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Theorem: Provided |T 0 | =
6 0 and |r00 | =
6 0, then |r00 × T 0 | = 0 if and only if
d 0
dt |r | = 0
By use of the division rule one can find:
T0 =
d 0
r0 dt
|r |
d r0
r00
=
−
0
0
0
dt |r |
|r |
|r |2
Cross product is distributive over addition, thus
r00 × T 0 =
d 0
1
00
00
dt |r |
· (r00 × r0 )
·
(r
×
r
)
−
|r0 |
|r0 |2
Simplifying further we reach
d 0
|r |
r00 × T 0 = − dt 0 2 · (r00 × r0 )
|r |
d 0
00
0
00
0
dt |r | = 0, then |r × T | = 0. Conversely if |r × T | = 0, then
d
0
|r |
d 0
as |T 0 × r0 | =
6 0, one can conclude − dt|r0 |2 = 0 and dt
|r | = 0.
If
1
|r00 × r0 | =
6 0
MathematicalRigor.pdf (PDF, 94.14 KB)
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