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

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13 Νοε 2013 (πριν από 3 χρόνια και 8 μήνες)

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Derivations for two-body kinematics
(both relativistic and non-relativistic)
Carl Wheldon
October 18,2013
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:
p
0
= p
2
cos(θ) +p
3
cos(φ)
0 = p
2
sin(θ) −p
3
sin(φ) (1)
Energy conservation:
E
0
+Q
0
= E
2
+E
3
+E
x
= E
tot.
+E
x
(2)
where Q
0
is the Q-value given by Q
0
= m
0
+m
1
−m
2
−m
3
,and E
x
is the excitation energy
of the particles after the reaction.Rearranging equations 1,gives
(p
0
−p
2
cos(θ))
2
= p
2
0
+p
2
2
cos
2
(θ) −2p
0
p
2
cos(θ) = p
2
3
cos
2
(φ)
p
2
2
sin
2
(θ) = p
2
3
sin
2
(φ) (3)
Adding Eqns.3 yields
p
2
0
+p
2
2
−2p
0
p
2
cos(θ) = p
2
3
(4)
Using energy conservation (Eqn.2) and that E =
p
2
2m
,implies
E
tot.

p
2
2
2m
2
=
p
2
3
2m
3
⇒ p
2
3
= 2m
3

E
tot.

p
2
2
2m
2
!
(5)
Substituting the above result into Eqn.4 yields
p
2
2

1 +
m
3
m
2

−p
2
(2p
0
cos(θ)) +

p
2
0
−2m
3
E
tot.

= 0 (6)
which can be solved as with any quadratic equation,using p
2
=
−b±

b
2
−4ac
2a
such that
1
p
2
=
2p
0
cos(θ) ±
r
4p
2
0
cos
2
(θ) −4

1 +
m
3
m
2

(p
2
0
−2m
3
E
tot.
)
2

1 +
m
3
m
2

(7)
Following on from this result the remaining quantities can be calculated.
E
2
=
p
2
2
2m
2
φ = arcsin(
p
2
p
3
sin(θ)) (8)
E
3
= E
tot.
−E
2
p
3
=
q
2m
3
E
3
1.1 Centre-of-mass frame (aka.centre-of-momentum frame)
Starting by calculating the centre-of-mass velocity from momentum conservation
(m
0
+m
1
)v
i
C
= m
0
v
0
(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,v
i
C
is the velocity of the centre-of-mass
in the initial frame and v
0
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 Q
0
value).Solving
Eqn.9 for v
i
C
gives the expression
v
i
C
=
v
0
m
0
(m
0
+m
1
)
(10)
The velocity of the initial particles is given by
v
C0
= v
0
−v
i
C
= v
0

1 −
m
0
(m
0
+m
1
)
!
=
v
0
m
1
m
0
+m
1
v
C1
= v
1
−v
i
C
= −v
i
C
=
−v
0
m
0
(m
0
+m
1
)
(11)
Momentum conservation before and after the collision can be used to calculate the centre-
of-mass velocity in the final frame,v
f
C
,
(m
0
+m
1
)v
i
C
= (m
2
+m
3
)v
f
C
v
f
C
=
(m
0
+m
1
)
(m
2
+m
3
)
v
i
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,
E
0
= E
Cin
+E
i
C
= E
C0
+E
C1
+E
i
C
⇒E
Cin
= E
C0
+E
C1
= E
0
−E
i
C
= E
C2
+E
C3
−Q = E
Cout
−Q
E
C0
=
m
0
v
2
C0
2
=

m
1
m
0
+m
1

2
m
0
v
2
0
2
=

m
1
m
0
+m
1

2
E
0
(13)
E
C1
=
m
1
v
2
C1
2
=

m
0
m
0
+m
1

2
m
1
v
2
0
2
=
m
0
m
1
(m
0
+m
1
)
2
E
0
2
Where E
i
C
=
(m
0
+m
1
)(v
i
C
)
2
2
.Note that the latter equations from Eqns.13 are obtained by
using the result from Eqns.11.Now calculating E
Cin
= E
C0
+E
C1
E
Cin
=

m
1
m
0
+m
1

2
E
0
+
m
0
m
1
(m
0
+m
1
)
2
E
0
=
m
2
1
+m
0
m
1
(m
0
+m
1
)
2
E
0
=
m
1
E
0
(m
0
+m
1
)
(14)
Furthermore,
E
Cout
= E
Cin
+Q−E
x
(15)
Now,considering the two out-going particles.The total momentumin the centre-of-mass
frame must be zero.From this point,the energies can be calculated,
p
C2
= p
C3
p
2
C2
= p
2
C3
m
2
2
v
2
C2
2m
2
=
m
2
3
v
2
C3
2m
2
(16)
E
C2
=
m
3
m
2
E
C3
Now an expression for E
Cout
can be obtained,
E
Cout
= E
C2
+E
C3
=
m
3
m
2
E
C3
+E
C3
= (1 +
m
3
m
2
)E
C3
=
(m
2
+m
3
)
m
2
E
C3
(17)
Finally the centre-of-mass angles can be calculated.Note that φ
C
= 180

− θ
C
since
p
C2
+p
C3
= 0.




C

C
f
2
C2
Figure 1:A velocity (v) vector diagram showing the laboratory and centre-of-mass frame
scattering angles,θ and θ
C
respectively.
From Fig.1,the equations for the centre-of-mass frame velocities for the outgoing particles
are
v
C2
cos(θ
C
) = v
2
cos(θ) −v
f
C
v
C2
sin(θ
C
) = v
2
sin(θ) (18)
From Eqns.18,
θ
C
= arctan

v
2
sin(θ)
v
2
cos(θ) −v
f
C
!
(19)
Alternatively,
θ
C
= arccos

v
2
cos(θ) −v
f
C
v
2
!
(20)
3
Further examination of Fig.1 reveals another way of extracting the centre-of-mass scat-
tering angle,θ
C
(often referred to as θ

).
θ
C
= θ

= arctan

p
2
(x)
p
C2
(z)
!
(21)
The quantity p
2
(x) = p
C2
(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
C2
(z) = v
2
(z) −v
f
C
(22)
Multiplying by m
2
yields the momentum:
p
C2
(z) = p
2
(z) −m
2
v
f
C
(23)
Equations 10 and 12 combined lead to an expression for v
f
C
,
v
f
C
=
(m
0
+m
1
)
(m
2
+m
3
)
v
0
m
0
(m
0
+m
1
)
=
v
0
m
0
(m
2
+m
3
)
=
p
0
(m
2
+m
3
)
(24)
From Eqn.23,the term,p
2
(z),can be rewritten by considering momentum conservation in
the laboratory frame as in Eqn.1,
p
0
= p
2
(z) +p
3
(z) ⇒ p
2
(z) = p
0
−p
3
(z) (25)
Substituting the expressions from Eqns.24 and 25 into Eqn.23 and manipulating gives,
p
C2
(z) = p
0
−p
3
(z) −m
2
p
0
(m
2
+m
3
)
p
C2
(z) = p
0

1 −
m
2
(m
2
+m
3
)
!
−p
3
(z) (26)
Revealing
p
C2
(z) =
m
3
p
0
(m
2
+m
3
)
−p
3
(z) (27)
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,
p
0
= p
2
cos(θ) +p
3
cos(φ)
0 = p
2
sin(θ) −p
3
sin(φ) (28)
4
Energy conservation is written as
E
0
+m
1
c
2
+Q = E
2
+E
3
+E
x
= E
tot.
+E
x
(29)
where E = T +mc
2
and E
2
= p
2
c
2
+m
2
c
4
.Kinetic energy is represented by T and E is the
total energy (kinetic + rest-mass energy).
Squaring and adding Eqns.28 gives,
(p
0
−p
2
cos(θ))
2
= p
2
0
+p
2
2
cos
2
(θ) −2p
0
p
2
cos(θ) = p
2
3
cos
2
(φ)
p
2
2
sin
2
(θ) = p
2
3
sin
2
(φ) (30)
⇒ p
2
0
c
2
+p
2
2
c
2
−2p
0
p
2
c
2
cos(θ) = p
2
3
c
2
Substituting E
2
3
= (E
tot.
−E
2
)
2
into the last line of Eqns.30 results in
p
2
0
c
2
+p
2
2
c
2
−2p
0
p
2
c
2
cos(θ) = (E
tot.
−E
2
)
2
−m
2
3
c
4
= E
2
tot.
+E
2
2
−2E
tot.
E
2
−m
2
3
c
4
p
2
0
c
2
+E
2
2
−m
2
2
c
4
−2p
0
p
2
c
2
cos(θ) = E
2
tot.
+E
2
2
−2E
tot.
E
2
−m
2
3
c
4
(m
2
3
c
4
−m
2
2
c
4
) −2p
0
p
2
c
2
cos(θ) +p
2
0
c
2
= E
2
tot.
−2E
tot.
E
2
(31)
2p
0
p
2
c
2
cos(θ) = (m
2
3
c
4
−m
2
2
c
4
) +p
2
0
c
2
+2E
tot.
E
2
−E
2
tot.
Squaring both sides of the last line in Eqns.31 leads to
4p
2
0
p
2
2
c
4
cos
2
(θ) = (m
2
3
c
4
−m
2
2
c
4
+p
2
0
c
2
+2E
tot.
E
2
−E
2
tot.
)
2
(E
2
2
−m
2
2
c
4
)4p
2
0
c
2
cos
2
(θ) = m
4
3
c
8
+m
4
2
c
8
+p
4
0
c
4
+E
4
tot.
+4m
2
3
c
4
E
2
tot.
+4m
2
3
c
4
E
2
E
tot.
−2m
2
2
c
4
p
2
0
c
2
+2m
2
2
c
4
E
2
tot.
−4m
2
3
c
4
E
2
E
tot.
−2p
2
0
c
2
E
2
tot.
+4p
2
0
c
2
E
2
E
tot.
−4E
3
tot.
E
2
(32)
Now the various terms of Eqns.32 can be grouped in terms of powers of E
2
which results in
the quadratic equation
E
2
2
(4p
2
0
c
2
cos
2
(θ) −4E
2
tot.
) +E
2
(4E
3
tot.
−4p
2
0
c
2
E
tot.
+4m
2
2
c
4
E
tot.
−4m
2
3
c
4
E
tot.
)
+(2p
2
0
c
2
E
2
tot.
−2m
2
2
c
4
E
2
tot.
+2m
2
2
c
4
p
2
0
c
2
+2m
2
3
c
4
E
2
tot.
−2m
2
3
c
4
p
2
0
c
2
(33)
+2m
2
3
c
4
m
2
2
c
4
−E
4
tot.
−p
4
0
c
4
−m
4
2
c
8
−m
4
3
c
8
−4m
2
2
c
4
p
2
0
c
2
cos
2
(θ)) = 0
which can be trivially solved using E
2
=
−b±

b
2
−4ac
2a
.
Once E
2
is known,the remaining quantities can be calculated,for example,
T
2
= E
2
−m
2
c
2
E
3
= E
tot.
−E
2
T
3
= E
3
−m
3
c
2
p
2
=
q
E
2
2
−m
2
2
c
4
(34)
p
3
=
q
E
2
3
−m
2
3
c
4
φ = arcsin

p
2
p
3
sin(θ)
!
5
The relativistic velocities are given by
v/c =
pc
E
(35)
This equation comes from combining E = γmc
2
with p = γmv where γ =
1
q
1−
(
v
c
)
2
.
2.1 Relativistic kinematics in the centre-of-mass frame
The centre-of-mass velocity in the initial frame is obtained from
v
i
C
=
p
0
+p
1
E
0
+E
1
=
p
0
E
tot.
(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
γ
i
C
(m
0
+m
1
)
v
i
C
c
= γ
f
C
(m
2
+m
3
)
v
f
C
c
γ
i
C
(m
0
+m
1
)v
i
C
(m
2
+m
3
)c
=
v
f
C
/c
s
1 −

v
f
C
c

2

γ
i
C
(m
0
+m
1
)v
i
C
(m
2
+m
3
)c
!
2
=
(v
f
C
/c)
2
1 −

v
f
C
c

2

v
f
C
c
!
2
=

γ
i
C
(m
0
+m
1
)v
i
C
(m
2
+m
3
)c

1 +

γ
i
C
(m
0
+m
1
)v
i
C
(m
2
+m
3
)c

(37)
v
f
C
c
=
v
u
u
u
u
u
t

γ
i
C
(m
0
+m
1
)v
i
C
(m
2
+m
3
)c

1 +

γ
i
C
(m
0
+m
1
)v
i
C
(m
2
+m
3
)c

The centre-of-mass total and kinetic energies can now be obtained
E
C0
= γ
i
C

E
0
−p
0
cv
i
C

T
C0
= E
C0
−m
0
c
2
E
C1
= γ
i
C

E
1
−p
1
cv
i
C

(38)
T
C1
= E
C1
−m
1
c
2
These equations are obtained using Lorentz’s transformation matrix,e.g.for particle 0:





p
C0
c
0
0
E
C0
c





=






γ
i
C
0 0 −γ
i
C
v
i
C
c
0 1 0 0
0 0 1 0
−γ
i
C
v
i
C
c
0 0 γ
i
C











p
0
c
0
0
E
0
c





(39)
6
Similar for the outgoing (final) frame particles 2 and 3,e.g.





p
C3
c cos(φ
C
)
p
C3
c sin(φ
C
)
0
E
C3
c





=







γ
f
C
0 0 −γ
f
C
v
f
C
c
0 1 0 0
0 0 1 0
−γ
f
C
v
f
C
c
0 0 γ
f
C












p
3
c cos(φ)
p
3
c sin(φ)
0
E
3
c





(40)
such that
E
C2
= γ
f
C

E
2
−p
2
c cos(θ)
v
f
C
c
!
T
C2
= E
C2
−m
2
c
2
E
C3
= γ
f
C

E
3
−p
3
c cos(φ)
v
f
C
c
!
(41)
T
C3
= E
C3
−m
3
c
2
Also,when considering relativistic velocities,Eqns.18 for example,take the form
v
C0
=
v
0
−v
i
C
1 −
v
0
v
i
C
c
2
v
C2
cos(θ
C
) =
v
2
cos(θ) −v
i
C
1 −
v
0
cos(θ)v
i
C
c
2
(42)
v
C2
sin(θ
C
) = v
2
sin(θ)
Alternatively,knowing the energy in the centre-of-mass frame
v
C
c
=
s
1 −
m
2
c
4
E
2
C
(43)
The above equation (43) is derived from Eqn.35,v/c =
pc
E
,as follows
v
C
c
=
s

pc
E

2
v
C
c
=
s
E
2
−m
2
c
4
E
2
v
C
c
=
v
u
u
t
1 −

m
2
c
4
E
!
2
(44)
Other equations relating to the energies in the centre-of-mass-frame are
T
Cin
= T
C0
+T
C1
T
Cout
= T
Cin
+Q
0
−E
x
E
Cin
= T
Cin
+m
0
c
2
+m
1
c
2
= E
C0
+E
C1
(45)
E
Cout
= T
Cout
+m
2
c
2
+m
2
c
2
= E
C2
+E
C3
7
Finally,from Eqn.40,θ
C
and φ
C
can be calculated.Various formulæ can be found.An
example is,
θ
C
= arctan



p
2
c sin(θ)
γ
f
C
(p
2
c cos(θ) −
v
f
C
c
E
2
)



φ
C
= 180

−θ
C
(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,
sin(θ
y
) =
y
r
,cos(φ) =
y
b
⇒sin(θ
y
) =
b cos(φ)
r
,(47)
further substituting sin(θ) =
b
r
leads to
sin(θ
y
) = sin(θ) cos(φ).(48)
Similarly,for θ
x
sin(θ
x
) =
x
a
,sin(φ) =
x
b
⇒sin(θ
x
) =
b sin(φ)
a
and sin(θ) =
b
r
,⇒sin(θ
x
) =
r
a
sin(θ) sin(φ).(49)
Further substitution of cos(θ
y
) =
a
r
,leads to
sin(θ
x
) =
sin(θ) sin(φ)
cos(θ
y
)
.(50)
This can be manipulated using cos(θ
y
)) =
q
(1 −sin
2

y
)) to yield
sin(θ
x
) =
sin(θ) sin(φ)
q
(1 −sin
2

y
))
⇒ sin(θ
x
) =
sin(θ) sin(φ)
q
(1 −sin
2
(θ) cos
2
(φ)
.(51)
Alternatively,
cos(θ
x
) =
z
a
,cos(θ) =
z
r
and cos(θ
y
) =
a
r
,(52)
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 (θ
x
) and out-of-plane (θ
y
)
angles.
cos(θ
x
) =
cos(θ)
cos(θ
y
)
.(53)
Knowing θ
x
and θ
y
,θ can be obtained by manipulating equation 53:
cos(θ) = cos(θ
x
) cos(θ
y
).(54)
Similarly,φ can be obtained by rearranging equation 48:
cos(φ) =
sin(θ
y
)
sin(θ)
.(55)
4 Frame rotation in two dimensions
Below,the transformation matrix connecting two frames,related by a positive rotation
through angle α,is derived.
9
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 =
q
(x
2
+y
2
).The projections in the rotated frame can
be expressed in term so of the projections in the original frame and the rotation angles,α,
x

r
= cos(θ −α) = cos(θ) cos(α) +sin(θ) sin(α) =
x
r
cos(α) +
y
r
sin(α) and
y

r
= sin(θ −α) = sin(θ) cos(α) −cos(θ) sin(α) =
y
r
cos(α) −
x
r
sin(α).(56)
These two expressions can be written in matrix form:

x

y

!
=

cos(α) sin(α)
−sin(α) cos(α)
!
x
y
!
.(57)
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
n
ecessary 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

= Xcos(φ) − Y sin(φ)
y

= Xsin(φ) + Y cos(φ) (58)
z

= Z
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
Secondly,considering a separate rotation through angle θ,
x

= X
y

= Y cos(θ) − Z sin(θ) (59)
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
′′
y
′′
z
′′


 =



1 0 0
0 cos(θ) −sin(θ)
0 sin(θ) cos(θ)






cos(φ) −sin(φ) 0
sin(φ) cos(φ) 0
0 0 1






X
Y
Z


 (60)
Leading to a matrix for the two rotations combined,



x
′′
y
′′
z
′′


 =



cos(φ) −sin(φ) 0
cos(θ) sin(φ) cos(θ) cos(φ) −sin(θ)
sin(θ) sin(φ) sin(θ) cos(φ) cos(θ)






X
Y
Z


 (61)
Since the above matrix (Eqn.61) 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
Y
Z



=



cos(φ) cos(θ) sin(φ) sin(θ) sin(φ)
−sin(φ) cos(θ) cos(φ) sin(θ) cos(φ)
0 −sin(θ) cos(θ)






x
′′
y
′′
z
′′



(62)
It is now possible to go from the Cartesian coordinates to spherical polar coordinates by
using the following expressions for x,y and z,
11
x = r sin(θ) sin(φ)
y = r sin(θ) cos(φ) (63)
z = r cos(θ)
Substituting Eqns.63 into the matrix Eqns.62 yields,
sin(θ) sin(φ) = sin(θ
R
) sin(φ
R
) cos(φ
F
) +sin(θ
R
) cos(φ
R
) cos(θ
F
) sin(φ
F
)
+cos(θ
R
) sin(θ
F
) sin(φ
F
)
sin(θ) cos(φ) = −sin(θ
R
) sin(φ
R
) sin(φ
F
) +sin(θ
R
) cos(φ
R
) cos(θ
F
) cos(φ
F
) (64)
+cos(θ
R
) sin(θ
F
) cos(φ
F
)
cos(θ) = −sin(θ
R
) cos(φ
R
) sin(θ
F
) +cos(θ
R
) cos(θ
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-
momentumis used as an example [taken fromWikipedia,“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 E
s
and E
0
is the energy
in the frame of the nucleus,i.e.unshifted.





E
s
p
x
p
y
p
z





=








γ γβ
x
′′
γβ
y
′′
γβ
z
′′
γβ
x
′′ 1 +(γ −1)
β
2
x
′′
β
2
(γ −1)
β
x
′′
β
y
′′
β
2
(γ −1)
β
x
′′
β
z
′′
β
2
γβ
y
′′
(γ −1)
β
y
′′ β
x
′′
β
2
1 +(γ −1)
β
2
y
′′
β
2
(γ −1)
β
y
′′ β
z
′′
β
2
γβ
x
′′
(γ −1)
β
z
′′ β
x
′′
β
2
(γ −1)
β
z
′′ β
y
′′
β
2
1 +(γ −1)
β
2
z
′′
β
2













E
0
p
x
′′
p
y
′′
p
z
′′





(65)
From the first line of Eqn.65,the Doppler shift equation can be obtained.
E
s
= γE
0
+γ [β
x
′′ p
x
′′ +β
y
′′ p
y
′′ +β
z
′′ p
z
′′ ]
⇒E
s
= γE
0

~
β ∙
~
p
′′
(66)
E
s
= γE
0
+γβp
′′
cos(θ
0
)
The angle,θ
0
is between the direction of β and the unshifted γ ray,i.e.between p
′′
and β.
Making a substitution into Eqn.66 for the magnitude of the momentum vector,p
′′
= E
0
for γ rays (p
′′
is in units of MeV in the four-vector notation).Therefore,the Doppler shift
formula is,
12
E
s
= γE
0
(1 +β cos(θ
0
))
or E
s
= E
0
(1 +β cos(θ
0
))

1 −β
2
.(67)
Clearly,knowing θ
0
,β and E
s
allows a Doppler correction to be performed via
E
0
= E
s

1 −β
2
(1 +β cos(θ
0
))
(68)
However,it is more common to use the inverse transformation using the Lorentz Boost
because it is usually the angle,θ
s
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.65 and making
a transformation from (E
s
,p
x
,p
y
,p
z
) to (E
0
,p
′′
x
,p
′′
y
,p
′′
z
),such that Eqn.66 becomes
E
0
= γE
s
−γ [β
x
p
x

y
p
y

z
p
z
]
⇒E
0
= γE
s
−γ
~
β ∙ ~p (69)
E
0
= γE
s
−γβp cos(θ
s
)
Making the substitution,p = E
s
gives the usual Doppler correction formula:
E
0
= γE
s
(1 −β cos(θ
s
))
or E
0
= E
s
(1 −β cos(θ
s
))

1 −β
2
.(70)
13