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For a constant magnetic field,is the symmetric gauge.

A uniform electric field perpendicular to a uniform magnetic field shifts each cyclotron oscillator's guiding centre and makes its energy depend linearly on the conserved momentum. The formerly degenerate Landau levels therefore become tilted bands.

For a spin-one-half particle with the Pauli magnetic term and gyromagnetic factor two, the orbital energies $\hslash {\omega }_c(n+1/2)$ split into $\hslash {\omega }_{c}n$ and $\hslash {\omega }_c(n+1)$. The lowest level has one spin branch; every positive level has two.

In a planar region of area $A$, each spinless Landau level has guiding-centre degeneracyMagnetic periodic boundary conditions require this flux count to be an integer.

A particle of charge $q$ and mass $m$ in a uniform magnetic field of magnitude $B$ has cyclotron angular frequency ${\omega }_c=|qB|/m$.

For a particle of charge $q$ in a vector potential $\mathbf{A}$, the canonical momentum in the free Hamiltonian is replaced by $\mathbf{p}-q\mathbf{A}$, giving

If $b(x)=B{x}^{-1/5}$, balancing the momentum flux and transverse viscosity gives

For a steady incompressible plume with localized vertical body force $b(x)\delta (y)$, decay of velocity and shear at transverse infinity gives

A Lagrangian with no explicit time dependence conserves $E={\sum }_i{\dot{q}}_i\partial L/\partial {\dot{q}}_i-L$.

Near a stable equilibrium, retaining quadratic Lagrangian terms gives a linear system $\mathsf{M}\ddot{z}+\mathsf{K}z=0$.

The resulting linearized motion decomposes into normal modes near the stable equilibrium.

For point masses with generalized coordinates, kinetic energy follows by differentiating their Cartesian position vectors and summing $m_i|{\dot{r}}_i{|}^2/2$.

A double pendulum consists of two pendula coupled because the second pivot moves with the first mass. It has two degrees of freedom: even when the upper pivot is fixed, moving the upper arm carries the lower pivot along. This makes the double pendulum a compact example of how mechanical constraints couple motion, and how Lagrangian mechanics converts geometry into dynamics.

Use two equal point masses $m>0$ and two massless rigid rods of equal length $\ell >0$, with joints that allow free rotation in one vertical plane. The upper pivot is fixed; gravity is uniform, with gravitational acceleration $g>0$. Neglect joint friction and drag force. Both angles ${\theta }_1,{\theta }_2$ are measured from the downward vertical, positive toward the right, in radians. In particular, ${\theta }_2$ is an absolute angle, not the bend between the rods: that bend is ${\theta }_2-{\theta }_1$.

The double pendulum Lagrangian below gives the exact dynamics of this ideal model. Near the downward stable equilibrium, normal modes of a double pendulum replace the coupled motion with two independent harmonic oscillators. Their fixed amplitude ratios explain which releases make the rods swing together and which make them swing in opposite directions.

If $L=|\dot{q}{|}^2/2-V(q)$ and $V$ depends only on $|q{|}^2$ and $(q\cdot n{)}^2$, then rotations about the axis $n$ leave $L$ invariant. The quantity supplied by Noether theorem is the axial angular momentumIf $L$ also has no explicit time dependence, the energy $|\dot{q}{|}^2/2+V(q)$ is conserved independently.

If a Lagrangian is independent of a coordinate, the conjugate momentum to that coordinate is conserved by the Euler--Lagrange equations.

The generalized, or conjugate, momentum associated with a coordinate $q_i$ is

In Lagrangian mechanics, the coordinate variable describes the configuration; its conjugate momentum need not equal the Cartesian mechanical momentum.

For electromagnetic scalar and vector potentials $\varphi$ and $A$,Withthe Euler--Lagrange equations give the Lorentz force $m\ddot{r}=q(E+\dot{r}\times B)$, while the canonical momentum is $p=m\dot{r}+qA$.

Replacing a Lagrangian byadds only the endpoint term $F(q(t_2),t_2)-F(q(t_1),t_1)$ to the action. Fixed-endpoint variations therefore give the same Euler--Lagrange equations.

The action of a path is the time integral of its Lagrangian. The principle of stationary action requires its first variation to vanish for every admissible fixed-endpoint variation.

For a topless rectangular box with dimensions $x,y,z$, volume $V$, and weighted face costan interior optimum equalizes the three terms:The same result and global minimality follow from the arithmetic-geometric mean inequality.

Forsuppose the optimizer and multiplier vary smoothly and use $L=f-\lambda (g-b)$. Differentiating the optimum value and using stationarity gives