HW #2
(1) t-decay
The electron or muon four-momentum in the t rest frame is
pl m = y
1
2
mt H1, sin q, 0, cos qL,
by suitably choosing the orientation of the H1, 2L plane. Here I've dropped "hat" on q, and renamed x-hat as y just to save time to
type. By boosting it to the lab frame (with g = mZ ê H2 mt L and taking the approximation b = 1), it is
pl m = y
1
2
mt Hg H1 + cos qL, sin q, 0, g H1 + cos qLL = y
Therefore,
2 pl 0
x=
mZ
=
1
2
mZ
4
JH1 + cos qL,
2 mt
mZ
sin q, 0, H1 + cos qLN
y H1 + cos qL.
Now it is compute the distribution in x by integrating over y and cos q with the delta function dIx dG
dx
= Ÿ0 d y Ÿ-1 d cos q dIx 1
= Ÿx d y
1
=
=
1
1
2
yH1 + cos qLM
2
5
2x
2 GF mt
B3 - 2 y ± J y
y 192 p3
GF 2 mt 5
192 p3
2
GF mt
192 p3
5
- 1N H2 y - 1LF y2
@3 - 2 y ± cos qH2 y - 1LD y2
2
Ÿx d y 2A3 y - 2 y ± H2 x - yL H2 y - 1LE
1
GF 2 mt 5 2
A3
96 p3
I1 - x3 M, 1 - 3 x2 + 2 x3 E
where the last expression is for + and - sign, respectively.
IntegrateA3 y - 2 y2 + H2 x - yL H2 y - 1L, 8y, x, 1<E
2
3
-
2 x3
3
IntegrateA3 y - 2 y2 - H2 x - yL H2 y - 1L, 8y, x, 1<E
1 - 3 x2 + 2 x3
We plot them:
1
2
yH1 + cos qLM.
2
HW2.nb
PlotB
2
3
I1 - x3 M, 8x, 0, 1<F
0.6
0.5
0.4
0.3
0.2
0.1
0.2
0.4
0.6
0.8
1.0
0.2
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0.8
1.0
0.4
0.6
0.8
1.0
PlotA1 - 3 x2 + 2 x3 , 8x, 0, 1<E
1.0
0.8
0.6
0.4
0.2
Show@%, %%D
1.0
0.8
0.6
0.4
0.2
0.2
The spectra resemble closely what we see in Fig. 13(a), (b) of the ALEPH paper, except for low x which is clealy affected by
acceptance bias. In order to make them appear even closer, we can plot them with the correct relative fraction fR = Isin2 qW M ,
2
fL = I -1
+ sin2 qW M .
2
2
HW2.nb
In[12]:=
PlotBI-0.5 + sW 2 M
2
2
3
I1 - x3 M ê. :sW Ø
0.23 >, 8x, 0, 1<F
0.04
0.03
Out[12]=
0.02
0.01
0.2
In[13]:=
0.4
0.6
0.8
PlotBIsW 2 M I1 - 3 x2 + 2 x3 M ê. :sW Ø
1.0
0.23 >, 8x, 0, 1<F
2
0.05
0.04
0.03
Out[13]=
0.02
0.01
0.2
In[14]:=
0.4
PlotBI-0.5 + sW 2 M
2
2
3
0.6
0.8
1.0
I1 - x3 M + IsW 2 M I1 - 3 x2 + 2 x3 M ê. :sW Ø
2
0.10
0.08
0.06
Out[14]=
0.04
0.02
0.2
0.4
0.6
0.8
1.0
0.23 >, 8x, 0, 1<F
3
4
HW2.nb
In[15]:=
Show@%, %%, %%%D
0.10
0.08
0.06
Out[15]=
0.04
0.02
0.2
0.4
0.6
0.8
1.0
This plot looks basically the same as that in the paper except for very low x. In the actual experiment, they fit the lepton spectrum
keeping sW free based on the predicted distributions for both left- and right-handed tau.
HW2.nb
5
(2) impact parameter distribution
Suppose the long-lived particle traveled over the distance l and decayed into a charged track along the polar angle q relative to the
original direction. There may of course be several of them, but we will use only one. We assume that the track is massless. We
define the coordinate system such that the original direction is the z-axis.
Ô
In the rest frame of the decaying particle, the charged track has the four-momentum p = EH1, sin q, 0, cos qL. (I omit the hat on q
just for saving time with typing.) Going go the lab frame, we use the boost factor b = v ê c and g = 1 ì
1 - b2 . The four-
momentum of the charged track then is p = EHg + g b cos q, sin q, 0, g cos q + g bL. Therefore, the track is pointing to the
direction n = Hsin q, 0, g cos q + g bL. If we extrapolate this vector from the decay point H0, 0, lL, the distance of the closest
Ø
bHtL = » H-t sin q, 0, l - t gHcos q + bLL »2 = t2 sin q + l2 - 2 l t gHcos q + bL + t2 g2 Hcos q + bL
approach
to
the
origin
(the
impact
2
parameter
b)
is
given
by
minimizing
2
2
= t2 Jsin2 q + g2 Hcos q + bL N - 2 l t gHcos q + bL + l2
2
The minimum is
2
b2 = l2 - Hl gHcos q + bLL2 í Jsin2 q + g2 Hcos q + bL N
= l2 B1 - g2 Hcos q + bL í Jsin2 q + g2 Hcos q + bL NF
2
2
The normalized distribution in the decay length is
¶
1
d l e-lêg c b t
g c b t Ÿ0
In addition, we assume the isotropic decay
1 1
d cos q
2 Ÿ-1
Therefore, in order to find the distribution in the impact parameter, we need to work out
¶
1
1
d l e-lêg c b t 12 Ÿ-1 d cos q
g c b t Ÿ0
Now we insert unity and use the expression,
¶
1
¶
2
2
1
d l e-lêg c b t 12 Ÿ-1 d cos q Ÿ0 d b2 dJb2 - l2 B1 - g2 Hcos q + bL í Jsin2 q + g2 Hcos q + bL NFN
g c b t Ÿ0
We can integrate over l easily because its solution for the delta function is
l2 = b2 í B1 - g2 Hcos q + bL í Jsin2 q + g2 Hcos q + bL NF.
2
2
The distribution is then given by
1
gc bt
2
Ÿ0 d l
¶
=
=
1
2l
e-lêg c b t 12 Ÿ-1 d cos q Ÿ0 d b2 dJl2 - b2 í B1 - g2 Hcos q + bL í Jsin2 q + g2 Hcos q + bL NFN
¶
1
¶
1
d
g c b t Ÿ0
¶
1
d
g c b t Ÿ0
b b Ÿ-1 d cos q
1
b Ÿ-1 d cos q
1
l2 1
b2 2 l
l
2b
2
e-lêg c b t
e-lêg c b t
where l is given by the above solution.
Let us consider the ultrarelativistic limit b Ø 1. In this case, the solution for l simplifies to
l2 = b2 í B1 - g2 Hcos q + 1L í Jsin2 q + g2 Hcos q + 1L NF.
2
2
= b2 Jsin2 q + g2 Hcos q + 1L N í sin2 q
2
In addition, we can ignore sin2 q in the numerator for g p 1, and
l = b g H1 + cos qL ê sin q.
Then the normalized distribution simplifies drastically to
¶
1
g H1+cos qL
1
d b Ÿ-1 d cos q 2 sin q e-b H1+cos qLê c t sin q
g c t Ÿ0
=
¶
1
d
2 c t Ÿ0
¶
1
b Ÿ-1 d cos q
1
1+cos q
sin q
1
1+cos q
e-b H1+cos qLê c t sin q
-b H1+cos qLê c t sin q
2
1
B1-g2 Hcos q+bL íJsin2 q+g2 Hcos q+bL NF
2
2
Let us consider the ultrarelativistic limit b Ø 1. In this case, the solution for l simplifies to
l2 = b2 í B1 - g2 Hcos q + 1L í Jsin2 q + g2 Hcos q + 1L NF.
2
2
= b2 Jsin2 q + g2 Hcos q + 1L N í sin2 q
6
HW2.nb
2
In addition, we can ignore sin2 q in the numerator for g p 1, and
l = b g H1 + cos qL ê sin q.
Then the normalized distribution simplifies drastically to
¶
1
g H1+cos qL
1
d b Ÿ-1 d cos q 2 sin q e-b H1+cos qLê c t sin q
g c t Ÿ0
=
=
=
¶
1
1+cos q
1
d b Ÿ-1 d cos q sin q e-b H1+cos qLê c t sin q
2 c t Ÿ0
¶
1
1+cos q
1
d b Ÿ-1 d cos q sin q e-b H1+cos qLê c t sin q
2 c t Ÿ0
¶
1
q
1
d b Ÿ-1 d cos q cot 2 e-b cot Hqê2Lê c t
2 c t Ÿ0
q
1+cos q
= sin q ,
2
¶
4t
t e-b tê c t
Ÿ0 d t
2
I1+t2 M
Using t = cot
¶
1
d
2 c t Ÿ0
b
IntegrateB
2
t2
c t I1 +
2
t2 M
E-b têHc tL , 8t, 0, ¶<, Assumptions -> ReB
MeijerGB99- 12 =, 8<=, 990,
MeijerGB99- 12 =, 8<=, 990,
c
1
c
c2
NB
t2
8<=,
b2
4 c2 t2
ct
F > 0F
F
p t
c
IntegrateB
1
, 12 =,
2
b
1
, 12 =,
2
8<=,
b2
4 c2 t2
F
, 8b, 0, ¶<F
p t
t
MeijerGB99- 12 =, 8<=, 990,
c
1
, 12 =,
2
8<=,
b2
4 c2 t2
F
ê. 8t Ø 1 ê c< ê. 8b Ø 0<F
p t
1.5708
plot¶ =
PlotB
MeijerGB99- 12 =, 8<=, 990,
c
1
, 12 =,
2
8<=,
b2
4 c2 t2
F
ê. 8t Ø 1 ê c<, 8b, 0, 5<, PlotRange Ø 80, 1.6<F
p t
1.5
1.0
0.5
0
1
2
3
4
5
are more focused as 1 ê g and hence the impact parameter distribution is approximately g-independent.
The result is clearly g-independent. This is because the large g would lead to a larger decay length as g, whie the decay products
For modest boots, we need to numerically integrate over cos q.
HW2.nb
are more focused as 1 ê g and hence the impact parameter distribution is approximately g-independent.
7
The result is clearly g-independent. This is because the large g would lead to a larger decay length as g, whie the decay products
For modest boots, we need to numerically integrate over cos q.
TogetherB1 -
HCos@qD + bL2
Sin@qD2 + HCos@qD + bL2
F
Sin@qD2
b2 + 2 b Cos@qD + Cos@qD2 + Sin@qD2
Simplify@%D
Sin@qD2
1 + b2 + 2 b Cos@qD
NIntegrateB
1
Sin@qD
L
gcbt 2b
:g Ø
E-LêHg c
b tL
ê. :L Ø b ì
SolveBb g ã bg ê. :b Ø
1 - g-2 >, gF
::g Ø -
1 + bg2 >, :g Ø
1 + bg2 >>
::g Ø -
1 + bg2 >, :g Ø
1 + bg2 >>
::g Ø -
1 + bg2 >, :g Ø
1 + bg2 >>
1-
1 - g-2 > ê. %@@2DD
1
1 + bg2
Simplify@%D
bg2
1 + bg2
HCos@qD + bL2
2
Sin@qD +
g2
HCos@qD + bL
2
> ê. :b Ø
1 + bg2 > ê. 8bg Ø 100< ê. 8c Ø 1, t Ø 1< ê. 8b Ø 0.001<, 8q, 0, p<F
1.55721
b ê. :b Ø
1 - g2
bg
1 + bg2
> ê.
8
HW2.nb
plot1000 = PlotBNIntegrateB
Sin@qD
1
L
gcbt 2b
:g Ø
E-LêHg c
b tL
ê. :L Ø b ì
1 - g2
HCos@qD + bL2
Sin@qD2 + g2 HCos@qD + bL2
> ê. :b Ø
1.0
0.5
1
2
3
4
5
Just to verify that the numerical integral for this large value of g is consistent with the analytic result for large g:
Show@plot¶, plot1000D
1.5
1.0
0.5
0
2
1 + bg
1 + bg2 > ê. 8bg Ø 1000< ê. 8c Ø 1, t Ø 1<, 8q, 0, p<F, 8b, 0, 5<, PlotRange Ø 80, 1.6<F
1.5
0
> ê.
bg
1
They are indistinguishable.
2
3
4
5
HW2.nb
plot10 = PlotBNIntegrateB
Sin@qD
1
L
gcbt 2b
:g Ø
E-LêHg c
b tL
ê. :L Ø b ì
1 - g2
HCos@qD + bL2
Sin@qD2 + g2 HCos@qD + bL2
> ê. :b Ø
> ê.
bg
2
1 + bg
1 + bg2 > ê. 8bg Ø 10< ê. 8c Ø 1, t Ø 1<, 8q, 0, p<F, 8b, 0, 5<, PlotRange Ø 80, 1.6<F
1.5
1.0
0.5
0
1
2
3
4
5
plot1 = PlotBNIntegrateB
Sin@qD
1
L
gcbt 2b
:g Ø
E-LêHg c
b tL
ê. :L Ø b ì
1 - g2
HCos@qD + bL2
Sin@qD + g2 HCos@qD + bL
2
> ê. :b Ø
bg
1 + bg2
1 + bg2 > ê. 8bg Ø 1< ê. 8c Ø 1, t Ø 1<, 8q, 0, p<F, 8b, 0, 5<, PlotRange Ø 80, 2.2<F
2.0
1.5
1.0
0.5
0
2
1
2
3
4
5
> ê.
9
10
HW2.nb
plot01 = PlotBNIntegrateB
Sin@qD
1
L
E-LêHg c
gcbt 2b
:g Ø
b tL
ê. :L Ø b ì
1 - g2
HCos@qD + bL2
Sin@qD2 + g2 HCos@qD + bL2
> ê. :b Ø
> ê.
bg
2
1 + bg
1 + bg2 > ê. 8bg Ø 0.1< ê. 8c Ø 1, t Ø 1<, 8q, 0, p<F, 8b, 0, 0.5<, PlotRange Ø 80, 16<F
15
10
5
0.0
0.1
0.2
0.3
0.4
0.5
plot001 =
PlotBNIntegrateBSin@qD
:b Ø
1
L
gcbt 2b
> ê. :g Ø
bg
E-LêHg c
b tL
ê. :L Ø b ì
1 - g2
Sin@qD2 + g2 HCos@qD + bL2
1 + bg2 > ê. 8bg Ø 0.001< ê.
2
1 + bg
8c Ø 1, t Ø 1<, 8q, 0, p<F, 9b, 0, 5 10-3 =, PlotRange Ø 80, 1600<F
1500
1000
500
0.000
0.001
0.002
0.003
(3) h Ø W + W - decay
0.004
0.005
HCos@qD + bL2
> ê.