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

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The superconduction gap equation simplifies in the weak coupling limit, where VN(0) is significantly less than 1. In this scenario, the equation reduces to lΔl = ωDexp(-1/VN(0)). The transformation is derived from the relationship arsinh(ωD/lΔl) = ln(ωD² + √(ωD² + lΔl²)), which is valid under the approximation that VN(0) is small. This leads to the negative exponent in the exponential term, illustrating the behavior of the superconducting gap under weak coupling conditions.

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The superconduction gap equation is:

1=VN(0)arsinh(ωD/lΔl).

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

lΔl = ωDexp(-1/VN(0))

But how is this obtained. We have:

arsinh(ωD/lΔl) = ln(ωD2+√(ωD2 + 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?
 
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hi aaaa202! :smile:
aaaa202 said:
arsinh(ωD/lΔl) = ln(ωD2+√(ωD2 + lΔl2))

erm :redface: … 

first write it ωD/lΔl = sinh(1/VN(0)) :wink:​
 
wow lol.. thanks
 

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