Self-gravitating Bose-Einstein Condensates

Abstract

Bose-Einstein condensates play a major role in condensed matter physics. Recently, it has been suggested that they could play an important role in astrophysics also. Indeed, dark matter halos could be gigantic quantum objects made of Bose-Einstein condensates. The pressure arising from the Heisenberg uncertainty principle or from the repulsive scattering of the bosons could stabilize dark matter halos against gravitational collapse and lead to smooth core densities instead of cuspy density profiles in agreement with observations. In order to reproduce the scales of dark matter halos, the mass of the bosons may range from $10^{-24}\, \mathrm{eV/c^2}$ to a few $\mathrm{eV/c^2}$ depending whether they interact or not. At the scale of galaxies, Newtonian gravity can be used so the evolution of the wave function is governed by the Gross-Pitaevskii-Poisson system. Self-gravitating Bose-Einstein condensates have also been proposed to describe boson stars. For these compact objects, one must use general relativity and couple the Klein-Gordon equation to the Einstein field equations. In that context, it has been proposed that neutron stars could be Bose-Einstein condensate stars due to their superfluid core. Indeed, the neutrons could form Cooper pairs and behave as bosons. In that case, the maximum mass of the neutron stars depends on the scattering length of the bosons and can be as large as $2M_{\odot }$. This could explain recent observations of neutron stars with a mass much larger than the Oppenheimer-Volkoff limit of $0.7M_{\odot }$ obtained by assuming that neutron stars are ideal fermion stars. Self-gravitating Bose-Einstein condensates may also find applications in the physics of black holes. For example, when the scattering length of the bosons is negative, a Newtonian self-gravitating Bose-Einstein condensate becomes unstable above a critical mass and undergoes a gravitational collapse leading ultimately to a singularity. On the other hand, stable boson stars with a positive scattering length could mimic supermassive black holes that reside at the center of galaxies. Finally, it has been proposed that microscopic quantum black holes could be Bose-Einstein condensates of gravitons. This contribution discusses fundamental aspects of the physics of self-gravitating Bose-Einstein condensates and considers recent applications in astrophysics, cosmology and black hole physics with promising perspectives.

Appendices

Appendix

6.7 Self-interaction Constant

We define the self-interaction constant by [15]:

$$\begin{aligned} \frac{\lambda }{8\pi }\equiv \frac{a}{\lambda _c}=\frac{a m c}{\hbar }, \end{aligned}$$

(6.87)

where $\lambda _c=\hbar /mc$ is the Compton wavelength of the bosons. This is a dimensionless parameter measuring the strength of the short-range interactions. It can be written as

$$\begin{aligned} \frac{\lambda }{8\pi }=5.07\, \frac{a}{1\,\mathrm{fm}}\frac{m}{1 \,\mathrm{GeV}/c^2}. \end{aligned}$$

(6.88)

Using this expression, we can express the results in terms of $\lambda$ and $m$ instead of $a$ and $m$.

6.8 Conservation of Energy

The total energy associated with the quantum Euler-Poisson system (6.14)–(6.16) is given by Eq. (6.37). According to Eq. (6.38), we have

$$\begin{aligned} \delta \Theta _c=\int \frac{\mathbf{u}^2}{2}\delta \rho \, d\mathbf{r}+\int \rho \mathbf{u}\cdot \delta \mathbf{u}\, d\mathbf{r}. \end{aligned}$$

(6.89)

On the other hand, using Eqs. (6.10) and (6.39), we find that

$$\begin{aligned} \delta \Theta _Q&= \frac{\hbar ^2}{m^2}\int \nabla \sqrt{\rho }\cdot \delta \nabla \sqrt{\rho }\, d\mathbf{r} =\frac{\hbar ^2}{m^2}\int \nabla \sqrt{\rho }\cdot \nabla \left( \frac{1}{2\sqrt{\rho }}\delta \rho \right) \, d\mathbf{r}\nonumber \\&=-\frac{\hbar ^2}{2 m^2}\int \frac{\varDelta \sqrt{\rho }}{\sqrt{\rho }}\delta \rho \, d\mathbf{r}=\frac{1}{m}\int Q \delta \rho \, d\mathbf{r}. \end{aligned}$$

(6.90)

Finally, according to Eqs. (6.40) and (6.41), we have

$$\begin{aligned} \delta U=\frac{4\pi a\hbar ^2}{m^3}\int \rho \delta \rho \, d\mathbf{r},\quad \delta W=\int \varPhi \delta \rho \, d\mathbf{r}. \end{aligned}$$

(6.91)

Taking the time derivative of the total energy $E_{tot}$ and using the previous relations, we get

$$\begin{aligned} \dot{E}_{tot}=\int \left( \frac{\mathbf{u}^2}{2}+\frac{Q}{m}+\frac{4\pi a\hbar ^2}{m^3}\rho +\varPhi \right) \frac{\partial \rho }{\partial t}\, d\mathbf{r}+\int \rho \mathbf{u}\cdot \frac{\partial \mathbf{u}}{\partial t}\, d\mathbf{r}. \end{aligned}$$

(6.92)

Using the equation of continuity (6.14), integrating by parts, using the Euler equation (6.15) and the identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla \left( {\mathbf{u}^2}/{2}\right) -\mathbf{u}\times (\nabla \times \mathbf{u})$, we obtain after simplification

$$\begin{aligned} \dot{E}_{tot}=\int \rho \mathbf{u}\cdot \left( \mathbf{u}\times (\nabla \times \mathbf{u})\right) \, d\mathbf{r}. \end{aligned}$$

(6.93)

Since $\mathbf{u}$ is a potential flow, we have $\nabla \times \mathbf{u}=\mathbf{0}$ yielding $\dot{E}_{tot}=0$. Actually, we note that this result remains valid even if $\mathbf{u}$ is not a potential flow since $\mathbf{u}\cdot (\mathbf{u}\times (\nabla \times \mathbf{u}))=0$.

6.9 Virial Theorem

In this Appendix, we establish the time-dependent tensorial Virial theorem associated with the quantum Euler-Poisson system (6.14)–(6.16).

Taking the time derivative of the moment of inertia tensor $I_{ij}=\int \rho x_i x_j \, d\mathbf{r}$ and using the equation of continuity (6.14), we obtain after an integration by parts

$$\begin{aligned} \dot{I}_{ij}=\int \rho (x_i u_j+x_j u_i)\, d\mathbf{r}. \end{aligned}$$

(6.94)

Taking the time derivative of Eq. (6.94), we get

$$\begin{aligned} \ddot{I}_{ij}=\int x_i \frac{\partial }{\partial t}(\rho u_j) \, d\mathbf{r} + (i\leftrightarrow j), \end{aligned}$$

(6.95)

where $\partial _t(\rho u_j)$ is given by Eq. (6.13). We need to evaluate four terms. The first term is

$$\begin{aligned} -\int x_i \partial _k(\rho u_j u_k) \, d\mathbf{r} =\int \rho u_i u_j\, d\mathbf{r}. \end{aligned}$$

(6.96)

The second term is

$$\begin{aligned} -\int x_i \frac{\partial p}{\partial x_j} \, d\mathbf{r} =\delta _{ij} \int p \, d\mathbf{r}. \end{aligned}$$

(6.97)

The third term is

$$\begin{aligned} -\int \rho x_i \frac{\partial \varPhi }{\partial x_j} \, d\mathbf{r} =W_{ij}, \end{aligned}$$

(6.98)

where $W_{ij}$ is the potential energy tensor. It is a simple matter to show that this tensor is symmetric: $W_{ij}=W_{ji}$ [1]. The fourth term is

$$\begin{aligned} -\int x_i \frac{\rho }{m} \frac{\partial Q}{\partial x_j} \, d\mathbf{r} =-\int x_{i}\partial _{k}P_{jk}\, d\mathbf{r}=\int P_{ij}\, d\mathbf{r}, \end{aligned}$$

(6.99)

where $P_{ij}$ is the quantum pressure tensor defined by Eq. (6.17). Substituting these results in Eq. (6.95), we obtain the tensorial Virial theorem

$$\begin{aligned} \frac{1}{2}\ddot{I}_{ij}=\int \rho u_iu_j\, d\mathbf{r}+\int P_{ij}\, d\mathbf{r} +\delta _{ij}\int p\, d\mathbf{r}+W_{ij}. \end{aligned}$$

(6.100)

Contracting the indices and using the fact that $\int P_{ii}\, d\mathbf{r}=2\Theta _Q$ [this can be obtained from Eqs. (6.17) and (6.39)] and $W_{ii}=-\int \rho \mathbf{r}\cdot \nabla \varPhi \, d\mathbf{r}=W$ [the Virial of the gravitational force in $d=3$ is equal to the potential energy [1]], we obtain

$$\begin{aligned} \frac{1}{2}\ddot{I}=2(\Theta _c+\Theta _Q)+3\int p\, d\mathbf{r}+W, \end{aligned}$$

(6.101)

where $I=\int \rho r^2\, d\mathbf{r}$ is the moment of inertia. For a steady state ($\ddot{I}_{ij}=0$ and $\mathbf{u}=\mathbf{0}$), we obtain the equilibrium tensorial Virial theorem

$$\begin{aligned} \int P_{ij}\, d\mathbf{r}+\delta _{ij}\int p\, d\mathbf{r}+W_{ij}=0 \end{aligned}$$

(6.102)

and the scalar Virial theorem

$$\begin{aligned} 2\Theta _Q+3\int p\, d\mathbf{r}+W=0. \end{aligned}$$

(6.103)

These results are valid for an arbitrary equation of state $p(\rho )$. For the equation of state (6.12), using $\int p\, d\mathbf{r}=U$, Eqs. (6.101) and (6.103) reduce to Eqs. (6.42) and (6.43).

6.10 Stress Tensor

The equation of continuity (6.14) may be written as

$$\begin{aligned} \frac{\partial \rho }{\partial t}+\nabla \cdot \mathbf{j}=0, \end{aligned}$$

(6.104)

where $\mathbf{j}=\rho \mathbf{u}$ is the density current. Using Eqs. (6.6) and (6.7), the density current can be expressed in terms of the wave function as

$$\begin{aligned} \mathbf{j}=\frac{N\hbar }{2i}\left( \psi ^*\nabla \psi -\psi \nabla \psi ^*\right) . \end{aligned}$$

(6.105)

On the other hand, the quantum Euler equation (6.13) may be written as

$$\begin{aligned} \frac{\partial \mathbf{j}}{\partial t}=-\nabla (\rho \mathbf{u}\otimes \mathbf{u}) -\nabla p-\rho \nabla \varPhi -\frac{\rho }{m}\nabla Q. \end{aligned}$$

(6.106)

Introducing the quantum pressure tensor, we find that the equation for the density current is given by

$$\begin{aligned} \frac{\partial {j_i}}{\partial t}=-\partial _j T_{ij}-\rho \partial _i\varPhi , \end{aligned}$$

(6.107)

where

$$\begin{aligned} T_{ij}=\rho u_i u_j+p\delta _{ij}+P_{ij} \end{aligned}$$

(6.108)

is the stress tensor. Using Eq. (6.17), we have

$$\begin{aligned} T_{ij}=\rho u_i u_j+p\delta _{ij} -\frac{\hbar ^2}{4m^2}\rho \partial _i\partial _j\ln \rho \end{aligned}$$

(6.109)

or, alternatively,

$$\begin{aligned} T_{ij}=\rho u_i u_j+\left( p-\frac{\hbar ^2}{4m^2}\varDelta \rho \right) \delta _{ij} +\frac{\hbar ^2}{4m^2}\frac{1}{\rho }\partial _i\rho \partial _j\rho . \end{aligned}$$

(6.110)

Using Eqs. (6.6) and (6.7), we find after straightforward algebra that

$$\begin{aligned} \frac{\hbar ^2}{4m^2}\frac{1}{\rho }\partial _i\rho \partial _j\rho = \frac{N\hbar ^2}{4m}\frac{1}{|\psi |^2}(\psi ^*\partial _i\psi +\psi \partial _i\psi ^*) (\psi ^*\partial _j\psi +\psi \partial _j\psi ^*) \end{aligned}$$

(6.111)

and

$$\begin{aligned} \rho u_i u_j=-\frac{N\hbar ^2}{4m}\frac{1}{|\psi |^2}(\psi ^*\partial _i\psi -\psi \partial _i\psi ^*) (\psi ^*\partial _j\psi -\psi \partial _j\psi ^*). \end{aligned}$$

(6.112)

Therefore

$$\begin{aligned} \rho u_i u_j+\frac{\hbar ^2}{4m^2}\frac{1}{\rho }\partial _i\rho \partial _j\rho =\frac{N\hbar ^2}{m}\mathrm{Re}\left( \frac{\partial \psi }{\partial x_i}\frac{\partial \psi ^*}{\partial x_j}\right) . \end{aligned}$$

(6.113)

Regrouping these results, the stress tensor can be expressed in terms of the wave function as

$$\begin{aligned} T_{ij}=\frac{N\hbar ^2}{m}\mathrm{Re}\left( \frac{\partial \psi }{\partial x_i}\frac{\partial \psi ^*}{\partial x_j}\right) +\left( \frac{2\pi a\hbar ^2}{m}N^2|\psi |^4-\frac{N\hbar ^2}{4m}\varDelta |\psi |^2\right) \delta _{ij}. \end{aligned}$$

(6.114)

Finally, we introduce the density of energy

$$\begin{aligned} e=\frac{\mathbf{u}^2}{2}+\frac{Q}{m}+\frac{2\pi a\hbar ^2}{m^3}\rho +\frac{\varPhi }{2}. \end{aligned}$$

(6.115)

Using the equation of continuity (6.14) and the quantum Euler equation (6.15), we obtain the energy equation

$$\begin{aligned} \frac{\partial }{\partial t}(\rho e)+\nabla \cdot (\rho e \mathbf{u})=-\nabla \cdot (p \mathbf{u})-\frac{1}{2}\rho \mathbf{u}\cdot \nabla \varPhi +\frac{\rho }{m}\frac{\partial Q}{\partial t} +\frac{1}{2}\rho \frac{\partial \varPhi }{\partial t}. \end{aligned}$$

(6.116)

The conservation of energy directly results from this equation. Taking the time derivative of the total energy $E_{tot}=\int \rho e\, d\mathbf{r}$, and using Eq. (6.116), we get

$$\begin{aligned} \dot{E}_{tot}=-\frac{1}{2}\int \rho \mathbf{u}\cdot \nabla \varPhi \, d\mathbf{r}+\int \frac{\rho }{m}\frac{\partial Q}{\partial t}\, d\mathbf{r} +\frac{1}{2}\int \rho \frac{\partial \varPhi }{\partial t}\, d\mathbf{r}. \end{aligned}$$

(6.117)

Using the Poisson equation (6.16) and the equation of continuity (6.14), and integrating by parts, we obtain

$$\begin{aligned} \frac{1}{2}\int \rho \frac{\partial \varPhi }{\partial t}\, d\mathbf{r}&= \frac{1}{8\pi G}\int \varDelta \varPhi \frac{\partial \varPhi }{\partial t}\, d\mathbf{r}=\frac{1}{2}\int \varPhi \frac{\partial \rho }{\partial t}\, d\mathbf{r}\nonumber \\&=-\frac{1}{2}\int \varPhi \nabla \cdot (\rho \mathbf{u})\, d\mathbf{r}=\frac{1}{2}\int \rho \mathbf{u}\cdot \nabla \varPhi \, d\mathbf{r}. \end{aligned}$$

(6.118)

On the other hand, using Eqs. (6.38) and (6.90), we find that

$$\begin{aligned} \int \rho \frac{\partial Q}{\partial t}\, d\mathbf{r}=m\frac{d\Theta _Q}{dt}-\int Q\frac{\partial \rho }{\partial t}\, d\mathbf{r} =\int Q\frac{\partial \rho }{\partial t}\, d\mathbf{r}-\int Q\frac{\partial \rho }{\partial t}\, d\mathbf{r}=0. \end{aligned}$$

(6.119)

Substituting these identities in Eq. (6.117), we obtain $\dot{E}_{tot}=0$.

6.11 Lagrangian and Hamiltonian

In this Appendix, we discuss the Lagrangian and Hamiltonian structure of the GP equation and of the corresponding hydrodynamic equations.

The Lagrangian of the GP equation (6.4) is

$$\begin{aligned}&L=\int \biggl \lbrace i\frac{\hbar }{2}N\left( \psi ^*\frac{\partial \psi }{\partial t}-\psi \frac{\partial \psi ^*}{\partial t}\right) -\frac{N\hbar ^2}{2m}|\nabla \psi |^2 -\frac{1}{2}Nm|\psi |^2\varPhi -\frac{2\pi a\hbar ^2}{m}N^2|\psi |^4\biggr \rbrace \, d\mathbf{r}.\nonumber \\&\end{aligned}$$

(6.120)

We can view the Lagrangian (6.120) as a functional of $\psi$, $\dot{\psi }$, and $\nabla \psi$. The action is $S=\int L\, dt$. The least action principle $\delta S=0$, which is equivalent to the Lagrange equations

$$\begin{aligned} \frac{\partial }{\partial t}\left( \frac{\delta L}{\delta \dot{\psi }}\right) +\nabla \cdot \left( \frac{\delta L}{\delta \nabla \psi }\right) -\frac{\delta L}{\delta \psi }=0 \end{aligned}$$

(6.121)

returns the GP equation (6.4). The Hamiltonian is obtained from the transformation

$$\begin{aligned} H=\int i\frac{\hbar }{2}N\left( \psi ^*\frac{\partial \psi }{\partial t}-\psi \frac{\partial \psi ^*}{\partial t}\right) \, d\mathbf{r}-L \end{aligned}$$

(6.122)

leading to

$$\begin{aligned} H=\int \biggl \lbrace \frac{N\hbar ^2}{2m}|\nabla \psi |^2 +\frac{1}{2}Nm|\psi |^2\varPhi + \frac{2\pi a\hbar ^2}{m}N^2|\psi |^4 \biggr \rbrace \, d\mathbf{r}. \end{aligned}$$

(6.123)

Of course, this expression coincides with the total energy (6.37) in the wavefunction representation. Using the Lagrange equations, one can show that the Hamiltonian is conserved. On the other hand, the GP equation (6.4) can be written as

$$\begin{aligned} i\hbar \frac{\partial \psi }{\partial t}=\frac{1}{N}\frac{\delta H}{\delta \psi ^*}. \end{aligned}$$

(6.124)

A stable stationary solution of the GP equation is a minimum of energy under the normalization condition. Writing the variational principle as $\delta H-\alpha N m\int |\psi |^2\, d\mathbf{r}=0$ where $\alpha$ is a Lagrange multiplier (chemical potential), we recover the time-independent GP equation (6.19) with $E=\alpha m$.

Using the Madelung transformation (see Sect. 6.2.2), we can rewrite the Lagrangian and the Hamiltonian in terms of hydrodynamic variables. According to Eqs. (6.6) and (6.7) we have

$$\begin{aligned} \frac{\partial S}{\partial t}=\frac{\hbar }{2i}\frac{1}{|\psi |^2}\left( \psi ^*\frac{\partial \psi }{\partial t}-\psi \frac{\partial \psi ^*}{\partial t}\right) \end{aligned}$$

(6.125)

and

$$\begin{aligned} |\nabla \psi |^2=\frac{1}{Nm\hbar ^2}\left[\rho (\nabla S)^2+\frac{\hbar ^2}{4\rho }(\nabla \rho )^2\right]. \end{aligned}$$

(6.126)

Substituting these identities in Eq. (6.120) we get

$$\begin{aligned} L=-\int \biggl \lbrace \frac{\rho }{m}\frac{\partial S}{\partial t}+\frac{\rho }{2m^2}(\nabla S)^2 +\frac{\hbar ^2}{8m^2}\frac{(\nabla \rho )^2}{\rho }+\frac{1}{2}\rho \varPhi +\frac{2\pi a\hbar ^2}{m^3}\rho ^2\biggr \rbrace \, d\mathbf{r}. \end{aligned}$$

(6.127)

We can view the Lagrangian (6.127) as a functional of $S$, $\dot{S}$, $\nabla S$, $\rho$, $\dot{\rho }$, and $\nabla \rho$. The Lagrange equations for the phase

$$\begin{aligned} \frac{\partial }{\partial t}\left( \frac{\delta L}{\delta \dot{S}}\right) +\nabla \cdot \left( \frac{\delta L}{\delta \nabla S}\right) -\frac{\delta L}{\delta S}=0 \end{aligned}$$

(6.128)

return the equation of continuity (6.8). The Lagrange equations for the density

$$\begin{aligned} \frac{\partial }{\partial t}\left( \frac{\delta L}{\delta \dot{\rho }}\right) +\nabla \cdot \left( \frac{\delta L}{\delta \nabla \rho }\right) -\frac{\delta L}{\delta \rho }=0 \end{aligned}$$

(6.129)

return the quantum Hamilton-Jacobi (or Bernouilli) equation (6.9) leading to the quantum Euler equation (6.15). The Hamiltonian is obtained from the transformation

$$\begin{aligned} H=-\int \frac{\rho }{m}\frac{\partial S}{\partial t}\, d\mathbf{r}-L \end{aligned}$$

(6.130)

leading to

$$\begin{aligned} H=\int \biggl \lbrace \frac{1}{2}\rho \mathbf{u}^2+\frac{\hbar ^2}{8m^2}\frac{(\nabla \rho )^2}{\rho } +\frac{1}{2}\rho \varPhi +\frac{2\pi a\hbar ^2}{m^3}\rho ^2 \biggr \rbrace \, d\mathbf{r}. \end{aligned}$$

(6.131)

Of course, this expression coincides with the total energy (6.37) in the hydrodynamical representation. Using the Lagrange equations, one can show that the Hamiltonian is conserved.