IPhO 2003, Теория — Задача 3. Theoretical Question 3

• (a) $E_{\max} = \frac{c^2}{2m_n}\left[m_n^2 + m_e^2 - (m_p + m_v)^2\right] \approx 1.29\ \mathrm{MeV}, \quad \frac{v_m}{c} = \frac{\sqrt{(m_n + m_e + M)(m_n + m_e - M)(m_n - m_e + M)(m_n - m_e - M)}}{m_n^2 - m_e^2 + M^2} \approx 0.00127, \quad M = m_p + m_v$

  Official values: E_max ≈ 1.292569 MeV ≈ 1.29 MeV; v_m/c ≈ 0.00126538 ≈ 0.00127.

• (b) $P = \frac{4mgcR^2}{(n-1)^2\delta^2}$

Solution to Theoretical Question 3

Part A

Neutrino Mass and Neutron Decay

(a) Let $(c^2E_e, c\vec{q}_e)$, $(c^2E_p, c\vec{q}_p)$, and $(c^2E_v, c\vec{q}_v)$ be the energy-momentum 4-vectors of the electron, the proton, and the anti-neutrino, respectively, in the rest frame of the neutron. Notice that $E_e, E_p, E_v, \vec{q}_e, \vec{q}_p, \vec{q}_v$ are all in units of mass. The proton and the anti-neutrino may be considered as forming a system of total rest mass $M_c$, total energy $c^2E_c$, and total momentum $c\vec{q}_c$. Thus, we have

Note that the magnitude of the vector $\vec{q}_c$ is denoted as $q_c$. The same convention also applies to all other vectors.

Since energy and momentum are conserved in the neutron decay, we have

When squared, the last equation leads to the following equality

From Eq. (A4) and the third equality of Eq. (A1), we obtain

With its second and third terms moved to the other side of the equality, Eq. (A5) may be divided by Eq. (A2) to give

As a system of coupled linear equations, Eqs. (A2) and (A6) may be solved to give

Using Eq. (A8), the last equality in Eq. (A4) may be rewritten as

Eq. (A8) shows that a maximum of $E_e$ corresponds to a minimum of $M_c^2$. Now the rest mass $M_c$ is the total energy of the proton and anti-neutrino pair in their center of mass (or momentum) frame so that it achieves the minimum

when the proton and the anti-neutrino are both at rest in the center of mass frame. Hence, from Eqs. (A8) and (A10), the maximum energy of the electron $E = c^2E_e$ is

When Eq. (A10) holds, the proton and the anti-neutrino move with the same velocity $v_m$ of the center of mass and we have

where the last equality follows from Eq. (A3). By Eqs. (A7) and (A9), the last expression in Eq. (A12) may be used to obtain the speed of the anti-neutrino when $E = E_{\max}$. Thus, with $M = m_p+m_v$, we have

[Alternative Solution]

Assume that, in the rest frame of the neutron, the electron comes out with momentum $c\vec{q}_e$ and energy $c^2E_e$, the proton with $c\vec{q}_p$ and $c^2E_p$, and the anti-neutrino with $c\vec{q}_v$ and $c^2E_v$. With the magnitude of vector $\vec{q}_\alpha$ denoted by the symbol $q_\alpha$, we have

Conservation of energy and momentum in the neutron decay leads to

When squared, the last two equations lead to

Subtracting Eq. (5A) from Eq. (4A) and making use of Eq. (1A) then gives

or, equivalently,

If $\theta$ is the angle between $\vec{q}_p$ and $\vec{q}_v$, we have $\vec{q}_p \cdot \vec{q}_v = q_pq_v\cos\theta \le q_pq_v$ so that Eq. (7A) leads to the relation

Note that the equality in Eq. (8A) holds only if $\theta = 0$, i.e., the energy of the electron $c^2E_e$ takes on its maximum value only when the anti-neutrino and the proton move in the same direction.

Let the speeds of the proton and the anti-neutrino in the rest frame of the neutron be $c\beta_p$ and $c\beta_v$, respectively. We then have $q_p = \beta_pE_p$ and $q_v = \beta_vE_v$. As shown in Fig. A1, we introduce the angle $\phi_v$ ($0 \le \phi_v < \pi/2$) for the antineutrino by

Similarly, for the proton, we write, with $0 \le \phi_p < \pi/2$,

Eq. (8A) may then be expressed as

The factor in parentheses at the end of the last equation may be expressed as

and clearly assumes its minimum possible value of 1 when $\phi_p = \phi_v$, i.e., when the anti-neutrino and the proton move with the same velocity so that $\beta_p = \beta_v$. Thus, it follows from Eq. (11A) that the maximum value of $E_e$ is

and the maximum energy of the electron $E = c^2E_e$ is

When the anti-neutrino and the proton move with the same velocity, we have, from Eqs. (9A), (10A), (2A) ,(3A), and (1A), the result

Substituting the result of Eq. (13A) into the last equation, the speed $v_m$ of the anti-neutrino when the electron attains its maximum value $E_{\max}$ is, with $M = m_p+m_v$, given by

Part B

Light Levitation

(b) Refer to Fig. B1. Refraction of light at the spherical surface obeys Snell’s law and leads to

Neglecting terms of the order $(\delta/R)^3$ or higher in sine functions, Eq. (B1) becomes

For the triangle $\Delta FAC$ in Fig. B1, we have

Let $f_0$ be the frequency of the incident light. If $n_p$ is the number of photons incident on the plane surface per unit area per unit time, then the total number of photons incident on the plane surface per unit time is $n_p\pi\delta^2$. The total power $P$ of photons incident on the plane surface is $(n_p\pi\delta^2)(hf_0)$, with $h$ being Planck’s constant. Hence,

The number of photons incident on an annular disk of inner radius $r$ and outer radius $r+dr$ on the plane surface per unit time is $n_p(2\pi rdr)$, where $r = R\tan\theta_i \approx R\theta_i$. Therefore,

The $z$-component of the momentum carried away per unit time by these photons when refracted at the spherical surface is

so that the $z$-component of the total momentum carried away per unit time is

where $\tan\theta_{im} = \dfrac{\delta}{R} \approx \theta_{im}$. Therefore, by the result of Eq. (B5), we have

The force of optical levitation is equal to the sum of the $z$-components of the forces exerted by the incident and refracted lights on the glass hemisphere and is given by

Equating this to the weight $mg$ of the glass hemisphere, we obtain the minimum laser power required to levitate the hemisphere as

Marking Scheme

Theoretical Question 3

Neutrino Mass and Neutron Decay

Light Levitation