Question

Help

Answer 1

@ganeshie8

Answer 2

@texaschic101

Answer 3

@Zarkon

Answer 4

@Michele_Laino

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

I'm thinking....

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

You mean isomorphic to our group?

Answer 17

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

Answer 18

I am not getting th cyclic part

Answer 19

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: \(J_z\)

Answer 20

when*

Answer 21

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

Answer 22

for example, the \(n-\)th roots of unity are a cyclic group

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

good question

Answer 29

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

Answer 30

ok

Answer 31

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

Answer 32

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

Answer 33

real* set

Answer 34

OK thanks...:)

Answer 35

:)