338
6 Special Theory of Relativity
Fig. 6.4 Decay of a charged
unstable particle into three
pions
Therefore, P = [
p(x)
2
+ (
p(y)
2 ]
1/2
= 856.4 MeV/c
The mass of the particle is given by
M =
E
2 −
p
2
1/2
= [(989)
2
− (856.4)
2 ]
1/2
= 494.7 MeV
It is a K meson
We can find the direction of K meson by calculating the resultant momenta
of the three pions and its orientation with respect to one of the pions.
By the vector addition of p 2 and p 3 we find the resultant P 23 = 576.6 MeV/c
inclined at angle α = 2.84
◦ above p 2 as in the Fig. 6.4. The angle inclined
between P 23 and p 1 is ϕ = 2.84 + 22.4 = 25.24
◦ .
When p 23 is combined with vector p we find that the resultant is inclined at
an angle of 16.7
◦ above p 1 .
6.3.2 Length, Time, Velocity
6.16 L = γ L 0 = (1 − β
2 )
1/2 L 0 = (1 − 0.8
2 )
1/2 L 0
= 0.6L 0
ΔL = L 0 − L = 0.4L 0
6.17 L = L 0 /γ = L 0 /2
γ = 2 → β = (γ
2
− 1)
1/2
/γ = (2
2
− 1)
1/2
/2 = 0.866
v = βc = 0.866 × 3 × 10
8
= 2.448 × 10
8 ms
−1
6.18 β = v/c = 30 km s
−1
/3 × 10
5 km s
−1
= 10
−4
1/γ = (1 − β
2 )
1/2
= (1 − 1/2 × β
2 )
ΔL = L 0 − L = L 0 − L 0 /γ = L 0 − L 0 (1−β
2 )
1/2
=
1
2
L 0 β
2
= 1/2×6,400×
10
−8 km = 3.2 cm
Thus the earth appears to be shrunk by 3.2 cm.
6.19 τ = γ τ 0
1.5 × 10
−5
= 2.2 × 10
−6
γ
γ = 6.818
β = (γ
2
− 1)
1/2
/γ = 0.9892
v = 0.9890c
6 Special Theory of Relativity
Fig. 6.4 Decay of a charged
unstable particle into three
pions
Therefore, P = [
p(x)
2
+ (
p(y)
2 ]
1/2
= 856.4 MeV/c
The mass of the particle is given by
M =
E
2 −
p
2
1/2
= [(989)
2
− (856.4)
2 ]
1/2
= 494.7 MeV
It is a K meson
We can find the direction of K meson by calculating the resultant momenta
of the three pions and its orientation with respect to one of the pions.
By the vector addition of p 2 and p 3 we find the resultant P 23 = 576.6 MeV/c
inclined at angle α = 2.84
◦ above p 2 as in the Fig. 6.4. The angle inclined
between P 23 and p 1 is ϕ = 2.84 + 22.4 = 25.24
◦ .
When p 23 is combined with vector p we find that the resultant is inclined at
an angle of 16.7
◦ above p 1 .
6.3.2 Length, Time, Velocity
6.16 L = γ L 0 = (1 − β
2 )
1/2 L 0 = (1 − 0.8
2 )
1/2 L 0
= 0.6L 0
ΔL = L 0 − L = 0.4L 0
6.17 L = L 0 /γ = L 0 /2
γ = 2 → β = (γ
2
− 1)
1/2
/γ = (2
2
− 1)
1/2
/2 = 0.866
v = βc = 0.866 × 3 × 10
8
= 2.448 × 10
8 ms
−1
6.18 β = v/c = 30 km s
−1
/3 × 10
5 km s
−1
= 10
−4
1/γ = (1 − β
2 )
1/2
= (1 − 1/2 × β
2 )
ΔL = L 0 − L = L 0 − L 0 /γ = L 0 − L 0 (1−β
2 )
1/2
=
1
2
L 0 β
2
= 1/2×6,400×
10
−8 km = 3.2 cm
Thus the earth appears to be shrunk by 3.2 cm.
6.19 τ = γ τ 0
1.5 × 10
−5
= 2.2 × 10
−6
γ
γ = 6.818
β = (γ
2
− 1)
1/2
/γ = 0.9892
v = 0.9890c