Superfluidity, Landau Criterion, and Accelerated Frames

The previous page used random lattices and matrix models to make geometry dynamical. We now turn in a different direction, but the conceptual issue is secretly similar: what do we mean by the vacuum when the time coordinate itself is changed?

In a Lorentz-invariant relativistic QFT the vacuum is invariant under boosts. A medium is different. A fluid at rest selects a preferred frame. A superfluid is especially sharp: it has a ground state with a macroscopic phase, gapless sound modes, and a remarkable metastability against slow motion. The same physical state can look stable in one frame and unstable in another, because the Hamiltonian in a moving frame is shifted by momentum.

This page develops three related ideas. First, the Landau criterion says that a moving body can radiate excitations only when its velocity exceeds

$$v_c=\inf_{\mathbf p\ne0}\,{\varepsilon(\mathbf p)\over |\mathbf p|},$$

where $\varepsilon(\mathbf p)$ is the quasiparticle dispersion in the rest frame of the fluid. Second, rotating systems obey the analogous rule with $H$ replaced by $H-\Omega L$, giving the same kinematic seed as superradiance. Third, accelerated observers use a different time coordinate, the Rindler time, so their natural Hamiltonian and mode functions differ from those of inertial observers. The Rindler calculation is not just curved-coordinate gymnastics; it is another example of the same theme: the notion of positive frequency depends on the time flow used to define energy.

We finish by returning to the superfluid order parameter. The low-energy field is the phase $\varphi$ of a condensate. Its stiffness produces sound, persistent flow, and a dynamical current response with a massless pole. This is the bridge to vortices and compact phases on the next page.

Frame-dependent vacua in a medium

For the Galilean discussion set $\hbar=1$. A state with energy $E$, momentum $\mathbf P$, and fixed total mass $M$ has energy in a frame moving with velocity $\mathbf v$

$$E'=E-\mathbf v\cdot\mathbf P+{1\over2}Mv^2.$$

The last term is common to all states in the fixed-mass sector and drops out of excitation energies.

A system with no medium has Poincaré symmetry. If the vacuum is unique, translation- and Lorentz-invariance strongly constrain it. A uniform fluid, however, is not Lorentz-invariant. It has a rest frame, a density, and a chemical potential. Even if the microscopic theory is Galilean-invariant, the state is not invariant under boosts. Boosting the state makes a moving fluid.

The Galilean transformation of energy is the basic calculation. Let the total mass be $M$ and the total momentum be $\mathbf P$. Up to an additive constant, the Hamiltonian in a frame moving with velocity $\mathbf v$ is

$$H_{\mathbf v}=H-\mathbf v\cdot\mathbf P.$$

For a quasiparticle above the ground state,

$$H|\mathbf p\rangle=(E_0+\varepsilon(\mathbf p))|\mathbf p\rangle, \qquad \mathbf P|\mathbf p\rangle=\mathbf p|\mathbf p\rangle,$$

so

$$H_{\mathbf v}|\mathbf p\rangle =\bigl(E_0+\varepsilon(\mathbf p)-\mathbf v\cdot\mathbf p\bigr)|\mathbf p\rangle.$$

Thus the excitation energy in the moving frame is

$$\boxed{\varepsilon_{\mathbf v}(\mathbf p)=\varepsilon(\mathbf p)-\mathbf v\cdot\mathbf p.}$$

If this quantity is negative for some $\mathbf p$, the moving state is not the lowest-energy state with respect to $H_{\mathbf v}$. The system can lower its energy by producing that excitation. This is the kinematic heart of superfluidity.

A normal gas has single-particle excitations with

$$\varepsilon(\mathbf p)={\mathbf p^2\over2m}.$$

Then

$${\varepsilon(p)\over p}={p\over2m}\to0 \qquad (p\to0),$$

so the kinematic critical velocity vanishes. Provided an impurity, boundary, or another subsystem can exchange momentum with the gas, arbitrarily slow motion has an allowed soft-excitation channel. A superfluid instead has a phonon branch

$$\varepsilon(p)=c_s p+O(p^3),$$

with sound speed $c_s$. If no other branch has smaller $\varepsilon(p)/p$, then $v_c=c_s$. In real helium there is also a roton minimum, and the infimum of $\varepsilon(p)/p$ can occur at finite momentum. The clean statement is therefore not “the critical velocity is the sound speed,” but rather “the critical velocity is the greatest lower bound of slopes from the origin to the dispersion curve.” When that bound is attained, it is an ordinary minimum.

The Landau velocity is the infimum of $\varepsilon(p)/p$. A line $vp$ with $v>v_c$ lies above the dispersion for some momentum, so $\varepsilon(p)-vp<0$ in the moving frame. For $v<v_c$, every quasiparticle has nonnegative moving-frame energy.

This criterion is only a stability criterion for creating elementary excitations. It is necessary, not always sufficient. Boundaries, impurities, vortex nucleation, finite-size effects, and heating can reduce the observed critical velocity. Still, the criterion captures a universal point: dissipation by quasiparticle emission requires an energetically allowed channel.

Derivation from phonon emission

Consider a heavy body of mass $M$ moving through the fluid. In the fluid rest frame it has initial momentum $\mathbf P$ and energy

$$E(\mathbf P)={\mathbf P^2\over2M}.$$

Suppose it emits a quasiparticle of momentum $\mathbf k$ and energy $\varepsilon(\mathbf k)$. Momentum conservation gives the final body momentum $\mathbf P-\mathbf k$, and energy conservation requires

$$E(\mathbf P)=E(\mathbf P-\mathbf k)+\varepsilon(\mathbf k).$$

Substituting the nonrelativistic kinetic energy,

$${\mathbf P^2\over2M} ={|\mathbf P-\mathbf k|^2\over2M}+\varepsilon(\mathbf k),$$

so

$${2\mathbf P\cdot\mathbf k-\mathbf k^2\over2M}=\varepsilon(\mathbf k).$$

Writing $\mathbf v=\mathbf P/M$,

$$\mathbf v\cdot\mathbf k =\varepsilon(\mathbf k)+{\mathbf k^2\over2M}.$$

For a macroscopic body $M$ is large, so recoil is negligible:

$$\mathbf v\cdot\mathbf k\simeq\varepsilon(\mathbf k).$$

Since $\mathbf v\cdot\mathbf k\le v|\mathbf k|$, emission is possible only if

$$v\ge {\varepsilon(\mathbf k)\over|\mathbf k|}$$

for some $\mathbf k$. Therefore the threshold is

$$\boxed{v_c=\inf_{\mathbf k\ne0}{\varepsilon(\mathbf k)\over |\mathbf k|}.}$$

For phonons,

$$\varepsilon(\mathbf k)=c_s|\mathbf k|,$$

and the threshold is $v_c=c_s$. For a finite emitted momentum and finite recoil, one needs $v>c_s$; the value $c_s$ is approached as $k\to0$. This is the direct analogue of Cherenkov radiation: a source moving faster than the wave speed can radiate waves.

A heavy body with velocity $\mathbf v=\mathbf P/M$ can emit a quasiparticle only when $\mathbf k\cdot\mathbf v$ can match $\varepsilon(\mathbf k)$ up to the small recoil term $\mathbf k^2/(2M)$. For phonons this gives the threshold $v=c_s$.

The same derivation can be phrased without the body. A uniformly moving superfluid is described by the frame Hamiltonian

$$H_{\mathbf v}=H-\mathbf v\cdot\mathbf P.$$

If every excitation has $\varepsilon(\mathbf p)-\mathbf v\cdot\mathbf p>0$, the moving state is locally stable. If some excitation has negative energy, it can be produced while lowering $H_{\mathbf v}$. The heavy body simply supplies a concrete mechanism for momentum exchange.

Rotating frames and superradiance

A rotating system gives a close cousin of the Landau criterion. In a frame rotating with angular velocity $\Omega$, the Hamiltonian is

$$H_\Omega=H-\Omega L_z.$$

A mode with energy $\omega$ and angular momentum quantum number $m$ has rotating-frame energy

$$\omega_\Omega=\omega-m\Omega.$$

For a positive-frequency co-rotating mode with $m>0$, if

$$0<\omega<m\Omega,$$

then the mode has negative energy with respect to the rotating Hamiltonian. If the rotating body, horizon, or medium has an absorptive channel, scattering can amplify the outgoing wave while extracting rotational energy. Negative rotating-frame energy is the kinematic condition; dissipation or horizon boundary conditions supply the physical channel. This is the same algebraic structure as

$$\varepsilon_{\mathbf v}(\mathbf p)=\varepsilon(\mathbf p)-\mathbf v\cdot\mathbf p.$$

The replacement is

$$\mathbf v\cdot\mathbf p \quad\longleftrightarrow\quad \Omega L_z.$$

In a rotating frame, a mode with quantum numbers $(\omega,m)$ has energy $\omega-m\Omega$. When this is negative and an absorptive channel is present, amplified emission can extract rotational energy. This is the rotating analogue of the Landau condition in a moving superfluid.

In hydrodynamic language, a point at radius $R$ on the rotating object has tangential speed $v=\Omega R$. A mode with angular quantum number $m$ has azimuthal wave number roughly $m/R$, so the condition $\omega<m\Omega$ says that the boundary moves faster than the phase velocity $\omega/(m/R)$. Again the logic is Cherenkov-like: a moving source radiates when it outruns a mode.

This observation is deliberately kinematic. A full superradiance calculation must impose boundary conditions, distinguish ingoing and outgoing positive-norm modes, and compute the absorptive flux. The invariant core is the sign of the conserved frequency associated with the co-rotating time flow.

Superfluid order parameter and broken Galilean symmetry

A superfluid is most economically described by a complex order parameter

$$\psi(x)=\rho(x)e^{i\varphi(x)}.$$

For a weakly interacting Bose gas, a Landau–Ginzburg energy functional has the form

$$F[\psi]=\int d^d x\, \left[{1\over2m}|\nabla\psi|^2-\mu|\psi|^2+{g\over2}|\psi|^4\right], \qquad g>0.$$

The potential is minimized at nonzero amplitude when $\mu>0$. Writing $\psi=\rho e^{i\varphi}$ gives

$${1\over2m}|\nabla\psi|^2 ={1\over2m}(\nabla\rho)^2+{\rho^2\over2m}(\nabla\varphi)^2.$$

Thus

$$F=\int d^d x\, \left[{1\over2m}(\nabla\rho)^2+U(\rho)+{\rho^2\over2m}(\nabla\varphi)^2\right],$$

where

$$U(\rho)=-\mu\rho^2+{g\over2}\rho^4.$$

At the minimum $\rho=\rho_0$, with

$$U'(\rho_0)=0,$$

the amplitude fluctuation is massive, while the phase is massless. Freezing the amplitude at long distances leaves

$$F_{\rm IR}={n_s\over2}\int d^d x\,(\nabla\varphi)^2.$$

Here $n_s$ denotes the phase stiffness, not necessarily the number density; at this mean-field level $n_s=\rho_0^2/m$. Authors who use $n_s$ for the superfluid number density instead write the coefficient as $n_s/(2m)$. The invariant statement is that gradients of the phase cost energy.

The condensate chooses a point on a circle of vacua. Radial fluctuations are massive, while angular fluctuations are the massless phase mode. The dynamical response has a sound pole; after the phase is integrated out, the static spatial kernel is transverse.

The superfluid velocity is proportional to the phase gradient,

$$\mathbf v_s={1\over m}\nabla\varphi$$

in the most common convention. A uniform phase gradient therefore represents a flowing condensate. An active Galilean boost that gives the condensate velocity $\mathbf u$ acts on the microscopic boson field as

$$\psi_{\mathbf u}(t,\mathbf x) =e^{im(\mathbf u\cdot\mathbf x-u^2t/2)} \psi(t,\mathbf x-\mathbf u t).$$

For a spatially uniform rest condensate, or after accounting for the shifted argument, this changes the phase gradient by

$$\nabla\varphi\mapsto\nabla\varphi+m\mathbf u.$$

So a boost changes the superfluid velocity. The ground state is not invariant under Galilean boosts; it is mapped to a different state with persistent flow. This is the symmetry statement behind the frame-dependent energy formula.

The same phase field gives the phonon. A time-dependent effective action has the schematic form

$$S_{\rm eff}=\int dt\,d^d x\, \left[{\chi\over2}(\partial_t\varphi)^2-{n_s\over2}(\nabla\varphi)^2+\cdots\right],$$

where $\chi$ is the compressibility. The equation of motion is

$$\chi\, \partial_t^2\varphi-n_s\nabla^2\varphi=0,$$

so the phase mode has

$$\omega=c_s|\mathbf k|, \qquad c_s^2={n_s\over\chi}.$$

This is the sound branch used in the Landau criterion.

Gauge response and the massless pole

Couple the conserved particle-number current to a background field $A_\mu$. At quadratic order the real-time phase action is

$$S[\varphi,A]=\frac12\int dt\,d^d x\, \left[ \chi(\partial_t\varphi-QA_0)^2 -n_s(\nabla\varphi-Q\mathbf A)^2 \right].$$

Here $Q$ is the charge coupled to the source, and $n_s$ retains the stiffness normalization defined above.

The phase propagator at $A_\mu=0$ has denominator

$$\chi(\omega+i0)^2-n_s k^2 =\chi\left[(\omega+i0)^2-c_s^2k^2\right].$$

Current correlators therefore have a Goldstone pole at $\omega^2=c_s^2k^2$. In a neutral superfluid this propagating pole is the phonon.

The zero-frequency spatial response is related but must be stated separately. Its free energy is

$$F[\varphi,A]={n_s\over2}\int d^d x\,(\nabla\varphi-Q\mathbf A)^2.$$

The phase is not optional: it adjusts itself to the longitudinal part of $\mathbf A$. Decompose

$$A_i=A_i^{\rm T}+\partial_i\chi, \qquad \partial_i A_i^{\rm T}=0.$$

Choosing

$$\varphi=Q\chi$$

cancels the longitudinal part. The remaining energy is

$$F[A]={n_s Q^2\over2}\int d^d x\,(A_i^{\rm T})^2.$$

In momentum space the transverse projection is

$$A_i^{\rm T}(\mathbf k)=\left(\delta_{ij}-{k_ik_j\over k^2}\right)A_j(\mathbf k),$$

so, up to the overall sign convention used to define the response kernel,

$$\Pi_{ij}(\mathbf k)=n_s Q^2 \left(\delta_{ij}-{k_ik_j\over k^2}\right)$$

for $\mathbf k\ne0$. The local $n_sQ^2\delta_{ij}$ contact term and exchange of the massless phase combine to cancel the longitudinal static response, leaving this transverse projector. Its nonanalytic $k_i k_j/k^2$ term is the zero-frequency remnant of the dynamical Goldstone pole; the full static kernel itself is transverse, not longitudinal. If $A_\mu$ is promoted from a background source to a dynamical electromagnetic field, the charged Goldstone mode is absorbed by the gauge field and the transverse stiffness produces the Meissner effect. In chiral symmetry breaking, the analogous stiffness is the pion decay constant:

$$n_s\quad\longleftrightarrow\quad f_\pi^2.$$

This analogy is valuable: phase stiffness in condensed matter and current algebra in relativistic QFT are two versions of the same infrared idea. A broken continuous symmetry gives a massless field, and the coefficient of its gradient energy controls current correlators.

Accelerated coordinates and Rindler time

We now shift from moving media to accelerated frames. The connection is conceptual rather than material: different time flows define different Hamiltonians, hence different notions of positive energy.

Specializing the site-wide $(+---)$ convention to $1+1$ dimensions gives $ds^2=dt^2-dx^2$. In the right Rindler wedge $x>|t|$, introduce

$$t=\rho\sinh\eta, \qquad x=\rho\cosh\eta, \qquad \rho>0.$$

Then

$$dt=\sinh\eta\,d\rho+\rho\cosh\eta\,d\eta,$$

and

$$dx=\cosh\eta\,d\rho+\rho\sinh\eta\,d\eta.$$

Therefore

$$ds^2=dt^2-dx^2 =\rho^2d\eta^2-d\rho^2.$$

Worldlines with fixed $\rho$ are hyperbolae,

$$x^2-t^2=\rho^2.$$

Their proper time is

$$d\tau_{\rm prop}=\rho\,d\eta,$$

and their proper acceleration is

$$a={1\over\rho}.$$

Thus Rindler time $\eta$ is the natural time coordinate for uniformly accelerated observers.

The right Rindler wedge is foliated by hyperbolae $\rho={\rm const}$ and rays $\eta={\rm const}$. After analytic continuation $\eta=-i\theta$, the metric is locally polar, $ds_E^2=d\rho^2+\rho^2d\theta^2$, so regularity at the origin requires $\theta$ to have period $2\pi$.

The Euclidean continuation is especially revealing. Set

$$\eta=-i\theta.$$

Then

$$ds_E^2=d\rho^2+\rho^2d\theta^2.$$

This is simply the flat Euclidean plane in polar coordinates. To avoid a conical singularity at $\rho=0$, the angle must have period

$$\theta\sim\theta+2\pi.$$

For an observer at fixed $\rho=a^{-1}$, proper Euclidean time is

$$\tau_E=\rho\theta={\theta\over a}.$$

The period in proper Euclidean time is therefore

$$\beta={2\pi\over a},$$

which corresponds to temperature

$$T={a\over2\pi}.$$

For the Minkowski vacuum, Euclidean correlation functions are regular at the origin and inherit this angular periodicity. The corresponding Kubo–Martin–Schwinger period is the Euclidean kernel of the Unruh effect. It does not say that every state in the wedge is thermal: the regularity and analyticity of the Minkowski vacuum are essential. What matters for the present page is that the accelerated Hamiltonian is not the inertial Hamiltonian. Positive Rindler frequency and positive Minkowski frequency are different decompositions of the same field.

Rindler modes and the exponential wall

For a scalar field in $1+1$ dimensions,

$$S={1\over2}\int dt\,dx\, \left[(\partial_t\phi)^2-(\partial_x\phi)^2-m^2\phi^2\right].$$

Using

$$ds^2=\rho^2d\eta^2-d\rho^2,$$

one can write the wave equation as

$$\left[{1\over\rho^2}\partial_\eta^2-{1\over\rho}\partial_\rho(\rho\partial_\rho)+m^2\right]\phi=0.$$

It is often cleaner to introduce a reference length $\rho_0$ and the dimensionless coordinate

$$z=\log {\rho\over\rho_0}, \qquad \rho=\rho_0e^z.$$

Then

$$ds^2=\rho_0^2e^{2z}(d\eta^2-dz^2).$$

The massless kinetic term is conformally simple in two dimensions, while the mass term is multiplied by the Weyl factor. The Rindler Hamiltonian becomes, up to the standard factor $1/2$,

$$H_R={1\over2}\int dz\, \left[\pi^2+(\partial_z\phi)^2+(m\rho_0)^2e^{2z}\phi^2\right].$$

So the Rindler problem is equivalent to a one-dimensional wave equation with an exponential potential wall. A mode of Rindler frequency $\omega$,

$$\phi(\eta,z)=e^{-i\omega\eta}f_\omega(z),$$

obeys

$$\boxed{ \left(\partial_z^2+\omega^2-(m\rho_0)^2e^{2z}\right)f_\omega(z)=0. }$$

The decaying solution at large $z$ is

$$f_\omega(z)\propto K_{i\omega}(m\rho_0e^z)=K_{i\omega}(m\rho),$$

where $K_\nu$ is the modified Bessel function. Near the horizon, $\rho\to0$ or $z\to-\infty$, the potential disappears and the solutions behave like plane waves in $z$:

$$f_\omega(z)\sim A(\omega)e^{i\omega z}+A^*(\omega)e^{-i\omega z}.$$

In the dimensionless coordinate $z=\log(\rho/\rho_0)$, a massive scalar mode in the Rindler wedge sees the potential $(m\rho_0)^2e^{2z}$. Near the horizon $z\to-\infty$ the mode is approximately a free wave; far from the horizon the modified Bessel function $K_{i\omega}(m\rho)$ decays through the exponential wall.

The fixed-frequency Green function is correspondingly built from these Rindler modes. For example, a Rindler-vacuum time-ordered Green function has the schematic form

$$G(\eta_1,\rho_1;\eta_2,\rho_2) \sim \int_0^\infty d\omega\, \mathcal N(\omega) K_{i\omega}(m\rho_1)K_{i\omega}(m\rho_2) e^{-i\omega|\eta_1-\eta_2|},$$

with normalization and an $i0$ prescription fixed by the Green function and mode normalization. The Minkowski-vacuum correlator restricted to the wedge is not obtained by using the same zero-temperature weights: it contains both frequency signs with Bose factors at dimensionless inverse temperature $2\pi$, in accord with the KMS condition above. The formula is less important than the lesson: the accelerated observer expands the same field in modes adapted to $\partial_\eta$, not $\partial_t$.

This is why accelerated frames belong naturally in the same discussion as moving superfluids. In both cases, the “energy” relevant for stability or occupation is the generator of the time flow used by the observer or medium:

$$H_{\mathbf v}=H-\mathbf v\cdot\mathbf P, \qquad H_\Omega=H-\Omega L, \qquad H_R=i\partial_\eta.$$

The differences are substantial — Rindler coordinates introduce horizons and thermal behavior — but the algebraic theme is common.

Summary

A medium chooses a rest frame. In a moving frame, excitation energies are shifted by momentum:

$$\varepsilon_{\mathbf v}(\mathbf p)=\varepsilon(\mathbf p)-\mathbf v\cdot\mathbf p.$$

A moving superfluid is stable against quasiparticle emission only if this remains nonnegative for all $\mathbf p$. This gives the Landau criterion

$$v_c=\inf_{\mathbf p\ne0}{\varepsilon(\mathbf p)\over |\mathbf p|}.$$

For a pure phonon dispersion, $v_c=c_s$. For a normal gas with $\varepsilon=p^2/(2m)$, $v_c=0$.

The rotating analogue uses

$$H_\Omega=H-\Omega L,$$

so a mode is superradiant when $\omega-m\Omega<0$. The accelerated analogue uses Rindler time, whose Euclidean continuation is polar angle. Regularity fixes the Euclidean period and leads to the temperature $T=a/(2\pi)$ for a uniformly accelerated observer.

The low-energy superfluid field is the phase of the condensate. Its stiffness

$$F_{\rm IR}={n_s\over2}\int d^d x\,(\nabla\varphi)^2$$

is the origin of the phonon, persistent flow, current response, and the analogy with current algebra in relativistic symmetry breaking.

Common pitfalls

Kinematic threshold versus observed breakdown. The Landau criterion is the threshold for creating quasiparticles in an idealized kinematic calculation, not a universal experimental critical velocity. Real flows can break down earlier through vortices, boundary roughness, turbulence, or heating.

The sound speed need not win. The sound speed is the answer only when the phonon branch gives the infimum of $\varepsilon(p)/p$. A roton-like minimum or another low-energy branch can lower $v_c$.

Boost symmetry does not leave the state fixed. A Galilean boost of the microscopic theory maps one superfluid ground state to another flowing state. This is why the phase gradient and the frame velocity are physically meaningful.

Rindler energy is boost energy. The Rindler Hamiltonian is not the ordinary Minkowski Hamiltonian written in strange coordinates. It is the generator of boosts restricted to a wedge, and its notion of positive frequency differs from inertial positive frequency.

The phase is compact. The local stiffness action treats $\varphi$ as a smooth real field, but the physical phase obeys $\varphi\sim\varphi+2\pi$. Vortices and winding sectors require that compactness; that is where the next page begins.

Exercises

A dispersion whose bound is not attained

Let a quasiparticle have dispersion

$$\varepsilon(p)=\sqrt{\Delta^2+c^2p^2}.$$

Compute the Landau critical velocity

$$v_c=\inf_{p>0}{\varepsilon(p)\over p}.$$

Interpret the result.

Solution

We have

$${\varepsilon(p)\over p}=\sqrt{{\Delta^2\over p^2}+c^2}.$$

For $\Delta>0$, this decreases monotonically as $p$ increases and approaches $c$ from above:

$$\lim_{p\to\infty}{\varepsilon(p)\over p}=c.$$

Thus the infimum is

$$v_c=c.$$

This result should be interpreted with care. The relativistic-looking dispersion has no finite-momentum minimum of $\varepsilon/p$; only its infimum is approached asymptotically. In a real condensed-matter system the high-momentum dispersion will eventually deviate from this form, and the actual minimum may occur elsewhere. The exercise shows both why the full dispersion matters and why the criterion is best written with an infimum.

Finite recoil in phonon emission

A heavy body of mass $M$ moves with velocity $\mathbf v$ through a superfluid. Keeping the recoil term, show that emission of a quasiparticle with momentum $\mathbf k$ requires

$$\mathbf v\cdot\mathbf k=\varepsilon(\mathbf k)+{\mathbf k^2\over2M}.$$

For phonons $\varepsilon(k)=c_s k$, find the infimum of speeds that permit emission at finite $M$, and state whether it is attained at nonzero $k$.

Solution

Energy conservation gives

$${\mathbf P^2\over2M}={|\mathbf P-\mathbf k|^2\over2M}+\varepsilon(\mathbf k).$$

Expanding the square,

$$|\mathbf P-\mathbf k|^2=\mathbf P^2-2\mathbf P\cdot\mathbf k+\mathbf k^2.$$

Therefore

$$0={-2\mathbf P\cdot\mathbf k+\mathbf k^2\over2M}+\varepsilon(\mathbf k),$$

or

$${\mathbf P\over M}\cdot\mathbf k=\varepsilon(\mathbf k)+{\mathbf k^2\over2M}.$$

Since $\mathbf v=\mathbf P/M$,

$$\mathbf v\cdot\mathbf k=\varepsilon(\mathbf k)+{\mathbf k^2\over2M}.$$

For phonons,

$$v k\cos\theta=c_s k+{k^2\over2M}.$$

The most favorable angle is $\cos\theta=1$, so

$$v=c_s+{k\over2M}.$$

For any finite $k$ this is larger than $c_s$, but by taking $k\to0$ one approaches

$$v_c=c_s.$$

Thus recoil does not change the infimum of the threshold in the ideal phonon theory, although no finite-$k$ emission occurs exactly at $v=c_s$ and recoil changes the kinematics at every fixed nonzero $k$.

Euclidean regularity of Rindler space

Derive the Rindler metric from

$$t=\rho\sinh\eta, \qquad x=\rho\cosh\eta.$$

Then analytically continue $\eta=-i\theta$ and show that regularity requires $\theta$ to have period $2\pi$.

Solution

Differentiate:

$$dt=\sinh\eta\,d\rho+\rho\cosh\eta\,d\eta,$$ $$dx=\cosh\eta\,d\rho+\rho\sinh\eta\,d\eta.$$

Then

$$dt^2-dx^2 =(\sinh^2\eta-\cosh^2\eta)d\rho^2 +\rho^2(\cosh^2\eta-\sinh^2\eta)d\eta^2,$$

while the cross terms cancel. Since

$$\cosh^2\eta-\sinh^2\eta=1,$$

we get

$$ds^2=\rho^2d\eta^2-d\rho^2.$$

Set $\eta=-i\theta$. The Euclidean metric is

$$ds_E^2=d\rho^2+\rho^2d\theta^2.$$

This is the flat plane in polar coordinates. At $\rho=0$, the coordinate $\theta$ is an angle. The plane is smooth only if

$$\theta\sim\theta+2\pi.$$

A different period would produce a conical singularity.

Rindler mode equation

Starting from the massive Klein–Gordon equation in Rindler coordinates,

$$\left[{1\over\rho^2}\partial_\eta^2-{1\over\rho}\partial_\rho(\rho\partial_\rho)+m^2\right]\phi=0,$$

set $\rho=\rho_0e^z$ and $\phi=e^{-i\omega\eta}f_\omega(z)$. Show that

$$\left(\partial_z^2+\omega^2-(m\rho_0)^2e^{2z}\right)f_\omega(z)=0.$$

Solution

Since $\rho=\rho_0e^z$,

$$\partial_\rho={1\over\rho}\partial_z.$$

Therefore

$$\rho\partial_\rho=\partial_z,$$

and

$${1\over\rho}\partial_\rho(\rho\partial_\rho) ={1\over\rho}\partial_\rho\partial_z ={1\over\rho^2}\partial_z^2.$$

Substituting into the wave equation gives

$$\left[{1\over\rho^2}\partial_\eta^2-{1\over\rho^2}\partial_z^2+m^2\right]\phi=0.$$

Multiplying by $\rho^2=\rho_0^2e^{2z}$,

$$\left[\partial_\eta^2-\partial_z^2+(m\rho_0)^2e^{2z}\right]\phi=0.$$

With $\phi=e^{-i\omega\eta}f_\omega(z)$,

$$\left[-\omega^2-\partial_z^2+(m\rho_0)^2e^{2z}\right]f_\omega(z)=0.$$

Multiplying by $-1$ gives

$$\left(\partial_z^2+\omega^2-(m\rho_0)^2e^{2z}\right)f_\omega(z)=0.$$

Static transverse response

Consider the phase-only superfluid free energy

$$F[\varphi,A]={n_s\over2}\int d^d x\,(\nabla\varphi-Q\mathbf A)^2.$$

Decompose $\mathbf A=\mathbf A^{\rm T}+\nabla\chi$ with $\nabla\cdot\mathbf A^{\rm T}=0$. Minimize over $\varphi$ and show that only the transverse field costs energy.

Solution

Substitute the decomposition:

$$\nabla\varphi-Q\mathbf A =\nabla\varphi-Q\nabla\chi-Q\mathbf A^{\rm T} =\nabla(\varphi-Q\chi)-Q\mathbf A^{\rm T}.$$

The transverse field is orthogonal to gradients under integration by parts:

$$\int d^d x\,\mathbf A^{\rm T}\cdot\nabla f =-\int d^d x\,f\,\nabla\cdot\mathbf A^{\rm T}=0,$$

assuming boundary terms vanish. Therefore

$$F={n_s\over2}\int d^d x\, \left[(\nabla(\varphi-Q\chi))^2+Q^2(\mathbf A^{\rm T})^2\right].$$

The minimum over $\varphi$ is obtained by choosing

$$\varphi=Q\chi+\text{constant},$$

so

$$F_{\rm min}[A]={n_s Q^2\over2}\int d^d x\,(\mathbf A^{\rm T})^2.$$

Thus the phase screens the longitudinal part of $\mathbf A$, while the transverse part measures the stiffness. This is the static origin of the superfluid or superconducting current response.

References

• Birrell, N. D., and Davies, P. C. W. Quantum Fields in Curved Space. Cambridge Monographs on Mathematical Physics. Cambridge: Cambridge University Press, 1982.

• Landau, L. D., and Lifshitz, E. M. Statistical Physics, Part 2. 3rd ed. Course of Theoretical Physics 9. Oxford: Pergamon Press, 1980.

• Leggett, A. J. Quantum Liquids: Bose Condensation and Cooper Pairing in Condensed-Matter Systems. Oxford: Oxford University Press, 2006.

• Pitaevskii, L., and Stringari, S. Bose–Einstein Condensation and Superfluidity. Oxford: Oxford University Press, 2016.

• Polyakov, A. M. Gauge Fields and Strings. Contemporary Concepts in Physics 3. Chur: Harwood Academic Publishers, 1987.

• Wald, R. M. Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics. Chicago Lectures in Physics. Chicago: University of Chicago Press, 1994.