I have what I suspect is a simple problem. Yet, I find myself stuck.
My equation describes the electric field at a point $\mathbf{r}$ due to cumulative polarization $\mathbf{P}(\mathbf{r^\prime})$ at points $\mathbf{r^\prime}$, and looks like this:
$$ \mathbf{E}(\mathbf{r}) = \int \left\{\frac{3(\mathbf{r}-\mathbf{r^\prime})}{|\mathbf{r}-\mathbf{r^\prime}|^5}[\mathbf{P}(\mathbf{r^\prime})\cdot(\mathbf{r}-\mathbf{r^\prime})] - \frac{\mathbf{P}(\mathbf{r^\prime})}{|\mathbf{r}-\mathbf{r^\prime}|^3}\right\}d\mathbf{r^\prime} $$
I want to eliminate the $\mathbf{r^\prime}$ by expressing $\mathbf{E}(\mathbf{r})$ as a convolution of two functions of the vector $\mathbf{r}$.
Wikipedia tells me that the convolution is defined as: $$ (f * g )(t) = \int_{-\infty}^\infty f(\tau) g(t - \tau) \, d\tau $$ So it seems there should be a way to express my equation as a convolution with respect to $\mathbf{r}-\mathbf{r^\prime}$. However, I'm not sure if this applies to vectors, or how to deal with the dot product in my equation above.
Any clues would be really appreciated!
Hugh
We begin with splitting the integral into two terms: $$ \mathbf{E}(\mathbf{r}) = \int \left\{\frac{3(\mathbf{r}-\mathbf{r^\prime})}{|\mathbf{r}-\mathbf{r^\prime}|^5}[\mathbf{P}(\mathbf{r^\prime})\cdot(\mathbf{r}-\mathbf{r^\prime})] - \frac{\mathbf{P}(\mathbf{r^\prime})}{|\mathbf{r}-\mathbf{r^\prime}|^3}\right\}d\mathbf{r^\prime} \\ = \int \frac{3(\mathbf{r}-\mathbf{r^\prime})}{|\mathbf{r}-\mathbf{r^\prime}|^5}[\mathbf{P}(\mathbf{r^\prime})\cdot(\mathbf{r}-\mathbf{r^\prime})] d\mathbf{r^\prime} - \int \frac{\mathbf{P}(\mathbf{r^\prime})}{|\mathbf{r}-\mathbf{r^\prime}|^3} d\mathbf{r^\prime} $$
The second integral can easily be written as a convolution: $$ \int \frac{\mathbf{P}(\mathbf{r^\prime})}{|\mathbf{r}-\mathbf{r^\prime}|^3} d\mathbf{r^\prime} = \mathbf{P}(\mathbf{r}) * \frac{1}{|\mathbf{r}|^3} $$
The first integral is more difficult. The $i$th coordinate is somewhat easier: $$ \left( \int \frac{3(\mathbf{r}-\mathbf{r^\prime})}{|\mathbf{r}-\mathbf{r^\prime}|^5}[\mathbf{P}(\mathbf{r^\prime})\cdot(\mathbf{r}-\mathbf{r^\prime})] d\mathbf{r^\prime} \right)_i = \int \frac{3(x_i-x_i^\prime)}{|\mathbf{r}-\mathbf{r^\prime}|^5}[\mathbf{P}(\mathbf{r^\prime})\cdot(\mathbf{r}-\mathbf{r^\prime})] d\mathbf{r^\prime} \\ = \int \mathbf{P}(\mathbf{r^\prime})\cdot\frac{3(x_i-x_i^\prime)(\mathbf{r}-\mathbf{r^\prime})}{|\mathbf{r}-\mathbf{r^\prime}|^5} d\mathbf{r^\prime} = \mathbf{P}(\mathbf{r}) \stackrel{\cdot}{*} \frac{3 x_i \mathbf{r}}{|\mathbf{r}|^5} $$ where the operator $\stackrel{\cdot}{*}$ denotes that both a convolution and a dot product are made: $$ \mathbf{P}(\mathbf{r}) \stackrel{\cdot}{*} \frac{3 x_i \mathbf{r}}{|\mathbf{r}|^5} = \sum_j P_j(\mathbf{r}) * \frac{3 x_i x_j}{|\mathbf{r}|^5} $$
Thus, $$ \int \frac{3(\mathbf{r}-\mathbf{r^\prime})}{|\mathbf{r}-\mathbf{r^\prime}|^5}[\mathbf{P}(\mathbf{r^\prime})\cdot(\mathbf{r}-\mathbf{r^\prime})] d\mathbf{r^\prime} = \sum_i \left( \mathbf{P}(\mathbf{r}) \stackrel{\cdot}{*} \frac{3 x_i \mathbf{r}}{|\mathbf{r}|^5} \right) \mathbf{e}_i = \sum_i \sum_j P_j(\mathbf{r}) * \frac{3 x_i x_j}{|\mathbf{r}|^5} \mathbf{e}_i $$ so $$ \mathbf{E}(\mathbf{r}) = \sum_i \sum_j P_j(\mathbf{r}) * \frac{3 x_i x_j}{|\mathbf{r}|^5} \mathbf{e}_i + \mathbf{P}(\mathbf{r}) * \frac{1}{|\mathbf{r}|^3} $$
Isn't the integrand the expansion of something? I think that I recognize it. The non-expanded expression might be easier to write as a pure convolution.