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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 ?


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) ?

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  • $\begingroup$ Because if the total energy was greater than zero the star would be unbound! $\endgroup$ – Rob Jeffries Jul 1 '17 at 19:26
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    $\begingroup$ @RobJeffries, the energy $E_{\text{magn}} + U_{\text{grav}}$ isn't the total energy ! It's just a part of it. There may be gaz pressure inside the star, of degeneracy fermi pressure, or even rotation kinetic energy contributing. So it doesn't explain why $E_{\text{magn}} + U_{\text{grav}}$ should cancels. $\endgroup$ – Cham Jul 1 '17 at 19:31
  • $\begingroup$ Or is it the worst case scenario, for which the star has no other internal energy except its own magnetic field sustaining the whole weight ? $\endgroup$ – Cham Jul 1 '17 at 19:35
  • $\begingroup$ Yes of course there is internal energy and that is positive. So the case you have there is the maximum that the magnetic field energy could be and still have a bound star. $\endgroup$ – Rob Jeffries Jul 1 '17 at 19:41
  • $\begingroup$ If it's the worst case scenario, the total energy \begin{equation}E = \frac{\mu_0 \, \mu^2}{4 \pi R^3} - \frac{3 G M^2}{5 R}\end{equation} admits a minimum value at some non-trivial $R$ (considering $\mu$ as a constant for a given star). $\endgroup$ – Cham Jul 1 '17 at 19:41
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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.

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  • $\begingroup$ I agree with this, but the total energy (magnetic field + gravitationnal energy) admits a minimum for some non-trivial $R$, if $\mu$ is considered as a constant. This gives a specific value for the magnetic field which is like (4), without the factor of 3 inside the square root. $\endgroup$ – Cham Jul 1 '17 at 19:48
  • $\begingroup$ @cham Your question asks for the justification of a maximum value. Of course there is a global minimum - that will be the equilibrium configuration for the star if the mass and magnetic dipole moment is fixed but the radius can change. This gives you the radius of a star supported (solely) by the magnetic energy density. $\endgroup$ – Rob Jeffries Jul 1 '17 at 19:51
  • $\begingroup$ There's something that I don't understand clearly about the variables and the constants. Is $\mu$ a constant, if the radius $R$ changes ? Or is $B_{\text{int}} \propto \mu/R^3$ a constant ? Or what else ? $\endgroup$ – Cham Jul 1 '17 at 19:57
  • $\begingroup$ @cham I don't know - it's your question! My answer in terms of energy considerations does not depend on what you consider to be fixed or variable. It is true in general, although I do need to make a small edit to be correct. $\endgroup$ – Rob Jeffries Jul 1 '17 at 19:58
  • $\begingroup$ Ahaa! Your edit is a crucial point, actually. So do you agree that \begin{equation}B_{\text{int max}} = \sqrt{\frac{3 \mu_0 \, G}{5 \pi}} \, \frac{M}{R^2} ? \end{equation} Then what could we say about $R$ ? I think my confusion is coming from here. $\endgroup$ – Cham Jul 1 '17 at 20:11

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