Papapetrou transformation: Conditions to be satisfied to achieve transformation

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SUMMARY

The discussion focuses on the Papapetrou transformation as outlined in Chapter 6, Section 52 of Chandrasekhar's "The Mathematical Theory of Black Holes" (1983). The transformation aims to convert the line element into diagonal form with equal coefficients for the variables (dx2)² and (dx3)². The necessary and sufficient conditions for this transformation are derived, leading to equations involving the derivatives of the new coordinates ρ and z. The participants question the differences between their derived conditions and those presented by Chandrasekhar, specifically regarding the conditions for achieving the transformation.

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julian
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TL;DR
The Papapetrou transformation. Conditions to be satisfied to achieve requirements of transformation. My conditions don't match Chandrasekhar's conditions.
I'm looking at the Papapetrou transformation in Ch. 6, ##\S 52## of Chandrasekhar's book. He cf's Ch. 2, ##\S##11.I understand Ch. 2, ##\S##11. There he considers a coordinate transformation,

\begin{align*}
{x'}^1 = \phi (x^1,x^2) \qquad \text{and} \qquad {x'}^2 = \psi (x^1,x^2)
\end{align*}

which will reduce the contravariant form of the line element

\begin{align*}
ds^2 = g^{11} (dx_1)^2 + 2 g^{12} dx_1 dx_2 + g^{22} (dx_2)^2
\end{align*}

to diagonal form with equal coefficients for ##(dx_1)^2## and ##(dx_2)^2##. For a transformation to achieve this it is necessary and sufficient that

\begin{align*}
g^{'12} = g^{11} \phi_{,1} \psi_{,1} + 2 g^{12} (\phi_{,1} \psi_{,2} + \phi_{,2} \psi_{,1}) + g^{22} \phi_{,2} \psi_{,2} = 0
\end{align*}

\begin{align*}
g^{'11} - g^{'22} = g^{11} ({\phi_{,1}}^2 - {\psi_{,1}}^2) + 2 g^{12} (\phi_{,1} \phi_{,2} - \psi_{,1} \psi_{,2}) + g^{22} ({\phi_{,2}}^2 - {\psi_{,2}}^2) = 0
\end{align*}

I get all of this.In Ch. 6, ##\S##52, (b) The Papapetrou transformation, he is wanting to perform a coordinate transformation

\begin{align*}
(x^2,x^3) \rightarrow (\rho , z)
\end{align*}

such that

\begin{align*}
e^{2 \mu} [(dx_2)^2 + (dx_3)^2] \rightarrow f (\rho , z) [(d \rho)^2 + (dz)^2]
\end{align*}Regarding the possibility of making such a coordinate transformation, he cf's Ch. 2 ##\S##11. So I was thinking I should write

\begin{align*}
{x'}^2 = \rho (x^2,x^3) \qquad \text{and} \qquad {x'}^3 = z (x^2,x^3)
\end{align*}

where ##\rho## and ##z## are to be chosen so that the metric remains in diagonal form and with equal coefficients for ##(d \rho)^2## and ##(dz)^2##. For a transformation to achieve this it is necessary and sufficient that

\begin{align*}
g^{'23} = g^{22} \rho_{,2} z_{,2} + 2 g^{23} (\rho_{,2} z_{,3} + \rho_{,3} z_{,2}) + g^{33} \rho_{,3} z_{,3} = 0
\end{align*}

\begin{align*}
g^{'22} - g^{'33} = g^{22} ({\rho_{,2}}^2 - {z_{,2}}^2) + 2 g^{23} (\rho_{,2} \rho_{,3} - z_{,2} z_{,3}) + g^{33} ({\rho_{,3}}^2 - {z_{,3}}^2) = 0
\end{align*}

As ##g^{23} = 0## and ##g^{22} = g^{33}##, the first condition requires
\begin{align*}
\rho_{,2} z_{,2} + \rho_{,3} z_{,3} = 0
\end{align*}

As ##g^{23} = 0## and ##g^{22} = g^{33}##, the second condition requires

\begin{align*}
{\rho_{,2}}^2 - {z_{,2}}^2 = - {\rho_{,3}}^2 + {z_{,3}}^2
\end{align*}However, Chandrasekhar gets these conditions instead:

\begin{align*}
{\rho_{,2}}^2 + {z_{,2}}^2 & = {\rho_{,3}}^2 + {z_{,3}}^2
\nonumber \\
\rho_{,2} \rho_{,3} + z_{,2} z_{,3} & = 0
\end{align*}

How does Chandrasekhar arrive at these conditions?Are my conditions not necessary and sufficient conditions for the transformation to achieve the requirements I stated? Does Chandrasekhar have other requirements in mind? Chandrasekhar notes that his conditions are satisfied by ##\rho_{,2} = +z_{,3}## and ##\rho_{,3} = - z_{,2}##. I notice that my conditions are satisfied by these choices as well.
 
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julian said:
[...] Chandrasekhar's book. [...]
Which book? He's written quite a few.

Edit: Oh, I guess you mean "The Mathematical Theory of Black Holes" (1983).
Geez, I hate Chandrasekhar's conventions for chapter/section numbering.... (sigh)

Alas, I don't have time to give a detailed answer right now. I'll try later this week if no one else jumps in first. :oldfrown:
 
Last edited:

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