rotational symmetry


by adkinje
Tags: rotational, symmetry
adkinje
adkinje is offline
#1
Jun6-09, 04:16 AM
P: 11
I've been following along with Lenny Susskinds lectures on modern classical mechanics on youtube.

at 34:30 he writes a few translation formulas on the board:
delta X = - epsilon Y
delta Y = epsilon X
http://www.youtube.com/watch?v=FZDy_Dccv4s

It's not obvious to me why these equations are true. I can't seem to find a derivation anywhere, nor can I work one out myself. Any help?
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EricAngle
EricAngle is offline
#2
Jul9-09, 11:46 PM
P: 4
I haven't watched the video, but if you perform a rotation by an angle [latex]\theta[/latex] about the [latex]z[/latex] axis on the vector [latex] {\bf r} = \left(x,y,z\right) [/latex], you get [latex] r' = r + \Delta r = \left(\cos \theta x - \sin \theta y, \sin \theta x + \cos \theta y ,z\right) [/latex]. For [latex]\theta = \epsilon[/latex] infinitesimal, this becomes [latex] r' = r + \delta r = \left(x - \epsilon y, \epsilon x + y ,z\right) = r + \left(- \epsilon y, \epsilon x ,0\right) [/latex], so that [latex] \delta x = - \epsilon y[/latex] and [latex] \delta y = - \epsilon x[/latex].
adkinje
adkinje is offline
#3
Jul14-09, 02:38 AM
P: 11
thanks, that's what I was looking for.


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