# What is the Constant of Motion in a Rotating Potential Field?

• etotheipi
In summary, the Lagrangian and equation of motion can be written as $\mathcal{L} = \frac{1}{2}m\dot{\mathbf{x}}^2 - V_0(R^{-\omega t}\mathbf{x})$ and $m\ddot{\mathbf{x}} = -\nabla_{\mathbf{x}} V_0(R^{-\omega t}\mathbf{x})$, respectively. The quantity $E - \omega L$ is shown to be a constant of motion by looking at $\frac{dE}{dt}$ and $\frac{d(\omega L)}{dt}$, which are found to be equal. To find the symmetries of the
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

Last edited by a moderator:
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.

kuruman said:
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}##

Last edited by a moderator:
Delta2
Yes, I see it now. Thanks.

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!

Last edited by a moderator:
etotheipi said:
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!

Last edited by a moderator:
Nice. That all looks good to me.

etotheipi
TSny said:
Nice. That all looks good to me.

Cool! Happy Christmas Eve, and thanks for the guidance

TSny

## 1. What is a rotating potential field?

A rotating potential field is a type of potential field in which the potential energy varies with both position and time. This means that the strength and direction of the field changes as an object moves through it, creating a rotating effect.

## 2. How is a rotating potential field created?

A rotating potential field can be created by applying a varying force or torque to an object, such as a rotating magnet or electric field. It can also occur naturally in systems where there is a combination of forces acting on an object, such as in the Earth's atmosphere or in fluid dynamics.

## 3. What are the applications of a rotating potential field?

Rotating potential fields have various applications in science and technology. They are commonly used in navigation systems, such as GPS, to determine the position and orientation of objects. They are also used in fluid dynamics to model and understand the behavior of fluids in rotating systems.

## 4. How is a rotating potential field different from a static potential field?

A rotating potential field differs from a static potential field in that it changes over time, while a static potential field remains constant. This means that the effects of a rotating potential field are dynamic and can cause objects to move or change direction, while a static potential field will not have this effect.

## 5. How is a rotating potential field measured and studied?

Rotating potential fields can be measured and studied using various techniques, such as mathematical modeling, computer simulations, and physical experiments. These methods allow scientists to understand the behavior and effects of rotating potential fields and their applications in different systems.

Replies
1
Views
1K
Replies
1
Views
609
Replies
1
Views
1K
Replies
8
Views
1K
Replies
26
Views
3K
Replies
44
Views
3K
Replies
6
Views
1K
Replies
5
Views
2K