Derivations for two-body kinematics
(both relativistic and non-relativistic)
Carl Wheldon
October 27, 2015
1
Non-relativistic two-body equations
Starting from energy and momentum conservation, the equations for non-relativistic twobody reactions are derived both in the laboratory frame and the centre-of-mass frame. The
projectile and target are particles 0 and 1 respectively. The ejectile (scattered projectile-like
species) and recoil (target-like) are particles 2 and 3 respectively.
Momentum conservation:
p0 = p2 cos(θ) + p3 cos(φ)
0 = p2 sin(θ) − p3 sin(φ)
(1)
Energy conservation:
E0 + Q0 = E2 + E3 + Ex = Etot. + Ex
(2)
where Q0 is the Q-value given by Q0 = m0 + m1 − m2 − m3 , and Ex is the excitation energy
of the particles after the reaction. Rearranging equations 1, gives
(p0 − p2 cos(θ))2 = p20 + p22 cos2 (θ) − 2p0 p2 cos(θ) = p23 cos2 (φ)
p22 sin2 (θ) = p23 sin2 (φ)
(3)
Adding Eqns. 3 yields
p20 + p22 − 2p0 p2 cos(θ) = p23
Using energy conservation (Eqn. 2) and that E =
p2
2m
(4)
, implies
p22
p23
Etot. −
=
2m2
2m3
!
2
p
2
2
⇒ p3 = 2m3 Etot. −
2m2
(5)
Substituting the above result into Eqn. 4 yields
p22
m3
1+
− p2 (2p0 cos(θ)) + p20 − 2m3 Etot. = 0
m2
which can be solved as with any quadratic equation, using p2 =
1
√
−b± b2 −4ac
2a
(6)
such that
p2 =
2p0 cos(θ) ±
r
4p20 cos2 (θ) − 4 1 +
2 1+
m3
m2
m3
m2
(p20 − 2m3 Etot. )
(7)
Following on from this result the remaining quantities can be calculated.
p22
E2 =
2m2
p2
φ = arcsin( sin(θ))
p3
E3 = Etot. − E2
p3 =
1.1
q
(8)
2m3 E3
Centre-of-mass frame (aka. centre-of-momentum frame)
Starting by calculating the centre-of-mass velocity from momentum conservation
(m0 + m1 )vCi = m0 v0
(9)
since the net overall momentum is now zero within the centre-of-mass frame, so only the
momentum of the frame must be calculated. Here, vCi is the velocity of the centre-of-mass
in the initial frame and v0 is the velocity of the beam in the laboratory frame. This will be
different to that in the final frame due to the mass change (i.e. non-zero Q0 value). Solving
Eqn. 9 for vCi gives the expression
v0 m 0
vCi =
(10)
(m0 + m1 )
The velocity of the initial particles is given by
vC0 = v0 −
vCi
vC1
!
v0 m 1
m0 + m1
−v0 m0
= v1 − vCi = −vCi =
(m0 + m1 )
= v0
m0
1−
(m0 + m1 )
=
(11)
Momentum conservation before and after the collision can be used to calculate the centreof-mass velocity in the final frame, vCf ,
(m0 + m1 )vCi = (m2 + m3 )vCf
(m0 + m1 ) i
vCf =
v
(m2 + m3 ) C
(12)
Using the above expressions it is now possible to write down the equations for the energies
of the particles in the centre-of-mass frame,
⇒ ECin
E0 = ECin + ECi = EC0 + EC1 + ECi
= EC0 + EC1 = E0 − ECi = EC2 + EC3 − Q = ECout − Q
2
2
2
m0 v02
m0 vC0
m1
m1
E0
=
=
EC0 =
2
m0 + m1
2
m0 + m1
2
2
m1 vC1
m1 v02
m0 m1
m0
EC1 =
=
=
E0
2
m0 + m1
2
(m0 + m1 )2
2
(13)
(m +m )(v i )2
Where ECi = 0 21 C . Note that the latter equations from Eqns. 13 are obtained by
using the result from Eqns. 11. Now calculating ECin = EC0 + EC1
ECin
m1
=
m0 + m1
2
E0 +
m0 m1
m21 + m0 m1
m1 E 0
E
=
E0 =
0
2
2
(m0 + m1 )
(m0 + m1 )
(m0 + m1 )
(14)
Furthermore,
ECout = ECin + Q − Ex
(15)
Now, considering the two out-going particles. The total momentum in the centre-of-mass
frame must be zero. From this point, the energies can be calculated,
pC2 = pC3
p2C2 = p2C3
2
m22 vC2
m2 v 2
= 3 C3
2m2
2m2
m3
EC2 =
EC3
m2
(16)
Now an expression for ECout can be obtained,
ECout = EC2 + EC3 =
m3
m3
(m2 + m3 )
EC3 + EC3 = (1 +
)EC3 =
EC3
m2
m2
m2
(17)
Finally the centre-of-mass angles can be calculated. Note that φC = 180◦ − θC since
pC2 + pC3 = 0.
ν2
θC
θ
νCf
νC2
Figure 1: A velocity (v) vector diagram showing the laboratory and centre-of-mass frame
scattering angles, θ and θC respectively.
From Fig. 1, the equations for the centre-of-mass frame velocities for the outgoing particles
are
vC2 cos(θC ) = v2 cos(θ) − vCf
vC2 sin(θC ) = v2 sin(θ)
(18)
From Eqns. 18,
Alternatively,
v2 sin(θ)
θC = arctan
v2 cos(θ) − vCf
!
(19)
v2 cos(θ) − vCf
θC = arccos
v2
!
(20)
3
Further examination of Fig. 1 reveals another way of extracting the centre-of-mass scattering angle, θC (often referred to as θ∗ ).
p2 (x)
θC = θ = arctan
pC2 (z)
∗
!
(21)
The quantity p2 (x) = pC2 (x) is the x component of momentum and is the same in both
the laboratory and centre-of-mass frames (Eqn. 18 and Fig. 1). The z component in the
centre-of-mass frame can be obtained from looking at the projections onto the beam-axis in
Fig. 1, such that,
vC2 (z) = v2 (z) − vCf
(22)
Multiplying by m2 yields the momentum:
pC2 (z) = p2 (z) − m2 vCf
(23)
Equations 10 and 12 combined lead to an expression for vCf ,
vCf =
(m0 + m1 ) v0 m0
v0 m 0
p0
=
=
(m2 + m3 ) (m0 + m1 )
(m2 + m3 )
(m2 + m3 )
(24)
From Eqn. 23, the term, p2 (z), can be rewritten by considering momentum conservation in
the laboratory frame as in Eqn. 1,
p0 = p2 (z) + p3 (z) ⇒ p2 (z) = p0 − p3 (z)
(25)
Substituting the expressions from Eqns. 24 and 25 into Eqn. 23 and manipulating gives,
pC2 (z) = p0 − p3 (z) − m2
pC2 (z) = p0
p0
(m2 + m3 )
!
m2
− p3 (z)
1−
(m2 + m3 )
(26)
Revealing
pC2 (z) =
2
m3 p0
− p3 (z)
(m2 + m3 )
(27)
Relativistic two-body equations
As with Section 1, the equations for velocities, scattering angles and energies will be derived,
but this time using the relativistic formalism. The equations for momentum conservation
remain the same as,
p0 = p2 cos(θ) + p3 cos(φ)
0 = p2 sin(θ) − p3 sin(φ)
4
(28)
Energy conservation is written as
E0 + m1 c2 + Q = E2 + E3 + Ex = Etot. + Ex
(29)
where E = T + mc2 and E 2 = p2 c2 + m2 c4 . Kinetic energy is represented by T and E is the
total energy (kinetic + rest-mass energy).
Squaring and adding Eqns. 28 gives,
(p0 − p2 cos(θ))2 = p20 + p22 cos2 (θ) − 2p0 p2 cos(θ) = p23 cos2 (φ)
p22 sin2 (θ) = p23 sin2 (φ)
⇒ p20 c2 + p22 c2 − 2p0 p2 c2 cos(θ) = p23 c2
(30)
Substituting E32 = (Etot. − E2 )2 into the last line of Eqns. 30 results in
2
p20 c2 + p22 c2 − 2p0 p2 c2 cos(θ) = (Etot. − E2 )2 − m23 c4 = Etot.
+ E22 − 2Etot. E2 − m23 c4
2
p20 c2 + E22 − m22 c4 − 2p0 p2 c2 cos(θ) = Etot.
+ E22 − 2Etot. E2 − m23 c4
2
− 2Etot. E2
(m23 c4 − m22 c4 ) − 2p0 p2 c2 cos(θ) + p20 c2 = Etot.
2 2
2
2
2 4
2 4
2p0 p2 c cos(θ) = (m3 c − m2 c ) + p0 c + 2Etot. E2 − Etot.
(31)
Squaring both sides of the last line in Eqns. 31 leads to
2
4p20 p22 c4 cos2 (θ) = (m23 c4 − m22 c4 + p20 c2 + 2Etot. E2 − Etot.
)2
4
2
(E22 − m22 c4 )4p20 c2 cos2 (θ) = m43 c8 + m42 c8 + p40 c4 + Etot.
+ 4m23 c4 Etot.
+ 4m23 c4 E2 Etot.
2
2
−2m22 c4 p20 c2 + 2m22 c4 Etot.
− 4m23 c4 E2 Etot. − 2p20 c2 Etot.
3
+4p20 c2 E2 Etot. − 4Etot.
E2
(32)
Now the various terms of Eqns. 32 can be grouped in terms of powers of E2 which results in
the quadratic equation
2
3
E22 (4p20 c2 cos2 (θ) − 4Etot.
) + E2 (4Etot.
− 4p20 c2 Etot. + 4m22 c4 Etot. − 4m23 c4 Etot. )
2
2
2
+(2p20 c2 Etot.
− 2m22 c4 Etot.
+ 2m22 c4 p20 c2 + 2m23 c4 Etot.
− 2m23 c4 p20 c2
(33)
2 4 2 4
4
4 4
4 8
4 8
2 4 2 2
2
+2m3 c m2 c − Etot. − p0 c − m2 c − m3 c − 4m2 c p0 c cos (θ)) = 0
√
2
which can be trivially solved using E2 = −b± 2ab −4ac .
Once E2 is known, the remaining quantities can be calculated, for example,
T2 = E 2 − m2 c2
E3 = Etot. − E2
T3 = E 3 − m3 c2
p2 =
p3 =
q
q
E22 − m22 c4
E32 − m23 c4
p2
φ = arcsin
sin(θ)
p3
5
!
(34)
The relativistic velocities are given by
v/c =
pc
E
(35)
This equation comes from combining E = γmc2 with p = γmv where γ = q
2.1
1
1−( vc )
2
.
Relativistic kinematics in the centre-of-mass frame
The centre-of-mass velocity in the initial frame is obtained from
vCi =
p0 + p1
p0
=
E0 + E1
Etot.
(36)
From the conservation of momentum in the initial and final frames (cf. Eqn. 12), the
outgoing centre-of-mass velocity can be obtained via
vCf
vCi
f
= γC (m2 + m3 )
+ m1 )
c
c
f
i
i
v /c
γC (m0 + m1 )vC
= s C 2
(m2 + m3 )c
vf
1 − cC
γCi (m0
γCi (m0 + m1 )vCi
(m2 + m3 )c
!2
vCf
c
!2
=
1−
f
vC
c
i (m +m )v i
γC
0
1 C
(m2 +m3 )c
1+
v
u
u
f
u
vC
u
=u
t
c
=
(vCf /c)2
1+
i (m +m )v i
γC
0
1 C
(m2 +m3 )c
i (m +m )v i
γC
0
1 C
(m2 +m3 )c
2
i (m +m )v i
γC
0
1 C
(m2 +m3 )c
(37)
The centre-of-mass total and kinetic energies can now be obtained
EC0 = γCi E0 − p0 cvCi
EC1
TC0 =
E − m0 c2
C0
= γCi E1 − p1 cvCi
(38)
TC1 = EC1 − m1 c2
These equations are obtained using Lorentz’s transformation matrix, e.g. for particle 0:





pC0 c
0
0
EC0
c






=



γCi
0
0
vi
−γCi cC
0
1
0
0
6
0 −γCi
0
0
1
0
0
γCi
i
vC
c






p0 c
0
0
E0
c





(39)
Similar for the outgoing (final) frame particles 2 and 3, e.g.





pC3 c cos(φC )
pC3 c sin(φC )
0
EC3
c


f
γCf





=



0
0
−γCf
f
vC
c
0 0
1 0
0 1
v
−γCf cC
0 0
γCf
0
0


p3 c cos(φ)

  p c sin(φ)
 3


0

E
3
c





(40)
such that
EC2 =
γCf
EC3 = γCf
vf
E2 − p2 c cos(θ) C
c
TC2 = EC2 − m2 c2
!
vCf
E3 − p3 c cos(φ)
c
TC3 = EC3 − m3 c2
!
(41)
Also, when considering relativistic velocities, Eqns. 18 for example, take the form
v0 − vCi
vC0 =
v vi
1 − 0c2C
v2 cos(θ) − vCi
vC2 cos(θC ) =
(42)
v cos(θ)v i
1 − 0 c2 C
vC2 sin(θC ) = v2 sin(θ)
Alternatively, knowing the energy in the centre-of-mass frame
vC
=
c
s
1−
m2 c4
EC2
The above equation (43) is derived from Eqn. 35, v/c =
vC
=
c
vC
=
c
vC
=
c
v
u
u
t
s
s
pc
E
(43)
pc
,
E
as follows
2
E 2 − m2 c4
E2
m2 c4
1−
E
!2
(44)
Other equations relating to the energies in the centre-of-mass-frame are
TCin = TC0 + TC1
TCout = TCin + Q0 − Ex
ECin = TCin + m0 c2 + m1 c2 = EC0 + EC1
ECout = TCout + m2 c2 + m2 c2 = EC2 + EC3
7
(45)
Finally, from Eqn. 40, θC and φC can be calculated. Various formulæ can be found. An
example is,

θC = arctan 

p2 c sin(θ)
γCf (p2 c cos(θ)
f
vC
E2 )
c
◦



−
φC = 180 − θC
(46)
The equations derived in this document are used in the program ckin.c for calculating
two-body kinematics.
3
Coordinate transformations
Below, the equations linking Cartesian and spherical polar angles are derived as defined in
Fig. 2. Starting from,
y
y
sin(θy ) = , cos(φ) =
r
b
b
further substituting sin(θ) = r leads to
⇒ sin(θy ) =
b cos(φ)
,
r
sin(θy ) = sin(θ) cos(φ).
(47)
(48)
Similarly, for θx
x
x
b sin(φ)
, sin(φ) =
⇒ sin(θx ) =
a
b
a
r
b
and sin(θ) = , ⇒ sin(θx ) = sin(θ) sin(φ).
r
a
sin(θx ) =
(49)
Further substitution of cos(θy ) = ar , leads to
sin(θx ) =
This can be manipulated using cos(θy )) =
sin(θ) sin(φ)
sin(θx ) = q
(1 − sin2 (θy ))
sin(θ) sin(φ)
.
cos(θy )
q
(50)
(1 − sin2 (θy )) to yield
⇒ sin(θx ) = q
sin(θ) sin(φ)
(1 − sin2 (θ) cos2 (φ)
.
(51)
Alternatively,
z
z
a
cos(θx ) = , cos(θ) =
and cos(θy ) = ,
a
r
r
yielding,
8
(52)
Figure 2: Angle definitions. The beam axis lies on the z-axis. Spherical polar angles θ and
φ are shown for the vector r, as well as the Cartesian in-plane (θx ) and out-of-plane (θy )
angles.
cos(θx ) =
cos(θ)
.
cos(θy )
(53)
Knowing θx and θy , θ can be obtained by manipulating equation 53:
cos(θ) = cos(θx ) cos(θy ).
(54)
Similarly, φ can be obtained by rearranging equation 48:
cos(φ) =
sin(θy )
.
sin(θ)
(55)
The spherical polar angles are obtained from the vector components by:
z
θ = arccos
r
!
x
φ = arctan
.
y
9
(56)
4
Frame rotation in two dimensions
Below, the transformation matrix connecting two frames, related by a positive rotation
through angle α, is derived.
Figure 3: Frame rotation definitions. The original frame (x,y) is rotated by a positive angle,
α, resulting in frame (x’,y’).
q
The magnitude of the vector is r = (x2 + y 2 ). The projections in the rotated frame can
be expressed in terms of the projections in the original frame and the rotation angles, α,
x
y
x′
= cos(θ − α) = cos(θ) cos(α) + sin(θ) sin(α) = cos(α) + sin(α) and
r
r
r
y
x
y′
= sin(θ − α) = sin(θ) cos(α) − cos(θ) sin(α) = cos(α) − sin(α).
r
r
r
(57)
These two expressions can be written in matrix form:
x′
y′
5
!
=
cos(α) sin(α)
− sin(α) cos(α)
!
x
y
!
.
(58)
Frame rotation in three dimensions
Consider the left-handed coordinate system shown in Fig. 2. The azimuthal angle, φ is defined as a rotation from Y towards X about the Z axis. Given this, to be consistent, it is
necessary to define the angle θ as a rotation from Z to Y about X. 1
1
Note, that defining θ in another way leads to inconsistencies when combining this with the φ rotation.
Since φ is defined such that it cycles ‘backwards’ from Y to X, then θ must similarly be defined as cycling
‘backwards’ from Z to Y , rather than, say, from Z to X, which would cycle ‘forwards’.
10
The new axis, x′ , y ′ and z ′ can now be defined. Firstly, following a rotation through angle
φ,
x′ = X cos(φ) − Y sin(φ)
y ′ = X sin(φ) + Y cos(φ)
z′ = Z
(59)
Secondly, considering a separate rotation through angle θ,
x′ = X
y ′ = Y cos(θ) − Z sin(θ)
z ′ = Y sin(θ) + Z cos(θ)
(60)
Figure 4: Frame rotation definitions in three dimensions. Left: the original frame (x, y, z)
is rotated by a positive angle, φ, about the z-axis resulting in frame (x′ , y ′ , z ′ ). Right: this
frame is subsequently rotated by a positive angle θ about the x′ axis resulting in frame
(x′′ , y ′′ , z ′′ ).
As these two rotations have been defined in a consistent manner (i.e. cyclically φ: y→x,
and θ: z→y), the resulting rotation matrices can be multiplied together,
X
cos(φ) − sin(φ) 0
x′′
1
0
0



 ′′ 

cos(φ) 0 
 Y 
 y  =  0 cos(θ) − sin(θ)   sin(φ)
Z
0
0
1
0 sin(θ)
cos(θ)
z ′′






(61)
Leading to a matrix for the two rotations combined,
X
cos(φ)
− sin(φ)
0
x′′



 ′′ 
 y  =  cos(θ) sin(φ) cos(θ) cos(φ) − sin(θ)   Y 
Z
sin(θ) sin(φ) sin(θ) cos(φ)
cos(θ)
z ′′





(62)
Since the above matrix (Eqn. 62) is unitary and all the elements are real, the inverse matrix,
to transform from the rotated frame (x′′ , y ′′ , z ′′ ) frame to the original, non-rotated frame
11
(X, Y , Z), can be written as the transpose,
x′′
cos(φ) cos(θ) sin(φ) sin(θ) sin(φ)
X
  ′′ 



 Y  =  − sin(φ) cos(θ) cos(φ) sin(θ) cos(φ)   y 
z ′′
0
− sin(θ)
cos(θ)
Z





(63)
It is now possible to go from the Cartesian coordinates to spherical polar coordinates by
using the following expressions for x, y and z,
x = r sin(θ) sin(φ)
y = r sin(θ) cos(φ)
z = r cos(θ)
(64)
Substituting Eqns. 64 into the matrix Eqns. 63 yields,
sin(θ) sin(φ) = sin(θR ) sin(φR ) cos(φF ) + sin(θR ) cos(φR ) cos(θF ) sin(φF )
+ cos(θR ) sin(θF ) sin(φF )
sin(θ) cos(φ) = − sin(θR ) sin(φR ) sin(φF ) + sin(θR ) cos(φR ) cos(θF ) cos(φF )
+ cos(θR ) sin(θF ) cos(φF )
cos(θ) = − sin(θR ) cos(φR ) sin(θF ) + cos(θR ) cos(θF )
(65)
where the subscript R denotes angles in the rotated frame and the F subscript indicates the
angles by which the frame itself has been rotated. The angles in the original, non-rotated
frame have no subscripts.
6
Lorentz transformation in an arbitrary direction.
Considering a frame moving with velocity, β, in an arbitrary direction to the observers
frame, but for which the x, y and z axes coincide (i.e. a non-rotated inertial frame), the
following Lorentz transformation matrix can be applied to any four-vector. Here, the fourmomentum is used as an example [taken from Wikipedia, “Lorentz Transformation” accessed
04/07/2013]. Note that for a ‘boost’ β is used, but the matrix equations used below are
for a transformation from the inertial frame (′′ ) to the observers frame and, therefore, a
substitution of −β has been made. The Doppler shifted energy is Es and E0 is the energy
in the frame of the nucleus, i.e. unshifted.





Es
px
py
pz


γ

 γβx′′



=


 γβy′′

γβx′′
γβx′′
β2
1 + (γ − 1) βx2′′
β ′′ β ′′
(γ − 1) y β 2 x
(γ − 1) βz′′ββ2 x′′
γβy′′
(γ −
γβz′′
β ′′ β ′′
1) x β 2 y
βy2′′
1 + (γ − 1)
(γ − 1)
12
β2
βz′′ βy′′
β2
(γ − 1) βx′′ββ2 z′′
β ′′ β ′′
(γ − 1) y β 2 z
β2
1 + (γ − 1) βz2′′


E0


  px′′

  p ′′
y


pz′′





(66)
From the first line of Eqn. 66, the Doppler shift equation can be obtained:
Es = γE0 + γ [βx′′ px′′ + βy′′ py′′ + βz′′ pz′′ ]
⇒ Es = γE0 + γ β~ · p~′′
(67)
′′
Es = γE0 + γβp cos(θ0 ).
The angle, θ0 is between the direction of β and the unshifted γ ray, i.e. between p′′ and β.
Making a substitution into Eqn. 67 for the magnitude of the momentum vector, p′′ = E0
for γ rays (p′′ is in units of MeV in the four-vector notation). Therefore, the Doppler shift
formula is,
Es = γE0 (1 + β cos(θ0 ))
or Es = E0
(1 + β cos(θ0 ))
√
.
1 − β2
(68)
Clearly, knowing θ0 , β and Es allows a Doppler correction to be performed via
√
1 − β2
.
E0 = Es
(1 + β cos(θ0 ))
(69)
However, it is more common to use the inverse transformation using the Lorentz Boost
because it is usually the angle, θs between the the shifted γ ray (p) and β that is known
(measured). Using the Lorentz boost involves changing the sign of β in Eqn. 66 and making
a transformation from (Es , px , py , pz ) to (E0 , p′′x , p′′y , p′′z ), such that Eqn. 67 becomes
E0 = γEs − γ [βx px + βy py + βz pz ]
⇒ E0 = γEs − γ β~ · p~
(70)
E0 = γEs − γβp cos(θs ).
Making the substitution, p = Es gives the usual Doppler correction formula:
E0 = γEs (1 − β cos(θs ))
or E0 = Es
7
(1 − β cos(θs ))
√
.
1 − β2
(71)
Triangle plots of three-alpha break-up.
Plotting relative or fractional
energies, εi , for three-break-up particles in the centre-of-mass
!
frame
εi = E i /
X
Ei
can be accomplished using a three-axis ’triangle’ plot, with axes
i
13
Figure 5: Left; A three-axis ’triangle’ plot. The three energy axes are labelled by εi . Right;
same plot with features used in the derivation highlighted. In the centre the three energies
are equal, contributing 1/3 each.
ε1,2,3 , as shown in Fig. 5. Each of the three parameters, εi , has a maximum of 1 at a vertex
and a minimum of 0 where it meets the side at the mid-point.
In order to make such a plot, expressions for the Cartesian coordinates, x and y, can be
derived. Taking the base of the triangle as y = 0,
y ε1 = ε 1
yε2 = −ε2 sin(30) = −1/2ε2
yε3 = −ε3 sin(30) = −1/2ε3 ,
(72)
where yε2 is denoted by the vertical blue-dashed line in Fig. 5 (left) and yε2 = yε3 . Summing
the three components together yields an expression for y,
y=
(2ε1 − ε2 − ε3 )
.
2
(73)
For x, zero lies at the centre of the base of the triangle. Expressions are needed for the
distances x1 and x2 on Fig. 5 (right). Firstly, the length, hb , on the red triangle on Fig. 5
(right) is,
√
1/3
3
1
(74)
= tan(30) =
⇒ hb = √ .
hb
3
3
The length, hb is half the length of the base of the large triangle and also the hypotenuse of
the blue triangle. This can be used to obtain the magnitudes of x1 and x2 :
1
|x1 | = hb cos(60) = √
2 3
1
1
1
|x2 | = hb − |x1 | = √ − √ = √ .
3 2 3
2 3
14
(75)
The three contributions to x are as follows:
x ε1 = 0
3
1
xε2 = ε2 cos(30) − |x2 | = ε2
− √
2
2 3
√
1
1
3
xε3 = |hb | − |x1 | − ε3 cos(30) = √ − √ − ε3
,
2
3 2 3
√
(76)
where the expression for xε3 makes use of the symmetry between xε3 and xε2 . Summing all
three contributions yields an expression for x,
x = (ε2 − ε3 )
15
√
3
.
2
(77)