IPHO 2003
Solution − Theoretical Question 3
(prelimary)
p.1/5
Solution- Theoretical Question 3
Part A
Neutrino Mass and Neutron Decay



(a) Let (c 2 Ee , cqe ) , (c 2 E p , cq p ) , and (c 2 Ev , cqv ) 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 , q e , q p , q 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 2 Ec , and total momentum cqc . Thus, we have
Ec  E p  Ev ,



qc  q p  qv ,
M c2  Ec2  qc2
(A1)

Note that the magnitude of the vector q c is denoted as qc. The same convention
also applies to all other vectors.
Since energy and momentum are conserved in the neutron decay, we have
E c  E e  mn
(A2)


(A3)
qc  qe
When squared, the last equation leads to the following equality
qc2  qe2  Ee2  me2
(A4)
From Eq. (A4) and the third equality of Eq. (A1), we obtain
Ec2  M c2  Ee2  me2
(A5)
With its second and third terms moved to the other side of the equality, Eq. (A5)
may be divided by Eq. (A2) to give
1
(A6)
Ec  Ee 
( M c2  me2 )
mn
As a system of coupled linear equations, Eqs. (A2) and (A6) may be solved to give
1
(A7)
Ec 
(mn2  me2  M c2 )
2mn
Ee 
1
(mn2  me2  M c2 )
2mn
(A8)
Using Eq. (A8), the last equality in Eq. (A4) may be rewritten as
qe 
1
2m n
1

2m n
(mn2  me2  M c2 ) 2  (2mn m e ) 2
(A9)
(mn  m e  M c )( mn  me  M c )( mn  me  M c )( mn  me  M c )
Eq. (A8) shows that a maximum of Ee corresponds to a minimum of M c2 .
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
IPHO 2003
Solution − Theoretical Question 3
(prelimary)
M  m p  mv
p.2/5
(A10)
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 = c2Ee is
c2
Emax 
mn2  me2  (m p  mv ) 2  1.292569 MeV  1.29 MeV (A11)*1
2mn


When Eq. (A10) holds, the proton and the anti-neutrino move with the same
velocity vm of the center of mass and we have
qp
vm
q
q
q
(A12)
 ( v ) | E  Emax  ( ) | E  Emax  ( c ) | E  Emax  ( e ) | M c m p mv
c
Ev
Ep
Ec
Ec
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 = Emax. Thus, with M = mp+mv, we have
(mn  me  M )( mn  me  M )( mn  me  M )( mn  me  M )
vm

c
mn2  me2  M 2
(A13)*
 0.00126538  0.00127
-----------------------------------------------------------------------------------------------------[Alternative Solution]
Assume that, in the rest frame of the neutron, the electron comes out with


momentum cqe and energy c2Ee, the proton with cq p and c 2 E p , and the


anti-neutrino with cqv and c 2 Ev . With the magnitude of vector q denoted by
the symbol q, we have
E 2p  m 2p  q 2p ,
Ev2  mv2  qv2 ,
Ee2  me2  qe2
(1A)
Conservation of energy and momentum in the neutron decay leads to
E p  Ev  mn  Ee



q p  qv   qe
(2A)
(3A)
When squared, the last two equations lead to
E 2p  Ev2  2 E p Ev  (mn  Ee ) 2
 
q 2p  qv2  2q p  qv  qe2  Ee2  me2
(4A)
(5A)
Subtracting Eq. (5A) from Eq. (4A) and making use of Eq. (1A) then gives
 
m 2p  mv2  2( E p Ev  q p  qv )  mn2  me2  2mn Ee
(6A)
IPHO 2003
Solution − Theoretical Question 3
(prelimary)
p.3/5
or, equivalently,
 
2mn Ee  mn2  me2  m 2p  mv2  2( E p Ev  q p  qv )
(7A)

 

If  is the angle between q p and q v , we have q p  qv  q p qv cos   q p qv so that
Eq. (7A) leads to the relation
2mn Ee  mn2  me2  m 2p  mv2  2( E p Ev  q p qv )
(8A)
Note that the equality in Eq. (8A) holds only if  = 0, i.e., the energy of the electron
c2Ee 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 p and c v , respectively. We then have q p   p E p and qv   v Ev . As
shown in Fig. A1, we introduce the angle  v ( 0   v   / 2 ) for the antineutrino by
qv  mv tan  v ,
Ev  mv2  qv2  mv sec  v ,
 v  qv / Ev  sin  v
(9A)
Ev
qv
Figure A1
v
mv
Similarly, for the proton, we write, with 0   p   / 2 ,
q p  m p tan  p ,
E p  m 2p  q 2p  m p sec  p ,
 p  q p / E p  sin  p
(10A)
Eq. (8A) may then be expressed as
1  sin  p sin  v
2mn Ee  mn2  me2  m 2p  mv2  2m p mv (
)
cos  p cos  v
(11A)
The factor in parentheses at the end of the last equation may be expressed as
1  sin  p sin  v
cos  p cos  v

1  sin  p sin  v  cos  p cos  v
cos  p cos  v
1 
1  cos( p   v )
cos  p cos  v
 1  1 (12A)
and clearly assumes its minimum possible value of 1 when  p =  v, i.e., when the
anti-neutrino and the proton move with the same velocity so that  p =  v. Thus, it
follows from Eq. (11A) that the maximum value of Ee is
( Ee ) max 
1
(mn2  me2  m 2p  mv2  2m p mv )
2m n
1

[mn2  me2  (m p  mv ) 2 ]
2m n
1
An equation marked with an asterisk contains answer to the problem.
(13A)*
IPHO 2003
Solution − Theoretical Question 3
(prelimary)
p.4/5
and the maximum energy of the electron E = c2Ee is
Emax  c 2 ( Ee ) max  1.292569 MeV  1.29 MeV
(14A)*
When the anti-neutrino and the proton move with the same velocity, we have,
from Eqs. (9A), (10A), (2A) ,(3A), and (1A), the result
qp
q p  qv
Ee2  me2
qv
qe
v   p 




E p E v E p  E v mn  E e
mn  E e
(15A)
Substituting the result of Eq. (13A) into the last equation, the speed vm of the
anti-neutrino when the electron attains its maximum value Emax is, with M = mp+mv,
given by
( Ee ) 2max  me2
vm
 (  v ) max Ee 

c
mn  ( Ee ) max

(mn2  me2  M 2 ) 2  4mn2 me2
2mn2  (mn2  me2  M 2 )
(mn  me  M )( mn  me  M )( mn  me  M )( mn  me  M )
(16A)*
mn2  me2  M 2
 0.00126538  0.00127
------------------------------------------------------------------------------------------------------
Part B
Light Levitation
(b) Refer to Fig. B1. Refraction of light at the spherical surface obeys Snell’s law and
leads to
n sin  i  sin  t
(B1)
Neglecting terms of the order ( /R)3or higher in sine functions, Eq. (B1) becomes
n i   t
(B2)
For the triangle FAC in Fig. B1, we have
   t   i  n i   i  (n  1) i
(B3)
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  2 . The total power P of photons
incident on the plane surface is (n p )( hf 0 ) ,
2
with h being Planck’s constant. Hence,
P
(B4)
np 
 2 hf 0
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rdr ) , where
z
F
t

A
i
n
i
C

Fig. B1
IPHO 2003
Solution − Theoretical Question 3
(prelimary)
p.5/5
r  R tan  i  R i . Therefore,
n p (2rdr )  n p (2R 2 ) i d i
(B5)
The z-component of the momentum carried away per unit time by these photons
when refracted at the spherical surface is
hf o
hf
2
(2rdr ) cos   n p 0 (2R 2 )(1 
) d
c
c
2 i i
hf
(n  1) 2 3
 n p 0 (2R 2 )[ i 
 i ] d i
c
2
dFz  n p
(B6)
so that the z-component of the total momentum carried away per unit time is
hf 0  im
(n  1) 2 3
Fz  2R n p (
) [ i 
 i ] d i
c 0
2
hf
(n  1) 2 2
2
 R 2 n p ( 0 ) im
[1 
 im ]
c
4
2
where tan  im 

R
Fz 
(B7)
  im . Therefore, by the result of Eq. (B5), we have
(n  1) 2  2
(n  1) 2  2
R 2 P hf 0  2
P
(
)
[
1

]

[
1

]
c
 2 hf 0 c R 2
4R 2
4R 2
(B8)
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
(n  1) 2  2
(n  1) 2  2 P
P
P P
(B9)
 ( Fz )   [1 
]
c
c c
c
4R 2
4R 2
Equating this to the weight mg of the glass hemisphere, we obtain the minimum
laser power required to levitate the hemisphere as
P
4mgcR 2
(n  1) 2  2
(B10)*