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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.
Answer
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
Dimensional analysis and mean lifetime
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
your considerations.
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?
Answer
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
lifetime:
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
the quantity γ using the particles energy, we start with.
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.
Answer
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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