Mathematical Physics Vol 1
4.6 Examples
157
Solution
Given that
r = r r 0 ,
it follows that
d r d t
d r 0 d t
d r d t
= v = r
r 0 .
+
From the previous Example, we have
d 2 r d t 2
γ M r 2
d v d t
r 0 ,
(4.182)
= −
=
aswell as
r × v = 2 H = h .
(4.183)
It follows that
h = r r 0 × r
r 0 = r
d r 0 d t
d r 0 d t
d r d t
2 r
(4.184)
0 ×
+
.
Let us form a vector product of both sides of the equation (4.182) with vector h , from the right. Using (4.184) and Example 5d on p. 58, we then obtain
r 0 × h = − γ M r 0 × r 0 ×
d t
γ M r 2
d v d t ×
d r 0
h = −
= − γ M r 0 ·
d t
d t
d r 0
d r 0
d r 0 d t
= γ M
r 0 − ( r 0 · r 0 )
(4.185)
,
d r 0
d t · r 0 = 0 (see Example 26 on p. 69). Given that h is a constant vector, it
because
follows that
d v d t ×
d d t
h =
( v × h ) ,
and then, based on (4.185),
d r 0 d t
d d t
( v × h )= γ M
(4.186)
.
Integration of the relation (4.186) yields
v × h = γ M r 0 + p , where p is an arbitrary constant vector. From here, by a scalar product with r , we obtan r · ( v × h )= γ M r · r 0 + r · p = = γ Mr + r r 0 · p = γ Mr + rp cos θ
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