## anonymous one year ago Help

1. anonymous

2. anonymous

@ganeshie8

3. anonymous

@texaschic101

4. anonymous

@Zarkon

5. anonymous

@Michele_Laino

6. Michele_Laino

yes the set of rotation of the $$x,y-$$plane is a group, for example, consider this rotation around the $$z-$$axis $\left\{ {\begin{array}{*{20}{c}} {x' = x\cos \alpha + y\sin \alpha } \\ {y' = - x\sin \alpha + y\cos \alpha } \\ {z' = z} \end{array}} \right.$

7. Michele_Laino

or, in a more compact form: $\left( {\begin{array}{*{20}{c}} {x'} \\ {y'} \\ {z'} \end{array}} \right) = {R_\alpha }\;\left( {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right) \Rightarrow R = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha }&0 \\ { - \sin \alpha }&{\cos \alpha }&0 \\ 0&0&1 \end{array}} \right)$

8. anonymous

HI... The group is already defined .. And the rotation is taken about (0,0)

9. Michele_Laino

even if the elements of a general group of rotation need not commute, namely: ${R_i}{R_j} \ne {R_j}{R_i}$

10. Michele_Laino

In your case, please try to verify the axioms of a group, using my formulas above

11. Michele_Laino

if you prefer you can consider the case when z=0, namely a rotation of the $$x,y-$$plane

12. anonymous

Yes... It will be a subgroup then. What about the cyclic part?

13. Michele_Laino

I'm thinking....

14. Michele_Laino

I think that the rotations of a polygon, around the origin are a cyclic group

15. anonymous

For a cyclic group we need a generator.. In this case the angles are takes from the set of all real numbers...

16. Michele_Laino

I know, nevertheless if we can find a cyclic group with respect to our group is a subgroup, we have the thesis

17. anonymous

You mean isomorphic to our group?

18. Michele_Laino

I mean if $$R_1$$ is a cyclic group, such that $$R$$ is included in $$R_1$$ then also $$R$$ is cyclic

19. anonymous

I am not getting th cyclic part

20. Michele_Laino

whe 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: $$J_z$$

21. Michele_Laino

when*

22. anonymous

I mean the group mentioned in the question.. HOw is it having a generator, that part I am not getting

23. Michele_Laino

for example, the $$n-$$th roots of unity are a cyclic group

24. anonymous

Yeah I know that its a cyclic group, I meant the given group is cyclic or not

25. Michele_Laino

and 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

26. Michele_Laino

If 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

27. Michele_Laino

so, if we define an infinitesimal rotation, namely $$\delta \theta$$, also our group is a cyclic group

28. anonymous

Then how can we represent any rotation which is less than it but greater than 0?

29. Michele_Laino

good question

30. Michele_Laino

I consider the matrix: $R = \left( {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right)$

31. anonymous

ok

32. Michele_Laino

Using the induction principle, we can write: ${R^n} = \left( {\begin{array}{*{20}{c}} {\cos \left( {n\alpha } \right)}&{\sin \left( {n\alpha } \right)} \\ { - \sin \left( {n\alpha } \right)}&{\cos \left( {n\alpha } \right)} \end{array}} \right)$

33. Michele_Laino

since 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

34. Michele_Laino

real* set

35. anonymous

OK thanks...:)

36. Michele_Laino

:)