Infinitesimal Rotation vector

[Kashmir]

chapter 4.8 of Goldstein’s classical mechanics 3rd edition that deals with infinitesimal rotations, and the following is the part I got stuck:
(p.166~167) :

"If $d\boldsymbol{\Omega}$ is to be a vector in the same sense as $\mathbf{r}$, it must transform under $\mathbf{B}$ in the same way. As we shall see, $d\boldsymbol{\Omega}$ passes most of this test for a vector, although in one respect it fails to make the grade. One way of examining the transformation properties of $d\boldsymbol{\Omega}$ is to find how the matrix $\boldsymbol{\epsilon}$ transforms under a coordinate transformation. The transformed matrix $\boldsymbol{\epsilon}’$ is obtained by a similarity transformation:
$\begin{equation} \boldsymbol{\epsilon}’=B\boldsymbol{\epsilon}B^{-1}\end{equation}$
As the antisymmetry property of a matrix is preserved under an orthogonal similarity transformation, $\boldsymbol{\epsilon}’$ consists of nonvanishing elements $d\Omega’_i$ such that $\begin{equation}d\Omega'_i=|B|b_{ij}d\Omega_j.\end{equation}"$"

I'm not able to understand what the author is trying to say. How does "If $d\boldsymbol{\Omega}$ is to be a vector in the same sense as $\mathbf{r}$, it must transform under $\mathbf{B}$ in the same way" prove that $d\boldsymbol{\Omega}$ is indeed a vector?
I appreciate your help.

 

[PeroK]

How does "If $d\boldsymbol{\Omega}$ is to be a vector in the same sense as $\mathbf{r}$, it must transform under $\mathbf{B}$ in the same way" prove that $d\boldsymbol{\Omega}$ is indeed a vector?

That's the (physical) definition of a vector: that it obeys the vector transformation rules. Not to be confused with the pure mathematical definition of a vector as an object in a vector space.

I wonder whether you have dived in at the deep end with Goldstein. There is perhaps merit in pushing yourself, but if you are missing too may elementary prerequisites, then you risk a descent into general confusion.

 

[vanhees71]

$$\newcommand{\uvec}[1]{\underline{#1}}$$
What is it specifically you don't understand?

A rotation is represented, given some fixed right-handed Cartesian basis by an SO(3) matrix $\hat{R}$ with $\hat{R}^{-1}=\hat{R}^{\text{T}}$ and $\mathrm{det} \hat{R}=1$.

An infinitesimal rotation is given by
$$R_{jk}=\delta_{jk} + \delta \tilde{\Omega}_{jk},$$
and from the orthonormality condition you get $\delta \tilde{\Omega}_{jk}=-\delta \tilde{\Omega}_{kj}$. Now you can write any antisymmetric matrix with help of the Levi-Civita symbol as
$$\delta \tilde{\Omega}_{jk}=\epsilon_{jkl} \delta \Omega_l.$$

If you now use another righthanded Cartesian basis $\vec{e}_k'$, Then there's an SO(3) matrix $\hat{B}$ such that
$$\vec{e}_{k}'=B_{jk} \vec{e}_j \; \Leftrightarrow \; \vec{e}_j=B_{jk} \vec{e}_{k}'.$$
For an arbitrary vector $\vec{V}$ you have
$$\vec{V}=V_j \vec{e}_j = V_j B_{jk} \vec{e}_k' \; \Rightarrow \; V_k'=B_{jk} V_j$$
or in matrix-vector notation
$$\uvec{V}'=\hat{B} \uvec{V} ; \Leftrightarrow \; \uvec{V} =\hat{B}^{\text{T}} \uvec{V}'$$
Now for the rotated vector components you have
$$\uvec{V}_R'=\hat{B} \hat{R} \uvec{V}=\hat{B}\hat{R} \hat{B}^{\text{T}} \uvec{V}'$$
or
$$\uvec{V}_R'=\hat{R}' \uvec{V}' \quad \text{with} \quad \hat{R} = \hat{B}\hat{R} \hat{B}^{\text{T}}.$$
For the infinitesimal rotation this implies
$$(\hat{1} + \delta \tilde{\Omega})' = \hat{B} (\hat{1}+\delta \tilde{\Omega}) \hat{B}^{\text{T}} = \hat{1} + \hat{B} \delta \tilde{\Omega} \hat{B}^{\text{T}}.$$
This implies
$$\delta \tilde{\Omega}' = \hat{B} \tilde{\Omega} \hat{B}^{\text{T}}.$$
In Components this reads
$$\delta \tilde{\Omega}_{jk}' = B_{jl} B_{km} \delta \tilde{\Omega}_{lm} = B_{jl} B_{km} \epsilon_{lmn} \delta \Omega_n=B_{jl} B_{km} \epsilon_{lmn} B_{no} \delta \Omega_{o}'.$$
Now because
$$B_{jl} B_{km} \epsilon_{lmn} B_{on}=\mathrm{det} \hat{B} \epsilon_{jko}=\epsilon_{jko}$$
you have
$$\delta \tilde{\Omega}_{jk}'=\epsilon_{jko} \delta{\Omega}_o'.$$
This means the rotation matrix wrt. the basis $\vec{e}_k'$ is given by the vector
$$\delta \uvec{\Omega}'=\hat{R}' \delta \uvec{\Omega},$$
i.e., $\delta \uvec{\Omega}$ transforms as vector components under SO(3) tranformations $\hat{B}$ of the basis, as claimed.

 

[Kashmir]

That's the (physical) definition of a vector: that it obeys the vector transformation rules. Not to be confused with the pure mathematical definition of a vector as an object in a vector space.

I wonder whether you have dived in at the deep end with Goldstein. There is perhaps merit in pushing yourself, but if you are missing too may elementary prerequisites, then you risk a descent into general confusion.

I self studied linear algebra from strang. and this definition of a vector was never presented. This is the first time I've seen that in physics a more restricted form is only allowed i.e objects that transform in a particular way.

Also since there is no way for me to clear my doubts i am most of the time uncertain to ask it here or to think about it myself which sometimes leads to many days and I've not got any ahead.

 

[PeroK]

I self studied linear algebra from strang. and this definition of a vector was never presented. This is the first time I've seen that in physics a more restricted form is only allowed i.e objects that transform in a particular way.

Also since there is no way for me to clear my doubts i am most of the time uncertain to ask it here or to think about it myself which sometimes leads to many days and I've not got any ahead.

That's my point. A less advanced textbook than Goldstein would explain more. Goldstein assumes you know more than you do, so you have to find another source to cover the gaps in your knowledge.

 

[Kashmir]

That's my point. A less advanced textbook than Goldstein would explain more. Goldstein assumes you know more than you do, so you have to find another source to cover the gaps in your knowledge.

Yes, that's correct.

 

[vanhees71]

I self studied linear algebra from strang. and this definition of a vector was never presented. This is the first time I've seen that in physics a more restricted form is only allowed i.e objects that transform in a particular way.

Also since there is no way for me to clear my doubts i am most of the time uncertain to ask it here or to think about it myself which sometimes leads to many days and I've not got any ahead.

Physicists often do not distinguish clearly between vectors and components of vectors wrt. a basis. That often can make the understanding a bit complicated ;-).

Here you have vectors in connection with the 3D Euclidean space as part of the Newtonian spacetime model, and there it makes sense to distinguish between polar and axial vectors. A polar vector is precisely what you learn in pure mathematicians' linear-algebra books. Additionally to the vector-space axioms you also have a positive definite bilinear form, i.e., a scalar product, which induces the Euclidean norm ("length") of vectors.

Typical polar vectors in mechanics are the position vectors. By definition, under an ("active") rotation they transform as $\vec{x}'=R \vec{x}$, where $R \in \mathrm{SO}(3)$, i.e., it is a linear map with determinant 1 fulfilling $(R \vec{a}) \cdot (R \vec{b})=\vec{a} \cdot \vec{b}$ for all (polar) vectors $\vec{a}$ and $\vec{b}$ (that makes $R$ and orthogonal linear map) and it has $\mathrm{det} R=1$ (this makes is a special orthogonal map). Since time is a scalar (in the sense of the Euclidean vector space) all time derivatives of the position vector are polar vectors too, particularly velocity $\dot{\vec{x}}$ and acceleration $\dot{\vec{v}}=\ddot{\vec{x}}$. Since also the mass of a particle is a scalar also its momentum $\vec{p}=m \vec{v}$ is a polar vector.

Now also the cross product is of great use in 3D Euclidean space. Now the vector product of two polar vectors, like e.g. angular momentum $\vec{L}=\vec{x} \times \vec{p}$ transforms under rotations like a vector, because $\vec{L}'=(R \vec{x}) \times (R \vec{p})=R (\vec{x} \times \vec{p})$.

However, for a spatial reflection, which is defined for polar (!) vectors as $P=-\text{id}$, it behaves slightly different: $\vec{L}'=(P \vec{x}) \times (P \vec{p}) = (-\vec{x}) \times (-\vec{p})=+\vec{x} \times \vec{p}=\vec{L}$, i.e., for a orthogonal map with determinant $-1$ it gets an additional sign, i.e., for $O \in \mathrm{O}(3)$ you generally have
$$\vec{L}'=\mathrm{det} O O(\vec{x} \times \vec{p}).$$
That's why physicists call $\vec{L}$ an axial vector.

 

[Kashmir]

Physicists often do not distinguish clearly between vectors and components of vectors wrt. a basis. That often can make the understanding a bit complicated ;-).

Here you have vectors in connection with the 3D Euclidean space as part of the Newtonian spacetime model, and there it makes sense to distinguish between polar and axial vectors. A polar vector is precisely what you learn in pure mathematicians' linear-algebra books. Additionally to the vector-space axioms you also have a positive definite bilinear form, i.e., a scalar product, which induces the Euclidean norm ("length") of vectors.

Typical polar vectors in mechanics are the position vectors. By definition, under an ("active") rotation they transform as $\vec{x}'=R \vec{x}$, where $R \in \mathrm{SO}(3)$, i.e., it is a linear map with determinant 1 fulfilling $(R \vec{a}) \cdot (R \vec{b})=\vec{a} \cdot \vec{b}$ for all (polar) vectors $\vec{a}$ and $\vec{b}$ (that makes $R$ and orthogonal linear map) and it has $\mathrm{det} R=1$ (this makes is a special orthogonal map). Since time is a scalar (in the sense of the Euclidean vector space) all time derivatives of the position vector are polar vectors too, particularly velocity $\dot{\vec{x}}$ and acceleration $\dot{\vec{v}}=\ddot{\vec{x}}$. Since also the mass of a particle is a scalar also its momentum $\vec{p}=m \vec{v}$ is a polar vector.

Now also the cross product is of great use in 3D Euclidean space. Now the vector product of two polar vectors, like e.g. angular momentum $\vec{L}=\vec{x} \times \vec{p}$ transforms under rotations like a vector, because $\vec{L}'=(R \vec{x}) \times (R \vec{p})=R (\vec{x} \times \vec{p})$.

However, for a spatial reflection, which is defined for polar (!) vectors as $P=-\text{id}$, it behaves slightly different: $\vec{L}'=(P \vec{x}) \times (P \vec{p}) = (-\vec{x}) \times (-\vec{p})=+\vec{x} \times \vec{p}=\vec{L}$, i.e., for a orthogonal map with determinant $-1$ it gets an additional sign, i.e., for $O \in \mathrm{O}(3)$ you generally have
$$\vec{L}'=\mathrm{det} O O(\vec{x} \times \vec{p}).$$
That's why physicists call $\vec{L}$ an axial vector.

Thank you once again sir. I'm studying your answer and I need some time to understand it :)