# Difference between revisions of "Masers"

### Cosmic Masers

${\displaystyle OH}$ and ${\displaystyle H_{2}0}$ masers can occur in dusty, star-forming regions which are cold enough for these molecules to form. The dust’s black-body radiation in the infrared band is absorbed by these molecules and a population inversion is established. When maser emission is caused by via stimulated emission, these clouds can get very bright (brightness temperatures ${\displaystyle \sim 10^{14}K}$). Temperatures this high cannot be thermal, so we know a maser when we see one.

Generally, masers are useful for tracing the galaxy’s magnetic field (emission lines are Zeeman split), and for following disks of gas and dust around stars in star-forming regions. In order to detect them, we need clouds which are moving uniformly together, and have velocity coherence both ${\displaystyle \perp ,\|}$ to our line-of-sight through a disk of rotation gas. In general, the intensity we observe depends on the path length through the masing cloud, so we like long path lengths with the same velocity.

### How masers Work

We will disucss population inversion a bit later, for now lets just assume that there are many more atoms in state 2 then state 1, where state 2 has higher energy than state 1. If a photon with an energy equal to E2-E1 comes along and interacts with an atom, it will thus be significantly more likely to produce stimulated emission (ie cause the atom to emit an identical photon with energy E2-E1 and drop from state 2 to 1) than to radiatively excite the atom (ie absorb the photon and transition from state 1 to 2). We will ignore the possibility of radiative excitation from state 1 to 2 entirely. The duplicate pair of photons produced by stimulated emission then go on to interact with other atoms leading to an exponential increase in the number of photons (and thus the intensity of radiation). Now onto the math. Consider a molecule with two rotational energy levels. Observing a homogeneous slab of this molecule, the intensity we receive is given by the familiar (see Radiative Transfer Equation):

${\displaystyle I_{\nu }=S_{\nu }(1-e^{-\tau _{\nu }})\,\!}$

where ${\displaystyle S_{\nu }={j_{\nu } \over \alpha _{\nu }}}$. ${\displaystyle j_{\nu }}$ and ${\displaystyle \alpha _{\nu }}$ are given by (see Einstein Coefficients):

{\displaystyle {\begin{aligned}j_{\nu }&=\overbrace {n_{2}{A_{21}}} ^{pertime}\overbrace {h\nu } ^{perE}\overbrace {\phi (\nu )} ^{per\ Hz}\overbrace {1 \over 4\pi } ^{persteradian}\\\alpha _{\nu }&={h\nu \over 4\pi }\phi (\nu )[\overbrace {n_{1}{B_{12}}} ^{abs}-\overbrace {n_{2}{B_{21}}} ^{stim \atop emis}]\\\end{aligned}}\,\!}

Thus, our source function looks like:

${\displaystyle S_{\nu }={n_{2}{A_{21}}\phi (\nu )h\nu {1 \over 4\pi } \over {h\nu \over 4\pi }\phi (\nu )[n_{1}{B_{12}}-n_{2}{B_{21}}]}\,\!}$

Then since ${\displaystyle g_{1}{B_{12}}=g_{2}{B_{21}}}$ and ${\displaystyle {{A_{21}} \over {B_{21}}}={2h\nu ^{3} \over c^{2}}}$, we have:

${\displaystyle S_{\nu }={2h\nu ^{3} \over c^{2}}{1 \over {n_{1}g_{2} \over g_{1}n_{2}}-1}\,\!}$

A population is said to be inverted when ${\displaystyle n_{1}g_{2} (not ${\displaystyle n_{1}). ${\displaystyle n_{1} \over g_{1}}$ is an expression for the population per degenerate sub-level in energy level 1. If ${\displaystyle {n_{1} \over g_{1}}<{n_{2} \over g_{2}}}$, then ${\displaystyle S_{\nu }<0}$, and we have a maser. Expressed in terms of the excitation temperature (${\displaystyle {n_{2} \over n_{1}}={g_{2} \over g_{1}}e^{-{h\nu \over kT_{ex}}}}$), we have:

${\displaystyle S_{\nu }={2h\nu ^{3} \over c^{2}}{1 \over e^{h\nu \over kT_{ex}}-1}\,\!}$

which is less than 0 when ${\displaystyle T_{ex}<0}$. We can express the optical depth of this slab to maser radiation as:

${\displaystyle \tau _{\nu }=\alpha _{\nu }L={h\nu \over 4\pi }\phi (\nu ){B_{21}}\left[{n_{1}g_{2} \over g_{1}}-n_{2}\right]L={h\nu \over 4\pi }\phi (\nu ){B_{21}}n_{2}\left[{n_{1}g_{2} \over g_{1}n_{2}}-1\right]L\,\!}$

If ${\displaystyle {n_{2} \over g_{2}}>{n_{1} \over g_{1}}}$, then ${\displaystyle \tau _{\nu }<0}$. Now you might think we’re talking nonsense with a negative source function and a negative optical depth, but we’re not. Only observable quantities need to be positive, since optical depth and the source function are just mathematical record keeping objects, they can take on any values. Recall that the intensity is:

${\displaystyle I_{\nu }=S_{\nu }(1-e^{-\tau })\,\!}$

If ${\displaystyle \tau \to -\infty }$, then ${\displaystyle I_{\nu }\to -S_{\nu }e^{\tau _{\nu }}}$, so ${\displaystyle I_{\nu }\gg 1}$. On the other hand, if ${\displaystyle \tau <0}$ and ${\displaystyle |\tau |\ll 1}$, then ${\displaystyle I_{\nu }\to S_{\nu }\tau _{\nu }}$, which is the product of two negatives = positive. This should be convincing you that ${\displaystyle I_{\nu }}$ is always positive, and therefore, actually manifested.

### Maser Species

${\displaystyle OH}$ mases at around 18 cm, and is found around AGB (asymptotic-giant-branch) stars in star-forming regions and around the galactic nucleus. AGB’s are important because they have lots of dust. The two masing transitions are from ${\displaystyle 1667\to 1612MHz}$ and ${\displaystyle 1720\to 1665MHz}$.

${\displaystyle H_{2}O}$ mases at ${\displaystyle 1.35cm}$ in a transition to its ground rotational state. Note that an order-of-magnitude calculation of ${\displaystyle \Delta E={\hbar \over 2I}}$ for ${\displaystyle H_{2}O}$ gives us an estimate of ${\displaystyle \sim 1mm}$, which is incorrect. The correct transitional energy is caused by a slight degeneracy in water molecules.

${\displaystyle SiO}$ mases at ${\displaystyle 3.4mm}$ in its ground vibrational state, and is typically found in star-forming regions and around AGB stars.

Other molecules mase, and there is even potential for detecting atomic lasers around massive stars (${\displaystyle L\sim 10^{5}L_{\odot }}$). Hydrogen transitions from ${\displaystyle 10\to 9}$ have been observed (at ${\displaystyle 55\mu m}$), and Stielnitski 1996 claims to have observed a population inversion in atomic H.

### Saturated vs. Unsaturated masers

There are two modes of operation for masers. For unsaturated masers, the gain is exponential with the path length, and for saturated ones, the gain only grows linearly with path length. The masers we find in the cosmos are typically saturated. The following is working toward understanding why there are two modes in masers. Before we start with the math lets imagine a sequence of events that might lead to saturation. Initially we have lots of atoms in state 2. If a photon that has energy E2-E1 interacts with one of these atoms we get two identical photons. Both of these photons can then interact with other atoms in state 2 giving 4 photons. This leads to an exponential amplification of the number of photons. If this increasing number of photons continues to propogate through the atoms, a point will eventually be reached where there are more photons with energy E2-E1 then there are atoms in state 2 for them to interact with. This means that we transition from an exponential amplification (from photon doubling) to a linear amplification (from continuing to travel through a medium with some density of atoms in state 2). This is called saturation. The graph given below shows intensity ${\displaystyle I_{\nu }}$ as a function of optical depth ${\displaystyle \tau }$. The discontinuity in the derivative on the graph is the saturation point.

Now let us prove this intuitive solution mathematically. We begin with the expression for specific intensity from emissivity (see Radiative Transfer Equation and Einstein Coefficients):

{\displaystyle {\begin{aligned}{dI_{\nu } \over dz}&=j_{\nu }-\alpha _{\nu }I_{\nu }\\&={h\nu \over 4\pi }\phi (\nu )n_{2}A_{21}-{h\nu \over 4\pi }\phi (\nu )[n_{1}{B_{12}}-n_{2}{B_{21}}]I_{\nu }\\\end{aligned}}\,\!}

In principle, the Line Profile Functions governing spontaneous emission and the line-profile governing stimulated emission might not have to be the same, but evidence seems to suggest they are. Let’s assume they are, and rewrite this:

{\displaystyle {\begin{aligned}{dI_{\nu } \over dz}&={h\nu \over 4\pi }\phi (\nu ){n_{2}{A_{21}} \over g_{2}}\cdot g_{2}-{h\nu \over 4\pi }\phi (\nu )g_{2}{B_{21}}\left[{n_{1} \over g_{1}}-{n_{2} \over g_{2}}\right]\end{aligned}}\,\!}

Defining ${\displaystyle N_{1}\equiv {n_{1} \over g_{1}}}$, ${\displaystyle N_{2}\equiv {n_{2} \over g_{2}}}$, ${\displaystyle A\equiv {A_{21}}g_{2}}$, and ${\displaystyle B={B_{21}}g_{2}}$, our equation looks like:

${\displaystyle {dI_{\nu } \over dz}={h\nu \over 4\pi }\phi (\nu )[N_{2}A+B(N_{2}-N_{1})I_{\nu }]\,\!}$

Now we’ll integrate over frequency. ${\displaystyle \phi (\nu )}$ is a sharply peaked function, so we’ll treat it as a ${\displaystyle \delta }$-function. Then substituting ${\displaystyle I\equiv \int {\phi (\nu )I_{\nu }d\nu }}$, we have:

${\displaystyle {\int {I_{\nu }d\nu } \over \Delta \nu }\equiv \int {I_{\nu }\phi (\nu )d\nu }\,\!}$

which is true by definition of ${\displaystyle \Delta \nu }$: ${\displaystyle \int {I_{\nu }d\nu }=I\Delta \nu }$. So now we have:

${\displaystyle {dI \over dz}={h\nu \over 4\pi \Delta \nu }[(N_{2}-N_{1})BI+N_{2}A]\,\!}$

This is an equation with three unknowns. To close this system, we’ll use the equations of statistical equilibrium, which say:

${\displaystyle {dN_{2} \over dt}=-N_{2}BJ+N_{1}BJ-N_{2}A+R_{2}(N-N_{1}-N_{2})-\Gamma _{2}N_{2}\,\!}$

where ${\displaystyle J\equiv \int {I_{\nu }\phi (\nu )d\nu }}$, and ${\displaystyle J_{\nu }={1 \over 4\pi }\int {I_{\nu }d\Omega }}$. ${\displaystyle R_{2}}$ is the “pumping rate”. It describes the rate at which molecules not in state 1 or 2 (counted by ${\displaystyle N-N_{1}-N_{2}}$) are radiatively or collisionally knocked into state 2. ${\displaystyle \Gamma _{2}}$ is the loss rate of molecules in state 2 into any state other than state 1. In the above equation, we’ve neglected two terms: collisional excitation ${\displaystyle 1\to 2}$ and ${\displaystyle 2\to 1}$. These are crucial terms, being tightly related to local thermal equilibrium. However, we will neglect them to simplify analysis. We also have a similar equation for state 1:

${\displaystyle {dN_{1} \over dt}=N_{2}BJ-N_{1}BJ+N_{2}A+R_{1}(N-N_{1}-N_{2})-\Gamma _{1}N_{1}\,\!}$

We’ll further simply matters by setting the loss-rates for the two populations equal to each other (${\displaystyle \Gamma _{1}=\Gamma _{2}=\Gamma }$). Now let’s solve for the steady-state solution (${\displaystyle {dN \over dt}=0}$).

${\displaystyle {d(N_{2}-N_{1}) \over dt}=-(N_{2}-N_{1})2BJ-2N_{2}A+(R_{2}-R_{1})(N-N_{12})-\Gamma (N_{2}-N_{1})\,\!}$

where ${\displaystyle N_{12}=N_{1}+N_{2}}$. Two terms in this equation are acting to reduce the population inversion of state 2 with respect to state 1: ${\displaystyle (N_{1}-N_{1})2BJ}$ and ${\displaystyle \Gamma (N_{2}-N_{1})}$. In the unsaturated regime where ${\displaystyle BJ\ll \Gamma }$, stimulated emission is a minor perturbation to the inverse. In this case we can solve the system of equations:

{\displaystyle {\begin{aligned}{d\Delta N \over dt}={d(N_{2}-N_{1}) \over dt}&=-2N_{2}A+(R_{2}-R_{1})(N-N_{12})-\Gamma \Delta N=0\\{dN_{12} \over dt}&=(R_{2}+R_{1})(N-N_{12})-\Gamma (N_{1}+N_{2})=0\\\end{aligned}}\,\!}

Then solving for ${\displaystyle N-N_{12}}$:

${\displaystyle N-N_{12}={\Gamma (N_{1}+N_{2}) \over R_{1}+R_{2}}={\Gamma N_{12} \over R_{1}+R_{2}}\,\!}$

Substituting this into the difference equation, which reads:

${\displaystyle -2N_{2}A={R_{2}-R_{1} \over R_{1}+R_{2}}\Gamma N_{12}-\Gamma \Delta N=0\,\!}$

the using ${\displaystyle N_{2}={N_{12}+\Delta N \over 2}}$ (this is just an identity), we get:

${\displaystyle \underbrace {\left(1+{\Gamma \over A}\right)} _{\equiv 2\beta }\Delta N=N_{12}\left[\underbrace {{R_{2}-R_{1} \over R_{1}+R_{2}}{\Gamma \over A}} _{\equiv {1 \over \alpha }}-1\right]\,\!}$

Thus:

${\displaystyle \Delta N={N_{12} \over 2\beta }\left[{1 \over \alpha }-1\right]={N_{12}(1-\alpha ) \over 1\alpha \beta }\,\!}$

Now we’ll define ${\displaystyle S\equiv {\alpha \beta \over 1-\alpha }}$. ${\displaystyle S}$ is meant to connote the source function here. The reason we might expect ${\displaystyle S}$ to be related to the source function is that ${\displaystyle S_{\nu }\propto {N_{2} \over N_{2}-N_{1}}\propto {N_{2} \over \Delta N}}$, so ${\displaystyle \Delta N\propto {N_{12} \over 2S}}$. This is why we use an ${\displaystyle S}$ here. Getting back to our original equation, we have:

${\displaystyle {4\pi \Delta \nu \over h\nu A}{dI \over dz}=\Delta N\underbrace {BI \over A} _{\equiv {\mathfrak {I}}}+N_{2}\,\!}$

Using our newly defined ${\displaystyle S}$, this becomes:

${\displaystyle {4\pi \Delta \nu \over h\nu B}{d{\mathfrak {I}} \over dz}={N_{12} \over 2S}{\mathfrak {I}}+{\frac {1}{2}}(N_{12}+\Delta N)\,\!}$

Finally, defining an unsaturated gain length ${\displaystyle L\equiv {4\pi \Delta \nu \over Bh\nu }{2S \over N_{12}}}$, we have:

${\displaystyle L={\mathfrak {I}}+S+{\frac {1}{2}}\,\!}$

Then using ${\displaystyle ds={dz \over L}}$, we have:

${\displaystyle {d{\mathfrak {I}} \over ds}={\mathfrak {I}}+{\frac {1}{2}}+S\,\!}$

To order of magnitude:

{\displaystyle {\begin{aligned}{\mathfrak {I}}&\equiv {B \over A}I={B \over A}B_{\nu }(T_{bright})\\&={c^{2} \over 2h\nu ^{3}}{2h\nu ^{3} \over c^{2}}{1 \over e^{h\nu \over kT_{bright}}-1}\\&={1 \over e^{h\nu \over kT_{bright}}-1}\\\end{aligned}}\,\!}

Now since ${\displaystyle 1\ll {kT_{bright} \over h\nu }}$ (typical maser wavelengths are in mm and ${\displaystyle T_{bright}}$ is typically several K), we can throw away our ${\displaystyle {\frac {1}{2}}}$ in our equation for ${\displaystyle {d{\mathfrak {I}} \over ds}}$:

${\displaystyle {d{\mathfrak {I}} \over ds}={\mathfrak {I}}+S\Rightarrow \int {d{\mathfrak {I}} \over {\mathfrak {I}}+S}=\int {ds}\Rightarrow {\mathfrak {I}}+S=De^{s}\,\!}$

Choosing ${\displaystyle {\mathfrak {I}}=0,s=0}$ (we’re assuming there is no background source), then ${\displaystyle D=S}$, so:

${\displaystyle {{\mathfrak {I}}=S(e^{s}-1)}\,\!}$

If ${\displaystyle s\ll 1}$, we have ${\displaystyle {\mathfrak {I}}=Ss}$ from spontaneous emission, and this is the saturated case. If ${\displaystyle s\gg 1}$, the ${\displaystyle {\mathfrak {I}}=Se^{s}}$ from stimulated emission, and this is the unsaturated case. Earlier we threw away the ${\displaystyle BJ}$ term, but we could have done all of this including that term, and the algebra would have been the same. If we do this, we get the answer:

${\displaystyle {{d{\mathfrak {I}} \over ds}={\beta ({\mathfrak {I}}+{\frac {1}{2}}) \over \beta +{\mathfrak {J}}}+S}\,\!}$

where ${\displaystyle {\mathfrak {J}}}$ is the non-dimensionalized, integrated flux, ${\displaystyle {\mathfrak {J}}={BJ \over A}}$.

### The Saturated Mode

Recall our masing equation:

${\displaystyle {d{\mathfrak {I}} \over ds}={\beta ({\mathfrak {I}}+{\frac {1}{2}}) \over \beta +{\mathfrak {J}}}+S\,\!}$

In the unsaturated case, ${\displaystyle {\beta \over \beta +{\mathfrak {J}}}\to 1\iff \beta \gg {\mathfrak {J}}}$. This is equivalent to saying:

${\displaystyle {\frac {1}{2}}\left(1+{\Gamma \over A}\right)\gg {BJ \over A}\,\!}$

And if ${\displaystyle \Gamma \gg A}$, then:

${\displaystyle A+\Gamma \gg BJ\iff \Gamma \gg BJ\,\!}$

In the saturated case, then ${\displaystyle \beta \ll {\mathfrak {J}}}$. In this case:

${\displaystyle {d{\mathfrak {I}} \over ds}={\beta {\mathfrak {I}} \over {\mathfrak {J}}}+S={4\pi \beta {\mathfrak {I}} \over \int {{\mathfrak {I}}ds}}+S\,\!}$

where we used that ${\displaystyle {\mathfrak {J}}\sim {1 \over 4\pi }\int {{\mathfrak {I}}d\Omega }}$. If we consider a single beam of photons through a cloud, then ${\displaystyle \int {{\mathfrak {I}}d\Omega }\approx {\mathfrak {I}}\Delta \Omega }$, so:

{\displaystyle {\begin{aligned}{d{\mathfrak {I}} \over ds}&={4\pi \beta {\mathfrak {I}} \over {\mathfrak {I}}\Delta \Omega }+S\\&={4\pi \beta \over \Delta \Omega }+S\\\end{aligned}}\,\!}

### How We Get Population Inversions

There is another dichotomy in masers: ones which are excited radiatively and those which are excited collisionally (see Collisional Excitation). We’ll discuss a cloud of molecules which have 3 energy states which are populated by simple collisional pumping. Before we dive into the math lets consider conceptually what is going on. The diagram shows a simple 3 state atom. We can only go between state E1 and E3 collisionally (ie they do not communicate radiatively). State E3 spontaneously decays radiatively to state E2 with a short lifetime. State E2 radiatively decays to state E1, but with a very long lifetime. This means that there can be a large number of atoms that are in state E2. Thus we have the requisite population inversion for masing to occur. Now onto the math. The rate of change of the population of energy state 1 in this case is given by (sources - sinks):

${\displaystyle {dn_{1} \over dt}=n_{2}{A_{21}}+n_{2}C_{21}+n_{3}C_{31}+n_{2}{B_{21}}J_{21}-n_{1}{B_{12}}J_{12}-n_{1}C_{12}-n_{1}C_{13}\,\!}$

Then ${\displaystyle J_{12}=J_{21}}$, and because of local collisional thermal equilibrium, ${\displaystyle C_{12}=C_{21}{g_{2} \over g_{1}}e^{-{E_{21} \over kT}}}$. Similarly, ${\displaystyle C_{13}=C_{31}{g_{3} \over g_{1}}e^{-{E_{21} \over kT}}}$. So dividing by ${\displaystyle g}$, and defining ${\displaystyle N_{1}\equiv {n_{1} \over g_{1}}}$, we have:

{\displaystyle {\begin{aligned}{dN_{1} \over dt}=&\underbrace {n_{2} \over g_{2}} _{N_{2}}\underbrace {{{A_{21}} \over g_{1}}g_{2}} _{A_{21}}+\underbrace {n_{2} \over g_{2}} _{N_{2}}\underbrace {{C_{21} \over g_{1}}g_{2}} _{\equiv C}+\underbrace {n_{2} \over g_{3}} _{N_{3}}\underbrace {{C_{31} \over g_{1}}g_{3}} _{\equiv C}+\underbrace {n_{2} \over g_{2}} _{N_{2}}\underbrace {{{B_{21}} \over g_{1}}g_{2}} _{B_{21}}J_{21}\\&-N_{1}\underbrace {B_{21}} _{{B_{21}}{g_{2} \over g_{1}}\equiv {B_{21}}}-N_{1}\underbrace {C_{21}{g_{2} \over g_{1}}} _{\equiv C}e^{-{E_{21} \over kt}}-N_{1}\underbrace {C_{31}{g_{3} \over g_{1}}} _{\equiv C}e^{-{E_{31} \over kT}}\\\end{aligned}}\,\!}

Phew. Notice that we set ${\displaystyle C_{21}=C_{31}}$. This is just to make our lives easier. We can do the same for ${\displaystyle dN_{2} \over dt}$, but omitting ${\displaystyle B_{23}}$, because we’re deciding not to have absorptions from ${\displaystyle 2\to 3}$ and no stimulated emission from ${\displaystyle 3\to 2}$. Then our total population is ${\displaystyle N=N_{1}+N_{2}+N_{3}}$. Without being careful, instinct tells us that in steady state, we’ll have a population inversion ${\displaystyle {N_{2} \over N_{1}}>1}$ if ${\displaystyle C\ll A_{32}}$. This instinct is correct, but let’s do this carefully. First we’ll make some assumptions:

• ${\displaystyle {A_{21}}\ll C}$. For ${\displaystyle H_{2}O}$:
${\displaystyle {A_{21}}\sim 10^{8}s^{-1}\left({1216\mathrm {\AA} \over 1.35cm}\right)^{3}\sim 10^{8}\cdot 10^{-15}s^{-1}\sim 10^{-7}s^{-1}\left({\mu \over ea_{0}}\right)^{2}\,\!}$

where ${\displaystyle \mu }$ is our way of accommodating the fact that the dipole moment for ${\displaystyle H_{2}O}$ might not be the same as for the fine structure of hydrogen. It turns out the answer is ${\displaystyle {A_{21}}\sim 2\cdot 10^{-9}s^{-1}}$.

Estimating C:

${\displaystyle C\sim n_{H_{2}}\sigma v_{rel}\sim n_{H_{2}}\cdot 10^{-15}\cdot (3{km \over s})\sim n_{H_{2}}\cdot 3\cdot 10^{-10}\,\!}$

which is ${\displaystyle \gg 10^{-7}s^{-1}\left({\mu \over ea_{0}}\right)^{2}}$ when ${\displaystyle n_{H_{2}}\gg 10^{3}cm^{-3}\left({\mu \over ea_{0}}\right)^{2}}$.

• Next we’ll assume ${\displaystyle E_{21}\ll kT}$. Define ${\displaystyle {E_{21} \over kT}\equiv \delta \ll 1}$.

Now we have a 2-step:

• Step 1: Since ${\displaystyle 1\to 3}$ are not linked by radiation,
${\displaystyle {N_{3} \over N_{1}}\approx e^{-{E_{31} \over kT}}\equiv \theta \leq 1\,\!}$
• Step 2: Radiative decays from ${\displaystyle 3\to 2}$. To get an inversion, we’ll argue that the sources into 2 are larger than the sinks out of 2 (this is a little weird because we’re solving our steady-state equations, but whatever):
${\displaystyle N_{3}A_{32}+N_{3}C+N_{1}Ce^{-{E_{21} \over kT}}>N_{2}Ce^{-{E_{32} \over kT}}+N_{2}C\,\!}$

We can rewrite this as:

{\displaystyle {\begin{aligned}\underbrace {N_{2} \over N_{1}} _{=\theta }{A_{32} \over C}&>\left[{N_{2} \over N_{1}}\underbrace {e^{-{E_{32} \over kT}}} _{E_{32}=E_{32}-E_{12}}-{N_{3} \over N_{2}}\right]+\left[{N_{2} \over N_{1}}-e^{-{E_{21} \over kT}}\right]\\{A_{32}\theta \over C}&>\left[\underbrace {N_{2} \over N_{1}} _{1}\theta (1+\delta )-\theta \right]+\left[\underbrace {N_{2} \over N_{1}} _{1}-(1-\delta )\right]\\&=\theta \delta +\delta =\delta (\theta +1)\\\end{aligned}}\,\!}

Thus, the ${\displaystyle \delta }$ really helps get masing started.

Now one last thing: we’d chosen to ignore stimulated radiative transfer between energy states 3 and 2. In general, this process will tend to reduce the population inversion. However, for optically thick clouds ${\displaystyle \tau \gg 1}$, photons only have a ${\displaystyle P\sim {1 \over \tau }}$ probability of escaping, so we can describe this by “diluting” the ${\displaystyle A_{32}}$ term by ${\displaystyle {1 \over \tau }}$.

#### The 4 state atom

While 3 state masers are possible, most actual masers involve 4 states. Lets discuss population inversion in a 4 state maser. We will relax the condition that collisions be responsible for the transition to the highest state. As shown in the diagram we require that there is some strong transition from state 1 to 4. This provides atoms in a state that can dexcite to our population-inverted state. Imagine that we want states 3 and 2 to be the population inverted states. To achieve this, we need a relatively strong transition from state 4 to 3 to populate state 3. Similarly we need a strong 2 to 1 transition to empty out state 2. We are thus left with lots of atoms in state 3, and almost none in 2. This provides the inversion necessary for masing. Our masing transition will be 3 to 2 (ie it will correspond to our population inverted states).