Question
The Lagrangian for a harmonic oscillator if given as:
The transformed coordinate under rotation about the Z-axis is given as:
Determine the corresponding y' so that the transformation leaves the Lagrangian invariant.
Answer :
Word Count : 476
We are asked to find (y') such that the Lagrangian [ L = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) + \frac{1}{2} m \omega^2 (x^2 + y^2) ] remains invariant under a rotation about the (z)-axis, with (x') given as [ x' = x \sin \theta + y \cos \theta. ] --- For a rotation about the (z)-axis, the standard 2D rotation transformation is: [ \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}. ] Notice carefully that in the question, (x') is given as [ x' = x \sin \theta + y \cos \theta. ] We want the Lagrangian to be invariant under this rotation, _________ _______ ________ _____ _______ _____ ______ ___ ____ ______ ____.
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We are asked to find (y') such that the Lagrangian [ L = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2) + \frac{1}{2} m \omega^2 (x^2 + y^2) ] remains invariant under a rotation about the (z)-axis, with (x') given as [ x' = x \sin \theta + y \cos \theta. ] --- For a rotation about the (z)-axis, the standard 2D rotation transformation is: [ \begin{pmatrix} x' \ y' \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}. ] Notice carefully that in the question, (x') is given as [ x' = x \sin \theta + y \cos \theta. ] We want the Lagrangian to be invariant under this rotation, _________ _______ ________ _____ _______ _____ ______ ___ ____ ______ ____.
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