IPhO 2005 · Theory problem 3

Neutrons in a Gravitational Field

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Th 3 NEUTRONS IN A GRAVITATIONAL FIELD

In the familiar classical world, an elastic bouncing ball on the Earth’s surface is an ideal example for perpetual motion. The ball is trapped: it can not go below the surface or above its turning point. It will remain bounded in this state, turning down and bouncing up once and again, forever. Only air drag or inelastic bounces could stop the process and will be ignored in the following.

A group of physicists from the Institute Laue - Langevin in Grenoble reported\(^{1}\) in 2002 experimental evidence on the behaviour of neutrons in the gravitational field of the Earth. In the experiment, neutrons moving to the right were allowed to fall towards a horizontal crystal surface acting as a neutron mirror, where they bounced back elastically up to the initial height once and again.

The setup of the experiment is sketched in Figure F-1. It consists of the opening W, the neutron mirror M (at height \(z = 0\)), the neutron absorber A (at height \(z = H\) and with length \(L\)) and the neutron detector D. The beam of neutrons flies with constant horizontal velocity component \(v_x\) from W to D through the cavity between A and M. All the neutrons that reach the surface of A are absorbed and disappear from the experiment. Those that reach the surface of M are reflected elastically. The detector D counts the transmission rate \(N(H)\), that is, the total number of neutrons that reach D per unit time.

Figure 1

F-1

Marks are indicated at the beginning of each subquestion, in parenthesis.

The neutrons enter the cavity with a wide range of positive and negative vertical velocities, \(v_z\). Once in the cavity, they fly between the mirror below and the absorber above.

  1. (1.5) Compute classically the range of vertical velocities \(v_z(z)\) of the neutrons that, entering at a height \(z\), can arrive at the detector D. Assume that \(L\) is much larger than any other length in the problem.

  2. (1.5) Calculate classically the minimum length \(L_c\) of the cavity to ensure that all neutrons outside the previous velocity range, regardless of the values of \(z\), are absorbed by A. Use \(v_x = 10\ \mathrm{m\,s^{-1}}\) and \(H = 50\ \mu\mathrm{m}\).

The neutron transmission rate \(N(H)\) is measured at D. We expect that it increases monotonically with \(H\).

  1. (2.5) Compute the classical rate \(N_c(H)\) assuming that neutrons arrive at the cavity with vertical velocity \(v_z\) and at height \(z\), being all the values of \(v_z\) and \(z\) equally probable. Give the answer in terms of \(\rho\), the constant number of neutrons per unit time, per unit vertical velocity, per unit height, that enter the cavity with vertical velocity \(v_z\) and at height \(z\).

\(^{1}\) V. V. Nesvizhevsky et al. “Quantum states of neutrons in the Earth’s gravitational field.” Nature, 415 (2002) 297. Phys Rev D 67, 102002 (2003).

The experimental results obtained by the Grenoble group disagree with the above classical predictions, showing instead that the value of \(N(H)\) experiences sharp increases when \(H\) crosses some critical heights \(H_1\), \(H_2\) ... (Figure F-2 shows a sketch). In other words, the experiment showed that the vertical motion of neutrons bouncing on the mirror is quantized. In the language that Bohr and Sommerfeld used to obtain the energy levels of the hydrogen atom, this can be written as: “The action \(S\) of these neutrons along the vertical direction is an integer multiple of the Planck action constant \(h\)”. Here \(S\) is given by

\[ S=\int p_z(z)\,dz=nh,\qquad n=1,2,3\ldots \qquad\text{(Bohr-Sommerfeld quantization rule)} \]

where \(p_z\) is the vertical component of the classical momentum, and the integral covers a whole bouncing cycle. Only neutrons with these values of \(S\) are allowed in the cavity.

Figure 2

F-2

  1. (2.5) Compute the turning heights \(H_n\) and energy levels \(E_n\) (associated to the vertical motion) using the Bohr-Sommerfeld quantization condition. Give the numerical result for \(H_1\) in \(\mu\mathrm{m}\) and for \(E_1\) in eV.

The uniform initial distribution \(\rho\) of neutrons at the entrance changes, during the flight through a long cavity, into the step-like distribution detected at D (see Figure F-2). From now on, we consider for simplicity the case of a long cavity with \(H < H_2\). Classically, all neutrons with energies in the range considered in question 1 were allowed through it, while quantum mechanically only neutrons in the energy level \(E_1\) are permitted. According to the time-energy Heisenberg uncertainty principle, this reshuffling requires a minimum time of flight. The uncertainty of the vertical motion energy will be significant if the cavity length is small. This phenomenon will give rise to the widening of the energy levels.

  1. (2.0) Estimate the minimum time of flight \(t_q\) and the minimum length \(L_q\) of the cavity needed to observe the first sharp increase in the number of neutrons at D. Use \(v_x = 10\ \mathrm{m\,s^{-1}}\).

Data:

Quantity Value
Planck action constant \(h = 6.63\cdot10^{-34}\ \mathrm{J\,s}\)
Speed of light in vacuum \(c = 3.00\cdot10^8\ \mathrm{m\,s^{-1}}\)
Elementary charge \(e = 1.60\cdot10^{-19}\ \mathrm{C}\)
Neutron mass \(M = 1.67\cdot10^{-27}\ \mathrm{kg}\)
Acceleration of gravity on Earth \(g = 9.81\ \mathrm{m\,s^{-2}}\)

If necessary, use the expression:

\[ \int(1-x)^{1/2}\,dx=-\frac{2(1-x)^{3/2}}{3} \]

Official English text of the problem.

Answers

  1. For \(0\le z<H\), the transmitted velocities satisfy \[ \boxed{-\sqrt{2g(H-z)}<v_z(z)<\sqrt{2g(H-z)}.} \] The endpoints reach the absorber and are therefore excluded.

  2. The minimum classical cavity length is \[ \boxed{L_c=2v_x\sqrt{\frac{2H}{g}} =6.39\times10^{-2}\ \mathrm{m}=6.39\ \mathrm{cm}.} \]

  3. The classical transmission rate is \[ \boxed{N_c(H)=\frac{4}{3}\rho\sqrt{2g}\,H^{3/2}.} \]

  4. Using precisely the quantization rule stated in the problem, \[ \boxed{H_n=\left(\frac{9n^2h^2}{32M^2g}\right)^{1/3} =H_1n^{2/3},} \] \[ \boxed{E_n=MgH_n =\left(\frac{9n^2h^2Mg^2}{32}\right)^{1/3} =E_1n^{2/3}.} \] Numerically, \[ \boxed{H_1=16.5\ \mu\mathrm{m},\qquad E_1=1.69\times10^{-12}\ \mathrm{eV}.} \]

  5. Using \(\Delta E\,t_q\gtrsim\hbar/2\) and the resolution scale \(\Delta E\sim E_1\), the uncertainty-limit estimates are \[ \boxed{t_q\gtrsim\frac{h}{4\pi E_1} =1.95\times10^{-4}\ \mathrm{s},\qquad L_q\gtrsim1.95\times10^{-3}\ \mathrm{m}.} \] A sharply resolved step requires longer flight. If the rough uncertainty estimate is instead written as \(\Delta E\,t_q\sim h\), the corresponding estimates are \[ \boxed{t_q\sim2.45\times10^{-3}\ \mathrm{s},\qquad L_q\sim2.45\times10^{-2}\ \mathrm{m}.} \] The numerical prefactor is not fixed without a quantitative sharpness criterion.

Solution

1. Classical range of transmitted vertical velocities

The horizontal motion is uniform and does not affect the vertical energy. Between reflections, conservation of mechanical energy gives

\[ E_z=\frac12 Mv_z^2+Mgz. \]

An elastic reflection at the mirror reverses \(v_z\) without changing \(E_z\). At the upper turning point \(z_{\mathrm{t}}\), the vertical velocity vanishes, so

\[ Mgz_{\mathrm{t}}=\frac12 Mv_z^2+Mgz, \]

and therefore

\[ z_{\mathrm{t}}=z+\frac{v_z^2}{2g}. \]

In a sufficiently long cavity, every neutron whose trajectory reaches \(z=H\) is absorbed. Transmission consequently requires \(z_{\mathrm{t}}<H\), or

\[ v_z^2<2g(H-z). \]

Thus

\[ \boxed{-\sqrt{2g(H-z)}<v_z<\sqrt{2g(H-z)},\qquad 0\le z<H.} \]

Both signs occur: an initially downward-moving neutron first reflects from the mirror, then rises to the same turning height. Equality is excluded because the problem states that every neutron reaching the absorber is absorbed.

2. Minimum classical cavity length

We must find the longest possible time before a neutron outside this range reaches the absorber.

Write its conserved vertical energy as \(E_z=Mga\), where \(a\ge H\) is the turning height it would have without the absorber. Its vertical speed at height \(z\) is

\[ |v_z(z)|=\sqrt{2g(a-z)}. \]

The time to travel upward from the mirror to the absorber is

\[ \begin{aligned} \tau(a) &=\int_0^H\frac{dz}{\sqrt{2g(a-z)}}\\ &=\sqrt{\frac{2}{g}} \left(\sqrt a-\sqrt{a-H}\right). \end{aligned} \]

This decreases as \(a\) increases, and its largest value is

\[ \tau(H)=\sqrt{\frac{2H}{g}}. \]

An initially upward-moving neutron reaches the absorber within one such upward passage. An initially downward-moving neutron can first descend to the mirror and then rise to the absorber. Its total time is at most \(2\tau(a)\).

The limiting longest trajectory starts just below the absorber, moving downward very slowly, with energy just sufficient to return to height \(H\). Hence the supremum of the absorption time is

\[ t_c=2\sqrt{\frac{2H}{g}}. \]

Since the horizontal flight time is \(L/v_x\), all such neutrons are removed when \(L/v_x\ge t_c\). Any smaller length allows some trajectories sufficiently close to the limiting one to escape absorption. Therefore

\[ \boxed{L_c=2v_x\sqrt{\frac{2H}{g}}.} \]

Numerically,

\[ t_c=2\sqrt{\frac{2(50\times10^{-6}\ \mathrm{m})} {9.81\ \mathrm{m\,s^{-2}}}} =6.39\times10^{-3}\ \mathrm{s}, \]

so

\[ \boxed{L_c=(10\ \mathrm{m\,s^{-1}})t_c =6.39\times10^{-2}\ \mathrm{m}.} \]

3. Classical transmission rate

By definition, the incident rate in an element \(dz\,dv_z\) is

\[ dN=\rho\,dz\,dv_z. \]

For a long cavity, only the velocity interval found in part 1 contributes. Integrating over this interval and over all entrance heights,

\[ N_c(H)= \rho\int_0^H \int_{-\sqrt{2g(H-z)}}^{+\sqrt{2g(H-z)}} dv_z\,dz. \]

Thus

\[ \begin{aligned} N_c(H) &=2\rho\sqrt{2g}\int_0^H(H-z)^{1/2}\,dz\\ &=2\rho\sqrt{2g}\,\frac23H^{3/2}. \end{aligned} \]

Therefore

\[ \boxed{N_c(H)=\frac43\rho\sqrt{2g}\,H^{3/2}.} \]

This predicts a smooth, increasing curve starting at \(N_c(0)=0\), with no steps.

4. Quantized turning heights and energies

For a neutron whose vertical turning height is \(H_n\),

\[ E_n=MgH_n. \]

The vertical momentum on the upward part of its trajectory is

\[ p_z(z)=\sqrt{2M(E_n-Mgz)} =M\sqrt{2g(H_n-z)}. \]

During descent, both \(p_z\) and \(dz\) are negative, so the downward contribution to the action equals the upward contribution. The instantaneous reflection at the mirror contributes no displacement. Consequently, the action over one full bounce is

\[ \begin{aligned} S &=2\int_0^{H_n}M\sqrt{2g(H_n-z)}\,dz\\ &=\frac43M\sqrt{2g}\,H_n^{3/2}. \end{aligned} \]

Applying the supplied Bohr-Sommerfeld rule \(S=nh\),

\[ \frac43M\sqrt{2g}\,H_n^{3/2}=nh. \]

Solving,

\[ \boxed{ H_n=\left(\frac{3nh}{4M\sqrt{2g}}\right)^{2/3} =\left(\frac{9n^2h^2}{32M^2g}\right)^{1/3}. } \]

Multiplication by \(Mg\) gives

\[ \boxed{ E_n=\left(\frac{9n^2h^2Mg^2}{32}\right)^{1/3}. } \]

Both quantities scale as \(n^{2/3}\).

For \(n=1\),

\[ H_1= \left[ \frac{3(6.63\times10^{-34})} {4(1.67\times10^{-27})\sqrt{2(9.81)}} \right]^{2/3} =1.653\times10^{-5}\ \mathrm{m}. \]

Therefore

\[ \boxed{H_1=16.5\ \mu\mathrm{m}.} \]

The corresponding energy is

\[ E_1=MgH_1 =2.708\times10^{-31}\ \mathrm{J}. \]

Using \(1\ \mathrm{eV}=1.60\times10^{-19}\ \mathrm{J}\),

\[ \boxed{ E_1=\frac{2.708\times10^{-31}}{1.60\times10^{-19}} =1.69\times10^{-12}\ \mathrm{eV}. } \]

5. Minimum quantum flight time and cavity length

The first onset of transmission occurs near \(MgH=E_1\). If the energy width of the first level is comparable to \(E_1\), this onset is substantially smeared. The relevant energy scale for an initial resolution estimate is therefore

\[ \Delta E\sim E_1. \]

Using the time-energy uncertainty estimate in the form

\[ \Delta E\,t_q\gtrsim\frac{\hbar}{2} =\frac{h}{4\pi}, \]

a necessary flight-time scale is

\[ t_q\gtrsim\frac{h}{4\pi E_1}. \]

Substituting the energy in joules,

\[ t_q\gtrsim \frac{6.63\times10^{-34}} {4\pi(2.708\times10^{-31})} =1.95\times10^{-4}\ \mathrm{s}. \]

The horizontal motion then gives

\[ \boxed{ L_q=v_xt_q\gtrsim (10)(1.95\times10^{-4}) =1.95\times10^{-3}\ \mathrm{m}. } \]

These values characterize an uncertainty bound, rather than guaranteeing a particular experimental sharpness. To make that distinction quantitative, suppose the permitted energy width is \(\Delta E=\epsilon E_1\). Since \(E=MgH\), this also corresponds to a relative height width \(\Delta H/H_1=\epsilon\). The estimates become

\[ t_q\gtrsim\frac{1.95\times10^{-4}}{\epsilon}\ \mathrm{s}, \qquad L_q\gtrsim\frac{1.95\times10^{-3}}{\epsilon}\ \mathrm{m}. \]

A sharp step, with \(\epsilon\ll1\), requires times and lengths appreciably larger than the basic bound.

For an order-of-magnitude treatment using \(\Delta E\,t_q\sim h\), one instead quotes

\[ t_q\sim\frac{h}{E_1} =2.45\times10^{-3}\ \mathrm{s}, \qquad \boxed{L_q\sim2.45\times10^{-2}\ \mathrm{m}=2.45\ \mathrm{cm}.} \]

The solution was checked against the official solution.