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anonymous
 one year ago
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anonymous
 one year ago
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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes the set of rotation of the \(x,y\)plane is a group, for example, consider this rotation around the \(z\)axis \[\left\{ {\begin{array}{*{20}{c}} {x' = x\cos \alpha + y\sin \alpha } \\ {y' =  x\sin \alpha + y\cos \alpha } \\ {z' = z} \end{array}} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1or, in a more compact form: \[\left( {\begin{array}{*{20}{c}} {x'} \\ {y'} \\ {z'} \end{array}} \right) = {R_\alpha }\;\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) \Rightarrow R = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha }&0 \\ {  \sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0HI... The group is already defined .. And the rotation is taken about (0,0)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1even if the elements of a general group of rotation need not commute, namely: \[{R_i}{R_j} \ne {R_j}{R_i}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1In your case, please try to verify the axioms of a group, using my formulas above

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1if you prefer you can consider the case when z=0, namely a rotation of the \(x,y\)plane

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes... It will be a subgroup then. What about the cyclic part?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I'm thinking....

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that the rotations of a polygon, around the origin are a cyclic group

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For a cyclic group we need a generator.. In this case the angles are takes from the set of all real numbers...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I know, nevertheless if we can find a cyclic group with respect to our group is a subgroup, we have the thesis

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You mean isomorphic to our group?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I mean if \(R_1\) is a cyclic group, such that \(R\) is included in \(R_1\) then also \(R\) is cyclic

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am not getting th cyclic part

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1whe we are in quantum mechanics we can speak about the generator of rotations, and if we consider the rotation around the \(z\)axis, then such generator is the \(z\)component of angular momentum of our mechanical system: \(J_z\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I mean the group mentioned in the question.. HOw is it having a generator, that part I am not getting

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1for example, the \(n\)th roots of unity are a cyclic group

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I know that its a cyclic group, I meant the given group is cyclic or not

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and as a generator I can take any one of that roots, of course the order of such group is \(n\) Yes I think so, nevertheless I'm not able to justify it

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1If we consider the set of symmetry rotations, then we have a cyclic group, since such group is isomorphic to the group of integers modulo n, where \(2\pi/n\) is the littlest rotation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so, if we define an infinitesimal rotation, namely \(\delta \theta\), also our group is a cyclic group

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then how can we represent any rotation which is less than it but greater than 0?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I consider the matrix: \[R = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ {  \sin \alpha }&{\cos \alpha } \end{array}} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1Using the induction principle, we can write: \[{R^n} = \left( {\begin{array}{*{20}{c}} {\cos \left( {n\alpha } \right)}&{\sin \left( {n\alpha } \right)} \\ {  \sin \left( {n\alpha } \right)}&{\cos \left( {n\alpha } \right)} \end{array}} \right)\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1since not all real numbers are in the form \(n\alpha\) because the real line is a vector space on the re4al set, then our group is not cyclic
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