Scalar product and generalised coordinates

[dyn]

TL;DR Summary

Scalar Product and generalized coordinates

Hi
If i have 2 general vectors written in Cartesian coordinates then the scalar product a.b can be written as a_ib_i because the basis vectors are an orthonormal basis.
In Hamiltonian mechanics i have seen the Hamiltonian written as H = p_iv_i - L where L is the lagrangian and v is the time derivative of position . I have also seen this written as H= p.v - L.
My question is ; how is p_iv_i equivalent to p.v ? Surely this only applies if p and v are written in an orthonormal basis but are p and v even vectors when written using generalised coordinates and generalised momenta ? Generalised coordinates q can sometimes just be angles and i have never seen any mention of basis vectors in this case.
So is it just a convention that p_iv_i is equivalent to p.v in Hamiltonian mechanics ?
Thanks

 

[ergospherical]

yes the inner product in $\mathbf{R}^n$ is defined $\mathbf{a} \cdot \mathbf{b} = \displaystyle{\sum_i} a_i b_i$. in the lagrangian and hamiltonian formalisms, the vector $\mathbf{q} = (q_1, \dots, q_n)$, the vector $\dot{\mathbf{q}} = (\dot{q}_1, \dots, \dot{q}_n)$, the vector $\mathbf{p} = (p_1, \dots, p_n)$, etc. are vectors in $\mathbf{R}^n$

 

[dyn]

Thanks. So is this just a convention ? As opposed to the 3-D case where it only applies in the case of orthonormal basis vectors ?

 

[ergospherical]

definitely not a convention - one does not need to even introduce a basis of $\mathbf{R}^n$ to determine the inner product of two of its elements

say you do have some arbitrary basis $\{ \mathbf{e}_1, \dots, \mathbf{e}_n \}$ of n-tuples in $\mathbf{R}^n$, this means any $\mathbf{v} = (v_1, \dots, v_n) =\sum_i \tilde{v}_i \mathbf{e}_i \in \mathbf{R}^n$ can be taken to some other n-tuple $(\tilde{v}_1, \dots, \tilde{v}_n)$, its "representation" in the basis

of course now in this representation $\mathbf{a} \cdot \mathbf{b} = \sum_i \sum_j \tilde{a}_i \tilde{b}_j \mathbf{e}_i \cdot \mathbf{e}_j \equiv \sum_i \sum_j g_{ij} \tilde{a}_i \tilde{b}_j$, and if the basis is not orthonormal then $g_{ij} \neq \delta_{ij}$

 

[dyn]

of course now in this representation $\mathbf{a} \cdot \mathbf{b} = \sum_i \sum_j \tilde{a}_i \tilde{b}_j \mathbf{e}_i \cdot \mathbf{e}_j \equiv \sum_i \sum_j g_{ij} \tilde{a}_i \tilde{b}_j$, and if the basis is not orthonormal then $g_{ij} \neq \delta_{ij}$

You have just stated that a.b = a_ib_i only if the basis is orthonormal .

So why in Hamiltonian mechanics is p.v written as p_iv_i without knowing if the basis is orthonormal or not. To be honest i don't even know if generalised coordinates such as angles have a basis

 

[ergospherical]

no you are missing the point! elements of $\mathbf{R}^n$ are lists of numbers and a priori you don't need any basis of any kind

you only worry about the metric when working with representations of elements of $\mathbf{R}^n$ with respect to some particular basis of tuples (orthonormal or not), as in post 4

 

[dyn]

I do not understand. When working with 3-D vectors a.b only equals a_ib_i if the basis vectors of a and b are orthonormal. Is that correct ?
So i do not understand how p.v can be written as p_iv_i without knowing anything about basis vectors and if they are orthonormal or not

 

[Mark44]

You have just stated that a.b = a_ib_i only if the basis is orthonormal .

No, that's not at all what @ergospherical is saying. He's not saying anything about whatever basis is involved.
Also, the equation you wrote is incorrect. The right side should be a summation; i.e., $\sum a_i b_i$, where i ranges from 1 to n in the case of vectors in $\mathbb R^n$.

So why in Hamiltonian mechanics is p.v written as p_iv_i

Same comments as above.

When working with 3-D vectors a.b only equals a_ib_i if the basis vectors of a and b are orthonormal. Is that correct ?

Again, no.

 

[hutchphd]

no you are missing the point! elements of Rn are lists of numbers and a priori you don't need any basis of any kind

you only worry about the metric when working with representations of elements of Rn with respect to some particular basis of tuples (orthonormal or not), as in post 4

I believe the issue of confusion here is what is the n in the Hamiltonian formulation. This is loosely the number of degrees of freedom of the problem (k particles in m dimensions would be $n=k*m$) and the sum is over all n . All the rest is dividing the baby.
Shoot me if I'm wrong...

 

[dyn]

#9 is telling me that the following statement is incorrect ;

In 3-D ; a.b = a_ib_i only when the basis vectors are orthonormal

Every textbook i have ever read says that my statement above is correct

 

[hutchphd]

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! )

 

[ergospherical]

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! )

@hutchphd I hate to say it but this is irrelevant, because the inner product of two elements of $R^n$ is defined independently of the notion of any set of basis vectors

$\mathbf{p}$ is a list of numbers, $\dot{\mathbf{q}}$ is a list of numbers, and $\mathbf{p} \cdot \dot{\mathbf{q}} = \displaystyle{\sum_i} p_i \dot{q}_i$ is their inner product

(you only worry about using the metric when you have a vector space isomorphism $V \rightarrow R^n$ from $V$ (any vector space of finite dimension $n$) to a coordinate space of $n$-tuples with elements in $R$ - this isomorphism is induced by selecting a basis of $V$. note that $V$ can also be $R^n$, in which case the vector and its representation are the same types of objects)

 

[hutchphd]

I like being contradicted, and I'm sure I may well have misstated this. I do understand your point, but I do not think I have said otherwise.
If I remember correctly the reason the Hamiltonian can be written in this form follows from the independence of the generalized coordinates (this is a physics constraint). In the end the choice of basis and representation for these coordinates is of crucial importance to make the problem tractable. These two processes were being intermingled in the OP, and my attempt is to describe this comingling in the way I understand it.
Please do correct as necessary (relevancy is in the eye of the beholder, however).

 

[ergospherical]

no the generalised coordinates are independent (by definition), but this has nothing to do with the ability to use vector notation as opposed to summation notation

to transform one set $(q,\dot{q})$ of independent variables to another set $(q, p)$ one effects a legendre transformation\begin{align*}
dL &= \sum_i \dfrac{\partial L}{\partial q^i} dq_i + \sum_i \dfrac{\partial L}{\partial \dot{q}^i} d\dot{q}^i \\

dL &= \sum_i \dot{p}_i dq^i + \sum_i p_i d\dot{q}^i \\

-\sum_i \dot{p}_i dq^i + \sum_i \dot{q}^i dp_i &= d(\sum_i p_i \dot{q}^i - L) \equiv dH

\end{align*}i.e. $H \equiv \sum_i p_i \dot{q}^i - L$ or equivalently $H \equiv \mathbf{p} \cdot \dot{\mathbf{q}} - L$ where these are n dimensional vectors

 

[hutchphd]

no the generalised coordinates are independent (by definition)

Perhaps I don't understand what independent means in this context.

 

[vanhees71]

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! )

Of course you are not bound to the special case, where the metric components are diagonal. The Lagrange formalism is completely invariant under general diffeomorphisms, the Hamilton formalism even under general symplectomorphisms (canonical transformations) of phase-space variables.

 

[hutchphd]

It seems to me that the way I was taught this was that there is

definitely not a convention - one does not need to even introduce a basis of $\mathbf{R}^n$ to determine the inner product of two of its elements

say you do have some arbitrary basis $\{ \mathbf{e}_1, \dots, \mathbf{e}_n \}$ of n-tuples in $\mathbf{R}^n$, this means any $\mathbf{v} = (v_1, \dots, v_n) =\sum_i \tilde{v}_i \mathbf{e}_i \in \mathbf{R}^n$ can be taken to some other n-tuple $(\tilde{v}_1, \dots, \tilde{v}_n)$, its "representation" in the basis

of course now in this representation $\mathbf{a} \cdot \mathbf{b} = \sum_i \sum_j \tilde{a}_i \tilde{b}_j \mathbf{e}_i \cdot \mathbf{e}_j \equiv \sum_i \sum_j g_{ij} \tilde{a}_i \tilde{b}_j$, and if the basis is not orthonormal then $g_{ij} \neq \delta_{ij}$

@hutchphd I hate to say it but this is irrelevant, because the inner product of two elements of $R^n$ is defined independently of the notion of any set of basis vectors

$\mathbf{p}$ is a list of numbers, $\dot{\mathbf{q}}$ is a list of numbers, and $\mathbf{p} \cdot \dot{\mathbf{q}} = \displaystyle{\sum_i} p_i \dot{q}_i$ is their inner product

(you only worry about using the metric when you have a vector space isomorphism $V \rightarrow R^n$ from $V$ (any vector space of finite dimension $n$) to a coordinate space of $n$-tuples with elements in $R$ - this isomorphism is induced by selecting a basis of $V$. note that $V$ can also be $R^n$, in which case the vector and its representation are the same types of objects)

You have forced me to blow the dust off my copy of Nomizu "Fundamentals of Linear Algebra" (guite a lot of dust!)

I see that this is mostly an issue with nomenclature. Nomizu, after defining addition and scalar multiplication, defines $\mathbf{R}^n$ as the n-dimensional standard real vector space. In addition he defines the standard basis as the set of n n-tuples $$ (1,0,0...,0)$$ $$(0,1,0,...,0)$$ $$etc$$

He then defines the inner product in terms of the standard (manifestly orthonormal) basis, and makes provision for how to change bases. I much prefer this (and hopefully @dyn will) to just saying " you don't need no stinkin' bases" **

Nomizu is a pretty good text!...

**Treasure of the Sierra Madre

 

[ergospherical]

again, you do not need any basis set to define the inner product of two vectors in $R^n$ (the space of n-tuples with real elements)

the "standard basis" $E = \{ (0,\dots, \underbrace{1}_{\mathrm{i^{\mathrm{th}}} \, \mathrm{position}}, \dots, 0) \}$ of $R^n$ is distinguished in that every $v \in R^n$ is equal to its representation $[v]_E$ with respect to this basis, i.e. $v = [v]_E$

see e.g. Halmos

 

[hutchphd]

Under what circumstance does this make a difference? Are we parsing words here or am I missing something important?

 

[ergospherical]

it's not important but its worth to understand the distinction

on $R^n$ (n-dimensional tuple space) there is a canonical inner product $\mathbf{x} \cdot \mathbf{y} = \sum x_i y_i$
where $\mathbf{x} = (x_1, \dots, x_n)$ and $\mathbf{y} = (y_1, \dots, y_n)$
[no mention of any basis set]

then, given any other finite $n$-dimensional vector space [including $R^n$], an inner product can be chosen by mapping each vector to $R^n$ (by choice of a basis), and using the canonical inner product on $R^n$

e.g. #1 choose a basis $B = (1,x,x^2)$ on the polynomial space $P_2(R)$, and let $\varphi$ be the standard representation with respect to $B$ i.e. $\varphi(a_0 + a_1 x + a_2 x^2) = (a_0,a_1,a_2)$, then you can define\begin{align*}
\langle a_0 + a_1 x + a_2 x^2, b_0 + b_1 x + b_2 x^2 \rangle \equiv (a_0,a_1,a_2) \cdot (b_0,b_1,b_2) = a_0 b_0 + a_1 b_1 + a_2 b_2
\end{align*}

e.g. #2 choose a basis $B = \{ (\cos{\theta}, \sin{\theta}), (-\sin{\theta}, \cos{\theta}) \} \equiv \{\mathbf{x}_1, \mathbf{x}_2 \}$ of $R^2$. one can define a map from vectors in $R^2$ to their representations in this rotated basis (which happen to also live in $R^2$)\begin{align*}
\mathbf{v} &\mapsto [\mathbf{v}]_B \\
(v_1, v_2) &\mapsto (v_1', v_2') = (v_1 \cos{\theta} + v_2 \sin{\theta}, -v_1 \sin{\theta} + v_2 \cos{\theta})
\end{align*}
given two vectors $\mathbf{u}, \mathbf{w} \in R^2$, one can find their inner product either directly,\begin{align*}
\mathbf{u} \cdot \mathbf{w} = u_1 w_1 + u_2 w_2
\end{align*}or by their representations in the basis\begin{align*}
\mathbf{u} \cdot \mathbf{w} = \sum_{i,j} (u_i' \mathbf{x}_i) \cdot (w_j' \mathbf{x}_j) = \sum u_i' w_j' \mathbf{x}_i \cdot \mathbf{x}_j = \sum u_i' w_i' = u_1' w_1' + u_2' w_2'
\end{align*}where it is simple in this case since $\mathbf{x}_i \cdot \mathbf{x}_j = \delta_{ij}$

 

[hutchphd]

I understand that is how you learned it (which is fine) but it is operationally identical I believe. One could call the standard basis the "canonical" basis and (other than the slightly religious overtones) nothing changes. For me it is a cleaner definition and makes @dyn question vanish. If I misunderstand some implication of this I would really like to know. Thanks.

 

[ergospherical]

I don't know how to explain this any more clearly because you are missing the point

it is by definition that $R^n$ is a space of $n$-tuples and the canonical inner product on $R^n$ is given by $u \cdot v =\displaystyle{ \sum_i} u_i v_i$. you may see any book, Halmos, Axler, Spivak, Rudin, etc. You do not need the additional structure of a basis [a set of linearly independent elements of $R^n$ itself which span $R^n$] in order to define this inner product

maybe it is easier to see like this: even if you introduce the standard basis [the set of tuples with $1$ in the $ith$ position and $0$ in the rest] you require a method to compute the inner products $\mathbf{e}_i \cdot \mathbf{e}_j$ of these vectors, which of course brings you back to the canonical definition

[the "standard" basis really is only distinguished by the property that $v = [v]_E$, other than that it is nothing special]

 

[hutchphd]

maybe it is easier to see like this: even if you introduce the standard basis [the set of tuples with 1 in the ith position and 0 in the rest] you require a method to compute the inner products ei⋅ej of these vectors, which of course brings you back to the canonical definition

So can we agree that a more useful answer to the question

You have just stated that a.b = a_ib_i only if the basis is orthonormal .

So why in Hamiltonian mechanics is p.v written as p_iv_i without knowing if the basis is orthonormal or not. To be honest i don't even know if generalised coordinates such as angles have a basis

might be the following: "One can always define the standard basis (manifestly orthonormal as discussed ) and so there is no contradiction."

Seems better than "we don't need no stinking basis"

 

[ergospherical]

that doesn't make any sense, because the standard basis [a concept only defined on vector spaces of type $F^n$] is nothing but that set of vectors with $1$ in the $ith$ place and $0$ in the rest, in otherwords it's trivially defined

Seems better than "we don't need no stinking basis"

is facetious

i'm sorry but you're simply wrong
why do you cling so strongly to the concept of bases?

 

[hutchphd]

This is a vector space of the type $R^n$ . I truly do not know what I have said that makes no sense...can you please be more specific?

 

[ergospherical]

this

"One can always define the standard basis (manifestly orthonormal as discussed ) and so there is no contradiction."

is contentless. of course you can "always" define the standard basis of $R^n$, it's simply $\{ (1,\dots, 0), (0,1,\dots,0), \dots \}$. of course it's orthonormal using the canonical inner product

you do not need any basis sets to define inner product in R^n

this whole thread is ridiculous
i will not answer more questions on this subject because my pprevious posts are sufficient

 

[hutchphd]

Arguments fundamentally about dogma always are.

 

[vanhees71]

In physics it's very important to distinguish vectors (tensors) from from their components wrt. a basis. It took me a lot of struggle to understand vectors, because many textbooks don't make this simple distinction. Particularly the also very important subject of basis changes and the various subgroups of the general linear group becomes very lucid.

 

[ergospherical]

In physics it's very important to distinguish vectors (tensors) from from their components wrt. a basis. It took me a lot of struggle to understand vectors, because many textbooks don't make this simple distinction. Particularly the also very important subject of basis changes and the various subgroups of the general linear group becomes very lucid.

Yes, and in particular when the underlying vectors are elements of the space $R^n$, since then both the vector $\mathbf{v} = (v_1, \dots, v_n)$ and its representation $[\mathbf{v}]_B = \varphi_B(\mathbf{v})$ with respect to $B$ are real $n$-tuples.

[where, if $B = \{ \mathbf{x}_i \}$ with $\mathbf{x}_i$ a set of linearly independent $n$-tuples which span $R^n$, then $\varphi_B$ is a linear map defined by $\varphi_B(\mathbf{x}_i) = (0,\dots, 1, \dots 0)$ with $1$ in the $ith$ position]

 

[hutchphd]

In physics it's very important to distinguish vectors (tensors) from from their components wrt. a basis.

I do think I understand this well enough. That being said I again sincerely invite criticism of my reply to the OP question about dot product and orthonormal basis:

So can we agree that a more useful answer to the question
might be the following: "One can always define the standard basis (manifestly orthonormal as discussed ) and so there is no contradiction."

Is this incorrect in some way?

 

[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.