d}{dt} \left( \sum_i \frac{\partial L}{\partial \dot{q}_i} \dot{q}_i - L \right) = -\frac{\partial L}{\partial t}$

The quantity $\displaystyle H=\sum_i\frac{\partial L}{\partial\dot{q}_i}\dot{q}_i-L$ is the Hamiltonian (or energy) of the system. This equation shows that if the Lagrangian does not explicitly depend on time ( $\displaystyle \frac{\partial L}{\partial t}=0$ ), then the Hamiltonian $\displaystyle H$ is a conserved quantity. This is the conservation of energy and is obtained as a consequence of time invariance of the Lagrangian.

Now, let’s consider a general continuous transformation $\displaystyle q_i\to q_i+\epsilon K_i(q,\dot{q},t)$ where $\displaystyle \epsilon$ is an infinitesimal parameter and $\displaystyle K_i$ are functions describing the transformation. If the Lagrangian is invariant under this transformation (meaning $\displaystyle \delta L=0$ ), we will see that there exists a conserved quantity $\displaystyle G$ . The derivation of $\displaystyle G$ involves the using generalized momentum $\displaystyle p_i=\frac{\partial L}{\partial\dot{q}_i}$ .

The variation of the action $\displaystyle \delta S$ must be zero. If the Lagrangian is invariant under a transformation, then $\displaystyle \delta L=0$ . This implies:

$\displaystyle \delta L=\sum_i\left(\frac{\partial L}{\partial q_i}\delta q_i+\frac{\partial L}{\partial\dot{q}_i}\delta\dot{q}_i\right)=0$

Substituting the Euler-Lagrange equations, $\displaystyle \frac{\partial L}{\partial q_i}=\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}_i}\right)$ , we get:

$\displaystyle \sum_i\left(\frac{d}{dt}\left(\frac{\partial L}{\partial\dot{q}_i}\right)\delta q_i+\frac{\partial L}{\partial\dot{q}_i}\frac{d}{dt}(\delta q_i)\right)=0$

This is precisely the total time derivative of $\displaystyle \sum_i\frac{\partial L}{\partial\dot{q}_i}\delta q_i$ :

$\displaystyle \frac{d}{dt}\left(\sum_i\frac{\partial L}{\partial\dot{q}_i}\delta q_i\right)=0$

Since $\displaystyle \delta q_i=\epsilon K_i$ . We can write:

$\displaystyle \frac{d}{dt}\left(\sum_i p_i K_i\right)=0$

Thus, the quantity $\displaystyle G=\sum_i p_i K_i$ is conserved. This is the general form of Noether’s theorem for a continuous symmetry.

Specific Examples:

Time-translation symmetry: If $\displaystyle \frac{\partial L}{\partial t}=0$ (Lagrangian doesn’t explicitly depend on time), then energy is conserved. The conserved quantity as seen above is the Hamiltonian $\displaystyle H = \sum_i p_i \dot{q}_i - L$ .
Spatial-translation symmetry: If $\displaystyle L$ is invariant under translation in a direction (e.g., $\displaystyle x \to x + \epsilon$ , so $\displaystyle K_x=1$ , others $\displaystyle K_j=0$ ), then the linear momentum component in that direction is conserved. $\displaystyle G = p_x K_x = p_x \cdot 1 = p_x$
Rotational symmetry: If $\displaystyle L$ is invariant under rotations about an axis (e.g., z-axis: $\displaystyle \delta x = -\epsilon y$ , $\displaystyle \delta y = \epsilon x$ ; thus $\displaystyle K_x=-y$ , $\displaystyle K_y=x$ , $\displaystyle K_z=0$ ), then the angular momentum about that axis is conserved. $\displaystyle G = p_x K_x + p_y K_y = p_x(-y) + p_y(x) = x p_y - y p_x = L_z$