The python package PyNeb could calculate the 'Emissivity' of an emission line at given electron temperature and density in unit "erg.s-1.cm3". But I don't know why Emissivity should in such an unit but not "erg.s-1.cm-3", what does this unit mean?
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2$\begingroup$ @JamesK The unit is correct, but the term "emissivity" is not. Not a typo, just sloppiness. $\endgroup$– pelaCommented Dec 2, 2021 at 11:06
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2 Answers
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This answer is not right, please see pela's answer or reply by Christophe Morisset (one of the owner of PyNeb) in https://github.com/Morisset/PyNeb_devel/issues/21
I decide to hold this answer for reference because sometimes "emissivity" does mean two different things .
Old but not right answer
I have checked the code and found the "emissivity" returned by this method ($\epsilon_{code}$) is not actually the "emissivity" in equation 3 of V. Luridiana , C. Morisset , and R. A. Shaw 2015 ($\epsilon_{paper}$). The relation between these two "emissivity" are: $\epsilon_{code} = \dfrac{\epsilon_{paper}}{n_e \cdot n_{ion}}$ So the unit of $\epsilon_{code}$ should be $\text{erg } \text{s}^{-1} \text{ cm}^3$.
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That's because of another incident of the sloppiness of astronomers:
Emissivity vs rate coefficient
The term "emissivity" ($\varepsilon$) is mostly used as you think: The amount of energy emitted (in a given line) per time, per volume. Hence, in cgs it's measured in $\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-3}$, as you expected.
In the above and other cases, however, it refers to the quantum mechanical factor that tells you how likely a given physical process is to happen, the so-called rate coefficient, multiplied by the energy of the photon emitted by the process.
If the physical process is the recombination of an ion and an electron, it's called the recombination coefficient and is often written $\alpha_\mathrm{B}$ (for "case B" recombination, which is usually the case in astrophysics). If the process is a collision between two particles, a $C$ or a $q$ is sometimes used: For instance, a collision between a free electron and neutral hydrogen atom (HI) in the ground state (1s), resulting in the atoms' electron being excited to the first excited state (2p), would be called $q_{1\mathrm{s}\rightarrow 2\mathrm{p}}$.
The number of emitted photons from the process then depends on this rate coefficient, as well as on the densities of both types of particles. The more particles you have, the more likely the process is to happen, so you must multiply by the densities of both types. The number of photons emitted from a volume of gas is measured in $\mathrm{s}^{-1}\,\mathrm{cm}^{-3}$. Hence, the units of the rate coefficient must be $\mathrm{s}^{-1}\,\mathrm{cm}^{+3}$, even though this unit may not have a physical interpretation in itself.
Rate coefficients are temperature-dependent, because the speed of the particles affect the probability of the process occurring (too slow particles rarely meet, while too fast particles may result in other processes, e.g. ionization).
Example
As an example, consider the energy emitted from a blob of gas in the Lyman $\alpha$ line. Two processes contribute to the Ly$\alpha$ luminosity: The first, and typically dominant, process is the recombination of free electrons with ionized hydrogen (HII; i.e. just a proton), the densities of which are $n_e$ and $n_\mathrm{HII}$, respectively. The second is the collision of free electrons with neutral hydrogen HI, with density $n_\mathrm{HI}$.
Not all recombinations result in the emission of a Ly$\alpha$ photon, only a fraction $P(\mathrm{Ly}\alpha)$ (which is quite high, roughly 68% with a small temperature dependence). Hence, the total Ly$\alpha$ emissivity is $$ \begin{eqnarray} \varepsilon_{\mathrm{Ly}\alpha} & = & \varepsilon_\mathrm{rec} + \varepsilon_\mathrm{coll} \\ & = & P(\mathrm{Ly}\alpha) h \nu_0 \, n_\mathrm{e} n_\mathrm{HII} \, \alpha_\mathrm{B}(T) + h \nu_0 \, n_\mathrm{e} n_\mathrm{HI} \, q_{1\mathrm{s} \rightarrow 2\mathrm{p}}(T), \end{eqnarray} $$ where $h \nu_0$ is the energy per photon.
For completeness, here's a plot showing you the $T$ dependence and the relative importance of the two rate coefficients. Although collisional excitation seems to dominate at all high temperaures, the amount of neutral hydrogen atoms quickly decreases for $T\gtrsim10^4\,\mathrm{K}$.
