24 Jun 2025
Lagrangian Dynamics and Applications
Hamilton's principle, Euler-Lagrange equations, oscillators, uniform gravity, and electromagnetic coupling.
Lagrange’s equations
Suppose the generalized force is
\[Q_j=-\frac{\partial V}{\partial q_j}+Q_j^{(\mathrm{nc})},\]where $Q_j^{(\mathrm{nc})}$ contains any non-conservative part. With the Lagrangian $L=T-V$, D’Alembert’s equation becomes
\[\boxed{ \frac{d}{dt}\frac{\partial L}{\partial\dot q_j} -\frac{\partial L}{\partial q_j}=Q_j^{(\mathrm{nc})}}.\]For a conservative system, the right-hand side is zero. These are Lagrange’s equations, also called the Euler-Lagrange equations.
Hamilton’s principle
Let the action between two fixed times be
\[S[q]=\int_{t_1}^{t_2}L(q_i,\dot q_i,t)\,dt.\]Compare the physical path with $q_i(t,\epsilon)=q_i(t)+\epsilon\eta_i(t)$. The configuration endpoints are fixed:
\[\eta_i(t_1)=\eta_i(t_2)=0.\]The first variation is
\[\delta S =\int_{t_1}^{t_2}\sum_i\left( \frac{\partial L}{\partial q_i}\eta_i +\frac{\partial L}{\partial\dot q_i}\dot\eta_i \right)dt.\]Integrating the second term by parts gives
\[\delta S =\left[\sum_i\frac{\partial L}{\partial\dot q_i}\eta_i\right]_{t_1}^{t_2} +\int_{t_1}^{t_2}\sum_i\left[ \frac{\partial L}{\partial q_i} -\frac{d}{dt}\left(\frac{\partial L}{\partial\dot q_i}\right) \right]\eta_i\,dt.\]The endpoint term vanishes. Hamilton’s principle, $\delta S=0$ for arbitrary interior variations $\eta_i$, therefore yields the Euler-Lagrange equations.
One-dimensional simple harmonic oscillator
For a mass $m$ attached to a spring of force constant $k$,
\[L=\frac12m\dot x^2-\frac12kx^2.\]Since $\partial L/\partial\dot x=m\dot x$ and $\partial L/\partial x=-kx$,
\[m\ddot x+kx=0.\]Thus
\[x(t)=A\cos(\omega t+\delta), \qquad \boxed{\omega=\sqrt{\frac{k}{m}}},\]where $\omega$ has units $\mathrm{s^{-1}}$.
Falling body in uniform gravity
Take $y$ positive upward and choose $V=mgy$. Then
\[L=\frac12m\dot y^2-mgy.\]The Euler-Lagrange equation gives
\[m\ddot y+mg=0, \qquad \boxed{y(t)=y_0+v_{0y}t-\frac12gt^2}.\]The sign follows from the stated upward-positive convention.
Two coupled oscillators
Consider two equal masses and three equal springs, with $x_1$ and $x_2$ measured from equilibrium:
\[L=\frac m2(\dot x_1^2+\dot x_2^2) -\frac k2\left[x_1^2+(x_2-x_1)^2+x_2^2\right].\]Differentiation gives
\[m\ddot x_1+2kx_1-kx_2=0, \qquad m\ddot x_2+2kx_2-kx_1=0.\]For a normal mode $x_j=A_j e^{i\omega t}$,
\[\begin{pmatrix} 2k-m\omega^2&-k\\ -k&2k-m\omega^2 \end{pmatrix} \begin{pmatrix}A_1\\A_2\end{pmatrix}=0.\]A non-zero amplitude requires
\[(2k-m\omega^2)^2-k^2=0.\]The two modes are therefore
\[\boxed{\omega_1^2=\frac{k}{m},\quad(A_1,A_2)\propto(1,1)},\]and
\[\boxed{\omega_2^2=\frac{3k}{m},\quad(A_1,A_2)\propto(1,-1)}.\]
Charged particle in an electromagnetic field
Introduce scalar and vector potentials by
\[\mathbf E=-\nabla\phi-\frac{\partial\mathbf A}{\partial t}, \qquad \mathbf B=\nabla\times\mathbf A.\]The Lagrangian of a particle of charge $q$ is
\[\boxed{ L=\frac12m\dot{\mathbf r}^{\,2} +q\dot{\mathbf r}\cdot\mathbf A(\mathbf r,t) -q\phi(\mathbf r,t)}.\]For the Cartesian coordinate $x_i$,
\[\frac{\partial L}{\partial\dot x_i}=m\dot x_i+qA_i,\] \[\frac{d}{dt}\frac{\partial L}{\partial\dot x_i} =m\ddot x_i +q\frac{\partial A_i}{\partial t} +q\dot x_j\frac{\partial A_i}{\partial x_j},\]and
\[\frac{\partial L}{\partial x_i} =q\dot x_j\frac{\partial A_j}{\partial x_i} -q\frac{\partial\phi}{\partial x_i}.\]Repeated Cartesian indices are summed. The Euler-Lagrange equation becomes
\[m\ddot x_i =q\left[ -\frac{\partial\phi}{\partial x_i} -\frac{\partial A_i}{\partial t} +\dot x_j\left( \frac{\partial A_j}{\partial x_i} -\frac{\partial A_i}{\partial x_j} \right) \right].\]The last term is $(\dot{\mathbf r}\times\mathbf B)_i$, so
\[\boxed{m\ddot{\mathbf r}=q(\mathbf E+\dot{\mathbf r}\times\mathbf B)}.\]
Discussion