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May 17th, 2015, 06:50 PM  #1 
Newbie Joined: May 2015 From: USA Posts: 3 Thanks: 0  Working Some Physics, Stuck on Algebra
I'm working on laser physics, specifically steady state inversion and I can't get past this one algebra part... Here are what is known: These two equations are inserted into When the values for N_1 and N_2 are substituted in, the answer the book gives is I do not know how they got to that equation, I'm stuck. When I substitute and rearrange the numbers I get: Anyway, I really can't figure this one out. Any help would be appreciated. If you are wondering, this is suppose to be the equation for Steadystate inversion, the condition necessary for a laser to work. N_1 and N_2 are the numbers of atoms in high and low states, g_1 and g_2 are their respective degeneracies, T1 and T2 is the time period they spend in those states and A_21 is a constant based on the atomic structure... Thanks you much! 
May 17th, 2015, 07:45 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
$\displaystyle \frac{R_2T_2}{g_2} > \frac{R_1T_1 + R_2T_2 A_{21}T_1}{g_1}$ $\displaystyle \frac{R_2T_2}{g_2} > \frac{R_1T_1}{g_1} + \frac{R_2T_2 A_{21}T_1}{g_1}$ $\displaystyle \frac{R_2T_2}{g_2}  \frac{R_2T_2 A_{21}T_1}{g_1} >\frac{R_1T_1}{g_1}$ Multiply both sides by $\displaystyle \frac{g_1}{R_1T_1}$ $\displaystyle \frac{R_2T_2g_1}{R_1T_1g_2}  \frac{R_2T_2A_{21}}{R_1} > 1$ Here comes a trick: I'm going to multiply the second term one the LHS by $\displaystyle 1 = \frac{g_1}{g_2} \cdot \frac{g_2}{g_1} \cdot \frac{T_1}{T_1}$. (How do I know to do this? I solved the problem backward. ) $\displaystyle \frac{R_2T_2g_1}{R_1T_1g_2}  \frac{R_2T_2 A_{21}T_1g_1g_2}{R_1T_1g_1g_2} > 1$ Now factor: $\displaystyle \frac{R_2T_2g_1}{R_1T_1g_2} \cdot \left ( 1  \frac{A_{21}T_1g_2}{g_1} \right ) > 1$ Dan  
May 18th, 2015, 06:17 AM  #3 
Newbie Joined: May 2015 From: USA Posts: 3 Thanks: 0 
Thanks a lot Dan! That helps a lot.

May 18th, 2015, 08:41 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,343 Thanks: 1024 
James, there are 7 variables used: A21, g1, g2, R1, R2, T1, T2 Can you give me an actual/typical value for each? Thanks. I'm trying to understand the book's equation (the (1  ?) portion): how the results can be > 1 after a multiplication by (1  ?). Btw, quite an interesting equation... 
May 18th, 2015, 01:33 PM  #5 
Newbie Joined: May 2015 From: USA Posts: 3 Thanks: 0 
So inside the square brakets: is called the "necessary but not sufficient condition" for achieving population inversion. Population inversion is when you have more atoms per unit volume in the high state rather than the low state. This condition can be rearranged to give: is called the Einstein A coefficient, it describe the rate at which atoms decay to the lower level. A typical value for A_21 would be 5*10^7 s^1. So together with N_1 (in atoms/unit volume), we get a typical value for T (or tau actually) is 30ns, this is the "flourescence lifetime", it describes how long an atom lasts in the high or low states. The g values are the degeneracy values for each energy state. They are integers. In quantum physics there are many different ways to get to the same energy levels, if there are three different ways, the degeneracy value is 3. The R values are the pump rates. In order for a laser to work, you have to continuously pump atoms into the high state, there's lots of ways to do it, but they all involve energizing the atoms in some way, unfortunately, I do not have a good typical value since I'm still a novice at this. But if I find one I will message you. Last edited by JamesBrady; May 18th, 2015 at 01:39 PM. Reason: Forgot g values 
May 18th, 2015, 04:13 PM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,343 Thanks: 1024  
May 18th, 2015, 08:52 PM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,343 Thanks: 1024 
Another way to get to the book's format. R1, T1 = a, m R2, T2 = b, n g1, g2 = u, v A21 = k So, right before Dan gets tricky, we're in this position: bnu / (amv)  bnk / a > 1 And we want to change above to this format: **** bnu / (amv)[1  x] > 1 So bnux / (amv) = bnk / a Simplify: x = mvk / u **** bnu / (amv)[1  mvk / u] > 1 ...same as Dan. Note to James: when working with longish equations with unwieldy variables as is the case here, I find that changing the variables as I did here saves a whole lot of writing (and paper!). Something to remember... 

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