Scalar product and generalised coordinates

[vanhees71]

I don't know, what you mean by "standard basis". We have an Euclidean vector space $V$ with a scalar product. A Cartesian basis $\vec{e}_i$ fulfills by definition the propery $\vec{e}_i \cdot \vec{e}_j=\delta_{ij}$, and each vector can be uniquely written in terms of linear combinations of these Cartesian basis vectors,
$$\vec{v}=v^j \vec{e}_j.$$
Here the Einstein summation convention is used, i.e., over equal indices you have to sum. The numbers $v^j$ are the components of the vector $\vec{v}$ wrt. the Cartesian basis.

For the scalar product of two vectors you get
$$\vec{v} \cdot \vec{w} = (v^j \vec{e}_j) \cdot (w^k \vec{e}_k)= v^j w^k \delta_{jk}=w^j v^j.$$
This holds for the components wrt. any Cartesian basis. There is not any special "standard basis".

 

[dyn]

For elements of a 3D vector space space your statement is correct.

The configuration space for the Hamiltonian is a more general beast.
Please read this carefully and then see if your questions are put in better context:
https://en.wikipedia.org/wiki/Generalized_coordinates

The generalized coordinates for the Hamiltonian can be various. I believe they are required to be independent which always allows the inner product to be written in the familiar-looking diagonal sum. (This is stuff I learned 50 yrs ago so anybody feel free to correct me! )

This hints that the basis vectors are orthogonal which helps.

If i have 2 vectors a = a_ie_i and b = b_ie_i then consider the basis vectors are not orthonormal then in the scalar product a.b i will end up with term like a₁b₂e₁.e₂ which means in this case a.b≠ a_ib_i

 

[hutchphd]

For the scalar product of two vectors you get

For the scalar product of two vectors you get
v→⋅w→=(vje→j)⋅(wke→k)=vjwkδjk=vjvj.
This holds for the components wrt. any Cartesian basis. There is not any special "standard basis".

I assume you mean $\omega _j$ in last

The "standard basis" is the name I was taught for the trivial set of n-tuples {(1,0,...,0),(0,1,...0),...(0,...,0,1)} associated with $R^n$ and the canonical coordinates.

 

[vanhees71]

Sure, I've corrected it in the posting above.