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Treat $t,T$ as independent variables, with $T={ϵ}^{2}t$. The chain rule givesTherefore the method of multiple scales isThe unspecified $O({ϵ}^4)$ term includes the higher correction in $a$. Only the displayed terms through order ${ϵ}^2$ are needed to determine the leading slow-amplitude system.

On the free streamline choose $q=e^{-i\theta }$ with $-\pi /2\le \theta <0$. The lower endpoint of this parameter interval is $D$, and $\theta \to 0^{-}$ approaches $C$. The fractional expression in the map isHence $e^{\pi w/a}=-{\csc }^2\theta$, and the branch on the upper edge of the potential strip isDifferentiate to obtain $dw/d\theta =-2a\cot \theta /\pi$. Since $dw/dz=q$, the chain rule givesThe sign and interval are consistent with the free surface moving to the right and downward from $D$.

Take downward distance $s$ and downward speed $v=ds/dt\ge 0$, with constant gravitational acceleration $g>0$ and quadratic drag coefficient $k>0$. The printed vector square has no standard vector meaning; the directionally consistent drag force is ${\mathbf{F}}_d=-mk|\mathbf{v}|\mathbf{v}$. For this downward fall, Newton's second law becomesThe terminal velocity is obtained by setting the net force to zero:From rest, the speed increases monotonically towards this limiting value. In fact $v(t)=v_{\infty }\tanh (\sqrt{gk}t)$; thus it is a supremum approached at infinite time, and a tower of finite height need not permit it to be reached.

For the required distance, use the chain rule $dv/dt=vdv/ds$. Writing $w=v^2$ gives a linear ordinary differential equation:Consequently,Here $k$ has units of inverse length, so the result has the required units of distance and is independent of the falling mass.

In the position representation in quantum mechanics, $\hat{\mathbf{p}}=-i\hslash \nabla$. Therefore the orbital angular momentum operator givesAt fixed $r,\theta$, the spherical coordinates obey ${\partial }_{\varphi }x=-y$, ${\partial }_{\varphi }y=x$, and ${\partial }_{\varphi }z=0$. The chain rule consequently gives ${\partial }_{\varphi }=-y{\partial }_x+x{\partial }_y$, henceThe derivative holds the other two spherical coordinates fixed; it is not a derivative along an arbitrary spatial curve.

Following the moving fluid. The Eulerian velocity field $\mathbf{u}(\mathbf{x},t)$ assigns a velocity to each position. A fluid particle satisfies $\dot{\mathbf{x}}=\mathbf{u}$, so the chain rule gives the material derivative, also called the convective derivative,Even in steady flow a particle can accelerate because it moves between regions with different velocity. For example, flow around a curved obstacle changes direction despite ${\partial }_t\mathbf{u}=0$. Mass conservation isFor a constant-density incompressible fluid, $\nabla \cdot \mathbf{u}=0$.

Euler equations for an inviscid fluid and Bernoulli's equation. Neglect viscosity and take the body force per unit mass to be $-\nabla \Phi$. Newton's second law for a fluid element givesFor steady flow, dot with displacement along a streamline. The tangential component of $\mathbf{u}\cdot \nabla \mathbf{u}$ is the derivative of $u^2/2$ along that line. For constant density, integration givesIn uniform gravity, $\Phi =gz$. Thus pressure, kinetic energy per unit mass and gravitational potential energy per unit mass exchange along a streamline. For a barotropic compressible fluid, replace $p/\rho$ by $\int dp/\rho (p)$. Different streamlines need not have the same Bernoulli constant.

Vorticity and potential flow. Vorticity is $\mathbf{\omega }=\nabla \times \mathbf{u}$. A small fluid element rotates locally with angular velocity $\mathbf{\omega }/2$ after its deformational part is separated. The vector identitygives $\nabla B=\mathbf{u}\times \mathbf{\omega }$ for steady constant-density inviscid flow. Thus an irrotational region has one common Bernoulli constant on each connected region. In a simply connected irrotational domain, introduce a velocity potential $\mathbf{u}=\nabla \varphi$. Incompressibility then gives Laplace's equation, ${\nabla }^2\varphi =0$. For unsteady irrotational flow,An impermeable stationary wall fixes ${\partial }_n\varphi =0$, but ideal potential flow does not enforce no slip. In a multiply connected domain, irrotational flow may still have nonzero circulation around an excluded vortex core, so a globally single-valued potential requires care.

For example, uniform two-dimensional flow past a circular cylinder has $\varphi =U(r+a^2/r)\cos \theta$ outside $r=a$. Its normal velocity vanishes at the cylinder, and its surface tangential velocity is $-2U\sin \theta$. Bernoulli's equation predicts a symmetric pressure distribution and zero drag. This is D'Alembert's paradox: the assumptions omit the viscous boundary layers, separation and wake responsible for real drag.

For an incompressible fluid of constant kinematic viscosity $\nu$, the curl of the motion equation givesThe first term stretches and tilts vortex lines; the second diffuses vorticity. In two-dimensional incompressible inviscid flow the stretching term is absent and vorticity is carried with the fluid. In a three-dimensional inviscid flow it can be amplified by stretching. Density gradients not parallel to pressure gradients provide an additional source in a general compressible fluid.

Viscous flow. A Newtonian fluid's shear stress is proportional to rate of strain. For constant viscosity and incompressibility the Navier-Stokes equation iswith dynamic viscosity $\eta$ and kinematic viscosity $\nu =\eta /\rho$. At an ordinary solid wall, the no-slip boundary condition sets fluid velocity equal to the wall velocity. Viscosity transfers momentum between neighbouring layers and converts mechanical energy into heat; unlike the pressure force, its shear contribution depends on velocity gradients.

For plane Couette flow between plates $y=0,h$, with the upper plate moving at $U$, the steady solution is $u_x=Uy/h$, with shear stress $\eta U/h$. For pressure-driven flow through a pipe of radius $a$, a constant pressure gradient gives the parabolic Poiseuille profileThe Reynolds number $\mathrm{Re}=\rho UL/\eta$ compares convective inertia with viscous force. Small Reynolds number often permits the Stokes-flow approximation; large Reynolds number can produce thin boundary layers and, under suitable conditions, turbulence. It does not justify neglecting the wall conditions. Bernoulli's inviscid constant is generally not conserved along a viscous streamline, and pipe flow instead shows a pressure loss corresponding to dissipation.

The momentum of a material volume $V(t)$ is ${\int }_{V(t)}\rho \mathbf{u}dV$. Newton's second law equates its time derivative to the pressure force and gravity:Apply the given Reynolds transport theorem componentwise to the left side, and the divergence theorem to the pressure term. ThenSince this holds for arbitrary material volumes and the fields are continuous, the integrand vanishes pointwise. Dividing by positive density gives the Euler equations for an inviscid fluidAt a fixed physical time $t_0$, a streamline is a curve tangent to $\mathbf{u}(\cdot ,t_0)$, which may be parametrized by $d\mathbf{x}/ds=\mathbf{u}(\mathbf{x}(s),t_0)$. The chain rule gives $df(\mathbf{x}(s),t_0)/ds=\mathbf{u}\cdot \nabla f=0$, proving constancy along it. The time is held fixed; a time-dependent streamline statement does not by itself assert material conservation along particle trajectories.

Let $P=F_{y^{\prime }}$ and $Q=F_{y^{\prime \prime }}$. For a smooth variation $y+\epsilon \eta$, the first variation isHere the last expression follows by two integrations by parts. Taking variations supported in the interior and applying the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation for a second-derivative functional:If both endpoint values and slopes are free, their independent variations force the natural boundary conditions for second-derivative functionalsIf an endpoint value or slope is fixed instead, its corresponding variation vanishes and the associated natural condition is not required.

For $H=y^{\prime \prime }Q-y^{\prime }(Q^{\prime }-P)-F$, differentiate using the chain rule:The Euler-Lagrange equation cancels the first term. Consequently $H^{\prime }=-F_x$, and if there is no explicit $x$ dependence,This is a Beltrami identity for a second-derivative functional.

Differentiability at $a$ means that there is a linear map $L:V\to V_1$ withThis map is the total derivative $D{f}_a$. In the Frobenius norm, $|AB|\le |A||B|$, by the Cauchy-Schwarz inequality applied to matrix entries. Expanding with the order of all factors preserved givesThe remainder after the terms linear in $H$ has norm at most $3|A||H{|}^2+|H{|}^3=o(|H|)$. HenceFor $|C|<1$, the Neumann series $I-C+C^2-\cdots$ converges absolutely in the finite-dimensional space of matrices. Multiplying its partial sums by $I+C$ and passing to the limit shows that it is $(I+C{)}^{-1}$. Its remainder after $I-C$ has norm at most $|C{|}^2/(1-|C|)=o(|C|)$, proving the required expansion.

If $A$ is an invertible matrix and $|H|<1/|A^{-1}|$, then $A+H=A(I+A^{-1}H)$ is an invertible matrix. Thus the general linear group is an open subset of the matrix space. The preceding expansion also givesApply the chain rule to $g(X)=(X^{-1}{)}^3$, using the cubic total derivative at $A^{-1}$ and this total derivative of inversion:The three terms generally cannot be combined by commuting $A$ and $H$.

The chain rule at the specified point isIn standard coordinates this becomesThe matrix dimensions are respectively $k\times n$, $k\times m$, and $m\times n$, matching the composition of the two linear derivative maps.

The derivative of the curve isIts image lies on the unit circle, so $f(g(t))=1/2$ and $\boxed{D(f\circ g)(t)=0}$. Meanwhileand multiplication by $J_g(t)$ gives $-\sin t\cos t/2+\cos t\sin t/2=0$. This verifies the chain rule explicitly.

Write ${\mathcal{D}}_0=\mu {\partial }_{\mu }+{\beta }_0(g_0){\partial }_{g_0}$. Allow a general bare coefficient ${\gamma }_0$ temporarily, so the starting equation is $({\mathcal{D}}_0-n{\gamma }_0/2){\hat{\Gamma }}^{(n)}=0$. Substitute ${\hat{\Gamma }}^{(n)}=Z_{\varphi }(g_0{)}^{-n/2}{\Gamma }^{(n)}(p,g(g_0),\mu )$ and apply the product and chain rules:Multiplying by $Z_{\varphi }^{n/2}$ identifiesThus the renormalization-group beta function transforms by the chain rule under a coupling reparametrization, while the field coefficient also records the scale dependence of the finite or divergent field normalization. For the actual bare parametrization here, substitute ${\beta }_0=-ϵg_0$ and ${\gamma }_0=0$.

Continue with $m=0$. Along a characteristic $x=x_0+t$, the chain rule givesThus the initial wave packet ${\psi }_R(0,x)=f(x)$ evolves as ${\psi }_R(t,x)=f(x-t)$. Similarly, along $x=x_0-t$, the left component is constant, givingA feature initially at $x_0$ therefore appears at $x_0+t$ for the right component and $x_0-t$ for the left component. The profiles translate without changing shape at the speed of light, which is 1 in these units. For positive-energy plane waves, the right component has $E=p>0$ and the left component has $E=-p>0$, another way to identify their propagation directions.

The conclusion concerns the massless Dirac equation. For $m\ne 0$, the components couple as in c, and a positive-energy packet has group velocity $p/\sqrt{p^2+m^2}$, rather than a universal speed $\pm 1$. For example, a massive positive-energy rest solution has $p=0$ and ${\psi }_R={\psi }_L$, so neither component alone describes an independently moving light-speed packet.

Put $x_P=(t,-x,y)$ and $U_P={\gamma }^1$. Since $U_P^{†}=-{\gamma }^1$ and ${\gamma }^0{\gamma }^1{\gamma }^0=-{\gamma }^1$, the Dirac adjoint transforms asThe chain rule gives ${\partial }_{\mu }\psi (x_P)=r_{\mu }({\partial }_{\mu }\psi )(x_P)$, with $(r_0,r_1,r_2)=(1,-1,1)$ and no sum in this formula. The Clifford algebra givesMultiplication by $r_{\mu }$ therefore restores each kinetic term:However, the Dirac scalar bilinear changes sign because $({\gamma }^1{)}^2=-I_2$:Changing variables in the action has unit absolute Jacobian determinant. HenceA nonzero Dirac mass term for this single two-component field breaks parity symmetry in quantum field theory with the specified transformation.

The specified time reversal includes complex conjugating $i{\gamma }^{\mu }{\partial }_{\mu }$; omitting that operation changes the answer. Put $x_T=(-t,x,y)$, $U_T={\gamma }^2$, and $(r_0,r_1,r_2)=(-1,1,1)$. As in c, the Dirac adjoint in the source's transformed bilinear is ${\bar{\psi }}^{\prime }(x)=\bar{\psi }(x_T){\gamma }^2$. The conjugated Dirac operator has coefficient matricesCombining these coefficients with the chain rule signs and the Clifford algebra gives, without summing over $\mu$,For example the time coefficient is ${\gamma }^2(-i{\gamma }^0){\gamma }^2(-1)=i{\gamma }^0$, and the second spatial coefficient is ${\gamma }^2(-i{\gamma }^2){\gamma }^2=i{\gamma }^2$. Thus the kinetic term is invariant. The real Dirac mass term again reverses sign since ${\bar{\psi }}^{\prime }{\psi }^{\prime }=\bar{\psi }({\gamma }^2{)}^2\psi =-\bar{\psi }\psi$. The spacetime integration has unit absolute Jacobian determinant, soEach operation reverses the sign of the mass, while their composition reverses it twice and preserves the kinetic term. In an operator formulation the complex conjugation of coefficients expresses the antiunitary operator character of time reversal; the transformation must not be treated as an ordinary unitary substitution with all $i$'s held fixed. These statements concern the stated free action.

In the scalar two-particle model, the fermionic exchange symmetry and bosonic exchange symmetry conditions areFor particles with spin, exchange acts on all their labels; the stated coordinate-only model fixes the spatial symmetry sector. Define $S\Psi (x_1,x_2)=\Psi (x_2,x_1)$. Equal masses make the kinetic operator invariant under exchange, and the assumed potential is invariant too, so $HS=SH$. Therefore $H\Psi =E\Psi$ implies $H(S\Psi )=E(S\Psi )$. The componentsare respectively symmetric and antisymmetric eigenfunctions whenever nonzero. Their sum is the original nonzero quantum state, so at least one is nonzero. An eigenspace may contain both sectors if the energy is degenerate.

The normal modes of two harmonically coupled identical particles provide the separation. The coordinate transformation to $y_{+},y_{-}$ is orthogonal, with absolute Jacobian one. Exchanging the particles leaves $y_{+}$ fixed and changes $y_{-}$ to $-y_{-}$. Hence bosonic quantum states are even and fermionic quantum states odd in the relative coordinate. The chain rule givesAlso $x_1^2+x_2^2=y_{+}^2+y_{-}^2$ and $(x_1-x_2{)}^2=2y_{-}^2$. Combining the two trapping potentials with the interaction therefore gives the separated Hamiltonian operatorThese are independent oscillators of frequencies $\omega$ and $\omega \cos \alpha$. Both frequencies are positive in the stated range. Their product eigenstates have energiesThe parity of an oscillator level is $(-1{)}^n$. Consequently bosons allow even $n_{-}$ and fermions odd $n_{-}$, while $n_{+}$ is unrestricted.

The bosonic ground state uses $n_{+}=n_{-}=0$, and the fermionic ground state uses $n_{+}=0,n_{-}=1$. The oscillator length in the relative mode isUsing the normalized Gaussian for each mode givesIndeed ${\varphi }_{\tau }(y)=\sqrt{2}y{\psi }_{\tau }(y)/\tau$ by differentiating the Gaussian, so it is the normalized odd first excited state. The ground energies are $E_{\mathrm{B}}=\hslash \omega (1+\cos \alpha )/2$ and $E_{\mathrm{F}}=\hslash \omega (1+3\cos \alpha )/2$.

Choose normalized factors for a normalized product quantum state; this can always be done by reciprocal rescaling of its factors. Since the coordinate Jacobian has absolute value one,If the supplied factors are not individually normalized, the general expression has an additional factor $\Vert \psi {\Vert }^2$; choosing normalized factors is the convention in the displayed formula.

For the ground states, put $\rho (y)=|{\psi }_{\tau }(y){|}^2=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$. Then $|{\varphi }_{\tau }(y){|}^2=2y^2\rho (y)/{\tau }^2$ and ${\rho }^{\prime }=-2y\rho /{\tau }^2$. Thus, for differentiable $f$ with suitable Gaussian integrability and vanishing boundary terms,This is integration by parts, and every polynomial satisfies the required endpoint conditions. The relative-coordinate moments in even and odd oscillator ground states follow by substituting $f(y)=y^n$:For odd $n$, both expectations vanish by parity, so this is an equality of zeros rather than an increase in signed mean separation. The even moments show the physical difference. In particular ${\mathbb{E}}_{\mathrm{B}}[y_{-}^2]={\tau }^2/2$ and ${\mathbb{E}}_{\mathrm{F}}[y_{-}^2]=3{\tau }^2/2$, so the fermionic root-mean-square particle separation is $\sqrt{3}$ times the bosonic value. The fermionic probability density vanishes at coincidence $y_{-}=0$, expressing Pauli exclusion principle in this spatial symmetry sector. This is an exchange constraint on the wave function, not an additional classical force. As $\alpha$ approaches $\pi /2$, the relative oscillator softens and its width diverges; the excluded endpoint would no longer have this normalizable oscillator ground state.

For a normalized state the position probability density is $\boxed{f=|\psi {|}^2}$. With a real potential, the Time-dependent Schrödinger equation and its conjugate giveThis is the continuity equation $\boxed{f_t+j_x=0}$. Integrating over the line yields $d\Vert \psi {\Vert }^2/dt=-j(\infty )+j(-\infty )=0$ under the stated decay. A real-valued wave function gives $\overline{\psi }{\psi }_x-\psi \overline{{\psi }_x}=0$, so its instantaneous current vanishes; the same is true up to a spatially constant phase.

Put $\alpha =m\omega /\hslash$. Complete the square in the overlap:The Gaussian integral is $\sqrt{\pi /\alpha }$, also for complex displacement by analytic continuation. HenceThe product of square roots fixes the branch convention implicit in the printed square root of $\overline{u}s$. At $u=s\ne 0$ the norm squared is $|N{|}^2|s|\sqrt{\pi /\alpha }{e}^{2\alpha |s{|}^2}$, which becomes one for the stated $N$. In particularThe centre is $-2\mathrm{Re}s$ and its position variance is $1/(2\alpha )$.

When differentiating in $s$, keep $N$ fixed on the chosen constant-radius orbit. ThenSubstitution gives $HG=\hslash \omega s{G}_s$. Since $\dot{s}=-i\omega s$, the chain rule gives $\boxed{i\hslash G_t=HG}$. Continue $\sqrt{s(t)}$ continuously along the time orbit; its sign after a full orbit supplies the usual oscillator zero-point phase. The case of zero displacement is obtained as a limit, since the separate printed $N$ and $\sqrt{s}$ expressions are singular at $s=0$.

Finally $j=(\hslash /m)\mathrm{Im}(\overline{G}{G}_x)=-2\omega \mathrm{Im}s|G{|}^2$. For real nonzero $a$ this isThe printed final current omits the Gaussian normalization factor $\sqrt{m\omega /(\pi \hslash )}$. Retaining that factor is required both by the normalized state and by the continuity equation. This current is the Gaussian probability density times the velocity of its moving centre, $2a\omega \sin \omega t$.

In the scalar two-particle model, the fermionic exchange symmetry and bosonic exchange symmetry conditions areFor particles with spin, exchange acts on all their labels; the stated coordinate-only model fixes the spatial symmetry sector. Define $S\Psi (x_1,x_2)=\Psi (x_2,x_1)$. Equal masses make the kinetic operator invariant under exchange, and the assumed potential is invariant too, so $HS=SH$. Therefore $H\Psi =E\Psi$ implies $H(S\Psi )=E(S\Psi )$. The componentsare respectively symmetric and antisymmetric eigenfunctions whenever nonzero. Their sum is the original nonzero quantum state, so at least one is nonzero. An eigenspace may contain both sectors if the energy is degenerate.

The normal modes of two harmonically coupled identical particles provide the separation. The coordinate transformation to $y_{+},y_{-}$ is orthogonal, with absolute Jacobian one. Exchanging the particles leaves $y_{+}$ fixed and changes $y_{-}$ to $-y_{-}$. Hence bosonic quantum states are even and fermionic quantum states odd in the relative coordinate. The chain rule givesAlso $x_1^2+x_2^2=y_{+}^2+y_{-}^2$ and $(x_1-x_2{)}^2=2y_{-}^2$. Combining the two trapping potentials with the interaction therefore gives the separated Hamiltonian operatorThese are independent oscillators of frequencies $\omega$ and $\omega \cos \alpha$. Both frequencies are positive in the stated range. Their product eigenstates have energiesThe parity of an oscillator level is $(-1{)}^n$. Consequently bosons allow even $n_{-}$ and fermions odd $n_{-}$, while $n_{+}$ is unrestricted.

The bosonic ground state uses $n_{+}=n_{-}=0$, and the fermionic ground state uses $n_{+}=0,n_{-}=1$. The oscillator length in the relative mode isUsing the normalized Gaussian for each mode givesIndeed ${\varphi }_{\tau }(y)=\sqrt{2}y{\psi }_{\tau }(y)/\tau$ by differentiating the Gaussian, so it is the normalized odd first excited state. The ground energies are $E_{\mathrm{B}}=\hslash \omega (1+\cos \alpha )/2$ and $E_{\mathrm{F}}=\hslash \omega (1+3\cos \alpha )/2$.

Choose normalized factors for a normalized product quantum state; this can always be done by reciprocal rescaling of its factors. Since the coordinate Jacobian has absolute value one,If the supplied factors are not individually normalized, the general expression has an additional factor $\Vert \psi {\Vert }^2$; choosing normalized factors is the convention in the displayed formula.

For the ground states, put $\rho (y)=|{\psi }_{\tau }(y){|}^2=e^{-y^2/{\tau }^2}/(\sqrt{\pi }\tau )$. Then $|{\varphi }_{\tau }(y){|}^2=2y^2\rho (y)/{\tau }^2$ and ${\rho }^{\prime }=-2y\rho /{\tau }^2$. Thus, for differentiable $f$ with suitable Gaussian integrability and vanishing boundary terms,This is integration by parts, and every polynomial satisfies the required endpoint conditions. The relative-coordinate moments in even and odd oscillator ground states follow by substituting $f(y)=y^n$:For odd $n$, both expectations vanish by parity, so this is an equality of zeros rather than an increase in signed mean separation. The even moments show the physical difference. In particular ${\mathbb{E}}_{\mathrm{B}}[y_{-}^2]={\tau }^2/2$ and ${\mathbb{E}}_{\mathrm{F}}[y_{-}^2]=3{\tau }^2/2$, so the fermionic root-mean-square particle separation is $\sqrt{3}$ times the bosonic value. The fermionic probability density vanishes at coincidence $y_{-}=0$, expressing Pauli exclusion principle in this spatial symmetry sector. This is an exchange constraint on the wave function, not an additional classical force. As $\alpha$ approaches $\pi /2$, the relative oscillator softens and its width diverges; the excluded endpoint would no longer have this normalizable oscillator ground state.

Define the reflected function $\varphi (x)=\psi (-x)$. The chain rule gives ${\varphi }^{\prime }(x)=-{\psi }^{\prime }(-x)$ and ${\varphi }^{\prime \prime }(x)={\psi }^{\prime \prime }(-x)$. Evaluating the time-independent Schrödinger equation for $\psi$ at $-x$ and using the even function identity $V(-x)=V(x)$ givesThus reflection preserves the energy eigenvalue. The same substitution $y=-x$ in the norm givesThis is the action of the parity operator $R\psi (x)=\psi (-x)$, which is a unitary operator. For a reflection-symmetric physical problem it preserves the Hamiltonian operator's operator domain as well as its differential expression.