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In Lagrangian mechanics, the energy balance for an autonomous Lagrangian uses $E={\sum }_i{\dot{q}}_i\partial L/\partial {\dot{q}}_i-L$. For the stationary vector potential the velocity-linear magnetic term cancels in this expression, leaving kinetic energy. In cylindrical coordinates,Time independence and the cyclic angular coordinate give two constants:The three Euler-Lagrange equations areThe supplied angular velocity is $\dot{\varphi }=-2eg(z_0)/m$. Set $r=r_0$, $z=z_0$ and use this constant angular velocity. The radial right-hand side is $m{r}_0({\dot{\varphi }}^2+2eg\dot{\varphi }/m)=0$, the angular momentum is constant, and the vertical equation vanishes when $g^{\prime }(z_0)=0$. These equations and the initial data therefore giveFor nonzero radius and charge this is the requested circular orbit.

For an autonomous Lagrangian, the Euler-Lagrange equation usesApply the chain rule to the Hamiltonian along an arbitrary smooth path, without assuming the unforced Euler-Lagrange equation:Consequently the energy balance for an autonomous Lagrangian isWith explicit time dependence, an additional $-{\partial }_{t}L$ enters $dH/dt$. For ideal Gaussian white noise, the identity is understood through smooth-noise regularization or the Stratonovich chain rule. The original PDF correctly differentiates $L$ with respect to $\dot{x}$ in $H$; the supplied TeX's derivative with respect to $x$, its endpoint $t^2$, and its ordinary derivative of $A$ are transcription errors.

At thermal equilibrium, microscopic reversibility equates the probabilities of a path and its reversed path when both include their Boltzmann distribution initial weights. Denote their endpoint states by $z_1,z_2$, including velocity if needed. Since the Hamiltonian is even under time reversal in classical mechanics,This detailed balance condition and the energy balance for an autonomous Lagrangian yieldThis is the fluctuation-dissipation relation for a Langevin particle: the strength of Gaussian white noise is fixed by the damping and temperature, with $k_B$ the Boltzmann constant.

For an equilibrium coarse-grained variable, the unresolved microscopic states contribute entropy; their statistical weight is encoded in the Helmholtz free energy, rather than in a single microscopic energy. Relative to the same reference measure, ${\rho }_{\mathrm{eq}}(x)\propto e^{-\beta F(x)}$, so microscopic reversibility becomesThis extension assumes an equilibrium coarse-grained description with reversible path statistics; externally driven dynamics need not obey this relation.