Springs, Normal Modes, and Center of Mass coordinates

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Homework Help Overview

The discussion revolves around the problem of analyzing a linear triatomic molecule using normal mode frequencies and center of mass coordinates. The original poster attempts to reduce the problem to two degrees of freedom by introducing specific coordinates and eliminating one variable to maintain the center of mass at rest.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the transformation of coordinates and the implications of the center of mass condition. There are attempts to express potential and kinetic energy in terms of the new coordinates, with some questioning the algebraic steps needed to eliminate certain variables.

Discussion Status

There has been a mix of progress reports and requests for clarification. Some participants have made attempts to transform the equations but express uncertainty about their methods. Others are seeking guidance on specific algebraic manipulations, indicating a collaborative effort to understand the problem better.

Contextual Notes

Participants note the challenge of transforming kinetic energy into the new coordinate system and the constraints imposed by the requirement of a stationary center of mass. There is also mention of a lack of responses to earlier posts, suggesting a need for more engagement on the topic.

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[SOLVED] Springs, Normal Modes, and Center of Mass coordinates

1.) Problem

The problem of the linear triatomic molecule can be reduced to one of two degrees of freedom by introducing coordinates y_{1} = x_{2} - x_{1}, y_{2} = x_{3} - x_{2}, and eliminating x_{2} by requiring that the center of mass remain at rest. Obtain the frequencies of the normal modes in these coordinates and show that they agree with the results of Section 6.4.

-Classical Mechanics, Goldstein, 3rd Edition, pg 272

2.) Useful Formulae and Context

I've attached a picture I drew of the set-up.

In section 6.4, the normal modes of the triatomic molecule are found by

1. writing out the potential and kinetic energy,
2. converting to coordinates relative to the equilibrium positions,
3. expressing them (V and T) as tensors, and
4. solving the characteristic equation |V - \omega^{2} T| = 0

Explicitly,

V = \frac{k}{2} (x_{2} - x_{1} - b)^{2} + \frac{k}{2} (x_{3} - x_{2} - b)^{2}

The coordinates relative to the equilibrium positions are introduced:

\eta_{i} = x_{i} - x_{0i}

where

x_{02} - x_{01} = b = x_{03} - x_{02}

So the potential energy becomes,

V = \frac{k}{2} (\eta_{2} - \eta_{1})^{2} + \frac{k}{2} (\eta_{3} - \eta_{2})^{2}
V = \frac{k}{2} (\eta_{1}^{2} + 2\eta_{2}^{2} + \eta_{3}^{2} - 2\eta_{1}\eta_{2} - 2\eta_{2}\eta_{3})

which can pretty easily be written in tensor form. A similar thing is done with kinetic energy.

3.) Attempt at the Solution

Goldstein writes that we should impose the constraint that "the center of mass remain stationary at the origin:"

m(x_{1} + x_{3}) + M x_{2} = 0

and that this equation should be used to eliminate one of the coordinates from V and T.

Clearly, this coordinate should be x_{2}, since it appears in both y_{1}, y_{2}, right? I've scribbled and rearranged these equations over and over, and can't figure out how express V and T only in terms of y_{1}, y_{2}.

The "Center of Mass" is described by:

R = \frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{m(x_{1} + x_{3}) + M x_{2}}{2m + M}

How does knowing this help me? If someone could just point the way, or give me the smallest hint, I'm sure I could push this through - I'm just having a block on this. Thanks!
 

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Let me just add specifically some steps I've taken:

R = \frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{m(x_{1} + x_{3}) + M x_{2}}{2m + M} = 0
\rightarrow m(x_{1} + x_{3}) + M x_{2} = 0
\rightarrow x_{2} = - \frac{m}{M} (x_{1} + x_{3})

Now using,

y_{1} = x_{2} - x_{1}
y_{2} = x_{3} - x_{2}

I should be able to find some combination of those that would give me:

x_{1} + x_{3}

so that I could insert it into the equation up above for x_{2}, thereby eliminating it. But I can't find such a combination.
 
Last edited:
Hi all - it's been a week, and I still haven't made any progress on this problem.

There must be some simple algebra trick I'm missing here. Or maybe my method is completely off? Any thoughts at all?
 
Progress Report

Alright, I was clearly looking at this incorrectly. (Perhaps that's why I've had 120+ views, and zero replies in almost 2 weeks.)

Again, the given equations:

Energies for Triatomic Model in Cartesian Coordinates:
V=\frac{k}{2}(x_{2}-x_{1}-b)^{2}+\frac{k}{2}(x_{3}-x_{2}-b)^{2}
T=\frac{m}{2}(\dot{x_{1}}^{2}+\dot{x_{3}}^{2})+\frac{M}{2}\dot{x_{2}}^{2}

Transformation to Internal Coordinates:
y_{1}=x_{2}-x_{1}
y_{2}=x_{3}-x_{2}

Stipulation of Stationary Center of Mass:
m(\dot{x_{1}}+\dot{x_{3}})+M(\dot{x_{2}})=0

Now, transforming V to internal coordinates is easy if you apply a new transformation:
y_{1}-b=\gamma_{1}
y_{2}-b=\gamma_{2}

So we have,
V=\frac{k}{2}(\gamma_{1}^{2}+\gamma_{2}^{2})

or, written as a tensor:
\bar{V}=\frac{k}{2}\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)

The Kinetic Energy is where I'm having trouble. Can anyone help me transform \dot{x_{1}},\dot{x_{2}},\dot{x_{3}} to \dot{\gamma_{1}},\dot{\gamma_{2}}? I have pages of work, and I've seem to be getting no where. If no one is replying because I'm breaking a rule, could someone let me know, too? Thanks much!
 
Alright, it took 5 more pages of algebra, but I got it.
 


Can u help me how u solved it? its due tomorrow, so please respond asap. how did u express T in terms of gammas?

thanks in advance
 


Solve for x3 and x1 in terms of y1, y2, and x2. Then plug into center of mass equation and solve for x2. You then can get x1, x2, and x3 in terms of y1 and y2. Then plug away...
 

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