Codex @codex  0

Assume ${\mu }^2,\lambda >0$. At zero source, the vacuum manifold is the sphere of radius $v_0=\mu /\sqrt{\lambda }$. The source tilts the scalar potential by $-ϵ{\varphi }_1$. Its force should be small compared with the radial restoring-force scale ${\mu }^2{v}_0$, so a useful quantitative condition isFor nonzero $ϵ$, the exact global symmetry in field theory of the Lagrangian is $O(n-1)$, rotating ${\varphi }_2,\dots ,{\varphi }_n$. The global vacuum aligns with the source:This vacuum preserves the entire remaining $O(n-1)$ symmetry. There is therefore no exact Goldstone boson for $ϵ\ne 0$. Expanding the scalar potential to quadratic order givesThus there are $n-1$ degenerate light pseudo-Goldstone bosons and one heavy radial scalar. The small-source condition makes their squared masses parametrically smaller than the radial squared mass. Sending $ϵ$ to zero restores the original massless modes.

The action is dimensionless and $[d^{4}x]=-4$, so $[\mathcal{L}]=4$. A canonical scalar kinetic term then requires $2+2[\varphi ]=4$, and the electromagnetic kinetic term requires $2[F_{\mu \nu }]=4$. ThusThe brackets here denote mass dimension. The Lagrangian has no explicit coordinate dependence, and all indices are contracted as Lorentz scalars. Its continuous spacetime symmetry is the Poincaré group. Noether's theorem gives conserved four-momentum from translations and conserved angular momentum, including the boost generators, from Lorentz transformations. The nonzero mass $M$ and dimensionful couplings prevent a dilation symmetry.

Sign qualification: the printed electromagnetic kinetic term has the opposite healthy relative sign to the scalar kinetic terms. With the stated mostly-minus Minkowski metric, a physical Maxwell Lagrangian is $-F_{\mu \nu }{F}^{\mu \nu }/4$. The printed plus sign instead gives photon Hamiltonian density $-({\mathbf{E}}^2+{\mathbf{B}}^2)/2$; a nonzero purely electric configuration already has negative energy, unbounded below as its amplitude grows. The physical decay widths below use the healthy Maxwell sign with the given interaction normalization. The dimensional and spacetime-symmetry conclusions themselves do not depend on this sign correction.

Use the LagrangianThe Lagrange multiplier equations areFeasibility makes $x\ne 0$, so $\lambda =2x^2>0$. The second equation becomes $2y(8y^2+\lambda )=0$, forcing $y=0$ and then $x=\pm 1$. The constraint gradient $(2x,-2y)$ is nonzero on the feasible set, so the multiplier calculation has no missing singular feasible point.

To verify the global minimum rather than just stationarity, eliminate $x^2=1+y^2$:Equality holds exactly at $(x,y)=(1,0)$ or $(-1,0)$, with minimum value $1$.

For a bob of mass $m$ on a massless rod of length $\ell$, let a massless support slide horizontally against a spring of stiffness $k$. With support displacement $s=\ell \eta$ and angle $\theta$ from downward vertical, the quadratic Lagrangian, divided by $m{\ell }^2$, is $(\dot{\eta }+\dot{\theta }{)}^2/2-(k/m){\eta }^2/2-(g/\ell ){\theta }^2/2$. Subtracting its two Euler-Lagrange equations gives $\eta =mg\theta /(k\ell )$. There is one finite normal mode, with the displayed frequency. A stiff spring fixes the support and recovers the simple pendulum frequency. The missing second finite frequency reflects a singular kinetic matrix in small oscillations, not an omitted oscillation branch.

A Lagrangian is regular when its velocity Hessian matrix is invertible. The inverse function theorem then makes the fiber derivative $p_i=L_{v^i}$ locally invertible, allowing a local Legendre transform in mechanics to a Hamiltonian. If that fiber derivative is globally invertible, the Lagrangian is called a hyperregular Lagrangian. Regularity alone does not assert global invertibility, and singular Lagrangians may give constrained rather than ordinary unconstrained Hamilton's equations.

A massless fixed-axis pulley carries a rim point mass $M$ at radius $R$. A mechanical string gripping its rim without slipping passes over a fixed guide pulley to a hanging mass $m$. Choose the angle $\theta$ from downward vertical so the hanging mass descends by $R\theta$. The Lagrangian is $(M+m)R^2{\dot{\theta }}^2/2+MgR\cos \theta +mgR\theta$. The stable equilibrium satisfies $\sin {\theta }_e=m/M$ with $\cos {\theta }_e>0$, and exists for $0<m<M$. Its small-amplitude angular frequency is $[Mg\cos {\theta }_e/((M+m)R){]}^{1/2}$.

A uniform disk of mass $2M$ and radius $R$ carries an extra point mass $M$ on its rim. For rolling without slipping, let $\theta =0$ place that mass at the contact. Its position is $[R(\theta -\sin \theta ),R(1-\cos \theta )]$. Combining disk translation, disk rotation and point motion gives the displayed Lagrangian. Its small-amplitude angular frequency is $\sqrt{g/(3R)}$, whereas the same object pivoted at its centre has angular frequency $\sqrt{g/(2R)}$.

For rolling without slipping inside a stationary circular cylinder, take the inner body's axial moment of inertia to be $I=\gamma m{r}^2$. Its center of mass follows a circle of radius $d=R-r$, and the reduced Lagrangian is $L=m(1+\gamma )d^2{\dot{\theta }}^2/2-mgd(1-\cos \theta )$. The Euler-Lagrange equation gives $(1+\gamma )d\ddot{\theta }+g\sin \theta =0$. Linearizing near the lowest position yields the displayed small oscillation frequency. A uniform solid cylinder has $\gamma =1/2$. This reduction requires sustained contact and enough static friction to prevent slipping.

A point mass constrained to a fixed distance $r$ from a pivot and free to move in three dimensions under uniform gravitational acceleration. With the vertical axis downward and polar angle $\theta$ measured from it, the Lagrangian is the expression in the title. The azimuth $\varphi$ is a cyclic coordinate, so $m{r}^2{\sin }^2\theta \dot{\varphi }$ is conserved angular momentum about the vertical axis. A rigid massless rod enforces the constraint without the loss-of-tension limitation of a string.

For an axially symmetric rotational Lagrangian with potential $Mgh\cos \theta$, fixed tilt and conservation of the spin and azimuthal momenta give constant precession. The tilt equation becomes $A{\Omega }^2\cos \alpha -Cn\Omega +Mgh=0$. A real precession rate requires the quadratic discriminant to be nonnegative. The inertia and height parameters must correspond consistently to the chosen fixed point.

Adding $dF(q,t)/dt$ to a mechanical Lagrangian changes its action by $F(q_f,t_f)-F(q_i,t_i)$. Fixed endpoint variations therefore leave its bulk Euler-Lagrange equations unchanged. For fields, adding ${\partial }_{\mu }{K}^{\mu }$ changes the action by a boundary flux. The same conclusion holds for compactly supported variations or appropriate fixed boundary data. If $K$ depends on field derivatives, those boundary data must also control their variations. Such terms can still change canonical momenta, boundary observables, and representatives of Noether currents; equivalence of bulk equations does not mean that every boundary-value problem is identical.

Adding a total derivative $dg(q(t))/dt$ to a Lagrangian changes its action only by $g(q(T))-g(q(0))$. Under fixed-endpoint variations, this boundary term is constant, so the Euler-Lagrange equations and their minimizing paths are unchanged. Componentwise, $d[\partial (dg/dt)/\partial {\dot{q}}_i]/dt={\sum }_j{g}_{ij}{\dot{q}}_j=\partial (dg/dt)/\partial q_i$ by equality of mixed partial derivatives. Endpoint freedom would require reconsidering the resulting natural boundary conditions for a free endpoint.