Environmental Engineering Reference
In-Depth Information
it follows that the kaon rest frame must also move with a speed
u
relative
to the lab (in the same direction as the pion) in order that the other pion
should be at rest in the lab frame. Drawing a diagram might help you see
this. Thus the speed of the moving pion in the lab frame is 2
u/(
1
u
2
/c
2
)
and
u
can be deduced using energy conservation in the kaon rest frame,
i.e.
E
π
=
+
m
K
c
2
/
2so
γ
0
.
982
c
which corresponds to
an energy of 746 MeV. Alternatively you could determine the energy and
momentum in the kaon rest frame and then make a Lorentz transformation
to the lab.
12.11 Isotropic decay means that d
P/
d
=
=
m
K
/(
2
m
π
)
and
u
=
d
(
cos
θ
)
d
φ
is an element of solid angle in the pion rest frame. The normalization
is fixed since integration over all solid angles must give a unit prob-
ability
1
/(
4
π)
where d
=
to
find
the
photon.
We
need
to
understand
the
transformation
from
(θ
,φ
)
to
(θ, φ)
.
Consider
a
photon
with
four-momentum
equal
to
(E, E
cos
θ,E
sin
θ
cos
φ,E
sin
θ
sin
φ)
.
The
Lorentz
transformation
equations tell us that
E
=
γ(E
−
βE
cos
θ),
E
cos
θ
=
γ(E
cos
θ
−
βE),
E
sin
θ
cos
φ
=
E
sin
θ
cos
φ,
E
sin
θ
sin
φ
=
E
sin
θ
sin
φ,
φ
whilst the first two
can be used to eliminate the energies, i.e. dividing them gives
where
β
=
u/c
. The last two equations tell us
φ
=
cos
θ
−
β
cos
θ
=
β
cos
θ
.
−
1
Thus d
(
cos
θ
)
β
2
)/(
1
β
cos
θ)
2
=
d
(
cos
θ)
×
(
1
−
−
and the result follows.
PROBLEMS 14
14.1 Write
p
γ(u)mu
and differentiate.
14.2 Equation of motion is
γ
3
m
=
u
˙
=−
γκm
, i.e. the required time
t
satisfies
0
d
u
=−
κt
1
−
u
2
/c
2
c
2
which can be integrated to give
t
=
(c
ln 3
)/(
2
κ)
.
γ
3
m
mc
2
L/x
2
14.3 Equation
of
motion
is
u
˙
=−
which
is
solved
by
x
=
A
cos
(ωt
+
φ)
provided
Aω/c
=
1
(in
which
case
γ
=
A/x
).
The angular frequency is
ω
2
c
2
L/A
3
=
and the required solution is obtained
upon setting
A
=
L
.
