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

  • Thread starter Thread starter bobibomi
  • Start date Start date
  • Tags Tags
    Chain
Click For Summary
SUMMARY

The discussion centers on the motion of atoms in a one-dimensional crystal lattice, modeled using spring forces between adjacent atoms. The equation of motion is given by m \ddot x_i = D(x_{i+1} + x_{i-1} - 2x_i), with a proposed solution of x_i(t) = \sin(aik) \sin(\omega t) + ai. Participants explore substituting this solution into the differential equation to derive the angular frequency \omega as a function of the spring constant D, mass m, lattice constant a, and wave number k. The final expression for \omega is \omega = 2\sqrt{C/M} \sin(ka/2), indicating the relationship between these parameters.

PREREQUISITES
  • Understanding of classical mechanics, particularly Newton's laws of motion.
  • Familiarity with wave mechanics and harmonic oscillators.
  • Knowledge of differential equations and their applications in physics.
  • Proficiency in using LaTeX for mathematical expressions.
NEXT STEPS
  • Study the derivation of normal modes in one-dimensional lattices.
  • Learn about the implications of the dispersion relation in crystal lattices.
  • Investigate the effects of varying the spring constant D on the system's frequency.
  • Explore the application of Fourier analysis in solving wave equations in physics.
USEFUL FOR

Students and researchers in condensed matter physics, particularly those studying crystal dynamics and wave propagation in solid-state systems.

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

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


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

Is this right?
 
Physics news on Phys.org
  • #62
bobibomi said:
ω = √- \stackrel{(D((2*cos(ak)- 2))}{(m)}

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


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

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

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


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

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

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


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

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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
Replies
3
Views
842
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K