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

[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.