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Rotational symmetry

by adkinje
Tags: rotational, symmetry
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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

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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Jul9-09, 11:46 PM
P: 4
I haven't watched the video, but if you perform a rotation by an angle [itex]\theta[/itex] about the [itex]z[/itex] axis on the vector [itex] {\bf r} = \left(x,y,z\right) [/itex], you get [itex] r' = r + \Delta r = \left(\cos \theta x - \sin \theta y, \sin \theta x + \cos \theta y ,z\right) [/itex]. For [itex]\theta = \epsilon[/itex] infinitesimal, this becomes [itex] r' = r + \delta r = \left(x - \epsilon y, \epsilon x + y ,z\right) = r + \left(- \epsilon y, \epsilon x ,0\right) [/itex], so that [itex] \delta x = - \epsilon y[/itex] and [itex] \delta y = - \epsilon x[/itex].
Jul14-09, 02:38 AM
P: 11
thanks, that's what I was looking for.

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