Introduction to Health Physics: Fourth Edition - Ruang Baca FMIPA UB

REVIEW OF PHYSICAL PRINCIPLES 9
Substitution of the value of f from Eq. (2.12) into Eq. (2.8) gives
W =
r
0
d (mv)
dt
dr . (2.13)
Since v = dr/dt, Eq. (2.13) can be written as
W =
t
0
d (mv)
dt
and substituting m =
W =
v
0
⎧
v dt =
⎪⎨
m0v
v d
⎪⎩ 1 − v2
c 2
mv
0
m0
v d (mv), (2.14)
1 − v2
c 2
1/2 , we have
1/2
Differentiating the term in the parenthesis gives
W = m0
v
0
⎧
⎫
⎪⎨
v
⎪⎩ 1 − v2
c 2
1/2 +
⎪⎬
. (2.15)
⎪⎭
v3 c 2
1 − v2
c 2
3/2
⎫
⎪⎬
dv. (2.16)
⎪⎭
Now, multiply the numerator and denominator of the first term in Eq. (2.16) by
1 − v 2 /c 2 to obtain
W = m0
= m0
v
0
v
0
⎧
⎪⎨ v −
⎪⎩
v3
c 2
1 − v2
c 2
3/2 +
v
1 − v2
c 2
3/2
dv = m0
v3 c 2
1 − v2
c 2
v
0
3/2
⎫
1
v 1 − v2
c 2
The integrand in Eq. (2.18) is almost in the form
b
a
⎪⎬
dv (2.17)
⎪⎭
3/2
dv
. (2.18)
u n du = un+1
b
n + 1
, (2.19)
a