The trajectory of the atoms

The change of angle of the light in the collimator has to be the same as the change of angle of the velocity of the atoms. If the collimator is working well, the angle the velocity vector $\vec v$ of the atoms makes with the centerline is zero at the end of the collimator, the atoms all leave the collimator parallel to each other and to the centerline of the collimator. With Eq. 1.14 we can calculate the total change of angle of the light in the collimator, which will give the angle $\beta_{\rm max}$ of the velocity of the atoms at the beginning of the collimator:

$$\beta_{\rm max} = \alpha_0 - \sqrt{\alpha_0^2 - 2 \gamma \frac{L}{D}}$$ (1.16)

Since the total change of angle of the velocity vector $\vec v$ is equal to the total change of angle of the laser light in the collimator, we can use eqn 1.9 to show:

$$\beta_{\rm max} = N \gamma$$ (1.17)

with $N$ the total number of reflections in the collimator. If $\gamma\frac{L}{D}\ll \alpha_0^2$, Eq. 1.16 can be approximated by:

$$\beta_{\rm max} = \frac{\gamma L}{\alpha_0 D}$$ (1.18)

In our experiments, $\frac{\gamma L}{D \alpha_0^ 2}\approx$ 0.3, so Eq. 1.18 gives a good approximation of the maximum capture angle $\beta_{\rm max}$.

Another way of finding the maximum capture angle, is looking at the force that can be exerted on the atom during the time it is in the collimator. This force should reduce the transverse velocity $v_{\perp}$ to zero, according to:

$$v_{\perp} - \frac{F}{M} t = 0$$ (1.19)

with $F$ the force exerted on the Helium atom, $M$ the mass of the Helium atom, and $t$ the interaction time. If we take half the maximum resonant radiation pressure $F = F_{\rm max}/2 = \hbar k \Gamma/4$ , and take $L/v_{\parallel}$ as the interaction time in the collimator, we get:

$$v\sin{\theta} - \frac{\hbar k \Gamma}{4M} \frac{L}{v \cos{\theta}} = 0$$ (1.20)

Solving this for $\theta$ gives:

$$\theta = \beta_{\rm max} = \frac{1}{2}\sin^{-1}{\frac{\hbar k \Gamma L}{2M v^2}}$$ (1.21)

Filling in the specific values $L = 0.2$ m, $k = 5.8 \cdot 10^6$ m$^{-1}$, $\Gamma =1.6$ MHz, $v = 1000$ m/s and $M = 6.65 \cdot 10^{-27}$ kg, gives an angle $\beta_{\rm max} = 0.055$ rad.

Now that we know the values for $\beta_{\rm max}$ and $\alpha_0$ we can find a value for $\gamma$ using Eq. 1.16, and filling in the values of $\beta_{\rm max}$ and $\alpha_0$. The value we found was $\gamma = 1.74$ mrad. Since $\beta_{\rm max} = N \gamma$, we find a total number of reflections $N = 31$.