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Power dependence of dynamic spatial replenishment of femtosecond pulses propagating in air

M. Mlejnek, E.M. Wright, and J.V. Moloney

Open Access Open Access

Abstract

Numerical simulations of nonlinear pulse propagation in air show an initial pulse formed, absorbed by plasma generation, and subsequently replenished by power from the trailing edge. Here we show that this scenario implies that the length scale for filament propagation is relatively insensitive to the peak input power beyond the threshold for filament generation.

©1999 Optical Society of America

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Supplementary Material (4)

Media 1: MOV (2232 KB)     
Media 2: MOV (2356 KB)     
Media 3: MOV (2558 KB)     
Media 4: MOV (2690 KB)     

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Figures (6)
Figure 1.
Figure 1. Global maximum over time of the on-axis intensity as a function of propagation distance z for peak input power of ~ 4Pcr . The Rayleigh range was in this case ~ 2 m. (Dashed line … n 2 without the Raman contribution.)

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Figure 2.
Figure 2. Global maximum over time of the on-axis intensity as a function of propagation distance z for peak input power of ~ 6Pcr . The Rayleigh range was in this case ~ 2 m. (Dashed line … n 2 without the Raman contribution.)

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Figure 3.
Figure 3. Global maximum over time of the on-axis intensity as a function of propagation distance z for peak input power of ~ 6.5Pcr . The Rayleigh range was in this case ~ 2 m. Dashed line … n 2 without the Raman contribution.)

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Figure 4.
Figure 4. Global maximum over time of the on-axis intensity as a function of propagation distance z for peak input power of ~ 7Pcr . The Rayleigh range was in this case ~ 2 m. (Dashed line … n 2 without the Raman contribution.)

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Figure 5.
Figure 5. The intensity of The field for the case shown in Fig. 1 (Raman response included) at the propagation distance of 1 m.

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Figure 6.
Figure 6. The intensity of the field for the case shown in Fig. 1 (Raman response included) at the propagation distance of 2.8 m.

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Equations (5)

Equations on this page are rendered with MathJax. Learn more.

∂ ε ∂ z = i 2 k ( ∂ 2 ∂ r 2 + 1 r ∂ ∂ r ) ε − ik ″ 2 ∂ 2 ε ∂ t 2 − σ 2 ( 1 + iωτ ) ρε − β ( K ) 2 ∣ ε ∣ 2 K − 2 ε
+ i k 0 ( 1 − f ) n 2 ∣ ε ∣ 2 ε + i k 0 f n 2 [ ∫ − ∞ ∞ dt ′ R ( t − t ′ ) ∣ ε ( t ′ ) ∣ 2 ] ε ,
R ( t ) = θ ( t ) Ω 2 e − Γ t / 2 sin ⁡ ( Λ t ) Λ , Λ = Ω 2 − Γ 2 / 4 ,
∂ ρ ∂ t = 1 n b 2 σ E g ρ ∣ ε ∣ 2 + β ( K ) ∣ ε ∣ 2 K Kħω − a ρ 2 .
ε ( r , 0 , t ) = 2 P in π ω 0 2 exp ⁡ ( − r 2 ω 0 2 − t 2 t p 2 ) ,
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