The movement of a nerve in a magnetic field: application to MRI Lorentz effect imaging

Abstract

Direct detection of neural activity with MRI would be a breakthrough innovation in brain imaging. A Lorentz force method has been proposed to image nerve activity using MRI; a force between the action currents and the static MRI magnetic field causes the nerve to move. In the presence of a magnetic field gradient, this will cause the spins to precess at a different frequency, affecting the MRI signal. Previous mathematical modeling suggests that this effect is too small to explain the experimental data, but that model was limited because the action currents were assumed to be independent of position along the nerve and because the magnetic field was assumed to be perpendicular to the nerve. In this paper, we calculate the nerve displacement analytically without these two assumptions. Using realistic parameter values, the nerve motion is <5 nm, which induced a phase shift in the MRI signal of <0.02°. Therefore, our results suggest that Lorentz force imaging is beyond the capabilities of current technology.

Appendix

1.1 Potential and current density along an axon

The calculation of potential and current density is identical to that presented previously [2, 19]. The transmembrane potential V _m(z) varies along the axon and can be expressed in terms of its Fourier transform V _m(k)

$$V_{\text{m}} (z) = \frac{1}{2\pi }\int\limits_{ - \infty }^{\infty } {V_{\text{m}} (k)e^{ - ikz} {\text{d}}k} ,$$

(1)

where

$$V_{\text{m}} (k) = \int\limits_{ - \infty }^{\infty } {V_{\text{m}} (z)e^{ikz} {\text{d}}z} ,$$

(2)

and k is the spatial frequency. The intracellular potential, V _i(r,z), and extracellular potential, V _e(r,z), obey Laplace’s equation. At the membrane, the difference between V _i(a,z) and V _e(a,z) is equal to the transmembrane potential, and the membrane current is continuous. Clark and Plonsey [2] showed that the Fourier transforms of the potentials are therefore

$$V_{\text{i}} \left( {r,k} \right) = \frac{{I_{0} \left( {\left| k \right|r} \right)}}{{\beta \left( {\left| k \right|a} \right)I_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right)$$

(3)

and

$$V_{\text{e}} \left( {r,k} \right) = \frac{{K_{0} \left( {\left| k \right|r} \right)}}{{\alpha \left( {\left| k \right|a} \right)K_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right),$$

(4)

where

$$\alpha \left( {\left| k \right|a} \right) = - \left( {1 + \gamma \left( {\left| k \right|a} \right)} \right),$$

(5)

$$\beta \left( {\left| k \right|a} \right) = 1 + \frac{1}{{\gamma \left( {\left| k \right|a} \right)}},$$

(6)

$$\gamma \left( {\left| k \right|a} \right) = \frac{{\sigma_{\text{e}} K_{1} \left( {\left| k \right|a} \right)I_{0} \left( {\left| k \right|a} \right)}}{{\sigma_{\text{i}} K_{0} \left( {\left| k \right|a} \right)I_{1} \left( {\left| k \right|a} \right)}}$$

(7)

and I ₀, I ₁, K ₀, and K ₁ are modified Bessel functions [5]. The intracellular (r < a) and extracellular (r > a) current densities in the r and z directions are

$$J_{iz} \left( {r,k} \right) = ik\sigma_{i} \frac{{I_{0} \left( {\left| k \right|r} \right)}}{{\beta \left( {\left| k \right|a} \right)I_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right),$$

(8)

$$J_{ir} \left( {r,k} \right) = - \left| k \right|\sigma_{\text{i}} \frac{{I_{1} \left( {\left| k \right|r} \right)}}{{\beta \left( {\left| k \right|a} \right)I_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right),$$

(9)

$$J_{ez} \left( {r,k} \right) = ik\sigma_{\text{e}} \frac{{K_{0} \left( {\left| k \right|r} \right)}}{{\alpha \left( {\left| k \right|a} \right)K_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right),$$

(10)

$$J_{er} \left( {r,k} \right) = \left| k \right|\sigma_{\text{e}} \frac{{K_{1} \left( {\left| k \right|r} \right)}}{{\alpha \left( {\left| k \right|a} \right)K_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right).$$

(11)

1.2 Magnetic field perpendicular to the axon

Assume that a magnetic field of strength B ₀ points in the x direction, perpendicular to the axon (Fig. 1)

$${\mathbf{B}}_{0} = B_{0} {\hat{\mathbf{x}}} = B_{0} \,\left( {{\hat{\mathbf{r}}}\cos \theta - {\hat{\mathbf{\theta }}}\sin \theta } \right) .$$

(12)

The Fourier transform of the Lorentz force per unit volume, ${\mathbf{F}} = {\mathbf{J}} \times {\mathbf{B}}$, is

$$F_{ir} \left( {r,k} \right) = ik\sigma_{\text{i}} B_{0} sin\theta \frac{{I_{0} \left( {\left| k \right|r} \right)}}{{\beta \left( {\left| k \right|a} \right)I_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right),$$

(13)

$$F_{i\theta } \left( {r,k} \right) = ik\sigma_{\text{i}} B_{0} cos\theta \frac{{I_{0} \left( {\left| k \right|r} \right)}}{{\beta \left( {\left| k \right|a} \right)I_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right),$$

(14)

$$F_{iz} \left( {r,k} \right) = \left| k \right|\sigma_{\text{i}} B_{0} sin\theta \frac{{I_{1} \left( {\left| k \right|r} \right)}}{{\beta \left( {\left| k \right|a} \right)I_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right),$$

(15)

$$F_{er} \left( {r,k} \right) = ik\sigma_{\text{e}} B_{0} sin\theta \frac{{K_{0} \left( {\left| k \right|r} \right)}}{{\alpha \left( {\left| k \right|a} \right)K_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right),$$

(16)

$$F_{e\theta } \left( {r,k} \right) = ik\sigma_{\text{e}} B_{0} cos\theta \frac{{K_{0} \left( {\left| k \right|r} \right)}}{{\alpha \left( {\left| k \right|a} \right)K_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right),$$

(17)

$$F_{ez} \left( {r,k} \right) = - \left| k \right|\sigma_{\text{e}} B_{0} sin\theta \frac{{K_{1} \left( {\left| k \right|r} \right)}}{{\alpha \left( {\left| k \right|a} \right)K_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right).$$

(18)

In our mechanical model, the tissue experiences pressure and undergoes displacements. The stress tensor, τ _ij , is [8]

$$\tau_{ij} = - p\delta_{ij} + 2\mu \varepsilon_{ij} + T_{ij} ,$$

(19)

where p is the hydrostatic fluid pressure, ε _ij is the strain tensor, μ is the tissue shear modulus, and T _ij is the Maxwell stress tensor [5] that results in the Lorentz force. We assume that the shear modulus is the same within the nerve and in the surrounding tissue.

The stress tensor specifies the stress in the tissue, but to determine the net force on an element of tissue, we must examine how the stress changes with position. For instance, if the stress is larger on the left side of an element of tissue than on the right that tissue element will experience a net force. Mathematically, the divergence of the stress tensor gives Navier’s equation that describes the elastic state of the medium in static equilibrium and says that the sum of the forces (elastic plus magnetic) is zero. Navier’s equation in polar coordinates is [7]

$$- \frac{{{\partial }p}}{\partial r} + 2\mu \left( {\frac{{\partial \varepsilon_{rr} }}{\partial r} + \frac{1}{r}\frac{{\partial \varepsilon_{r\theta } }}{\partial \theta } + \frac{{\partial \varepsilon_{rz} }}{\partial z} + \frac{{\varepsilon_{rr} - \varepsilon_{\theta \theta } }}{r}} \right) + F_{r} = 0,$$

(20)

$$- \frac{1}{r}\frac{\partial p}{\partial \theta } + 2\mu \left( {\frac{{\partial \varepsilon_{r\theta } }}{\partial r} + \frac{1}{r}\frac{{\partial \varepsilon_{\theta \theta } }}{\partial \theta } + \frac{{\partial \varepsilon_{\theta z} }}{\partial z} + \frac{{2\varepsilon_{r\theta } }}{r}} \right) + F_{\theta } = 0$$

(21)

$$- \frac{\partial p}{\partial z} + 2\mu \left( {\frac{{\partial \varepsilon_{rz} }}{\partial r} + \frac{1}{r}\frac{{\partial \varepsilon_{\theta z} }}{\partial \theta } + \frac{{\partial \varepsilon_{zz} }}{\partial z} + \frac{{\varepsilon_{rz} }}{r}} \right) + F_{z} = 0.$$

(22)

In the linear approximation, the displacement of the tissue, u = (u _r , u _θ , u _z ), is related to the strain tensor by [7]

$$\varepsilon_{rr} = \frac{{\partial u_{r} }}{\partial r},\quad \varepsilon_{\theta \theta } = \frac{1}{r}\frac{{\partial u_{\theta } }}{\partial \theta } + \frac{{u_{r} }}{r},\quad \varepsilon_{zz} = \frac{{\partial u_{z} }}{\partial z},\quad \varepsilon_{\theta z} = \frac{1}{2}\left( {\frac{1}{r}\frac{{\partial u_{z} }}{\partial \theta } + \frac{{\partial u_{\theta } }}{\partial z}} \right),$$

$$\varepsilon_{zr} = \frac{1}{2}\left( {\,\frac{{\partial u_{r} }}{\partial z} + \frac{{\partial u_{z} }}{\partial r}} \right),\quad \varepsilon_{r\theta } = \frac{1}{2}\left( {\frac{{\partial u_{\theta } }}{\partial r} - \frac{{u_{\theta } }}{r} + \frac{1}{r}\frac{{\partial u_{r} }}{\partial \theta }} \right).$$

(23)

We assume that the tissue is incompressible (∇ · u = 0), which implies that the displacement can be specified by a stream function ψ, a scalar function whose spatial derivatives give the components of the displacement vector [7]

$$u_{r} = - \frac{1}{r}\frac{\partial \psi }{\partial \theta },\quad u_{\theta } = \frac{\partial \psi }{\partial r},\quad u_{z} = 0.$$

(24)

These relationships allow the solution of Navier’s equation to be stated in terms of two scalar functions: the pressure and the stream function. At the boundary of the nerve (r = a), the displacement and radial components of the stress tensor are continuous.

We have found an analytical solution to Navier’s equation subject to these boundary conditions. Inside (i) and outside (e) of the nerve, the Fourier transform of the stream function and pressure are given by

$$\psi_{\text{i}} \left( {r,k} \right) = - \frac{i}{2\mu k\left| k \right|}\frac{{\sigma_{\text{i}} B_{o} }}{{\beta \left( {\left| k \right|a} \right)I_{0} \left( {\left| k \right|a} \right)}}cos\theta V_{\text{m}} \left( k \right)\left[ {r\left| k \right|I_{0} \left( {\left| k \right|r} \right) - \left( {2 + a\left| k \right|\frac{{K_{0} \left( {\left| k \right|a} \right)}}{{K_{1} \left( {\left| k \right|a} \right)}}} \right)I_{1} \left( {\left| k \right|r} \right)} \right],$$

(25)

$$p_{i} \left( {r,k} \right) = \frac{ik}{\left| k \right|}\frac{{\sigma_{\text{i}} B_{o} }}{{\beta \left( {\left| k \right|a} \right)I_{0} \left( {\left| k \right|a} \right)}}I_{1} \left( {\left| k \right|r} \right)sin\theta V_{\text{m}} \left( k \right),$$

(26)

$$\psi_{\text{e}} \left( {r,k} \right) = - \frac{i}{2\mu k\left| k \right|}\frac{{\sigma_{\text{e}} B_{o} }}{{\alpha \left( {\left| k \right|a} \right)K_{0} \left( {\left| k \right|a} \right)}}\cos \theta V_{\text{m}} \left( k \right)\left[ {r\left| k \right|\,K_{0} \left( {\left| k \right|r} \right) + \left( {2 - a\left| k \right|\frac{{I_{0} \left( {\left| k \right|a} \right)}}{{I_{1} \left( {\left| k \right|a} \right)}}} \right)K_{1} \left( {\left| k \right|r} \right)} \right],$$

(27)

$$p_{\text{e}} \left( {r,k} \right) = - \frac{ik}{\left| k \right|}\frac{{\sigma_{e} B_{o} }}{{\alpha \left( {\left| k \right|a} \right)K_{0} \left( {\left| k \right|a} \right)}}K_{1} \left( {\left| k \right|r} \right)sin\theta V_{\text{m}} \left( k \right).$$

(28)

These expressions are complicated, but become simpler if |k|a ≪ 1, which means that the spatial extent of the action potential (and more specifically, of the rising phase of the action potential) is much greater than the radius. The intracellular stream function then becomes

$$\psi_{\text{i}} \left( {r,k} \right) = - \frac{i}{\mu k}\sigma_{i} B_{\text{o}} rcos\theta V_{\text{m}} \left( k \right)\left( \frac{ka}{2} \right)^{2} ln\left( {\frac{\left| k \right|a}{2}} \right).$$

(29)

This stream function is consistent with a displacement in the y direction of

$$u_{iy} = - \frac{i}{\mu k}\sigma_{\text{i}} B_{o} \,V_{\text{m}} \left( k \right)\left( \frac{ka}{2} \right)^{2} \ln \left( {\frac{\left| k \right|a}{2}} \right).$$

(30)

In the limit of |k|a ≪ 1, the current density in Eq. (8) becomes J _iz  = ikσ _i V _m, so the displacement is

$$u_{iy} = - \frac{{B_{o} J_{iz} }}{4\mu }\,a^{2} \,ln\left( {\frac{\left| k \right|a}{2}} \right).$$

(31)

1.3 Magnetic field parallel to the axon

Now assume that the magnetic field points in the z direction, parallel to the axon (Fig. 2)

$${\mathbf{B}}_{0} = B_{0} {\hat{\mathbf{z}}}.$$

(32)

In this case, the Fourier transform of the Lorentz force is

$$F_{i\theta } \left( {r,k} \right) = \left| k \right|\sigma_{\text{i}} B_{0} \frac{{I_{1} \left( {\left| k \right|r} \right)}}{{\beta \left( {\left| k \right|a} \right)I_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right),$$

(33)

$$F_{e\theta } \left( {r,k} \right) = - \left| k \right|\sigma_{\text{e}} B_{0} \frac{{K_{1} \left( {\left| k \right|r} \right)}}{{\alpha \left( {\left| k \right|a} \right)K_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right).$$

(34)

The force is entirely in the θ direction, is independent of θ, and causes a twisting motion about the axis of the axon.

Using the same mechanical model as described earlier, the pressure is zero and the Fourier transform of the stream function is given by

$$\psi_{\text{i}} \left( {r,k} \right) = \frac{1}{{2\mu k^{2} }}\frac{{\sigma_{\text{i}} B_{\text{o}} }}{{\beta \left( {\left| k \right|a} \right)\,I_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right)\left[ {\left( {a\left| k \right|\frac{{K_{0} \left( {\left| k \right|a} \right)}}{{K_{1} \left( {\left| k \right|a} \right)}} + 2} \right)I_{0} \left( {\left| k \right|r} \right) - r\left| k \right|I_{1} \left( {\left| k \right|r} \right)} \right],$$

(35)

$$\psi_{\text{e}} \left( {r,k} \right) = \frac{1}{{2\mu k^{2} }}\frac{{\sigma_{\text{e}} B_{\text{o}} }}{{\alpha \left( {\left| k \right|a} \right)K_{0} \left( {\left| k \right|a} \right)}}V_{\text{m}} \left( k \right)\left[ {\left( { - a\left| k \right|\frac{{I_{0} \left( {\left| k \right|a} \right)}}{{I_{1} \left( {\left| k \right|a} \right)}} + 2} \right)K_{0} \left( {\left| k \right|r} \right) + r\left| k \right|K_{1} \left( {\left| k \right|r} \right)} \right].$$

(36)

1.4 Numerical methods

To perform the numerical calculations, we performed the Fourier transforms and Bessel function calculations in MATLAB. We approximated the z axis with a grid of 121 points, using a distance between points of 0.2 mm. For the transmembrane potential, we used the three-Gaussian approximation to the action potential used by Clark and Plonsey [2].