How does the superconduction gap equation simplify in the weak coupling limit?

[aaaa202]

The superconduction gap equation is:

1=VN(0)arsinh(ω_D/lΔl).

My book says that in the weak coupling limit VN(0)<<1 this will reduce to:

lΔl = ω_Dexp(-1/VN(0))

But how is this obtained. We have:

arsinh(ω_D/lΔl) = ln(ω_D²+√(ω_D² + lΔl²)), but what has the approximation of this got to do with the prefactor being small? Also I don't see how you get a negative exponent?

 

[tiny-tim]

hi aaaa202!

arsinh(ω_D/lΔl) = ln(ω_D²+√(ω_D² + lΔl²))

erm … 

first write it ω_D/lΔl = sinh(1/VN(0)) ​

 

[aaaa202]

wow lol.. thanks