Finding the gain coefficient at line centre for a HeNe Laser

AI Thread Summary
The discussion focuses on calculating the gain coefficient for a Helium Neon (HeNe) laser, emphasizing the importance of understanding the line profile width and the density of the upper lasing state. The gain coefficient is defined in terms of light intensity and propagation distance, with specific formulas provided for its calculation. The spontaneous emission rate coefficient plays a crucial role in determining the decay time of the population in the upper state. Participants express challenges in finding comprehensive resources on the topic, indicating that gain coefficient discussions are often overlooked in educational materials. The thread concludes with an acknowledgment of the complexity of the subject and a call for further exploration of the underlying principles.
MarkNormand99
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Homework Statement
The upper lasing transition of neon in a helium-neon laser has a spontaneous
radiative decay to the lower laser level with a transition probability of 3.39 × 10^7
s^−1
Determine the gain coefficient at line centre if the density of the upper lasing state for the helium neon laser is 10^18 m−3.
Assume the lower states population is negligible
Relevant Equations
There were no relevant equations given in the question
I have found the value for the full width at half maximum line profile width , I know the the density of the upper lasing state for the helium neon laser is 10^18 m−3, and I am assuming I have to use dN_2/dt = -A_21N_2, which I can turn into N_2(t)=N_2(t=0)exp(-A_21*t), and I have calculated the mean lifetime of the state to be t=1/A_21. I`m not too sure where i should go on from here, any advice would be appreciated, thanks.
 
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I looked at the Füchbauer–Ladeberg formula, the Einstein Coefficients, the four level laser, etc, etc couldn't find an answer... interesting problem... maybe look into the details of the Helium Neon Laser. Not sure that Boltzmann is the right track- but could be wrong... this seems to be Post Graduate Level stuff... so there is perhaps a lack of foundational information... not surprised you haven't got any responses. Good luck- let us know if you come up with any interesting sources. Spontaneous decay (emission)(should perhaps be the same as absorption from Einstein)- 10^18 m-3 perhaps photonic density of lasing (or transparency??). ?
 
It seems this question slipped through the cracks and the OP is probably long gone. I think this one isn't so bad. I'd just like to clear it up for everyone else reading this thread.

There's a very skeletal discussion of the gain coefficient here. The gist of it is that the gain coefficient is defined as ##\gamma(\nu) = \frac{\partial I / \partial z}{I}## where ##I## is the intensity of the light in the gain medium and ##z## is the distance propagated. It also shows that ##\gamma(\nu) = n\frac{\lambda^2}{8\pi t_{sp}} g(\nu)## where ##\lambda## is the lasing wavelength (633nm for HeNe's), ##t_{sp}## is the spontaneous emission lifetime aka ##\frac{1}{A}## in terms of the spontaneous emission rate coefficient, and ##g(\nu)## is a unit-normalized Lorentzian lineshape representing the lasing transition. The unit-normalzied lineshape is given by $$g(\nu) = \frac{\Delta \nu / 2\pi}{(\nu - \nu_0)^2 + (\Delta \nu / 2)^2}$$, and if I'm not mistaking the notation of the linked pdf then ##\Delta \nu = \frac{A}{2\pi}##. So on resonance ##g(\nu_0) = \frac{1}{\pi} \times \frac{2}{A/2\pi} = \frac{4}{A}##. From this, it follows that $$\gamma(\nu_0) = 10^{18} m^{-3} \frac{(633*10^{-9}m)^2 * 3.39*10^7 s^{-1}}{8\pi} \frac{4}{3.39*10^7 s^{-1}} = 63772$$

I may very well have some rogue factors of ##2\pi## in there, but it gets the point across.
 
Yes after commenting I realized the publish date. Thanks Twigg for the explanation. I couldn't follow the pdf- but your maths looks pretty fine- I'll have to look further at the background. Not sure how you slotted the transition probability into the equation- I'll have to review it again... Thanks Twigg.
 
The spontaneous emission rate coefficient ##A## comes in because it's related to the decay time. Since with no applied light the rate equation is $$\dot{N_e} = -A N_e$$ the population decays as $$N(t) = N(0) e^{-At}$$. In other words, the spontaneous emission decay time (##t_{sp}##), aka the time it takes for the population to decay to 1/e-th of it's initial level, is ##t_{sp} = \frac{1}{A}##.

Edit: sorry about the pdf being so poorly explained. It seemed like everything I found that discussed the gain coefficient was lecture slides. I think it's kind of a topic that no one really spends much time on. I've only heard "gain coefficient" used once or twice outside of class, after working on 20ish lasers in the time since. It's just an obscure term.
 
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