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