University of Jyväskylä
Helsinki Institute of Physics
All you need to know about π 0 decay
kinematics
Authors:
J. Rak
December 21, 2009
Contents
1 Opening angle between the two decay photons
2
2 Photons in π 0 rest frame
3
3 Momentum distribution of decay photons
4
4 Direct photons, isolation cut, fake rate due to π 0 decay
7
4.1
Phase space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
A APPENDIX
9
11
A.1 Note on the Taylor expansion in powers of 2/γ 2 . . . . . . . . . . . . . . . .
11
A.2 Random variable rxF according given distribution F (x)
11
. . . . . . . . . . .
B Bit more about kinematics
12
B.1 Beam Rapidity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
B.2 Spherical geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
1
1
Opening angle between the two decay photons
The Lorentz boost of the decay photon can be from the π 0 rest frame RF into LAB can
be decomposed into two components:
mπ
(cos θ∗ ± β)
2
mπ
=
sin θ∗
2
mπ
=
(1 ± β cos θ∗ )
2
LAB
E±,||
= γ(E||RF ± βE RF ) = γ
E⊥LAB = E⊥RF
E±LAB
mπ
=
2
q
γ 2 (cos θ∗ ± β)2 + sin2 θ∗
(1)
where E|| and E⊥ refer to a longitudinal and perpendicular component of the photon energy
with respect to the π 0 momentum. It is useful to define an asymmetry parameter
E+ − E− α = E+ + E− → α = β cos θ∗
from (1)
(2)
It is easy to see that from Eq. (1) and Eq. (2) follows
1
E+ E− = Eπ (1 − α2 )
4
mπ
E+=
2
(3)
E+
π0
*
θ
θL
EFigure 1: The π 0 → 2γ decay kinematics. The γ momenta were calculated for Eπ = 250
MeV according Eq. (1).
Let us evaluate θL , the opening angle between two decay photons emitted from the decay
of π 0 of the energy Eπ . The opening angle can be derived from the usual four-momentum
algebra:
2
π→ 2γ opening angle [rad]
π→ 2γ opening angle [deg]
α = 0.0
α = 0.5
α = 0.7
α = 0.9
0.4
0
0.35
10
0.3
0.25
pi energy
1.5 GeV
2.0 GeV
3.0 GeV
5.0 GeV
15.0 GeV
2/γ
0.2
0.15
0.1
0.05
1
0
2
4
6
0
0
8 10 12 14 16 18 20
pTπ [GeV/c]
0.2
0.4
0.6
asymmetry α
0.8
1
Figure 2: The π 0 → 2γ opening angle as a function π 0 total energy (at y = 0 Eπ = pT π ) for
different values of asymmetry α (left) and π 0 energy (right).
m2π = (G+ + G− )2 = 2E+ E− − 2E+ E− cos θL → cos θL =
2E+ E− − m2π
2E+ E−
(4)
From this equation and taking into account Eq. (1) one can write
cos θL =
γ 2 (1 − α2 ) − 2
γ 2 (β 2 − α2 ) − 1
Eπ2 (1 − α2 ) − 2m2π
=
=
γ 2 (1 − α2 )
γ 2 (1 − α2 )
Eπ2 (1 − α2 )
(5)
where the γ 2 − β 2 γ 2 = 1 was used. For an opening angle distributions evaluated for various
α and Eπ see Fig. 2. It is worthwhile to notice, that for the small values of α one can expand
p
cos−1 expression Eq. (5) in powers of 2/γ 2 and find (see App. A.1)
θL |α→0 ≈
2
2
γ
(6)
Photons in π 0 rest frame
For the π 0 decay study one has to generate two photons of energy mπ /2 with a random
3D angular distribution. It means, one has to generate two random angles 0 < θ < π and
0 < φ < 2π uniformly covering the mπ /2 sphere. One way of generating a random point
3
on the sphere is to generate two uniform random numbers: −R < z < R and 0 < φ < 2π .
The second angle θ can be then evaluated as θ = cos−1 (z/R). The four-vectors of the two
photons in the π 0 rest frame can be written as
mπ
(1, sin θ cos φ, sin θ sin φ, cos θ)
2
mπ
=
(1, − sin θ cos φ, − sin θ sin φ, − cos θ)
2
GRF
=
+
GRF
−
which, obviously, obeys (G+ + G− )2 = m2π
3
Momentum distribution of decay photons
The θ∗ decay angle (see Fig. 1) is randomly distributed. However, when taking into ac-
count also the azimuthal angle φ (2D→3D) then the number of photons per θ∗ gets Jacobian
dNγ = sin θ∗ dθ∗
and from Eq. (1) it is easy to see that
dEγ
1
= ∓ Eπ β sin θ∗
∗
dθ
2
Using the chain rule we see that
dNγ dNγ dθ∗
2
=
=
∗
dEγ y=0
dθ dEγ
pT π
(7)
According Eq. (7) the decay photon distribution is flat. Since asymmetry α = β cos θ∗ it
is quite easy to see that dNγ /dα is also flat (constant with α). This uniform α distribution
is truncated from the upper side if the the Eπ m0 doesn’t hold (the negative photon
momentum component wrt π 0 for small θ∗ is comparable to m0 /2) or if there is minimum
Eγ which can be detected (always the case ).
4
10
1
0.9
1
10-1
10-2
π, n=7.03±0.00
0.8
γ , n=7.04±0.01
0.7
[a.u.]
1/Nev dN/dpT (GeV/c)-1
y=0, input n=7.0
-3
0.6
0.5
Rγ ,π
10
10-4
-5
10
0.4
0.3
0.2
-6
10
0.1
-7
10 2
4
6
8
10
p [GeV/c]
12
0
0
14
2
4
T
6
8
10
p [GeV/c]
12
14
T
2
0.3
0.25
1
0.15
0.1
Rγ ,π
pT π
0.2
dN/dEγ |
1.5
2/pT,π
[a.u.]
(GeV/c)-1
Figure 3: The π 0 distribution generated according pT ∝ p−7
T and the decay photon distribution (left panel). The ratio of decay photon to the mother π 0 distribution, Rγ,π (right panel).
The solid line is the 0th-order polynomial fit, the dashed line represents an expectation value
according Eq. (9).
pT,π
0.5
2/7.0
0.05
0
0
2
4
6
8
0
0
10 12 14 16 18 20
p
T
[GeV/c]
2
4
6
8
10
p [GeV/c]
12
14
T
Figure 4: Decay photon energy distribution for Gaussian rapidity distribution −1 < y < 1 of
the width σy =2 and fixed pT π0 =10 GeV/c (left). The ratio of decay photon to the mother
π 0 distribution, Rγ,π for −1 < y < 1 (right panel). The solid line is the 0th-order polynomial
fit, the dashed line represents an expectation value according Eq. (9).
5
1/Nev dN/dpT (GeV/c)-1
10
-0.35<y0.35; σy=2, input n=7.0
1
π, n=7.01±0.02
γ , n=7.00±0.04
-1
[a.u.]
10
1
10-2
Rγ ,π
10-3
10-4
0.5
-5
10
2/7.0
10-6
10-72
4
6
8
10
p [GeV/c]
12
0
0
14
2
T
4
6
8
10
p [GeV/c]
12
14
T
Figure 5: The π 0 distribution generated according pT ∝ p−7
T and Gaussian rapidity of the
width σy =2 together the decay photon distribution (left panel). The ratio of decay photon
to the mother π 0 distribution, Rγ,π (right panel). The solid line is the 0th-order polynomial
fit, the dashed line represents an expectation value according Eq. (9).
Knowing Eq. (7) one can evaluate the decay photon distribution provided the π 0 pT
distribution has power law form dNπ /dpT ≈ p−n
T
Z √s/2
2
2 1
dNγ =
dpT π0 = Eγ−n
n
dEγ π0
n
pT =Eγ pT π pT π
(8)
and thus the double ratio of π 0 decay photon to π 0 distribution
Rγ,π0 =
dNγ /dEγ
2
= .
dNπ0 /dpπ0
n
(9)
Comment: Power n usually refers to the invariant cross section dσ/pT dpT ∝ pT−n . In
the analysis about we used the probability distribution dN/dpT ∝ pT−n+1 of power of n − 1.
So in literature (e.g. [1]) you will find
Rγ,π0 =
2
n−1
instead of Eq. (9). An example for n = 7 distribution is shown on Fig. 3.
6
(10)
It is important to notice that this is valid only for midrapidity distribution where Eπ =
pT π . For the more general case where π 0 is not emitted parallel to the beam axis the simple
scaling Eq. (9) doesn’t hold. Decay photon energy distribution corresponding to the more
realistic Gaussian distribution −1 < y < 1 of the width σy =2 and fixed pT π0 =10 GeV/c is
shown on the left panel of Fig. 4. Below the π 0 threshold the energy distribution of decay
photons is flat as it is in the case of y = 0. However, there is tail above the π 0 pT coming
from π 0 decays with significant longitudinal component where the Eπ > pT π and the Lorentz
factor in Eq. (1) is larger then those corresponding the π 0 pT . The right panel of Fig. 4
displays the Rγ,π ratio for the π 0 rapidity distribution −1 < y < 1 and pT ∝ p−7
T .
In the case of small rapidity acceptance calorimeters as PHENIX and PHOS in ALICE
Eq. (9) serves a good approximation (see Fig. 5)
4
Direct photons, isolation cut, fake rate due to π 0
decay
Collinear π 0 decays (α → 0) are the source of the “fake” direct photons signal. The
anti-collinear photon has a vanishing negative momentum component wrt π 0 momentum
and cannot be detected by a calorimeter with minimum energy threshold Emin typically of
the order of 100-200 MeV. According Eq. (7) the energy distribution of decay photon for
fixed π 0 momentum is constant and hence the fraction of single-detected decay photons Rfake
is equal to
Rfake =
Emin
pT π
(11)
Folded with the π 0 power law distribution p−n
T we obtained
dN
Emin
= n+1
dEfake
pT π
(12)
Simple simulation of π 0 decay and detection of single decay photon is shown on Fig. 6
where the decay photons were counted only when the second photon had a energy below
7
Emin . These decay photon cannot be distinguished from the decay photons.
10
-0.00<y<0.00; σy=2, input n=7.0
1
1/Nev dN/dp (GeV/c)-1
1/Nev dN/dp (GeV/c)-1
10
π, n=7.03±0.01
single decay γ Eγ >0.2
10-1
Eγ ,min/p n+1
T,pi
-2
10-1
Eγ ,min/p n+1
T,pi
-2
10
-3
10-3
10
10-4
10-4
-5
10
-5
10
10-6
10-70
π, n=7.03±0.01
single decay γ Eγ >0.1
T
T
10
-0.35<y<0.35; σy=2, input n=7.0
1
10-6
2
4
6
8
10
p [GeV/c]
12
10-70
14
T
2
4
6
8
10
p [GeV/c]
12
14
T
Figure 6: Monte carlo simulated of π 0 decay of p−7
T distribution (filled triangles) and the
photon pT distribution for decays when only one decay photon had energy larger then 200
MeV (open circles) for midrapidity (left panel). The dashed line corresponds to calculation
according Eq. (12). Left panel shows the same simulation compared to Eq. (12) for the
PHENIX acceptance and lower energy threshold Emin = 100 MeV.
Knowing the “fake” rate Rfake and using the theoretical NLO estimation of direct photon
to π 0 yield [2] (left panel of Fig. 7) one can then evaluate the yield of decay photons when
the second photons is not detected (E− < Emin ) per one direct photon (right panel of Fig. 7)
as
Emin NLO π 0
Nisolated decay γ
=
(13)
direct γ
pT NLO direct γ
√
One can see that for p + p s=200 GeV at photon momentum 6 GeV/c the probability
that the isolated photon is a direct one is 50% and to get below reasonable fake rate 10%
the isolated photon momentum must be larger than 13 GeV/c for Emin =200 MeV or larger
than 10 GeV/c for Emin =100 MeV.
For the LHC case where the relative ratio of direct photon to π 0 is smaller as compared
to RHIC, the 50% probability of isolated photon being the direct one is around 10-15 GeV/c
and in order to get below 10% fake rate on has to study a isolated photons of momenta
above 20-40 GeV/c depending on the Emin settings.
8
single decay γ /direct γ
10
1
10-1
10-2
10-3
p+p
p+p
p+p
p+p
10-4
s=200 GeV, E =0.2 GeV
min
s=200 GeV, Emin=0.1 GeV
s=14 TeV, E =0.2 GeV
min
s=14 TeV, Emin=0.1 GeV
10
p [GeV/c]
102
T
Figure 7: NLO estimation of direct photon to π 0 yield [2] (left panel). Number of isolated
decay
photon per one direct photon calculated according Eq. (13) for p + p collisions at
√
s=200 GeV and 14 TeV and two energy threshold Emin =100 and 200 MeV.
4.1
Phase space
There s always possibility asymmetric π 0 decay where one decay photon misses the detector or its energy falls below the detection threshold. From Eq. (1) it is easy to see the
opening angle between π 0 and decay photon, θ, can be expressed as
cos θ =
cos θ∗ + β
β − cos θ
→ cos θ∗ =
∗
1 + β cos θ
β cos θ − 1
(14)
Using the same arguments as in section 3 one sees that
∗
dNγ
d cos θ∗
(1 − β 2 ) sin θ
∗ dθ
= sin θ
=−
=
dθ
dθ
dθ
(β cos θ − 1)2
An example of opening angle distributions is shown Fig. 8
9
(15)
102
π0
π0
π0
π0
10
1
mom = 1 GeV/c
mom = 3 GeV/c
mom = 10 GeV/c
mom = 30 GeV/c
dN/dθ
10-1
10-2
10-3
10-4
10-5
10-6
0
0.5
1
1.5
2
θ=π∠γ
Figure 8:
10
2.5
3
A
APPENDIX
A.1
Note on the Taylor expansion in powers of 2/γ 2
Opening angle for small asymmetry can be written (see Eq. (5))
2
θL α→0 = arccos 1 − 2
γ
(16)
when one try to expand arccos(1 − x) for x → 0 it is not possible because
1
1
dx arccos(1 − x) = p
≈ √ →∞
2
2x
1 − (1 − x)
for
x→0
However, one can expand
√
2x
2
dx arccos(1 − x2 ) = p
≈√
= 2
2 − x2
1 − (1 − x2 )2
and since x =
for
x→0
p
2/γ 2 the first non-zero Taylor term is 2/γ.
An easier way is to expand directly cos θL = 1 − θL2 /2 and since
2
cos θL α→0 = 1 − 2
γ
(17)
it is easy to see that θL α→0 = 2/γ.
A.2
Random variable rxF according given distribution F (x)
If the F (x) function has an analytic primitive function then the random variable rxF can
be found by solving the integral equation
R rxF
R0,1 = aR b
a
F (x0 )dx0
F (x0 )dx0
where R0,1 is the uniform random number 0 < R0,1 < 1.
11
Exponential function F (x) = e−αx and a < x < b
1
rxexp = − log e−αa − R0.1 (e−αa − e−αb )
α
Power law function F (x) = x−n
1/(1−n)
rxpowerlaw = a1−n + R0,1 ∗ (b1−n − a1−n )
(18)
Example: Let’s assume one has to generate a realistic particle distribution. Usually
some given rapidity and pT distributions are required. One can then generate a pT according
e.g. Eq. (18) and y according Gaussian. The remaining component p|| can be evaluated as
p|| = mT sinh y.
B
Bit more about kinematics
Let us prove, for the record
E = mT cosh y
p|| = mT sinh y
where mT =
p
m20 + p2T → m0 for particles traveling along the beam. First of all, let us
remind an algebra of hyperbolic functions (see Fig. 9):
1
sinh2 x = (e2x − 2 + e−2x ) =
4
1
2
cosh x = (e2x + 2 + e−2x ) =
4
12
1
(cosh 2x − 1)
2
1
(cosh 2x + 1)
2
This implies important equalities:
sinh2 x + cosh2 x = cosh 2x
cosh2 x − sinh2 x = 1
(19)
1
1
sinh x cosh x = (e2x − e−2x ) = sinh 2x
4
2
5
hypergeom. fcn
4
3
2
1
0
cosh(x)
sinh(x)
acosh(x)
asinh(x)
ln(2x)
-1
-2
-1
0
1
2
3
4
5
x
Figure 9: Hypergeometric function and their inverse. Dotted black line shows the log(2x)
x → ∞ approximation of acosh(x) and asinh(x).
One can easily derive a relation between the particle energy E and rapidity y
p
p
(E + p|| )2 + (E − p|| )2
2E
q
=
exp(y) + exp(−y) =
= 2 cosh(y)
mT
E 2 − p2||
13
or alternatively and more complicated
exp(2y) + exp(−2y) =
1
(2m2T + 4p2|| ) = 2 cosh(2y)
2
mT
and thus
2p2||
= 2 sinh2 y + 1 ⇒ p|| = mT sinh y
q
q
2
2
E = mT + p|| = mT 1 + sinh2 y = mT cosh y
cosh(2y) = 1 +
B.1
m2T
(20)
(21)
Beam Rapidity
Some people prefer to express a beam rapidity for given center of mass energy
e.g. [3]) as
√
s (see
√
yb = acosh
s
.
2
(22)
It follows from simple consideration
yb =
√
Eb + pb
Eb + pb
1
log
= log
≈ log 2Eb = log s.
2
Eb − pb
m0
(23)
With help of identities Eq. (19) it is easy to see that
p
cosh + sinh = exp(x) ⇒ x = log(cosh + cosh2 −1)
(24)
and thus
acosh(x) = log(x +
√
x2 − 1) and asinh(x) = log(x +
√
x2 + 1)
acosh(x) ≈ asinh(x) ≈ log(2x) for x 1
√
√
√
s
s
yb ≈ acosh
≈ asinh
≈ log s
2
2
14
(25)
where log
√
s is the best approximation for yb quantity because it follows directly from
Eq. (23).
B.2
Spherical geometry
Some frequently used identities are listed in the appendix.
Figure 10: http://mathworld.wolfram.com/SphericalCoordinates.html
p
x2 + y 2 + z 2 x = r cos θ sin φ
y = r sin θ sin φ
θ = tan−1 xy
φ = cos−1 zr
z = r cos φ
r =
Since we always define the z-axes to be collinear with the beam, we just need to swap
θ ↔ φ and thus we will use:
p
x2 + y 2 + z 2 x = r sin θ cos φ
y = r sin θ sin φ
θ = cos−1 zr
φ = tan−1 xy
z = r cos θ
r =
The line element
ds = drr̂ + rdφφ̂ + r sin φdθθ̂
15
The area element
da = r2 sin φdθdφr̂
The volume element
dV = r2 sin φdφdθdr
Jacobian
∂(x, y, z) 2
∂(r, φ, θ) = r sin φ
References
[1] T. Ferbel and W. R. Molzon, Rev. Mod. Phys. 56, 181 (1984).
[2] F. Arleo et al., (2004), hep-ph/0311131.
[3] D. G. d’Enterria, (2007), 0708.0551.
16