A simple model of a laser ( )

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A simple model of a laser (URGENT!!!!!!!)

Homework Statement



Need to turn this in 2 hrs from now. Yes, I'm a slacker.

(a) Suppose that N evolves much more quickly than n so that we can make the quasi-static approximation N ≈ 0. Given this approximation, re-write our equations as a 1-D dynamical system for n.

(b) Show that for p > pc, the "trivial" fixed point n* = 0 is unstable, and that for p < pc, n* = 0 is stable. No need to explicitly determine pc.

(c) What type of bifurcation occurs as the pump strength increases past the laser threshold pc?

(d) For what range in the parameters (G, k, f, p) do you expect the quasi-static approximation to be good?

Homework Equations



"This will give some good intuition and background information on the physics involved. The problem we will consider is as follows. Let N be the number of excited atoms in our laser, and let n be the number of photons (particles of light) that ... [etc.] ....

n'(t) = GnN - kn
N'(t) = -GnN - fN + p,

where G is the gain coefficient .... [etc.] ....



The Attempt at a Solution



Part (a) is easy: I have n'(t) = Gnp/(Gn+f) - k. Note that this means n'(t) = 0 if n = 0 or n = p/k - f/g.

So then (n'(t))'(n) = Gpf/(Gn+f)2 - k. Plugging in n*=0 to that gives Gp/f - k. I'm assuming that the pc in the problem means pc = kf/G ????????? Because then we have a situation where p > pc makes Gp/f - k > 0, and p < pc makes Gp/f - k < 0.

I've noticed also that p = fk/g makes (n'(t))'(n) = 0 when I plug in the other fixed point, n* = p/k - f/g. I'm still trying to figure out the significance of this.
 

Answers and Replies

  • #2
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I'm assuming that the pc in the problem means pc = kf/G ?
Right, as this point is the border between a stable an an unstable point n*=0.

I've noticed also that p = fk/g makes (n'(t))'(n) = 0 when I plug in the other fixed point, n* = p/k - f/g. I'm still trying to figure out the significance of this.
Saddle point? Could be stable.
 
  • #3
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Right, as this point is the border between a stable an an unstable point n*=0.

That's why I hate these deceptive homework problems. When it says "No need to explicitly determine pb", I get the impression that pc is a value that's extremely difficult to solve for.
 
  • #4
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Saddle point? Could be stable.

I think it would be considered a saddle point since there's only one fixed point when p = pc, but then another appears out of nowhere when p ≠ pc.

EDIT: It would be a "transcritical bifurcation" since the stability at n=0 changes as p is varied from p=pc to p>pc.
 

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