Our way to get to Eqs (7.11) without having to go through Chapter 6.
To go with generalized coordinates qi we want to set basis vectors (not necessarily orthonormal) such that components of certain vectors in this basis are particularly simple.
For example, the gradient. The following is true just by the chain rule, as you can verify component by component:
ÑU =
å
i
¶U
¶qi
Ñqi,
(1)
teaching us that the components of ÑU are simply the partial derivatives ¶U/¶qi (without correction factors like those in Eq (4.74)) in a basis whose ith basis vector is Ñqi. (Instead of
Ñqi we will often just write dqi, considering "d" as another notation for Ñ.) So we want to use this new basis in which vectors have "generalized" or "covariant" components.
If I have any vector v whose Cartesian components are va, what are its components vi in this new basis? (Greek subscripts will mean Cartesian components, latin ones the generalized components; Cartesian coordinates are xa, generalized coordinates are qi.) Since
v =
å
a
va dxa =
å
a, i
va
¶xa
¶qi
dqi =
å
i
vi dqi
the generalized components are
vi =
å
a
va
¶xa
¶qi
In particular for Newton's law we have
m
××
x
i
= m
å
a
××
x
a
¶xa
¶qi
= -
¶U
¶qi
(2)
where the components on the right hand side are already familiar from Eq (1).
The task is now to write the LHS in a similarly simple way. To do this we rewrite it, similar to a step in the derivation of the angular momentum equations, as a total time derivative plus a correction:
å
a
××
x
a
¶xa
¶qi
=
å
a
d
dt
æ è
×
x
a
¶xa
¶qi
ö ø
-
å
a
×
x
a
¶
×
x
a
¶qi
The second term has the desired simple form, namely it is a partial
qi-derivative (of 1/2åa x·a x·a = 1/2v2 = T/m). I will explain in lecture that [(¶xa/¶qi = ¶x·a)/(¶q·i)] so that the first term can also be written as a partial derivative of T,
m
å
a
d
dt
æ ç
è
×
x
a
¶
×
x
a
¶
×
q
i
ö ÷
ø
=
d
dt
æ ç
è
¶T
¶
×
q
i
ö ÷
ø
.
When we put it all together this gives us an equation equivalent to (7.11).