Discrete dipole approximation: Difference between revisions

Revision as of 21:23, 29 July 2025

Revision as of 21:50, 29 July 2025

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Each matrix-vector multiplication <math>\mathbf{G}_{\alpha\beta} \mathbf{P}_\beta</math> can be computed as a convolution when the dipoles are arranged on a regular grid, allowing the use of Fast Fourier Transforms (FFTs) to accelerate the solution.

Let <math>\boldsymbol{\beta}_j = \boldsymbol{\alpha}_j^{-1}</math> denote the inverse polarizability tensor for dipole <math>j</math>. Each <math>\boldsymbol{\beta}_j</math> is a complex-valued <math>3 \times 3</math> matrix. This gives:

The component-wise action of the inverse polarizability tensor in the rearranged dipole vector format is:

<math>

(\boldsymbol{\alpha}^{-1beta} \mathbf{P})_y_x &=

\begin{aligned}

(\boldsymbolmathrm{\alphadiag}^(\beta_{-1,xx}, \dots, \beta_{N,xx}) \, \mathbf{P})_x &= +

\mathrm{diag}(\alpha_beta_{1,xx}^{-1xy}, \dots, \alpha_beta_{N,xx}^{-1xy}) \, \mathbf{P}_x_y +

\mathrm{diag}(\alpha_beta_{1,xy}^{-1xz}, \dots, \alpha_beta_{N,xy}^{-1xz}) \, \mathbf{P}_y +_z

\mathrm{diag}(\alpha_{1,xz}^{-1}, \dots, \alpha_{N,xz}^{-1}) \, \mathbf{P}_z \\

(\boldsymbol{\alpha}^{-1} \mathbf{P})_y &=

\mathrm{diag}(\alpha_{1,yx}^{-1}, \dots, \alpha_{N,yx}^{-1}) \, \mathbf{P}_x +

\mathrm{diag}(\alpha_{1,yy}^{-1}, \dots, \alpha_{N,yy}^{-1}) \, \mathbf{P}_y +

\mathrm{diag}(\alpha_{1,yz}^{-1}, \dots, \alpha_{N,yz}^{-1}) \, \mathbf{P}_z \\

(\boldsymbol{\alpha}^{-1} \mathbf{P})_z &=

\mathrm{diag}(\alpha_{1,zx}^{-1}, \dots, \alpha_{N,zx}^{-1}) \, \mathbf{P}_x +

\mathrm{diag}(\alpha_{1,zy}^{-1}, \dots, \alpha_{N,zy}^{-1}) \, \mathbf{P}_y +

\mathrm{diag}(\alpha_{1,zz}^{-1}, \dots, \alpha_{N,zz}^{-1}) \, \mathbf{P}_z

\end{aligned}

</math>

<math>

For isotropic and homogeneous particle this reduces to simple multiplication by the inverse polarizability.

(\boldsymbol{\beta} \mathbf{P})_y =

\mathrm{diag}(\alpha_beta_{1,yx}^{-1}, \dots, \alpha_beta_{N,yx}^{-1}) \, \mathbf{P}_x +

\mathrm{diag}(\alpha_beta_{1,yy}^{-1}, \dots, \alpha_beta_{N,yy}^{-1}) \, \mathbf{P}_y +

\mathrm{diag}(\alpha_beta_{1,xz}^{-1yz}, \dots, \alpha_beta_{N,xz}^{-1yz}) \, \mathbf{P}_z \\

</math>

<math>

(\boldsymbol{\alpha}^{-1beta} \mathbf{P})_z &=

\mathrm{diag}(\alpha_beta_{1,zx}^{-1}, \dots, \alpha_beta_{N,zx}^{-1}) \, \mathbf{P}_x +

\mathrm{diag}(\alpha_beta_{1,zy}^{-1}, \dots, \alpha_beta_{N,zy}^{-1}) \, \mathbf{P}_y +

\mathrm{diag}(\alpha_beta_{1,yz}^{-1zz}, \dots, \alpha_beta_{N,yz}^{-1zz}) \, \mathbf{P}_z \\

</math>

In the special case of an isotropic and homogeneous particle, the polarizabilities <math>\boldsymbol{\alpha}_j</math> are identical for all dipoles and proportional to the identity matrix: <math>\boldsymbol{\alpha}_j = \alpha \, \mathbf{I}</math>. Then, the inverse becomes <math>\boldsymbol{\beta}_j = \alpha^{-1} \, \mathbf{I}</math>, all off-diagonal elements vanish, and the expressions reduce to a simple element-wise division:

<math>

(\boldsymbol{\beta} \mathbf{P}) = \frac{1}{\alpha} \, \mathbf{P}

</math>