Transmembrane potential generated by a magnetically induced transverse electric field in a cylindrical axonal model

Abstract

During the electrical stimulation of a uniform, long, and straight nerve axon, the electric field oriented parallel to the axon has been widely accepted as the major field component that activates the axon. Recent experimental evidence has shown that the electric field oriented transverse to the axon is also sufficient to activate the axon, by inducing a transmembrane potential within the axon. The transverse field can be generated by a time-varying magnetic field via electromagnetic induction. The aim of this study was to investigate the factors that influence the transmembrane potential induced by a transverse field during magnetic stimulation. Using an unmyelinated axon model, we have provided an analytic expression for the transmembrane potential under spatially uniform, time-varying magnetic stimulation. Polarization of the axon was dependent on the properties of the magnetic field (i.e., orientation to the axon, magnitude, and frequency). Polarization of the axon was also dependent on its own geometrical (i.e., radius of the axon and thickness of the membrane) and electrical properties (i.e., conductivities and dielectric permittivities). Therefore, this article provides evidence that aside from optimal coil design, tissue properties may also play an important role in determining the efficacy of axonal activation under magnetic stimulation. The mathematical basis of this conclusion was discussed. The analytic solution can potentially be used to modify the activation function in current cable equations describing magnetic stimulation.

Appendices

Appendix 1: Expression of vector potential $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {A}$ in cylindrical coordinates (r′, θ′, z′)

In cylindrical coordinates (r′, θ′, z′) centered at O′, we have

$$ \nabla \times \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}}{A} = \left( {{\frac{1}{{{r^{\prime}}}}}{\frac{{\partial A_{{z}^{\prime}}}}{\partial \theta^{\prime}}} - {\frac{{\partial A_{\theta ^{\prime}} }}{\partial z^{\prime}}}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}}{{r}^{\prime}} + \left( {{\frac{{\partial A}_{{r}^{\prime}}}{\partial z^{\prime}}} - {\frac{{\partial A_{z^{\prime}} }}{\partial r^{\prime}}}} \right)\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {{\theta}^{\prime}} + \left[ {{\frac{1}{r^{\prime}}}{\frac{\partial }{\partial r^{\prime}}}(r^{\prime}A_{\theta^{\prime}} ) - {\frac{1}{r^{\prime}}}{\frac{{\partial A_{r^{\prime}} }}{\partial \theta ^{\prime}}}} \right]\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {{z^\prime}}$$

(13)

The magnetic vector potential$\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {A}$ induced by an external current is

$$ \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {A} = {\frac{\mu }{4\pi }}\iiint {{\frac{{\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {J} {\text{d}}v}}{R}}} $$

(14)

where $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {J}$ is the current density; R is the distance between the current element and the target tissue where $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {A}$ is evaluated; dv is the volume of the element carrying current density $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {J}$; μ is the magnetic conductivity; $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {A}$ is in the direction of -$\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {\theta } ^{\prime}$ and is symmetrical about the O′Z′ axis.

$$ A_{r^{\prime}} = 0,\;A_{\theta ^{\prime}} = A,\;A_{z^{\prime}} = 0 $$

(15)

Substituting (15) into (13), we have

$$ \nabla \times \overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {A} = {\frac{1}{r^{\prime}}}{\frac{\partial }{\partial r^{\prime}}}(r^{\prime}A_{\theta ^{\prime}} )\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {z} ^{\prime} $$

(16)

Also

$$ \nabla \times\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}}{A} =\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}}{B} = -\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}}{{z} ^{\prime}}B_{0} e^{j\omega t} $$

(17)

From (16) and (17), we have

$$ A_{\theta ^{\prime}} = - \frac{1}{2}r^{\prime}B_{0} e^{j\omega t} + {\frac{CO}{r^{\prime}}} $$

(18)

where CO is a constant. As the current is symmetrical about the Z′ axis. $A_{{\theta^{\prime}(r^{\prime} = 0)}} \; = \;0.$ Therefore CO = 0. Thus,

$$ A_{\theta ^{\prime}} = - \frac{1}{2}r^{\prime}B_{0} e^{j\omega t} $$

(19)

Appendix 2: Coordinate transformation from cylindrical coordinates (r′, θ′, z′) to (r, θ, z) in the computation of the magnetic vector potential $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {A}$

In cylindrical coordinates (r′, θ′, z′) with origin at o′, from Eq. 19, we know

$$ A(r^{\prime}) = 0 $$

(20)

$$ A(\theta ^{\prime}) = - {\frac{{r^{\prime}B_{0} }}{2}} $$

(21)

$$ A(z^{\prime}) = 0 $$

(22)

Here, we have omitted the time factor e ^jωt to simplify the calculation, which allows us to investigate the “instant” polarization of the axon. The vector potential can be expressed in a Cartesian basis using a matrix transformation

$$ A\left( {x^{\prime},y^{\prime},z^{\prime}} \right) = \left[\begin{array}{lll} {\cos \theta^{\prime}} & - \sin \theta^{\prime} & 0 \\ {\sin \theta^{\prime}} & {\cos \theta^{\prime}} &0 \\ 0 & 0 & 1 \\ \end{array} \right]\left[ \begin{array}{l} 0\\ { - {\frac{{r^{\prime}B_{0} }}{2}}} \\ 0 \\ \end{array}\right] = \left[ \begin{array}{l} {{\frac{{B_{0} r^{\prime}\sin\theta^{\prime}}}{2}}} \\ {{\frac{{B_{0} r^{\prime}\cos\theta^{\prime}}}{2}}} \\ 0 \\\end{array} \right] = \left[ \begin{array}{l}{{\frac{{B_{0} y^{\prime}}}{2}}} \\ {{\frac{{B_{0}x^{\prime}}}{2}}} \\ 0 \\ \end{array} \right] $$

(23)

In Fig. 1, since x′ = x − C, y′ = y, z′ = z, the vector potential $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {A}$ can be expressed in (x, y, z) coordinates

$$ A(x,y,z) = \left[\begin{array}{cc} {{\frac{{B_{0} y}}{2}}}\\ { - {\frac{{B_{0} (x - C)}}{2}}} \\ 0 \\ \end{array} \right]$$

(24)

Expressed in cylindrical coordinates(r, θ, z)

$$ \left[ {\begin{array}{lll} {A_{\rm or} } \\ {A_{{\rm o}\theta }}\\ {A_{\rm oz} } \\ \end{array} } \right] = \left[{\begin{array}{lll} {\cos \theta } & {\sin \theta } & 0 \\ { -\sin \theta } & {\cos \theta } & 0 \\ 0 & 0 & 1 \\ \end{array} }\right]\left[ {\begin{array}{c} {{\frac{{B_{0} y}}{2}}} \\ { -{\frac{{B_{0} (x - C)}}{2}}} \\ 0 \\ \end{array} } \right] =\left[ {\begin{array}{c} {{\frac{{B_{0} }}{2}}C\sin \theta }\\ { - {\frac{{B_{0} }}{2}}(r - C\cos \theta )} \\ 0 \\\end{array} } \right] $$

(25)

Appendix 3: Determining unknown coefficients C _n , D _n in Eq. 11 using boundary conditions (A)–(D)

As V is bounded at r = 0 and r → ∞ and from Eq. 11, we have

$$ D_{0} = 0 $$

$$ A_{2} = 0 $$

Therefore, expressions for the potential distribution in the extracellular media, the membrane, and in the cytoplasm are

$$ V_{0} = {\frac{{A_{0} }}{r}}\sin \theta $$

(26)

$$ V_{1} = \left( {{\frac{{A_{1} }}{r}} + D_{1} r} \right)\sin \theta $$

(27)

$$ V_{0} = D_{2} r\sin \theta $$

(28)

We substituted A _0r (Eq. 6) and the $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\rightharpoonup}$}} {r}$ components of ∇V in the three regions into (9) to yield the expressions of the normal components of the electric fields in the three regions:

$$ E_{0r} = - {\frac{{j\omega B_{0} C}}{2}}\sin \theta + {\frac{{A_{0} }}{{r^{2} }}}\sin \theta $$

(29)

$$ E_{1r} = - {\frac{{j\omega B_{0} C}}{2}}\sin \theta + \left( {{\frac{{A_{1} }}{{r^{2} }}} - D_{1} } \right)\sin \theta $$

(30)

$$ E_{2r} = - {\frac{{j\omega B_{0} C}}{2}}\sin \theta - D_{2} \sin \theta $$

(31)

Following boundary condition (A), V is continuous at the extracellular media/membrane (r = R ₊) and membrane/intracellular cytoplasm interfaces (r = R ₋),

$$ {\frac{{A_{0} }}{{R_{ + } }}} = {\frac{{A_{1} }}{{R_{ + } }}} + D_{1} R_{ + } $$

(32)

$$ {\frac{{A_{1} }}{{R_{ - } }}} + D_{1} R_{ - } = D_{2} R_{ - } $$

(33)

We then used the boundary condition (B), that the normal components of the current densities are continuous between two different media (Eqs. 1, 2), to obtain the following equations:

$$ S_{0} \left( { - {\frac{{j\omega B_{0} C}}{2}} + {\frac{{A_{0} }}{{R_{ + }^{2} }}}} \right) = S_{1} \left( { - {\frac{{j\omega B_{0} C}}{2}} + {\frac{{A_{1} }}{{R_{ + }^{2} }}} - D_{1} } \right) $$

(34)

$$ S_{1} \left( { - {\frac{{j\omega B_{0} C}}{2}} + {\frac{{A_{1} }}{{R_{ - }^{2} }}} - D_{1} } \right) = S_{2} - \left( {{\frac{{j\omega B_{0} C}}{2}} - D_{2} } \right) $$

(35)

We solved (32), (33), (34) and (35) to get the last four unknown coefficients:

$$ A_{0}\,=\,{\frac{{j\omega B_{0} C}}{2}}{\frac{{R_{ + }^{2} \left[ {\left( {S_{0} + S_{1} } \right)\left( {S_{1} - S_{2} } \right)R_{ - }^{2} + \left( {S_{0} - S_{1} } \right)\left( {S_{1} + S_{2} } \right)R_{ + }^{2} } \right]}}{{\left( {S_{0} - S_{1} } \right)\left( {S_{1} - S_{2} } \right)R_{ - }^{2} + \left( {S_{0} + S_{1} } \right)\left( {S_{1} + S_{2} } \right)R_{ + }^{2} }}} $$

$$ A_{1}\,=\,{\frac{{j\omega B_{0} C}}{2}}{\frac{{2R_{ - }^{2} R_{ + }^{2} S_{0} \left( {S_{1} - S_{2} } \right)}}{{\left( {S_{0} - S_{1} } \right)\left( {S_{1} - S_{2} } \right)R_{ - }^{2} + \left( {S_{0} + S_{1} } \right)\left( {S_{1} + S_{2} } \right)R_{ + }^{2} }}} $$

$$ D_{1}\,=\,-{\frac{{j\omega B_{0} C}}{2}}{\frac{{\left( {S_{0} - S_{1} } \right)\left[ {\left( {S_{1} - S_{2} } \right)R_{ - }^{2} - \left( {S_{1} + S_{2} } \right)R_{ + }^{2} } \right]}}{{\left( {S_{0} - S_{1} } \right)\left( {S_{1} - S_{2} } \right)R_{ - }^{2} + \left( {S_{0} + S_{1} } \right)\left( {S_{1} + S_{2} } \right)R_{ + }^{2} }}} $$

$$ D_{2} = - {\frac{{j\omega B_{0} C}}{2}}{\frac{{\left( {S_{0} - S_{1} } \right)\left( {S_{1} - S_{2} } \right)R_{ - }^{2} + \left[ {S_{0} \left( {S_{2} - 3S_{1} } \right) + S_{1} \left( {S_{1} + S_{2} } \right)} \right]R_{ + }^{2} }}{{\left( {S_{0} - S_{1} } \right)\left( {S_{1} - S_{2} } \right)R_{ - }^{2} + \left( {S_{0} + S_{1} } \right)\left( {S_{1} + S_{2} } \right)R_{ + }^{2} }}} $$