Can Magnetic Traps Achieve Bose-Einstein Condensate?

[Kalrag]

Lately I've been wondering how magnetic traps work. So can anyone tell me how they work, How long they can hold atoms and if they can achieave Bose-Einstein Condensate? Also I have been wondering if they could hold several grams of atoms and then create the Bose-Einstein Effect. Can anyone help me?

 

[Dickfore]

Magnetic traps work by essentially creating a gradient in the magnetic field. As you may know, atoms possesses a magnetic moment. This is why they have a potential energy:

$$U = -\mathbf{\mu} \cdot \mathbf{B}$$

The negative gradient of this potential energy determines the force:

$$\mathbf{F} = -\nabla U = \nabla(\mathbf{\mu} \cdot \mathbf{B}) = \nabla(\mathbf{\mu} \cdot \stackrel{\downarrow}{\mathbf{B}})$$

In the adiabatic approximation, we neglect the possible precession of the magnetic moment of the atoms and assume that they are always aligned along the magnetic field line, i.e.

$$\mathbf{\mu} = \mu \, \hat{\mathbf{b}}$$

However, using:

$$\nabla B = \nabla (B^{2})^{\frac{1}{2}} = \nabla (\mathbf{B} \cdot \mathbf{B})^{\frac{1}{2}} = \frac{1}{2} \, (\mathbf{B} \cdot \mathbf{B})^{-\frac{1}{2}} \nabla (\mathbf{B} \cdot \mathbf{B}) = \frac{1}{B} \nabla (\mathbf{B} \cdot \stackrel{\downarrow}{\mathbf{B}}) = \nabla(\hat{\mathbf{b}} \cdot \stackrel{\downarrow}{\mathbf{B}})$$

we may write the force as:

$$\mathbf{F} = \mu \nabla B$$

If $\mu > 0$, then this force is in the same direction as the gradient of the intensity of the magnetic field. This means it is directed in the direction where the intensity increases. Such atoms are called high-field seekers. On the contrary, atoms with $\mu < 0$ are called low field seekers. This means that atoms can have stable equilibrium in points where the intensity of the magnetic field is maximal or minimal.

There is a theorem in Classical Electrodynamics that states that the magnetic field cannot have a local maximum in intensity in a space with no currents present. Thus, high-field seekers cannot be trapped.

For low-field seekers, we can produce a local minimum if we create a point where $\mathbf{B} = \mathbf{0}$. This may be achieved by using two anti-Helmholtz coils (the currents are directed in opposite directions, so that the fields cancel in the region between the two). This kind of field configuration is known as a quadrupole magnetic field. Near the point where the field vanishes (assuming axial symmetry), the magnetic field is approximately:

$$\mathbf{B} = B' \left\langle x, y, -2 z \right\rangle$$

and the magnitude then becomes:

$$B = B' \left(x^{2} + y^{2} + 4 z^{2}\right)^{\frac{1}{2}}$$

The potential energy is:

$$U = \mu \, B' \, \left(x^{2} + y^{2} + 4 z^{2}\right)^{\frac{1}{2}}$$

 

[Kalrag]

Thanks. Now that I know how they work, can they create a fair amount of the Bose-Einstein Condensate for a long period of time?

 

[Dickfore]

One of the pioneering papers on the subject (Science 269, 198 (1995)) reported that a condensate fraction of ⁸⁷Rb atoms first appeared at 170 nK with a central density $2.5 \times 10^{12} \, \mathrm{cm}^{-3}[/ite x] and they managed to contain it for 15 s. They had about 2,000 atoms in the condensate.$

 

[Kalrag]

Now is it possible to take a gram or more of atoms and then create the Bose-Einstein condensate? I guess what I am trying to say is if you can freeze not just 2 thousand atoms but 5 grams worth of atoms.

 

[Dickfore]

I don't know what the current record is. The problem is that during evaporative cooling, you lose a large fraction of your initial atoms.