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For the Weyl scalars of all spacetimes, at any point, possess one special structure, the so called principal null directions. Consider a general null tetrad $\{ l_a,n_a,m_a,\overline{m}_a \}$, we would like to perform a linear transformation that preserves the vector $n_a$. The most general one is given by $l_a→l_a+\overline{b}m_a + b \overline{m}_a+|b|^2 n_a$ and $m_a→m_a+bn_a$, where $b$ is a complex parameter that defines the transformation. Now the Weyl scalar $\Psi_0$ will, under this change of basis, transform as

$$\Psi_0→\Psi_0+4b\Psi_1+6b^2\Psi_2+4b^3\Psi_3+b^4\Psi_4$$

Since the transformation is quartic in $b$ there are exactly four roots for the equation $\Psi_0=0$. Therefore, in any given point in the spacetime, there are four null directions $l_a$ such that $\Psi_0=0$. These are called principal null directions. A Petrov classification is given via:

  1. Petrov I - all roots are distinct
  2. Petrov II - two roots coincide
  3. Petrov III - three roots coincide
  4. Petrov D - two distinct double roots
  5. Petrov N - all roots coincide

as given for example in "The Mathematical Theory of Black Holes" by Chandrasekhar.

However in the book "Advanced General Relativity" by John Stewart the Petrov classification is given in terms of the spinor representation of the Weyl tensor as follows. The Weyl tensor in terms of spinors reads

$$C_{ABCDA'B'C'D'} = \Psi_{ABCD} \epsilon_{A'B'} \epsilon_{C'D'} + \overline{\Psi}_{A'B'C'D'} \epsilon_{AB} \epsilon_{CD}$$

where $\Psi_{ABCD}$ is totally symmetric. As it is symmetric, it can be shown that

$$\Psi_{ABCD} = \alpha_{(A} \beta_B \gamma_C \delta_{D)}$$

Now the spinor $\alpha_A$ defines a null vector by $\alpha_a \sim \alpha_A \overline{\alpha}_{A'}$, same with $\beta, \gamma, \delta$ and a Petrov classification scheme can be defined in terms of them and their "repeatedness". This is how it is presented in Stewart's book. These supposedly are the same as the principal null vectors as described above, and this is what you need to prove to show the two schemes are equivalent.

Just a bit more background - the first part above which was written in terms of null tetrads, can be written in terms of spinors...You have a spinor basis $(o,i)$ and

$$l_a \sim o_A \overline{o}_{A'} , \;\; n_a \sim i_A \overline{i}_{A'} , \;\; m_a \sim o_A \overline{i}_{A'} , \;\; \overline{m}_a \sim i_A \overline{o}_{A'}$$

and in this basis

\begin{align}\Psi_0 &= \Psi_{ABCD} o^A o^B o^C o^D\\ \Psi_1 &= \Psi_{ABCD} o^A o^B o^C i^D\\ \Psi_2 &= \Psi_{ABCD} o^A o^B i^C i^D\\ \Psi_3 &= \Psi_{ABCD} o^A i^B i^C i^D\\ \Psi_4 &= \Psi_{ABCD} i^A i^B i^C i^D\end{align}

Now the transformation given above for null tetrads is the same as doing:

$$(\hat{o} , \hat{i}) \mapsto (o + b i , i)$$

Obviously this will result in the same transform as above, namely:

$$\Psi_0→\Psi_0+4b\Psi_1+6b^2\Psi_2+4b^3\Psi_3+b^4\Psi_4$$

The principal null directions, as defined above in terms of null tetrads, are now equivalently defined in terms of spinors:

$$o_A + b_1 i_A , \;\; o_A + b_2 i_A , \;\; o_A + b_3 i_A, \;\; o_A + b_4 i_A$$

where $b_1,b_2,b_3,b_4$ are the roots of $\Psi_0 = 0$.

After all this background, my question boils down to this: to show the two schemes coincide we need to show that

$$\Psi_{ABCD} = \alpha_{(A} \beta_B \gamma_C \delta_{D)}$$

where

$$\alpha_A = o_A + b_1 i_A , \;\; \beta_A = o_A + b_2 i_A , \;\; \gamma_A = o_A + b_3 i_A, \;\; \delta_A = o_A + b_4 i_A$$

What does this reduce to showing?

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EDIT I can now include the other cases, not just when the roots are distinct.

Under the transformation $(o,i) \mapsto (o + b i , i)$, we have

$\hat{\Psi}_0 (b) = \Psi_0 + 4 b \Psi_1 + 6 b^2 \Psi_2 + 4 b^3 \psi_3 + b^4 \Psi_4 \\ \hat{\Psi}_1 (b) = \Psi_1 + 3 b \Psi_2 + 3 b^2 \Psi_3 + b^3 \psi_4 \\ \hat{\Psi}_2 (b) = \Psi_2 + 2 b \Psi_3 + \Psi_4 \\ \hat{\Psi}_3 (b) = \Psi_3 + b \Psi_4 \\ \hat{\Psi}_4 (b) = \Psi_4 . $

We are interested in the roots of $\hat{\Psi}_4 (b) = 0$ which is quartic in $b$ and so can be written,

$ \hat{\Psi}_0 (b) = \Psi_4 (b-b_1) (b-b_2) (b-b_3) (b-b_4) . $

Below we will often use this together with the obvious formula:

$ \hat{\Psi}_1 (b) = {1 \over 4} {d \over d b} \hat{\Psi}_0 (b) \\ \hat{\Psi}_2 (b) = {1 \over 3} {d \over d b} \hat{\Psi}_1 (b) \\ \hat{\Psi}_3 (b) = {1 \over 2} {d \over d b} \hat{\Psi}_2 (b) \\ \hat{\Psi}_4 (b) = {d \over d b} \hat{\Psi}_3 (b) = \Psi_4 . $

CASE (1) First we consider the case where the four roots $b_1, b_2, b_3, b_4$ are distinct. Write

$\rho_1^A := o^A + b_1 i^A \\ \rho_2^A := o^A + b_2 i^A \\ \rho_3^A := a^A + b_3 i^A \\ \rho_4^A := a^A + b_4 i^A$

Then $\hat{\Psi}_0 = 0$ implies the 4 equations for $\alpha , \beta , \gamma, \delta$:

$\alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_1^A \rho_1^B \rho_1^C \rho_1^D = 0 \\ \alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_2^A \rho_2^B \rho_2^C \rho_2^D = 0 \\ \alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_3^A \rho_3^B \rho_3^C \rho_3^D = 0 \\ \alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_4^A \rho_4^B \rho_4^C \rho_4^D = 0$

which reduce to the 4 equations:

$(\alpha_A \rho_1^A) (\beta_B \rho_1^B) (\gamma_C \rho_1^C) (\delta_D \rho_1^D) = 0 \quad Eq 1 \\ (\alpha_A \rho_2^A) (\beta_B \rho_2^B) (\gamma_C \rho_2^C) (\delta_D \rho_2^D) = 0 \quad Eq 2 \\ \\ (\alpha_A \rho_3^A) (\beta_B \rho_3^B) (\gamma_C \rho_3^C) (\delta_D \rho_3^D) = 0 \quad Eq 3 \\ (\alpha_A \rho_4^A) (\beta_B \rho_4^B) (\gamma_C \rho_4^C) (\delta_D \rho_4^D) = 0 \quad Eq 4 $

Now we use that for spinors, $\alpha_A \rho^A = 0$ if and only if $\alpha$ is proportional to $\rho$ (we write $\alpha_A = \lambda \rho_A$).

We are considering the case where all the roots $b_1,b_2,b_3,b_4$ are all different and as such that the spinors $\rho_1 , \rho_2 , \rho_3 , \rho_4$ are not proportional to each other. Then Eq 1 is zero if and only if one of at least one of the brackets vanish. Say the first bracket is one that vanishes, so we can say $\alpha_A = \lambda_1 \rho_{1 A} = \lambda_1 (o_A + b_1 i_A)$. The first bracket in Eq 2 cant then vanish because $\rho_1$ is not proportional to $\rho_2$, and so one of the other brackets must vanish. Say the second bracket is one that vanishes, and so $\beta_A = \lambda_2 \rho_{2A} = \lambda_2 (o_A + b_2 i_A)$. The first two brackets of Eq 3 cant vanish, so at least one of the other two vanish, say it is the 3rd bracket then $\gamma_A = \lambda_3 \rho_{3A} = \lambda_3 (o_A + b_3 i_A)$. The first 3 brackets of Eq 4 can't vanish and so it must be the last bracket that vanishes, and so $\delta_A = \lambda_4 \rho_{4A} = \lambda_4 (o_A + b_4 i_A)$. And So $\Psi_{ABCD} = \alpha_{(A} \beta_B \gamma_C \delta_{D)}$ where the spinors $\alpha_A , \beta_A , \gamma_A , \delta_A$ are all distinct and each representing a principal null direction.

CASE (2) We consider the case where just two roots coincide, say $b_1=b_2$. As $\rho_1 = \rho_2$ we have three independent equations from $\hat{\Psi}_0 = 0$:

$(\alpha_A \rho_1^A) (\beta_B \rho_1^B) (\gamma_C \rho_1^C) (\delta_D \rho_1^D) = 0 \quad Eq 5 \\ (\alpha_A \rho_3^A) (\beta_B \rho_3^B) (\gamma_C \rho_3^C) (\delta_D \rho_3^D) = 0 \quad Eq 6 \\ (\alpha_A \rho_4^A) (\beta_B \rho_4^B) (\gamma_C \rho_4^C) (\delta_D \rho_4^D) = 0 \quad Eq 7 $

Then Eq 5 is zero if and only if one of at least one of the brackets vanish. Say the first bracket is one that vanishes, so we can say $\alpha_A = \lambda_1 \rho_{1 A} = \lambda_1 (o_A + b_1 i_A)$. The first bracket in Eq 6 cant then vanish because $\rho_1$ is not proportional to $\rho_3$, and so one of the other brackets must vanish. Say the second bracket is one that vanishes, and so $\beta_A = \lambda_3 \rho_{3A} = \lambda_3 (o_A + b_3 i_A)$. The first two brackets of Eq 3 cant vanish, so at least one of the other two vanish, say it is the 3rd bracket then $\gamma_A = \lambda_4 \rho_{4A} = \lambda_4 (o_A + b_4 i_A)$.

It can easily be shown that with parameter $b = b_1 (= b_2)$ $\hat{\Psi}_0$ and $\hat{\Psi}_1$ will vanish. So we also have the equation

$\alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_1^A \rho_1^B \rho_1^C i^D = 0$

or

$\rho_{1(A} \rho_{3B} \rho_{4C} \delta_{D)} \; \rho_1^A \rho_1^B \rho_1^C i^D = 0$

which reduce to

$(\rho_{1A} i^A) (\rho_{3B} \rho_1^B) (\rho_{4C} \rho_1^C) (\delta_{D} \rho_1^D) = 0$

implying $\delta_A = \rho_{1A} = \lambda_1 (o_A + b_1 i_a)$. So now we have that $\Psi_{ABCD} = \alpha_{(A} \beta_B \gamma_C \delta_{D)}$ where the spinors $\alpha_A , \beta_A , \gamma_A , \delta_A$ each represent a principal null direction with two directions coinciding.

CASE (3) Three roots coincide and $b = b_1 (= b_2 = b_3)$. As $\rho_1 = \rho_2 = \rho_3$ we have two independent equations from $\hat{\Psi}_0 = 0$:

$(\alpha_A \rho_1^A) (\beta_B \rho_1^B) (\gamma_C \rho_1^C) (\delta_D \rho_1^D) = 0 \quad Eq 8 \\ (\alpha_A \rho_4^A) (\beta_B \rho_4^B) (\gamma_C \rho_4^C) (\delta_D \rho_4^D) = 0 \quad Eq 9 $

Then Eq 8 is zero if and only if one of at least one of the brackets vanish. Say the first bracket is one that vanishes, so we can say $\alpha_A = \lambda_1 \rho_{1 A} = \lambda_1 (o_A + b_1 i_A)$. The first bracket in Eq 9 cant then vanish because $\rho_1$ is not proportional to $\rho_4$, and so one of the other brackets must vanish. Say the second bracket is one that vanishes, and so $\beta_A = \lambda_4 \rho_{4A} = \lambda_4 (o_A + b_4 i_A)$.

It is easily shown that with parameter $b = b_1 (= b_2 = b_3)$ $\hat{\Psi}_0$, $\hat{\Psi}_1$ and $\hat{\Psi}_2$ will vanish simultaneously. So we also have the equations

$\alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_1^A \rho_1^B \rho_1^C i^D = 0 \quad Eq 10 \\ \alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_1^A \rho_1^B i^C i^D = 0 \quad Eq 11 $

Eq 10 is

$\rho_{1(A} \rho_{4B} \gamma_C \delta_{D)} \; \rho_1^A \rho_1^B \rho_1^C i^D = 0 $

which reduces to

$(\rho_{1A} i^A) (\rho_{4B} \rho_1^B) (\gamma_C \rho_1^C) (\delta_{D} \rho_1^D) = 0 $

This implies that $\gamma_A = \lambda_1 \rho_{1A} = \lambda_1 (o_A + b_1 i_A)$. Eq 11 then reads

$ \rho_{1(A} \rho_{4B} \rho_{1C} \delta_{D)} \; \rho_1^A \rho_1^B i^C i^D = 0 $

which reduces to

$ (\rho_{1A} i^A) (\rho_{4B} \rho_1^B) (\rho_{1C} i^C) (\delta_{D} \rho_1^D) = 0 $

which means that $\delta_{D} = \rho_{1D} = \lambda_1 (o_a + b_1 i_A)$. So now we have that $\Psi_{ABCD} = \alpha_{(A} \beta_B \gamma_C \delta_{D)}$ where the spinors $\alpha_A , \beta_A , \gamma_A , \delta_A$ each represent a principal null direction with three directions coinciding.

CASE(4) Two distinct double roots $b_1$ and $b_2$. As $\rho_1 = \rho_3$ and $\rho_2 = \rho_4$ we have two independent equations from $\hat{\Psi}_0 = 0$:

$(\alpha_A \rho_1^A) (\beta_B \rho_1^B) (\gamma_C \rho_1^C) (\delta_D \rho_1^D) = 0 \quad Eq 12 \\ (\alpha_A \rho_2^A) (\beta_B \rho_2^B) (\gamma_C \rho_2^C) (\delta_D \rho_2^D) = 0 \quad Eq 13 $

Then Eq 12 is zero if and only if one of at least one of the brackets vanish. Say the first bracket is one that vanishes, so we can say $\alpha_A = \lambda_1 \rho_{1 A} = \lambda_1 (o_A + b_1 i_A)$. The first bracket in Eq 13 cant then vanish because $\rho_1$ is not proportional to $\rho_4$, and so one of the other brackets must vanish. Say the second bracket is one that vanishes, and so $\beta_A = \lambda_2 \rho_{2A} = \lambda_2 (o_A + b_2 i_A)$.

It is easily shown that with parameter $b = b_1$ we have $\hat{\Psi}_1 = 0$

$\alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_1^A \rho_1^B \rho_1^C i^D = 0 $

or

$\rho_{1(A} \rho_{2B} \gamma_C \delta_{D)} \; \rho_1^A \rho_1^B \rho_1^C i^D = 0$

which reduces to

$(\rho_{1A} i^A) (\rho_{2B} \rho_1^B) (\gamma_C \rho_1^C) (\delta_D \rho_1^D) = 0$

So that at least one of the last two brackets vanish. Say the third bracket vanishes, then $\gamma_A = \lambda_1 \rho_{1A} = \lambda_1 (o_A + b_1 i_A)$.

It is easily shown that with parameter $b = b_2$ we have $\Psi_1 = 0$

$\alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_2^A \rho_2^B \rho_2^C i^D = 0 $

or

$\rho_{1(A} \rho_{2B} \rho_{1C} \delta_{D)} \; \rho_2^A \rho_2^B \rho_2^C i^D = 0$

which reduces to

$(\rho_{1A} \rho_2^A) (\rho_{2B} i^B) (\rho_{1C} \rho_2^C) (\delta_D \rho_2^D) = 0$

So that at least one of the last two brackets vanish. Say the third bracket vanishes, then $\delta_A = \lambda_2 \rho_{2A} = \lambda_2 (o_A + b_2 i_A)$. So now we have that $\Psi_{ABCD} = \alpha_{(A} \beta_B \gamma_C \delta_{D)}$ where the spinors $\alpha_A , \beta_A , \gamma_A , \delta_A$ each represent a principal null direction with two different pairs repeated.

CASE(5) All roots coincide and we have for $b=b_1$ that $\hat{\Psi}_0 = \hat{\Psi}_1 = \hat{\Psi}_2 = \hat{\Psi}_3 = 0$. We have the equations:

$\alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_1^A \rho_1^B \rho_1^C \rho_1^D = 0 \quad Eq 14 \\ \alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_1^A \rho_1^B \rho_1^C i^D = 0 \quad Eq 15 \\ \alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_1^A \rho_1^B i^C i^D = 0 \quad Eq 16 \\ \alpha_{(A} \beta_B \gamma_C \delta_{D)} \; \rho_1^A i^B i^C i^D = 0 \quad Eq 17$

Eq 14 reduces to

$(\alpha_A \rho_1^A) (\beta_B \rho_1^B) (\gamma_C \rho_1^C) (\delta_D \rho_1^D) = 0$

At least one of the brackets vanish, say the first. So that $\alpha_A = \lambda_1 \rho_{1A} = \lambda_1 (o_A + b_1 i_A)$. Eq 15 reduces to

$(\rho_{1A} i^A) (\beta_B \rho_1^B) (\gamma_C \rho_1^C) (\delta_D \rho_1^D) = 0$.

At least one the last three brackets vanish, say the second bracket vanishes. Then $\beta_A = \lambda_1 \rho_{1A} = \lambda_1 (o_a + b_1 i_A)$. Eq 16 reduces to

$(\rho_{1A} i^A) (\rho_{1B} i^B) (\gamma_C \rho_1^C) (\delta_D \rho_1^D) = 0$.

At least one the last two brackets vanish, say the third bracket vanishes. Then $\gamma_A = \lambda_1 \rho_{1A} = \lambda_1 (o_a + b_1 i_A)$. Eq 17 reduces to

$(\rho_{1A} i^A) (\rho_{1B} i^B) (\rho_{1C} i^C) (\delta_D \rho_1^D) = 0$.

The last bracket must vanish, therefore $\delta_A = \lambda_1 \rho_{1A} = \lambda_1 (o_a + b_1 i_A)$. So now we have that $\Psi_{ABCD} = \alpha_{(A} \beta_B \gamma_C \delta_{D)}$ where the spinors $\alpha_A , \beta_A , \gamma_A , \delta_A$ each represent a principal null direction with all four directions coinciding.

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