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Assume $\omega \ne 0$. With $h=0$, the method of characteristics gives $f(t,\Phi (t,0,z))=f_0(z)$. HenceLet $K=\mathrm{supp}{f}_0$ and $E_{\ast }={\max }_K{H}_{\omega }$; if $f_0=0$, the conclusion is immediate. By the Hamiltonian energy balance, every image point of $K$ remains in the energy sublevelThis is a fixed compact ellipsoid in phase space. Therefore the support bound is uniform in time:This uniform support bound from a coercive conserved energy requires no explicit solution of the curves.

The nonzero-frequency qualification is necessary. At $\omega =0$ the energy does not control $x$, and the equation is free transport equation. Choose a smooth compactly supported $f_0$ that is nonzero at $(x_0,v_0)$ with $v_0\ne 0$. Its transported value at $(x_0+t{v}_0,v_0)$ stays nonzero, so the union of the supports is unbounded. Each individual support is compact, but there is no fixed compact support for all times.

For positive smooth $f$ with the stated decay, integration by parts givesand, using unit mass,For zero values of $f$, these calculations can be made with the positive unit-mass approximation $(f+\epsilon \gamma )/(1+\epsilon )$ and then passed to the limit whenever the quantities are finite. Positive-time solutions also have the usual Gaussian smoothing.

Since $-\log \gamma =v^2/2+\frac{1}{2}\log (2\pi )$, mass conservation givesThe derivative of the first term is $\int {\partial }_t{f}_t\log f_t$, because $\int {\partial }_t{f}_t=0$. The two identities above and the energy equation in (b), with $d=M=1$, yieldNow expand the relative Fisher information:where $\int v{f}^{\prime }=-1$. Thus the entropy dissipation identity for Ornstein-Uhlenbeck flow is

Use the nearly conserved oscillator energy $E=({\dot{x}}^2+x^2)/2$. The differential equation givesFor a leading-order orbit $x=A\cos \theta$, $\dot{x}=-A\sin \theta$, the cycle averages are $⟨{\dot{x}}^4⟩=3A^4/8$, $⟨x^3\dot{x}⟩=0$, $⟨x^4{\dot{x}}^2⟩=A^6/16$, and $⟨{\dot{x}}^2⟩=A^2/2$. The energy balance method therefore givesThe two nonzero simple roots give the leading limit cycles $x\simeq \cos (t+{\theta }_0)$ and $x\simeq \sqrt{2}\cos (t+{\theta }_0)$. Below amplitude $1$ the energy increases, between $1$ and $\sqrt{2}$ it decreases, and above $\sqrt{2}$ it increases. Hence the amplitude-$1$ cycle is stable and the amplitude-$\sqrt{2}$ cycle unstable for positive small $\epsilon$.

The linearized averaged radial exponents are $-\epsilon /2$ and $\epsilon$, respectively, consistent with nontrivial Floquet multipliers $e^{-\pi \epsilon +O({\epsilon }^2)}$ and $e^{2\pi \epsilon +O({\epsilon }^2)}$. The nonzero derivatives at these averaged zeros also justify persistence of nearby cycles for sufficiently small $\epsilon$.

The Bose-Einstein distribution gives $n_i=[\exp ((E_i-\mu )/(k_{B}T))-1{]}^{-1}$, with chemical potential below the lowest energy when its discrete occupation is finite. In the continuum approximation,If the zero-energy ground state is macroscopically occupied, its discrete occupation $N_0$ must be added to the first integral; it contributes no energy to the second. This distinction is essential when applying the formulas to Bose-Einstein condensation.

Let $f(r)=1-r_s/r$ and parametrize the timelike geodesic by proper time $\tau$. Time-translation and rotational symmetry conserve $E=f\dot{t}$ and $h=r^2\dot{\varphi }$ in the equatorial plane. The normalized four-velocity obeysSubstitution and multiplication by $f$ give ${\dot{r}}^2+f(1+h^2/r^2)=E^2$, orThe dot denotes a proper-time derivative, and $E,h$ are energy and angular momentum per unit rest mass.

The linear elastic wave equation is $\rho {\mathbf{u}}_{tt}=(\lambda +\mu )\nabla (\nabla \cdot \mathbf{u})+\mu \Delta \mathbf{u}$. For a shear-horizontal wave, take $\mathbf{u}=(0,v(x,z,t),0)$ with no $y$ dependence. Its divergence vanishes, andRigid fixed boundaries require $v=0$ at $z=0,h$. The modes are $v=A\sin (q_{n}z)\cos (kx-{\omega }_{n}t)$ with $q_n=n\pi /h$, $n=1,2,\dots$, andThere is no $n=0$ displacement mode with these clamped boundary conditions. For $k>0$, the phase velocity and group velocity areThe phase velocity decreases from infinity to $c_s$, while the group velocity increases from zero to $c_s$; their product is $c_s^2$.

For a mode of frequency $\omega$, the appropriate time and cross-sectional average is $⟨A⟩=\omega /(2\pi h){\int }_0^{2\pi /\omega }{\int }_0^{h}Adzdt$, per unit transverse width. Its average kinetic energy density is $\rho A^2{\omega }^2/8$. The only nonzero strains are $e_{xy}=e_{yx}=v_x/2$ and $e_{yz}=e_{zy}=v_z/2$, while $e_{kk}=0$. Thus the elastic energy density is $\mu (v_x^2+v_z^2)/2$, whose average is $\mu A^2(k^2+q_n^2)/8$. The dispersion relation proves equality of average kinetic and elastic energies.

For the ray asymptotics of a clamped elastic waveguide mode, expand the localized initial displacement in transverse sine modes and Fourier transform in $x$. For a displacement released from rest, each mode evolves with $\cos ({\omega }_n(k)t)$. At $x=Vt$ the oscillatory phases are $t(kV\pm {\omega }_n(k))$. If $0<|V|<c_s$, one phase has a stationary point withThe stationary phase method gives a generic oscillatory amplitude proportional to $t^{-1/2}$ for each contributing mode. If the initial Fourier coefficient vanishes at the stationary point, that mode can decay faster; smooth localized data make the modal sum well behaved. If $|V|>c_s$, there is no stationary point. Repeated integration by parts gives rapid decay for smooth rapidly decaying data; for initially compactly supported displacement, finite propagation speed makes the displacement on that ray exactly zero after a sufficiently long time. These are the requested subsonic and super-shear-speed regimes.

The two cyclic Euler angles give conserved canonical momentaHere $L_3$ is the body-axis angular momentum and $L_z$ the vertical angular momentum. Thus ${\omega }_3=L_3/I_3$ is constant. Since the Lagrangian has no explicit time dependence, the energy is also conserved:These are the three independent conserved quantities furnished by the continuous symmetries of the heavy symmetric top.

Eliminate the cyclic velocities using $\dot{\varphi }=(L_z-L_3\cos \theta )/(I_1{\sin }^2\theta )$. The conserved energy becomesDifferentiating the effective potential giveswhich is also the $\theta$ Euler-Lagrange equation. Hence $I_1\ddot{\theta }=-V_{\mathrm{eff}}^{\prime }(\theta )$. Deriving it directly from the Euler-Lagrange equation ensures it also holds at turning points, where division by $\dot{\theta }$ in an energy argument would be invalid.

The work done by a force is $W=\int \mathbf{F}\cdot d\mathbf{r}$. By the work-energy theorem, this has the same physical dimension as energy. ThusAlthough torque and energy have the same physical dimension, they represent different physical quantities.

Now $m$ denotes the instantaneous rest mass of the rocket, including its fuel and stored internal energy, and $v$ is measured in a fixed inertial frame. Assume the exhaust is expelled backwards with speed $0<u<c$ in the rocket's instantaneous rest frame. By the relativistic velocity-addition formula, material emitted at time $\tau$ has inertial velocityIt coasts at this velocity after emission.

The crucial change is that the expelled material's rest mass is not simply $-dm$: some of the rocket's rest energy becomes exhaust motion. Instead, use conservation of relativistic energy. Put $q(t)=m(t)\gamma (t)$, so the rocket's energy is $c^{2}q(t)$. The exhaust emitted during $d\tau$ carries energy $-c^2{q}^{\prime }(\tau )d\tau$. For material moving at velocity $w$, the relativistic energy-momentum relation gives momentum $wE/c^2$, since $E={\gamma }_{\mathrm{ex}}\mu c^2$ and $p={\gamma }_{\mathrm{ex}}\mu w$. Hence the modified total momentum isMeanwhile the total relativistic energy is $c^2[q(t)-{\int }_0^t{q}^{\prime }(\tau )d\tau ]=c^{2}q(0)$. These expressions consistently account for both energy and momentum of the expelled material.

Differentiate the momentum formula. The momentum conservation law givesTo simplify it, useMultiplying by ${\gamma }^2(1-uv/c^2)$, the equation becomesThe two terms containing $uv/c^2$ cancel. Dividing by $\gamma >0$ proves the relativistic rocket equation

For this question use the mostly-plus Dirac convention: the Minkowski metric is ${\eta }_{ab}=\mathrm{diag}(-1,1,1,1)$ andThis convention matches the printed plane-wave phase and final identity. It is related to the usual mostly-minus Dirac equation by reversing the metric and taking the negatives of the usual gamma matrices. In particular, the resulting Dirac action and its Dirac adjoint describe the same physical massive field. The Dirac gamma matrices are four complex $4\times 4$ matrices representing the spacetime Clifford algebra with quadratic form $-\eta$. The irreducible complex representation has dimension four. A convenient explicit choice is the negative of the standard Dirac representation of the gamma matrices:where ${\sigma }^i$ are the Pauli matrices. Their multiplication law verifies the displayed anticommutator.

In this representation the gamma matrix adjoint and transpose identities areandThe invariant way to express the latter pattern uses the charge-conjugation matrix:Individual transpose signs depend on the basis. More generally, a similarity transformation ${\gamma }^{\prime a}=M{\gamma }^a{M}^{-1}$ changes the Hermitizing matrix to $H^{\prime }=M^{-†}{\gamma }^0{M}^{-1}$ and the charge-conjugation matrix to $C^{\prime }=MC{M}^T$. Then ${\gamma }^{\prime a†}=H^{\prime }{\gamma }^{\prime a}{H}^{\prime -1}$ and $C^{\prime -1}{\gamma }^{\prime a}{C}^{\prime }=-{\gamma }^{\prime aT}$. Thus the simple formula with ${\gamma }^{\prime 0}$ itself presumes a compatible Hermitian basis, rather than an arbitrary nonunitary similarity transformation.

Applying $(i{\gamma }^a{\partial }_a-m)$ to the Dirac equation gives $(\square -m^2)\Psi =0$. Its mass shell is $p^2=-m^2$, so the frequencies are $p^0=\pm \sqrt{{\mathbf{p}}^2+m^2}$. The Dirac spinor transforms in the four-component Spinor representation of the Lorentz group. Under spatial rotations, the two upper and the two lower components each transform as a two-component spin-$1/2$ representation: the spin angular momentum matrices are $\mathrm{diag}({\sigma }^i,{\sigma }^i)/2$. At rest the positive-energy equation selects the upper two components, giving two independent spin polarizations, and the negative-frequency equation selects the lower two.

In the quantum theory a mode expansion isWith the mostly-plus Dirac convention, $e^{ip\cdot x}=e^{-iEt+i\mathbf{p}\cdot \mathbf{x}}$ is positive frequency. The negative-frequency coefficient obeys $(/p+m)v_s(p)=0$. The fermionic annihilation operators $b_s$ and $d_s$ satisfy the canonical anticommutation relations, with their respective fermionic creation operators. The $b_s^{†}$ excitations are particles; the $d_s^{†}$ excitations are antiparticles with the same positive energy, mass and spin-$1/2$ but opposite charge. After normal ordering, the Hamiltonian operator contains positive multiples of $b_s^{†}{b}_s+d_s^{†}{d}_s$. Reinterpreting the negative-frequency part as antiparticle creation supplies a spectrum bounded below rather than a physical tower of negative-energy particles.

For the printed $e^{ip\cdot x}$ wave, ${\partial }_a\Psi =i{p}_a\Psi$. Substitution givesThe spin label $s$ indexes the two states of a spin-$1/2$ particle, rather than varying the particle's total spin. For real on-shell $p$, Hermitian conjugation and ${\gamma }^{a†}={\gamma }^0{\gamma }^a{\gamma }^0$ giveThese are right and left null-vector equations for the same on-shell matrix.

To obtain the Gordon identity, take both external Dirac spinors to have the same real mass $m$. Their two equations implyDefine ${\gamma }^{ab}=[{\gamma }^a,{\gamma }^b]/2$. The Clifford algebra relation yieldsThereforeMultiplying the previous null-vector equation by $-1$ proves the required formula exactly:For $m\ne 0$ it can be solved for the vector-current matrix element, separating a momentum term from the antisymmetric Dirac spinor term. The identity before division also holds at $m=0$. The signs depend jointly on the metric, Clifford relation, Dirac mass term and plane-wave phase. In the mostly-minus convention of Questions 1 and 3, the printed phase instead gives $(/p+m)u=0$; the corresponding identity uses $(p_b-p_b^{\prime })$ in the antisymmetric term. Mixing that convention with the formula proved here would produce an apparent sign error.

Use ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$ and $\hslash =1$. The free real scalar field has actionThe canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$, and the Hamiltonian operator is obtained by the Legendre transform of the density:Canonical quantization promotes $\varphi$ and $\pi$ to Hermitian operator-valued fields and imposes the equal-time canonical commutation relationsTheir Heisenberg equation of motion gives $\dot{\varphi }=\pi$, $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, hence the Klein-Gordon equation $(\square +m^2)\varphi =0$.

Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. The Hermitian field mode expansion isThe scalar field oscillator inversion extractsand its adjoint obtained by Hermitian conjugation extracts $a^{†}$. Substitute these expressions into the equal-time canonical commutation relation. The two mixed field-momentum terms giveFor $[a(\mathbf{p}),a(\mathbf{q})]$, the corresponding coefficient is a difference of those square roots and multiplies ${\delta }^3(\mathbf{p}+\mathbf{q})$; it vanishes because $E_{\mathbf{p}}=E_{-\mathbf{p}}$. Taking the adjoint gives $[a^{†},a^{†}]=0$. Thus these are bosonic annihilation operators and creation operators. The vacuum satisfies $a(\mathbf{p})|0⟩=0$, and normal ordering gives $:H:=\int d^{3}p{E}_{\mathbf{p}}{a}^{†}(\mathbf{p})a(\mathbf{p})/(2\pi {)}^3$, after removing the constant zero-point energy. A one-particle excitation has energy $E_{\mathbf{p}}$ and spin zero.

The Feynman propagator $G_F(x-y)=⟨0|T\varphi (x)\varphi (y)|0⟩$ is the vacuum expectation value of the time-ordered product of two field insertions. For $x^0>y^0$, the field at $y$ creates a one-particle excitation from the vacuum and the field at $x$ annihilates it; the opposite time ordering reverses the roles. It is a propagation amplitude and correlation function of vacuum fluctuations, rather than a transition probability. From the oscillator expansion, only the $a$-$a^{†}$ contraction survives, so with $t=x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$,Here $\theta$ is the Heaviside step function, and changing $\mathbf{p}$ to $-\mathbf{p}$ in the second term gives the last line.

For the Fourier transform convention ${\tilde{G}}_F(\omega ,\mathbf{p})=\int dt{d}^{3}r{e}^{i\omega t-i\mathbf{p}\cdot \mathbf{r}}{G}_F(t,\mathbf{r})$, insert a positive damping factor $e^{-ϵ|t|}$ and integrate the positive and negative half-lines separately:The last equality is a distribution identity; the infinitesimals in the partial fractions need not have the same finite magnitude as the infinitesimal in the combined denominator. The Feynman i-epsilon prescription means that the positive-frequency pole is just below the real energy axis and the negative-frequency pole is just above it. Closing the contour integral below for $t>0$, and above for $t<0$, reproduces the oscillator result by the residue theorem. This prescription fixes which homogeneous solutions are added to the Green function. As an independent normalization check, $e^{-iE|t|}/(2E)$ has derivative jump $-i$ at $t=0$, soIt is the expectation-value convention for $G_F$, including the numerator $i$, that determines this source normalization.

The vacuum is a centered free Gaussian state. The Wick theorem expresses the four-field time-ordered product as the sum of all pair Wick contractions plus terms containing a normal-ordered product. The latter terms have zero vacuum expectation value. There are exactly three complete pairings, with no fermionic signs for this bosonic field. Thus the free scalar four-point function isThese formulas describe canonical quantization of a real scalar field; in particular, a real scalar field uses a single oscillator family rather than independent charged-particle and antiparticle families.

Work away from $x=0$, where the inverse-square potential is singular. Variation of the phase-space action givesThe last equation is the constraint imposed by the Lagrange multiplier $e$. The canonical momentum conjugate to $t$ is $-E$, not $E$. Consequently the nonzero canonical Poisson brackets areIn particular,With the convention $\delta z=\alpha \{z,Q\}$, the Noether charge generating $\delta t=\alpha$ is $Q=-E$. Equivalently, the conserved physical energy is $E$; a convention that calls the energy the time-translation charge absorbs this minus sign into its parameter.

For the dilation charge,on the constraint surface. Its infinitesimal canonical transformations areFor constant $\alpha$, extending this by ${\delta }_{D}e=-\alpha e$ preserves the action: the kinetic terms are invariant and $E-H$ scales oppositely to $e$. Thus the dilation acts on time and position with their nonrelativistic relative scaling. Its Poisson bracket with the energy is

Put $C=E-H$. Direct variation of the Hamiltonian givessoThe supplied transformations are generated by the special conformal charge of inverse-square mechanicsChooseThen $\delta (eC)=0$ even for a time-dependent parameter. Expanding the kinetic-term variation, keeping the terms in $\dot{\beta }$, givesHenceFor a constant parameter only the boundary term remains, identifying $K$ as the Noether charge. Independently, the equations of motion giveUsing the same canonical Poisson bracket convention,Together, $\{D,E\}=-E$, $\{D,K\}=K$ and $\{E,K\}=2D$ form the sl2R Lie algebra of conformal mechanics. These are time translation, dilation and special conformal transformation. A useful normalization check is the Casimir elementon the constraint surface.