Finite rotations and infinitesimal rotations

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SUMMARY

The discussion centers on the non-commutative nature of finite rotations versus the commutative property of infinitesimal rotations, as outlined in Kleppner's "An Introduction to Mechanics." Participants clarify that while infinitesimal rotations can be added together and commute, finite rotations do not share this property due to the underlying structure of the rotation group and its Lie algebra. The conversation emphasizes that the mathematical representation of rotations, particularly through matrix exponentials, reveals the complexities of their commutation properties.

PREREQUISITES
  • Understanding of rotation groups and Lie algebras
  • Familiarity with matrix exponentials and their applications in physics
  • Basic knowledge of vector calculus and linear transformations
  • Concept of commutation in mathematical operations
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  • Study the properties of Lie groups and Lie algebras in the context of physics
  • Learn about the matrix representation of infinitesimal rotations
  • Explore the implications of non-commutative operations in quantum mechanics
  • Investigate the mathematical foundations of rotation groups in advanced mechanics
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Physics students, mathematicians, and educators seeking a deeper understanding of rotational dynamics and the mathematical structures that describe them.

  • #31
vanhees71 said:
With Lie-group theory that's very easy to understand.
Oh! Now I see what you call Lie group theory. This is not yet Lie group theory this is just some trivial argument. Lie group theory is a serious mathematical branch. By the way, your formulas are too cumbersome. Particularly there is no need to refer on coordinate representations every time. Everything is much simpler.

Let ##u## be any vector frozen into a rigid body. The rigid body moves so that ##u=u(t)##. Define a linear operator $$A(t)u(0)=u(t).\qquad (!)$$ It is a unitary operator it's clear:
$$AA^*=I,\qquad (*)$$
Actually, formulas (*) and (!) are the definition of a rigid body.

Differentiate formula (!):
$$\dot u(t)=\dot A(t)u(0)=\dot A(t)A^{-1}(t)u(t)=\dot A(t)A^*(t)u(t).\qquad (**)$$
The operator ##\Omega=\dot AA^*## is skew-symmetric: ##\Omega^*=-\Omega##. Indeed, just take ##\frac{d}{dt}## from both sides of (*).

And finally, any skew-symmetric operator in ##\mathbb{R}^3## can unequally be presented with the help of (pseudo) vector such that ##\Omega x=\omega\times x,## and correspondingly formula (**) takes the form ##\dot u(t)=\omega\times u(t)##
 
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  • #32
vanhees71 said:
No, it's very simple!
Simple for you, but probably not to the OP! Personally, I think talk of Lie groups, Lie algebra, and the like is too advanced for an I-level thread.
 
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