the ring and the sum is over all particles in the ring. Similar to the surface density, we can construct a Fourier series for the vertical displacement:

$$\label{eq:displacement}
Z\left  (R,\,\phi,\,t\right ) = {\rm Re}\left \{ \sum_{m=0}^\infty
Z_m\left (R,\,t\right ) e^{-im\phi}\right \}~,$$

where we define

$$\label{eq:bendmodefouriercoef}
Z_m\left (R,\,t\right ) \equiv A_\alpha^{-1} \sum_{j\in\alpha} 
\frac{z_j \, m_j }{\Sigma\left (R_j,\,\phi_j,\,t\right )} \, e^{im\phi_j}~.$$

The surface density at particle $j$, $\Sigma\left
  (R_j,\,\phi_j,\,t\right )$, is estimated by using the Fourier series for $\Sigma$ (equation \[eq:surfdens\]) truncated at the 5th order terms.

![Fourier mode decomposition of $Z(R, \phi, t \sim 4\,{\rm Gyr})$ for $m=0,\,1,\,2,$ and $3$, as indicated, for Model 1. Dotted concentric circles indicate increments of 5 kpc in radius. The rotation of the disc is counter-clockwise. The colour scale is in units of kpc, and differs between the upper and lower panels in order to highlight the extremal values for each $m$. The factor of $1/2$ for the zeroth order term in the Fourier Series is reflected in the $m = 0$ panel. From the figure it is clear that the (leading) $m = 1$ term is the dominant one in the decomposition. \[fig:Zm\_maps\]](fig4.pdf){width="\columnwidth"}

In Fig. \[fig:Zm\_maps\] we show the $m=0,\,1,\,2,$ and $3$ Fourier modes 