What does a retarding field energy analyzer measure? According to [1], p.58, the current collected by a retarding field analyzer is:
$$I= qA \int^\infty_{v_{min}} vf(v)dv $$
where $q$ is the electron charge magnitude (here singly charged ions are assumed), $A$ is the effective collecting area of the probe, $v$ is the velocity of an ion perpendicular to the plane of the RFEA head, $f(v)$ is the velocity distribution function of the ions (IVDF), and $v_{min}$ is the
minimum ion velocity given by:
$$v_{min} = \sqrt{\frac{2qV_D}{M_i}}$$
with $M_i$ the ion mass, and $V_D$ the discriminator voltage.
Now, according to [2], the collected current is:
$$I = -\frac{q^2n_iA}{M} \int^\infty_{qV_D} g(E)dE$$
Where $g(E)$ is the ion energy distribution function (IEDF).
So we have the same instrument, some say it measures the IVDF, others the IEDF. I know the two are related (because $E = \frac{1}{2} M_i v^2$ = qV), but I just don't understand:


*

*How are these expressions of the current first obtained?

*Why is the density of the ions $n_i$ present in the second equation and not the first?

*How do I go from the IEDF to the IVDF (and reciprocally)?


[1] Lafleur T., Helicon Wave Propagation in Low Diverging Magnetic Fields, https://openresearch-repository.anu.edu.au/handle/1885/8676
[2] Ferda B., Retarding Potential Analyzer Theory and Design, http://w3.pppl.gov/ppst/docs/ferda.pdf
 A: 
How are these expressions of the current first obtained?

The expression for current density, $\mathbf{j}$, is a well known thing, which in the macroscopic form is given by:
$$
\mathbf{j} = \sum_{s} \ q_{s} \ n_{s} \ \mathbf{v}_{s}
$$
where $q_{s}$ is the charge of species $s$, $n_{s}$ is the number density of species $s$, and $\mathbf{v}_{s}$ is the bulk flow velocity of species $s$, i.e., the first velocity moment of the particle velocity distribution function (e.g., Maxwellian).
A retarding potential analyzer (RPA) is basically a mixture of a Faraday Cup and a Langmuir probe.  It measures energy per charge, much like a Faraday cup.  A detailed derivation explaining how one finds current densities can be found in the link to J. Kasper's PhD thesis on the Wind spacecraft's NASA webpage.
The first integral expression will yield a speed times a density, i.e., the density is included within the expression for $f\left(v\right)$.  The form of $g\left(E\right)$ in the second integral expression should really be in terms of energy per charge, but one can differentiate it to find:
$$
g\left(\frac{E}{q}\right) = - \frac{ M }{ q^{2} \ n_{i} \ A } \ \frac{ dI }{ dV }
$$

Why is the density of the ions $n_{i}$ present in the second equation and not the first?

As I said above, $n_{i}$ is contained within $f\left(v\right)$.  From a simple dimensional analysis, one can see this must be true for the units to work out.

How do I go from the IEDF to the IVDF (and reciprocally)?

You just use the chain rule and know that:
$$
f\left(E\right) \ dE = f\left(\mathbf{p}\right) d^{3}p
$$
where $\mathbf{p}$ is the three vector momentum and in the nonrelativistic limit, $E = \tfrac{p^{2}}{2 \ m}$.
There are detailed steps given in the Wikipedia article on the Maxwell-Boltzmann distribution that show how to go from one form to another.  However, be careful as I do not think the general form of $f\left(v\right)$ is the same as is implied by $g\left(E\right)$.
