Whipple's First Equilibrium

Our initial populations can be described by a phase density for electrons and ions as follows:

$$f_e = n \left( \frac{m_e}{2\pi kT}\right)^{\frac{3}{2}} e^{\left( -\frac{K}{kT} \right)}\\$$ (4)

$$f_i = (n - n_b) \left( \frac{m_e}{2\pi kT}\right)^{\frac{3}{2}} e^{\left(- \frac{K}{kT} \right)} +$$ (5)

$$n_b/(\pi B_0)\left( \frac{m_e}{2} \right)^{\frac{3}{2}} K_b^{\frac{1}{2}}\delta (\mu - \mu_b) \delta (K - K_b)$$

where we specify the density of the beam as $n_b$, and the density of the cold plasma as $n$; the first invariant is $\mu$; the beam value is $\mu_b$; the total energy $K$; the beam energy $K_b$; and the cold plasma temperature, $kT$. Generalizing from Whipple's equation 18, we allow $\mu_b$ to take on values other than zero.

If we carry out the zeroeth moment integral to find the quasi-neutrality condition according to the Whipple prescription, we arrive at a similar equation,

$$x\equiv q\Phi/kT$$ (6)

$$K_{\vert\vert}\equiv K_b - \mu_b B$$ (7)

$$r\equiv n_b/n$$ (8)

$$e^x= (1 - r) e^{-x} + \frac{rB}{B_0}\sqrt{\frac{K_{\vert\vert}}{K_{\vert\vert}- x kT}}$$ (9)

By the appropriate asymptotic expansions, we can characterize the solution as a function of $x$. Let $x$ be small and positive, corresponding to a few volt (few $kT$) potential needed to shift the thermal plasma such that the ion beam is neutralized.

$$e^{x}+ (r-1)e^{-x}= \frac{rB}{B_0\sqrt{1 - \frac{x kT}{K_{\vert\vert}}}}$$ (10)

$$e^{2x}+2r-2+(r-1)^2e^{-2x}= \frac{r^2B^2}{B^2_0(1-\frac{x kT}{K_{\vert\vert}})}$$ (11)

Taylor expanding the exponential and truncating gives:

$${\left(1 - {\frac{xkT}{K_{\vert\vert}}}\right)(1 - 2x + 4x/r)\frac{B^2_0}{B^2}} \cong 1$$ (12)

$${\frac{B^2/B^2_0 - 1}{4/r - 2 - \frac{kT}{K_{\vert\vert}}}} \cong x$$ (13)

From this expansion, we see that when we are at the equator, $B/B_0=1$ and the potential is zero, as defined by Whipple. Furthermore, since $r < 1$ and $kT/(K_b-\mu_b B)\ll 1$ is positive, the denominator is always positive definite. Thus the potential quadratically increases with $B/B_0$ away from the equator, which means a proton will be confined to the equator, but an electron will be accelerated away from the equator. This solution produces the well-known potential of several $kT$ along the field line [15] as documented by Whipple [5]. Heuristically, this is all the voltage needed to shift massive numbers of cold electrons to the ion mirror point and shield the ion charge.