Lagrangian Mechanics
Lagrangian mechanics reformulates dynamics in terms of generalized coordinates $q_i(t)$ and a single scalar function $L(q_i,\dot{q}_i,t)$, the Lagrangian. Rather than starting from forces, one derives the equations of motion by demanding that the physical trajectory makes a functional called the action stationary among all nearby paths with the same endpoints.
\[S[q]=\int_{t_1}^{t_2} L(q_i,\dot{q}_i,t)\,dt\]The principle of stationary action is
\[\delta S=0\]Action Principle and Euler–Lagrange Equations
Consider a variation of the path $q_i(t)\to q_i(t)+\delta q_i(t)$ with fixed endpoints $\delta q_i(t_1)=\delta q_i(t_2)=0$. The first variation of the action is
\[\delta S=\int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q_i}\delta q_i+\frac{\partial L}{\partial \dot{q}_i}\delta \dot{q}_i\right)dt\]Using integration by parts on the term involving $\delta\dot{q}_i$,
\[\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot{q}_i}\delta \dot{q}_i\,dt = \left[\frac{\partial L}{\partial \dot{q}_i}\delta q_i\right]_{t_1}^{t_2} - \int_{t_1}^{t_2}\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)\delta q_i\,dt\]The boundary term vanishes because the endpoints are fixed. Therefore,
\[\delta S=\int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q_i}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}\right)\delta q_i\,dt\]Since the variations $\delta q_i(t)$ are otherwise arbitrary, the integrand must vanish, yielding the Euler–Lagrange equations,
\[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)-\frac{\partial L}{\partial q_i}=0\]These equations replace Newton’s laws in the Lagrangian framework and are valid for any coordinates and any Lagrangian that correctly encodes the dynamics.
Structure of Lagrangians and Generalized Momentum
For conservative systems in simple coordinates, a common and useful choice is
\[L=T-V\]where $T$ is kinetic energy and $V$ is potential energy. This form is not fundamental; it is a special case that works when forces derive from a potential and velocity-dependent interactions are absent. In many important systems, $L$ contains mixed velocity terms (e.g. $\dot{x}\dot{y}$) or velocity–position couplings (e.g. $x\dot{y}$), which naturally describe constraints, electromagnetic interactions, and effective descriptions.
A general quadratic Lagrangian that captures many mechanical and field-motivated examples is
\[L=\frac{1}{2}a_{ij}(q)\dot{q}_i\dot{q}_j+b_i(q)\dot{q}_i-V(q)\]In this form:
- $a_{ij}(q)$ acts like a configuration-space metric, allowing kinetic coupling between coordinates.
- $b_i(q)\dot{q}_i$ introduces velocity–position coupling (often called a “gyroscopic” or “magnetic-type” term).
- $V(q)$ is the potential.
The generalized momentum conjugate to $q_i$ is defined by
\[p_i=\frac{\partial L}{\partial \dot{q}_i}\]Unlike Newtonian momentum, $p_i$ can depend on multiple velocities when the kinetic term is coupled. For instance, for $L=\dot{x}\dot{y}$ one finds
\[p_x=\dot{y},\quad p_y=\dot{x}\]showing that the coordinates are dynamically coupled at the level of canonical momenta.
Worked Examples, Conservation Laws, and Nonstandard Lagrangians
A minimal example illustrating coupled velocities is
\[L=\dot{x}\dot{y}\]Then
\[\frac{\partial L}{\partial \dot{x}}=\dot{y},\quad \frac{\partial L}{\partial x}=0\]so the Euler–Lagrange equation for $x$ gives
\[\frac{d}{dt}(\dot{y})=0 \Rightarrow \ddot{y}=0\]and similarly,
\[\ddot{x}=0\]Velocity–position couplings encode forces without introducing a potential. A standard “magnetic-type” term is
\[L=\frac{m}{2}(\dot{x}^2+\dot{y}^2)+qB(x\dot{y}-y\dot{x})\]which yields Lorentz-force-like equations,
\[m\ddot{x}=qB\dot{y},\quad m\ddot{y}=-qB\dot{x}\]Symmetry principles organize conserved quantities. If $L$ has no explicit time dependence, $\frac{\partial L}{\partial t}=0$, then the energy-like quantity
\[E=\sum_i \dot{q}_i\frac{\partial L}{\partial \dot{q}_i}-L\]is conserved. More generally, invariance of $L$ under a continuous transformation implies a conserved quantity (Noether’s theorem), with key correspondences:
- time translation $\to$ energy conservation,
- spatial translation $\to$ momentum conservation,
- rotation $\to$ angular momentum conservation.
Not every physical system admits a simple $L=T-V$ representation. Examples where nonstandard constructions arise include dissipative systems (often requiring enlarged variable sets), PT-symmetric models, effective theories, and constrained systems. The correct viewpoint is that the Lagrangian is any function that reproduces the correct equations of motion through the Euler–Lagrange equations.
Exercises
Q1. Derive the Euler–Lagrange equation for
\[L=\frac{1}{2}m\dot{x}^2-V(x)\]Solution:
\[\frac{d}{dt}(m\dot{x})-\left(-V'(x)\right)=0 \Rightarrow m\ddot{x}+V'(x)=0\]Q2. For
\[L=\dot{x}\dot{y}\]find equations of motion and conserved quantities.
Solution:
\[\ddot{x}=0,\quad \ddot{y}=0\]and
\[p_x=\dot{y},\quad p_y=\dot{x}\]If $L$ has no explicit time dependence, then
\[E=\dot{x}p_x+\dot{y}p_y-L=\dot{x}\dot{y}\]Q3. Show that
\[L=\frac{1}{2}m\dot{x}^2+qA(x)\dot{x}\]leads to modified momentum.
Solution:
\[p=\frac{\partial L}{\partial \dot{x}}=m\dot{x}+qA(x)\]Q4. Derive equations of motion for
\[L=\frac{1}{2}(\dot{x}^2+\dot{y}^2)+\alpha x\dot{y}\]Solution:
\[\frac{d}{dt}(\dot{x})-\alpha\dot{y}=0 \Rightarrow \ddot{x}=\alpha\dot{y}\]and
\[\frac{d}{dt}(\dot{y}+\alpha x)=0 \Rightarrow \ddot{y}+\alpha\dot{x}=0\]Q5. Show that adding a total time derivative
\[L'=L+\frac{dF}{dt}\]does not change the equations of motion.
Solution:
\[\int_{t_1}^{t_2}\frac{dF}{dt}\,dt=F(t_2)-F(t_1)\]so its variation vanishes for fixed endpoints:
\[\delta\int_{t_1}^{t_2}\frac{dF}{dt}\,dt=0\]Q6. For two coupled oscillators,
\[L=\dot{x}\dot{y}-\omega^2xy\]find normal modes.
Solution: the equations are
\[\ddot{x}+\omega^2 y=0,\quad \ddot{y}+\omega^2 x=0\]Define
\[z_1=x+y,\quad z_2=x-y\]Then
\[\ddot{z}_1+\omega^2 z_1=0,\quad \ddot{z}_2-\omega^2 z_2=0\]Additional Problem
We consider a two-degree-of-freedom Lagrangian in coordinates $x(t),y(t)$ with velocity couplings typical of planar systems in a rotating frame. The goal is to (i) write the Euler–Lagrange equations cleanly and (ii) Obtain the solution to the equation of motion of the form $e^{\omega t}(k_1+k_2 t)$ for the homogeneous differential equation it solves.
Correct Form of the Lagrangian and Notation
To preserve the intended structure (quadratic potential, Coriolis-type coupling, kinetic term, and an $xy$ coupling), we write the Lagrangian as
\[L=-\frac{\varepsilon}{2}(x^2+y^2)+\omega\left(x\dot y-y\dot x\right)+\dot x\,\dot y-\omega^2xy.\]The velocities are
\[\dot x=\frac{dx}{dt}, \qquad \dot y=\frac{dy}{dt}.\]Euler–Lagrange Equations and the Coupled Dynamics
For the coordinate $x$, the Euler–Lagrange equation is
\[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot x}\right)-\frac{\partial L}{\partial x}=0.\]Compute the derivatives:
\[\frac{\partial L}{\partial \dot x}=-\omega y+\dot y,\]hence
\[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot x}\right)=-\omega\dot y+\ddot y.\]Next,
\[\frac{\partial L}{\partial x}=-\varepsilon x+\omega\dot y-\omega^2y.\]Substituting into Euler–Lagrange gives
\[\left(-\omega\dot y+\ddot y\right)-\left(-\varepsilon x+\omega\dot y-\omega^2y\right)=0,\]so
\[\ddot y-2\omega\dot y+\omega^2y+\varepsilon x=0.\]Similarly, for the coordinate $y$:
\[\ddot x+2\omega\dot x+\omega^2x+\varepsilon y=0.\]Thus, the Lagrangian generates a coupled linear system for $(x,y)$ with constant coefficients; the $\varepsilon$ terms couple the coordinates, while the $\pm 2\omega$ terms represent the velocity coupling.
Solving the Homogeneous ODE and Interpreting the Result
Solve
\[\ddot y-2\omega\dot y+\omega^2y=0\]by the trial form
\[y=e^{rt}.\]Then
\[\dot y=re^{rt}, \qquad \ddot y=r^2e^{rt}.\]Substitution yields
\[r^2e^{rt}-2\omega re^{rt}+\omega^2e^{rt}=0.\]Since $e^{rt}\neq 0$,
\[r^2-2\omega r+\omega^2=0,\]which factors as
\[(r-\omega)^2=0.\]Therefore the characteristic root is repeated:
\[r=\omega.\]For a double root, the two linearly independent solutions are $e^{\omega t}$ and $t e^{\omega t}$, so the general solution is
\[y(t)=e^{\omega t}(k_1+k_2 t).\]The structural reason for the $t e^{\omega t}$ term is the repeated root of the characteristic polynomial: the second independent solution must be multiplied by $t$ to remain linearly independent of the first.
\[y(t)=e^{\omega t}(k_1+k_2 t)\]