# A rotating potential field

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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## Answers and Replies

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How should one interpret the RHS of ##V(\mathbf{x}, t) = V_0 (R^{-\omega t} \mathbf{x})##? It looks suspiciously like a vector to me. Maybe I am not familiar with your notation.

How should one interpret the RHS of ##V(\mathbf{x}, t) = V_0 (R^{-\omega t} \mathbf{x})##? It looks suspiciously like a vector to me. Maybe I am not familiar with your notation.

Here the potential ##V(\mathbf{x}, t) = V_0(\mathbf{x}_0(\mathbf{x}, t)) = V_0(R^{-\omega t} \mathbf{x})## is a scalar field which is a function of ##\mathbf{x}_0(\mathbf{x}, t)##, which is itself a function of the position vector ##\mathbf{x}## and time ##t##. So ##V_0## is a function ##V_0 : \mathbb{R}^3 \rightarrow \mathbb{R}##

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Delta2
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Yes, I see it now. Thanks.

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The notation here tends to give me a headache. (Not your fault.) I resorted to getting my hands dirty by writing things out explicitly in terms of components. Thus, let ##(x, y)## be the Cartesian coordinates of ##\mathbf x##. So, ##V(\mathbf x, t)## may be written ##V(x, y, t)##. Let ##(u, v)## be the Cartesian coordinates of the rotated vector ##R^{-\omega t} \mathbf x##. So, ##V_0(R^{-\omega t} \mathbf x)## can be written as ##V_0(u, v)## where

## u = x \cos \omega t + y \sin \omega t##
## v = -x \sin \omega t + y \cos \omega t##

Then, ##V(x, y, t) = V_0(u, v)##.

So, you can work out ##\frac {\partial V}{\partial t}## beginning with

##\large \frac {\partial V}{\partial t} = \frac {\partial V_0}{\partial u} \frac {\partial u}{\partial t} + \frac {\partial V_0}{\partial v} \frac {\partial v}{\partial t}##

Likewise, for the torque ##\large \frac{dL}{dt}##, you can begin with

##\large \frac{dL}{dt} = y (\frac {\partial V}{\partial x}) - x (\frac {\partial V}{\partial y})##

where ##\large \frac {\partial V}{\partial x} = \frac {\partial V_0}{\partial u} \frac{\partial u}{\partial x} + \frac {\partial V_0}{\partial v} \frac{\partial v}{\partial x} ##, etc.

Not a pretty way to do it, but it seems to work out.

Delta2 and etotheipi
Awesome, thanks! I was a bit hesitant to try to break everything down, but how you reformulated the functions makes it a lot easier to understand. So we have$$\frac{dE}{dt} = \frac{\partial V_0}{\partial t} = \frac{\partial V_0}{\partial u} \left( \omega y \cos{\omega t} - \omega x \sin{\omega t} \right) + \frac{\partial V_0}{\partial v} \left( -\omega x \cos{\omega t} - \omega y \sin{\omega t} \right)$$Whilst for the second part$$\omega \frac{dL}{dt} = \omega {\varepsilon^{i}}_j x^j \partial_i V = \omega y \left[ \frac{\partial V_0}{\partial u}\cos{\omega t} - \frac{\partial V_0}{\partial v}\sin{\omega t} \right] - \omega x \left[ \frac{\partial V_0}{\partial u}\sin{\omega t} + \frac{\partial V_0}{\partial v} \cos{\omega t} \right]$$and these two are the same, so ##\frac{d}{dt} \left( E- \omega L \right) = 0##, which is neat! Now for the second part about finding the symmetries of this Lagrangian, as far as I know, we're going to look for transformations that maybe look like$$\bar{t} = t + \varepsilon T, \quad \bar{\mathbf{x}} = \mathbf{x} + (\delta \theta) \mathbf{z} \times \mathbf{x}$$that are pseudosymmetries which result in ##\delta L = \varepsilon\frac{dF}{dt}## and ##\delta L = \delta \theta \frac{dF}{dt}## respectively, with ##F## being a function of ##\mathbf{x}## or ##t##, which then implies a conserved quantity$$Q = \mathcal{H}T - \frac{\partial \mathcal{L}}{\partial \dot{\mathbf{x}}} \cdot (\mathbf{z} \times \mathbf{x}) + F$$where ##\mathbf{z}## is the unit vector out of the plane. So we can take ##T = 1## and ##F = 0##, show that the resulting variations in the Lagrangian are correct, and then write down$$Q = \mathcal{H} - m \dot{\mathbf{x}} \cdot \mathbf{z} \times \mathbf{x} = \mathcal{H} - \mathbf{z} \cdot \mathbf{x} \times m\dot{\mathbf{x}} = \mathcal{H} - mL_z$$which is our conserved quantity. Does that approach look okay to you?

Thanks a bunch!

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Awesome, thanks! I was a bit hesitant to try to break everything down, but how you reformulated the functions makes it a lot easier to understand. So we have$$\frac{dE}{dt} = \frac{\partial V_0}{\partial t} = \frac{\partial V_0}{\partial u} \left( \omega y \cos{\omega t} - \omega x \sin{\omega t} \right) + \frac{\partial V_0}{\partial v} \left( -\omega x \cos{\omega t} - \omega y \sin{\omega t} \right)$$Whilst for the second part$$\omega \frac{dL}{dt} = {\varepsilon^{i}}_j x^j \partial_i V = \omega y \left[ \frac{\partial V_0}{\partial u}\cos{\omega t} - \frac{\partial V_0}{\partial v}\sin{\omega t} \right] - \omega x \left[ \frac{\partial V_0}{\partial u}\sin{\omega t} + \frac{\partial V_0}{\partial v} \cos{\omega t} \right]$$and these two are the same, so ##\frac{d}{dt} \left( E- \omega L \right) = 0##, which is neat!
Looks good.

Now for the second part about finding the symmetries of this Lagrangian, as far as I know, we're going to look for transformations that maybe look like$$\bar{t} = t + \varepsilon T, \quad \bar{\mathbf{x}} = \mathbf{x} + (\delta \theta) \mathbf{z} \times \mathbf{x}$$that are pseudosymmetries which result in ##\delta L = \varepsilon\frac{dF}{dt}## and ##\delta L = \delta \theta \frac{dF}{dt}## respectively, with ##F## being a function of ##\mathbf{x}## or ##t##

You need to specify how ##\delta \theta## is related to ##\epsilon## in order for ##L## to be invariant under the combined transformations of ##t## and ##\mathbf x##. You are looking for one symmetry transformation parameterized by ##\epsilon##. So, the transformations of both ##t## and ##\mathbf x## should be written in terms of ##\epsilon##.

which then implies a conserved quantity$$Q = \mathcal{H}T - \frac{\partial \mathcal{L}}{\partial \dot{\mathbf{x}}} \cdot (\mathbf{z} \times \mathbf{x}) + F$$where ##\mathbf{z}## is the unit vector out of the plane.
There is something missing in the second termof ##Q##. This has to do with how ##\delta \theta## is related to ##\epsilon##.

So we can take ##T = 1## and ##F = 0##, show that the resulting variations in the Lagrangian are correct, and then write down$$Q = \mathcal{H} - m \dot{\mathbf{x}} \cdot \mathbf{z} \times \mathbf{x} = \mathcal{H} - \mathbf{z} \cdot \mathbf{x} \times m\dot{\mathbf{x}} = \mathcal{H} - mL_z$$which is our conserved quantity. Does that approach look okay to you?
The conserved quantity should be ##\mathcal{H} - \omega L_z## rather than ##\mathcal{H} - mL_z##.

etotheipi
Ah, yeah that's funny I didn't notice about the ##m## instead of the ##\omega## (it's also shouldn't be there anyway, since the ##m## should have been absorbed into the ##\mathbf{p} = m\dot{\mathbf{x}}##... whoops!). I'm not sure exactly what I'm doing, but here's my second go at it!

Under the transformations ##\bar{t} = t + \varepsilon T##, and ##\bar{\mathbf{x}} = \mathbf{x} + \delta \theta \mathbf{z} \times \mathbf{x}##, our transformed Lagrangian is

$$\mathcal{L}(\bar{\mathbf{x}}, \dot{\bar{\mathbf{x}}}, \bar{t}) = \frac{1}{2}m \dot{\bar{\mathbf{x}}}^2 - V_0(R^{-\omega(t+\varepsilon T)} \bar{\mathbf{x}})$$To first order in ##\delta \theta##,$$\dot{\bar{\mathbf{x}}}^2 = \dot{\mathbf{x}}^2 + 2\delta \theta \dot{\mathbf{x}} \cdot \mathbf{z} \times \dot{\mathbf{x}} + \mathcal{O}(\delta \theta^2) = \dot{\mathbf{x}}^2 + \mathcal{O}(\delta \theta^2)$$whilst to first order in ##\delta \theta## and ##\varepsilon## - ignoring the cross term including ##\varepsilon \delta \theta## - we have for the argument of the potential energy$$R^{-\omega(t+\varepsilon T)} \bar{\mathbf{x}} \approx R^{-\omega t} [\mathbf{x} + (\delta \theta - \omega \varepsilon T) \mathbf{z} \times \mathbf{x} ]$$We can make the variation in the argument of the potential energy zero by setting ##\delta \theta = \omega \varepsilon T##, which means we now can express the transformation as a one-parameter transformation, as is required. The variation in the lagrangian under this transformation is$$\delta \mathcal{L} = \mathcal{L}(\bar{\mathbf{x}}, \dot{\bar{\mathbf{x}}}, \bar{t}) - \mathcal{L}(\mathbf{x}, \dot{\mathbf{x}}, t) \approx 0 \overset{!}{=} \varepsilon\frac{dF}{dt}$$so the transformation is a perfect symmetry, i.e. ##F=0##. Taking ##T=1##, the conserved quantity is then$$Q = \mathcal{H} T - \mathbf{p} \cdot \mathbf{K} = \mathcal{H} - m \dot{\mathbf{x}} \cdot (\omega \mathbf{z} \times \mathbf{x}) = \mathcal{H} - \omega L_z$$since the rotation transformation is ##\bar{\mathbf{x}} = \mathbf{x} + \varepsilon (\omega T \mathbf{z} \times \mathbf{x})## with a corresponding generator of rotations ##\mathbf{K} = \omega T \mathbf{z} \times \mathbf{x}##.

Phew! Does that look good? Thanks for your help, you're amazing!

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Nice. That all looks good to me.

etotheipi
Nice. That all looks good to me.

Cool! Happy Christmas Eve, and thanks for the guidance 🥳

TSny