- #1
sebpinski
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Hi all,
I'm having a few problems with crystal dynamics of a simple monatomic chain.
Taking the dispersion relation:
[itex]\omega^2 = \frac{4k}{m}\left(\sin^2 \left( \frac{\kappa a}{2}\right)\right)[/itex]
Where k=spring constant, m=mass, [itex]\kappa[/itex]=wavevector, a= lattice constant and [itex]\omega[/itex]= frequency.
Now assume we have a periodic chain made up of four ions and applying the boundary conditions we get:
[itex]\kappa = \frac{2 \pi n}{a L}[/itex]
where n is just an integer and L is the length of the chain (here L=4).
Substituting this into the dispersion relation and assuming mass and springs are unity we can calculate the four frequencies for the four admissible modes as [itex]\omega^2=[/itex]0,2,2 and 4.
Now solving numerically, we construct the dynamical matrix using the same values for masses and springs as:
(2, -1, 0, -1; -1, 2, -1, 0; 0, -1, 2, -1; -1, 0, -1, 2)
Diagonalizing this gives the eigenvalues ([itex]\omega^2[/itex]) identical to the ones above calculated analytically and the following eigenvectors (extensions from equilibrium):
(0.5, 0.5, 0.5, 0.5)
(0.7071, 0, -0.7071, 0)
(0, 0.7071, 0, -0.7071)
(-0.5, 0.5, -0.5, 0.5)
I've tried analytically to obtain these eigenvectors but I just can't see how. I recall the equation for extension from equilibrium as:
[itex] u_s = U exp[\frac{i 2 \pi n s}{L}] [/itex]
Again L=4, U is the amplitude and for normalization it is [itex]L^{-1/2}[/itex], [itex]u_s[/itex] is the extension of site s. If I just substitute the values site numbers 1 to 4 in along with mode numbers n= 1 to 4. I do not reproduce the eigenvectors found from diagonalization.
Thanks for any help in advance.
Sebastian
I'm having a few problems with crystal dynamics of a simple monatomic chain.
Taking the dispersion relation:
[itex]\omega^2 = \frac{4k}{m}\left(\sin^2 \left( \frac{\kappa a}{2}\right)\right)[/itex]
Where k=spring constant, m=mass, [itex]\kappa[/itex]=wavevector, a= lattice constant and [itex]\omega[/itex]= frequency.
Now assume we have a periodic chain made up of four ions and applying the boundary conditions we get:
[itex]\kappa = \frac{2 \pi n}{a L}[/itex]
where n is just an integer and L is the length of the chain (here L=4).
Substituting this into the dispersion relation and assuming mass and springs are unity we can calculate the four frequencies for the four admissible modes as [itex]\omega^2=[/itex]0,2,2 and 4.
Now solving numerically, we construct the dynamical matrix using the same values for masses and springs as:
(2, -1, 0, -1; -1, 2, -1, 0; 0, -1, 2, -1; -1, 0, -1, 2)
Diagonalizing this gives the eigenvalues ([itex]\omega^2[/itex]) identical to the ones above calculated analytically and the following eigenvectors (extensions from equilibrium):
(0.5, 0.5, 0.5, 0.5)
(0.7071, 0, -0.7071, 0)
(0, 0.7071, 0, -0.7071)
(-0.5, 0.5, -0.5, 0.5)
I've tried analytically to obtain these eigenvectors but I just can't see how. I recall the equation for extension from equilibrium as:
[itex] u_s = U exp[\frac{i 2 \pi n s}{L}] [/itex]
Again L=4, U is the amplitude and for normalization it is [itex]L^{-1/2}[/itex], [itex]u_s[/itex] is the extension of site s. If I just substitute the values site numbers 1 to 4 in along with mode numbers n= 1 to 4. I do not reproduce the eigenvectors found from diagonalization.
Thanks for any help in advance.
Sebastian