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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.

Let $Q$ be the configuration smooth manifold, and let $L(t,q,v)$ be a smooth Lagrangian on its tangent bundle. For a path with fixed endpoints, its action isThe principle of stationary action requires the first variation to vanish for every fixed-endpoint variation. In a coordinate chart take $q_s=q+s\eta$, with $\eta (a)=\eta (b)=0$. Differentiation under the integral and integration by parts giveRepeated coordinate indices are summed. The endpoint term vanishes. Since compactly supported variations can be chosen independently in each coordinate, the fundamental lemma of the calculus of variations yields the Euler-Lagrange equationsConversely these equations make the displayed first variation zero for every fixed-endpoint variation. Variations localized in coordinate charts establish the same assertion for paths on $Q$. Thus the Euler-Lagrange equations are exactly the stationary-path condition, not necessarily a condition for an action minimum.

Use natural units and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. A real scalar field assigns a real variable $\varphi (t,\mathbf{x})$ to each spatial point. Its Lagrangian density can be taken to beThe principle of stationary action gives the Euler-Lagrange equation $\square \varphi +V^{\prime }(\varphi )=0$. With $V(\varphi )=m^2{\varphi }^2/2$, this is the Klein-Gordon equation. An additional nonlinear part of $V$ describes interactions.

The canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. The Legendre transform in mechanics gives the canonical Hamiltonian density of a real scalar fieldThe Hamiltonian equations $\dot{\varphi }=\pi$ and $\dot{\pi }={\nabla }^2\varphi -V^{\prime }(\varphi )$ recover the same field equation. In canonical quantization, the fields become operators satisfying the equal-time canonical commutation relationsA spatial lattice makes the analogy with many coupled quantum-mechanical coordinates precise. Each lattice field value is a coordinate, with its own conjugate momentum. The path integral is another representation of the same quantum evolution.

To see its origin, first consider one coordinate with $H=p^2/(2M)+V(q)$. Split a time interval into $N$ steps of length $\epsilon$ and insert position and momentum resolutions of the identity. The short-time kernel isMultiplying the kernels and integrating over intermediate positions gives the phase-space path integralThe endpoints of $q$ are fixed. The momentum integrals are Gaussian integrals; completing the square produces the configuration-space path integralAt finite slicing its normalization contains $(M/(2\pi i\epsilon ){)}^{N/2}{\prod }_{j=1}^{N-1}d{q}_j$. This fixes the composition law and the initial delta-function kernel. One sums over all paths, not merely solutions of the classical equation. Restoring $\hslash$ replaces the weight by $e^{iS/\hslash }$; stationary phase explains the emergence of classical trajectories.

For the field, use scalar field configuration eigenstates $|\phi ⟩$, satisfying $\hat{\varphi }(\mathbf{x})|\phi ⟩=\phi (\mathbf{x})|\phi ⟩$. Insert their completeness relations on every time slice. This givesThe endpoint field configurations are fixed. Integrating the Gaussian momentum variables leaves the scalar field path integral $\mathcal{N}\int \mathcal{D}\varphi e^{iS[\varphi ]}$. The functional measure means a regulated product over the field variables. A spacetime lattice or another ultraviolet cutoff makes this product finite before the continuum limit; interacting continuum calculations may require renormalization. The oscillatory Minkowski weight is an amplitude, not a positive probability density.

For vacuum expectation values, the boundaries must select the vacuum rather than arbitrary field configurations. Long imaginary-time evolution suppresses excited states: $e^{-TH}|n⟩=e^{-T{E}_n}|n⟩$, so after normalization only the lowest-energy component remains as $T\to \infty$. This is vacuum projection by imaginary time. The corresponding Feynman i-epsilon prescription in the real-time integral specifies the vacuum boundary conditions and the poles of the propagator. With $t=-i\tau$, the Euclidean path integral has the weight $e^{-S_E}$, whereIt is often a useful regulated starting point; analytic continuation returns the vacuum time-ordered quantities.

Introduce a classical source $J(x)$ and define the normalized vacuum generating functionalwith the same vacuum prescription in numerator and denominator. A functional derivative brings down $i\varphi (x)$. The order of the time slices makes the operator insertion time-ordered. Thus source differentiation inserts time-ordered field operators:The denominator removes vacuum diagrams and gives normalized expectation values. It is essential that these are time-ordered products; differentiating this vacuum functional does not directly give every possible operator ordering.

The free theory illustrates the method. Its quadratic kernel is $K=-\square -m^2$ with the vacuum pole prescription, and completing the square gives the Gaussian evaluation of a free scalar generating functionalTwo source derivatives give ${\Delta }_F=⟨\Omega |T\{\hat{\varphi }(x)\hat{\varphi }(y)\}|\Omega ⟩$. Higher derivatives give all pairings, the content of Wick theorem. For an interaction $S_{\mathrm{int}}$, one may use path-integral perturbation by source derivatives:Expanding this expression generates Feynman diagrams and their Wick contractions. The connected generating functional $W[J]=-i\log Z[J]$ retains connected contributions; in particular $\delta W/\delta J=⟨\varphi {⟩}_J$. These functionals turn the computation of field-operator expectations into source differentiation of an ordinary regulated integral.

Use gamma matrices satisfying the Clifford algebra relation $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, with ${\gamma }^{0†}={\gamma }^0$ and ${\gamma }^{i†}=-{\gamma }^i$. The Dirac adjoint is $\bar{\Psi }={\Psi }^{†}{\gamma }^0$. Take the Dirac actionIt differs from the manifestly Hermitian form with $(i/2)\bar{\Psi }{\gamma }^{\mu }{\overleftrightarrow{\partial }}_{\mu }\Psi$ only by a boundary term. Treat $\Psi$ and $\bar{\Psi }$ as independent variables when applying the principle of stationary action. Varying $\bar{\Psi }$ givesVarying $\Psi$ and integrating by parts gives the adjoint Dirac equation, $i({\partial }_{\mu }\bar{\Psi }){\gamma }^{\mu }+m\bar{\Psi }=0$. Multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ also gives $(\square +m^2)\Psi =0$, so its dispersion relation is Lorentz invariant.

For a Lorentz transformation $x^{\prime }=\Lambda x$, the field transforms in the Spinor representation of the Lorentz group:Since ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$, this identity gives the Lorentz covariance of the Dirac operator:Thus every solution is carried to another solution. The field is a spinor rather than a four-vector; the transformation of the gamma matrices supplies the necessary covariance. The Dirac action is Lorentz invariant because its integrand is a scalar and $d^{4}x$ is invariant.

For electric charge $q$, use the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$. The minimal electromagnetic coupling of a Dirac field isThe local gauge transformations in this convention areDirect substitution gives $D_{\mu }^{\prime }{\Psi }^{\prime }=e^{-iq\alpha }{D}_{\mu }\Psi$, which proves gauge covariance of the charged Dirac equation. The mass and kinetic terms are invariant, and $F_{\mu \nu }^{\prime }=F_{\mu \nu }$, so the full action has gauge invariance. The conserved current is $j^{\mu }=q\bar{\Psi }{\gamma }^{\mu }\Psi$. It transforms as a four-vector under Lorentz transformations, and varying $A_{\mu }$ gives ${\partial }_{\nu }{F}^{\nu \mu }=j^{\mu }$.

To make the infinitesimal spinor transformation explicit, write ${\Lambda }^{\mu }{}_{\nu }={\delta }^{\mu }{}_{\nu }+{\omega }^{\mu }{}_{\nu }$ with ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$. DefineThe Lorentz generators from gamma-matrix commutators have precisely the needed algebra: $[{\sigma }^{\mu \nu },{\gamma }^{\rho }]=2i({\eta }^{\nu \rho }{\gamma }^{\mu }-{\eta }^{\mu \rho }{\gamma }^{\nu })$, which verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }$ to first order. The infinitesimal transformation of a Dirac field is thereforeAt the same coordinate argument, the orbital change must also be included:Both formulas describe the same transformation; their arguments differ.

Finally, the gamma adjoint identities imply ${\sigma }^{\mu \nu †}={\gamma }^0{\sigma }^{\mu \nu }{\gamma }^0$ and hence the pseudo-unitarity of the spinor Lorentz representation, $S^{†}{\gamma }^{0}S={\gamma }^0$. It follows thatThus the Dirac scalar bilinear $\bar{\Psi }\Psi$ is a Lorentz scalar. Infinitesimally the spin terms in $\delta (\bar{\Psi }\Psi )$ cancel; at fixed coordinates only the ordinary scalar orbital transformation remains.

Choose boundary states with nonzero overlap with the lowest-energy state in the vacuum sector $Q=0$ and in the one-kink topological sector $Q=1$. In a finite spatial box, a fixed scalar field configuration eigenstate may be used formally, with temporal endpoints equal to the chosen configuration. More generally, smear the endpoints with wavefunctionals ${\Psi }_Q$. Their unitary time evolution kernels areThe inner scalar field path integral remains in the chosen topological sector, with $\varphi (+\infty )-\varphi (-\infty )=2\pi Q$. A zero-total-momentum projection can be included to select the rest state; alternatively, the translational prefactor does not change the large-time exponential. After a Wick rotation to physical Euclidean time $\tau$, the energy eigenstate expansion gives $K_Q^E(\tau )\sim C_Q{e}^{-E_Q\tau }$. Thus the exact vacuum-subtracted soliton mass isEquivalently it is $i{T}^{-1}\log (K_1/K_0)$ at large time with a damping prescription. The ratio subtracts the vacuum energy; the Hamiltonian operator and action here are the fully regulated and renormalized ones, not merely their classical approximations.

With the dimensionless coordinates of this paper, the correctly normalized classical action isFor the static Sine-Gordon kink, ${\varphi }_K^{\prime }=2\mathrm{sech}x$ and $1-\cos {\varphi }_K=2{\mathrm{sech}}^{2}x$, so $M_{\mathrm{cl}}=8m/{\beta }^2$. Write $\eta =\delta \varphi$. Expanding and integrating by parts givesThe first variation is ${\beta }^{-2}\int ({\varphi }_K^{\prime \prime }-\sin {\varphi }_K)\eta$, plus boundary terms. It vanishes because the kink satisfies the Euler-Lagrange field equation and the fluctuations have the prescribed temporal endpoints and admissible spatial boundary behavior. This is the principle of stationary action, not a symmetry assumption about $\eta$.

The printed expansion omits ${\beta }^{-2}$ despite the stated Lagrangian density. Its displayed form is the expansion of ${\beta }^{2}S$; it is not the physical $S$ at arbitrary coupling. Alternatively, writing $\eta =\beta \chi$ puts the quadratic term in canonical normalization, while leaving the classical term $-8mT/{\beta }^2$ unchanged. This normalization repair does not change ${\Delta }_x$ or the physical fluctuation frequencies.

The Sine-Gordon kink fluctuation operator has the useful factorizationIt is nonnegative, and $A{\psi }_0=0$ gives the normalized translational zero mode of a sine-Gordon kink:A displacement $a$ changes ${\varphi }_K(x-a)$ by $-a{\varphi }_K^{\prime }(x)$. Thus the zero mode in field theory is the position collective coordinate of the kink, reflecting translation invariance. It has no restoring force and no oscillator zero-point energy. Integrate that collective coordinate separately rather than inserting a zero factor into the Gaussian functional determinant.

In the Gaussian fluctuation approximation, the formal oscillator contribution to the one-loop soliton mass correction, before adding any counterterm, isUse a common regulator for the two sums, include all discrete modes, and treat the translation mode as above. The vacuum sum is essential: subtracting only classical vacuum energy would leave an extensive oscillator energy. A periodic fluctuation and its derivative are matched at the two ends of the large box; the one-kink background lies in the twisted topological sector, and its infinite-line profile is accurate up to exponentially small boundary corrections.

For a continuum scattering wavefunction, equality of its two asymptotic values gives the periodic-box phase-shift quantizationAway from the threshold, expand at a matched mode number:Since consecutive free wave numbers are separated by $2\pi /L$, replacing the continuum-mode sum by an integral proves the displayed continuum contribution:The ultraviolet cutoff is retained until the counterterm is added.

There is a finite threshold issue if this expression is identified with the complete oscillator correction. It can be settled directly using the factorization: a continuum eigenfunction is $f_k\propto (\tanh x-ik)e^{ikx}$. Its transmission phase obeyswith the odd phase branch that tends to zero at large $|k|$. For $k>0$, the continuum roots have labels $n=1,2,\dots$; there is no periodic continuum root at $k=0$, since the limiting eigenfunction $\tanh x$ has opposite signs at the two ends. The bound state at ${\omega }_0=0$ replaces the free $k=0$ oscillator with ${\omega }_0^{(0)}=1$. Consequently, in mode-number regularization of soliton masses,The PDF's continuum-only formula misses this finite $-m/2$ under this standard phase and mode-counting convention. It has the correct logarithmic ultraviolet divergence, but the missing term is not suppressed by large $L$. Changing the phase branch requires changing the mode labels and endpoint terms consistently; it cannot erase a physical mode from the formal spectrum sum.

At high momentum, $\delta (k)=2/k+O(k^{-3})$, so both expressions have divergent part $-(m/\pi )\log \Lambda$. The canonical field $\phi =\varphi /\beta$ has a quartic interaction with coupling $-m^2{\beta }^2$. Its vacuum tadpole diagram shifts the squared mass by $-m^2{\beta }^{2}J(\Lambda )/4$, whereThe Sine-Gordon vacuum tadpole counterterm has $\delta m^2=+m^2{\beta }^{2}J/4$ and adds potential energy density $\delta m^2(1-\cos \varphi )/{\beta }^2$. Since $\int (1-\cos {\varphi }_K)dx=4$, its vacuum-subtracted kink energy iswhich cancels the logarithmic ultraviolet divergence. Finite parts require a specified renormalization condition. As a consistency check, with this tadpole subtraction and matched mode-number cutoff, integration by parts yieldsThe complete semiclassical soliton mass is then $8m/{\beta }^2-m/\pi +O(m{\beta }^2)$ in that convention. This last finite result uses the bound-mode term and is additional to the requested ultraviolet cancellation.

The principle of stationary action says that the physical path makes the action stationary under smooth variations $q+\eta h$ with fixed endpoints $h(t_1)=h(t_2)=0$. It need not minimize the action. Differentiating and integrating by parts givesThe fundamental lemma of the calculus of variations then gives the Euler-Lagrange equation

The principle of stationary action, also called Hamilton's principle, requires $\delta {\int }_{t_1}^{t_2}Ldt=0$ for fixed-endpoint path variations. Here the original PDF has ${\dot{r}}^2$ in the radial kinetic energy; the TeX has lost that dot. With $m>0$ and $r>0$, the Lagrangian isThe Euler-Lagrange equation for each coordinate givesThe two conserved quantities are angular momentum $h=m{r}^2\dot{\varphi }$ and energy $E=\frac{m}{2}{\dot{r}}^2+\frac{m}{2}{r}^2{\dot{\varphi }}^2+km{r}^2$. By Noether's theorem, they arise respectively from rotational invariance and time-translation invariance of the action. The coordinate $\varphi$ is a cyclic coordinate, and $L$ has no explicit time dependence.

The conjugate momentum for each coordinate is $p_r=m\dot{r}$, $p_{\varphi }=m{r}^2\dot{\varphi }$. The Legendre transform $H=p_r\dot{r}+p_{\varphi }\dot{\varphi }-L$ gives the HamiltonianHamilton's equations areSetting $p_{\varphi }=h$ yields $m\ddot{r}=-V_{\mathrm{eff}}^{\prime }(r)$ with the effective potential $V_{\mathrm{eff}}=h^2/(2m{r}^2)+km{r}^2$.

For $k>0$ and $h\ne 0$, this potential tends to infinity both as $r\downarrow 0$ and as $r\to \infty$, and has its unique stationary point atThe effective potential stability criterion proves a stable circular orbit in a quadratic central potential, with $|\dot{\varphi }|=\sqrt{2k}$. For a small radial displacement $\eta$, $\ddot{\eta }+8k\eta =0$ to first order, so radial perturbations oscillate rather than grow. The diagram shows the centrifugal and quadratic contributions and their sum in dimensionless units:

The printed paper only says that $k$ is constant. Its claimed positive-radius stable circular orbit needs the additional attractive-force assumption $k>0$ and nonzero angular momentum. For $k<0$ there is no such minimum; for $k=0,h\ne 0$ the effective potential is strictly decreasing. With $h=0,k>0$, the minimum is at the origin and is not a positive-radius circular orbit. These cases are not represented by the diagram.

Use natural units $\hslash =c=1$ and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ throughout. An Abelian gauge transformation is $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$. For a sufficiently regular $\chi$, commuting the partial derivatives gives $F_{\mu \nu }\mapsto F_{\mu \nu }$, so the Maxwell Lagrangian is invariant pointwise.

Vary the action with respect to the electromagnetic four-potential, using variations of compact support or vanishing on the boundary. The antisymmetry of the electromagnetic field tensor givesAfter integration by parts, $\delta S=\int d^{4}x({\partial }_{\mu }{F}^{\mu \nu })\delta A_{\nu }$. The principle of stationary action therefore gives the source-free Maxwell equations,In the Lorenz gauge, ${\partial }_{\mu }{A}^{\mu }=0$, this becomes $\square A_{\nu }=0$ with $\square ={\partial }_{\mu }{\partial }^{\mu }$. The PDF's name “Lorentz gauge” denotes the usual Lorenz gauge, named after Lorenz. Residual gauge transformations preserve it when $\square \chi =0$; this condition is not itself a complete removal of the gauge freedom.

Use the first-order actionVarying $q$ with fixed endpoints and varying $p$ freely givesThe principle of stationary action therefore yields Hamilton's equations

For $H=p^2+q^{-2}$ these become $\dot{q}=2p$ and $\dot{p}=2q^{-3}$. Along their Hamiltonian flow, $H=E$ is conserved andThus $F=pq-ctH$ is an integral of motion exactly when $c=2$.

Write its constant value as $F_0$. The surfaces $pq-2tE=F_0$ in extended $(q,p,t)$-space giveUsing $E{q}^2=p^2{q}^2+1$,The sign of $q$ is fixed. In phase space, each energy orbit is one of the two branchesEach branch approaches $p=-\sqrt{E}$ and $p=+\sqrt{E}$ as $|q|\to \infty$ and turns at $(p,q)=(0,\pm E^{-1/2})$.

Write the action density asUnder a compactly supported variation $\varphi \mapsto \varphi +ϵ\eta$,Integrating the first term by parts in time and the second in space, and discarding the boundary terms, givesThe principle of stationary action requires this to vanish for every $\eta$. Sincedivision by $-a^3$ yields the field Euler-Lagrange equationThis is the equation generated by the inflaton action in an expanding universe.

The Euler-Lagrange equation is $m\ddot{x}=0$. The unique path with the prescribed endpoints is thereforeIt is a minimum of the Euclidean action and a stationary point of the real-time action. Its classical action isThe principle of stationary action consequently fixes the position-dependent phase of the semiclassical propagator asBecause the action is quadratic, the stationary-phase evaluation of the path integral is exact. Composition of propagators, or the Van Vleck determinant, gives $C(T)=\sqrt{m/(2\pi i\hslash T)}$, reproducing part i.

The chain rule andgiveApplying the same rule to $L_1$ and $L_2$ in the Euler-Lagrange field equation yields the exact modulated equation

Multiply this equation by ${\Phi }_{\theta }$. Since $k$ and $\omega$ are independent of the fast phase,The product rule then rearranges the field equation into the exact modulated-wave first integralAverage this identity over one $2\pi$ period in $\theta$. Periodicity kills the first term, while differentiation of the averaged Lagrangian givesConsequently

The same two equations follow directly from the modulated variational principle. Varying $\Phi$ giveswhich is the exact field equation above. For a variation of $\Theta$, use $\delta k={\partial }_X\delta \Theta$ and $\delta \omega =-{\partial }_T\delta \Theta$. Integration by parts in $X$ and $T$ gives the averaged equation. Thus variation with respect to the periodic profile reproduces the local wave equation, whereas variation with respect to its slow phase gives the Whitham modulation equation for wave action.

Sincethe determinant variation givesIntegrating by parts and taking $\delta X^{\mu }$ to vanish on the boundary yieldsThe principle of stationary action therefore gives