Fourier transform of particles on a loaded string

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SUMMARY

The discussion focuses on the Fourier transform of the quantity \(\Sigma (x_{j+1} - x_j)^2\) for a loaded string, where \(x_j\) is defined as \(x_j = \Sigma A_k \cdot \exp(-iqkj)\). The Fourier transform is evaluated under periodic boundary conditions, leading to the expression \(\Sigma_j (\Sigma_k A_k \cdot \exp(-iqkj) (\exp(-iqk) - 1))^2\). The final result simplifies to \(\Sigma_k A_k \cdot A_{-k} \cdot (2\sin(qk/2))^2\), utilizing the identity \(e^{iqk/2} - e^{-iqk/2} = 2\sin(qk/2)\) for simplification.

PREREQUISITES
  • Understanding of Fourier transforms and their applications in physics.
  • Familiarity with complex exponentials and periodic boundary conditions.
  • Knowledge of trigonometric identities, particularly involving sine functions.
  • Basic principles of string theory and particle dynamics.
NEXT STEPS
  • Study the derivation of Fourier transforms in the context of string theory.
  • Learn about periodic boundary conditions and their implications in physical systems.
  • Explore trigonometric identities and their applications in simplifying complex expressions.
  • Investigate the role of sine functions in wave mechanics and particle interactions.
USEFUL FOR

Students and researchers in physics, particularly those focusing on wave mechanics, string theory, and Fourier analysis. This discussion is beneficial for anyone looking to deepen their understanding of Fourier transforms in the context of particle dynamics on strings.

Mrbluesky323
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Homework Statement


Can someone tell me how to Fourier transform this quantity:
[tex]\Sigma[/tex] (x_(j+1) - x_j)^2

where the sum is from j=1 to N

Homework Equations


Define the Fourier transform as
x_j = [tex]\Sigma[/tex] A_k *exp(-iqkj)
**Where i is sqrt(-1)
**The Sum is from k=0 to (N-1)
**q = (2*pi)/N
**Assume periodic boundary conditions (i.e. x_(N+1) = x_1)


The Attempt at a Solution



I get to

[tex]\Sigma[/tex]j ([tex]\Sigma[/tex]k Ak* exp(-iqkj) (exp(-iqk) - 1))^2

I'm told the answer one arrives at ought to be:

[tex]\Sigma[/tex]k Ak* A-k*(2sin(qk/2))^2
I'm not sure how to get there from where i am.
Any help would be appreciated.
 
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As a first step, factor out [itex]e^{-iqk/2}[/itex] from the expression [itex]e^{-iqk}-1[/itex] and note the identity:
[tex]e^{iqk/2} - e^{-iqk/2} = 2\sin{qk/2}[/tex]
 

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