Loading [MathJax]/jax/output/HTML-CSS/jax.js
Synchro
Fast degradation limit
français | english
Navigation
Home
Sitemap
This wiki
This page

If the period is much bigger than the half-life, ie. the degradation rate is much bigger than the frequency : γΩ ie. δ<<1, we can make some simplifications.

 

Let us recall the transcendental equation for the frequency :

 

Ω=ω+S2(δ2+1)14Im(ei2(arctan(ωγ)arctan(δ))(11exp((δ2+1)14λei2arctan(δ)d+iβ)+11exp((δ2+1)14λei2arctan(δ)diβ)1))

We do not know if Ωγ is negligible in comparison to ωγ so we will keep this term and write  ei(arctan(ωγ)arctan(Ωγ))=A+iB 

If Ω=ω+ϵ,ϵ<<1 then A=1 and B=ϵ In a simulation, we find that Ωω103 for example.

It is useful to define the three following independant parameters :

 

λ=Dγ (coupling range in meters)

S=αDγ (coupling strength in 1/seconds)

 

δ=Ωγ

 

 

 

At first order in δ :

 

Ω=ω+S2Im((A+iB)(11e(1+iδλ+iβ)+11e(1+iδλiβ)1))=ω+S2Im((A+iB)(11e(1λ+iβ)+11e(1λiβ)1iδ2λ(1e(1λiβ)1+1e(1λ+iβ)1)))

 

A common choice of parameters (alpha=50; omega=10; gamma=100.; D=1)would give δ0.1 and B0.001, so we will choose to take the zeroth order in B (consequently, A=1)and the first order in δ.  Since the sum of two complex conjugates is real, taking the imaginary part leads to :

 

 

Ω=ωS4δλ2Re(1e(1λ+iβ)1)

 

 

(1e(1λ+iβ)1)=e1λcos(β)1ie1λsin(β)|e(1λ+iβ)1|2 so Ω=ωSδ2λe1λcos(β)1|e(1λ+iβ)1|2

 

Considering that λ=Dγ, in many cases where γ is big, we have λ small, and we can approximate the previous expression taking the leading order in e1λ when β is not far from 0:

 

Ω=ωSδ2λe1λcos(β)112e1λcos(β)+e2λωSδ2λe1λcos(β)

 

We can finally solve the equation for Ω, recalling that δ=Ωγ :

 

 

Ω=ω(1+S2γλe1λcos(β))1ω(1e1λS2γλcos(β)), or putting the distance between the oscillators back in the formula : Ω=ω(1edλSd2γλcos(β))
 
Conclusion : three hypotheses were necessary : 
  1. The variation of  Ω around the frequency ω is not too big (which translates to α not being to great).
  2. The degradation rate γ is much bigger than the frequency Ω
  3. The range of the coupling is small (D/d2γ (NB: this condition is less stringent than the previous one, because the approximation is much smaller)

 

 

Search
Share