Shifted momentum from a total-derivative modification: $p^{\mathrm{\prime }}=p+F_x$, $H^{\mathrm{\prime }}=H-F_t$ and unchanged motion on explicit data

Declare the one-label model of the force branch of Adopted force model: force maps, the equation of motion, additive composition of force contributions and frame covariance with $N=1$, mass $m>0$, time window $I_0=\mathbb{R}$, configuration domain $Q={\mathbb{R}}^3$, a $C^2$ potential $U:\mathbb{R}\times Q\to \mathbb{R}$ and no residual forces; write $L(t,x,v)=\frac{1}{2}m\mid v{\mid }_2^2-U(t,x)$ for its mechanical Lagrangian (Potential presentations of a force model, potential and mechanical energy, closed versus exact force fields, local potentials and equipotential level sets, The action functional of a force model with a potential presentation: the mechanical Lagrangian, its chart instance and its dimensions). Fix a real parameter $\alpha$ and the smooth function $F(t,x):=\alpha t{x}_1$ on the full time-configuration projection $\mathbb{R}\times Q$, and define

$L^{\mathrm{\prime }}(t,x,v):=L(t,x,v)+F_t(t,x)+F_x(t,x)\cdot v=L(t,x,v)+\alpha x_1+\alpha t{v}_1\mathrm{.}$

The units are $[m]=\mu$, $[x]=\mathrm{\ell }$, $[t]=\tau$, $[\alpha ]=\mu \mathrm{\ell }{\tau }^{-2}$ (force, newtons), and $[F]=\mu {\mathrm{\ell }}^2{\tau }^{-1}$ (action). The free initial speed $q$ below has dimension $\mathrm{\ell }{\tau }^{-1}$ and is read in metres per second. Then:

1. $L^{\mathrm{\prime }}$ is a fibre-full $C^2$ time-dependent Lagrangian chart with velocity gradient $L_v^{\mathrm{\prime }}(t,x,v)=mv+\alpha t{e}_1$, $e_1=(1,0,0)$; the fibre map $v\mapsto L_v^{\mathrm{\prime }}(t,x,v)$ is a $C^2$ bijection of ${\mathbb{R}}^3$ onto itself with inverse $v=(p^{\mathrm{\prime }}-\alpha t{e}_1)\mathrm{/}m$, so $L^{\mathrm{\prime }}$ is hyperregular on its generic time-dependent chart (The Legendre map of a time-dependent Lagrangian: regularity and hyperregularity) and its fibre derivative is the shifted momentum $p^{\mathrm{\prime }}=p+\alpha t{e}_1,p=mv,$ which differs from the mechanical momentum $mv$ whenever $\alpha t\mathrm{\ne }0$.
2. the Hamiltonian of the branch of $L^{\mathrm{\prime }}$ is $H^{\mathrm{\prime }}(t,x,p^{\mathrm{\prime }})=\frac{\mid p^{\mathrm{\prime }}-\alpha t{e}_1{\mid }_2^2}{2m}+U(t,x)-\alpha x_1=H(t,x,p^{\mathrm{\prime }}-\alpha t{e}_1)-\alpha x_1,$ where $H$ is the Hamiltonian of the presentation (The Hamiltonian of a hyperregular potential presentation: the Legendre transform of the mechanical Lagrangian, its domain and its dimensions, Every potential presentation is hyperregular in its charts, with inverse branch $v={\mathsf{M}}^{-1}p$ (or $G(q{)}^{-1}p$) and the explicit kinetic-plus-potential Hamiltonian);
3. the Hamilton systems of $H$ and $H^{\mathrm{\prime }}$ are equivalent under the shift $p^{\mathrm{\prime }}=p+\alpha t{e}_1$: $\dot{x}=H_{p^{\mathrm{\prime }}}^{\mathrm{\prime }}=(p^{\mathrm{\prime }}-\alpha t{e}_1)\mathrm{/}m$ and ${\dot{p}}^{\mathrm{\prime }}=-H_x^{\mathrm{\prime }}=-{\mathrm{\nabla }}_{x}U+\alpha e_1$ hold exactly when $\dot{x}=p\mathrm{/}m$ and $\dot{p}=-{\mathrm{\nabla }}_{x}U$ hold, so both systems have the same $x$-curves, namely the motions of the model (Hamilton's equations of the mechanical presentation are equivalent to the model's equation of motion, with residual forces as momentum sources and the coordinate form on every chart, Total-derivative Lagrangians, the potential representative and scaling: invariant stationary curves with endpoint terms); the Euler-Lagrange expressions coincide, $E_{L^{\mathrm{\prime }}}(x)=E_L(x)$, and the actions differ by the endpoint value $\alpha (b{x}_1(b)-a{x}_1(a))$;
4. on free data $U\equiv 0$, with $m>0$, $\alpha \mathrm{\ne }0$, initial position $x(0)=0$ and initial velocity $v(0)=(q,0,0)$ with $q>0$, the curves $x(t)=(qt,0,0),p^{\mathrm{\prime }}(t)=(mq+\alpha t,0,0)$ solve the shifted Hamilton system, and along them $H^{\mathrm{\prime }}(t,x(t),p^{\mathrm{\prime }}(t))=\frac{1}{2}m{q}^2-\alpha qt,K(t)+U(t,x(t))=\frac{1}{2}m{q}^2\mathrm{.}$ Thus $H^{\mathrm{\prime }}$ differs from the mechanical energy for $t\mathrm{\ne }0$ while the motion is unchanged. For the numerical instance $m=1\mathrm{k}\mathrm{g}$ and $\alpha =1\mathrm{N}$ the same dimensionally explicit formulas apply.

All quantities are model values of the declared idealization; no measurement is reported.

• The shift is exactly the gauge derivative. The general pattern is $p^{\mathrm{\prime }}=p+F_x(t,x)$ and $H^{\mathrm{\prime }}(t,x,p^{\mathrm{\prime }})=H(t,x,p^{\mathrm{\prime }}-F_x(t,x))-F_t(t,x)$: the momentum is translated by the gradient of the added function and the Hamiltonian is corrected by its time derivative. The modified Lagrangian is hyperregular but is not the mechanical Lagrangian of the presentation, and $p^{\mathrm{\prime }}$ is not the momentum $\mathsf{M}v$ of The momentum of a potential presentation: the fibre derivative of the mechanical Lagrangian, the Cartesian momentum Mv and generalized momenta; it is the fibre derivative of the modified chart, which is why the Hamiltonian of the modified chart is not the mechanical energy.

• Source check. Tong's Section 2.5.7 records the invariance of the equations under addition of a total derivative, and his Section 4.1.3 electromagnetic example exhibits a canonical momentum differing from the mechanical one by a gradient term; Stewart's Section 1.2 moving-cart computation exhibits the same pattern with a constant gradient and shows that the shifted Hamiltonian can be conserved while the mechanical energy is not. The computation above is the model instance of those statements; no electromagnetic interpretation of the parameter $\alpha$ is claimed.