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

[bobibomi]

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

 

[haruspex]

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

 

[bobibomi]

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?

 

[haruspex]

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?

 

[bobibomi]

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?

 

[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]