Introduction to Particles and Fields solution # 1 Contents Ira Wolfson, 040558926

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Introduction to Particles and Fields solution # 1 Contents Ira Wolfson, 040558926
```Introduction to Particles and Fields solution # 1
Ira Wolfson, 040558926
April 10, 2016
Contents
1 Dimensional analysis
2
2 Dimensional analysis and mean lifetime
4
3 Dilation transformation
6
1
1
Dimensional analysis
In this assignment we will put together some basic units for particle physics.
You are given the following quanta:
• ~ - reduced Planck constant.
• G - Newton’s gravitational constant.
• c - Speed of light.
• KB - Boltzmann constant.
Construct the following units from these quanta alone:
1) Time unit.
2) Distance unit.
3) Mass unit.
4) Energy unit.
5) Temperature unit.
6) Force unit.
7) Momentum unit.
First off let’s note the units for the quantities given , and represent them in
vector form such that:


α
 β 

M α Lβ T γ K δ → 
 γ 
δ
2
Where M, L, T, K are Mass, Length, Time and temperature respectively.


1
2
 2 
ML

[~] = Energy · T =
=
 −1 
T
0


−1
 3 
L3

=
[G] =
 −2 
2
MT
0


0
L 
1 

[c] =
=


−1
T
0


1
 2 
Energy
M L2

= 2 =
[KB ] =
 −2 
K
T K
−1
So, Now that we have the 4 basis vectors for this vector space, let’s see how we
get the standard basis êi from these:

 
 
 

0
1
−1
1
r
 0  1  2   3   1 
 ⇒ mpl = ~c
+
  = 
−
 0  2  −1   −2   −1 
G
0
0
0
0
By the same token:
 

0
 1  1 
  = 
 0  2 
0

0
r

 1 
~G

 − 2

 −1  ⇒ lpl =
c3
0
 
−1
1
 3
2 
+
−1   −2
0
0


 
1
−1
 3
2 
+
−1   −2
0
0


And:



0
 0  1 
  = 
 1  2 
0

0
r

 1 
~G
 − 5


 −1  ⇒ tpl =
c5
0
Finally the Plank temperature can be given by
units of energy from the basic Planck units:
r
2
mpl · lpl
~c5
=
Epl =
2
tpl
G
3
Epl
KB
thus we can first compose
And so the Planck temperature would be:
s
~c5
Planck temperature = Tpl =
2
GkB
We now have just force and momentum to construct:
mpl lpl
c4
=
2
tpl
G
r
mpl
~c3
Ppl =
=
c
G
Fpl =
2
In this assignment we will work out some important ”thumb-rules” in particle
physics.
1) The characteristic mean lifetime of a particle is denoted by τ . Use dimensional analysis, to find an expression for the mean lifetime of a particle,
depending only on the characteristics of the particle. Explain in detail
2) Assuming c = ~ = 1 simplify the expression found in the previous part.
3) The mean lifetime of a tauon (τ − , a heavier variant of the electron) is
2.906·10−13 sec. What is the maximum distance from a τ − creation event
we can place a detector, and expect to pick up an appreciable signal, if
the outgoing tauon is a 10 GeV particle?
1) Using the Heisenberg uncertainty principle we state that ∆E∆t ≥ ~2 .
Since ~2 is a constant (except maybe on Tuesdays), we can deduce That
the lifetime of a particle scales as its inverse energy:
~
E
Now, in its own frame of reference, a massive particle has no momentum,
thus we are left with:
~
τ∝
m 0 c2
τ∝
2) Setting ~ = c = 1 we end up with the known thumb rule for particle
1
τ'
m
Where here the mass of the particle denotes the rest mass of the particle.
4
3) In its own reference frame the tauon has a mean lifetime of 2.906·10−13 sec.
In a reference frame that has been boosted by a measure of γ the observed
lifetime will be prolonged and is given by τ → γτ . So we are left to find
Given Lorentz transformations:
 E 
 E0  
γ
−βγ
c
c
  Px 
 P 0   −βγ
γ


 x =
  Py 
 P0  
1
y
1
Pz
Pz0
And setting the original system to be the rest frame of the particle we are
left with:
E 0 = γE ⇒ γ =
E
m0 c2
In the detector frame of reference the distance D is simply given by:
D = v∆t = βc∆t
But since we have time dilation such that ∆t0 = γ∆t, meaning the observed time interval is always longer by factor γ than the time interval in
the rest frame one can see that:
t = γτ
Thus we have:
D = βγcτ
Now let’s play a little bit with the βγ expression to get a function of γ
alone:
p
p
γ2 − 1
1
γ≡p
⇒β=
⇒ γβ = γ 2 − 1
2
γ
1−β
So in general we have:
s
D=
p
γ2
− 1cτ =
E2
− 1 · cτ
m20 c4
In regions where E 2 m20 c4 , or differently put, in the kinematic region
(where virtually all of the energy is in momentum) we can simplify this
to be:
D = γcτ
5
Thus for a 10 GeV Tau lepton, with rest mass of mτ = 1.77682 GeV /c2 ,
we have:
1020
E2
=
= 31.704
E02
1.7762 · 1018
If we feel that is not deep enough into the kinematic region we use the
general result to get:
√
D = 31.704 − 1 · 3 · 1010 · 2.9 · 10−13 = 0.0482 cm
The approximate result (with the kinematic region approximation) is given
by D = 0.0489 cm which is about a 2% deviation.
In CERN they go all the way up to 14 TeV, so if we take a 10 T eV
tau lepton we get:
D = 5.630 · 103 · 3 · 1010 · 2.9 · 10−13 = 48.98 cm
3
Dilation transformation
Given the dilation transformation:
x0 = αx
t0 = αt
Find the Generators for this transformation.
And find their commutation relations with boost transformations.
We want to find the generators of this transformation. we do this by doing the
following:
x → (1 + δα) x
t
t→
≈ (1 − δα) t
(1 + δα)
(1)
(2)
Therefore the generator is given by:
Gdil = x
d
d
−t
dx
dt
(3)
∂
∂
Setting Px = −i ∂x
and E = i ∂t
we get:
Gdil = i (XPx − T E)
(4)
Alternatively, in matrix representation we can set:
 0   1


t
t
α
 x0  
 x 
α
 0 =


 y  
 y 
1
z0
z
1
6
so the transformation is given by the matrix above. We can apply G =
to get:


−1


1

Gdil = 


0
0
dT dα α=1
And we already saw in TA class that boost transformation generator is given
by:
Gboost = x
d
d
+t
= −i (XE − T Px )
dt
dx
or in matrix form:

0
 −1
GΛx = 



−1
0



0
(5)
0

GΛy = 
 −1
0
−1


0




GΛz = 

0
0
0
−1
There are three ways to do this, first we will do it explicitly, later we will do
it by using the Poincaré group commutation relations: [Xµ , Pν ] = iηµν , lastly
we will do this in matrix representation, and show commutation relations with
different axis boosts as well.
∂
∂
∂
∂
Gdil Gboost ψ = x
−t
x +t
ψ
∂x
∂t
∂t
∂x
∂
∂ = x
−t
xψ̇ + tψ 0 = x2 − t2 ψ̇ 0 + ψ 00 − ψ̈ xt + xψ̇ − tψ 0
∂x
∂t
Gboost Gdil ψ =
∂
∂
∂
∂
x
−t
ψ
x +t
∂t
∂x
∂x
∂t
= x2 − t2 ψ̇ 0 + xt ψ 00 − ψ̈ − xψ̇ + tψ 0
And so,
[Gdil , Gboost ] ψ = Gdil Gboost ψ − Gboost Gdil ψ = 2 xψ̇ − tψ 0
∂
∂
⇒ [Gdil , Gboost ] = 2 x − t
∂t
∂x
7




0
Finding the commutation relations
First method:
−1
0
0
Second method:
[Gdil , Gboost ] = [XP + T E, XE − T P ] =
[XP, XE] + [T E, XE] − [XP, T P ] − [T E, T P ]
= EX [P, X] + X [T, E] E − T [X, P ] P − P T [E, T ]
∂
∂
= i (−2EX − 2P T ) = −2i (T P + XE) = 2 x − t
∂t
∂x
Matrix representation
−1
[Gdil , Gboost ] =
0
−
−1
0
−1
−1
1
−1
0
−1
0
1
=
2
=
−2
Where the indices 2,3 are suppressed in the interest of brevity.
In the case of other indices (for instance the y direction) we have:



−1
0
−1


1
0
[Gdil , Gboost ] = 
0
−1
0



0
−1
−1

=
0
1
−
−1
0
0


0 0 1
= 0 0 0 
−1 0 0
8
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