How Does Atom Interaction Influence Motion in a One-Dimensional Crystal Lattice?

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Homework Help Overview

The discussion revolves around the interaction of atoms in a one-dimensional crystal lattice, modeled as a system of coupled oscillators. The original poster presents equations of motion for the atoms, which involve spring forces and seeks to explore the nature of standing wave solutions within this context.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formulation of the equation of motion and the appropriate form of the solution involving standing waves. There are attempts to substitute various forms into the equations and questions about the validity of these substitutions.

Discussion Status

Some participants have offered guidance on how to approach the problem, suggesting that the original poster substitute the proposed solution into the differential equation. There is ongoing confusion regarding the use of exponential versus trigonometric forms and the interpretation of indices in the equations.

Contextual Notes

There are indications of miscommunication regarding the notation used in the equations, particularly concerning the representation of indices and the distinction between real and imaginary components. Participants express uncertainty about how to proceed with the analysis of the equations given the constraints of the problem.

  • #61
ω = √- [itex]\stackrel{(D((2*cos(ak)- 2))}{(m)}[/itex]

ω = √[itex]\stackrel{(-D((2*cos(ak)- 2))}{(m)}[/itex]


-(A-B) = (-A - -B) = (-A+B) = (B-A)
ω = √-[itex]\stackrel{(D((2*cos(ak)- 2))}{(-m)}[/itex] = √[itex]\stackrel{(-D((2*cos(ak)- 2))}{(--m)}[/itex] = √[itex]\stackrel{(-D((2*cos(ak)- 2))}{(+m)}[/itex] = √[itex]\stackrel{(-D((2*cos(ak)- 2))}{(m)}[/itex]

Is this right?
 
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  • #62
bobibomi said:
ω = √- [itex]\stackrel{(D((2*cos(ak)- 2))}{(m)}[/itex]

ω = √[itex]\stackrel{(-D((2*cos(ak)- 2))}{(m)}[/itex]


-(A-B) = (-A - -B) = (-A+B) = (B-A)
ω = √-[itex]\stackrel{(D((2*cos(ak)- 2))}{(-m)}[/itex] = √[itex]\stackrel{(-D((2*cos(ak)- 2))}{(--m)}[/itex] = √[itex]\stackrel{(-D((2*cos(ak)- 2))}{(+m)}[/itex] = √[itex]\stackrel{(-D((2*cos(ak)- 2))}{(m)}[/itex]

Is this right?
Move the minus sign in the numerator inside the parentheses, using the steps I showed with A and B.
 
  • #63
ω = √- [itex]\stackrel{(D((2*cos(ak)- 2))}{(m)}[/itex]

ω = √[itex]\stackrel{(-D((2*cos(ak)- 2))}{(m)}[/itex]


-(A-B) = (-A - -B) = (-A+B) = (B-A)
ω = √[itex]\stackrel{(-D((2*cos(ak)- 2))}{(m)}[/itex] = √[itex]\stackrel{(D((-2*cos(ak)- - 2))}{(m)}[/itex] = √[itex]\stackrel{(D((-2*cos(ak) + 2))}{(+m)}[/itex] = √[itex]\stackrel{(D((2-2*cos(ak)))}{(m)}[/itex]

Is this right? I have no other idea
 
  • #64
bobibomi said:
ω = √- [itex]\stackrel{(D((2*cos(ak)- 2))}{(m)}[/itex]

ω = √[itex]\stackrel{(-D((2*cos(ak)- 2))}{(m)}[/itex]


-(A-B) = (-A - -B) = (-A+B) = (B-A)
ω = √[itex]\stackrel{(-D((2*cos(ak)- 2))}{(m)}[/itex] = √[itex]\stackrel{(D((-2*cos(ak)- - 2))}{(m)}[/itex] = √[itex]\stackrel{(D((-2*cos(ak) + 2))}{(+m)}[/itex] = √[itex]\stackrel{(D((2-2*cos(ak)))}{(m)}[/itex]

Is this right? I have no other idea
Yes, that's right (but why do you persist in using \stackrel instead \frac?).
Note that 2-2*cos() = 2*(1-cos()) which is never negative. Therefore the square root always has a real solution.
 

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