Question

Consider one dimensional chain consists of many atoms with identical masses m and separated by a distance of a (lattice parameter). At equilibrium the abscissa of the nth atom is xᵃₙ = na . the atoms can move only in a direction parallel to the chain and we note uₙ the displacement with respect to its equilibrium position. for harmonic approximation. uₙ = a e⁻¹⁽ᵈˣ⁻ᵏˣ ᵃ/ⁿ⁾ we consider the first and second nearest neighbours’ interaction.

Determine the dispersion relation ω(k) for normal modes

Answer 1

In a one-dimensional chain of atoms, we can use the equation uₙ = a e⁻¹⁽ᵈˣ⁻ᵏˣ ᵃ/ⁿ⁾ to represent the displacement of atoms from their equilibrium position. By considering the interactions with first and second nearest neighbors, we can determine the dispersion relation ω(k) for normal modes.

Step-by-step explanation:

In this one-dimensional chain of atoms, the displacement of the nth atom from its equilibrium position can be represented as uₙ = a e⁻¹⁽ᵈˣ⁻ᵏˣ ᵃ/ⁿ⁾. To determine the dispersion relation ω(k) for normal modes, we consider the first and second nearest neighbors' interactions. The dispersion relation can be found by solving the equation of motion for the uₙ's and finding the frequencies ω(k) that satisfy the equation.

The total potential of the system would be influenced by the positions of all interacting neighbors given the interactions are not limited to just adjacent atoms.
In a harmonic approximation, these inter-atomic forces can be approximated as springs obeying Hooke's law, which states that the force exerted by a spring is proportional to the displacement from the equilibrium. The equation uₙ = a e⁻¹(ᴅₓ⁻ᴋₓ ᵃ/ₙ) is a typical representation of a displacement that varies exponentially with the position along the chain.