Are Bose-Einstein condensates affected by gravitational attraction? In sufficient quantity, do Bose-Einstein condensates exert a gravitational force?
[Physics] Bose-Einstein condensates and gravity
bose-einstein-condensategravityquantum mechanics
Related Solutions
1) Some of the assumptions of the Gross-Pitaevskii equation (GPE) are:
- all atoms are in the same condensate wave function,
- the condensate is at $T=0$,
- collisions between atoms are sufficiently low energy that the interactions can be well described by the $s$-wave scattering length, so that the interaction can be written $g\delta(\mathbf{x}_i-\mathbf{x}_j)$.
Generalized GPEs can also be solved, allowing for thermal and quantum depletion (some atoms not in the condensate) and allowing for other forms of interaction, such as dipolar.
2) The interaction term, $g|\Psi(\mathbf{x})|^2$, is in addition to the external potential $V_\mathrm{ext}(\mathbf{x})$, the effective potential is the sum of both: $V_\mathrm{ext}(\mathbf{x})+g|\Psi(\mathbf{x})|^2$. The condensate density is $n_0(\mathbf{x})=|\Psi(\mathbf{x})|^2$, so the interaction term is $gn_0(\mathbf{x})$ which is the potential due to interaction with the condensate itself.
More detail in response to the OP's comment:
The interaction potential between two atoms can usually be written as $V(\mathbf{r}_{ij})$ where $\mathbf{r}_{ij} =\mathbf{x}_i-\mathbf{x}_j$. For neutral atoms without a significant magnetic dipole moment, the dominant interaction is van der Waals so $V(\mathbf{r}_{ij})\propto r_{ij}^{-6}$.
When considering the scattering between two atoms, we can do a partial wave expansion (matching incoming and outgoing wave functions and expanding in terms of Legendre polynomials, e.g. "Quantum Mechanics", Ch. 17, Landau and Lifshitz). For slow particles with van der Waals interaction, the $s$-wave term is dominant and the interaction can be simplified to $V(\mathbf{r}_{ij}) = g \delta(\mathbf{r}_{ij})$ where $g=4\pi\hbar^2 a_s/m$ and $a_s$ is the $s$-wave scattering length. To get a feel for the scattering length, in the the $s$-wave approximation, the cross section is $\sigma=4\pi a_s^2$, so $a_s$ is a length scale for the interaction.
The interaction potential in the GPE can be written $$\int d\mathbf{x'} V(\mathbf{x}'-\mathbf{x})|\Psi(\mathbf{x'})|^2$$ When $V(\mathbf{x}'-\mathbf{x})=g\delta(\mathbf{x}'-\mathbf{x})$, this simplifies to $$\int d\mathbf{x'} g\delta(\mathbf{x}'-\mathbf{x})|\Psi(\mathbf{x'})|^2 = g|\Psi(\mathbf{x})|^2$$
3) The external potential $V_\mathrm{ext}(\mathbf{x})$ is generally due to applied optical or magnetic fields, and is often approximately a harmonic oscillator. The oscillator strength may be very strong in some directions creating quasi one or two dimensional confinement. A particle in a box is not possible yet (the atoms would interact with the "walls"), but the external potential may be locally approximately uniform near the center of the trap. Lattice potentials are also common, where (in addition to harmonic confinement) the atoms are trapped in a standing wave created by counterpropogating lasers resulting in a periodic potential. Many other shapes are possible, such as toroids.
A good reference is the book "Bose-Einstein Condensation in Dilute Gases" by Pethick and Smith. This slightly dated review is also good (free arXiv version here): section III is relevant to your question 2.
For a BEC, you want atoms to be in the same quantum state, not necessarily at the same position.
For a BEC, the temperature is low enough so that the de Broglie wavelength $\lambda_{\mathrm{dB}} \propto 1/\sqrt{T}$ is larger than the interatomic spacing $\propto n^{-1/3}$, $n$ being the density. This means that the wave nature of the atoms is large enough for it to be felt by other atoms, in other words atoms "see" each other even without exactly sitting on top of each other. This is just to further justify the claim that you don't need atoms at the same position. Actually, if you had a perfect box potential of side $L$, and you reached BEC, then the atoms will macroscopically occupy the ground state $ |\Psi|^2 \propto \sin^2(x/L)$ which is very much extended. If you let $L\rightarrow \infty$, the atomic distribution becomes flat. So, again, very much atoms not at the same positions.
Ok, so now interactions and collapse.
First of all, BEC is a non-interacting effect. It is not driven by a competition of interaction terms, but solely by Bose-Einstein statistics. It is experimentally interesting that BEC seems to exist also in interacting systems, though there is no general theoretical proof. By BEC in an interacting system I mean macroscopic occupation of the ground state + Off-Diagonal Long-Range Order (ODLRO) — so not all superfluids are BECs. Let me also point out that you need interactions to reach a BEC as you need to reach thermal equilibrium.
The interaction strength among weakly interacting Bose-condensed bosons is quantified by a $g n$ term in the Hamiltonian, where $g$ is $4\pi\hbar^2 a/m$ (Gross-Pitaevski equation). You can make this interaction attractive with $a<0$ and repulsive with $a>0$, where $a$ is the scattering length and it is given by $a(B) = a_0 f(B)$, where $a_0$ is the background scattering length in the presence on no external magnetic field $B$ ($f$ is some function).
The pressure of a weakly interacting Bose-condensed gas is (at $T=0$): $$ P = -\frac{\partial E}{\partial V} = \frac{1}{2}gn^2.$$
Because $n^2$ is always positive, the condition for stability (i.e. not to collapse) is $P>0$ and hence $g>0 \Rightarrow a>0$ i.e. a repulsive system. With a positive pressure, the gas expands until it hits a wall (e.g. the confining potential). But if $P<0$ then the system is intrinsically unstable and collapses.
Rb-87 is "easy" because its background scattering length is positive and therefore trivially allows for a stable BEC. K-39, on the other hand, has a negative background scattering length so its "BEC" would collapse (and eventually explode). But its scattering length can be made repulsive by the use of a Feshbach resonance (applying a field $B$ to change $a$) so that it can undergo BEC.
Best Answer
Yes, Bose-Einstein condensates are affected by gravity. Most condensates are formed in laser traps and often (especially in the early experiments) the lasers must be turned off to get a good image of the condensate, with the consequence that many images of condensates (again, especially from the early experiments) show them falling.
An example (source):