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Suppose $f$ is an absolutely continuous function on every finite interval $[0,T]$, both $e^{-st}f(t)$ and $e^{-st}{f}^{\prime }(t)$ are Lebesgue integrable functions, and $e^{-sT}f(T)\to 0$. Then integration by parts on $[0,T]$ followed by a limit givesHere $f(0+)$ is the right one-sided limit. Exponential order bounds on $f$ and $f^{\prime }$ suffice when $\mathrm{Re}s$ exceeds both bounds. For $f(t)=e^{-2t}$, the identity reads $-2/(s+2)=s/(s+2)-1$ on $\mathrm{Re}s>-2$. This is the ordinary derivative on nonnegative time; differentiating a causal extension by zero at its jump instead introduces a Dirac delta function. ^[ref].

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.

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.