Intermediate Modern Physics  Review Problems
Fall, 2009
1. A proton moving with
β
= 0
.
99 will appear like a moving disk. (a) Compute the thickness of
the disk assuming that in the proton’s rest frame the proton’s diameter is 1 fm. (b) Compute
the proton’s
kinetic energy
in MeV, assuming its mass is 938 MeV/c
2
. The proton collides
and fuses with a stationary proton to create a new particle. (c) What is the mass of the new
particle? (d) How fast will the new particle be moving?
solution
(a) The proton will be Lorentzcontracted by the factor
γ
= 1
/
p
1

β
2
= 7
.
09, so its
thickness is 1fm / 7.09 = 0.14 fm.
(b) According to Einstein, the total energy of a (free) particle is
E
=
γmc
2
and
mc
2
is the
rest energy. The kinetic energy is therefore
K
=
E

mc
2
= (
γ

1)
mc
2
. Given
γ
= 7
.
09 and
mc
2
= 938 MeV, we ﬁnd
K
= 5
.
7 GeV.
(c) Since 4momentum is conserved, we can write
P
1
+
P
2
=
P
3
,
where
P
1
and
P
2
are the 4momenta of the incident and stationary protons, respectively, and
P
3
is the 4momentum of the new particle. The rest energy of the new particle, and therefore
its mass, is given by
P
2
3
= (
m
3
c
2
)
2 1
, that is, by
P
2
3
=
P
2
1
+
P
2
2
+ 2
P
1
P
2
,
= 2(
mc
2
)
2
+ 2
P
1
P
2
,
= 2(
mc
2
)
2
+ 2(
γmc
2
, ~
pc
)(
mc
2
,
~
0)
,
= 2(
mc
2
)
2
(1 +
γ
)
.
(1)
Therefore,
m
3
c
2
=
mc
2
p
2(1 +
γ
) = 3
.
77 GeV.
(d) The total energy of the new particle is
E
3
=
E
1
+
E
2
=
mc
2
(
γ
+ 1), so its gamma
factor is
γ
3
=
E
3
/m
3
c
2
; that is,
mc
2
(1 +
γ
)
/mc
2
p
2(1 +
γ
) =
p
(1 +
γ
)
/
2 = 2
.
01. Therefore,
β
3
=
p
γ
2
3

1
/γ
3
= 0
.
87.
1
Since
P
= (
γmc
2
, ~
pc
), we have that
P
2
=
γ
2
(
mc
2
)
2

(
γmvc
)
2
, that is,
P
2
=
γ
2
(
mc
2
)
2
(1

β
2
) = (
mc
2
)
2
.
1