Physics 417G : Solutions for Problem set 11
Due : April 22, 2016
Please show all the details of your computations including intermediate steps.
1
Problem 1
a) Consider a Lorentz boost transformation

γi
−βi γi 0
 −βi γi
γi
0
Λi = 

0
0
1
0
0
0
along x− coordinate in the vector xµ = (ct, x, y, z)T ,

0
vi
1
0 
 ,
,
βi =
γi = p
.
2
0 
c
1 − βi
1
By doing two successive boost transformations explicitly Λ1 Λ2 , obtain the resulting velocity and gamma
factor.
b) Find the matrix element describing a Lorentz transformation with βx along the x− axis followed
by a Lorentz transformation with velocity βz along z− coordinate. Does it matter in what order the
transformations are carried out?
c) An event happens at xµa = (15, 5, 0, 4) and another event happens at xµb = (5, 10, 0, 9) both in
system S.
(i) What is the invariant interval I = (xµa − xµb )(xaµ − xbµ ) = (xa − xb )µ (xa − xb )µ between a
and b?
(ii) Is there an inertial system in which they occur simultaneously? If so, find its velocity
(magnitude and direction) relative to S.
(iii) Is there an inertial system in which they occur at the same point? If so, find its velocity
relative to S.
d) Repeat the analysis of c) for the two events xµc = (1, 2, 1, 4) and another event happens at
µ
xd = (3, 10, 1, 4) both in system S.
e) Construct a 4×4 Lorentz transformation matrix Λφ that represent the counter clockwise rotation
by angle φ in xy plane.
f) The parallel between rotations and Lorentz transformations is even more striking if we introduce
the rapidity, θ = tanh−1 (v1 /c). Express the Lorentz transformation matrix Λ1 given in Problem 1, a)
in terms of theta.
g) Express the Einstein velocity addition law studied in Problem 1, a) in terms of rapidity. Check
that the rapidity ranges from ∞ to ∞, instead of −c to +c. Furthermore, check that rapidities add,
whereas velocities do not.
h) Compare the Lorentz transformation matrices, Λφ and Λi written in terms of the rapidity.
Remark the similarities and differences.
i) Exercise Λi Λφ − Λφ Λi .
Sol: a) The two Lorentz transformations give


 
γ1
−β1 γ1 0 0
γ2
−β2 γ2 0 0
γ1 γ2 (1 + β1 β2 ) −(β1 + β2 )γ1 γ2
 −β1 γ1
  −β2 γ2
  −(β1 + β2 )γ1 γ2 γ1 γ2 (1 + β1 β2 )
γ
0
0
γ
0
0
1
2

=
Λ1 Λ2 = 

0
0
1 0 
0
0
1 0  
0
0
0
0
0 1
0
0
0 1
0
0

0 0
0 0 
 ,
1 0 
0 1
which gives γ 0 = γ1 γ2 (1 + β1 β2 ) and β 0 γ 0 = (β1 + β2 )γ1 γ2 . The latter gives β 0 γ1 γ2 (1 + β1 β2 ) =
(β1 + β2 )γ1 γ2 , and thus
γ 0 = γ1 γ2 (1 + β1 β2 ) = γ1 γ2 (1 +
v1 v2
),
c2
b) A direct computations give


γx
−βx γx 0 0
γz
 −βx γx


γ
0
0
0
x

Λx Λz = 

0
0
1 0 
0
0
0
0 1
−βz γz
β0 =
0 0
1 0
0 1
0 0
(β1 + β2 )
(1 + β1 β2 )
→
v0 =
v1 + v2
.
(1 + v1 v2 /c2 )
 
−βz γz
γx γz
  −βx γx γz
0
=
 
0
0
γz
−βz γz
−βx γx
γx
0
0
0
0
1
0

−βz γx γz
γ x γ z βx βz 
 ,

0
γz

γx γz
 −βx γx
Λz Λx = 

0
−γx γz βz

0 −βz γz

0
 .

1
0
0
γz
−βx γx γz
γx
0
−γx γz βx βz
Thus the matrices do not commute.
c) (i,ii,iii) The invariant interval is −(10)2 + 52 + 52 = −50. Thus these two events are separated by
non-zero time ∆t 6= 0, meaning that they can not occur at the same time. In a frame S̄, the condition
is met by −(c∆t)2 = −50. One can solve this problem using the following transformation matrix


 √  
10
γ
−βx γ 0 −βz γ
5 2


 0   −βx γ
γ
0
0 
  −5  ,
 

 0 = 0
0
1
0  0 
−5
−βz γ
0
0
γ
0
where βx = vx /c, βz = vz /c, γ = √
1
.
2 +β 2 )
1−(βx
z
There are three equations. Two of the three relations
are −10βz γ − 5γ = 0 and −10βx γ − 5γ = 0, giving us βx = βz = −1/2. Thus we have ~v = −c/2(x̂ + ẑ).
d) (i,ii,iii) The invariant interval is −(2)2 + 82 = 40. Thus these two events are separated by non-zero
space interval ∆x 6= 0, meaning that they can not occur at the same position. In a frame S̄, the
condition is met by (∆x)2 = 40. One can solve this problem using the following transformation matrix


 

−2
γ
−βx γ 0 0
√0


 2 10   −βx γ
γ
0 0 
  −8  ,
 

 0 = 0
0
1 0  0 
0
0
0 1
0
0
where βx = vx /c, γ = √
1
.
2
1−βx
There are two equations. One of them are 8βx γ − 2γ = 0, giving us
βx = 1/4. Thus we have ~v = c/4x̂.
e) The matrix has the following form

1
 0
Λφ = 
 0
0
f,h) From the relations tanh θ =
√
cosh θ
cosh2 θ−sinh2 θ
2
sinh θ
cosh θ , cosh
0
cos φ
sin φ
0

0
0

− sin φ
 .
cos φ 0 
0
1
θ − sinh2 θ = 1, we have γ = √
1
1−v 2 /c2
= √
1
1−tanh2 θ
=
= cosh θ. Furthermore, γβ = sinh θ. This gives

cosh θ
 − sinh θ
Λθ = 

0
0
− sinh θ
cosh θ
0
0

0 0
0 0 
 .
1 0 
0 1
One can compare the similarity with the rotation matrix given in e). It is easy to see that the rapidity
ranges from ∞ to ∞, instead of −c to +c.
0
v1 /c+v2 /c
g) It is clear that for the velocity it does not add from a). The addition rule vc = (1+v
for the
2
1 v2 /c )
two Lorentz transformations in rapidity form give
tanh θ0 =
tanh θ1 + tanh θ2
= tanh(θ1 + θ2 )
(1 + tanh θ1 tanh θ2 )
→
θ 0 = θ1 + θ2 .
i) The computation Λφ Λθ − Λθ Λφ goes


 
1
0
0
0
cosh θ − sinh θ 0 0
cosh θ − sinh θ 0
 0 cos φ − sin φ
  − sinh θ cosh θ 0 0   − sinh θ cosh θ 0


−
 0 sin φ cos φ 0  
0
0
1 0  
0
0
1
0
0
0
1
0
0
0 1
0
0
0


0
(cos φ − 1) sinh θ
− sin φ sinh θ
0
 (1 − cos φ) sinh θ

0
(cosh
θ
−
1)
sin
φ
 .
=
 − sin φ sinh θ
(cosh θ − 1) sin φ
0
0 
0
0
0
0

0
1
0
 0 cos φ
0 

0   0 sin φ
1
0
0

0
0

− sin φ

cos φ 0 
0
1
2
Problem 2
Proper
ptime τ of a particle world line is defined by an observer’s time and measured velocity ~u as
dτ = 1 − u2 /c2 dt.
x
a) Starting from an ordinary four vector xµ = (ct, ~x) and ~u = d~
dt for the observer, construct the
µ
proper four velocity U µ = dx
dτ .
~.
b) From the result of a), invert the relation to obtain the formula for ~u in terms of U
An air plane is traveling along the 45o counter clockwise from x− coordinate in (x, y) plane (in the
reference frame S) with speed √25 c.
c) Find the components ux and uy of the velocity. What are the corresponding components Ux and
Uy of the proper velocity?
d) Find the time component of the proper 4-velocity, U 0 .
q
System S 0 is moving in the x− direction with speed 25 c, relative to S.
e) By using the appropriate transformation laws, find the velocity components u0x and u0y in S 0 .
f) Find the proper velocity components Ux0 and Uy0 in S 0 .
~ 0 = √ ~u0
g) As a consistency check, verify U
.
02
2
1−u /c
Sol: a) From dt/dτ = γ(~u), we get
Uµ =
d
dxµ
T
=
(ct, ~x) = γ(c, ~u)T .
dτ
dτ
where T represent the transpose.
~ 2 = ~u2 , and thus ~u = √ U~ .
b) By squaring the relation we get (1 − ~u2 )U
~2
1+U
q
q
√
u
2
2
1
c) ux = uy = u cos 45o = 5 c. And Ux = Uy = √ x2 2 = 5 c/ √5 = 2c.
1−u /c
√
d) U 0 = γc = 5c. √
√2
√2
q
q
2
uy
ux −v
2
2
1
0
5 c−
5c
5c
=
=
1
−
e) u0x = 1−u
=
0
and
u
=
=
2
y
γ 1−uy v/c2
5 1− 25
3 c.
1− 25
x v/c
q
q
√
√
√
f) Ux0 = γ(Ux − βU 0 ) = 1 − 25 ( 2c − 25 5c) = 0 and Uy0 = Uy = 2c.
√
√
(0, 23 c)T
~ 0 = √ ~u0
√
=
= (0, 2c)T .
g) U
02
2
1−u /c
3
1−2/3
Problem 3
a) In classical mechanics Newton’s law can be written in the more familiar form F~ = m~a. Compute
the corresponding relativistic expression for F~ = d~
p/dt in terms of ~u and ~a = d~u/dt.
µ
2
µ
d x
Now define proper acceleration as αµ = dU
dτ = dτ 2 .
µ
b) Find α in terms of ~u and ~a.
c) Compute αµ αµ in terms of ~u and ~a. Also compute αµ U µ .
µ
µ
µ
d) Compute K µ = dp
dτ in terms of α . What is K Uµ ?
e) Compute K µ Kµ in terms of an angle between ~u and F~ .
Sol: a) The force is given by
d
m~u
p
F~ = d~
p/dt =
=m
dt 1 − u2 /c2
p
1
d
u
dt ~
− u2 /c2
d
2~u · dt
~u/c2
−
2(1 − u2 /c2 )3/2
!
=p
~u(~u · ~a)
~a + 2
.
c − u2
1 − u2 /c2
m
b) Similar computation gives
dU 0
dU 0 dt
1
d
c
(~u · ~a)
p
=
=p
=
,
dτ
dt dτ
c(1 − u2 /c2 )2
1 − u2 /c2 dt 1 − u2 /c2
~
~ dt
dU
dU
1
~u(~u · ~a)
α
~=
=
=
~
a
+
.
dτ
dt dτ
(1 − u2 /c2 )
(c2 − u2 )
α0 =
c) The invariant can be computed in a straightforward manner
(~u · ~a)2
1
2
µ
0 2
a + 2
αµ α = −(α ) + α
.
~ ·α
~=
(1 − u2 /c2 )2
(c − u2 )
αµ U µ can be directly computed using the explicit expressions. Using the invariant U µ Uµ = −c2 is
1
(U µ Uµ ) = 2αµ Uµ = 0.
easier. By taking a derivative dτ
µ
µ
µ
µ
d) K µ = dp
dτ = mα . Thus K Uµ = mα Uµ = 0. 2
m(~
u·~
a)
u·~
a)
0
= (1−u
e) From a) we have F~ · ~u = F u cos θ = √ m 2 2 ~u · ~a + uc2(~
= mα0 =
2 /c2 )3/2 . Thus K
−u2
1−u /c
c
√uF cos2 θ
1−u /c2
~ =√
and K
~
F
1−u2 /c2
~ ·K
~ =−
K µ Kµ = −(K 0 )2 + K
4
u2 F 2 cos2 θ
F 2 (1 − cos2 θu2 /c2 )
F2
=
.
+
2
2
2
2
− u /c ) 1 − u /c
1 − u2 /c2
c2 (1
Problem 4
a) A pion at rest decays into a muon and a neutrino. Find the energy of the outgoing muon, in terms
of the two masses, mπ and mµ (assume that the neutrino has no mass mv = 0). Find the velocity of
the muon.
b) A particle of mass m whose total energy is three times its rest energy collides with an identical
particle at rest. If they stick together, what is the mass of the resulting composite particle? What is
its velocity?
c) A photon of energy E0 ”bounces” off an electron, initially at rest. Find the energy E of the
outgoing photon, as a function of the scattering angle θ.
2
Sol: a) Energy
q and momentum conservation tells that mπ c = Eµ + Eν , p~µ + p~ν = 0. The latter gives
|~
pν | = |~
pµ | =
Eµ2 − m2µ c4 /c. Using Eν = |~
pν |c, we get
mπ c2 = Eµ + |~
pν |c = Eµ +
q
Eµ2 − m2µ c4 /c
→
Eµ =
(m2π + m2µ )c2
.
2πµ
The velocity of the muon can be computed
Eµ =
(m2π + m2µ )c2
mµ c2
=p
2πµ
1 − v 2 /c2
→
v=
m2π − m2µ
c.
m2π + m2µ
2
b) We solve the problem with energy and momentum conservation. Initial energy
√ is 4mc and initial
2
2 2
2 2
2 2 2
2 2
2 2
momentum is p = E /c − m c = (3mc ) /c − m c = 8m c . Thus p = 2 2mc. For the final
state, the energy conservation gives Ef2 − p2f c2 = (4mc2 )2 − 8m2 c4 = 8m2 c4 = Mf2 c4 . Thus the
√
2
mγvc2
resulting final mass is Mf = 2 2m. For the final velocity, one can use pc
E = mγc2 = v. Thus
p c2
2
√
3
2mc
vf = Ef f = pEi ci = 2 4mc
= √c2 .
2
c) Conservation of momentum in the vertical direction gives pe sin φ = pp sin θ = E/c × sin θ with pp =
p
E/c. For horizontal direction, we get E0 /c = pe cos φ+pp cos θ = E/c×cos θ +pe 1 − (sin θE/(pe c))2 ,
2
2
2
which gives p2e c2 = (E0 − E cos θ)2 +
E cos θ + E 2 . Now the energy conservation
pE sin θ = E0 − 2E0p
2
2
4
2
2
gives E0 + mc = E + Ee = E + m c + pe c = E + m2 c4 + E02 − 2E0 E cos θ + E 2 . This can
1
hc
be solved for E to get E = (1−cos θ)/mc
2 +1/E . In terms of photon wavelength E = hν = λ , we get
0
h
λ = λ0 + mc
(1 − cos θ).