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 😮

 

[kuruman]

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 $V_0 : \mathbb{R}^3 \rightarrow \mathbb{R}$

 

[kuruman]

Yes, I see it now. Thanks.

 

[TSny]

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.

 

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!

 

[TSny]

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 $\mathcal{H} - mL_z$.

 

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!

 

[TSny]

Nice. That all looks good to me.

 

Nice. That all looks good to me.

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