338 6 Special Theory of Relativity
Fig. 6.4Decay of a charged
unstable particle into three
pions
Therefore,P=[∑
p(x)^2 +(∑
p(y)^2 ]^1 /^2 = 856 .4MeV/c
The mass of the particle is given by
M=[(∑
E
) 2
−
∣
∣∑p∣
∣^2
] 1 / 2
=[(989)^2 −(856.4)^2 ]^1 /^2
= 494 .7MeV
It is aKmeson
We can find the direction ofKmeson by calculating the resultant momenta
of the three pions and its orientation with respect to one of the pions.
By the vector addition ofp 2 andp 3 we find the resultantP 23 = 576 .6MeV/c
inclined at angleα= 2. 84 ◦abovep 2 as in the Fig. 6.4. The angle inclined
betweenP 23 andp 1 isφ= 2. 84 + 22. 4 = 25. 24 ◦.
Whenp 23 is combined with vectorpwe find that the resultant is inclined at
an angle of 16. 7 ◦abovep 1.6.3.2 Length, Time, Velocity .........................
6.16 L=γL 0 =(1−β^2 )^1 /^2 L 0 =(1− 0. 82 )^1 /^2 L 0
= 0. 6 L 0
ΔL=L 0 −L= 0. 4 L 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 × 108 = 2. 448 × 108 ms−^1
6.18β=v/c=30 km s−^1 / 3 × 105 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 =^12 L 0 β^2 = 1 / 2 × 6 , 400 ×
10 −^8 km= 3 .2cm
Thus the earth appears to be shrunk by 3.2 cm.
6.19τ=γτ 0
- 5 × 10 −^5 = 2. 2 × 10 −^6 γ
γ= 6. 818
β=(γ^2 −1)^1 /^2 /γ= 0. 9892
v= 0. 9890 c