How to justify the maximal value of a star's magnetic field at surface?

Question

In many lectures, it is stated that the maximal value $B_{\text{max}}$ of the magnetic field at the surface of a star can be found in Newton's gravitation theory by equating the gravitational potential energy with the magnetic field energy. For a sphere of mass $M$ and radius $R$, of uniform density and uniformly magnetized : \begin{equation}\tag{1} |\, U_{\text{grav}}| = \frac{3 G M^2}{5 R} = E_{\text{magn}} = \frac{\mu_0 \, \mu^2}{4 \pi R^3}, \end{equation} where $\mu$ is the sphere's dipolar magnetic moment. The right hand part is the total energy stored in the magnetic field of a dipole : \begin{align} E_{\text{magn}} = \int \frac{B^2}{2 \mu_0} \, d^3 x &= \int_0^R \frac{B_{\text{int}}^2}{2 \mu_0} \, 4 \pi r^2 \, dr + \int_R^{\infty} \frac{B_{\text{ext}}^2(r, \theta)}{2 \mu_0} \, r^2 \, dr \, \sin{\theta} \, d\theta \, d\varphi \\[12pt] &= \frac{\mu_0 \, \mu^2}{4 \pi R^3}. \tag{2} \end{align} Since the sphere is uniformly magnetized in its volume, the internal magnetic field is a constant : \begin{equation}\tag{3} B_{\text{int}} = \frac{2}{3} \, \mu_0 \, M = \frac{\mu_0 \, \mu}{2 \pi R^3} \quad \Rightarrow \quad \mu = \frac{2 \pi B_{\text{int}} \, R^3}{\mu_0}. \end{equation} Substituting this magnetic moment into equ. (1) gives the maximal field strength inside and at the surface of the sphere : \begin{equation}\tag{4} B_{\text{int max}} = \sqrt{\frac{3 \mu_0 \, G}{5 \pi}} \, \frac{M}{R^2}. \end{equation} So for a star of mass $M = 0.6 \, M_{\odot}$ and radius $R = 10^4 \, \mathrm{km}$ (a typical white dwarf), this give \begin{equation} B_{\text{int max}} \approx 5 \times 10^7 \, \mathrm{tesla} = 5 \times 10^{11} \, \mathrm{gauss}. \end{equation}

But how can we justify equation (1) ? Can it be made more rigorous ? Why should we have $E_{\text{magn}} + U_{\text{grav}} = 0$ for the maximal field strength ?

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EDIT : In the case of a canonical neutron star of radius $R \approx 10 \, \mathrm{km}$ and mass $M \approx 1,44 \, M_{\odot}$, equation (4) gives \begin{equation}\tag{5} B_{\text{int max NS}} \approx 10^{14} \, \mathrm{tesla} = 10^{18} \, \mathrm{gauss}, \end{equation} which is pretty excessive (AFAIK). The strongest known magnetars have at most a field of about $10^{15} \, \mathrm{gauss}$. So is there a theoretical way in reducing the value (5) ?

Answer 1

The total energy of a star must be less than zero for it to be a gravitationally bound object.

The total energy is the sum of negative gravitational potential energy (your expression assumes a star of uniform density) and positive terms due to gas pressure, turbulence, rotation and of course magnetic fields.

The maximum magnetic energy can be found by equating the total energy to zero and making the other positive terms zero. If they are $>0$ (which they are in a real star) then of course the largest magnetic energy possible will be smaller.

You then have to decide how you want to relate that to the magnetic field and dipole moment, since the magnetic energy density will depend on the equilibrium size of the star, though I would have thought you should make $B_{\rm int} R^2$ a constant, because magnetic flux through the surface is conserved when the size changes.