(Oxford) Solutions for B2: Symmetry and Relativity,

B2: Uno
cial Past Paper Solutions (2011-2016)
Toby Adkins


November 21, 2017




Disclaimer: These solutions are produced for student use, and have no connection to the
department what-so-ever. As such, they cannot be guaranteed to be correct, nor have any
bearing on the actual mark schemes that were used in the exams. Furthermore, they are
almost literally word for word my practise solutions, so they are far from 'model' solu-
tions. The formatting may suer somewhat as a result. As to the question of why I typed
them, as supposed to just scanning my solutions; anyone that knows me personally will
attest to the dire state of my handwriting. For queries and corrections, please email me at


2011
Question 1
Proper time - The time as de
ned by a clock following the coordinate of a particular object.
In terms of the metric:
Z
1 p
τ= −gµν dxµ dxν
c
Rapidity - De
ned as
β = tanh ρ, γ = cosh ρ

Proper acceleration - The rate of change of the velocity of an object as measured in its
instantaneous rest frame.

Consider the four acceleration:
dU d
U̇ = = (γ(c, u))
dτ dτ
Now,
dγ dt dγ dγ dγ du dγ
= =γ =γ · = γa ·
dτ dτ dt dt du dt du
Now,
dγ γ3u
= 2
du c
dγ u·a
= γ3 3
dt c
Thus,
u · a u·a 2 
U̇ = γ 2 γ2, γ u + a
c2 c2

1

,Toby Adkins B2 Past Papers


Equating invariants, we have that
γ6
a20 = γ 4 a2 + (u · a)2
c2
Now,
(u · a)2 = u2 a2 − (u × a)2

such that
(u × a)2
 
a20 62
=γ a −
c2

If the acceleration is in the same direction as the motion of the object, u × a = 0. Thus,
a0
a0 = γ 3 a2 −→ a=
γ2
as required.

Now,
dβ dρ d 1 dρ 1 dρ
= (tanh ρ) = 2 dτ = γ 2 dτ
dτ dτ dρ cosh ρ
dβ dβ γa a0
=γ = = 2
dτ dτ c γ c
Equating these two expressions, it follows that
dρ a0 a0 τ
= ←→ ρ=
dτ c c
This means that rapidity is Lorentz invariant (a0 , c and τ are all themselves invariant),
and that it is additive for velocities along a straight line. This makes it quite useful in
calculations treating proper motion.
dM
= −αM
dτ
We want to
nd the proper acceleration. By the conservation of momentum in the instan-
taneous rest frame of the rocket:
M dv = −cdM = αcM dt

such that a0 = αc, ρ = ατ . In terms of the proper time, the speed is given by
β = tanh ρ = tanh ατ

However, we need to
nd this in terms of the coordinate time t:
Z Z
1
t= dτ γ = dτ cosh ατ = sinh ατ
α
so
αt
tanh ατ =
c



2

,Toby Adkins B2 Past Papers


Thus,
αt v αt
β=p −→ =p
1 − β2 c 1 + (αt)2
In the second case, we are considering a photon powered rocket. In the rest frame of the
launchpad, the photons have 4-vectors:
 
1
dP = αM0 cdt
1
In the rest frame of the rocket,
 
0 γ(1 − β)
dP = ΛdP = αM0 cdt
γ(1 − β)
The three momentum delivered to the rocket in its rest frame is
dp0 dt dt
= 2αM0 c γ(1 − β) = M0 a0 −→ a0 = 2αc γ(1 − β)
dτ dτ dτ
Doppler shift due to photons:
s
dt 1−β
=
dτ 1+β
such that
 
1−β
a0 = 2αc
1+β
As β → 1, a0 → 0, meaning that the
rst spacecraft will clearly catch it up.

Question 2
Adopt the metric ηµν = diag(−1, 1, 1, 1). Then, consider the 4-momenta in the CM frame
(c ≡ 1):
P00 = (m0 , 0)
P01 = (E10 , p0 )
P02 = (E20 , −p0 )
By the conservation of 4-momentum,
P00 = P01 + P02
(P00 − P01 )2 = P202
−m20 − m21 − 2P00 · P01 = −m22
−m20 − m21 − 2(−m0 E10 ) = −m22
m20 + m21 − m22
→ E10 =
2m0
Consider E10 = m21 + p02 :
p

E12 = m21 + p2
p2 = E12 − m21
(m20 + m21 − m22 )2 − 4m21 m20
=
4m20
1
1/2
→p= (m21 + m22 − m20 )2 − 4m21 m22
2m0

3

, Toby Adkins B2 Past Papers


The Lorentz factor is given by
E0
γ=
m0
so
s  2
m0
β= 1−
E0

In the CM frame, the second decay product has:
m20 + m22 − m21
E20 =
2m0
1
1/2
p0 = (m21 + m22 − m20 )2 − 4m21 m22
2m0
This means that the lab frame quantities are given by:
     0
E γ βγ 0 0 E
px  βγ γ 0 0  p0x 
 =
0 1 0  p0y 
 
py   0
0 0 0 0 1 0

where - without loss of generality - we have assumed that the decay product occurs in a
plane. So:
E = γ(E 0 + βp0x )
px = γ(p0x + βE 0 )
py = p0y

Note that we have assumed that the original particle moves along the x-axis.
• Parallel decay, p0y = 0:

px = γ(p0 + βE 0 )
E = γ(E 0 + βp0 )

If the second particle is approximately massless,
1
1/2 m21 − m20
p0 ≈ (m21 − m20 )2 =
2m0 2m0

• Perpendicular decay, p0x = 0:

py = p0
px = βγE 0
E = γE 0

This means that the decay product is emitted at an angle of
p0
   
−1 py −1
θ = tan = tan
px βγE 0




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