Higher-than-ballistic conduction of viscous electron flows

Significance

Free electron flows through constrictions in metals are often regarded as an ultimate high-conduction charge transfer. We predict that electron fluids can flow with a resistance that is much smaller than the fundamental quantum mechanical ballistic limit for nanoscale electronics. The “superballistic” low-dissipation transport is particularly striking for the flow through a viscous point contact, a constriction exhibiting the quantum mechanical ballistic transport at zero temperature but governed by viscous electron hydrodynamics at a higher temperature. Unlike other mechanisms of low-dissipation transport, for example, superconductivity, the viscous electron flows can be realized at elevated temperatures, granting a new route for the low-power electronics research.

Abstract

Strongly interacting electrons can move in a neatly coordinated way, reminiscent of the movement of viscous fluids. Here, we show that in viscous flows, interactions facilitate transport, allowing conductance to exceed the fundamental Landauer’s ballistic limit $G_{\text{ball}}$. The effect is particularly striking for the flow through a viscous point contact, a constriction exhibiting the quantum mechanical ballistic transport at $T=0$ but governed by electron hydrodynamics at elevated temperatures. We develop a theory of the ballistic-to-viscous crossover using an approach based on quasi-hydrodynamic variables. Conductance is found to obey an additive relation $G=G_{\text{ball}}+G_{\text{vis}}$, where the viscous contribution $G_{\text{vis}}$ dominates over $G_{\text{ball}}$ in the hydrodynamic limit. The superballistic, low-dissipation transport is a generic feature of viscous electronics.

Free electron flow through constrictions in metals is often regarded as an ultimate high-conduction charge transfer mechanism (1–5). Can conductance ever exceed the ballistic limit value? Here we show that superballistic conduction is possible for strongly interacting systems in which electron movement resembles that of viscous fluids. Electron fluids are predicted to occur in quantum-critical systems and in high-mobility conductors, so long as momentum-conserving electron–electron (ee) scattering dominates over other scattering processes (6–9). Viscous electron flows feature a host of novel transport behaviors (10–22). Signatures of such flows have been observed in ultraclean GaAs, graphene, and ultrapure PdCoO₂ (23–26).

We will see that electrons in a viscous flow can achieve through cooperation what they cannot accomplish individually. As a result, resistance and dissipation of a viscous flow can be markedly smaller than that for the free-fermion transport. As a simplest realization, we discuss viscous point contact (VPC), where correlations act as a ”lubricant” facilitating the flow. The reduction in resistance arises due to the streaming effect illustrated in Fig. 1, wherein electron currents bundle up to form streams that bypass the boundaries, where momentum loss occurs. This surprising behavior is in a clear departure from the common view that regards electron interactions as an impediment for transport.

Fig. 1.

(A) Current streamlines (black) and potential color map for viscous flow through a constriction. Velocity magnitude is proportional to the density of streamlines. Current forms a narrow stream, avoiding the boundaries where dissipation occurs and allowing the resistance, Eq. 1, to drop below the ballistic limit value. (B) Current distribution in the constriction for different carrier collision mean-free-path values. The distribution evolves from a constant in the ballistic regime to a semicircle in the viscous regime, Eq. 10, illustrating the interaction-induced streaming effect. Parameters used are $L=3w$ and $b={10}^{5}v$. The distributions are normalized to unit total current. A Fourier space filter was used to smooth out the Gibbs phenomenon.

A simplest VPC is a 2D constriction pictured in Fig. 1A. The interaction effects dominate in constrictions of width $w$ exceeding the carrier collision mean free path $l_{\text{ee}}$ (and much greater than the Fermi wavelength ${\lambda }_F$). The VPC conductance, evaluated in the absence of impurity scattering, scales as a square of the width $w$ and inversely with the electron viscosity $\eta$,

$$G_{\text{vis}}(w)=\frac{\pi n^2{e}^2{w}^2}{32\eta },\hspace{1em}w\gg l_{\text{ee}},$$ [1]

where $n$ and $e$ are the carrier density and charge. In the opposite limit, $l_{\text{ee}}\gg w$, the ballistic free-fermion model (1, 5) predicts the conductance $G_{\text{ball}}=2e^2/hN$, where $N\approx 2w/{\lambda }_F$ is the number of Landauer’s open transmission channels. The conductance $G_{\text{vis}}$ grows with width faster than $G_{\text{ball}}$. Therefore, for large enough $w$, viscous transport yields $G$ values above the ballistic bound.

Conveniently, both regimes are accessible in a single constriction, because transport is expected to be viscous at elevated temperatures and ballistic at $T=0$. The crossover temperature can be estimated in terms of the ee scattering mean free path as

$$l_{\text{ee}}(T)/w={\pi }^2/16\approx \mathrm{0.62.}$$ [2]

This relation is found by setting $R_{\text{vis}}=R_{\text{ball}}$ and expressing viscosity as $\eta =\nu nm=\frac{1}{4}{v}_F{l}_{\text{ee}}nm$, where $m$ is the carrier mass and the kinetic viscosity $\nu$ is estimated in Eq. S21. The condition Eq. 2 can be readily met in micron-size graphene junctions.

Several effects of electron interactions on transport in constrictions were discussed recently. Refs. 14 and 15 study junctions with spatially varying electron density and, using the time-dependent current density functional theory, predict a suppression of conductance. A hydrodynamic picture of this effect was established in ref. 16. In contrast, here we study junctions in which, in the absence of applied current, the carrier density is approximately position-independent. This situation was analyzed in ref. 27 perturbatively in the ee scattering rate, finding a conductance enhancement that resembles our results.

The relation Eq. 1 points to a simple way to measure viscosity by the conventional transport techniques. Precision measurements of viscosity in fluids date as far back as the 19th century (28). They relied, in particular, on measuring resistance of a viscous fluid discharged through a narrow channel or an orifice, a direct analog of our constriction geometry. Further, viscosity-induced electric conduction has a well-known counterpart in the kinetics of classical gases, where momentum exchange between atoms results in a slower momentum loss and a lower resistance of gas flow. Viscous effects are responsible, in particular, for a dramatic drop in the hydrodynamic resistance upon a transition from Knudsen to Poiseuille regime. For a viscous flow through scatterers spaced by a distance $L$, the typical time of momentum transfer is $\tau \sim L^2/\nu \approx L^2/v_T\mathrm{\ell }$, whereas, for an ideal gas, this time is $\tau \prime =L/v_T$, where $v_T$ is thermal velocity and $\mathrm{\ell }$ is the mean free path. For $\mathrm{\ell }\ll L$, the viscous time $\tau$ is much longer than the ballistic time $\tau \prime$.

The peculiar correlations originating from fast particle collisions in proximity to scatterers can be elucidated by a spatial argument: Particle collisions near a scatterer reduce the average velocity component normal to the scatterer surface, $v_{\perp }$, which slows down the momentum loss rate per particle, $m\text{𝐯}{v}_{\perp }/L$. Momentum exchange makes particles flow collectively, on average staying away from scatterers and thus lowering the resistance.

The viscosity-induced drop in resistance can be used as a vehicle to overcome the quantum ballistic limit for electron conduction. Indeed, we can compare the values $R_{\text{vis}}$ and $R_{\text{ball}}$ by putting them in a Drude-like form $R=m/n{e}^2\tau$, with $m$ as the carrier mass and $\tau$ as a suitable momentum relaxation time. Eq. 1 can be modeled in this way using the time of momentum diffusion across the constriction $\tau =w^2/\nu$, whereas $R_{\text{ball}}$ can be put in a similar form with $\tau \prime =w/v_F$ as the flight time across the constriction. Estimating $\nu =\frac{1}{4}{v}_F{l}_{\text{ee}}$, we see that Eq. 1 predicts resistance below the ballistic limit values so long as $\tau \gtrsim \tau \prime$, i.e., in the hydrodynamic regime $w\gtrsim l_{\text{ee}}$.

Understanding the behavior at the ballistic-to-viscous crossover is a nontrivial task. Here, to tackle the crossover, we use a kinetic equation with a simplified ee collision operator chosen in such a way that the relaxation rates for all nonconserved harmonics of momentum distribution are the same. This model provides a closed-form solution for transport through VPC for any ratio of the length scales $w$ and $l_{\text{ee}}$, predicting a simple additive relation

$$G_{\text{VPC}}=G_{\text{ball}}+G_{\text{vis}}.$$ [3]

This dependence, derived from a microscopic model, interpolates between the ballistic and viscous limits, $w\ll l_{\text{ee}}$ and $w\gg l_{\text{ee}}$, in which the terms $G_{\text{ball}}$ and $G_{\text{vis}}$, respectively, dominate.

The Hydrodynamic Regime

We start with a simple derivation of the VPC resistance in Eq. 1 using the model of a low-Reynolds electron flow that obeys the Stokes equation (29).

$$(\eta {\nabla }^2-{(ne)}^2\rho )\text{𝐯}(\text{𝐫})=ne\nabla \varphi (\text{𝐫}).$$ [4]

Here $\varphi (r)$ is the electric potential, $\eta$ is the viscosity, and the second term describes ohmic resistivity due to impurity or phonon scattering. Our analysis relies on a symmetry argument and invokes an auxiliary electrostatic problem. We model the constriction in Fig. 1A as a slit $-\frac{w}{2}<x<\frac{w}{2}$, $y=0$. The $y\to -y$ symmetry ensures that the current component $j_y$ is an even function of $y$ whereas both the component $j_x$ and the potential $\varphi$ are odd in $y$. As a result, the quantities $j_x$ and $\varphi$ vanish within the slit at $y=0$. This observation allows us to write the potential in the plane as a superposition of contributions due to different current elements in the slit,

$$\varphi (x,y)={\int }_{-\frac{w}{2}}^{\frac{w}{2}}dx\prime \mathcal{R}(x-x\prime ,y)j(x\prime ),$$ [5]

where the influence function $\mathcal{R}(x,y)=\frac{\beta (y^2-x^2)}{{(x^2+y^2)}^2}$ describes potential in a half-plane due to a point-like current source at the edge, obtained from Eq. 4 with no-slip boundary conditions and $\rho =0$ (30). Here $\beta =\frac{2\eta }{\pi {(en)}^2}$, and, without loss of generality, we focus on the $y>0$ half-plane.

Crucially, rather than providing a solution to our problem, the potential-current relation Eq. 5 merely helps to pose it. Indeed, a generic current distribution would yield a potential that is not constant inside the slit. We must therefore determine the functions $j(x)$ and $\varphi (x,y)$ self-consistently, in a way that ensures that the resulting $\varphi (x,y)$ vanishes on the line $y=0$ inside the slit. Namely, Eq. 5 must be treated as an integral equation for an unknown function $j(x)$. Denoting potential values at the half-plane $y\ge 0$ edge as ${\varphi }_{+0}(x)=\varphi {(x,y)}_{y=+0}$, we can write the relation Eq. 5 as

$${\varphi }_{+0}(x)=-\frac{\beta }{2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx\prime \left[\frac{j(x\prime )}{{(x-x\prime +i0)}^2}+\frac{j(x\prime )}{{(x-x\prime -i0)}^2}\right],$$ [6]

where $j(x)$ is the current $y$ component, which is finite inside and zero outside the interval $[-\frac{w}{2},\frac{w}{2}]$.

A solution of this integral equation such that ${\varphi }_{+0}(x)$ vanishes for all $-\frac{w}{2}<x<\frac{w}{2}$ can be obtained from a 3D electrostatic problem for an ideal-metal strip of width $w$ placed in a uniform external electric field $E_0=\lambda \widehat{\text{𝐱}}$. The strip is taken to be infinite, zero thickness, and positioned in the $Y=0$ plane such that

$$-\frac{w}{2}<X<\frac{w}{2},\hspace{1em}Y=0,\hspace{1em}-\mathrm{\infty }<Z<\mathrm{\infty }$$ [7]

(for clarity, we denote 3D coordinates by capital letters). Potential ${\mathrm{\Phi }}_{3D}(X,Y)$ is a harmonic function, constant on the strip and behaving asymptotically as $-E_{0}X$. It is easily checked that the 3D electrostatic problem translates to the 2D viscous problem as

$$\begin{array}{cccc}3D,Y=0:\hfill & \hfill X\hfill & \hfill \sigma (X)\hfill & \hfill E_x(X)\hfill \\ & \hfill \downarrow \hfill & \hfill \downarrow \hfill & \hfill \downarrow \hfill \\ 2D,y=+0:\hfill & \hfill x\hfill & \hfill -\frac{\beta }{2}\partial j/\partial x\hfill & \hfill {\varphi }_{+0}(x)\hfill \end{array}.$$ [8]

This mapping transforms Coulomb’s charge–field relation between the electric field at $Y=0$ and the surface charge density, $E_x(X)=2{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sigma (X\prime )dX\prime }{X-X\prime }$, into the 2D viscous relation in Eq. 6. Potential ${\mathrm{\Phi }}_{3D}$, obtained through a textbook application of conformal mapping, then equals

$${\mathrm{\Phi }}_{3D}(X,Y)=-\text{Re}\lambda \sqrt{{\zeta }^2-\frac{w^2}{4}},\hspace{1em}\zeta =X+iY.$$ [9]

Eq. 9 describes the net contribution of the external field $E_0$ and the charges $\sigma (X)$ induced on the strip. The field component $E_x(X)=-{\partial }_X{\mathrm{\Phi }}_{3D}$ vanishes on the strip $-\frac{w}{2}<x<\frac{w}{2}$ and equals $\lambda$ far outside. We can therefore identify $\lambda$ with $V/2$ in the viscous problem (Fig. 1A).

Charge density on the strip, found from 9 with the help of Gauss’s law, $\sigma (X)=\frac{\lambda X}{2\pi \sqrt{w^2/4-X^2}}$, under the mapping Eq. 8, gives a semicircle current distribution,

$$j(|x|<\frac{w}{2})=\frac{\lambda }{\pi \beta }\sqrt{\frac{w^2}{4}-x^2},\hspace{1em}j(|x|>\frac{w}{2})=0.$$ [10]

The potential map in Fig. 1A is then obtained by plugging this result into Eq. 5. The flow streamlines are obtained from a similar relation for the stream function (see ref. 30). Evaluating the current $I={\int }_{-\frac{w}{2}}^{\frac{w}{2}}j(x)dx=\lambda w^2/8\beta$ and setting $\lambda =V/2$ yields $R=V/I=16\beta /w^2$, which is Eq. 1. The inverse-square scaling $R\propto w^{-2}$ is distinct from the $w^{-1}$ scaling found in the ballistic free-fermion regime. The scaling, as well as the lower-than-ballistic $R$ values, can serve as a hallmark of a viscous flow.

Potential, inferred from the 2D/3D correspondence, is

$$\varphi {(x)}_{y=\pm 0}=\{\begin{array}{cc}\hfill \frac{V|x|}{2\sqrt{x^2-w^2/4}}\text{sgn}y\hfill & \hfill |x|>\frac{w}{2}\hfill \\ \hfill 0\hfill & \hfill |x|<\frac{w}{2}\hfill \end{array},$$ [11]

where $\text{sgn}y$ corresponds to the upper and lower sides, $y=\pm 0$. Potential grows toward the slit, diverging at the end points $x=\pm \frac{w}{2}$. This interesting behavior, representing an up-converting DC current transformer, arises due to the electric field pointing against the current near the viscous fluid edge as discussed in ref. 29.

Crossover to the Ballistic Regime

Our next goal is to develop a theory of the ballistic-to-viscous crossover for a constriction. Because we are interested in the linear response, we use the kinetic equation linearized in deviations of particle distribution from the equilibrium Fermi step (assuming $k_{B}T\ll E_F$),

$$\left({\partial }_t+\text{𝐯}{\nabla }_{\text{𝐱}}\right)f(\theta ,\text{𝐱},t)=I_{\text{ee}}(f)+I_{\text{bd}}(f),$$ [12]

where $\theta$ is the angle parameterizing particle momentum at the 2D Fermi surface. Here, $I_{\text{ee}}$ and $I_{\text{bd}}$ describe momentum-conserving carrier collisions and momentum-nonconserving scattering at the boundary, respectively.

In the presence of momentum-conserving collisions, transport is succinctly described by “quasi-hydrodynamic variables” defined as deviations in the average particle density and momentum from local equilibrium (31). These quantities can be expressed as angular harmonics of the distribution $f(\theta ,\text{𝐱},t)$,

$$f_0={⟨f(\theta )⟩}_{\theta },\hspace{1em}{f}_{\pm 1}={⟨e^{\mp i\theta }f(\theta )⟩}_{\theta },$$ [13]

where we introduced notation ${⟨\mathrm{\dots }⟩}_{\theta }=\oint \mathrm{\dots }\frac{d\theta }{2\pi }$. The quantities $f_0$ and $f_{\pm 1}$, conserved in the ee collisions, represent the zero modes of $I_{\text{ee}}$. For suitably chosen $I_{\text{ee}}$, the task of solving the kinetic equation in a relatively complicated constriction geometry is reduced to analyzing a self-consistency equation for the variables $f_0$ and $f_{\pm 1}$. We will derive a linear integral equation for these quantities and solve it to obtain the current density, potential, and conductance.

To facilitate the analysis, we model $I_{\text{ee}}$ by choosing a single relaxation rate for all nonconserved harmonics,

$$I_{\text{ee}}(f)=-\gamma (f-Pf),\hspace{1em}P=\sum _{m=0,\pm 1}|m⟩⟨m|,$$ [14]

where $\gamma$ represents the ee collision rate, with $l_{\text{ee}}=v/\gamma$, and $P$ is a projector in the space of angular harmonics of $f(\theta )$ that selects the harmonics conserved in ee collisions. Here we introduced Dirac notation for $f(\theta )$ with the inner product $⟨f_1|f_2⟩=\oint \frac{d\theta }{2\pi }{\overline{f}}_1(\theta )f_2(\theta )$. Namely,

$$⟨\theta |m⟩=e^{im\theta },\hspace{1em}Pf(\theta )=\sum _{m=0,\pm 1}\oint \frac{d\theta \prime }{2\pi }{e}^{im(\theta -\theta \prime )}f(\theta \prime ).$$

As in quantum theory, the Dirac notation proves to be a useful bookkeeping tool to account, on equal footing, for the distribution function position and wavenumber dependence, as well as the angle dependence.

To simplify our analysis, we replace the constriction geometry by that of a full plane, with a part of the line $y=0$ made impenetrable through a suitable choice of $I_{\text{bd}}(f)$. Scattering by disorder at the actual boundary conserves $f_0$ but not $f_{\pm 1}$. We can therefore model momentum loss due to collisions at the boundary using

$$I_{\text{bd}}(f)=-\alpha (x)P\prime f,\hspace{1em}\alpha (\text{𝐱})=\{\begin{array}{cc}\hfill 0,\hfill & \hfill |x|<\frac{w}{2}\hfill \\ \hfill b\delta (y),\hfill & \hfill |x|\ge \frac{w}{2}\hfill \end{array},$$ [15]

where $P\prime$ is a projector defined in a manner similar to $P$, projecting $f$ on the harmonics $m=\pm 1$. The term $\alpha (x)$ describes momentum relaxation on the line $y=0$, equal to zero within the slit and to $b$ outside. The parameter $b>0$ with the dimension of velocity, introduced for mathematical convenience, describes a partially transparent boundary. An impenetrable no-slip boundary, which corresponds to the situation of interest, can be modeled by taking the limit $b\to \mathrm{\infty }$.

We will analyze the flow induced by a current applied along the $y$ direction, described by a distribution,

$$f(\theta ,\text{𝐱})=f^{(0)}(\theta )+\delta f(\theta ,\text{𝐱}),\hspace{1em}{f}^{(0)}(\theta )=2j\sin \theta .$$ [16]

Here, $f^{(0)}$ and $\delta f$, which we will also write as $|f^{(0)}⟩$ and $|\delta f⟩$, represent a uniform current-carrying state and its distortion due to scattering at the $y=0$ boundary. The quantity $j$ is the current density. Once found, the spatial distribution $f(\theta ,\text{𝐱})$ will allow us to determine the resulting potential and resistance.The kinetic equation, Eq. 12, reads

$$[{\partial }_t+K+\alpha (\text{𝐱})P\prime ]|f⟩=0,\hspace{1em}K=\text{𝐯}\nabla +\gamma \widehat{1}-\gamma P$$ [17]

(from now on, we suppress the coordinate and angle dependence of $f$ and switch to the Dirac notation). Plugging $f=f^{(0)}+\delta f$, we rewrite Eq. 17 as $\left(K+\widehat{\alpha }\right)|\delta f⟩=-\widehat{\alpha }|f^{(0)}⟩$, where, for conciseness, we absorbed the projector $P\prime$ into $\widehat{\alpha }$ and set ${\partial }_{t}f=0$ for a steady state. We write a formal operator solution as

$$|\delta f⟩=-{(1+G\widehat{\alpha })}^{-1}G\widehat{\alpha }|f^{(0)}⟩,$$ [18]

where $G=K^{-1}$ is the Green’s function. Performing analysis in momentum representation, we treat the scattering term in Eq. 15 as an operator,

$$⟨\text{𝐤}|\widehat{\alpha }|\text{𝐤}\prime ⟩=P\prime {\alpha }_{k_1-{k\prime }_1},\hspace{1em}{\alpha }_k=2\pi b\delta (k)-bw\text{sinc}\frac{kw}{2},$$ [19]

where $\text{sinc}x=\frac{\sin x}{x}$. The two terms in ${\alpha }_k$ describe scattering at the $y=0$ line minus the slit contribution.

Next, we derive a closed-form integral equation for quasi-hydrodynamic variables by projecting the quantities in Eq. 18 on the $m=0,\pm 1$ harmonics, Eq. 13. Acting on Eq. 18 with $P$ gives $|P\delta f⟩=-{(1+\stackrel{\sim }{G}\widehat{\alpha })}^{-1}\stackrel{\sim }{G}\widehat{\alpha }|f^{(0)}⟩$, where $\stackrel{\sim }{G}=PGP$ is a $3\times 3$ matrix in the $m=0,\pm 1$ space (here we used the identity $\widehat{\alpha }=P\widehat{\alpha }P$, which follows from $PP\prime =P\prime P=P\prime$). The integral equation is obtained by acting on both sides with the operator $1+\stackrel{\sim }{G}\widehat{\alpha }$, giving

$$(1+\stackrel{\sim }{G}\widehat{\alpha })|\stackrel{\sim }{f}⟩=|f^{(0)}⟩.$$ [20]

Here, we defined $\stackrel{\sim }{f}=f^{(0)}+P\delta f$, the full distribution function projected on the $m=0,\pm 1$ harmonics.

The quantity $\stackrel{\sim }{f}$ represents an unknown function that can be found, in principle, by inverting the integral operator $1+\stackrel{\sim }{G}\alpha$ in Eq. 20. However, rather than attempting to invert $1+\stackrel{\sim }{G}\alpha$ directly in 2D, it is more convenient to proceed in two steps: First, analyze Eq. 20 in 1D, on the line $y=0$, and then extend the solution into 2D.

We start with finding $\stackrel{\sim }{G}$. As a first step, we evaluate the $3\times 3$ matrix $S=\gamma P{G}_{0}P$, where $G_0=1/(i\text{𝐤𝐯}+\gamma )$. The quantity $G_0$ is an auxiliary Green’s function describing transport in which all harmonics, including $m=0,\pm 1$, relax at a rate $\gamma$. Direct calculation gives matrix elements (here $m,m\prime =0,\pm 1$, $\mathrm{\Delta }m=m\prime -m$),

$$S_{mm\prime }={⟨\frac{\gamma e^{i(m\prime -m)\theta }}{\gamma +i\text{𝐤𝐯}}⟩}_{\theta }=\text{tanh}\beta \frac{e^{i{\theta }_k\mathrm{\Delta }m}}{{\left(i{e}^{\beta }\right)}^{|\mathrm{\Delta }m|}},$$ [21]

where we denote $\text{sinh}\beta =\frac{\gamma }{kv}$ and ${\theta }_k=\text{arg}(k_1+i{k}_2)$.

The matrix $\stackrel{\sim }{G}$ can now be expressed through the matrix $S$ by expanding the actual Green’s function as $G=1/(G_0^{-1}-\gamma P)=G_0+G_0\gamma P{G}_0+\mathrm{\dots }$, which gives

$$G=G_0+G_{0}T{G}_0,\hspace{1em}T=\frac{\gamma P}{1-\gamma P{G}_{0}P}.$$ [22]

Here, we resummed the series, expressing the result in terms of a $3\times 3$ matrix $T$ in a manner analogous to the derivation of the Lippmann–Schwinger $T$ matrix for quantum scattering with a finite number of “active” channels. We note that $\gamma P{G}_{0}P$ is nothing but the matrix $S$ in Eq. 21. Plugging Eq. 22 into $\stackrel{\sim }{G}=PGP$ and performing a tedious but straightforward matrix inversion, we obtain

$$\stackrel{\sim }{G}=\frac{{\gamma }^{-1}S}{1-S}=\frac{\text{sinh}\beta }{\gamma }\left(\begin{array}{ccc}\hfill e^{\beta }\hfill & \hfill -i{\overline{z}}_k\hfill & \hfill -e^{\beta }{\overline{z}}_k^2\hfill \\ \hfill -i{z}_k\hfill & \hfill e^{-\beta }\hfill & \hfill -i{\overline{z}}_k\hfill \\ \hfill -e^{\beta }{z}_k^2\hfill & \hfill -i{z}_k\hfill & \hfill e^{\beta }\hfill \end{array}\right),$$ [23]

where $z_k=e^{i{\theta }_k}$, and the basis vectors are ordered as $|+1⟩$, $|0⟩$, and $|-1⟩$.

In what follows, it will be convenient to transform $|\pm 1⟩$ to the even/odd basis $|c⟩=\frac{|+1⟩+|-1⟩}{\sqrt{2}}$, $|s⟩=\frac{|+1⟩-|-1⟩}{\sqrt{2}i}$. In this basis, $\stackrel{\sim }{G}$ reads

$$\left(\begin{array}{ccc}\hfill {\stackrel{\sim }{G}}_{00}\hfill & \hfill {\stackrel{\sim }{G}}_{0c}\hfill & \hfill {\stackrel{\sim }{G}}_{0s}\hfill \\ \hfill {\stackrel{\sim }{G}}_{c0}\hfill & \hfill {\stackrel{\sim }{G}}_{cc}\hfill & \hfill {\stackrel{\sim }{G}}_{cs}\hfill \\ \hfill {\stackrel{\sim }{G}}_{s0}\hfill & \hfill {\stackrel{\sim }{G}}_{sc}\hfill & \hfill {\stackrel{\sim }{G}}_{ss}\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill \frac{R_{-}}{\gamma {\kappa }^2}\hfill & \hfill \frac{-i\sqrt{2}{\kappa }_1}{\gamma {\kappa }^2}\hfill & \hfill \frac{-i\sqrt{2}{\kappa }_2}{\gamma {\kappa }^2}\hfill \\ \hfill \frac{-i\sqrt{2}{\kappa }_1}{\gamma {\kappa }^2}\hfill & \hfill \frac{2{\kappa }_2^2{R}_{+}}{\gamma {\kappa }^4}\hfill & \hfill \frac{-2{\kappa }_1{\kappa }_2{R}_{+}}{\gamma {\kappa }^4}\hfill \\ \hfill \frac{-i\sqrt{2}{\kappa }_2}{\gamma {\kappa }^2}\hfill & \hfill \frac{-2{\kappa }_1{\kappa }_2{R}_{+}}{\gamma {\kappa }^4}\hfill & \hfill \frac{2{\kappa }_1^2{R}_{+}}{\gamma {\kappa }^4}\hfill \end{array}\right),$$ [24]

where the basis vectors are ordered as $|0⟩$, $|c⟩$, and $|s⟩$ and we defined $R_{\pm }(\kappa )=\sqrt{{\kappa }^2+1}\pm 1$ and ${\kappa }_{\mathrm{1,2}}=\frac{v}{\gamma }{k}_{\mathrm{1,2}}$, $\kappa =\sqrt{{\kappa }_1^2+{\kappa }_2^2}$. The quantities $G$ and $\stackrel{\sim }{G}$ represent, through their dependence on k, translationally invariant integral operators in position representation and diagonal operators in momentum representation.

Next, we evaluate the matrix that represents the operator $\stackrel{\sim }{G}$ restricted to the line $y=0$,

$$D(k_1)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{d{k}_2}{2\pi }\stackrel{\sim }{G}(k_1,k_2).$$ [25]

The matrix elements ${\stackrel{\sim }{G}}_{0c}$ and ${\stackrel{\sim }{G}}_{0s}$ are odd in $k_2$ and therefore give zero upon integration in Eq. 25; as a result we obtain a block diagonal matrix,

$$D(k_1)=\left(\begin{array}{ccc}\hfill D_{00}(k_1)\hfill & \hfill D_{0c}(k_1)\hfill & \hfill 0\hfill \\ \hfill D_{c0}(k_1)\hfill & \hfill D_{cc}(k_1)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill D_{ss}(k_1)\hfill \end{array}\right).$$ [26]

The quantity $D_{ss}(k_1)$ will play a central role in our analysis. Indeed, because the flow of interest is symmetric under $y\to -y$ and $x\to -x$, the ${\stackrel{\sim }{f}}_0$ and ${\stackrel{\sim }{f}}_c$ components vanish on the $y=0$ line. As a result, the distribution function at $y=0$ is of a pure $|s⟩$ form, that is, $\stackrel{\sim }{f}(\theta ,x)=g(x)\sqrt{2}\sin \theta$.

Evaluating the integral over $k_2$ in Eq. 25, we obtain

$$D_{ss}(k)=\frac{\frac{\pi }{2}\text{sgn}\kappa +\kappa +({\kappa }^2+1){\cot }^{-1}\kappa }{\pi \kappa v},$$ [27]

where $\kappa =kv/\gamma$. This expression defines an even function of $k$ with the asymptotics

$$D_{ss}(|k|v\ll \gamma )=\frac{\gamma }{|k|v^2},\hspace{1em}{D}_{ss}(|k|v\gg \gamma )=\frac{2}{\pi v}.$$ [28]

Because the matrix element $D_{ss}$ is an eigenvalue of $D$ for the eigenvector $|s⟩$, the $\theta$ dependence can be factored out of Eq. 20, giving $(1+D\alpha )|g⟩=|g^{(0)}⟩$. Finally, multiplying by $D^{-1}$, we obtain the “central equation,”

$$D_{ss}^{-1}(k)g_k+\int \frac{dk\prime }{2\pi }{\alpha }_{k-k\prime }{g}_{k\prime }=2\pi \mu \delta (k),$$ [29]

where $\mu$ is an unspecified number, akin to a Lagrange multiplier, which fixes the total current value. Here, we wrote the relation $(D^{-1}+\alpha )|g⟩=\mu |k=0⟩$ as an integral equation, replacing $k_1$ with $k$ for clarity.

The origin of the $\mu$-term in Eq. 29, and its relation with the properties of the operator $D$, is simplest to understand using a discretized momentum representation. Letting $k_1=\frac{2\pi }{L}n$ and replacing

$$\int d{k}_1\mathrm{\dots }\to \frac{2\pi }{L}\sum _n\mathrm{\dots },\hspace{1em}2\pi \delta (k)\to L{\delta }_{k,0},$$ [30]

that is, putting the problem on a cylinder of circumference $L$, we see that the values ${\stackrel{\sim }{G}}_{ss}(k_1,k_2)$ vanish for $k_1=0$ and any $k_2$. This observation implies that the quantity $D_{ss}(k_1)$ also vanishes for $k_1=0$, and thus the operator $D$ does not have an inverse. In this case, caution must be exercised when multiplying by $D^{-1}$. Namely, the quantities $D^{-1}|f⟩$ are defined modulo a null vector of $D$, which is the $k_1=0$ mode with an unspecified coefficient, represented by the $\mu$-term. We note, parenthetically, that discretization has no impact on the values $D_{ss}(k_1\ne 0)$ given in Eqs. 27 and 28.

We obtain current distribution by solving, numerically, Eq. 29, discretized as in Eq. 30, and subsequently Fourier-transforming $g_k$ to position space (Supporting Information). A large value $b={10}^{5}v$ was used to ensure that current vanishes outside the interval $[-\frac{w}{2},\frac{w}{2}]$. The resulting distribution, shown in Fig. 1B, features interesting evolution under varying $\gamma$: Flat at small $\gamma$, the distribution gradually bulges out as $\gamma$ increases, peaking at $x=0$ and dropping to zero near $x=\pm \frac{w}{2}$. In the limit $\gamma \gg v/w$, it evolves into a semicircle coinciding with the hydrodynamic result, Eq. 10. Current suppression near the constriction edges is in agreement with the streaming picture discussed in the Introduction.

The solution on the line $y=0$ can now be used to determine the solution in the bulk. For example, to obtain the density $f_0(\text{𝐱})$, we project the relation Eq. 20 on $m=0$ harmonic, taking into account that both $f^{(0)}$ and $\alpha \stackrel{\sim }{f}$ are of an $|s⟩$ form. This procedure yields an expression for the 2D density of the form $f_0(\text{𝐱})=-\int dx\prime {\stackrel{\sim }{G}}_{0s}(\text{𝐱},x\prime )\alpha (x\prime )g(x\prime )$, with x being a 2D coordinate and $-\mathrm{\infty }<x\prime <\mathrm{\infty }$. To avoid handling the $b\to \mathrm{\infty }$ limit in $\alpha$, we write this relation, using Eq. 29, as

$$f_0(\text{𝐱})=-\int dx\prime {\stackrel{\sim }{G}}_{0s}(\text{𝐱},x\prime )\left[\mu -\left(D_{ss}^{-1}g\right)(x\prime )\right].$$ [31]

Plugging in ${\stackrel{\sim }{G}}_{0s}(\text{𝐤})=\frac{-i\sqrt{2}{k}_2}{v(k_1^2+k_2^2)}$, Fourier-transforming, and carrying out the $k_2$ integral by the residue method, $\int d{k}_2{e}^{i{k}_{2}y}\frac{i{k}_2}{k_1^2+k_2^2}=-\pi e^{-|y{k}_2|}\text{sgn}y$, we obtain

$$f_0(\text{𝐱})=\frac{\text{sgn}y}{\sqrt{2}v}\int \frac{d{k}_1}{2\pi }{e}^{i{k}_{1}x-|k_{1}y|}[D_{ss}^{-1}(k_1)g_{k_1}-2\pi \mu \delta (k_1)].$$ [32]

The resulting distributions, shown in Fig. 2, are step-like. At large $y$, the $\mu$-term dominates, giving $f_0(|\mathbf{\text{x}}|\gg w)\approx -\frac{\mu }{\sqrt{2}v}\text{sgn}y$. Therefore, the step height equals $\frac{\sqrt{2}}{v}\mu$ regardless of the parameter values used.

Fig. 2.

Potential distribution induced by a unit current through a constriction (A) at the crossover, $l_{\text{ee}}\approx w$, and (B) in the viscous regime, $l_{\text{ee}}\gg w$. The spikes at the constriction edges in B are a signature of a hydrodynamic behavior (Eq. 11 and accompanying text). Plotted is particle density deviation from equilibrium, $f_0(\text{𝐱})$, which is proportional to potential (see discussion preceding Eq. 33). Parameters used are (i) $\gamma =v/w$ and (ii) $\gamma =15v/w$; other parameter values are the same as in Fig. 1B.

This relation provides a route to evaluate resistance. Namely, because of charge neutrality, the density $f_0$ obtained from a noninteracting model translates directly into potential distribution $\varphi (\text{𝐱})=\frac{1}{e{\nu }_0}{f}_0(\text{x})$, where ${\nu }_0$ is the density of states. Dividing the potential difference $V=\frac{\sqrt{2}\mu }{e{\nu }_{0}v}$ by the total current $I=\int dxg(x)⟨ev\sin \theta |s⟩=\frac{ev}{\sqrt{2}}{g}_{k_1=0}$ yields a simple expression for resistance,

$$R=\frac{\mu {\rho }_{\ast }}{v{g}_{k=0}},\hspace{1em}{\rho }_{\ast }=\frac{2}{e^{2}v{\nu }_0},$$ [33]

where $g_{k=0}=\int g(x)dx$, and ${\rho }_{\ast }$ is a quantityof dimension $\text{Ohm}\cdot \text{cm}$. Because $g\propto \mu$, the resulting $R$ values are $\mu$-independent. Fig. 3A shows $R$ plotted vs. $\gamma$. As expected, $R$ decreases as $\gamma$ increases, i.e., carrier collisions enhance conduction.

Fig. 3.

(A) The resistance $R$, Eq. 33, plotted vs. $\gamma$. Upon rescaling $R\to Rw$, $\gamma \to \gamma w$, all of the curves collapse on one curve, confirming that the only relevant parameter is the ratio $w/l_{\text{ee}}=w\gamma /v$. (B) Scaled conductance $G=1/(Rw)$ vs. $\gamma w$. All curves collapse onto a single straight line, which can be fitted with $(0.694+0.378\gamma w){\rho }_{\ast }^{-1}$. This dependence matches Eq. 3 which corresponds to $(2/\pi +\pi /8\gamma w){\rho }_{*}^{-1}$. Parameters used are $b={10}^{6}v$, the number of sampling points within the constriction $\approx 160$, and the length unit $w_0=\frac{1}{30}L$.

As a quick sanity check on Eq. 33, we consider the near-collisionless limit $\gamma \ll v/w$. In this case, $D_{ss}(k)\approx 2/\pi v$, and the integral Eq. 29 turns into an algebraic equation, which is solved by a step-like distribution,

$$g(|x|>w/2)=\frac{2\mu }{\pi v+2b},\hspace{1em}g(|x|<w/2)=\frac{2\mu }{\pi v}.$$ [34]

In the limit $b\to \mathrm{\infty }$, the total current is $I=\frac{ev}{\sqrt{2}}\frac{2w\mu }{\pi v}$. Taking the 2D density of states ${\nu }_0=\frac{Nm}{2\pi {\mathrm{\hslash }}^2}$ (here $N$ is spin–valley degeneracy, e.g., $N=4$ for graphene), we find

$$R=\frac{V}{I}=\frac{1}{N}\frac{h}{e^2}\frac{{\lambda }_F}{2w},\hspace{1em}{\lambda }_F=\frac{2\pi }{k_F}.$$ [35]

This result coincides with the collisionless Landauer conductance value. Spatial dependence can be obtained by plugging $g(x)$ into Eq. 32. Integrating and taking the limit $b\to \mathrm{\infty }$ gives

$$f_0(\text{𝐱})=-\frac{\text{sgn}y}{\sqrt{2}v}\mu \left[1-\frac{1}{\pi }\theta (\text{𝐱})\right],$$ [36]

where $\theta (\text{𝐱})={\tan }^{-1}\frac{|y|w}{x^2+y^2-\frac{1}{4}{w}^2}$ is the angle at which the interval $[-\frac{w}{2},\frac{w}{2}]$ is seen from the point $\text{𝐱}=(x,y)$. This result confirms the value $\frac{\mu }{\sqrt{2}v}$ for the step height.

The dependence $R$ vs. $\gamma$ shows several interesting features, some expected and some unexpected. First, on general grounds, we expect that the dependence on $\gamma$ is controlled solely by the ratio $w/l_{\text{ee}}$. Indeed, plotting the rescaled quantity $Rw$ vs. $\gamma w$, we find a family of curves that all collapse on one curve (this universality only holds at large b, cf. Fig. S1). Second, quite remarkably, inverting this quantity and plotting $1/(Rw)$ vs. $\gamma w$, we find a nearly perfect straight line with a positive offset at $\gamma =0$; see Fig. 3B. The straight line, which is identical for all $w$ values, is described by ${\rho }_{\ast }/(Rw)=a_1+a_2\gamma w$. This dependence translates into a simple addition rule for conductance, $G=G_{\text{ball}}+G_{\text{vis}}$. The term $G_{\text{ball}}$ describes a $\gamma$-independent ballistic contribution that scales linearly with $w$, whereas $G_{\text{vis}}$ describes a viscous contribution proportional to $\gamma$ that scales as $w^2$; the two terms yield values $a_1=2/\pi$ and $a_2=\pi /8$, respectively. This finding is in good agreement with the values $a_1=0.694$ and $a_2=0.378$ obtained from a best fit to the data in Fig. 3B.

Fig. S1.

Conductance scaled by constriction width vs. $\gamma w$. Plots are obtained at $w_0=\frac{1}{30}L$, $b=50v$, with the number of sampling points within the constriction of about 160. Unlike Fig. 3B, here different curves do not collapse on one curve, indicating that the universality fails for not-too-large $b$.

The additive behavior of conductance at the ballistic-to-viscous crossover comes as a surprise and, to the best of our knowledge, is not anticipated on simple grounds. It is also a stark departure from the Matthiessen’s rule that mandates an additive behavior for resistivity in the presence of different scattering mechanisms, as observed in many solids (32). This rule is, of course, not valid if the factors affecting transport depend on each other, because individual scattering probabilities cannot be summed unless they are mutually independent. The independence is certainly out of question for momentum-conserving ee collisions that do not, by themselves, result in momentum loss but can only impact momentum relaxation due to other scattering mechanisms. Furthermore, the addition rule for conductance, Eq. 3, describes a striking “anti-Matthiessen” behavior: Rather than being suppressed by collisions, conductance exceeds the collisionless value.

SI Integral Equation on a Circle

The integral Eq. 29, which describes current distribution in the constriction, is defined on a line $-\mathrm{\infty }<x<\mathrm{\infty }$ in position representation. It reads

$$\alpha (x)g(x)+{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx\prime D_{ss}^{-1}(x-x\prime )g(x\prime )dx\prime =\mu ,$$ [S1]

$$D_{ss}^{-1}(x-x\prime )=\int \frac{dk}{2\pi }\frac{e^{ik(x-x\prime )}}{D_{ss}(k)},\hspace{1em}\alpha (x)=\{\begin{array}{c}\hfill b,|x|>\frac{w}{2}\hfill \\ \hfill 0,|x|<\frac{w}{2}\hfill \end{array}.$$

Before we proceed to discuss the general solution, it is instructive to consider Eq. S1 in the collisionless limit $l_{\text{ee}}\gg w$ and in the hydrodynamic limit $w\gg l_{\text{ee}}$. These regimes are described by the large-$k$ and small-$k$ limits of $D_{ss}(k)$, given in Eq. 28.

In the first case, $\gamma =0$ and $D_{ss}(k)=2/\pi v$, and the integral Eq. S1 turns into an algebraic equation. This equation is solved by

$$f(|x|>w/2)=\frac{\mu }{1+\stackrel{\sim }{b}},\hspace{1em}f(|x|<w/2)=\frac{2}{\pi }\mu ,$$ [S2]

where $\stackrel{\sim }{b}=2b/\pi v$. Taking the limit $b\to \mathrm{\infty }$, describing a nontransparent boundary, we obtain a box-like solution that vanishes outside the slit $|x|<w/2$, which agrees with the current distribution in the ballistic limit $\gamma =0$.

In the second case, $\gamma \gg v/w$ and $D_{ss}(k)=\frac{\gamma }{|k|v^2}$, we have

$$D_{ss}^{-1}(x-x\prime )={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{dk}{2\pi }\frac{|k|v^2}{\gamma }{e}^{ik(x-x\prime )}=-\frac{v^2}{2\pi \gamma }\left[\frac{1}{{(x-x\prime +i0)}^2}+\frac{1}{{(x-x\prime -i0)}^2}\right].$$ [S3]

This quantity coincides, up to a numerical factor, with the kernel in Eq. 6. We will now show that the integral Eq. S1, in the limit $b\to \mathrm{\infty }$, is satisfied by a semicircle solution identical to that found by an electrostatic method. The analysis is facilitated by representing the semicircle solution, with a yet-undetermined normalization factor, as

$$g(x)=a\sqrt{1-\frac{4x^2}{w^2}}=\text{Im}{f}_{+}(z)-\text{Im}{f}_{-}(z),\hspace{1em}z=\frac{2x}{w},$$

where $f_{\pm }(z)$ are given by $\frac{a}{2}(\sqrt{z^2-1}-z)$ continued from large $z$ to $-1<z<1$ through the upper or lower complex $z$ half-plane, respectively. Using this representation and the expression in Eq. S3, we can carry out the integral in Eq. S1 by the method of residues, closing the integration path through the upper half-plane for $f_{+}(z)$ and the lower half-plane for $f_{-}(z)$. The contributions of large $z$ drop out because the functions $f_{\pm }(z)$ vanish at infinity, giving

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}dx\prime D_{ss}^{-1}(x-x\prime )g(x\prime )dx\prime =\text{Im}\left[\frac{2i{v}^2}{\gamma w}({f\prime }_{+}(z)-{f\prime }_{-}(z))\right]=\frac{2v^{2}a}{\gamma w}.$$ [S4]

Here we have taken $x$ to be in the interval $[-\frac{w}{2},\frac{w}{2}]$. Inserting this result in Eq. S1, we determine the normalization factor $a=\frac{\gamma w}{2v^2}\mu$. The resistance is obtained by evaluating

$$g_{k=0}={\int }_{-\frac{w}{2}}^{\frac{w}{2}}g(x)dx=\frac{\pi }{4}wa$$

and plugging it into Eq. 33. The resulting resistance value is

$$R=\frac{8v{\rho }_{\ast }}{\pi \gamma w^2}=\frac{16}{\pi e^2\gamma w^2{\nu }_0}.$$

Writing the 2D density of states as ${\nu }_0=\frac{Nm}{2\pi {\mathrm{\hslash }}^2}=\frac{2n}{m{v}^2}$ and expressing $\gamma$ through viscosity $\eta =nm{v}^2/4\gamma$ (Eq. S21), we find

$$R=\frac{8m{v}^2}{\pi e^2\gamma w^{2}n}=\frac{32\eta }{\pi e^2{n}^2{w}^2}.$$

This expression coincides with the hydrodynamic result given in Eq. 1.

Next, to facilitate numerical analysis, we put our 2D problem on a cylinder, choosing a large enough cylinder circumference $L$ to provide a good approximation to the 2D problem. Closing the $x$ axis into a circle does not impact in any way the $2D\to 1D$ reduction, which yields an integral equation defined in the domain $[-\frac{1}{2}L,\frac{1}{2}L]$,

$$\alpha (x)g(x)+{\int }_{-L/2}^{L/2}dx\prime {\stackrel{\sim }{D}}_{ss}^{-1}(x-x\prime )g(x\prime )dx\prime =\mu ,$$ [S5]

$${\stackrel{\sim }{D}}_{ss}^{-1}(x-x\prime )=\sum _{m=-\mathrm{\infty }}^{\mathrm{\infty }}{D}_{ss}^{-1}(x-x\prime -mL),$$

with periodic boundary conditions, $g(x\pm L)=g(x)$. It may seem that the problem defined by Eq. S5 is identical to that in Eq. S1, because any function $g(x)$ satisfying Eq. 5, after being continued periodically outside the domain $[-\frac{1}{2}L,\frac{1}{2}L]$, satisfies also Eq. S1. We note, however, that such a prescription generates functions that are nonzero not only in the constriction interval $[-\frac{w}{2},\frac{w}{2}]$ ($w<L$) but also in the intervals $[Lm-\frac{w}{2},Lm+\frac{w}{2}]$ where the solution of the original problem, Eq. 1, must vanish in the limit $b\to \mathrm{\infty }$. Physically, this procedure is equivalent to replacing one slit with an infinite array of slits of width $w$ each, and periodicity $L$. The behavior near one slit will not be affected by other slits so long as $L\gg w$. In our numerical study, taking $L$ equal to a few times $w$ was found sufficient to provide a reasonably good approximation.

To handle the $L$-periodic boundary conditions, we write Eq. S5 in momentum space, with momentum taking discrete values

$$k=\frac{2\pi n}{L},$$ [S6]

where $n$ is an integer. We transform Eq. S5 by inserting a resolution of identity $\frac{1}{L}{\sum }_k|k⟩⟨k|=1$, and using $⟨x|k⟩=\exp (ikx)$, $D|k_x⟩=D_{ss}(k_x)|k_x⟩$, and $⟨k|g(x)⟩=g_k$, where

$$g_k={\int }_{-L/2}^{L/2}dx{e}^{-ikx}g(x).$$ [S7]

Finally, we obtain

$$\sum _{k_1}{\alpha }_{k-k_1}{g}_{k_1}+D_{ss}^{-1}(k)g_k=\mu L{\delta }_{k,0},$$ [S8]

where ${\alpha }_k=b[{\delta }_{k,0}-\frac{w}{L}\text{sinc}(kw/2)]$. In numerical calculation, the values $n$ in Eq. S6 are limited by $-\frac{1}{2}N\le n<\frac{1}{2}N$, where $N$ is a suitably chosen large number. Restricting $n$ to a finite range corresponds to discretizing functions $f(x)$ in position space by using an $N$-point mesh $x_i=\frac{iL}{N}$, $i=-\frac{N}{2},-\frac{N}{2}+1,\mathrm{\dots },\frac{N}{2}-1$ in the interval $[-\frac{1}{2}L,\frac{1}{2}L]$.

We solve Eq. S8 numerically to obtain current distributions pictured in Fig. 1B: the calculation was done by first finding the distribution $f_k$ in momentum space, and then Fourier-transforming to position space. We used $L=3w$, and a large value $b={10}^{5}v$ to ensure that current vanishes outside the interval $|x|<\frac{w}{2}$. A Fourier space filter was used to smooth out the Gibbs phenomenon near the points $x=\pm \frac{w}{2}$ where current distribution drops abruptly to zero.

In the plots, the value $\mu$ was chosen such that the net current is normalized to unity. The resulting current distribution evolves in an interesting way upon $\gamma$ increasing: The distribution is a flat step at small $\gamma$, as expected in the ballistic case, and then gradually bulges, forming a peak at $x=0$ and gradually dropping to zero near $x=\pm \frac{w}{2}$. In the extreme hydrodynamical limit $\gamma \gg v/w$, the distribution evolves into a semicircle, which coincides with the result obtained from hydrodynamic equations in the main text.

Using the solution $g_k$, resistance $R$ can be calculated from Eq. 33, giving the conductance $G=1/R$ shown in Fig. 3 and Fig. S1. For large $b={10}^{6}v$, the conductance plots $G$ vs. $\gamma$, obtained for different constriction widths $w$, collapse on one curve when rescaled to $G/w$ vs. $\gamma w$. This “universality” confirms that the only relevant parameter in the problem is the ratio $w/l_{\text{ee}}$. This scaling stops working already for not very large $b$, as illustrated in Fig. 1. The breakdown of scaling is not alarming, because physically meaningful results are expected only in the limit $b\to \mathrm{\infty }$. Interestingly, however, the dependence $G/w$ vs. $\gamma w$ is well fitted by a perfectly straight line for both large and not-too-large $b$. The linear dependence $G$ vs. $\gamma$, along with the scaling, indicates that the conductance at the crossover is described by the addition formula $G=G_{\text{ball}}+G_{\text{vis}}$, as discussed in the main text.

SI The Hydrodynamic Regime

As an illustration, here we use the approach developed in the main text to solve for the 2D potential distribution, current flow, and conductance in the hydrodynamic regime $\gamma \gg v/w$. In this case, the solution of the integral Eq. S1 is a semicircle,

$$g(x)=a\sqrt{1-\frac{4x^2}{w^2}},$$ [S9]

where $a$ is a normalization factor that, for the time being, we will leave undetermined. In the Fourier domain,

$$g(k)={\int }_{-\frac{w}{2}}^{\frac{w}{2}}g(x)e^{-ikx}\mathrm{d}x=\frac{\pi wa}{2}\frac{J_1(|kw/2|)}{|kw/2|},$$ [S10]

where $J_1$ is the Bessel Function. The solution of the 1D problem can be used to obtain the 2D flow by using the same procedure as the one used to obtain the density distribution, Eq. 31, with ${\stackrel{\sim }{G}}_{0s}$ replaced with ${\stackrel{\sim }{G}}_{cs}$ and ${\stackrel{\sim }{G}}_{ss}$, respectively. Using the values ${\stackrel{\sim }{G}}_{cs}$ and ${\stackrel{\sim }{G}}_{ss}$ given in Eq. 24, and approximating $D_{ss}^{-1}\approx |k_x|v^2/\gamma$, the $|c⟩$ and $|s⟩$ components of the flow are given by

$$f(k_x,k_y)=\frac{4\pi a}{v^2}{J}_1(|k_{x}w/2|)\left(\begin{array}{c}\hfill \frac{-k_x{k}_y}{k^4}\hfill \\ \hfill \frac{k_x^2}{k^4}\hfill \end{array}\right),$$ [S11]

where the two entries represent the $x$ and $y$ momentum components, respectively. In Eq. S11, the $\mu$-term does not contribute because the quantities ${\stackrel{\sim }{G}}_{cs}$ and ${\stackrel{\sim }{G}}_{ss}$, given by Eq. 24, vanish at $k_1=0$. The next step is to perform Fourier transform to obtain the real-space flow distribution $f(x,y)=\int \frac{\mathrm{d}{k}_x}{2\pi }\frac{\mathrm{d}{k}_y}{2\pi }f(k_x,k_y)e^{i{k}_{x}x+i{k}_{y}y}$. After the $k_y$ integral is calculated by the residue method, we have

$$f(x,y)=\int \frac{\mathrm{d}{k}_x}{2\pi }\frac{\pi a{e}^{i{k}_{x}x-|k_{x}y|}}{v^2}{J}_1(|k_{x}w/2|)\left(\begin{array}{c}\hfill -iy\hfill \\ \hfill \frac{1+|k_{x}y|}{|k_x|}\hfill \end{array}\right)=\text{Re}{\int }_0^{\mathrm{\infty }}\mathrm{d}{k}_x\frac{a{e}^{i{k}_{x}x-|k_{x}y|}}{v^2}{J}_1(|k_{x}w/2|)\left(\begin{array}{c}\hfill -iy\hfill \\ \hfill \frac{1+|k_{x}y|}{|k_x|}\hfill \end{array}\right).$$ [S12]

The $k_x$ integral can be evaluated using the identity

$${\int }_0^{\mathrm{\infty }}\text{d}x{e}^{-\alpha x}{J}_{\nu }(\beta x)=\frac{{\beta }^{-\nu }{(\sqrt{{\alpha }^2+{\beta }^2}-\alpha )}^{\nu }}{\sqrt{{\alpha }^2+{\beta }^2}}.$$ [S13]

This gives the flow velocity components,

$$f_x=-\frac{2a}{w{v}^2}\text{Im}\left(y\frac{|y|-ix}{Z}\right),f_y=\frac{2a}{w{v}^2}\text{Re}\left(Z-|y|\frac{|y|-ix}{Z}\right),Z=\sqrt{{(w/2)}^2+{(|y|-ix)}^2}.$$ [S14]

The resulting flow is shown in Fig. 1. Using Eq. 32 we can compute the density distribution,

$$f_0(x,y)=\frac{\text{sgn}y}{\sqrt{2}v}\left[\frac{2v^{\mathit{2}}a}{\gamma w}\text{Re}\left(1-\frac{|y|-ix}{Z}\right)-\mu \right].$$ [S15]

The value of $\mu$ is determined by $f_0$ continuity at the constriction, giving a relation $2v^{2}a=\gamma w\mu$ identical to that found above from Eq. S1 (see discussion following Eq. S4). Evaluating conductance as in the main text, we find $G{\rho }^{\ast }/w=v{g}_0/\mu w=(\pi /8)\gamma w\sim 0.39\gamma w$. This agrees both with the analytic result in the main text (Eq. 1) and the numerical results in Fig. 3. The latter yield the best-fit slope $\sim 0.378$, which matches the analytic result above.

SI Hydrodynamic Modes

Here we derive hydrodynamic modes using the method of quasi-hydrodynamic variables, developed in the main text; this will allow us to relate the collision rate $\gamma$ and viscosity. To that end, we consider Eq. 12 in the absence of boundary scattering, $I_{\text{bd}}=0$. In this case, Eq. 12 takes the form

$$(\widehat{K}-\gamma P)f=0,\hspace{1em}\widehat{K}={\partial }_t+\text{𝐯}{\nabla }_{\mathbf{\text{x}}}+\gamma \widehat{1}.$$ [S16]

Because $f_0$ and $f_{\pm 1}$ are zero modes of the collision operator $I_{\text{ee}}$, they dominate at low frequencies and long wavelengths. Accordingly, we can obtain hydrodynamic modes from plane–wave solutions, $f(\theta ,\text{𝐱},t)\sim f(\theta )e^{-i\omega t+i\mathbf{\text{kx}}}$. Solving Eq. S16 as $f=\gamma {\widehat{K}}^{-1}Pf$, we project $f$ on the harmonics $f_0$ and $f_{\pm 1}$. This gives three coupled equations,

$$f_m=g_{mm\prime }{f}_{m\prime },\hspace{1em}{g}_{mm\prime }=⟨m|\gamma P{\widehat{K}}^{-1}P|m\prime ⟩.$$ [S17]

Direct calculation gives

$$g_{mm\prime }={⟨\frac{\gamma e^{i(m\prime -m)\theta }}{{\gamma }_{\omega }+i\text{𝐤𝐯}}⟩}_{\theta }=\text{tanh}\beta \frac{\gamma e^{i{\theta }_k\mathrm{\Delta }m}}{{\gamma }_{\omega }{\left(i{e}^{\beta }\right)}^{|\mathrm{\Delta }m|}},$$ [S18]

where ${⟨\mathrm{\dots }⟩}_{\theta }=\oint \mathrm{\dots }\frac{\theta }{2\pi }$. Here ${\gamma }_{\omega }=\gamma -i\omega$, $\text{sinh}\beta =\frac{{\gamma }_{\omega }}{kv}$, $\mathrm{\Delta }m=m\prime -m$, where $m,m\prime =0,\pm 1$. The integral over $\theta$ in Eq. S18 is performed by writing $\text{𝐤𝐯}=kv\cos \stackrel{\sim }{\theta }$, where $\stackrel{\sim }{\theta }=\theta -{\theta }_k$ is the angle between particle velocity v and wavevector k, and integrating over $\stackrel{\sim }{\theta }$.

As we now show, the equations $f_m=g_{mm\prime }{f}_{m\prime }$ generate an acoustic and a viscous mode. Because the acoustic and viscous modes are longitudinal and transverse, respectively, it is convenient to do the analysis by performing an orthogonal transformation to the even/odd basis,

$$|0⟩,\hspace{1em}|c⟩=\frac{|1_k⟩+|-1_k⟩}{\sqrt{2}}\hspace{1em},|s⟩=\frac{|1_k⟩-|-1_k⟩}{\sqrt{2}i},$$ [S19]

where we use notation $|m_k⟩=e^{-im{\theta }_k}|m⟩$. The modes $|c⟩$ and $|s⟩$ correspond to normalized angular harmonics $f_c(\theta )=\sqrt{2}\cos \stackrel{\sim }{\theta }$ and $f_s(\theta )=\sqrt{2}\sin \stackrel{\sim }{\theta }$.

This transformation brings the $3\times 3$ matrix $g_{mm\prime }$ to a block diagonal form,

$$\left(\begin{array}{ccc}\hfill g_{00}\hfill & \hfill g_{0c}\hfill & \hfill 0\hfill \\ \hfill g_{c0}\hfill & \hfill g_{cc}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill g_{ss}\hfill \end{array}\right).$$ [S20]

For the odd-mode $1\times 1$ block, we find $g_{ss}=\frac{\gamma }{{\gamma }_{\omega }}\text{tanh}\beta (1+e^{-2\beta })$. Writing the dispersion relation $1=g_{ss}$ and Taylor-expanding in small $\omega$ and $k$ yields a viscous mode dispersing as

$$\omega =-i\nu k^2,\hspace{1em}\nu =v^2/4\gamma .$$ [S21]

Here $\nu$ is the viscosity defined so that the dispersion in Eq. S21 agrees with that obtained from the linearized Navier–Stokes equation $({\partial }_t-\nu {\nabla }^2)\text{𝐯}=-\nabla P$.

The acoustic mode can be obtained from the even-mode $2\times 2$ block,

$$\left(\begin{array}{cc}\hfill g_{00}\hfill & \hfill g_{0c}\hfill \\ \hfill g_{c0}\hfill & \hfill g_{cc}\hfill \end{array}\right)=\frac{\gamma \text{tanh}\beta }{{\gamma }_{\omega }}\left(\begin{array}{cc}\hfill 1\hfill & \hfill -i\sqrt{2}{e}^{-\beta }\hfill \\ \hfill -i\sqrt{2}{e}^{-\beta }\hfill & \hfill 1-e^{-2\beta }\hfill \end{array}\right).$$ [S22]

The dispersion relation $\text{det}(1-g)=0$ gives

$$\left(\frac{{\gamma }_{\omega }}{\gamma \text{tanh}\beta }-1\right)\left(\frac{{\gamma }_{\omega }}{\gamma \text{tanh}\beta }-1+e^{-2\beta }\right)+2e^{-2\beta }=0.$$ [S23]

Plugging in $\text{sinh}\beta =\frac{\gamma }{kv}$, simplifying and Taylor-expanding in $\omega$ and $k$, yields a damped acoustic mode,

$$\omega =\frac{1}{\sqrt{2}}kv-\frac{i}{2}\nu k^2,$$ [S24]

where we expressed damping through viscosity $\nu$, evaluated in Eq. S21.