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Rotational symmetry 
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#1
Jun609, 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? 


#2
Jul909, 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].



#3
Jul1409, 02:38 AM

P: 11

thanks, that's what I was looking for.



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