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

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.

  • #31
##m\ddot x_i=D(x_{i+1}+x_{i−1}−2x_i)##
##\ddot x_i = ## -sin(aik)sin(ωt)*ω2
xi-1 = sin(a (i-1) k) sin(ωt) + a (i-1)
xi+1 = sin(a (i+1) k) sin(ωt) + a (i+1)
xi = sin(a (i) k) sin(ωt) + a (i)

m * -sin(aik)sin(ωt)*ω2 = D(sin(a (i+1) k) sin(ωt) + a (i+1) + sin(a (i-1) k) sin(ωt) + a (i-1) - 2*(sin(a (i) k) sin(ωt) + a (i))

Is this right?
I´m not sure, what i should do now.
 
Physics news on Phys.org
  • #32
bobibomi said:
##m\ddot x_i=D(x_{i+1}+x_{i−1}−2x_i)##
##\ddot x_i = ## -sin(aik)sin(ωt)*ω2
xi-1 = sin(a (i-1) k) sin(ωt) + a (i-1)
xi+1 = sin(a (i+1) k) sin(ωt) + a (i+1)
xi = sin(a (i) k) sin(ωt) + a (i)

m * -sin(aik)sin(ωt)*ω2 = D(sin(a (i+1) k) sin(ωt) + a (i+1) + sin(a (i-1) k) sin(ωt) + a (i-1) - 2*(sin(a (i) k) sin(ωt) + a (i))

Is this right?
I´m not sure, what i should do now.
yes, that's the right equation. i hope you understand the process up to this point, and why the other avenues you took were wrong.
Now you need to look for cancellations. E.g. collect the non-trig terms on the RHS together. For the trig terms, you need to expand those with a plus or minus sign inside them. Do you know how to expand sin(a+b)?
 
  • #33
I´m sorry but I don´t understand your tip.
Could you please try to explain it easier for me.
What should I do with the equation?
 
  • #34
bobibomi said:
I´m sorry but I don´t understand your tip.
Could you please try to explain it easier for me.
What should I do with the equation?
Regroup the terms on the RHS of the equation so that the tms involving sines are together and the oth three terms are together. Looking at the three terms that do not involve sines, try to add them into a single term. What do you get?
 
  • #35
m * -sin(aik)sin(ωt)*ω2 = D(sin(a (i+1) k) sin(ωt) + a (i+1) + sin(a (i-1) k) sin(ωt) + a (i-1) - 2*(sin(a (i) k) sin(ωt) + a (i))

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(a (i+1) k) + sin(a (i-1) k)) + a (i+1) + a (i-1) - 2*(sin(a (i) k) sin(ωt) + a (i))

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(a (i+1) k) + sin(a (i-1) k)) + 2ai - 2*(sin(a (i) k) sin(ωt) + a (i))

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k) sin(ωt))

m * -sin(aik)sin(ωt)*ω2 =D(sin(ωt)(2sin(aik)) - 2*(sin(a (i) k) sin(ωt))

m * -sin(aik)sin(ωt)*ω2 =D(-sin(ωt)(2sin(aik))


Is this right? Could you please show me what I have done wrong?
 
  • #36
bobibomi said:
m * -sin(aik)sin(ωt)*ω2 = D(sin(a (i+1) k) sin(ωt) + a (i+1) + sin(a (i-1) k) sin(ωt) + a (i-1) - 2*(sin(a (i) k) sin(ωt) + a (i))

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(a (i+1) k) + sin(a (i-1) k)) + a (i+1) + a (i-1) - 2*(sin(a (i) k) sin(ωt) + a (i))

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(a (i+1) k) + sin(a (i-1) k)) + 2ai - 2*(sin(a (i) k) sin(ωt) + a (i))

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k) sin(ωt))
Fine up to here but the next step is wrong
m * -sin(aik)sin(ωt)*ω2 =D(sin(ωt)(2sin(aik)) - 2*(sin(a (i) k) sin(ωt))
An immediate simplification is to cancel the sin(ωt) since it is a factor in every term on both sides of the equation. Next, you need to expand the sin(a (i+1) k) and sin(a (i-1) k) terms. Do you know a formula for expanding sin(A+B)?
 
  • #37
m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k) sin(ωt))

m * -sin(aik)*ω2 = D(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k)

I´m sorry but I don´t know a formula for expanding sin(A+B).
 
  • #38
bobibomi said:
m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k) sin(ωt))

m * -sin(aik)*ω2 = D(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k)

I´m sorry but I don´t know a formula for expanding sin(A+B).

Sin(A+B) = sin(A)cos(B)+cos(A)sin(B)
 
  • #39
m * -sin(aik)*ω2 = D(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k)

Sin(A+B) = sin(A)cos(B)+cos(A)sin(B)

m * -sin(aik)*ω2 = D(sin(a (sin(i)cos(1)+cos(i)sin(1)) k) + sin(a (sin(i)cos(1)+cos(i)+sin(1)) k)) - 2*(sin(a (i) k)

Is this right?
 
  • #40
bobibomi said:
m * -sin(aik)*ω2 = D(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k)

Sin(A+B) = sin(A)cos(B)+cos(A)sin(B)

m * -sin(aik)*ω2 = D(sin(a (sin(i)cos(1)+cos(i)sin(1)) k) + sin(a (sin(i)cos(1)+cos(i)+sin(1)) k)) - 2*(sin(a (i) k)

Is this right?
No. Multiply out a(i+1)k. You should get a sum of two terms. Apply the sin(A+B) rule to those two terms.
 
  • #41
1)a(i+1)k = aki+ak
2)a(i-1)k = aki- ak

is the first step right?
 
  • #42
bobibomi said:
1)a(i+1)k = aki+ak
2)a(i-1)k = aki- ak

is the first step right?
Yes.
 
  • #43
m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(a (i+1) k) + sin(a (i-1) k)) - 2*(sin(a (i) k) sin(ωt))

1) a(i+1)k = aki+ak
2) a(i-1)k = aki - ak


m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(aki+ak) + sin(aki - ak)) - 2*(sin(aik)sin(ωt))

Sin(A+B) = sin(A)cos(B)+cos(A)sin(B)

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(aki)cos(ak)+cos(aki)sin(ak) + sin(aki)cos(ak)-cos(aki)sin(ak) - 2*(sin(aik)sin(ωt))

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(sin(aki)cos(ak)+ sin(aki)cos(ak)- 2*(sin(aik)sin(ωt))

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(2*sin(aki)cos(ak)- 2*(sin(aik)sin(ωt))

m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(2*cos(ak)- 2*(sin(ωt))

m * -sin(aik)sin(ωt)*ω2 = D(-sin(ωt)(2*cos(ak)

m * -sin(aik)*ω2 = D(2*cos(ak)

Is this right ?
 
  • #44
bobibomi said:
m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(2*sin(aki)cos(ak)- 2*(sin(aik)sin(ωt))
Fine to there, but let's cancel that sin(ωt):
m * -sin(aik)*ω2 = D(2*sin(aki)cos(ak)- 2*sin(aik))
m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(2*cos(ak)- 2*(sin(ωt))

Is this right ?
No. How do you turn sin(aki)cos(ak)- sin(aik) into cos(ak)? That's like saying 3*4-3=4.
 
  • #45
m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(2*sin(aki)cos(ak)- 2*(sin(aik)sin(ωt))

Ok, this was the last right step, but I don´t know how I should go on
 
  • #46
bobibomi said:
m * -sin(aik)sin(ωt)*ω2 = D(sin(ωt)(2*sin(aki)cos(ak)- 2*(sin(aik)sin(ωt))

Ok, this was the last right step, but I don´t know how I should go on
Why are you not cancelling out the sin(ωt)? I thought we'd done that earlier, but it has crept back in. Every term, on both sides of the equation has that factor, and it cannot be identically zero, so cancel it out:
m * -sin(aik)*ω2 = D((2*sin(aki)cos(ak)- 2*(sin(aik))
Can you now see another factor that can be canceled out?
 
  • #47
m * -sin(aik)*ω2 = D((2*sin(aki)cos(ak)- 2*(sin(aik))

I think:
sin(aik) we can canceled out:
m *ω2 = D(2*sin(aki)cos(ak)- (sin(aik))

Is this right?
and then?
 
  • #48
bobibomi said:
m * -sin(aik)*ω2 = D((2*sin(aki)cos(ak)- 2*(sin(aik))

I think:
sin(aik) we can canceled out:
m *ω2 = D(2*sin(aki)cos(ak)- (sin(aik))

Is this right?
and then?
Yes, that is the factor to cancel out, but you've done it incorrectly on both sides.
I'm concerned that you seem to have some very basic misunderstandings about how to manipulate algebraic expressions. Try this for me: cancel out the A term in the equation B*(-A) = 2A+2A*C
 
  • #49
B*(-A) = 2A+2A*C
-AB = 2A + 2ACSorry bu I don´t know how to go on.
 
  • #50
bobibomi said:
B*(-A) = 2A+2A*C
-AB = 2A + 2AC


Sorry bu I don´t know how to go on.
Each of the terms in the above equation includes a factor A. So you can divide each of them by A (as long as A is not zero). If you divide everything in an equation by the same thing the equation is still true.
 
  • #51
B*(-A) = 2A+2A*C
-AB = 2A + 2AC
divide by A:
-B = 2 + 2C

so in the real equation:
m * -sin(aik)*ω2 = D((2*sin(aki)cos(ak)- 2*(sin(aik))

divide by sin(aik):
m * -1*ω2 = D((2*cos(ak)- 2)

Is this right?
 
  • #52
bobibomi said:
B*(-A) = 2A+2A*C
-AB = 2A + 2AC
divide by A:
-B = 2 + 2C

so in the real equation:
m * -sin(aik)*ω2 = D((2*sin(aki)cos(ak)- 2*(sin(aik))

divide by sin(aik):
m * -1*ω2 = D((2*cos(ak)- 2)

Is this right?
Yes! Next you have to find omega as a function of D, m, a and k.
 
  • #53
m * -1*ω2 = D((2*cos(ak)- 2)

omega as a function of D,m,a,k
maybe this:
ω = -√\stackrel{D((2*cos(ak)- 2)}{m}

is this right?
 
  • #54
bobibomi said:
m * -1*ω2 = D((2*cos(ak)- 2)

omega as a function of D,m,a,k
maybe this:
ω = -√\stackrel{D((2*cos(ak)- 2)}{m}

is this right?
Almost, but you've made a mistake with that minus sign. Try it again, doing one step at a time, and post all the steps.
Btw, for a fraction use \frac, not \stackrel.
 
  • #55
m * -1*ω2 = D((2*cos(ak)- 2)

omega as a function of D,m,a,k
maybe this:
-1*ω2 = \stackrel{D((2*cos(ak)- 2)}{m}
-ω = √ \stackrel{D((2*cos(ak)- 2)}{m}
ω = - √ \stackrel{D((2*cos(ak)- 2)}{m}

Where is my mistake?
 
  • #56
bobibomi said:
-1*ω2 = \stackrel{D((2*cos(ak)- 2)}{m}
-ω = √ \stackrel{D((2*cos(ak)- 2)}{m}
The square root of -A is not -√A. What would you get if you squared -√A?
 
  • #57
ω = √- \stackrel{D((2*cos(ak)- 2)}{m}

Is this right?
 
  • #58
bobibomi said:
ω = \sqrt{-\frac{D((2*cos(ak)- 2)}{m}}

Is this right?
Yes, but you can simplify it slightly. Take that minus sign inside the parentheses.
 
  • #59
ω = √- \stackrel{D((2*cos(ak)- 2)}{m}

I don´t know how do you mean that. Maybe you can show it
 
  • #60
bobibomi said:
\omega = \sqrt{-\frac{D((2*cos(ak)- 2)}{m}}

I don´t know how do you mean that. Maybe you can show it
First, g the minus sign into the numerator: -(A/B) = (-A/B); then take it inside the parentheses: -(A-B) = (-A - -B) = (-A+B) = (B-A).
Use this form for the LaTex:
[ itex]\omega = \sqrt{-\frac{D((2*cos(ak)- 2)}{m}}[/itex]
 

Similar threads

Energy of a crystal in thermal equilibrium
  • · Replies 1 ·
Replies
1
Views
2K
Temperature as a function of vibration energy quanta
  • · Replies 1 ·
Replies
1
Views
2K
How Does Charge and Elasticity Affect Hamiltonian Dynamics?
  • · Replies 8 ·
Replies
8
Views
1K
How to Calculate Amplitude Decrease in Damped Harmonic Motion After 10 Cycles?
  • · Replies 3 ·
Replies
3
Views
895
Equation of motion of magnetic dipole chain
  • · Replies 9 ·
Replies
9
Views
2K
Solving Oscillating Systems Homework: Angular Frequency & Velocity
  • · Replies 2 ·
Replies
2
Views
2K
Determining angular frequency and amplitude for SHM
  • · Replies 6 ·
Replies
6
Views
3K
[PoM] Electrons Fermi level in a crystal
  • · Replies 8 ·
Replies
8
Views
2K
Time period for a Simple Harmonic Oscillator to go from 0-1m
  • · Replies 2 ·
Replies
2
Views
3K
Kleppner Mechanics: Disk and coil spring
  • · Replies 6 ·
Replies
6
Views
2K