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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.

Vary $F_i$ and $F_i^{\ast }$ independently. The auxiliary fields have algebraic, rather than differential, Euler-Lagrange equations:Alternatively complete the square:Substitution therefore yields the on-shell auxiliary fields and nonnegative scalar potentialThe sign in the Lagrangian is minus the F-term scalar potential; changing the metric convention does not change this algebraic result. A supersymmetric vacuum requires all $W_i=0$, namely F-flatness. For a noncanonical positive Kähler metric $K_{i\bar{j}}$, the purely bosonic elimination instead gives $V_F=K^{i\bar{j}}{W}_i\overline{W_j}$; fermionic connection terms must also be included in a complete nonlinear component action.

Complexifying the Clifford algebra removes the signature dependence of its representation dimension. Pair the $2k$ gamma matrices into operators $a_i,a_i^{†}$ with $\{a_i,a_j^{†}\}={\delta }_{ij}$ and $\{a_i,a_j\}=0$. The $k$ commuting occupation operators have eigenvalues zero or one. Starting from a Clifford vacuum, applying every subset of the $k$ fermionic creation operators produces $2^k$ independent states. Conversely, these operators construct matrices on a $2^k$-dimensional space, and their products span its full matrix algebra. Thus an irreducible complex Clifford algebra module in $D=2k$ hasA general Clifford module is a direct sum of these; the statement concerns the minimal spinor.

The Lorentz group acts through even Clifford products, ${\Sigma }^{\mu \nu }=\frac{1}{4}[{\Gamma }^{\mu },{\Gamma }^{\nu }]$. More precisely this is a representation of its double cover, the Spin group. The normalized product ${\Gamma }_{\ast }$ of all $2k$ matrices squares to one and anticommutes with each ${\Gamma }^{\mu }$, while commuting with every ${\Sigma }^{\mu \nu }$. Its eigenspaces are the two Weyl spinor representations, so the Dirac spinor is reducible under the connected spin group:Each summand is irreducible. The Clifford module itself remains irreducible because an individual ${\Gamma }^{\mu }$ interchanges the summands. In $D=2k+1$ one can use the old ${\Gamma }_{\ast }$ as the additional matrix, with an appropriate phase for its signature. No independent chirality matrix remains: the product of all $2k+1$ matrices is central and fixed to a scalar on an irreducible module. There are two choices of this central sign for the complex Clifford algebra, but each restricts to the same irreducible $2^k$-dimensional spinor representation of the connected spin group. An odd-dimensional spinor has no Weyl splitting. Majorana spinor reality conditions depend additionally on dimension and signature; they must be imposed separately from this complex dimension count.

For the massless field counts, use the rotational part $SO(D-2)$ of the little group and real physical polarizations. A massless p-form gauge field has only transverse, antisymmetric components after its gauge redundancy and equations of motion are imposed, giving $\left(\genfrac{}{}{0ex}{}{D-2}{p}\right)$ polarizations. A graviton has a symmetric traceless transverse tensor, giving $(D-2)(D-1)/2-1=D(D-3)/2$. A massless fermion obeys a Dirac equation which halves its off-shell spinor components. In ten dimensions a Majorana-Weyl spinor has sixteen real components off shell and eight physical components; in nine dimensions a Majorana spinor has sixteen off shell and eight physical components; in eleven dimensions a Majorana spinor has thirty-two off shell and sixteen physical components. A gravitino is a transverse vector-spinor with its gamma trace removed, leaving $(D-3)$ times the corresponding physical spinor dimension. Thus the ten-dimensional Majorana-Weyl gravitino has $7\times 8=56$ polarizations, the nine-dimensional Majorana gravitino has $6\times 8=48$, and the eleven-dimensional Majorana gravitino has $8\times 16=128$.

The common Neveu-Schwarz sector of type IIA supergravity and type IIB supergravity consists of a metric tensor, the Kalb–Ramond field $B_2$, and the dilaton $\varphi$. Their counts are respectively $35$, $\left(\genfrac{}{}{0ex}{}{8}{2}\right)=28$, and $1$. The Ramond-Ramond sectors and fermions distinguish the theories. In type IIA supergravity the independent bosonic potentials and counts areThere are two Majorana-Weyl gravitini of opposite chirality, giving $56+56=112$, and two Majorana-Weyl dilatini of opposite chirality, giving $8+8=16$. Their total is $128$. With gravitini labelled ${\psi }^{+}$ and ${\psi }^{-}$, the corresponding dilatini have signs $-$ and $+$ respectively. Type IIA is nonchiral.

In type IIB supergravity, the independent bosonic potentials and counts areThe five-form field strength associated with $C_4$ is a self-dual differential form, so only half the seventy transverse four-form components are independent. In Lorentzian ten dimensions the Hodge star operator on five-forms squares to one, making a real self-duality condition possible. The two Majorana-Weyl gravitini have the same chirality and contribute $112$ states. Both Majorana-Weyl dilatini have the opposite chirality to those gravitini and contribute $16$ states. Type IIB is chiral, with $128+128$ bosonic and fermionic polarizations.

For dimensional reduction, retain the massless zero modes on a flat circle, with no flux, gauging or fermion twist. A metric splits into a lower-dimensional metric, one vector and one scalar. An ordinary $p$-form splits into a $p$-form and a $(p-1)$-form, according as it has no compact index or one. This preserves counts by Pascal's identity, $\left(\genfrac{}{}{0ex}{}{D-2}{p}\right)=\left(\genfrac{}{}{0ex}{}{D-3}{p}\right)+\left(\genfrac{}{}{0ex}{}{D-3}{p-1}\right)$.

Reducing type IIA supergravity to nine dimensions gives the following decompositions; each number is a count of physical states:Thus the bosonic maximal nine-dimensional supergravity multiplet contains one graviton, three massless vectors, two massless two-forms, one massless three-form and three real scalars, withEach ten-dimensional Majorana-Weyl gravitino becomes one nine-dimensional Majorana gravitino and one Majorana spin-one-half field, giving $56=48+8$. Each ten-dimensional dilatino gives one more Majorana spin-one-half field. The fermionic multiplet therefore has two gravitini and four spin-one-half fields, with $2(48)+4(8)=128$. The fields belong to one maximal gravity supermultiplet, not separate interacting matter multiplets.

Reducing type IIB supergravity gives $g_{\mu \nu },A_{\mu }^{\mathrm{metric}},\rho$ from the metric; one two-form and one vector each from $B_2$ and $C_2$; and the two scalars $\varphi ,C_0$. The four-form potential yields a nine-dimensional four-form and three-form, but the ten-dimensional self-dual differential form condition relates their field strengths. Retaining either one gives $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=\left(\genfrac{}{}{0ex}{}{7}{4}\right)=35$ independent polarizations, not seventy. The same bosonic count, three scalars, three vectors, two two-forms and one three-form, follows. The fermionic reduction likewise gives two gravitini and four spin-one-half fields. These are the same maximal nine-dimensional supergravity spectrum, as expected from T-duality. The spectrum is nonchiral in nine dimensions, since the odd-dimensional spinor representation has no independent Weyl chirality; the ten-dimensional distinction is lost on restriction to nine-dimensional Lorentz symmetry.

Eleven-dimensional supergravity has a metric with $11(8)/2=44$ states, a three-form with $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84$, and a Majorana gravitino with $128$. For direct reduction on a flat two-torus, write the internal indices as $i=1,2$. The metric gives $g_{\mu \nu }$, two vectors $g_{\mu i}$ and three symmetric components $g_{ij}$; the three-form gives $A_{\mu \nu \rho }$, two two-forms $A_{\mu \nu i}$ and one vector $A_{\mu 12}$. HenceThe eleven-dimensional Majorana spinor restricts to two nine-dimensional Majorana spinors. The vector-spinor consequently yields two gravitini and four spin-one-half fields, giving $128=2(48)+4(8)$. This is again the same $128+128$ maximal nine-dimensional multiplet.

For reduction of massless type IIA supergravity to four dimensions, use a flat six-torus and retain all zero modes, without flux or projections. Internal indices $i=1,\dots ,6$ giveThe multiplicities $21,15,20$ come respectively from a symmetric pair of six internal indices, $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$, and $\left(\genfrac{}{}{0ex}{}{6}{3}\right)$. Before dualizing, this gives one graviton, $28$ vectors, $63$ scalars, seven two-forms and one three-form. The seven two-forms provide seven scalar polarizations, whereas the three-form has no local propagating polarization. Thus the final bosonic spectrum is one graviton, $28$ vectors and $70$ scalars, with $2+56+70=128$ states.

A ten-dimensional Majorana-Weyl spinor decomposes into four four-dimensional Majorana spinors. Each ten-dimensional gravitino supplies four four-dimensional gravitini and twenty-four spin-one-half fields, with $56=4(2)+24(2)$. Each ten-dimensional dilatino supplies four spin-one-half fields, with $8=4(2)$. Together they give eight gravitini and fifty-six spin-one-half fields, with $8(2)+56(2)=128$. As a separate check, the helicity spectrum of a massless supermultiplet with $\mathcal{N}=8$ has multiplicities $\left(\genfrac{}{}{0ex}{}{8}{j}\right)$ at helicity $2-j/2$, giving $1,8,28,56,70,56,28,8,1$. The reduced theory has the field content of four-dimensional N=8 supergravity, with $128+128$ physical states, exactly as in ten and eleven dimensions. Other compact manifolds or projections can reduce the number of preserved supercharges; the flat-torus assumption is essential to this spectrum.

Here is the local two-form scalar duality including its coupling dependence. For this calculation use signature $(-+++)$, $H_{\mu \nu \rho }=3{\partial }_{[\mu }{B}_{\nu \rho ]}$, and the contravariant volume tensor ${ϵ}^{0123}=+1$. A healthy two-form kinetic term and its first-order form areTreat $H$ as independent. Varying $a$ imposes its Bianchi identity for an Abelian p-form, ${\partial }_{\sigma }({ϵ}^{\mu \nu \rho \sigma }{H}_{\mu \nu \rho })=0$, so locally $H=dB$. Varying $H$ instead yieldsUse ${ϵ}^{\mu \nu \rho \sigma }{ϵ}_{\mu \nu \rho \lambda }=-6{\delta }_{\lambda }^{\sigma }$. The original kinetic term becomes $+(g_B^2/2)(\partial a{)}^2$, and the multiplier term becomes $-g_B^2(\partial a{)}^2$. Both terms must be substituted; replacing $H$ only in the original kinetic term would produce the wrong sign. ThusAt fixed normalization of the Bianchi multiplier, the dual kinetic coefficient is the inverse of the original coefficient: if the scalar convention is $-\frac{1}{2g_a^2}(\partial a{)}^2$, then $g_a=1/g_B$. A canonical scalar is $g_{B}a$ for constant $g_B$, but that rescaling hides the formal coupling inversion and changes any assigned scalar periodicity. For a scalar-dependent positive kinetic matrix ${\mathcal{G}}_{IJ}$ of several two-forms, the same calculation gives $-\frac{1}{2}({\mathcal{G}}^{-1}{)}^{IJ}\partial a_I\partial a_J$. Additional topological couplings modify the multiplier's derivative terms but do not change this basic inversion of the two-form kinetic matrix. The Hodge star operator exchanges the two-form Euler-Lagrange equations and Bianchi identities for an Abelian p-form with those of the dual scalars; the duality is local and global flux sectors require separate treatment.

A four-dimensional three-form instead has $F_4=d{C}_3=f{\mathrm{vol}}_4$. Its equation of motion, $d(g_3^{-2}\ast F_4)=0$, makes $f/g_3^2$ a spacetime constant. For constant coupling the first-order expression in this convention can be writtenHere the first-order construction includes a multiplier $q(F_4-d{C}_3)$: varying $C_3$ imposes $dq=0$, and the displayed density retains the term fixing the flux normalization. Eliminating $f$ gives $f=g_3^{2}q$ and ${\mathcal{L}}_{\mathrm{eff}}=-g_3^2{q}^2/2$. Thus it can encode a constant-flux contribution to the vacuum energy, but no local massless particle or scalar wave. On the zero-flux perturbative vacuum it contributes no state. The seven two-forms are scalar duals; the three-form is nondynamical, which completes the four-dimensional field count without discarding a possible global flux parameter.

The product of the magnetic field with the velocity in this question is the three-dimensional cross product. Fix ${ϵ}_{123}=1$ and write $K^i{}_j={ϵ}^i{}_{kj}{B}^k$, so $K\mathbf{v}=\mathbf{B}\times \mathbf{v}$. In coordinates $y^0=t$, $y^i=x^i$ on ${\mathbb{R}}^3\times \mathbb{R}$, prescribe an affine connection by the following Christoffel symbols:with all other Christoffel symbols zero. These are smooth globally in the given Cartesian coordinates. The lower-index symmetry makes this a torsion-free connection. It is the geometrization of the Lorentz force by an affine connection; no condition on the curl of $\mathbf{E}$ or the divergence of $\mathbf{B}$ is required for this construction.

An affinely parametrized geodesic, with primes denoting differentiation with respect to $\lambda$, satisfiesOn the branch $t^{\prime }=c\ne 0$, use $t$ itself as an affine parameter. Dividing the spatial equation by $c^2$ givesConversely, each solution of this equation makes $\lambda =t$, $y(t)=(t,\mathbf{x}(t))$ an affinely parametrized geodesic of the constructed affine connection. Therefore its image is also an unparametrized geodesic. A general change of parameter adds a term proportional to the tangent in the geodesic equation, leaving the curve unchanged. The branch $t^{\prime }=0$ is not a trajectory with time as parameter.

For the metric realization, assume $\mathbf{E}=-\nabla U$ and $\nabla \cdot \mathbf{B}=0$. Introduce the differential formThen $dF=(\nabla \cdot \mathbf{B})d{x}^1\wedge d{x}^2\wedge d{x}^3=0$. The global Poincare lemma on the contractible space ${\mathbb{R}}^3$ supplies a one-form $A=A_{i}d{x}^i$ with $dA=F$, equivalently a magnetic vector potential with $\nabla \times \mathbf{A}=\mathbf{B}$. An explicit radial-gauge potential for a divergence-free magnetic field isThis is the radial homotopy formula for a closed two-form and is smooth even at the origin. For a constant magnetic field it gives $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.

Consider the Eisenhart-Duval lift with the Lorentzian metricThe independent one-forms $d{x}^1,d{x}^2,d{x}^3,dt,du-2A_{i}d{x}^i-Udt$ exhibit three positive directions and a two-dimensional block with one positive and one negative direction. Thus the metric is nondegenerate, of signature $(4,1)$. Its coefficients do not depend on $u$, and $g_{uu}=0$, so $\xi ={\partial }_u$ is a null Killing vector field. Indeed $g_{Au}$ are constant, so ${\Gamma }^A{}_{Bu}=0$ for the Levi-Civita connection and $\xi$ is parallel.

The geodesic Lagrangian for an affine parameter $\lambda$ isThe cyclic coordinate $u$ gives the conserved quantity $p_u=\partial L/\partial u^{\prime }=t^{\prime }$. Work at a nonzero value of $p_u$ and rescale the affine parameter to set $t^{\prime }=1$. This is the essential step in the null Kaluza-Klein reduction of a stationary force: one fixes the momentum along the null isometry and projects its geodesics, rather than dividing by $g_{uu}$.

The spatial Euler-Lagrange equations, before setting $t^{\prime }=1$, areSince $t^{\prime \prime }=0$ and $F_{ij}{v}^j=-(\mathbf{B}\times \mathbf{v}{)}_i$, their reduction isEquivalently, after taking $t$ as the affine parameter, the term $\dot{u}$ is a total derivative and the reduced Lagrangian is $L_{\mathrm{red}}=\frac{1}{2}|\dot{\mathbf{x}}{|}^2-2\mathbf{A}\cdot \dot{\mathbf{x}}-U$. Its Euler-Lagrange equations give the same sign and factor two.

There is also an explicit converse using null geodesics. For any physical trajectory, setThis makes its five-dimensional tangent null. The conserved momentum of the cyclic coordinate $t$ becomesThe quantity in parentheses is conserved, because its derivative is $\dot{\mathbf{x}}\cdot (2\mathbf{B}\times \dot{\mathbf{x}})=0$. Hence the $t$ equation, as well as the spatial and $u$ Euler-Lagrange equations, is satisfied. Every trajectory admits a null geodesic lift with $p_u=1$, and every such lift projects to the required trajectory.

Finally, a time-independent gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ is absorbed by $u\mapsto u+2\chi$. The one-form $du-2A_{i}d{x}^i-Udt$ and the five-dimensional metric are unchanged. These gauge transformations of an Eisenhart-Duval lift alter the reduced Lagrangian only by the total derivative $-2d\chi /dt$.

First let $L$ be a bounded linear operator on $H$ satisfying the symmetric positive-definiteness convention in part (a). For any direction $w\in H$, expansion of the quadratic functional givesThus vanishing of the first variation in every direction is precisely the weak equation $⟨Lu,w⟩=⟨f,w⟩$ for every $w$. In this whole-space bounded-operator setting it is equivalent to $Lu=f$, the Euler-Lagrange equation.

If $u$ solves that equation, set $v=u+w$. The linear terms cancel:with equality only when $v=u$. Hence the weak solution is the unique global minimizer. Conversely, any minimizer has zero first variation, and therefore solves the weak equation. For a symmetric bounded bilinear form $B$ on a form space $V$, exactly the same calculation gives $I(v)=B(v,v)-2\ell (v)$ and $B(u,w)=\ell (w)$ for every $w\in V$; it does not require an unbounded differential operator to map every $v\in V$ into $H$.

Existence for every $f$ requires an extra hypothesis if “positive definite” means only strict positivity. The coercive operator bound makes the form coercive, so the Lax-Milgram theorem supplies existence and uniqueness. Without that bound, the diagonal operator on sequence space $L(x_n)=(x_n/n)$ on ${\ell }^2$ is symmetric and strictly positive, but $f=(1/n)$ belongs to ${\ell }^2$ and its formal inverse $(1,1,\dots )$ does not. Thus strict positivity alone proves uniqueness and the minimizing property of a solution when one exists, not existence for all $f$.

For one small transverse displacement, the elastic energy of the elastic filament iswhere $A$ is the filament bending modulus. Its first variation isThe clamped boundary conditions remove the left boundary terms. A force-free and torque-free tip permits independent $\delta h(L)$ and $\delta h^{\prime }(L)$, so the natural boundary conditions for a free endpoint are $h^{\prime \prime }(L)=h^{\prime \prime \prime }(L)=0$: zero bending moment and transverse shear.

There is an important distinction between an energy minimum and a normal mode. The unloaded higher-order Euler-Lagrange equation is $h^{\prime \prime \prime \prime }=0$, whose only solution with these four boundary conditions is $h=0$. A general fluctuating shape is a sum of modes, not one of the stated sinusoidal/hyperbolic functions. To obtain the clamped--free bending modes, extremize the bending Rayleigh quotient, or equivalently $U[W]-\frac{1}{2}A{k}^4\int W^{2}dx$ with a fixed norm. Its Euler-Lagrange equation isThis is the constrained variational characterization of bending modes. Its four characteristic roots are $\pm k,\pm ik$. Clamping gives $F=-D$ and $E=-B$, henceThe free-tip conditions reduce toIts determinant is $-2(1+\cos q\cosh q)$, so nontrivial modes requireNo zero mode exists, since a cubic satisfying the homogeneous clamp/free conditions is zero. The clamped-free bending spectrum starts with $q_1\in (\pi /2,2)$; bisection or Newton iteration gives $q_1\simeq 1.875104$, so $k_1\simeq 1.875104/L$. Choosing $D=N(\sin q_n+\sinh q_n)$ and $B=-N(\cos q_n+\cosh q_n)$ givesThe second boundary condition follows from the root equation, and $N$ is arbitrary until a normalization is chosen.

The bending operator with these boundary conditions is positive and self-adjoint. Twice integrating by parts givesSymmetry implies orthogonality when $n\ne m$. Expand $h(x)={\sum }_n{a}_n{W}_n(x)$ and use $I_n=L^{-1}{\int }_0^L{W}_n^{2}dx$. Thenby the equipartition theorem. The supplied endpoint identity therefore yields the thermal bending fluctuations of a clamped filament:To evaluate the sum, apply a tip force $f$ and minimize $U-fh(L)$. The modified free-end conditions are $h^{\prime \prime }(L)=0$ and $-A{h}^{\prime \prime \prime }(L)=f$, and $h^{\prime \prime \prime \prime }=0$ in the interior. Integration gives $h(x)=f{x}^2(3L-x)/(6A)$, so the tip-force compliance of a cantilever is $L^3/(3A)$. On the other hand, minimizing the modal energy minus $f{\sum }_n{a}_n{W}_n(L)$ gives $a_n=f{W}_n(L)/(AL{k}_n^4{I}_n)$ and henceComparison evaluates the fourth inverse-power sum of the cantilever spectrum without truncating the modes:The boxed variance is for the specified single transverse direction. An independent equilibrium check follows by differentiating the Gaussian partition function with respect to $f$: $\partial ⟨h(L)⟩/\partial f=⟨h(L{)}^2⟩/(k_{B}T)$ at zero load, reproducing the same result from the static compliance. Keeping only the first normal mode gives about $0.32356k_{B}T{L}^3/A$, roughly $97.1\%$ of the exact variance. With two independent equivalent transverse directions, their summed variance is twice the boxed result. The small-slope model requires $k_{B}TL/A\ll 1$, or $L$ small compared with the usual three-dimensional persistence length $A/(k_{B}T)$.

Let $s$ be arclength, $\mathbf{t}={\partial }_s\mathbf{r}$ the unit tangent, and $\mathbf{v}$ the local filament velocity relative to the fluid. The resistive-force theory force per unit arclength, exerted by the fluid on the filament, isThe parallel and perpendicular drag coefficients of a slender filament are proportional to dynamic viscosity. For radius $a$ and a relevant axial cutoff length $\ell \gg a$, the anisotropic slender-filament drag from Stokeslet integration gives the leading logarithmic formsThe cutoff may be set by filament length or deformation wavelength. End shape and the choice of cutoff change order-one constants inside the logarithms. Slender-body theory also includes nonlocal interactions that the local resistive-force theory approximation omits. Typically ${\xi }_{\perp }>{\xi }_{\parallel }$, with ratio tending to two in the slender limit; this drag anisotropy enables propulsion.

Write $C={\xi }_{\perp }-{\xi }_{\parallel }>0$. For a small-amplitude planar deformation, $\mathbf{t}=(1,y_x)+O(y_x^2)$ and the leading transverse velocity is $y_t$. Exact inextensibility requires the quadratic longitudinal displacement of an inextensible planar filament: if $X(s,t)=s+X_2(s,t)$ and $Y(s,t)=y(s,t)$, then $X_{2,s}=-y_s^2/2$ to this order. Its velocity is quadratic and has zero time average for a periodic material cycle without mean axial drift. Thus the fixed material abscissa in the question is leading-order notation; this correction does not change the averaged propulsive term below. An imposed mean translation would contribute its separate ordinary axial drag.

To leading nonzero order,The time-averaged deformation-induced propulsive force of a periodic planar filament and power of a periodic planar filament are consequentlyThese are the leading quadratic functionals. Their signs refer to force on the filament. For example, $y=a\sin (kx-\Omega t)$ gives negative $F_x$, opposing the direction of wave propagation.

Let $\eta (x,t)$ be an arbitrary smooth variation periodic in both variables. The first variation of the constrained functional isUsing integration by parts in $x$ and $t$, every boundary term cancels by periodic boundary conditions. ThereforeThe Euler-Lagrange equation isFor a nontrivial stroke with nonzero thrust, $\Gamma \ne 0$. Set $c=C/(\Gamma {\xi }_{\perp })$ and $v=y_t$. Then $v_t+c{v}_x=0$, whose method of characteristics solution is $v=V(x-ct)$. Integrating in time gives the travelling-wave stationarity of planar filament propulsion:Here $g$ is an arbitrary static periodic component. It contributes neither power nor mean propulsion because the time average of $y_t$ vanishes. It can be set to zero when describing the active stroke. The wave must be compatible with both specified periods: $f(x-cT)=f(x)$.

For a pure travelling wave, its leading force and power areThus positive $x$ propulsion uses a wave travelling in the negative $x$ direction. Any smooth interior maximizer satisfying only the stated power constraint must have this travelling active deformation. If the drag were isotropic, $C=0$ and there would be no leading propulsion.

There is a qualification to the printed maximization claim: the variation proves stationarity, not the existence of an unrestricted global maximum. The bandwidth limitation in fixed-power filament optimization can be seen directly. Set $k=2\pi /\lambda$, $\Omega =2\pi /T$, and consider $y_m=a\sin (mkx+\Omega t)$. The leading expressions giveAt fixed power the formal quadratic objective increases with $m$. Eventually that sequence violates the small-slope approximation, but the approximation supplies no numerical slope or wavelength cutoff with which to define the global optimization. Even within a strict small-slope neighbourhood, a sufficiently small admixture of a higher spatial harmonic can improve a candidate interior maximum while keeping power fixed. For example, $y=a[\sqrt{1-{\eta }^2}\sin (kx+\Omega t)+\eta \sin (mkx+\Omega t)]$ retains the fundamental periods and exactly the same leading power, while its force is multiplied by $1+(m-1){\eta }^2$. For fixed $m>1$, taking $a$ and $\eta$ small keeps every slope small.

With an explicit admissible Fourier bandwidth or an additional geometric constraint, the intended conclusion can be made precise. For a finite set of nonstatic Fourier series modes proportional to $e^{i(k_{j}x-{\Omega }_{j}t)}$, with wavenumbers $k_j$ and frequencies ${\Omega }_j$,This is a weighted average of $-C{k}_j/({\xi }_{\perp }{\Omega }_j)$. A largest allowed ratio is attained by a single travelling mode, or by modes sharing the same phase speed; their superposition is still a travelling wave. Thus the Euler-Lagrange calculation yields the requested travelling-wave form, while a global maximum requires a specified admissible shape class.