Environmental Engineering Reference
In-Depth Information
Solution 7.2.3
sfasfd
mc
2
,i.e.
(a) We can compute the rest energy using E
rest
=
10
−
27
)
10
8
)
2
10
−
10
J
E
rest
=
×
·
×
×
(
1
.
67
(
3
1
.
5
940
MeV.
(b) To get the speed we use
γ mc
2
=
7000
GeV
10
12
7
×
⇒
γ
7400
.
10
6
940
×
This is vastly greater than unity so we are certainly in the highly relativistic
limit. We could have seen that right away since the proton mass is several
thousand times smaller than its kinetic energy. Using
v
c
=
1
γ
2
1
2
γ
2
.
1
−
1
−
In order to avoid having to evaluate the square root on a calculator (which
might be a problem on some calculators) we instead have performed a Taylor
expansion, which should be a good approximation since
1
/γ
2
1
. Putting in
the numbers gives
v
c
10
−
9
.
1
−
9
.
0
×
(c) We can determine the proton momentum using p
=
γmv.Sincev/c is so close
to unity (from part (b)) we can approximate p
E/c where
E
=
E
rest
+
7000
GeV
7001
GeV.
Hence the momentum is approximately equal to 7000GeV/c. If we did want to
re-express this result in SI units then we would need to evaluate
10
12
)
10
−
19
)
(
7
×
·
(
1
.
6
×
10
−
15
Ns.
p
=
Ns
3
.
7
×
3
×
10
8
(d) The LHC protons collide head on with equal and opposite momentum, i.e. the
total momentum for any given collision is zero. This means that all of the incom-
ing energy can be used to create new particles, with all of them at rest. (If the
total momentum were not zero then momentum conservation would force some
of the outgoing particles to have some motion.) The total energy before the
collision is
2
7000
GeV and each proton has a mass of 940 MeV, therefore
the collision is capable of producing some
×
15000
new pro-
tons. This number of protons is energetically possible but in reality there are
other laws of physics which prevent 15000 protons being produced (not least
the conservation of electric charge). Nevertheless, it is the case that thousands
of new particles can easily be produced in any given proton-proton collision at
the LHC.
14000
/
0
.
94
