Widdekind Posted August 8, 2011 Posted August 8, 2011 (edited) At a given pressure, inside stars, for a given inward gravity, electrons, being less massive, experience less inward pull, despite their similar thermal energies, and higher velocities. I read in a book, that this means stars shed a few electrons, until their slight positive charge affords electrons a balancing EM attraction. The following equations attempt to mathematically model said effect. First, from the Virial Theorem [math]<U> = -2 <K>[/math], applied separately to the electronic, and ionic, components, of the stellar plasma, w.h.t. [math]-\frac{G M m_e}{R} - \frac{Q e}{4 \pi \epsilon_0 \, R} = -2 \left( \frac{3}{2} k_B T_e \right)[/math] [math]-\frac{G M <A> m_p}{R} + \frac{Q <Z> e}{4 \pi \epsilon_0 \, R} = -2 \left( \frac{3}{2} k_B T_i \right)[/math] Now, assuming thermalization of electrons & ions, s.t. [math]T_e \approx T_i[/math], we can equate these expressions, s.t.: [math]-\frac{G M m_e}{R} - \frac{Q e}{4 \pi \epsilon_0 \, R^2} = -\frac{G M <A> m_p}{R} + \frac{Q <Z> e}{4 \pi \epsilon_0 \, R}[/math] which, after algebraic manipulations, yields (in Coulombs): [math]Q = \left( \frac{4 \pi \epsilon_0}{e} \right) \left( G M m_p \right) \frac{<A>}{<Z>+1} \approx 300 \left( \frac{M}{2 M_{\odot}} \right) \frac{<A>}{<Z>+1}[/math] since NS are [math]\approx 2 M_{\odot}[/math], characteristically (Kirshner. Extravagent Universe, p.36). For a pure H plasma, [math]A = Z = 1[/math], so that the 'plasma fraction' is ~1/2; whereas for a pure (and fully ionized) Fe plasma, characteristic of massive pre-collapse star cores (at [math]T \approx 300 keV[/math]) w.h.t. [math]<A> \approx 2 <Z>[/math], so that the 'plasma fraction' is ~2. Even in such a case, that's only 0.6 m-mol of electrons, out of an entire NS worth of mass. Now, a spinning charged sphere generates a magnetic moment, and dipole-like magnetic field, of characteristic surface-strength [math]B \approx \frac{\mu_0}{4 \pi} \frac{m}{R^3}[/math], where the magnetic moment [math]m \approx \frac{Q}{2 M} L = \frac{1}{5} Q R^2 \omega[/math]. Thus, w.h.t. (in Tesla): [math]B \approx \frac{1}{5} \frac{\mu_0}{4 \pi} \frac{Q \omega}{R} \approx 8 \times 10^{-6} \left( \frac{M}{2 M_{\odot}} \right) \left( \frac{R}{7 km} \right)^{-1} \left( \frac{\omega}{716 Hz} \right)[/math] even using the most rapidly rotating pulsar observed to date. Thus, even a 1ms pulsar, would not be expected, according to these equations, to generate a surface magnetic field, in excess of 0.1G, through this method. Edited August 8, 2011 by Widdekind
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