Orbital Angular Momentum as Generator for Circular Translations

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Homework Help Overview

The discussion revolves around the concept of orbital angular momentum as presented in Binney's text, specifically focusing on the definitions and implications of rotation vectors in relation to the plane of rotation.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definitions of the rotation vector ##\vec{\alpha}## and its relationship to the unit normal vector ##\hat{n}##. Questions arise regarding the implications of these definitions on the direction of rotation and the nature of the angle rotated.

Discussion Status

Some participants have provided clarifications regarding the notation and definitions used in the text, while others express confusion about the implications of these definitions on the physical interpretation of the rotation vector. Multiple interpretations of the notation are being explored.

Contextual Notes

There is mention of specific equations from Binney's text, and references to video explanations and additional resources that may provide further context to the discussion.

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Homework Statement



Taken from Binney's Text, pg 143.

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Homework Equations





The Attempt at a Solution



From equation (7.36): we see that ##\delta a## is in the direction of the angle rotated, ##\vec{x}## is the position vector, and ##\vec{n}## is the unit normal to the plane of rotation.

51dctx.png


But at the last line of (7.37), we see that ##\vec{\alpha} = \alpha \vec{n}##. Why is the angle rotated now the same direction as the unit normal to the plane?
 
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I'm not sure if I'm understanding your question. But, I think it's just a matter of definition. The symbol ##\vec{\alpha}## is defined to be ##\alpha \hat{n}##. Note how Binney uses the definition symbol (or equivalence symbol) [itex]\equiv[/itex]. This definition allows the rotation operator ##U## to be written more compactly. An angle of rotation has a certain magnitude about a certain axis. In this case ##\alpha## is the angle of rotation and ##\hat{n}## is the direction of the axis of rotation. The notation ##\vec{\alpha} = \alpha \hat{n}## is just a way to express that.
 
TSny said:
I'm not sure if I'm understanding your question. But, I think it's just a matter of definition. The symbol ##\vec{\alpha}## is defined to be ##\alpha \hat{n}##. Note how Binney uses the definition symbol (or equivalence symbol) [itex]\equiv[/itex]. This definition allows the rotation operator ##U## to be written more compactly. An angle of rotation has a certain magnitude about a certain axis. In this case ##\alpha## is the angle of rotation and ##\hat{n}## is the direction of the axis of rotation. The notation ##\vec{\alpha} = \alpha \hat{n}## is just a way to express that.

If ##\alpha## is the angle of rotation and ##\hat{n}## is the direction vector of rotation, then wouldn't ##\delta a = \delta \alpha## x ##x## be into the plane, perpendicular to the plane of rotation?
 
##\hat{n}## is a unit vector perpendicular to the plane of rotation. It points along the axis of rotation. So, ##\hat{n} \times \vec{x}## lies in the plane of rotation.

Since ##\vec{\alpha} = \alpha \hat{n}##, ##\vec{\alpha}## also lies along the axis of rotation (perpendicular to the plane of rotation). For example, a rotation in the x-y plane of ##\pi## radians could be written ##\vec{\alpha} = \pi \hat{z}##
 
TSny said:
For example, a rotation in the x-y plane of ##\pi## radians could be written ##\vec{\alpha} = \pi \hat{z}##

That doesn't make sense at all..Usually ##\vec{\alpha}## is the direction vector in which the system moves - I would say it's more like ##\alpha_x \vec{i} + \alpha_y \vec{j}##
 
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