Proof that Xp|k is a type (1, 1) tensor
We already know from the notes that DXp/dt is a type (1, 0) tensor, so we write down its transformation rule:
dt |
+ |
|
r | dt |
= | xp |
dt |
+ | h k |
Xh | dt |
Using the chain rule, we can rewrite this as
a |
+ |
|
r | a |
dt |
= | xp |
xb |
+ | h k |
Xh | xb |
a |
dt |
Now, this is true for any path whatsoever, entitling us to cancel the terms da/dt. If you cancel these terms, and stare at what is left, you will see
j|a | = | Xp|b | xp |
a |
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