Physics Asked by ivan44 on March 6, 2021
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
$$Psi_0→Psi_0+4bPsi_1+6b^2Psi_2+4b^3Psi_3+b^4Psi_4$$
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:
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^Dend{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:
$$Psi_0→Psi_0+4bPsi_1+6b^2Psi_2+4b^3Psi_3+b^4Psi_4$$
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?
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.
Answered by ivan44 on March 6, 2021
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP