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

[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.