Environmental Engineering Reference
In-Depth Information
PROBLEMS 11
mc
2
cosh
η
and
cp/E
mc
2
sinh
η
. Thus
11.1 We can write
E
=
=
tanh
η
,i.e.
cp
=
E
+
cp
cp
=
exp
(
2
η)
E
−
and the result follows.
11.2
10
−
.
G
(u)
=
u
1
Matrix
multiplication
gives
G
(u)
G
(v)
=
G
(u
+
v)
.
Two
Lorentz
transformations
lead
to
the
quoted
result
upon
using
the
identities:
cosh
η
1
cosh
η
2
+
sinh
η
1
sinh
η
2
=
cosh
(η
1
+
η
2
)
and
cosh
η
1
sinh
η
2
+
η
2
)
.
11.3 The left hand side can be written as
ε
ij k
R
jα
V
α
R
kβ
W
β
which can be fur-
ther written as
ε
abc
R
ia
R
jb
R
kc
R
jα
R
kβ
V
α
W
β
.Now
R
is an orthogonal matrix
which means that
R
T
sinh
η
1
cosh
η
2
=
sinh
(η
1
+
R
−
1
(R
−
1
)
ji
. We can therefore write
=
and so
R
ij
=
R
jb
R
jα
=
δ
cβ
which allows us to simplify the left hand
side to
ε
aαβ
R
ia
V
α
W
β
and this is equal to the right hand side.
δ
bα
and
R
kc
R
kβ
=
PROBLEMS 12
12.1 We write the initial four-momentum of the rocket as
m
i
(c,
0
)
,thefinal
four-momentum as
γm
f
(c, u)
and the total four-momentum of the photons
is
γm
f
(u,
u)
. The latter holds by virtue of the conservation of momen-
tum and the fact that each photon is massless. Equating the energy com-
ponents gives
m
i
c
−
=
γm
f
(c
+
u)
, as required. Solving for
u
gives
u/c
=
m
f
)
.
12.2 (a) In the zero momentum frame, the electron and positron must carry the
same energy,
E
e
. Similarly the proton and anti-proton carry the same energy,
E
p
. Energy conservation dictates that
E
e
(m
i
m
f
)/(m
i
−
+
E
p
. The minimum of this energy
occurs when the proton and anti-proton are produced at rest, i.e. the min-
imum kinetic energy is
m
p
c
2
=
m
e
c
2
937
.
8 MeV. (b) In the stationary
electron frame we exploit the fact that
(
P
e
+
+
−
=
P
p
)
2
is Lorentz
invariant. Thus we can compute it in the zero momentum frame where, at
threshold, the proton and anti-proton are produced at rest, i.e.
(
P
e
+
+
P
e
−
)
2
=
(
P
p
+
P
e
−
)
2
=
4
m
p
c
2
. We now work out the same quantity in the electron rest frame where
P
e
+
=
m
e
c)
2
p
2
4
m
p
c
2
. This
(E/c,
p
)
and
P
e
−
=
(m
e
c,
0
)
,i.e.
(E/c
+
−
=
(
2
m
p
−
m
e
)c
2
/m
e
=
can be re-arranged to give
E
=
3
.
45 TeV.
−
i
c
p
i
2
. We can compute it
in a frame of reference of our choice and, as so often, the zero momentum
frame is convenient. In that case
Mc
2
=
i
E
i
2
12.3 Invariant mass satisfies
M
2
c
4
=
i
E
CM
and this is minimized when
=
i
m
i
. Note that this proof assumes the
existence of the zero momentum frame, i.e. it assumes that the particles are
i
the particles are at rest, i.e.
M
