# Notation of a vector containing equation in a paper

I'm trying to implement Coulomb long range interactions into a molecular simulation program using a particle-particle/particle-mesh Ewald solver. The following equation from the paper "How to mesh up Ewald sums. I." (http://arxiv.org/abs/cond-mat/9807099) is problematic:

$\widetilde{g}(\boldsymbol{k}) = \cfrac{4\pi}{k^2}$ (above eq. (9))

How are the values of vector $\boldsymbol{k}$ used in this equation? My assumption is, I can implement it like this:

$\widetilde{g}(\boldsymbol{k}) = \cfrac{4\pi}{k_x^2} \cfrac{4\pi}{k_y^2} \cfrac{4\pi}{k_z^2}= \cfrac{(4\pi)^3}{k_x^2 k_y^2 k_z^2}$

Is this correct? (Same problem with $\widetilde{\gamma}(\boldsymbol{k}) = \exp{\frac{-k^2 }{4 \alpha^2}}$)

I come to this assumption beacuse eq. (18) in this paper is

$\widetilde{W}(k) = h\left( \cfrac{\sin(kh/2)}{kh/2} \right)^P$

and in a second paper (http://arxiv.org/abs/cond-mat/9807100, eq. (6)) they write

$\widetilde{W}(\boldsymbol{k}) = h^3 \left( \cfrac{\sin(\frac{1}{2}k_xh) \sin(\frac{1}{2}k_yh) \sin(\frac{1}{2}k_zh)}{\frac{1}{2}k_xh \quad \frac{1}{2}k_yh \quad \frac{1}{2}k_zh} \right)^P$

(note the difference in $\boldsymbol{k}$ and $k$).

The notation $k^2$ for vector $\boldsymbol{k}$ means the square length/magnitude of the vector; if $\boldsymbol{k} = (k_x,k_y,k_z)$ in Cartesian coordinates then $$k^2 = \left|\boldsymbol{k}\right|^2 = k_x^2+k_y^2+k_z^2$$
$$\widetilde{g}(\boldsymbol{k}) = \frac{4\pi}{k_x^2+k_y^2+k_z^2}$$
• But in eq. (31) they use $|\widetilde{\boldsymbol{D}}(\boldsymbol{k})|^2$ for the length of a vector, do you think they mixed up notations? – Jakob May 11 '16 at 13:59
• @Jakob It is not unusual for $\boldsymbol{a}^2=a^2=|\boldsymbol{a}|^2$ to coexist as notations in a paper. In this specific case it probably refers to the specific requirement of taking the sums of absolute values, $$|\widetilde{\boldsymbol{D}}(\boldsymbol k)|^2=|\widetilde{D}_x(\boldsymbol k)|^2+|\widetilde{D}_y(\boldsymbol k)|^2+|\widetilde{D}_z(\boldsymbol k)|^2,$$ since the Fourier transforms $\widetilde{D}_j$ are in general complex-valued (whereas $\boldsymbol k$ would be real). – Emilio Pisanty May 11 '16 at 14:05
• @EmilioPisanty Is $\widetilde{D_x}(\boldsymbol{k})$ in this case the same as $\widetilde{D}(k_x)$? – Jakob May 11 '16 at 14:20
• @Jakob Not at all. $\widetilde{\boldsymbol{D}}$ is a vector-valued function of a vector argument; you work the rest out. – Emilio Pisanty May 11 '16 at 15:10