Superfluid Dark Matter Lasha Berezhiani Department of Physics, Princeton University May 16, 2017 It faces some challenges at galactic scales. Introduction ΛCDM works remarkably well on large scales. Introduction ΛCDM works remarkably well on large scales. It faces some challenges at galactic scales. 4 vc Mb = GNa0 MOND fails on cosmological scales. MOND: Milgrom ’83 ( aN aN a0 a = p aNa0 aN a0 a0 ∼ H0 MOND fails on cosmological scales. MOND: Milgrom ’83 ( aN aN a0 a = p aNa0 aN a0 a0 ∼ H0 4 vc Mb = GNa0 MOND: Milgrom ’83 ( aN aN a0 a = p aNa0 aN a0 a0 ∼ H0 4 vc Mb = GNa0 MOND fails on cosmological scales. LB, Justin Khoury ’15 Unification can be achieved through superfluidity. 2M2 3=2 pl h 2i LMOND = − (r~ ') − ϕρb 3a0 vs (r~ ')2 L = P(X ) with X ≡ µ +' _ − superfluid 2m P depends on the properties of the superfluid. Unified Framework 2M2 3=2 pl h 2i LMOND = − (r~ ') − ϕρb 3a0 vs (r~ ')2 L = P(X ) with X ≡ µ +' _ − superfluid 2m P depends on the properties of the superfluid. Unified Framework LB, Justin Khoury ’15 Unification can be achieved through superfluidity. vs (r~ ')2 L = P(X ) with X ≡ µ +' _ − superfluid 2m P depends on the properties of the superfluid. Unified Framework LB, Justin Khoury ’15 Unification can be achieved through superfluidity. 2M2 3=2 pl h 2i LMOND = − (r~ ') − ϕρb 3a0 Unified Framework LB, Justin Khoury ’15 Unification can be achieved through superfluidity. 2M2 3=2 pl h 2i LMOND = − (r~ ') − ϕρb 3a0 vs (r~ ')2 L = P(X ) with X ≡ µ +' _ − superfluid 2m P depends on the properties of the superfluid. Condensation takes place when 1 λdB = p > ` mT At this point the particles become indistinguishable and the classical statistics breaks down. The critical temperature for this phase transition is estimated to be n2=3 T = c m Bose-Einstein Condensate Bose-Einstein Condensate Condensation takes place when 1 λdB = p > ` mT At this point the particles become indistinguishable and the classical statistics breaks down. The critical temperature for this phase transition is estimated to be n2=3 T = c m At finite temperature we have a mixture of 2 fluids: BEC + Normal fluid BEC exhibits superfluidity if it is a condensate of interacting (repulsive) particles. Bose-Einstein Condensate At T < Tc N T 3=2 k=0 = 1 − Ntot Tc BEC exhibits superfluidity if it is a condensate of interacting (repulsive) particles. Bose-Einstein Condensate At T < Tc N T 3=2 k=0 = 1 − Ntot Tc At finite temperature we have a mixture of 2 fluids: BEC + Normal fluid Bose-Einstein Condensate At T < Tc N T 3=2 k=0 = 1 − Ntot Tc At finite temperature we have a mixture of 2 fluids: BEC + Normal fluid BEC exhibits superfluidity if it is a condensate of interacting (repulsive) particles. For the contact 2-point interactions with coupling constant λ λn c2 = s m3 Superfluid The sound waves of the superfluid (so called ’phonons’) propagate with the dispersion relation k4 !2 ' c2k2 + k s 4m2 Superfluid The sound waves of the superfluid (so called ’phonons’) propagate with the dispersion relation k4 !2 ' c2k2 + k s 4m2 For the contact 2-point interactions with coupling constant λ λn c2 = s m3 Consider an impurity moving through the fluid with velocity v. The dissipation happens through the process Impurity ! Impurity + phonon Landau’s criterion for superfluidity v < cs Superfluidity For small momenta !k = cs k The dissipation happens through the process Impurity ! Impurity + phonon Landau’s criterion for superfluidity v < cs Superfluidity For small momenta !k = cs k Consider an impurity moving through the fluid with velocity v. Landau’s criterion for superfluidity v < cs Superfluidity For small momenta !k = cs k Consider an impurity moving through the fluid with velocity v. The dissipation happens through the process Impurity ! Impurity + phonon Superfluidity For small momenta !k = cs k Consider an impurity moving through the fluid with velocity v. The dissipation happens through the process Impurity ! Impurity + phonon Landau’s criterion for superfluidity v < cs e.g. λ L = −|@Ψj2 − m2jΨj2 − jΨj4 2 We are looking for a state with large number of particles in k = 0 state. This corresponds to the homogeneous classical solution. Superfluid It is a degenerate state of interacting bosons. We are looking for a state with large number of particles in k = 0 state. This corresponds to the homogeneous classical solution. Superfluid It is a degenerate state of interacting bosons. e.g. λ L = −|@Ψj2 − m2jΨj2 − jΨj4 2 This corresponds to the homogeneous classical solution. Superfluid It is a degenerate state of interacting bosons. e.g. λ L = −|@Ψj2 − m2jΨj2 − jΨj4 2 We are looking for a state with large number of particles in k = 0 state. Superfluid It is a degenerate state of interacting bosons. e.g. λ L = −|@Ψj2 − m2jΨj2 − jΨj4 2 We are looking for a state with large number of particles in k = 0 state. This corresponds to the homogeneous classical solution. This is the solution as long as µ~2 − m2 − λv 2 = 0 v is fixed by the concentration n = 2µ~v 2 Superfluid Let’s try Ψ = veiµ~t v is fixed by the concentration n = 2µ~v 2 Superfluid Let’s try Ψ = veiµ~t This is the solution as long as µ~2 − m2 − λv 2 = 0 Superfluid Let’s try Ψ = veiµ~t This is the solution as long as µ~2 − m2 − λv 2 = 0 v is fixed by the concentration n = 2µ~v 2 Easy to show that mρ = 2m and m' = 0 We can integrate out ρ to get (r~ ')2 L/ X 2 with X ≡ µ +' _ − 2m For generic potential we get a generic function of X . Fluctuations Ψ = (v + ρ)ei(~µt+') We can integrate out ρ to get (r~ ')2 L/ X 2 with X ≡ µ +' _ − 2m For generic potential we get a generic function of X . Fluctuations Ψ = (v + ρ)ei(~µt+') Easy to show that mρ = 2m and m' = 0 For generic potential we get a generic function of X . Fluctuations Ψ = (v + ρ)ei(~µt+') Easy to show that mρ = 2m and m' = 0 We can integrate out ρ to get (r~ ')2 L/ X 2 with X ≡ µ +' _ − 2m Fluctuations Ψ = (v + ρ)ei(~µt+') Easy to show that mρ = 2m and m' = 0 We can integrate out ρ to get (r~ ')2 L/ X 2 with X ≡ µ +' _ − 2m For generic potential we get a generic function of X . Unified Framework 3=2 h 2i LMOND ∼ (r~ ') vs (r~ ')2 L = P(X ) with X ≡ µ +' _ − superfluid 2m P depends on the properties of the superfluid. Unitary Fermi Gas is described by 5=2 LUFG ∼ X Fractional Powers? 3=2 Lsuperfluid ∼ X Fractional Powers? 3=2 Lsuperfluid ∼ X Unitary Fermi Gas is described by 5=2 LUFG ∼ X 2 2 2 6 L = −|@µΦj − m jΦj − λjΦj Upon BEC, it exhibits superfluidity. EFT for phonons is L ∼ X 3=2 X 3=2? Upon BEC, it exhibits superfluidity. EFT for phonons is L ∼ X 3=2 X 3=2? 2 2 2 6 L = −|@µΦj − m jΦj − λjΦj X 3=2? 2 2 2 6 L = −|@µΦj − m jΦj − λjΦj Upon BEC, it exhibits superfluidity. EFT for phonons is L ∼ X 3=2 Thermalization requirement −1 Γ >∼ tdyn BEC of Dark Matter For degeneracy 1 m1=3 λdB ∼ > ` ≡ mv ρ BEC of Dark Matter For degeneracy 1 m1=3 λdB ∼ > ` ≡ mv ρ Thermalization requirement −1 Γ >∼ tdyn Therefore, the region above some critical density will transform into the superfluid. Density Profile At virialization we expect NFW profile 4 3 2 1 0.5 1.0 1.5 2.0 As density increases, it becomes easier to satisfy the condensation criteria. Density Profile At virialization we expect NFW profile 4 3 2 1 0.5 1.0 1.5 2.0 As density increases, it becomes easier to satisfy the condensation criteria. Therefore, the region above some critical density will transform into the superfluid. σ=m 1=4 Galaxies: Rcore > R=3 ) m < 1:5 eV 0:5 cm2=g σ=m 1=4 Clusters: Rcore < R=10 ) m > 0:75 eV 0:5 cm2=g Density Profile NFW Superfluid σ=m 1=4 Clusters: Rcore < R=10 ) m > 0:75 eV 0:5 cm2=g Density Profile NFW Superfluid σ=m 1=4 Galaxies: Rcore > R=3 ) m < 1:5 eV 0:5 cm2=g Density Profile NFW Superfluid σ=m 1=4 Galaxies: Rcore > R=3 ) m < 1:5 eV 0:5 cm2=g σ=m 1=4 Clusters: Rcore < R=10 ) m > 0:75 eV 0:5 cm2=g DM particles are generated when H ∼ m, corresponding to p Tbaryons ∼ mMpl ∼ 50 TeV As soon as it is generated, DM becomes superfluid with T T ' 10−28 vs ' 10−2 Tc cosmo Tc MW Cosmology: DM with eV mass must be produced out of equilibrium. For instance, through axion-like vacuum displacement mechanism. As soon as it is generated, DM becomes superfluid with T T ' 10−28 vs ' 10−2 Tc cosmo Tc MW Cosmology: DM with eV mass must be produced out of equilibrium. For instance, through axion-like vacuum displacement mechanism. DM particles are generated when H ∼ m, corresponding to p Tbaryons ∼ mMpl ∼ 50 TeV T vs ' 10−2 Tc MW Cosmology: DM with eV mass must be produced out of equilibrium. For instance, through axion-like vacuum displacement mechanism. DM particles are generated when H ∼ m, corresponding to p Tbaryons ∼ mMpl ∼ 50 TeV As soon as it is generated, DM becomes superfluid with T ' 10−28 Tc cosmo Cosmology: DM with eV mass must be produced out of equilibrium.