- #1
jonny81
- 2
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Hello,
I want to prove that the function [itex]\mathcal{A}[/itex] in the 1D case satisfy
[itex]\mathcal{A}=\frac{48}{m}\sum_{j=1}^\infty \frac{\sin^2(qj/2)}{j^5}=\frac{12}{m}\left[2\zeta(5)-\text{Li}_5(e^{iq})-\text{Li}_5(e^{-iq})\right],[/itex]
with [itex]\text{Li}_n(z)[/itex] the polylogarithm function, and the matrix [itex]\mathcal{A}[/itex] in the 2D case satisfy
[itex]\mathcal{A}=\frac{3}{m}\sum_{j\neq 0}\left[1-\frac{5}{|\vec{r}_j^0|^2}\begin{pmatrix}(x_j^0)^2&x_jy_j\\x_jy_j&(y_j^0)^2\end{pmatrix}\right]\frac{\left(1-e^{i\vec{q}\vec{r}_j^0}\right)}{|\vec{r}_j^0|^5}[/itex]
Can somebody help me, please, to do this? In 2D the equilibrium positions [itex]\vec{r}_i^0[/itex] form a triangular lattice with basic lattice vectors [itex]a_1 = (1, 0)[/itex] and [itex]a_2 = (1,\sqrt{3})/2[/itex].
Starting point:
[itex]m\vec{\ddot x}_i=\frac{3}{2}\sum_{i\neq j}^{N}\left[\frac{5(\vec{r}_i^0 - \vec{r}_j^0)^2\left(\vec{x}_j-\vec{x}_i\right)}{|\vec{r}_i^0 - \vec{r}_j^0|^7}+\frac{\left(\vec{x}_i-\vec{x}_j\right)}{|\vec{r}_i^0 - \vec{r}_j^0|^5}\right][/itex]
with the ansatz
[itex]\vec{x}_i(t)=\epsilon_\lambda(\vec{q})e^{i\left(\vec{q}\vec{r}_i^0-\omega_\lambda(\vec{q})t\right)}[/itex]
[itex]\vec{\ddot x}_i(t)=-\omega^2\epsilon_\lambda(\vec{q})e^{i\left(\vec{q}\vec{r}_i^0-\omega_\lambda(\vec{q})t\right)}[/itex]
follows
[itex]-\omega^2_\lambda(\vec{q})\epsilon_\lambda(\vec{q})=\mathcal{A}\epsilon_\lambda(\vec{q})[/itex]
Thanks!
I want to prove that the function [itex]\mathcal{A}[/itex] in the 1D case satisfy
[itex]\mathcal{A}=\frac{48}{m}\sum_{j=1}^\infty \frac{\sin^2(qj/2)}{j^5}=\frac{12}{m}\left[2\zeta(5)-\text{Li}_5(e^{iq})-\text{Li}_5(e^{-iq})\right],[/itex]
with [itex]\text{Li}_n(z)[/itex] the polylogarithm function, and the matrix [itex]\mathcal{A}[/itex] in the 2D case satisfy
[itex]\mathcal{A}=\frac{3}{m}\sum_{j\neq 0}\left[1-\frac{5}{|\vec{r}_j^0|^2}\begin{pmatrix}(x_j^0)^2&x_jy_j\\x_jy_j&(y_j^0)^2\end{pmatrix}\right]\frac{\left(1-e^{i\vec{q}\vec{r}_j^0}\right)}{|\vec{r}_j^0|^5}[/itex]
Can somebody help me, please, to do this? In 2D the equilibrium positions [itex]\vec{r}_i^0[/itex] form a triangular lattice with basic lattice vectors [itex]a_1 = (1, 0)[/itex] and [itex]a_2 = (1,\sqrt{3})/2[/itex].
Starting point:
[itex]m\vec{\ddot x}_i=\frac{3}{2}\sum_{i\neq j}^{N}\left[\frac{5(\vec{r}_i^0 - \vec{r}_j^0)^2\left(\vec{x}_j-\vec{x}_i\right)}{|\vec{r}_i^0 - \vec{r}_j^0|^7}+\frac{\left(\vec{x}_i-\vec{x}_j\right)}{|\vec{r}_i^0 - \vec{r}_j^0|^5}\right][/itex]
with the ansatz
[itex]\vec{x}_i(t)=\epsilon_\lambda(\vec{q})e^{i\left(\vec{q}\vec{r}_i^0-\omega_\lambda(\vec{q})t\right)}[/itex]
[itex]\vec{\ddot x}_i(t)=-\omega^2\epsilon_\lambda(\vec{q})e^{i\left(\vec{q}\vec{r}_i^0-\omega_\lambda(\vec{q})t\right)}[/itex]
follows
[itex]-\omega^2_\lambda(\vec{q})\epsilon_\lambda(\vec{q})=\mathcal{A}\epsilon_\lambda(\vec{q})[/itex]
Thanks!
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