- #1

etotheipi

- Homework Statement
- A particle in 2D configuration space is subject to the potential energy $$V(\mathbf{x}, t) = V_0 (R^{-\omega t} \mathbf{x})$$ with ##R^{-\omega t}## being a rotation by angle ##- \omega t##. Show that ##E - \omega L## is a constant of motion and find the infinitesimal symmetry transformations corresponding to this quantity.

- Relevant Equations
- N/A

I'm getting a bit stuck here, the Lagrangian and equation of motion is$$\mathcal{L} = \frac{1}{2} m \dot{\mathbf{x}}^2 - V_0(R^{-\omega t} \mathbf{x}) \implies m\ddot{\mathbf{x}} = -\nabla_{\mathbf{x}} V_0(R^{-\omega t}\mathbf{x})$$as expected. To try and verify that the quantity ##E - \omega L## is a constant of motion I started by looking at ##E##,$$\frac{dE}{dt} = \frac{d}{dt} \left( \frac{1}{2} m \dot{\mathbf{x}}^2 + V_0(R^{-\omega t} \mathbf{x})\right) = m\dot{\mathbf{x}} \cdot \ddot{ \mathbf{x}} + \dot{\mathbf{x}} \cdot \nabla_{\mathbf{x}} V_0 (R^{-\omega t} \mathbf{x}) + \frac{\partial V_0 (R^{-\omega t} \mathbf{x})}{\partial t} = \frac{\partial V_0 (R^{-\omega t} \mathbf{x})}{\partial t}$$Then looking at ##L##,$$\frac{d (\omega L)}{dt} = -\omega \varepsilon_{ij} x^i (\nabla_{\mathbf{x}} V_0 (R^{-\omega t} \mathbf{x}))^j = -x^1 (\nabla_{\mathbf{x}} V_0(R^{-\omega t} \mathbf{x}))^2 + x^2 (\nabla_{\mathbf{x}} V_0(R^{-\omega t} \mathbf{x}))^1 $$I don't know what to do next

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