A rotating potential field

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
  • #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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  • #2
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.
 
  • #3
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}##
 
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  • #4
Yes, I see it now. Thanks.
 
  • #5
The notation here tends to give me a headache. :headbang: (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.
 
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  • #6
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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  • #7
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##.
 
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  • #8
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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  • #9
Nice. That all looks good to me.
 
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  • #10
TSny said:
Nice. That all looks good to me.

Cool! Happy Christmas Eve, and thanks for the guidance 🥳
 
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What is a rotating potential field?

A rotating potential field is a type of force field where the direction of the force continuously changes as an object moves within the field. This creates a rotating or swirling motion, similar to a whirlpool.

How is a rotating potential field created?

A rotating potential field is created by a combination of forces, such as gravity and electromagnetic forces, acting on an object. These forces can be generated by the rotation of a central object, such as a planet or star, or by the movement of multiple objects in a system.

What are some real-world examples of rotating potential fields?

Some examples of rotating potential fields include hurricanes, tornadoes, and whirlpools in bodies of water. These phenomena are all created by the rotation of air or water molecules, which creates a swirling motion.

How does a rotating potential field affect objects within it?

Objects within a rotating potential field experience a constantly changing force, which can cause them to rotate or spiral inwards towards the central source of the field. This effect is often seen in orbiting objects, such as planets around a star.

What are the applications of studying rotating potential fields?

Studying rotating potential fields can help us understand and predict natural phenomena, such as weather patterns and ocean currents. It also has practical applications in fields such as astrophysics, where it plays a crucial role in understanding the behavior of celestial objects.

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