Potential field

The magnetic field is based on the Altschuler and Newkirk (1969) and Zhang et al. (2019) model, where a potential field is defined

\[\Phi(r, θ, φ) = R_\odot \sum_{ℓ=0}^\infty \sum_{m=0}^ℓ P_ℓ^m(\cos θ) \left[g_{ℓ m} \cos(mφ) + h_{ℓ m} \sin(m φ)\right] \frac{ \displaystyle\left(\frac{R_\odot}{r}\right)^{ℓ+1} - \left(\frac{R_\odot}{R_{ss}}\right)^{ℓ+1} \left(\frac{r}{R_{ss}}\right)^ℓ }{ \displaystyle ℓ + 1 + ℓ \left(\frac{R_\odot}{R_{ss}}\right)^{2ℓ+1} }\]

and the magnetic field,

\[\mathbf{B}(r, θ, φ) = B_r(r, θ, φ) \hat{\mathbf{r}} + B_θ(r, θ, φ) \hat{\boldsymbol{θ}} + B_φ(r, θ, φ) \hat{\boldsymbol{φ}},\]

is the negative gradient of the potential field:

\[\mathbf{B}(r, θ, φ) = - \grad\Phi(r, θ, φ) = - \pdv{\Phi}{r} \hat{\mathbf{r}} - \frac{1}{r} \pdv{\Phi}{θ} \hat{\boldsymbol{θ}} - \frac{1}{r \sin θ} \pdv{\Phi}{φ} \hat{\boldsymbol{φ}}.\]

Note, here $P_ℓ^m$ is the associated Legendre Polynomial under quasi-Schmidt normalization.

For notational convenience, we define the potential function as a product solution, where each function is defined as

\[\begin{align*} F_ℓ(r) &= \frac{ \displaystyle \left(\frac{R_\odot}{r}\right)^{ℓ+1} - \left(\frac{R_\odot}{R_{ss}}\right)^{ℓ+1} \left(\frac{r}{R_{ss}}\right)^ℓ }{ \displaystyle ℓ + 1 + ℓ\left(\frac{R_\odot}{R_{ss}}\right)^{2ℓ+1} } \\ G_{ℓm}(θ) &= P_ℓ^m(\cos θ) \\[1ex] H_{ℓm}(φ) &= g_{ℓm} \cos(mφ) + h_{ℓ m} \sin(mφ) \end{align*}\]

Note that F, G, and H all implicitly depend on ℓ and m (only ℓ in F's case).

So, the potential function is instead

\[\Phi(r, θ, φ) = R_\odot \sum_{ℓ=0}^\infty F_ℓ(r) \sum_{m=0}^ℓ G_{ℓm}(θ) H_{ℓm}(φ)\]

Derivatives of potential field

We need the first and second derivatives of each single-variable function for use in the Jacobian of the magnetic field.

\[\begin{align*} \dv{}{r} F_ℓ(r) &= \frac{ \displaystyle -(ℓ+1) \frac{R_\odot^{ℓ+1}}{r^{ℓ+2}} - ℓ \left(\frac{R_\odot}{R_{ss}}\right)^{ℓ+1} \frac{r^{ℓ-1}}{R_{ss}^ℓ} }{ \displaystyle ℓ + 1 + ℓ \left(\frac{R_\odot}{R_{ss}}\right)^{2ℓ+1} } \\[1ex] &= \frac{ \displaystyle -\frac{ℓ+1}{r} \left(\frac{R_\odot}{r}\right)^{ℓ+1} - \frac{ℓ}{r} \left(\frac{R_\odot}{R_{ss}}\right)^{ℓ+1} \left(\frac{r}{R_{ss}}\right)^ℓ }{ \displaystyle ℓ + 1 + ℓ \left(\frac{R_\odot}{R_{ss}}\right)^{2ℓ+1} } \\[2ex] \frac{d^2}{dr^2} F_ℓ(r) &= \frac{\displaystyle (ℓ+1)(ℓ+2) \frac{R_\odot^{ℓ+1}}{r^{ℓ+3}} - ℓ(ℓ-1) \left(\frac{R_\odot}{R_{ss}}\right)^{ℓ+1} \frac{r^{ℓ-2}}{R_{ss}^ℓ} }{ \displaystyle ℓ + 1 + ℓ \left(\frac{R_\odot}{R_{ss}}\right)^{2ℓ+1} } \\[1ex] &= \frac{\displaystyle \frac{(ℓ+1)(ℓ+2)}{r^2} \left(\frac{R_\odot}{r}\right)^{ℓ+1} - \frac{ℓ(ℓ-1)}{r^2} \left(\frac{R_\odot}{R_{ss}}\right)^{ℓ+1} \left(\frac{r}{R_{ss}}\right)^ℓ }{ \displaystyle ℓ + 1 + ℓ \left(\frac{R_\odot}{R_{ss}}\right)^{2ℓ+1} } \end{align*}\]

\[\begin{align*} \frac{d}{dθ} G_{ℓm}(θ) &= - \sin(θ) ~ \frac{dP_ℓ^m(\cosθ)}{d(\cosθ)} \\[1ex] \frac{d^2}{dθ^2} G_{ℓm}(θ) &= \sin^2(θ) ~ \frac{d^2 P_ℓ^m(\cosθ)}{d(\cosθ)^2} - \cos(θ) ~ \frac{dP_ℓ^m(\cosθ)}{d(\cosθ)} \end{align*}\]

\[\begin{align*} \frac{d}{dφ} H_{ℓm}(φ) &= m \left[-g_{ℓ m} \sin(mφ) + h_{ℓ m} \cos(mφ)\right] \\[1ex] \frac{d^2}{dφ^2} H_{ℓm}(φ) &= -m^2 [g_{ℓ m} \cos(mφ) + h_{ℓ m} \sin(mφ)] = -m^2 H_ℓ^m(φ) \end{align*}\]

Jacobian of magnetic field

The Jacobian for an $\mathbb{R}^3$ is a 3×3 matrix. For B, it is

\[\begin{align*} J\mathbf{B}(r, θ, φ) &= \begin{bmatrix} \displaystyle -\frac{∂^2 \Phi}{∂ r^2} & \displaystyle -\frac{1}{r}\frac{∂^2 \Phi}{∂ θ \, ∂ r} & \displaystyle -\frac{1}{r \sin θ} \frac{∂^2 \Phi}{∂ φ \, ∂ r} \\[0.5ex] \displaystyle -\frac{1}{r} \frac{∂^2 \Phi}{∂ r \, ∂ θ} & \displaystyle -\frac{1}{r^2} \frac{∂^2 \Phi}{∂ θ^2} & \displaystyle -\frac{1}{r^2 \sinθ} \frac{∂^2 \Phi}{∂φ \, ∂θ} \\[0.5ex] \displaystyle -\frac{1}{r \sin θ} \frac{∂^2 \Phi}{∂ r \, ∂ φ} & \displaystyle -\frac{1}{r^2 \sin θ} \frac{∂^2 \Phi}{∂θ \, ∂φ} & \displaystyle -\frac{1}{r^2 \sin^2θ} \frac{∂^2 \Phi}{∂ φ^2} \end{bmatrix} \\[2ex] &= \begin{bmatrix} \longleftarrow & \grad^\TT B_r & \longrightarrow \\[0.5ex] \longleftarrow & \grad^\TT B_θ & \longrightarrow \\[0.5ex] \longleftarrow & \grad^\TT B_φ & \longrightarrow \end{bmatrix} \end{align*}\]

In terms of the SHTC expansion, the Jacobian is

\[J\mathbf{B}(r, θ, φ) = R_\odot \sum_{ℓ=0}^\infty \sum_{m = 0}^ℓ \begin{bmatrix} \displaystyle -\frac{d^2F_ℓ}{dr^2} G_{ℓm} H_{ℓm} & \displaystyle -\frac{1}{r} \dv{F_ℓ}{r} \dv{G_{ℓm}}{θ} H_{ℓm} & \displaystyle -\frac{1}{r\sinθ} \dv{F_ℓ}{r} G_{ℓm} \dv{H_{ℓm}}{φ} & \\[2ex] \displaystyle -\frac{1}{r} \dv{F_ℓ}{r} \dv{G_{ℓm}}{θ} H_{ℓm} & \displaystyle -\frac{1}{r^2} F_ℓ \frac{d^2G_{ℓm}}{dθ^2} H_{ℓm} & \displaystyle -\frac{1}{r^2\sinθ} F_ℓ \dv{G_{ℓm}}{θ} \dv{H_{ℓm}}{φ} & \\[2ex] \displaystyle -\frac{1}{r\sinθ} \dv{F_ℓ}{r} G_{ℓm} \dv{H_{ℓm}}{φ} & \displaystyle -\frac{1}{r^2\sinθ} F_ℓ \dv{G_{ℓm}}{θ} \dv{H_{ℓm}}{φ} & \displaystyle -\frac{1}{r^2\sin^2θ} F_ℓ G_{ℓm} \frac{d^2H_{ℓm}}{dφ^2} \end{bmatrix}\]

Factoring out the parts that appear in Φ, we have

\[J\mathbf{B}(r, θ, φ) = \sum_{ℓ=0}^\infty \sum_{m = 0}^ℓ - R_\odot F_ℓ(r) G_{ℓm}(θ) H_{ℓm}(φ) \begin{bmatrix} \displaystyle \frac{1}{F_ℓ} \frac{d^2F_ℓ}{dr^2} & \displaystyle \frac{1}{r} \frac{1}{F_ℓ G_{ℓm}} \dv{F_ℓ}{r} \dv{G_{ℓm}}{θ} & \displaystyle \frac{1}{r\sinθ} \frac{1}{F_ℓ H_{ℓm}} \dv{F_ℓ}{r} \dv{H_{ℓm}}{φ} \\[2ex] \displaystyle \frac{1}{r} \frac{1}{F_ℓ G_{ℓm}} \dv{F_ℓ}{r} \dv{G_{ℓm}}{θ} & \displaystyle \frac{1}{r^2} \frac{1}{G_{ℓm}} \frac{d^2G_{ℓm}}{dθ^2} & \displaystyle \frac{1}{r^2\sinθ} \frac{1}{G_{ℓm} H_{ℓm}} \dv{G_{ℓm}}{θ} \dv{H_{ℓm}}{φ} \\[2ex] \displaystyle \frac{1}{r \sinθ} \frac{1}{F_ℓ H_{ℓm}} \dv{F_ℓ}{r} \dv{H_{ℓm}}{φ} & \displaystyle \frac{1}{r^2\sinθ} \frac{1}{G_{ℓm} H_{ℓm}} \dv{G_{ℓm}}{θ} \dv{H_{ℓm}}{φ} & \displaystyle \frac{1}{r^2\sin^2θ} \frac{1}{H_{ℓm}} \frac{d^2H_{ℓm}}{dφ^2} \end{bmatrix}\]

Note that since B is a gradient field, the Jacobian is symmetric (negative Hessian of Φ).

Gradient of the field strength

∇B = ∇|B| can be determined from the Jacobian of B. What follows is a proof using index notation:

\[\pdv{B}{x_j} = \pdv{}{x_j} \left(B_i B_i\right)^{1/2} = \frac{1}{2} \left(B_i B_i\right)^{-1/2} \left(2 B_i \pdv{B_i}{x_j}\right) = \frac{B_i \pdv{B_i}{x_j}}{\left(B_k B_k\right)^{1/2}} = \frac{B_i}{B} \pdv{B_i}{x_j}\]

which implies $∇B = \hat{\mathbf{b}} ⋅ J\mathbf{B}$

Desired quantities

The following are the quantities we want to calculate

• The potential field

  \[Φ = \sum_{ℓ=0}^\infty \sum_{m=0}^ℓ R_\odot F_ℓ(r) G_{ℓm}(θ) H_{ℓm}(φ) = \sum_{ℓ,m} \Phi_ℓ^m,\]

  where $\Phi_ℓ^m = R_\odot F_ℓ(r) G_{ℓm}(θ) H_{ℓm}(φ)$

• The magnetic field

  \[\mathbf{B} = - \sum_{ℓ,m} R_\odot F_ℓ(r) G_{ℓm}(θ) H_{ℓm}(φ) \left(\frac{\hat{\mathbf{r}}}{F_ℓ} \dv{F_ℓ}{r} + \frac{\hat{\boldsymbol{θ}}}{r G_{ℓm}} \dv{G_{ℓm}}{θ} + \frac{\hat{\boldsymbol{φ}}}{r \sin(θ) H_{ℓm}} \frac{dH_{ℓm}}{d\varphi}\right)\]

• The magnetic field strength, B = |B|

• The gradient of the magnetic field strength, ∇B

  \[\begin{align*} \grad B &= \grad \sqrt{B_r^2 + B_θ^2 + B_φ^2} \\[1ex] &= \frac{1}{B} \begin{bmatrix} \longleftarrow & \dpdv{\mathbf{B}^\TT}{r} & \longrightarrow \\ \longleftarrow & \dfrac{1}{r} \dpdv{\mathbf{B}^\TT}{θ} & \longrightarrow \\ \longleftarrow & \dfrac{1}{r \sin θ} \dpdv{\mathbf{B}^\TT}{φ} & \longrightarrow \end{bmatrix} \mathbf{B} \\[1ex] &= \frac{1}{B} \begin{bmatrix} \uparrow & \uparrow & \uparrow \\ \grad B_r & \grad B_θ & \grad B_φ \\ \downarrow & \downarrow & \downarrow \end{bmatrix} \mathbf{B} \\[1ex] &= \frac{1}{B} (J\mathbf{B})^\TT \mathbf{B} \end{align*}\]

References