Paper259Part33-5

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Part 3.3
Curvature in the Single Polarization ECE Vacuum
To be a Beltrami field, the vector 𝒖 must satisfy the following [11]
𝒖×∇×𝒖= 𝟎
(175)
This is equivalent to [12]
∇×𝒖 =π‘˜π’–
(176)
A Beltrami field of the first kind [12] is a potential field where
𝒖 = ∇πœ“
(177)
with
π‘˜=0.
(178)
A Beltrami field of the second kind [12] occurs when
π‘˜ = π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘
(179)
Meaning that
∇βˆ™π’–=0.
(180)
A Beltrami field of the third kind [12] occurs when π‘˜ a function of the coordinate variables and
equations (175) or (176) apply.
The ECE vacuum for a single state of polarization is given by
𝑩= ∇×𝑨−𝝎×𝑨 = 𝟎
𝑬
𝟐
(181)
πœ•π‘¨
= −∇πœ™ + πŽπœ™ = −πœ”0 𝑨 − πœ•π‘‘ = 0
(182)
For the sake of simplicity, the vacuum state label has been omitted from the variables.
For any scalar potential πœ™
∇ × ∇πœ™ = 0 .
(183)
Therefore the field given by ∇πœ™ is a Beltrami field of the first kind.
From (182) therefore, since πŽπœ™ = ∇πœ™ , it is a Beltrami field of the first kind, i.e.
∇ × πŽπœ™ = 0
(184)
Using the standard vector expansion, equation (184) can be written
∇ × πŽπœ™ = πœ™∇ × πŽ + ∇πœ™ × πŽ = 𝟎
(185)
But from equation (182)
∇πœ™ × πŽ = ∇πœ™ ×
∇πœ™
=0
πœ™
Thus equation (185) becomes
∇ × πŽπœ™ = πœ™∇ × πŽ = 𝟎
(186)
making 𝝎 a Beltrami field of the first kind for the single polarization vacuum, as shown earlier.
Since, the single polarization curvature is given by
𝑹𝑠𝑝𝑖𝑛 = ∇ × πŽ
(187)
We note that the single polarization vacuum requires from equation (186)
𝑹𝑠𝑝𝑖𝑛 = 0 .
(188)
Also for a single polarization, from equations (8) and (15) we have that
π‘Ήπ‘œπ‘Ÿπ‘ × π‘¨ = 𝟎
(189)
π‘Ήπ‘œπ‘Ÿπ‘ βˆ™ 𝑨 = 0
(190)
This leads to the conclusion that
π‘Ήπ‘œπ‘Ÿπ‘ = 0
(191)
From equation (182)
πœ•π‘¨
−πœ”0 𝑨 =
(192)
πœ•π‘‘
so that
1 πœ•π΄π‘–
πœ”0 = − 𝐴
𝑖 πœ•π‘‘
πœ•
= − πœ•π‘‘ πΏπ‘œπ‘”(𝐴𝑖 ) .
(193)
Using
π‘Ήπ‘œπ‘Ÿπ‘ = −∇πœ”π‘œ −
πœ•πŽ
πœ•π‘‘
=0
(194)
We see that
πœ•
𝝎 = ∇ ∫ πœ•π‘‘ πΏπ‘œπ‘”(𝐴𝑖 ) 𝑑𝑑 = ∇πΏπ‘œπ‘”(𝐴𝑖 ) .
Therefore
(195)
𝝎 × π‘¨ = ∇πΏπ‘œπ‘”(𝐴𝑖 ) × π‘¨
(196)
A typical term for this cross product can be written
1
𝐴𝑖
πœ•π΄
πœ•π΄
(𝐴𝑗 πœ•π‘˜ 𝑖 − π΄π‘˜ πœ•π‘˜ 𝑖 )
π‘˜
𝑗
which in [13] was shown to be zero for the single polarization vacuum.
Therefore
𝝎×𝑨=∇×𝑨=𝟎 .
(197)
Thus 𝑨 and by equation (192), πœ”0 𝑨 are Beltrami fields of the first kind.
In summary, we have for the single polarization vacuum, that 𝑹𝑠𝑝𝑖𝑛 and π‘Ήπ‘œπ‘Ÿπ‘ are zero and that
𝑨 , 𝝎, ∇πœ™, πœ”0 𝑨 , and πŽπœ™ are Beltrami fields of the first kind.
References
[11] http://mathworld.wolfram.com/BeltramiField.html
[12] Thahar Amari,et. al., “Computing Beltrami Fields”,
https://www.ljll.math.upmc.fr/~boulmezaoud/PDFS/ARTIC/Amari_Boulbe_Boulmezaoud.pdf
[13] Myron W. Evans, Horst Eckardt, Douglas W. Lindstrom, “Generally Covariant Unified
Field Theory”, Ch. 16,Vol. VII, Abramis Publishing, UK , 2011
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