Defining the Centre of the Volume

In the above formula it is assumed that the volume of integration is centred about $x = 0$. If this is not the case (when the origin of the world coordinates is not at the centre of the brain) then the matrix $M$ can be simply modified to include a general centre.

That is, let $x_c$ be the desired centre, and define a new coordinate

$$\tilde{x} = x - x_c.$$ (33)

This new coordinate is now zero when $x = x_c$, so that the above derivation is valid for $\tilde{x}$. Furthermore, this translation can be expressed in matrix form (using homogeneous coordinates) as $\tilde{x} = M_c x$ where

$$M_c = \left[ \begin{array}{cc} I & -x_c \\ 0 \; 0 \; 0 & 1 \end{array} \right].$$

Therefore, $\Delta x = M \, M_c^{-1} \tilde{x}$, giving $\tilde{M} = M \, M_c^{-1}$. This implies that $\tilde{A} = A$ and $\tilde{t} = t + A \, x_c$ so that

$$E_{RMS}^2 = \frac{1}{5} R^2 \ensuremath{\mathrm{Trace}}(A^\top A) + (t+A \, x_c)^\top (t+A \, x_c).$$ (34)