Curvature in the Single Polarization ECE Vacuum
Proof that for single polarization ECE fields that the only non-zero curvature solutionsd to the ECE
vacuum equations are Beltrami.
The ECE vacuum for a single state of polarization is given by
𝑩 = ∇×𝑨−𝝎×𝑨 = 𝟎
𝑬
𝟐
(1)
𝜕𝑨
= −∇𝜙 + 𝝎𝜙 = −𝜔0 𝑨 − 𝜕𝑡 = 0
(2)
For the sake of simplicity, the vacuum state label has been omitted from the variables.
Let us look first at non-Beltrami solutions.
The single polarization curvature is
𝑹𝑠𝑝𝑖𝑛 = ∇ × 𝝎
𝑹𝑜𝑟𝑏 = −∇𝜔𝑜 −
(3)
𝜕𝝎
𝜕𝑡
(4)
From (2)
𝝎=
∇𝜙
(5)
𝜙
Substituting this into (3)
𝑹𝑠𝑝𝑖𝑛 = ∇ ×
∇𝜙
(6)
𝜙
This can be written
𝑹𝑠𝑝𝑖𝑛 = ∇ × ∇𝐿𝑜𝑔(𝜙) = 0
(7)
The only possible non-zero solution for 𝑹spin is a Beltrami solution.
To see if this is true for 𝑹𝑜𝑟𝑏 , we take the dot product of (4) with 𝑨 to give
𝑹𝑜𝑟𝑏 ∙ 𝑨 = −∇𝜔𝑜 ∙ 𝑨 −
𝜕𝝎
𝜕𝑡
∙𝑨
(8)
Now
−∇𝜔𝑜 ∙ 𝑨 = −∇ ∙ (𝜔𝑜 𝑨) − 𝜔𝑜 ∇ ∙ 𝑨
(9)
Using (2)
−∇𝜔𝑜 ∙ 𝑨 =
𝜕∇∙𝑨
𝜕𝑡
− 𝜔𝑜 ∇ ∙ 𝑨
or
𝜕
𝜕𝑡
−∇𝜔𝑜 ∙ 𝑨 = (
− 𝜔𝑜 ) ∇ ∙ 𝑨
(10)
Further
𝜕𝝎
𝜕𝑡
∙𝑨=
𝜕(𝝎∙𝑨)
−
𝜕𝑡
𝜕𝑨
𝝎 ∙ 𝜕𝑡
(11)
which from (2) becomes
𝜕𝝎
𝜕𝑡
∙𝑨=
𝜕(𝝎∙𝑨)
−
𝜕𝑡
𝜕
𝜔𝑜 𝝎 ∙ 𝑨 = (𝜕𝑡 − 𝜔𝑜 ) 𝝎 ∙ 𝑨
(12)
Therefore
𝜕
𝑹𝑜𝑟𝑏 ∙ 𝑨 = − (𝜕𝑡 − 𝜔𝑜 ) (∇ ∙ 𝑨 + 𝝎 ∙ 𝑨)
(13)
From the tetrad postulate, this is the sum of the spatial terms along the diagonal for the 𝜞 connection, so
that because of antisymmetry of 𝜞
∇∙𝑨+𝝎∙𝑨=𝟎
(14)
Therefore
𝑹𝑜𝑟𝑏 ∙ 𝑨 = 𝟎
(15)
We can take the cross product of equation (4) with 𝑨
𝑹𝑜𝑟𝑏 × 𝑨 = −∇𝜔𝑜 × 𝑨 −
𝜕𝝎
𝜕𝑡
×𝑨
(16)
This can be rewritten as
𝑹𝑜𝑟𝑏 × 𝑨 = −∇𝜔𝑜 × 𝑨 −
𝜕𝝎×𝑨
𝜕𝑡
+𝝎×
𝜕𝑨
𝜕𝑡
(17)
Using equations (1) and (2), this is
𝑹𝑜𝑟𝑏 × 𝑨 = −∇ × 𝜔𝑜 𝑨 + 𝜔𝑜 ∇ × 𝑨 −
𝜕𝝎×𝑨
𝜕𝑡
− 𝜔𝑜 𝝎 × 𝑨 = −∇ ×
𝜕∇×𝑨
𝜕𝑨
+
𝜕𝑡
𝜕𝑡
=0
(18)
Examining equations (15) and (18) we see that if 𝑹𝑜𝑟𝑏 is both parallel and perpendicular , therefore 𝑹𝑜𝑟𝑏
must be zero if is not a Beltrami solution.
We will confirm now that 𝑹𝒐𝒓𝒃
and 𝑹𝒔𝒑𝒊𝒏 are Beltrami.
By equation (5) and the fact that ∇𝜙 is Beltrami (see note at end)
∇ × 𝑹𝑠𝑝𝑖𝑛 = ∇ × ∇ × 𝝎 = ∇ × ∇ ×
∇𝜙
𝜙
1
= ∇ × (𝜙 ∇ × ∇𝜙 +
∇𝜙
𝜙
1
× ∇𝜙) = 𝜅∇ × 𝜙 ∇ × ∇𝜙
(19)
This reduces to
∇ × 𝑹𝑠𝑝𝑖𝑛 = 𝜅∇ × 𝝎 = 𝜅𝑹𝑠𝑝𝑖𝑛
So that 𝑹𝑠𝑝𝑖𝑛 is Beltrami.
If we take the curl of equation (4)
(20)
−∇ × 𝑹𝑜𝑟𝑏 = ∇ × (−∇𝜔𝑜 −
𝜕𝝎
)
𝜕𝑡
= (−∇ × ∇𝜔𝑜 −
𝜕∇×𝝎
𝜕𝑡
)
(21)
But, we can write
𝜕∇×𝝎
𝜕𝑡
𝜕
= 𝜅 𝜕𝑡 𝝎
(22)
so that
−∇ × 𝑹𝑜𝑟𝑏 = −𝜅
𝜕
𝝎
𝜕𝑡
− ∇ × ∇𝜔𝑜
(23)
By examining equation (27) we see that if ∇𝜔𝑜 is Beltrami, then so is 𝑹𝒐𝒓𝒃.
We note that
(∇ × ∇𝜔𝑜 ) ∙ 𝑨 = ∇ ∙ (∇𝜔𝑜 × 𝑨) + ∇𝜔𝑜 ∙ ∇ × 𝑨
(24)
and that
∇ × 𝜔𝑜 𝑨 = (∇𝜔𝑜 × 𝑨) + 𝜔𝑜 ∇ × 𝑨
(25)
If we take the divergence of this we have
∇ ∙ (∇𝜔𝑜 × 𝑨) = −∇ ∙ (𝜔𝑜 ∇ × 𝑨)
(26)
so that equation (24) becomes noting that 𝑨 has a Beltrami solution,
(∇ × ∇𝜔𝑜 ) ∙ 𝑨 = −∇ ∙ (𝜔𝑜 ∇ × 𝑨) + ∇𝜔𝑜 ∙ ∇ × 𝑨 = −∇ ∙ (𝜔𝑜 𝜅𝑨) + ∇𝜔𝑜 ∙ 𝜅𝑨
(27)
From equation (2) and that
∇∙𝑨=0
We
𝜕𝑨
∇ ∙ (𝜔𝑜 𝑨) = −∇ ∙ 𝜕𝑡 = 0
(28)
So that equation (27) becomes
(∇ × ∇𝜔𝑜 ) ∙ 𝑨 = ∇𝜔𝑜 ∙ 𝜅𝑨
(29)
or
(∇ × ∇𝜔𝑜 − 𝜅∇𝜔𝑜 ) ∙ 𝑨 = 𝟎
This suggests that if 𝑨 is not perpendicular to ∇𝜔𝑜 then ∇𝜔𝑜 satisfies the Beltrami equation.
To see that this is the case, we expand
∇ × (𝜔𝑜 ∇ × 𝑨) = ∇𝜔𝑜 × ∇ × 𝑨 + 𝜔𝑜 ∇ × ∇ × 𝑨
Since A is Beltrami, this can be rewritten
(30)
∇ × (𝜅𝜔𝑜 𝑨) = ∇𝜔𝑜 × 𝜅𝑨 + 𝜔𝑜 ∇ × 𝜅𝑨 = ∇𝜔𝑜 × 𝜅𝑨 + 𝜔𝑜 𝜅 2 𝑨
Since 𝑨 is Beltrami, so is its time derivative, and then by equation (2), so is 𝜔𝑜 𝑨 thus
∇𝜔𝑜 × 𝜅𝑨 = 𝟎
This proving that
There ∇𝜔𝑜 is parallel to 𝑨 ∇𝜔𝑜 satisfies the Beltrami equation.
Conclusions:
We conclude therefore that both 𝑹𝒔𝒑𝒊𝒏 and 𝑹𝒐𝒓𝒃 satisfy the Beltrami equation, and that the only
non-zero curvature solution to the ECE vacuum equations are Beltrami.
Proof that 𝛁𝝓 is Beltrami:
From (2)
∇ × ∇𝜙 = ∇ × 𝝎𝜙
But using standard vector expansion,
∇ × 𝝎𝜙 = 𝜙∇ × 𝝎 + ∇𝜙 × 𝝎
Again, using equation (2)
∇𝜙 × 𝝎 = ∇𝜙 ×
∇𝜙
𝜙
=0
Thus
∇ × 𝝎𝜙 = 𝜙∇ × 𝝎
We have from previous papers that
∇ × 𝝎 = 𝜅𝝎
We thus have that
∇ × 𝝎𝜙 = 𝜅𝜙𝝎
and so by equation (2)
∇ × 𝝎𝜙 = 𝜅∇𝜙 .
Q.E.D.