Let φ(τ ) = φ̄(α1 , α2 , α3 ), where
αj = j −1 tr τ j ,
j = 1, 2, 3.
(1)
We can use these three invariants instead of β1 , β2 , β3 since φ is isotropic.
The advantage is that the αj are explicit functions of τ , whereas the βA are
not so explicit (they can be expressed as roots of a cubic etc. but that is
a nightmare). We’ll return to the βA at the end. Also, we’ll ignore q for
simplicity.
So,
3
∂φ X ∂ φ̄ ∂αj
=
∂τ
∂αj ∂τ
j=1
=
3
X
∂ φ̄ j−1
τ
∂α
j
j=1
(2)
Then use the spectral form of τ along with
τ
j−1
=
3
X
βAj−1 nA ⊗ nA
(3)
A=1
to get
3
3
∂φ X X ∂ φ̄ j−1 =
β
nA ⊗ nA
∂τ A=1 j=1 ∂αj A
=
3
3
X
X
∂ φ̄ ∂αj nA ⊗ nA
∂α
∂β
j
A
j=1
A=1
3
X
∂ φ̂
nA ⊗ nA
=
∂βA
A=1
(4)
The last bit follows from
αj = j
−1
3
X
βAj
⇒
A=1
1
∂αj
= βAj−1
∂βA
(5)