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Self-focusing of an optical beam in
cold plasma
Gio Chanturia
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Intro
What do we have?
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A laser beam (ultra short). Cold plasma (collisionless).
Self-focusing of a beam in certain mediums
Due to non-linearity of medium, the beam focuses itself.
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Cold plasma and short pulse as our model
There are reasons, we use these models:
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Cool plasma model:
Ponderomotive effect;
Due to electromagnetic field.
Relativistic effect;
Due to free electrons in plasma.
NO thermal effect.
Short laser pulse model:
Short time scale;
No self-focusing process for quasineutral plasma.
Massive ions do not have time to respond
and therefore stay immobile.
Describing our system mathematically
What do we need to describe our system?
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Maxwell’s equations and equation of motion for a relativistic electrons:
Plasma current:
Electron velocity:
𝑱 = −𝑒𝑛𝑒𝒗 = −𝜔𝑝2
4𝜋𝑐
𝑁𝑒
1 + 𝐼𝑛𝑨
𝒗 =𝑷
𝑚𝛾=
𝜖
𝑚𝑐
𝑨
1 + 𝐼𝑛
Amplitude: 𝑨 = 𝑎𝑛 𝒓, 𝑡 𝑒𝑖 𝑘0𝑧−𝜔0𝑡−𝜓 𝒓,𝑡 𝒙 + 𝑖 𝒚
𝑁𝑒 = 1 +𝛿𝑛𝑒𝑛𝑜
Assumptions to deal with our calculations
Assumptions for amplitude and phase equations:
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Firstly, as we deal with axial symmetry.
That is, when none of the functions depend on 𝜃 and we’re left with only two variables:
𝑎 𝑟, 𝑧 = 𝑎 𝑟
𝜓 𝑟, 𝑧 = 𝑓 𝑧 + 𝑔(𝑟)
𝑥, 𝑦, 𝑧 → (𝑟, 𝜃, 𝑧)
Secondly, we assume, that amplitude doesn’t vary towards 𝑧 direction.
We allow phase to modulate and seek for solution as a sum of individual functions.
The first simplification
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Applying previous assumptions and separating variables, we get:
1
𝑎
𝑑2𝑎
𝑑𝑟2−
𝑑𝑔
𝑑𝑟
2
−1
𝜆𝑐2
𝑁𝑒
1 + 𝑎2= 𝐶1
𝑑2𝑔
𝑑𝑟2+
1
𝑎2𝑑(𝑎2)
𝑑𝑟
𝑑𝑔
𝑑𝑟= 0
These equations still look tricky. So, let us apply the slab limit:
{separation constant}
𝑦
𝑥
𝑦 → 0𝑟 → 𝑥
Slab limit results
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Within the slab limit we have:
1
𝑎
𝑑2𝑎
𝑑𝑥2−𝐶42
𝑎4−
1
𝜆𝑐2
𝑁𝑒
1 + 𝑎2= 𝐶1
𝑁𝑒 = 1 + 𝜆𝑐2 𝑑2
𝑑𝑥21 + 𝑎2
Which combines into:
1
𝑎
𝑑2𝑎
𝑑𝑥2−𝐶42
𝑎4−
1
𝜆𝑐2 1 + 𝑎2
−1
1 + 𝑎2
𝑑2
𝑑𝑥21 + 𝑎2 = 𝐶1
Lagrangian analogy
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The last equation can be written in a form:
We know from the least action principle, that Lagrangian of a particle is written like this:
𝐿 = 𝑔 𝑎𝑎′ 2
2− 𝑉(𝑎)
Which by Nother’s theorem gives:
𝑔 𝑎 𝑎′′ +1
2
𝑑𝑔
𝑑𝑎𝑎′ 2 −
𝜕𝑉
𝜕𝑎= 0 ≡ 𝐺(𝑎, 𝑎′, 𝑎′′)
1
𝑎
𝑑2𝑎
𝑑𝑥2−𝐶42
𝑎4−
1
𝜆𝑐2 1 + 𝑎2
−1
1 + 𝑎2
𝑑2
𝑑𝑥21 + 𝑎2 − 𝐶1 = 0 ≡ 𝐹(𝑎, 𝑎′, 𝑎′′)
Lagrangian analogy
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If for some integrating factor 𝜇 𝑎 :
Then we will be able to find the “potential” of amplitude and therefore describe the behavior of it.
𝐹 𝑎, 𝑎′, 𝑎′′ ∙ 𝜇 𝑎 = 𝐺(𝑎, 𝑎′, 𝑎′′)
Making calculations in this manner and flattening the metric by transformation
𝑎 = sinh(𝑦)
We obtain:
𝑉 𝑦 =𝐶42
2 [sinh 𝑦 ]2− cosh 𝑦 −
𝐶12[sinh 𝑦 ]2
(written in dimensionless transverse coordinate 𝜉 = 𝑥/𝜆𝑐)
Analyzing “potential”
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𝑉 𝑦 =𝐶42
2 [sinh 𝑦 ]2− cosh 𝑦 −
𝐶12[sinh 𝑦 ]2
𝐶42=0
-1<𝐶1<0
Analyzing “potential”
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𝑉 𝑦 =𝐶42
2 [sinh 𝑦 ]2− cosh 𝑦 −
𝐶12[sinh 𝑦 ]2
𝐶42=0
𝐶1>0
Analyzing “potential”
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𝑉 𝑦 =𝐶42
2 [sinh 𝑦 ]2− cosh 𝑦 −
𝐶12[sinh 𝑦 ]2
𝐶42=0
-1>𝐶1
Analyzing “potential”
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𝑉 𝑦 =𝐶42
2 [sinh 𝑦 ]2− cosh 𝑦 −
𝐶12[sinh 𝑦 ]2
1≫ 𝐶42 >0
-1<𝐶1<0
Analyzing “potential”
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𝑉 𝑦 =𝐶42
2 [sinh 𝑦 ]2− cosh 𝑦 −
𝐶12[sinh 𝑦 ]2
𝐶1>0
1≫ 𝐶42 >0
Analyzing “potential”
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𝑉 𝑦 =𝐶42
2 [sinh 𝑦 ]2− cosh 𝑦 −
𝐶12[sinh 𝑦 ]2
-1>𝐶1
1≫ 𝐶42 >0
Analyzing “potential”
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-1>𝐶1
1≫ 𝐶42 >0𝐶4
2=0
Summary:
𝐶1>0
-1<𝐶1<0
-1>𝐶1
𝐶1>0
-1<𝐶1<0
Physical values
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𝑉 𝑦 = −cosh 𝑦 −𝐶12[sinh 𝑦 ]2
휀 𝑦 > −3
2cosh 𝑦 − 𝐶1[sinh 𝑦 ]2
𝑁𝑒 > 0
Not every point of our potential corresponds to a
physical value.
To find out, a meaningful (meaning, useful for us in
this particular problem) values, we have to
remember condition:
Electron density can not be negative.
Exact solutions
𝑎 =2𝜅sech(𝜅𝜉)
1 − 𝜅2sech2(𝜅𝜉)
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𝜅2 = 1 + 𝜆𝑐2𝐶1 𝜉 = 𝑥/𝜆𝑐
Thank you!
• T.Kurki-Suonio, P.J. Morrison, T.Tajima –
“Self-focusing of an optical beam in plasma”;
• Stockholm’s Royal Institute of Technology
– “Nonlinear Optics 5A5513 (2003)”;
• Wolfram’s Mathematica (plots);
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Sources:
