Codex @codex  0

Write the poloidal magnetic flux function and toroidal amplitude explicitly in spherical components:Dotting the magnetostatic equilibrium equation with $\mathbf{B}$ gives $\mathbf{B}\cdot \nabla p=0$. Since all fields are axisymmetric, the poloidal magnetic field is tangent to the contours of $\chi$ in a meridional section. Thus, locally on regular connected flux surfaces, $p=p(\chi )$. This is a local flux-surface statement; disconnected surfaces with the same numerical value of $\chi$ need not have the same pressure without further global assumptions.

Taking the azimuthal component of the Lorentz force givesIts numerator is the meridional Jacobian determinant of $\psi$ and $\chi$. Vanishing therefore implies locally $\psi =\psi (\chi )$ on the same regular surfaces. DefineA direct curl calculation now givesSince ${\mathbf{e}}_{\varphi }\times {\mathbf{B}}_{\mathrm{pol}}=\nabla \chi /(r\sin \theta )$, the remaining force isConsequently the spherical magnetostatic pressure balance equation following from the stated physical force balance isThe PDF's displayed scalar equation has the wrong sign and omits ${\mu }_0$ for its stated pressure p. This is not corrected by setting ${\mu }_0=1$, which leaves the sign discrepancy. For example, $\chi =r^4{\sin }^2\theta$, $\psi =0$, $p=p_0-10\chi /{\mu }_0$ satisfies the stated vector force balance and the corrected scalar equation, because ${\Delta }_{\ast }\chi =10r^2{\sin }^2\theta$; it fails the printed scalar equation. One could obtain the printed form by replacing its pressure variable by $-{\mu }_0{p}_{\mathrm{physical}}$, but that is not the variable in the given force balance.

Now set $\psi =\alpha \chi$, $p=p_0+P\chi$, and $\chi =F(r){\sin }^2\theta$. The angular operator gives $-2F{\sin }^2\theta /r^2$, soFor the trigonometric form in the question, direct substitution yieldsFor $\alpha \ne 0$, this is a solution providedThe same coefficient relation makes the trigonometric pair regular at the origin: $\cos \alpha r-\sin \alpha r/(\alpha r)=-{\alpha }^2{r}^2/3+O(r^4)$. Thus all components of the magnetic field are finite there. With $p_0=0$, this explicitly constructs the requested proportional-pressure family; adding $p_0$ does not change force balance. The coefficient c is the one called C in the final sentence of the PDF.

For a confined linear-pressure spherical magnetic equilibrium, the condition that the entire magnetic field vanish on $r=a$ is precisely $F(a)=F^{\prime }(a)=0$. In components these conditions readThey eliminate both the radial and meridional components; the toroidal component vanishes too because $\psi =\alpha \chi$. For a nontrivial solution with $A\ne 0$, substituting $B=-A/\alpha$ givesThe last equation selects the allowed nonzero values of $\alpha a$. Equivalently, when division is legitimate, $\tan x=3x/(3-x^2)$; the undivided equation also handles vanishing denominators correctly. Since $\chi (a,\theta )=0$, $\boxed{p(a,\theta )=p_0}$ is independent of latitude. In the strictly proportional convention $p=P\chi$, this is zero. The zero-field solution is a trivial possibility; the displayed conditions describe the nontrivial confined family.