Derivations for Two-Body Kinematics (Both Relativistic and Non-Relativistic)

Derivations for two-body kinematics (both relativistic and non-relativistic)

Carl Wheldon June 13, 2019

1 Non-relativistic two-body equations

Starting from energy and momentum conservation, the equations for non-relativistic two- body 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= p sin(θ) p sin(φ) (1) 2 − 3 Energy conservation: E0 + Q0 = E2 + E3 + Ex = Etot. + Ex (2) where Q is the Q-value given by Q = m + m m m , and E is the excitation energy 0 0 0 1 − 2 − 3 x of the particles after the reaction. Rearranging equations 1, gives

2 2 2 2 2 2 (p0 p2 cos(θ)) = p0 + p2 cos (θ) 2p0p2 cos(θ)= p3 cos (φ) − − 2 2 2 2 p2 sin (θ)= p3 sin (φ) (3) Adding Eqns. 3 yields p2 + p2 2p p cos(θ)= p2 (4) 0 2 − 0 2 3 p2 Using energy conservation (Eqn. 2) and that E = 2m , implies

2 2 p2 p3 Etot. = − 2m2 2m3 2 2 p2 p3 =2m3 Etot. (5) ⇒ − 2m2 ! Substituting the above result into Eqn. 4 yields m p2 1+ 3 p (2p cos(θ)) + p2 2m E =0 (6) 2 m − 2 0 0 − 3 tot. 2 b √b2 4ac which can be solved as with any quadratic equation, using p2 = − ± 2a − such that

1 2p cos(θ) 4p2 cos2(θ) 4 1+ m3 (p2 2m E ) 0 ± 0 − m2 0 − 3 tot. p2 = r (7) m3 2 1+ m2 Following on from this result the remaining quantities can be calculated. 2 p2 E2 = 2m2 p φ = arcsin( 2 sin(θ)) (8) p3 E = E E 3 tot. − 2 p3 = 2m3E3 q 1.1 Centre-of-mass frame (aka. centre-of-momentum frame) Starting by calculating the centre-of-mass velocity from momentum conservation i (m0 + m1)vC = m0v0 (9) since the net overall momentum is now zero within the centre-of-mass frame, so only the i momentum of the frame must be calculated. Here, vC 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 i Eqn. 9 for vC gives the expression i v0m0 vC = (10) (m0 + m1) The velocity of the initial particles is given by

i m0 v0m1 vC0 = v0 vC = v0 1 = − − (m0 + m1)! m0 + m1 i i v0m0 vC1 = v1 vC = vC = − (11) − − (m0 + m1) Momentum conservation before and after the collision can be used to calculate the centre- f of-mass velocity in the ﬁnal frame, vC , i f (m0 + m1)vC =(m2 + m3)vC

f (m0 + m1) i vC = vC (12) (m2 + m3) 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, i i E0 = ECin + EC = EC0 + EC1 + EC i ECin = EC0 + EC1 = E0 EC = EC2 + EC3 Q = ECout Q ⇒ 2 − 2 2 − −2 m0vC0 m1 m0v0 m1 EC0 = = = E0 (13) 2 m0 + m1 2 m0 + m1 2 2 2 m1vC1 m0 m1v0 m0m1 EC1 = = = 2 E0 2 m0 + m1 2 (m0 + m1) 2 i 2 i (m0+m1)(vC ) Where EC = 2 . Note that the latter equations from Eqns. 13 are obtained by using the result from Eqns. 11. Now calculating ECin = EC0 + EC1

2 2 m1 m0m1 m1 + m0m1 m1E0 ECin = E0 + 2 E0 = 2 E0 = (14) m0 + m1 (m0 + m1) (m0 + m1) (m0 + m1) Furthermore, E = E + Q E (15) Cout Cin − x 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 2 2 pC2 = pC3 m2v2 m2v2 2 C2 = 3 C3 (16) 2m2 2m2 m3 EC2 = EC3 m2

Now an expression for ECout can be obtained,

m3 m3 (m2 + m3) ECout = EC2 + EC3 = EC3 + EC3 =(1+ )EC3 = EC3 (17) m2 m2 m2

Finally the centre-of-mass angles can be calculated. Note that φ = 180◦ θ since C − C pC2 + pC3 = 0.

ν 2 ν C2 θ θ C νf C

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

v cos(θ )= v cos(θ) vf C2 C 2 − C vC2 sin(θC )= v2 sin(θ) (18) From Eqns. 18, v2 sin(θ) θC = arctan f (19) v cos(θ) v ! 2 − C Alternatively, f v2 cos(θ) vC θC = arccos − (20) v2 !

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 (21) pC2(z) !

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,

v (z)= v (z) vf (22) C2 2 − C Multiplying by m2 yields the momentum:

p (z)= p (z) m vf (23) C2 2 − 2 C f Equations 10 and 12 combined lead to an expression for vC ,

f (m0 + m1) v0m0 v0m0 p0 vC = = = (24) (m2 + m3) (m0 + m1) (m2 + m3) (m2 + m3)

From Eqn. 23, the term, p2(z), can be rewritten by considering momentum conservation in the laboratory frame as in Eqn. 1,

p = p (z)+ p (z) p (z)= p p (z) (25) 0 2 3 ⇒ 2 0 − 3 Substituting the expressions from Eqns. 24 and 25 into Eqn. 23 and manipulating gives,

p0 pC2(z)= p0 p3(z) m2 − − (m2 + m3)

m2 pC2(z)= p0 1 p3(z) (26) − (m2 + m3)! −

Revealing

m3p0 pC2(z)= p3(z) (27) (m2 + m3) −

2 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= p sin(θ) p sin(φ) (28) 2 − 3

4 Energy conservation is written as

2 E0 + m1c + Q = E2 + E3 + Ex = Etot. + Ex (29) where E = T + mc2 and E2 = p2c2 + m2c4. Kinetic energy is represented by T and E is the total energy (kinetic + rest-mass energy). Squaring and adding Eqns. 28 gives,

2 2 2 2 2 2 (p0 p2 cos(θ)) = p0 + p2 cos (θ) 2p0p2 cos(θ)= p3 cos (φ) − − 2 2 2 2 p2 sin (θ)= p3 sin (φ) (30) p2c2 + p2c2 2p p c2 cos(θ)= p2c2 ⇒ 0 2 − 0 2 3 Substituting E2 =(E E )2 into the last line of Eqns. 30 results in 3 tot. − 2 p2c2 + p2c2 2p p c2 cos(θ)=(E E )2 m2c4 = E2 + E2 2E E m2c4 0 2 − 0 2 tot. − 2 − 3 tot. 2 − tot. 2 − 3 p2c2 + E2 m2c4 2p p c2 cos(θ)= E2 + E2 2E E m2c4 0 2 − 2 − 0 2 tot. 2 − tot. 2 − 3 (m2c4 m2c4) 2p p c2 cos(θ)+ p2c2 = E2 2E E (31) 3 − 2 − 0 2 0 tot. − tot. 2 2p p c2 cos(θ)=(m2c4 m2c4)+ p2c2 +2E E E2 0 2 3 − 2 0 tot. 2 − tot. Squaring both sides of the last line in Eqns. 31 leads to

4p2p2c4 cos2(θ) = (m2c4 m2c4 + p2c2 +2E E E2 )2 0 2 3 − 2 0 tot. 2 − tot. (E2 m2c4)4p2c2 cos2(θ) = m4c8 + m4c8 + p4c4 + E4 +4m2c4E2 +4m2c4E E 2 − 2 0 3 2 0 tot. 3 tot. 3 2 tot. 2m2c4p2c2 +2m2c4E2 4m2c4E E 2p2c2E2 − 2 0 2 tot. − 3 2 tot. − 0 tot. +4p2c2E E 4E3 E (32) 0 2 tot. − tot. 2

Now the various terms of Eqns. 32 can be grouped in terms of powers of E2 which results in the quadratic equation

E2(4p2c2 cos2(θ) 4E2 )+ E (4E3 4p2c2E +4m2c4E 4m2c4E ) 2 0 − tot. 2 tot. − 0 tot. 2 tot. − 3 tot. +(2p2c2E2 2m2c4E2 +2m2c4p2c2 +2m2c4E2 2m2c4p2c2 (33) 0 tot. − 2 tot. 2 0 3 tot. − 3 0 +2m2c4m2c4 E4 p4c4 m4c8 m4c8 4m2c4p2c2 cos2(θ)) = 0 3 2 − tot. − 0 − 2 − 3 − 2 0

b √b2 4ac which can be trivially solved using E2 = − ± 2a − . Once E2 is known, the remaining quantities can be calculated, for example, T = E m c2 2 2 − 2 E = E E 3 tot. − 2 T = E m c2 3 3 − 3 p = E2 m2c4 (34) 2 2 − 2 q p = E2 m2c4 3 3 − 3 q p φ = arcsin 2 sin(θ) p3 !

5 The relativistic velocities are given by pc v/c = (35) E

2 1 This equation comes from combining E = γmc with p = γmv where γ = 2 . 1 v −( c ) q 2.1 Relativistic kinematics in the centre-of-mass frame The centre-of-mass velocity in the initial frame is obtained from

i p0 + p1 p0 vC = = (36) E0 + E1 Etot. From the conservation of momentum in the initial and ﬁnal frames (cf. Eqn. 12), the outgoing centre-of-mass velocity can be obtained via

vi vf γi (m + m ) C = γf (m + m ) C C 0 1 c C 2 3 c γi (m + m )vi vf /c C 0 1 C = C (m2 + m3)c f 2 vC 1 c s − γi (m + m )vi 2 (vf /c)2 C 0 1 C = C f 2 (m2 + m3)c ! vC 1 c − γi (m0+m1)vi f 2 C C v (m2+m3)c C = i i (37) c ! γC (m0+m1)vC 1+ (m2+m3)c γi (m0+m1)vi f C C v (m2+m3)c C v = u i i c u γC (m0+m1)vC u1+ u (m2+m3)c u t The centre-of-mass total and kinetic energies can now be obtained

E = γi E p cvi C0 C 0 − 0 C T = E m c2 C0 C0 − 0 E = γi E p cvi (38) C1 C 1 − 1 C T = E m c2 C1 C1 − 1 These equations are obtained using Lorentz’s transformation matrix, e.g. for particle 0:

i i i vC pC0c γC 0 0 γC c p0c 0 0 10− 0 0   =     (39) 0  0 01 0  0  EC0   vi   E0     i C i     c   γC c 0 0 γC   c     −    6 Similar for the outgoing (ﬁnal) frame particles 2 and 3, e.g.

f f f vC pC c cos(φC ) γ 0 0 γ p c cos(φ) 3 C − C c 3 p c sin(φ )  0 10 0  p c sin(φ)  C3 C  =  3  (40) 0  0 01 0  0    f     EC3   f v f   E3   c   γ C 0 0 γ   c     C c C     −  such that

f f vC EC2 = γC E2 p2c cos(θ) − c ! 2 TC2 = EC2 m2c − f f vC EC3 = γC E3 p3c cos(φ) (41) − c ! T = E m c2 C3 C3 − 3 Also, when considering relativistic velocities, Eqns. 18 for example, take the form

i v0 vC vC0 = − i v0vC 1 c2 − i v2 cos(θ) vC vC2 cos(θC )= − i (42) v0 cos(θ)vC 1 2 − c vC2 sin(θC )= v2 sin(θ)

Alternatively, knowing the energy in the centre-of-mass frame

2 4 vC m c = 1 2 (43) c s − EC

pc The above equation (43) is derived from Eqn. 35, v/c = E , as follows

v pc 2 C = s c E v E2 m2c4 C = − c s E2

2 4 2 vC m c = v1 (44) c u − E ! u t Other equations relating to the energies in the centre-of-mass-frame are

TCin = TC0 + TC1

TCout = TCin + Q0 Ex 2 2 − ECin = TCin + m0c + m1c = EC0 + EC1 (45) 2 2 ECout = TCout + m2c + m2c = EC2 + EC3

7 Finally, from Eqn. 40, θC and φC can be calculated. Various formulæ can be found. An example is,

p2c sin(θ) θC = arctan  f  f vC γ (p2c cos(θ) E2)  C − c   φ = 180◦ θ (46) C − C 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 b cos(φ) sin(θ )= , cos(φ)= sin(θ )= , (47) y r b ⇒ y r b further substituting sin(θ)= r leads to

sin(θy) = sin(θ) cos(φ). (48)

Similarly, for θx x x b sin(φ) sin(θ )= , sin(φ)= sin(θ )= x a b ⇒ x a b r and sin(θ)= , sin(θ )= sin(θ) sin(φ). (49) r ⇒ x a a Further substitution of cos(θy)= r , leads to

sin(θ) sin(φ) sin(θx)= . (50) cos(θy)

This can be manipulated using cos(θ )) = (1 sin2(θ )) to yield y − y q sin(θ) sin(φ) sin(θ) sin(φ) sin(θx)= sin(θx)= . (51) 2 2 2 (1 sin (θy)) ⇒ (1 sin (θ) cos (φ) − − q q Alternatively, z z a cos(θ )= , cos(θ)= and cos(θ )= , (52) x a r y r yielding,

8 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(θ) cos(θx)= . (53) cos(θy)

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:

sin(θ ) cos(φ)= y . (55) sin(θ)

The spherical polar angles are obtained from the vector components by:

z x θ = arccos φ = arctan . (56) r y !

9 Note that it is clear from Fig. 2 that the distances x, y and z can be found from

a x = cos(θ ), = sin(θ ) x = r sin(θ ) cos(θ ), r y a x ⇒ x y y = r sin(θy)and (57) z a = cos(θ ), = cos(θ ) z = r cos(θ ) cos(θ ). a x r y ⇒ x x 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’).

The magnitude of the vector is r = (x2 + y2). The projections in the rotated frame can be expressed in terms of the projectionsq in the original frame and the rotation angles, α, x x y ′ = cos(θ α) = cos(θ) cos(α) + sin(θ) sin(α)= cos(α)+ sin(α) and r − r r y y x ′ = sin(θ α) = sin(θ) cos(α) cos(θ) sin(α)= cos(α) sin(α). (58) r − − r − r These two expressions can be written in matrix form:

x cos(α) sin(α) x ′ = . (59) y′ sin(α) cos(α) y ! − ! !

5 Frame rotation in three dimensions

Consider the left-handed coordinate system shown in Fig. 2. The azimuthal angle, φ is de- ﬁned as a rotation from Y towards X about the Z axis. Given this, to be consistent, it is

10 necessary to deﬁne the angle θ as a rotation from Z to Y about X. 1

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) (60)

z′ = Z Secondly, considering a separate rotation through angle θ,

x′ = X

y′ = Y cos(θ) Z sin(θ) (61) − z′ = Y sin(θ) + Z cos(θ)

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′′ 10 0 cos(φ) sin(φ) 0 X − y′′ = 0 cos(θ) sin(θ) sin(φ) cos(φ) 0 Y (62)    −      z 0 sin(θ) cos(θ) 0 0 1 Z  ′′                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  (63) z sin(θ) sin(φ) sin(θ) cos(φ)− cos(θ) Z  ′′            1Note, 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’.

11 Since the above matrix (Eqn. 63) 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 (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′′ (64)    −    Z 0 sin(θ) cos(θ) z      ′′     −    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) (65) z = r cos(θ) Substituting Eqns. 65 into the matrix Eqns. 64 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(θ ) sin(φ ) sin(φ ) + sin(θ ) cos(φ ) cos(θ ) cos(φ ) (66) − R R F R R F F + cos(θR) sin(θF ) cos(φF ) cos(θ) = sin(θ ) cos(φ ) sin(θ ) + cos(θ ) cos(θ ) − R R F R F 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 four- momentum 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.

γ γβx′′ γβy′′ γβz′′ 2 Es β ′′ ′′ E0 x′′ βx βy βx′′ βz′′  γβx′′ 1+(γ 1) 2 (γ 1) 2 (γ 1) 2  px β β β px′′   − − β2 −   = βy′′ βx′′ y′′ βy′′ βz′′ (67) py  ′′  py′′  γβy (γ 1) β2 1+(γ 1) β2 (γ 1) β2    2    − −β ′′ β ′′ − β  ′′  pz   βz′′ βx′′ z y z′′   pz     γβx′′ (γ 1) 2 (γ 1) 2 1+(γ 1) 2       β β β     − − −  12 From the ﬁrst line of Eqn. 67, the Doppler shift equation can be obtained:

Es = γE0 + γ [βx′′ px′′ + βy′′ py′′ + βz′′ pz′′ ] E = γE + γβ~ p~ (68) ⇒ s 0 · ′′ 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. 68 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))

(1 + β cos(θ )) or E = E 0 . (69) s 0 √1 β2 −

Clearly, knowing θ0, β and Es allows a Doppler correction to be performed via √1 β2 E0 = Es − . (70) (1 + β cos(θ0)) 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. 67 and making a transformation from (Es,px,py,pz) to (E0,px′′,py′′,pz′′), such that Eqn. 68 becomes

E = γE γ [β p + β p + β p ] 0 s − x x y y z z E = γE γβ~ ~p (71) ⇒ 0 s − · E = γE γβp cos(θ ). 0 s − s

Making the substitution, p = Es gives the usual Doppler correction formula:

E = γE (1 β cos(θ )) 0 s − s

(1 β cos(θs)) or E = Es − . (72) 0 √1 β2 −

7 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 = Ei/ Ei can be accomplished using a three-axis ‘triangle’ plot, with axes ! Xi 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,

yε1 = ε1 y = ε sin(30) = 1/2ε (73) ε2 − 2 − 2 y = ε sin(30) = 1/2ε , ε3 − 3 − 3 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, (2ε ε ε ) y = 1 − 2 − 3 . (74) 2

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 = tan(30) = hb = . (75) 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 = = . (76) | | −| | √3 − 2√3 2√3

14 The three contributions to x are as follows:

xε1 =0 √3 1 xε2 = ε2 cos(30) x2 = ε2 (77) −| | 2 − 2√3 1 1 √3 xε3 = hb x1 ε3 cos(30) = ε3 , | |−| | − √3 − 2√3 − 2 where the expression for xε3 makes use of the symmetry between xε3 and xε2 . Summing all three contributions yields an expression for x,

√3 x =(ε ε ) . (78) 2 − 3 2