Moment calculations

This note defines the quantities used in moment calculations.

Here we assume that (1) the look direction (angle) of a bin does not depend on energy; (2) the spacecraft potential only changes the particle’s speed/energy, not its direction, i.e. no electrostatic lensing/trajectory bending. So the the measured look direction $\hat{\mathbf e}(\Omega)$ is unchanged when mapping to $\mathbf v_\infty$ via $\mathbf v_\infty = \sqrt{2E_\infty/m}\,\hat{\mathbf e}(\Omega)$.

Since $E_\infty = E + q\Phi_{sc}$, we have $dE_\infty = dE$.

\[d^3v_\infty = \sqrt{\frac{2E_\infty}{m^3}}\, dE_\infty\, d\Omega_\infty = \sqrt{\frac{2E_\infty}{m^3}}\, dE\, d\Omega.\]

Number density

The ambient number density is given by

\[n = \int f(\mathbf v_∞) d^3v_\infty\]

where $f(\mathbf v_∞)$ is the ambient phase-space distribution function.

The differential number flux, which is measured by particle instruments, related to the velocity distribution by

\[j(E,\Omega) dE d\Omega = v f(\mathbf v_∞) d^3v .\]

Where the measured energy $E$ is related to the measured speed $v$ by $E=m v^2 / 2$. We get

\[j(E,\Omega) = \frac{2E}{m^2} f(\mathbf v_∞).\]

The differential energy flux therefore satisfies

\[f_E(E,\Omega) = E j(E,\Omega) = \frac{2E^2}{m^2} f(\mathbf v_∞)\]

so $f(\mathbf v_\infty) = \frac{m^2}{2E^2} f_E$. Substituting the relation between $f$ and $f_E$ into $n$ yields

\[n = \sqrt{\frac{m}{2}} \int f_E(E,\Omega) \frac{\sqrt{E_∞}}{E^2} dE d\Omega.\]

And after discretization in energy and angle,

\[n = \sqrt{\frac{m}{2}} \sum_i f_{E,i}\, a_i\, \frac{Δ E_i}{E_i ^ 2} \sqrt{E_{∞,i}} Δ \Omega_i\]

where the smooth cutoff/acceptance weight $a_i$ (physics/instrument correction factor applied per energy bin near the spacecraft potential to mitigate the effect of photoelectrons that may show up below the value of spacecraft potential.) is

\[a_i = \mathrm{clamp}\!\left(\frac{E_i + q\Phi_{sc}}{ Δ E_i} + \frac{1}{2},\; 0,\; 1\right).\]

Flux

The particle flux is the first velocity-space moment (vector):

\[\mathbf F = \int \mathbf v_\infty\, f(\mathbf v_\infty)\, d^3v_\infty, \qquad F_j = \int v_{\infty,j}\, f(\mathbf v_\infty)\, d^3v_\infty.\]

We obtain

\[F_j = \int \left(\sqrt{\frac{2E_\infty}{m}}\, \hat e_j\right) \left(\sqrt{\frac{2E_\infty}{m^3}}\right) \frac{m^2}{2E^2} f_E \, dE d\Omega = \int f_E \, \frac{E_\infty}{E^2}\, \hat e_j\, dE\, d\Omega.\]

After discretization in energy and angle, this becomes

\[F_j = \sum_i f_{E,i}\, a_i\, \frac{\Delta E_i}{E_i ^2}\, E_{\infty,i}\; \hat e_{j,i}\, Δ \Omega_i \qquad j \in \{x,y,z\}\]

Compared to density, the integrand acquires one extra factor of $E_\infty/E$ and uses the directional solid-angle weights.

Energy flux

The energy flux is the energy-weighted first moment:

\[\mathbf F_E = \int E_\infty\, \mathbf v_\infty\, f(\mathbf v_\infty)\, d^3v_\infty, \qquad F_{E,j} = \int E_\infty\, v_{\infty,j}\, f(\mathbf v_\infty)\, d^3v_\infty.\]

Using the same substitutions as above yields

\[F_{E,j} = \int f_E\, \frac{E_\infty^2}{E^2}\, \hat e_j\, dE\, d\Omega.\]

After discretization in energy and angle,

\[F_{E,j} = \sum_i f_{E,i}\, a_i\, \frac{\Delta E_i}{E_i^2}\, E_{\infty,i}^2\; \hat e_{j,i}\, Δ\Omega_i, \qquad j \in \{x,y,z\}.\]

Momentum flux

The (symmetric) momentum flux tensor is the second velocity-space moment (rank-2 tensor):

\[\mathrm{MF}_{jk} = \int m v_{\infty,j}\, v_{\infty,k}\, f(\mathbf v_\infty)\, d^3 v_\infty.\]

Using the relation between (1) $f$ and $f_E$, (2) $d^3v_\infty$ and $dE\ d\Omega$, we obtain

\[\mathrm{MF}_{jk} = \int m \left(\sqrt{\frac{2E_\infty}{m}}\,\hat e_j\right) \left(\sqrt{\frac{2E_\infty}{m}}\,\hat e_k\right) \left(\sqrt{\frac{2E_\infty}{m^3}}\right) \frac{m^2}{2E^2} f_E \, dE\, d\Omega = \sqrt{2 m} \int f_E \, \frac{E_\infty^{3/2}}{E^2}\, \hat e_j\,\hat e_k\, dE\, d\Omega.\]

After discretization in energy and angle, and defining $\Omega_{jk,i} \equiv \hat e_{j,i}\,\hat e_{k,i}\, Δ\Omega_i$, this becomes

\[\mathrm{MF}_{jk} = \sqrt{2 m}\sum_i f_{E,i}\, a_i\, \frac{\Delta E_i}{E_i^2}\, E_{\infty,i}^{3/2}\; \Omega_{jk,i}, \qquad (j,k) \in \{x,y,z\}^2.\]

In code and many data products, the symmetric tensor is stored as $(xx, yy, zz, xy, xz, yz)$.

Pressure tensor

The pressure tensor is the second central moment (momentum flux minus the bulk-flow/ram contribution):

\[P_{jk} = \mathrm{MF}_{jk} - V_j\,F_k\, m,\]

Heat flux

The heat flux is defined as the third velocity-space moment of the thermal (bulk-frame) velocity $\mathbf w_\infty \equiv \mathbf v_\infty - \mathbf V$:

\[\mathbf q = \int \tfrac{1}{2} m\,|\mathbf v_\infty - \mathbf V|^2\,(\mathbf v_\infty - \mathbf V)\, f(\mathbf v_\infty)\, d^3v_\infty, \qquad q_j = \int \tfrac{1}{2} m\,|\mathbf v_\infty - \mathbf V|^2\,(v_{\infty,j} - V_j)\, f(\mathbf v_\infty)\, d^3v_\infty.\]

Here $\mathbf V$ is the bulk velocity ($\mathbf V = \mathbf F/n$) and the thermal energy per bin is defined as the kinetic energy in the bulk-flow frame: $E_\mathrm{th}(E,\Omega) \equiv \tfrac{1}{2} m\, |\mathbf w_\infty|^2.$

The heat flux vector is

\[q_j = \int E_\mathrm{th}\, w_{\infty,j}\, f(\mathbf v_\infty)\, d^3v_\infty = \sqrt{\frac{m}{2}}\int f_E\, E_\mathrm{th}\, w_{\infty,j} \frac{\sqrt{E_\infty}}{E^2}\, dE\, d\Omega.\]

After discretization in energy and angle,

\[q_j = \sqrt{\frac{m}{2}}\sum_i f_{E,i}\, a_i\, \frac{\Delta E_i}{E_i^2}\, E_{\mathrm{th},i}\, w_{\infty,j,i}\, \sqrt{E_{\infty,i}}\; Δ \Omega_i \qquad j \in \{x,y,z\}.\]