Rotational Symmetry: Deriving Delta X & Delta Y

[adkinje]

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?

 

[EricAngle]

I haven't watched the video, but if you perform a http://mathworld.wolfram.com/RotationMatrix.html" by an angle \theta about the z axis on the vector {\bf r} = \left(x,y,z\right), you get r' = r + \Delta r = \left(\cos \theta x - \sin \theta y, \sin \theta x + \cos \theta y ,z\right). For \theta = \epsilon infinitesimal, this becomes r' = r + \delta r = \left(x - \epsilon y, \epsilon x + y ,z\right) = r + \left(- \epsilon y, \epsilon x ,0\right), so that \delta x = - \epsilon y and \delta y = - \epsilon x.

 

[adkinje]

thanks, that's what I was looking for.