*4.4. Extrinsic Curvature*

The expression for extrinsic curvature for our problem is given in (38). Similar to previous sections, we define

$$L\_{\mathbf{x}}(\mathbf{x}) := \log\_{10}(\mathbf{x}(\mathbf{x})).\tag{94}$$

Taking the log of (94), we have

$$\begin{aligned} L\_{\mathbf{x}}(\mathbf{x}) &= \quad \log\_{10}(\mathbf{x}(\mathbf{x})) \\ &= \quad \log\_{10}\left[an(n-1)\mathbf{x}^{n-2}\right] - \frac{3}{2}\log\_{10}\left[1 + (an\mathbf{x}^{n-1})^2\right] \\ &\approx \quad \log\_{10}\left[an(n-1)\mathbf{x}^{n-2}\right] - \frac{3}{2}\log\_{10}\left[(an\mathbf{x}^{n-1})^2\right] \\ &= \quad \log\_{10}\left[\frac{n-1}{(an)^2}\right] - (2n-1)\log\_{10}\mathbf{x}. \end{aligned} \tag{95}$$

Note that the second last expression is a valid approximation for *x* > 2. From (95) and (84) it is easy to establish the affine mapping

$$L\_K(\mathbf{x}) = \gamma\_1^K L\_\pi(\mathbf{x}) + \gamma\_0^K,\tag{96}$$

where

$$\begin{aligned} \gamma\_1^K &= \frac{n}{2n - 1}, \\ \gamma\_0^K &= \log\_{10}\left(\frac{n - 1}{na}\right) - \frac{n}{2n - 1} \log\_{10}\left[\frac{n - 1}{(an)^2}\right] \end{aligned} \tag{97}$$

Similarly, from (95) and (91) we can establish the affine relationship

$$L\_{\beta}(\mathbf{x}) = \gamma\_1^{\beta} L\_{\kappa}(\mathbf{x}) + \gamma\_0^{\beta} \, \mathrm{s} \tag{98}$$

where

$$\begin{aligned} \gamma\_1^\beta &= \frac{n}{2n - 1}, \\ \gamma\_0^\beta &= \log\_{10} \left[ \frac{(n - 1)\sigma\sqrt{2}}{na\sqrt{N\pi}} \right] - \frac{n}{2n - 1} \log\_{10} \left[ \frac{n - 1}{(an)^2} \right]. \end{aligned} \tag{99}$$

Using similar arguments used in previous sections, we infer that the extrinsic curvature and parameter-effects curvature have the same sign of the non-zero slopes. Similarly, the extrinsic curvature and direct parameter-effects curvature have the same non-zero slopes.
