Intro. Relativistic Heavy
Ion Collisions
Kinematic Variables
Manuel Calderón de la Barca Sánchez
Following C.Y. Wong’s Intro. to Rel. Heavy Ion
Collisions
Which follows: Bjorken & Drell, “Relativistic
Quantum Mechanics”.
Natural units:
= c =1
Space-time 4-position: xm = (x , x , x , x ) = (t, x) = (t, x, y, z)
Energy-momentum, 4-momentum:
0
1
2
3
pm = ( p0 , p1, p2 , p 3 ) = (E, p) = (E, pT , pz ) = (E, px , py , pz )
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In many processes, a detected particle can be
identified as originating from a parent particle.
Example: reaction a +b ® c + X
c: detected particle
b: beam, a: target
If detected particle c can be described as
originating from the beam particle b:
projectile fragmentation reaction
Region of momentum of c where this reaction is
dominant: “projectile fragmentation region”
Similar considerations for “target fragmentation
region”, if c originates from a.
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Beam axis: longitudinal axis, z-axis.
Use same symbol to represent particle and its 4-momentum
c = (c0 ,cT ,cz )
Define the forward light-cone momentum: c+ º c0 + cz
And the backward light-cone momentum: c º c - c
0
z
For an energetic particle traveling in
forward direction, along beam (+z-axis): c+ is large, c- small
backward direction (-z-axis):c- is large, c+ small
If the daughter particle c is considered to be fragmenting
from the parent particle b, define the forward light cone
variable:
c+ c0 + cz
x+ º
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b+
=
b0 + bz
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Under a Lorentz boost:
æ
ö æ
¢
ç c0 ÷ = ç g
ç c ÷ ç -bg
è 1 ø è
So:
-bg öæ c0
֍
g ÷øçè cz
ö
÷
÷
ø
c+¢ = c¢0 + c¢z = (g - bg )(c0 + cz ) = g (1- b )c+
The forward-light cone momentum transforms under a boost
by multiplying by a constant factor.
Hence:
c+¢ g (1- b )c+ c+
¢
x+ = =
= = x+
b+¢ g (1- b )b+ b+
The forward light-cone variable is Lorentz invariant
It also has the property: 0 < x+ < 1
The RHS is valid for daughter-parent relationship
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Similarly for the backward light-cone variable, for a daughter
particle
c whose parent is the target a.
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If the particle c is detected as a free particle in the detector
Suffers no additional interactions
4-momentum components obey the mass-shell condition:
Free particle is said to be “on mass shell”
Energy and momentum obey the usual special relativistic equation for
the invariant mass. c 2 = c02 - c 2 = mc2
4-momentum only has 3 degrees of freedom, e.g. if b is known: (x+ ,cT )
Other cases: the particle b can be considered as a composite
system, containing particle c, plus many others.
Particle c is not a free particle in this case
It is subject to interactions with other parts X.
4-momentum will not obey the mass-shell condition
Particle c is said to be “off the mass shell”, or “off-shell”.
4-momentum has 4 degress of freedom, e.g. (x ,c 2 ,c
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+
T
)
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Express c0 and cz in terms of x+ and c2.
We have c0 + cz = x+ (b0 +bz )
(c0 + cz )(c0 - cz ) c02 - cz2 c 2 + cT2
And:
c0 - cz =
=
=
c0 + cz
c0 + cz
c0 + cz
Add the two equations, divide by 2:
1æ
c 2 + cT2 ö
c0 = ç x+ (b0 + bz )+
÷
2è
x+ (b0 + bz ) ø
Subtract the two equations, divide by 2:
1æ
c 2 + cT2 ö
cz = ç x+ (b0 + bz ) ÷
2è
x+ (b0 + bz ) ø
With these, we can express anything in terms of x+, c2, cT
Note: For large, positive boosts: x+ ~ 1, and then x- ~ 0.
Vice versa for large, negative boosts.
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But if we use c0 and cz, these are both large.
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Express the relation between the differential
elements dc0dcT dcz and dx+dcT dc2
The Jacobian is:
é
ê
ê
J =ê
ê
êë
¶c0
¶x+
¶cz
¶x+
é
2
2
c
+
c
¶c0 ù
T
ê (b0 + bz ) - 2
ú
x+ (b0 + bz )
¶c 2 ú 1 ê
=
ê
¶cz ú 2 ê
c 2 + cT2
ú
(b + b )+
¶c 2 úû
ê 0 z x+2 (b0 + bz )
ë
ù
1
ú
x+ (b0 + bz ) ú
ú
-1
ú
x+ (b0 + bz ) úû
Using the Jacobian determinant, we get
1
dc0 dcz =| J | dx+ dc =
4
2
æ
ö æ
öæ
c 2 + cT2 öæ
-1
1
c 2 + cT2 ö
ç(b0 + bz ) - 2
֍
÷-ç
֍(b0 + bz )+ 2
÷
x
(b
+
b
)
x
(b
+
b
)
x
(b
+
b
)
x+ (b0 + bz ) ø
è
+
0
z øè +
0
z ø è +
0
z øè
1æ 1
c 2 + cT2
1
c 2 + cT2 ö 1 dx+ 2
= ç- + 3
- dc
÷=
4 è x+ x+ (b0 + bz )2 x+ x+3 (b0 + bz )2 ø 2 x+
Useful for change of variables to x+, c2.
Lorentz invariant quantities.
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Definition of xF:
czCM
xF º CM
cz (max)
Recall, we are considering reactions like: a +b ® c + X
Working in the CM system, let’s express czCM (max) in terms
of masses and the CM collision energy, √s.
Important: maximum longitudinal momentum happens when
the undetected component ‘X’ is just a single particle, with
mass mX, that can take the lowest value consistent with the
conservation laws.
CM
For this case: c
= -x CM and s = c + x = c (max) + m + c (max) + m
CM
0
Solve for czCM(max):
CM
z
c
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(max) =
CM
0
s 2 + mc4 + mx4 - 2(smc2 + smx2 + mc2 mx2 )
2 s
CM
z
º
2
2
c
CM
z
2
2
x
l (s, mc , mx )
2 s
9
Plug in this formula into our definition of xF:
czCM = xF
l (s, mc , mx )
2 s
To construct x+, we need cz and also the energy, c0.
CM
0
c
= m +c +c
2
c
2
T
CM
z
xF2 l 2 (s, mc2 , mx2 )
= m +c +
4s
2
c
2
T
We also need b0 and bz. (From the parent, beam particle)
In the CM system: b CM = -aCM and s = b + a = (b ) + m + (b ) + m
Solving for bzCM, as before: b = l (s,2m s, m )
Using this to find (b ) = (b ) + (b ) + m = (s + m4s- m )
CM
0
CM
z
CM 2
0
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CM 2
T
CM 2
z
a
2
b
CM
0
CM 2
z
2
b
CM 2
z
2
a
b
2
b
2 2
a
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Now we can write the forward light-cone variable
c0CM + czCM
x+ = CM CM
b0 + bz
In the high-energy limit, where
s
and
also
c (max) »
2
s
ma ,mb , mc , mx
CM
z
CM
0
b
+b
CM
z
=
s + mb2 - ma2 + s 2 + ma4 + mb4 - 2(sma2 + smb2 + ma2 mb2 )
2 s
»
s + mb2 - ma2 + s 1- 2ma2 / s - 2mb2 / s
2 s
æ ma2 ö
s + mb2 - ma2 + s - ma2 - mb2 2s - 2ma2
»
=
= s ç1- ÷
s ø
2 s
2 s
è
We can write
2
æ mc2 + mx2 ö 4mcT
æ mc2 + mx2 ö
2
xF ç1+ xF ç1÷ +
÷
s
s
s
è
ø
è
ø
1- ma2 / s
2
c0CM + czCM
x+ = CM CM »
b0 + bz
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For very high energies: x+ » xF
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Remarks:
The forward light-cone variable is always positive
xF can be zero or negative
So the two variables can differ substantially.
Only when the two are close to 1 are they
approximately equal.
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1 æ p0 + pz ö 1 æ p+ ö
y = ln ç
÷ = ln ç ÷
2 è p0 - pz ø 2 è p- ø
Definition:
Advantages of rapidity:
v
c
Non-relativistic limit:
Nice Properties under Lorentz boosts
y»
If we know the rapidity y in frame S, then the rapidity y’
in frame S’ moving with velocity b in the z-direction is:
1 æç p+¢ ö÷ 1 æ g (1- b )p+ ö 1 æ p+ ö 1 æ 1- b ö
y¢ = ln ç ÷ = ln ç
÷ = ln ç ÷ + ln ç
÷
2 è p-¢ ø 2 è g (1+ b ) p- ø 2 è p- ø 2 è 1+ b ø
1 æ p+ ö 1 æ 1+ b ö
= ln ç ÷ - ln ç
÷ = y - yb
2 è p- ø 2 è 1- b ø
A particle traveling at velocity b in frame S would
have rapidity yb.
So we can call yb the rapidity of frame S’
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Consider a particle moving only in the z direction
Energy: p0 = g m , momentum: pz = bg m
Its Rapidity is therefore:
1 æ p0 + pz ö 1 æ g m + bg m ö 1 æ 1+ b ö
y = ln ç
÷ = ln ç
÷ = ln ç
÷
2 è p0 - pz ø 2 è g m - bg m ø 2 è 1- b ø
=b+
b3 b5
3
+
5
+
b7
7
+ O(b 9 ) » b
So in the non-relativistic limit, the rapidity is the
same as the velocity, in units of the speed of light
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From the definition of rapidity:
e =
y
p0 + pz
and e-y =
p0 - pz
p0 - pz
p0 + pz
Add the equations, divide by 2:
æ
ö
p0 - pz ö 1 ç p0 + pz + p0 - pz ÷ æç
ey + e-y 1 æ p0 + pz
p0
÷÷ =
= çç
+
=ç
2
2
÷
2
2 è p0 - pz
p0 + pz ø 2 çè
p02 - pz2
ø è m + pT
ö
÷
÷
ø
Therefore: p0 = m2 + pT2 cosh y º mT cosh y
Subtract the equations, divide by 2:
æ
ö
p0 - pz ö 1 ç p0 + pz - p0 + pz ÷ æç
pz
ey - e- y 1 æ p0 + pz
÷÷ =
= çç
=ç
2
2
÷
2
2 è p0 - pz
p0 + pz ø 2 çè
p02 - pz2
ø è m + pT
Therefore
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ö
÷
÷
ø
pz = mT sinh y
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Beam 4-momentum: b
Target 4-momentum: a
By convention, the beam axis is z-axis, so beam and target
particles have no transverse momentum
Hence:
bz = mb sinh yb Þ yb = sinh-1 (bz / mb )
az = ma sinh ya Þ ya = sinh -1 (az / ma )
Also:
b0 + bz = mb (cosh yb + sinh yb ) = mb e
yb
b0 - bz = mb (cosh yb - sinh yb ) = mb e
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-yb
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Write the rapidity of the CM in terms of the rapidities of the
projectile and target:
Need to boost the projectile and target by a velocity such
that resulting longitudinal momenta are equal and opposite
Therefore:
azCM = -bzCM
g CM (az - bCM a0 ) = -g CM (bz - bCM b0 )
Solve for bCM: bCM = az + bz
a0 + b0
So the rapidity of the CM frame is:
yCM
1 æ 1+ bCM ö 1 æ (a0 + b0 )+ (az + bz ) ö 1 æ a0 + az + b0 + bz ö
= ln ç
÷ = ln ç
÷
÷ = ln ç
2 è 1- bCM ø 2 è (a0 + b0 ) - (az + bz ) ø 2 è a0 - az + b0 - bz ø
1 æ ma e ya + mb e yb e ya e yb ö 1 æ ma e ya + mb e yb ya yb ö 1 æ ma e ya + mb e yb ö ya + yb
= ln ç
×
× e e ÷ = ln ç
+
÷ = ln ç
y
y ÷
2 è ma e-ya + mb e-yb e ya e yb ø 2 è ma e yb + mb e ya
2
ø 2 è ma e b + mb e a ø
When ma=mb :
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yCM =
ya + yb
2
and
1
ybCM = yb - yCM = (yb - ya )
2
1
yaCM = ya - yCM = - (yb - ya ) = -ybCM
2
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Relationship between x+ and y
x+ =
mcT y-yb
e
and y = yb + ln x+ + ln(mb / mcT )
mb
projectile fragmentation
Best to use x+
target fragmentation
Best to use x-
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CM rapidity = midrapidity (if ma = mb)
Best to use y
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If we only measure the angle with respect to the beam axis,
define the pseudorapidity:
æ 1+ cosq ö 1 æ | p | +p ö
æ qö
z
h º -ln ç tan ÷ = ln ç
÷ = ln ç
÷
è 2ø
è 1- cosq ø 2 è | p | - pz ø
Does not require us to identify the particle.
When | p |» p0 then h » y
From the definition, we obtain:
And get formulae for pT and pz:
eh =
| p | + pz
| p | - pz
and e-h =
| p | - pz
| p | +pz
| p |= pT coshh and pz = pT sinhh
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Rapidity in terms of pseudorapidity
Just insert the relations for p0 and pz in terms of h and pT.
2
2
2
1 æç pT cosh h + m + pT sinhh ö÷
y = ln ç
2 è pT2 cosh 2 h + m 2 - pT sinh h ÷ø
The converse can also be found in a similar way:
2
2
2
1 æç mT cosh y - m + mT sinh y ö÷
h = ln ç 2
2 è mT cosh 2 h - m 2 - mT sinh y ÷ø
If particles have a distribution d2N/dydpT, then the
distribution in the pseudorapidity variable is
d2N
m2
d2N
= 1- 2
dhdpT
mT cosh 2 y dydpT
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Remarks:
d2N
m2
d2N
= 1- 2
dhdpT
mT cosh 2 y dydpT
For large y, the two distributions are approximately the
same.
In the region y~0, there is a depression (dip) in the
pseudorapidity distribution dN/dh relative to the dN/dy
Due to the Jacobian of the transformation
In the CM frame, the peak of the distribution is located at
y~0 and also at h~0
The peak value of dN/dh is smaller than the peak value of dN/dy by
approximately 1- m 2 mT2
In the Lab frame, the peak is located around h » yb / 2
The Jacobian is closer to unity, so dN/dh is similar to dN/dy.
e.g. yb~6 (proton at pz~100 GeV/c), pion detected with pT~0.35 GeV
yields a Jacobian of 0.993 : very close to unity.
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Unidentified charged particle multiplicity
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The form of the Jacobian is
commonly given in terms of
y and mT.
But we usually plot dN/dh
vs. h
And for unidentified
particles, we don’t know m
It is better to use the
Jacobian in terms of (h,pT)
¶y
( pT ,h ) =
¶h
( pT / m) coshh
2
( pT / m)
2
cosh 2 h +1
Fitting the pseudorapidity
distribution, one can obtain
dN/dy.
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dN hdy
dN h+
dy
= 295 ±18
= 304 ±18
MCBS, Ph.D. Thesis
23
We are now equipped with some of the basic
kinematic variables used in high energy, relativistic
heavy ion collisions.
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