# One-Dimensional chain

## Homework Statement

The atoms having in each case, the mass m and are located at positions with x i i ∈ Z. The rest position of the ith atom is i · a where a is the lattice constant of the crystal.
The interaction between the atoms can be modeled in a simple approximation, as a spring force of the spring constant D between adjacent atoms. The i-th atom causes therefore the (i - 1) th power of the atom has a size Fi→i−1 = D (xi − xi−1 − a).

(1) Set the equation of motion for the position of the ith atom in the crystal lattice and show that the equations of motion of the atoms are solved by standing waves of the form
xi (t) x = sin (a i k) sin (ω t) + a i. Look for the solution ω as a function of D, m, a and k, and determine the maximum value of ω as a function of the parameters of the crystal. Also sketch the history of ω as a function of k.

|k | = 2 π/λ
ω = 2 π f

## The Attempt at a Solution

Equation of motion for the position of the i-th atom:
Mx ̈i = C(xi+1 + xi-1 2xi)

I don´t know how i should go on.

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isn´t there anyone who could help me??
it is quite important for me to solve this problem.
i just can´t go on, i have no idea

haruspex
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isn´t there anyone who could help me??
it is quite important for me to solve this problem.
i just can´t go on, i have no idea
You are more likely to get help if you take the trouble to use LaTex to lay out your equations properly.
I assume your equation is supposed to read ##m \ddot x = D(x_{i+1}+x_{i-1}-2 x_i)##. Plug in the solution form you are given and see what you get.

Do you mean this is the right right equation of motion?

But how should i solve this equation with standing waves?

How should i go on?

haruspex
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Do you mean this is the right right equation of motion?

But how should i solve this equation with standing waves?

How should i go on?
You are given a form of solution and asked to show that it is a solution of the equation (if the parameters are set appropriately). So substitute the given form for xi into the differential equation and deduce the values of the parameters.

xi = Xo*e^i(kia-ωt)

xi-1 = xi*e^(-ika)
xi+1 = xi*e^(ika)

-Mω^2*xi =C(e^(ika) +e^(-ika) - 2)xi

ω^2 =2*C/M*(1-coska) = 4C/M*sin^2 * ka/2

ω = 2*√C/M * sin(ka/2)

is this right?
what have i done wrong?
what should i do know?

sorry, but can nobody say me if this is right?
i have no idea how i should go on...

thank for all your great help :) :)

haruspex
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Sorry, but I had no net access for a few days. Your failure to use LaTex makes it very hard to follow what you are doing. Sometimes i should be a subscript (perhaps), other times it stands for root of minus 1. I'm not even sure what the harmonic solution suggested in the OP is saying. Is that really ai in two places? If you repost in LaTex I'll take another look.

xi = x0 * ei*(k*i*a-ω*t)

xi-1 = xi*e(-ika)
xi+1 = xi*e(ika)

-Mω2xi=C(e(ika)+e(-ika)- 2)xi

ω2 =2*C/M*(1-coska) = 4C/M*sin2 * ka/2

ω = 2*√C/M * sin(ka/2)

I hope you can help me now

haruspex
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xi = x0 * ei*(k*i*a-ω*t)

xi-1 = xi*e(-ika)
xi+1 = xi*e(ika)

-Mω2xi=C(e(ika)+e(-ika)- 2)xi

ω2 =2*C/M*(1-coska) = 4C/M*sin2 * ka/2

ω = 2*√C/M * sin(ka/2)

I hope you can help me now
Well, it's still a bit confusing because you appear to be using i for two roles. How about using j for one of them?
But I don't understand how you have converted the given form of solution, a product of two sine functions, into the exponential form above. The original is purely real, whereas your form has an imaginary component.
Or maybe the i in your form is always referring to the index i, as in the original, and not sometimes meaning the square root of minus 1? In that case I don't understand how you changed sine into exp at all.

Here, k is the wave number (with dimension m^-1), x0 the amplitude and ω is the angular frequency, a is the distance between nearest the neighbors and i*a the rest position of the atom and index i.

Eigenmodes can be written to as plane waves are propagating along the chain:
xi = x0 * ei*(k*i*a-ω*t)

The newly introduced intrinsic modes no longer refer to individual atoms but on the totality of the atoms. They are characterized by their harmonic time dependence and are also called normal coordinates.
xi-1 = xi*e(-ika)
xi+1 = xi*e(ika)

By inserting into the equation of motion, we obtain:
-Mω2xi=C(e(ika)+e(-ika)- 2)xi

We divide by -Mxi and replace the exponential functions by trigonometric and receive:
ω2 =2*C/M*(1-coska) = 4C/M*sin2 * ka/2

From this we obtain the natural frequency:
ω = 2*√C/M * sin(ka/2)

I hope you can help me now

haruspex
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I fail to see what these exponential functions have to do with answering the given question. You are given a form of solution involving a product of two sine functions. No exponentials. Why not just substitute that form into the differential equation and demonstrate that it is a solution (when the parameters are set appropriately)?
If the exponential form you introduced is right then there's something about it that puzzles me. In the exponent you appear to have a quadratic in i. Is that correct or does one of the two i's represent the square root of -1? I keep asking this question in different ways and you have still not answered it.

xi = x0 * ei*(k*i*a-ω*t)

xi-1 = xi*e(-ika)
xi+1 = xi*e(ika)

-Mω2xi=C(e(ika)+e(-ika)- 2)xi

you think I should stop right here.?
no the is stand for is.

can you please show me what i should do

haruspex
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We don't seem to be making much progress. I have the feeling that you are quoting some textbook theory (the source of these equations with exponentials) but the xi terms in them do not correspond to those in the OP, so those equations are not relevant.
Is this possibly the case?

Okay, this equations are not the solution for this task.
But how should i go on? I have no other idea at the moment.
Or can you give me a tip where I should start to find the right solution.

haruspex
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Okay, this equations are not the solution for this task.
But how should i go on? I have no other idea at the moment.
Or can you give me a tip where I should start to find the right solution.
I have suggested another approach three times - posts #3, #5, and #12. Do you understand what I suggested?

I have got the equation of movement :
mx = D(xi+1 + xi-1 - 2xi)

and the equation for standing waves:
xi(t) = x * sin (a * i * k) * sin (ω * t) + a * i

Plug in the solution form you are given and see what you get.

So substitute the given form for xi into the differential equation and deduce the values of the parameters.

I am not sure if I understand you right.

Maybe:
x = (D(xi+1 + xi-1 - 2xi) / m

and this in (2)
xi(t) = (D(xi+1 + xi-1 - 2xi) / m * sin (a * i * k) * sin (ω * t) + a * i

I think that this is not the right way.

haruspex
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I have got the equation of movement :
mx = D(xi+1 + xi-1 - 2xi)

and the equation for standing waves:
xi(t) = x * sin (a * i * k) * sin (ω * t) + a * i
Need to get those equations right first:
##m \ddot x_i = D(x_{i+1} + x_{i-1}-2x_i)##
##x_i = sin(a i k) sin(\omega t) + a i##
OK?
Try again with those.

##m \ddot x_i = D(x_{i+1} + x_{i-1}-2x_i)##
##x_i = sin(a i k) sin(\omega t) + a i##

These are the two given equations and now i should solve this problem:

Look for the solution ω as a function of D, m, a and k, and determine the maximum value of ω as a function of the parameters of the crystal. Also sketch the history of ω as a function of k.

And I really don´t know what I should do with these two equations to get a solution.
Should I equate the formulas, or transform to a variable?
I do not know.

haruspex
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These are the two given equations and now i should solve this problem:

Look for the solution ω as a function of D, m, a and k, and determine the maximum value of ω as a function of the parameters of the crystal. Also sketch the history of ω as a function of k.

And I really don´t know what I should do with these two equations to get a solution.
Should I equate the formulas, or transform to a variable?
I do not know.
Use the second formula to substitute for all the xi, xi-1 and xi+1 references in the first.

(1)##m \ddot x_i = D(x_{i+1} + x_{i-1}-2x_i)##
(2)##x_i = sin(a i k) sin(\omega t) + a i##

xi+1=sin(a i k) sin(ωt) + a i +1
xi-1=sin(a i k) sin(ωt) + a i -1

Is this right, how i transformed the two equations?

then in (1):
mxi = D(sin(a i k) sin(ω t) + a i +1 + sin(a i k) sin(ωt) + a i -1 -2*(sin(a i k) sin(ωt) + a i)

Is this right?

haruspex
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(1)##m \ddot x_i = D(x_{i+1} + x_{i-1}-2x_i)##
(2)##x_i = sin(a i k) sin(\omega t) + a i##

xi+1=sin(a i k) sin(ωt) + a i +1
xi-1=sin(a i k) sin(ωt) + a i -1

Is this right, how i transformed the two equations?
No, I'm sorry but that is wrong. Take the equation ##x_i = sin(a i k) sin(\omega t) + a i## then change every instance of i to (i-1).
then in (1):
mxi = D(sin(a i k) sin(ω t) + a i +1 + sin(a i k) sin(ωt) + a i -1 -2*(sin(a i k) sin(ωt) + a i)

Is this right?
No. The LHS is not ##m x_i ## , it is ##m \ddot x_i ##. Do you understand what that means?

(1)##m \ddot x_i = D(x_{i+1} + x_{i-1}-2x_i)##
(2)##x_i = sin(a i k) sin(\omega t) + a i##

(2) xi-1 = sin(a (i-1) k) sin(ωt) + a (i-1)
xi+1 = sin(a (i+1) k) sin(ωt) + a (i+1)

For the second question:
No I don´t know the differences.

haruspex
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(1)##m \ddot x_i = D(x_{i+1} + x_{i-1}-2x_i)##
(2)##x_i = sin(a i k) sin(\omega t) + a i##

(2) xi-1 = sin(a (i-1) k) sin(ωt) + a (i-1)
xi+1 = sin(a (i+1) k) sin(ωt) + a (i+1)
yes
For the second question:
No I don´t know the differences.
Each dot represents differentiation wrt time. So the double dot indicates acceleration. You need to take the proposed solution for xi and differentiate it twice wrt t.

(1)##m \ddot x_i = D(x_{i+1} + x_{i-1}-2x_i)##
(2)##x_i = sin(a i k) sin(\omega t) + a i##

(2) xi-1 = sin(a (i-1) k) sin(ωt) + a (i-1)
xi+1 = sin(a (i+1) k) sin(ωt) + a (i+1)

I think I need to form the second derivation:

(2)
x_i = -sin(a i k) * -sin(ωt)
xi-1 = -sin(a (i-1) k) * -sin(ωt)
xi+1 = -sin(a (i+1) k) * -sin(ωt)

is this right? Can I insert it in the first equation?

haruspex
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I think I need to form the second derivation:

(2)
x_i = -sin(a i k) * -sin(ωt)
I assume you meant ##\ddot x_i =##, but the RHS is wrong. The ##\sin(aik)## term is just a constant factor, not a function of time, so that doesn't change. You correctly got a minus sign from differentiating sin(ωt) twice, but you forgot something to do with the ω. try it again.

(2) xi = sin(a i k) * -sin(ωt) * ω2

Is this right?

haruspex
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(2) xi = sin(a i k) * -sin(ωt) * ω2

Is this right?
Yes, sort of. More correctly ##\ddot x_i = - \sin(aik) \sin(\omega t) \omega^2 ##
Now write out your original DE with all the substitutions in place.

(1)##m \ddot x_i = D(x_{i+1} + x_{i-1}-2x_i)##

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

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

Is this right? How should i go on?

haruspex
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(1)##m \ddot x_i = D(x_{i+1} + x_{i-1}-2x_i)##

(2)
xi=-sin(aik)sin(ωt)*ω2
No no no! ##\ddot x_i = ## -sin(aik)sin(ωt)*ω2. Do not confuse ##x_i ## and ##\ddot x_i ## .
xi-1=-sin(a(i-1)k)sin(ωt)*ω2
xi+1=-sin(a(i+1)k)sin(ωt)*ω2
No. You had correct expressions for xi-1 and xi+1 before. There is no differentiation for these, so no ω2 term.
How should i go on?
In the differential equation ##m\ddot x_i=D(x_{i+1}+x_{i−1}−2x_i)## use ##\ddot x_i = ## -sin(aik)sin(ωt)*ω2 to substitute for ##\ddot x_i## and
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)
to substitute for xi-1, xi+1 and xi on the RHS.