Modeling Skull Electrical Properties

Abstract

Accurate representations and measurements of skull electrical conductivity are essential in developing appropriate forward models for applications such as inverse EEG or Electrical Impedance Tomography of the head. Because of its layered structure, it is often assumed that skull is anisotropic, with an anisotropy ratio around 10. However, no detailed investigation of skull anisotropy has been performed. In this paper we investigate four-electrode measurements of conductivities and their relation to tissue anisotropy ratio (ratio of tangential to radial conductivity) in layered or anisotropic biological samples similar to bone. It is shown here that typical values for the thicknesses and radial conductivities of individual skull layers produce tissue with much smaller anisotropy ratios than 10. Moreover, we show that there are very significant differences between the field patterns formed in a three-layered isotropic structure plausible for bone, and those formed assuming that bone is homogeneous and anisotropic. We performed a measurement of conductivity using an electrode configuration sensitive to the distinction between three-layered and homogeneous anisotropic composition and found results consistent with the sample being three-layered. We recommend that the skull be more appropriately represented as three isotropic layers than as homogeneous and anisotropic.

Appendices

Appendix A

Assume a brick-shaped structure such as that shown in Fig. A, consisting of two layers with respective conductivities σ₁ and σ₂ (or an arbitrary number of layers each one having conductivity either σ₁ or σ₂, for example the three layer structure shown in Fig. 2a). The resistances observed in response to homogeneous electric fields applied in the radial and tangential directions, respectively, are

$$ R_r =\frac{t_1 }{\sigma_1 lw}+\frac{t_2 }{\sigma_2 lw}\hbox{ and} $$

(A.1)

$$ R_t =\frac{l}{w}\frac{1}{\sigma_2 t_2 +\sigma_1 t_1 }.$$

(A.2)

Figure A

Brick-shaped structure with width w, length l and total thickness T, composed of two layers having conductivity σ₁ and thickness t ₁, and conductivity σ₂ and thickness t ₂, respectively. Homogeneous fields are applied through the faces with dimension l x w to obtain radial resistances and through faces having dimension w  x  T to obtain tangential resistances

We also have that

$$ k=\frac{\sigma_t }{\sigma_l }=\frac{R_l }{R_t }\frac{l^{2}}{T^{2}} $$

(A.3)

where T = t ₁ + t ₂, since $\sigma_t =\frac{1}{\rho_t}=\frac{l}{TwR_t}\hbox{ and }\sigma_r =\frac{1}{\rho_r }=\frac{T}{lwR_r}$ Combining Eqs. (A.1), (A.2) and (A.3), we obtain

$$ k=\frac{\sigma_t }{\sigma_r }=\left( \frac{1}{\sigma_1 }\left( 1-\frac{1}{\alpha_t +1} \right)+\frac{1}{\left( \alpha_t +1 \right)\sigma_2} \right)\left( \frac{\sigma_2}{\alpha_t +1}+\sigma_1 \left( 1-\frac{1}{\alpha_t +1} \right) \right) $$

(A.4)

where we have substituted $\alpha_t =\frac{t_1 }{t_2 }$ .

Taking the derivative of (A.4) with respect to α _t , we find a maximum at α _t = 1, or that the maximum value of k is at

$$ \begin{aligned} k_{\rm max } &=\left( \frac{1}{2\sigma_1 }+\frac{1}{2\sigma_2} \right)\left( \frac{\sigma_2}{2}+\frac{\sigma_1}{2}\right) \\ &=\frac{1}{4}\left( \frac{\sigma_2 }{\sigma_1 }+\frac{\sigma_1}{\sigma_2 }+2 \right)=\frac{1}{4\sigma_1 \sigma_2}\left( \sigma_1^2 +\sigma_2^2 +2\sigma_1 \sigma_2 \right)=\frac{\left( \sigma_1 +\sigma_2 \right)^{2}}{4\sigma_1 \sigma_2} \end{aligned} $$

(A.5)

Appendix B

Analytic expressions describing the dependence of diagonal and orthogonal resistivity or conductivity calculations on anisotropy, electrode separation and sample thickness can be derived using Livshitz et al.18 in conjunction with variable spatial scaling to account for anisotropy, similar to that described in Rush.28 The geometry used by Livshitz is shown in Fig. B.

Figure B

Configuration for single semi-infinite slab observations, after Livshitz et al.18 Conductivities σ₁ and σ₃ are zero. σ₂ may be anisotropic. Sources (x,y,z′) are positioned at A and D for diagonal observations, and voltage differences are calculated between B and C. For orthogonal observations, sources are positioned at A and C and voltage differences are calculated between B and D

The voltage within the a single semi-infinite slab (region 2) is described by

$$ V_2 \left( z,r \right)=\beta \mathop{\int}\limits_0^\infty {T_1 \left( \lambda \right)T_2 \left( \lambda \right)\left\{ e^{-\lambda \left| z-{z}^{\prime} \right|}+R_i \left( \lambda \right)e^{\lambda \left( z+{z}^{\prime} \right)} \right\}} J_o \left( {\lambda r} \right)d\lambda $$

(B.1)

where J ₀ is the zeroth order Bessel function, and β = I/4π σ₁ . The parameters z′ and r are the z coordinates of the source, and the xy distance between source and observation point respectively.

The functions T ₁ and T ₂ are transmission functions across the upper and lower boundaries, and are defined as

$$ T_i \left( \lambda \right)=\frac{1+K_{i-1}}{1+K_{i-1} R_i \left( \lambda \right)e^{2\lambda z_{i-1}}} $$

(B.2)

and

$$ R_i \left( \lambda \right)=\frac{K_i +R_{i+1} e^{2\lambda z_i}}{1+K_i R_{i+1} \left( \lambda \right)e^{2\lambda z_{i-1} }}e^{-2\lambda z_i} $$

(B.3)

The parameters K _i are defined as

$$ K_i =\frac{\sigma_i -\sigma_{i+1} }{\sigma_i +\sigma_{i+1} }\hbox{ and }K_0 =0 $$

(B.4)

Therefore, choosing z ₁ = 0 and z ₂ = h obtains

$$ \begin{array}{ll} {K_1 =-1}& {K_2 =1} \\ {R_2 =e^{-2\lambda h}}& {R_1 =K_1 +\frac{\left( 1-K_1 \right)\left( 1+K_1 \right)e^{-2\lambda h}}{1+K_1 e^{-2\lambda h}}} \\ {T_1 =1}& {T_2 =\frac{1+K_1 }{1+K_1 e^{-2\lambda h}}} \\ \end{array} $$

(B.5)

because σ₁ =  0 (the conductivity outside the slab), the relation (from Eq. B.4) is used to calculate β via

$$ \frac{1+K_1}{\sigma_1}=\frac{1-K_1}{\sigma_2} ;\,\mathop {\lim}\limits_{\sigma_1 \rightarrow 0} \frac{1+K_1 }{\sigma_1 }=\frac{2}{\sigma_2 } $$

(B.6)

Then, using the Weber–Lipschitz integral,

$$ \mathop{\int}\limits_0^\infty {e^{-\lambda \left| z \right|}J_0 \left( \lambda r \right)} d\lambda =\left( r^{2}+z^{2} \right)^{-\frac{1}{2}} $$

(B.7)

expanding the expression for R ₁ as a Taylor Series in K ₁ e ^−2λh and incorporating anisotropic scaling, we obtain the expression

$$ V_a =\frac{I}{2\pi \left( {\sigma_r \sigma_t } \right)^{1/2}}\left[ \begin{array}{l} \left( \alpha ^{2}r^{2}+\left| {z-{z}^{\prime}} \right|^{2} \right)^{-1/2}+\left( \alpha ^{2}r^{2}+\left( 2\left(-z-{z}^{\prime} \right)+h \right)^{2} \right)^{-1/2} \\ +\mathop{\sum}\limits_{m=1}^\infty {\left( \alpha ^{2}r^{2}+\left( 2mh+\left| z-{z}^{\prime} \right| \right)^{2} \right)^{-1/2}} \\ +\mathop{\sum}\limits_{m=1}^\infty {\left( \alpha ^{2}r^{2}+\left( \left( 2m+1 \right)h+\left| z-{z}^{\prime} \right| \right)^{2} \right)^{-1/2}} \end{array} \right] $$

(B.8)

where $\alpha =\sqrt{\sigma_r/\sigma_t}\le 1$.

For the case of diagonal observations, voltage is measured at B, positioned at $(x,y,z)=(a,0,0)=({\mathbf r},0)$. This voltage is the sum of contributions from current sources positioned at A =  (x′,y′,z′) =  (0,0,0) and D =  (x′,y′,z′) =  (a,0,h), respectively. In the case of a source at A

$$ V_{BA} =\frac{I}{2\pi \left( \sigma_r \sigma_t \right)^{1/2}}\left\{ \begin{array}{l} \frac{1}{\alpha a}+\frac{1}{\left( \alpha ^{2}a^{2}+h^{2} \right)^{1/2}} \\ +\mathop{\sum}\limits_{m=1}^\infty {\left[ \alpha ^{2}a^{2}+\left( 2mh \right)^{2} \right]^{-1/2}} +\mathop{\sum}\limits_{m=1}^\infty {\left[ \alpha ^{2}a^{2}+\left( 2m+1 \right)^{2}h^{2} \right]^{-1/2}} \end{array} \right\} $$

(B.9)

and for a source at D

$$ V_{BD} =\frac{I}{2\pi \left( \sigma_r \sigma_t \right)^{1/2}}\left\{ \frac{2}{h}+\mathop{\sum}\limits_{m=1}^\infty {\left( \frac{1}{\left( 2m+1 \right)h}+\frac{1}{2mh} \right)} \right\} $$

(B.10)

So the voltage on the electrode at B becomes

$$ V_B =V_{BA} -V_{BD} =\frac{I}{2\pi \left( \sigma_r \sigma_t \right)^{1/2}}\left\{ \begin{array}{l} \frac{1}{\alpha a}+\frac{1}{\left( \alpha ^{2}a^{2}+h^{2} \right)^{1/2}}-\frac{2}{h} \\ +\mathop{\sum}\limits_{m=1}^\infty {\left[ \begin{array}{l} \frac{1}{\left( \alpha ^{2}a^{2}+\left( {2mh} \right)^{2} \right)^{1/2}}+\frac{1}{\left( \alpha ^{2}a^{2}+\left( 2m+1 \right)^{2}h^{2} \right)^{1/2}} \\ -\frac{1}{\left( 2m+1 \right)h}-\frac{1}{2mh} \end{array} \right]} \end{array} \right\} $$

(B.11)

By symmetry, the difference between voltages at electrodes at B and C = (x,y,z) = (0,0,h) will be twice this quantity.

For the case of orthogonal observations, V _BA will be the same as for the diagonal case, but the voltage contribution from the source at C will be

$$ V_{BC} =\frac{I}{2\pi \left( {\sigma_r \sigma_t } \right)^{1/2}}\left\{ {\begin{array}{l} \frac{2}{\left( {\alpha ^{2}a^{2}+h^{2}} \right)^{1/2}} \\ +\mathop{\sum}\limits_{m=1}^\infty {\left[ {\left( {\alpha ^{2}a^{2}+\left( {2mh} \right)^{2}} \right)^{{-1}/2}+\left( {\alpha ^{2}a^{2}+\left( {2m+1} \right)^{2}h^{2}} \right)^{{-1}/2}} \right]} \end{array}} \right\} $$

(B.12)

and the voltage difference V _B will be

$$ V_B =V_{BA} -V_{BC} =\frac{I}{2\pi \left( \sigma_r \sigma_t \right)^{1/2}}\left[ \frac{1}{\alpha a}-\frac{1}{\left( {\alpha ^{2}a^{2}+h^{2}} \right)^{1/2}} \right]. $$

(B.13)