The collapse is perpendicular to the initial field, so magnetic flux freezing preserves the mass-to-flux ratio of each material flux tube:

\frac{B _{z}}{ρ} = \frac{B _{0}}{ρ _{0}}, B_{z} = \frac{B _{0}}{ρ _{0}} ρ .

\frac{d}{d R} (P + \frac{B _{z}^{2}}{8 π}) = - ρ \frac{d Φ}{d R} .

Define the effective polytropic constant

K_{m e ff} = K + \frac{B _{0}^{2}}{8 π ρ _{0}^{2}} .

P + B_{z}^{2} / (8 π) = K_{eff} ρ^{2}

. Dividing equilibrium by

ρ

, differentiating, and using the cylindrical Poisson equation

\frac{1}{R} \frac{d}{d R} (R \frac{d Φ}{d R}) = 4 π Gρ

gives

\frac{1}{R} \frac{d}{d R} (R \frac{d ρ}{d R}) + \frac{2 π G}{K _{eff}} ρ = 0 .

Put

a^{2} = K_{eff} / (2 π G)

. This is the order-zero Bessel differential equation, and regularity on the axis together with

ρ (0) = ρ_{1}

yields

ρ (R) = ρ_{1} J_{0} (R / a) .

ii

Solution

The physical filament ends when the mass density first reaches zero. If

x_{1} ≃ 2.4

is the first positive zero of the Bessel function

J_{0}

, then

R_{f} = x_{1} a = x_{1} \frac{K + B _{0}^{2} / ( 8 π ρ _{0}^{2} )}{2 π G} .

The first zero must be used because continuing to the next lobe would make the density negative.

λ = 2 π \int_{0}^{R_{f}} ρ (R) R d R .

Using

d [x J_{1} (x)] / d x = x J_{0} (x)

and

J_{1} (x) = - J_{0}^{'} (x)

gives

λ = 2 π ρ_{1} a^{2} x_{1} J_{1} (x_{1}) = - \frac{ρ _{1} K _{eff}}{G} x_{1} J_{0}^{'} (x_{1}) .

Since

x_{1} ≃ 2.4

and

J_{0}^{'} (x_{1}) ≃ - 0.52

λ ≃ 1.25 \frac{ρ _{1}}{G} (K + \frac{B _{0}^{2}}{8 π ρ _{0}^{2}}) .

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