Violation of bulk-edge correspondence in a hydrodynamic model Gian - PowerPoint PPT Presentation

Violation of bulk-edge correspondence in a hydrodynamic model Gian Michele Graf ETH Zurich PhD School: September 16-20, 2019 @Universit` a degli Studi Roma Tre

Violation of bulk-edge correspondence in a hydrodynamic model Gian Michele Graf ETH Zurich PhD School: September 16-20, 2019 @Universit` a degli Studi Roma Tre based on joint work with Hansueli Jud, Cl´ ement Tauber

Outline A hydrodynamic model Topology by compactification The Hatsugai relation Violation What goes wrong?

The Great Wave off Kanagawa (by K. Hokusai, ∼ 1831)

A hydrodynamic model Topology by compactification The Hatsugai relation Violation What goes wrong?

The model (take it or leave it) ◮ The Earth is rotating.

The model (take it or leave it) ◮ The Earth is rotating. Sure

The model (take it or leave it) ◮ The Earth is rotating. Sure ◮ The Earth is flat.

The model (take it or leave it) ◮ The Earth is rotating. Sure ◮ The Earth is flat. Well, locally yes

The model (take it or leave it) ◮ The Earth is rotating. Sure ◮ The Earth is flat. Well, locally yes ◮ The Sea covers the Earth.

The model (take it or leave it) ◮ The Earth is rotating. Sure ◮ The Earth is flat. Well, locally yes ◮ The Sea covers the Earth. Don’t despair. We’ll sight land

The model (take it or leave it) ◮ The Earth is rotating. Sure ◮ The Earth is flat. Well, locally yes ◮ The Sea covers the Earth. Don’t despair. We’ll sight land ◮ The Sea is shallow.

The model (take it or leave it) ◮ The Earth is rotating. Sure ◮ The Earth is flat. Well, locally yes ◮ The Sea covers the Earth. Don’t despair. We’ll sight land ◮ The Sea is shallow. Compared to wavelength

The model (take it or leave it) ◮ The Earth is rotating. Sure ◮ The Earth is flat. Well, locally yes ◮ The Sea covers the Earth. Don’t despair. We’ll sight land ◮ The Sea is shallow. Compared to wavelength Incompressible, shallow water equations (preliminary): ∂η ∂ t = − h ∇ · v ∂ v ∂ t = − g ∇ η − f v ⊥ ◮ fields (dynamic): velocity v = v ( x , y ), height above average η = η ( x , y ) ◮ parameters: gravity g , average depth h , angular velocity f / 2

A quick derivation Starting point: Euler equations for an incompressible fluid in dimension 3.

A quick derivation Starting point: Euler equations for an incompressible fluid in dimension 3. ρ D � v � g − ρ� v − � ∇ · � v = 0 , Dt = ρ� f ∧ � ∇ p p = 0 at z = η ( x , y ) D η Dt = v ◮ fields: velocity � v = � v ( x , y , z ) =: ( v , v ), pressure p = p ( x , y , z ) ◮ parameters: density ρ ; gravity in z -direction

A quick derivation Starting point: Euler equations for an incompressible fluid in dimension 3. ρ D � v � g − ρ� v − � ∇ · � v = 0 , Dt = ρ� f ∧ � ∇ p p = 0 at z = η ( x , y ) D η Dt = v ◮ fields: velocity � v = � v ( x , y , z ) =: ( v , v ), pressure p = p ( x , y , z ) ◮ parameters: density ρ ; gravity in z -direction Steps: (a) Linearization, (b) (2 + 1)-split, and (c) dimensional reduction

A quick derivation Starting point: Euler equations for an incompressible fluid in dimension 3. ρ D � v � g − ρ� v − � ∇ · � v = 0 , Dt = ρ� f ∧ � ∇ p p = 0 at z = η ( x , y ) D η Dt = v Steps: (a) Linearization, (b) (2 + 1)-split, and (c) dimensional reduction v · � (a) η ≪ h , � ∇ ≪ ∂/∂ t . Hence D / Dt ≈ ∂/∂ t

A quick derivation Starting point: Euler equations for an incompressible fluid in dimension 3. ρ D � v � g − ρ� v − � ∇ · � v = 0 , Dt = ρ� f ∧ � ∇ p p = 0 at z = η ( x , y ) D η Dt = v Steps: (a) Linearization, (b) (2 + 1)-split, and (c) dimensional reduction v · � (a) η ≪ h , � ∇ ≪ ∂/∂ t . Hence D / Dt ≈ ∂/∂ t (b) � v = � v ( x , y , z ) =: ( v , v )

A quick derivation Starting point: Euler equations for an incompressible fluid in dimension 3. ρ D � v � g − ρ� v − � ∇ · � v = 0 , Dt = ρ� f ∧ � ∇ p p = 0 at z = η ( x , y ) D η Dt = v Steps: (a) Linearization, (b) (2 + 1)-split, and (c) dimensional reduction v · � (a) η ≪ h , � ∇ ≪ ∂/∂ t . Hence D / Dt ≈ ∂/∂ t (b) � v = � v ( x , y , z ) =: ( v , v ); g = (0 , − g ), � f = (0 , f ), hence � v = ( f v ⊥ , ∗ ) � f ∧ �

A quick derivation Starting point: Euler equations for an incompressible fluid in dimension 3. ρ D � v � g − ρ� v − � ∇ · � v = 0 , Dt = ρ� f ∧ � ∇ p p = 0 at z = η ( x , y ) D η Dt = v Steps: (a) Linearization, (b) (2 + 1)-split, and (c) dimensional reduction v · � (a) η ≪ h , � ∇ ≪ ∂/∂ t . Hence D / Dt ≈ ∂/∂ t (b) � v = � v ( x , y , z ) =: ( v , v ); g = (0 , − g ), � f = (0 , f ), hence � v = ( f v ⊥ , ∗ ); to leading order � f ∧ � ρ g + ∂ p /∂ z = 0 , p = ρ g ( η − z ) , ∇ p = − ρ g ∇ η (c) Replace v by its average over 0 ≤ z ≤ h :

A quick derivation Starting point: Euler equations for an incompressible fluid in dimension 3. ρ D � v � g − ρ� v − � ∇ · � v = 0 , Dt = ρ� f ∧ � ∇ p p = 0 at z = η ( x , y ) D η Dt = v Steps: (a) Linearization, (b) (2 + 1)-split, and (c) dimensional reduction v · � (a) η ≪ h , � ∇ ≪ ∂/∂ t . Hence D / Dt ≈ ∂/∂ t (b) � v = � v ( x , y , z ) =: ( v , v ); g = (0 , − g ), � f = (0 , f ), hence � v = ( f v ⊥ , ∗ ); to leading order � f ∧ � ρ g + ∂ p /∂ z = 0 , p = ρ g ( η − z ) , ∇ p = − ρ g ∇ η (c) Replace v by its average over 0 ≤ z ≤ h : ❀ v = v ( x , y ) ∂η ρ∂ v ∂ t = − ρ f v ⊥ − ρ g ∇ η ∂ t = − h ∇ · v ,

A quick derivation Starting point: Euler equations for an incompressible fluid in dimension 3. ρ D � v � g − ρ� v − � ∇ · � v = 0 , Dt = ρ� f ∧ � ∇ p p = 0 at z = η ( x , y ) D η Dt = v Steps: (a) Linearization, (b) (2 + 1)-split, and (c) dimensional reduction v · � (a) η ≪ h , � ∇ ≪ ∂/∂ t . Hence D / Dt ≈ ∂/∂ t (b) � v = � v ( x , y , z ) =: ( v , v ); g = (0 , − g ), � f = (0 , f ), hence � v = ( f v ⊥ , ∗ ); to leading order � f ∧ � ρ g + ∂ p /∂ z = 0 , p = ρ g ( η − z ) , ∇ p = − ρ g ∇ η (c) Replace v by its average over 0 ≤ z ≤ h : ❀ v = v ( x , y ) ∂η ρ∂ v ∂ t = − ρ f v ⊥ − ρ g ∇ η ∂ t = − h ∇ · v ,

A hydrodynamic model Topology by compactification The Hatsugai relation Violation What goes wrong?

A convenient extension Momentum equations (in dimension 2): ρ Dv Dt = b + ∇ · σ body forces � b , stress tensor σ . To σ ij = − p δ ij (Euler) add either ( v i , j := ∂ v i /∂ x j ): ◮ even viscosity (Navier-Stokes) � 2 v 1 , 1 v 1 , 2 + v 2 , 1 � σ = − η , ∇ · σ = η ∆ v v 1 , 2 + v 2 , 1 2 v 2 , 2 ◮ odd viscosity (Avron) � � − ( v 1 , 2 + v 2 , 1 ) v 1 , 1 − v 2 , 2 ∇ · σ = − η ∆ v ⊥ σ = − η , v 1 , 2 + v 2 , 1 v 1 , 1 − v 2 , 2

The model (final form) Equations of motion ∂η ∂ t = − h ∇ · v ∂ v ∂ t = − g ∇ η − f v ⊥ − ν ∆ v ⊥ with ν = η/ρ .

The model (final form) Equations of motion ∂η ∂ t = − h ∇ · v ∂ v ∂ t = − g ∇ η − f v ⊥ − ν ∆ v ⊥ with ν = η/ρ . After rescaling ( gh = 1) ∂η ∂ t = −∇ · v ∂ v ∂ t = −∇ η − ( f + ν ∆) v ⊥

The model (final form) Equations of motion ∂η ∂ t = −∇ · v ∂ v ∂ t = −∇ η − ( f + ν ∆) v ⊥ In Hamiltonian form ( v =: ( u , v ), p x := − i ∂/∂ x ) i ∂ψ ∂ t = H ψ     η 0 p x p y  , i ( f − ν p 2 )  = H ∗ ψ = H = p x 0 u   − i ( f − ν p 2 ) v p y 0

The model as a spin 1 bundle By translation invariance (momentum k ∈ R 2 ), H reduces to fibers � 0 k x k y � H = i ( f − ν k 2 ) 0 k x k y − i ( f − ν k 2 ) 0

The model as a spin 1 bundle By translation invariance (momentum k ∈ R 2 ), H reduces to fibers � 0 k x k y � = � d · � � d ( k ) = ( k x , k y , f − ν k 2 ) H = i ( f − ν k 2 ) S , 0 k x k y − i ( f − ν k 2 ) 0 where � S is an irreducible spin 1 representation � 0 1 0 � 0 0 1 � 0 0 0 � � � S 1 = , S 2 = , S 3 = 1 0 0 0 0 0 0 0 i 0 0 0 1 0 0 0 − i 0

The model as a spin 1 bundle By translation invariance (momentum k ∈ R 2 ), H reduces to fibers � 0 k x k y � = � d · � � d ( k ) = ( k x , k y , f − ν k 2 ) H = i ( f − ν k 2 ) S , 0 k x k y − i ( f − ν k 2 ) 0 where � S is an irreducible spin 1 representation � 0 1 0 � 0 0 1 � 0 0 0 � � � S 1 = , S 2 = , S 3 = 1 0 0 0 0 0 0 0 i 0 0 0 1 0 0 0 − i 0 Eigenvalues d ( k ) | = ± ( k 2 + ( f − ν k 2 ) 2 ) 1 / 2 ω ± ( k ) = ±| � ω 0 ( k ) = 0 ,

The model as a spin 1 bundle H = � d · � � d ( k ) = ( k x , k y , f − ν k 2 ) S , Eigenvalues d ( k ) | = ± ( k 2 + ( f − ν k 2 ) 2 ) 1 / 2 ω ± ( k ) = ±| � ω 0 ( k ) = 0 ,

The model as a spin 1 bundle H = � d · � � d ( k ) = ( k x , k y , f − ν k 2 ) S , Eigenvalues d ( k ) | = ± ( k 2 + ( f − ν k 2 ) 2 ) 1 / 2 ω ± ( k ) = ±| � ω 0 ( k ) = 0 , Left: ω + as a function of k Right: projected along k y as a function of k x Remark: Gap is f > 0