Our way to get to Eqs (7.11) without having to go through Chapter 6.
To go with generalized coordinates q_{i} 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
¶q_{i}
Ñq_{i},
(1)
teaching us that the components of ÑU are simply the partial derivatives ¶U/¶q_{i} (without correction factors like those in Eq (4.74)) in a basis whose ith basis vector is Ñq_{i}. (Instead of
Ñq_{i} we will often just write dq_{i}, 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 v_{a}, what are its components v_{i} in this new basis? (Greek subscripts will mean Cartesian components, latin ones the generalized components; Cartesian coordinates are x_{a}, generalized coordinates are q_{i}.) Since
v =
å
a
v_{a} dx_{a} =
å
a, i
v_{a}
¶x_{a}
¶q_{i}
dq_{i} =
å
i
v_{i} dq_{i}
the generalized components are
v_{i} =
å
a
v_{a}
¶x_{a}
¶q_{i}
In particular for Newton's law we have
m
××
x
i
= m
å
a
××
x
a
¶x_{a}
¶q_{i}
= 
¶U
¶q_{i}
(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
¶x_{a}
¶q_{i}
=
å
a
d
dt
æ è
×
x
a
¶x_{a}
¶q_{i}
ö ø

å
a
×
x
a
¶
×
x
a
¶q_{i}
The second term has the desired simple form, namely it is a partial
q_{i}derivative (of ^{1}/_{2}å_{a} x^{·}_{a} x^{·}_{a} = ^{1}/_{2}v^{2} = T/m). I will explain in lecture that [(¶x_{a}/¶q_{i} = ¶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).