29. From
γ
=1+
K/mc
2
(see Eq. 3849) and
v
=
βc
=
c
p
1
−
γ
−
2
(see Eq. 388), we get
v
=
c
s
1
−
µ
1+
K
mc
2
¶
−
2
.
Therefore, for the Σ
∗
0
particle,
v
=(2
.
9979
×
10
8
m
/
s)
s
1
−
µ
1+
1000 MeV
1385 MeV
¶
−
2
=2
.
4406
×
10
8
m
/
s
,
and for Σ
0
,
v
0
=(2
.
9979
×
10
8
m
/
s)
s
1
−
µ
1+
1000 MeV
1192
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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