Fourier transform for loaded string with periodic boundary conditions.

[RufusDarkstar]

Homework Statement

So we have a string of N particles connected by springs like so:
*...*...*...*...*

A corresponding Hamiltonian that looks like:
H= 1/2* $$\Sigma$$ P_j^2 + (x_j - x_(j+1) )^2

Where x is transverse position of the particle as measured from the equilibrium position, and P is the momentum (essentially x-dot in this case)

Assume mass is 1.

The boundary condition is as follows:
x_0 = x_n
p_0 = p_n.

Assume the initial state of the system can be defined as:
$$x_{j}$$ = sin($$\pi$$*j/N)
$$p_{j}$$= 0

The object of the problem is to Fourier transform the coordinates and derive the equations of motion in momentum space.

Homework Equations

**$$A_{k}$$ = $$\frac{1}{n}$$ * $$\Sigma^{N}_{1}$$ $$x_{i}$$*$$e^{i*w*k*j}$$

**$$x_{j}$$ = $$\Sigma^{N-1}_{0}$$ $$A_{k}$$*$$e^{-i*w*k*j}$$

where w = 2*$$\pi$$/N.

** sin^2(x)= (1 - cos(2x))/2) (identity)

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The Attempt at a Solution

So, my first goal is to evaluate $$\Sigma$$($$x_{j}$$ - $$x_{j+1}$$))^2

which looks like:
$$\Sigma^{N}_{1}$$
(($$\Sigma^{N-1}_{0}$$ $$A_{k}$$*$$e^{-i*w*k*j}$$) - ($$\Sigma^{N-1}_{0}$$ $$A_{k}$$*$$e^{-i*w*k*j+1}$$))^2

=$$\Pi^{2}_{1}$$ $$\Sigma^{N-1}_{0}$$ $$A_{k}$$*$$e^{-i*w*k*j}$$* (1-$$e^{-i*w*k}$$)

= $$\Pi^{2}_{1}$$ $$\Sigma^{N-1}_{0}$$ $$A_{k}$$$$e^{-i*w*k*j}$$*2*sin^2(w*k)

This is as far as i can get. I'm not sure how to evaluate the Pi operator.

N.B. When i hit preview, this message looks like insanity. I'm hoping that that's just a glitch in the preview code.

 

[mark726]

Hi there,

Thank you for your post. I'm happy to help you with your problem. Let's start by looking at the Hamiltonian equation:

H= 1/2* \Sigma P_j^2 + (x_j - x_(j+1) )^2

We can simplify the equation by using the boundary condition x_0 = x_n and p_0 = p_n:

H= 1/2* \Sigma P_j^2 + (x_j - x_n )^2

Next, we can use the given initial state of the system to calculate x_j:

x_{j} = sin(\pi*j/N)

We can also use the identity given in the homework equations to simplify the equation:

sin^2(x)= (1 - cos(2x))/2)

Therefore, we can rewrite the Hamiltonian as:

H= 1/2* \Sigma P_j^2 + (sin(\pi*j/N) - sin(\pi*(j+1)/N) )^2

Now, let's focus on evaluating the sum of (x_j - x_(j+1))^2. We can rewrite it as:

(x_j - x_(j+1))^2 = (sin(\pi*j/N) - sin(\pi*(j+1)/N) )^2

= (sin(\pi*j/N))^2 + (sin(\pi*(j+1)/N))^2 - 2*sin(\pi*j/N)*sin(\pi*(j+1)/N)

= (1/2 - 1/2*cos(2\pi*j/N)) + (1/2 - 1/2*cos(2\pi*(j+1)/N)) - sin(\pi*j/N)*sin(\pi*(j+1)/N)

= 1 - 1/2*cos(2\pi*j/N) - 1/2*cos(2\pi*(j+1)/N) - sin(\pi*j/N)*sin(\pi*(j+1)/N)

= 1 - 1/2*cos(2\pi*j/N) - 1/2*cos(2\pi*j/N + 2\pi/N) - sin(\pi*j/N)*sin(\pi*(j+1)/N)

= 1 - 1/2*cos(2\pi*j/N) - 1/2*cos(2\pi*j