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

[haruspex]

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)?

 

[bobibomi]

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).

 

[haruspex]

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)

 

[bobibomi]

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?

 

[haruspex]

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.

 

[bobibomi]

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

is the first step right?

 

[haruspex]

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

is the first step right?

Yes.

 

[bobibomi]

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 ?

 

[haruspex]

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.

 

[bobibomi]

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

 

[haruspex]

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?

 

[bobibomi]

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?

 

[haruspex]

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

 

[bobibomi]

B*(-A) = 2A+2A*C
-AB = 2A + 2ACSorry bu I don´t know how to go on.

 

[haruspex]

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.

 

[bobibomi]

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?

 

[haruspex]

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.

 

[bobibomi]

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?

 

[haruspex]

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.

 

[bobibomi]

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?

 

[haruspex]

-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?

 

[bobibomi]

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

Is this right?

 

[haruspex]

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.

 

[bobibomi]

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

I don´t know how do you mean that. Maybe you can show it

 

[haruspex]

$\omega = \sqrt{-\frac{D((2*cos(ak)- 2)}{m}}$

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]

 

[bobibomi]

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?

 

[haruspex]

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.

 

[bobibomi]

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

 

[haruspex]

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.