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  Derivation of the Ernst Lagrangian

+ 2 like - 0 dislike

In 1968 [1], Ernst derived his famous equation for a single complex variable (which acts as a potential for the metric components) for the Einstein equations. The Lagrangian he writes down, just above equation (4) in [1], seems to be very non-trivial to derive. I had thought it was just the Einstein-Hilbert Lagrangian $\sqrt{-g} R$, but it is not since $R$ contains second order derivatives which do not appear in Ernst's Lagrangian. However, they are suspiciously similar (the terms involving $\omega$ are exactly the same).

Can someone please explain how the derivation goes? Is it the Einstein-Hilbert Lagrangian but modified by using the actual field equations themselves to remove some terms? Or do some other tricky manipulations take place?

Given Ernst's Lagrangian I can derive the rest of his equations quite nicely, following through the Euler-Lagrange approach. It is just the actual Lagrangian itself I cannot reproduce!


[1] Ernst, 1968, "New formulation of the axially symmetric gravitational field problem"

asked Dec 14, 2015 in Theoretical Physics by anonymous [ revision history ]
reshown Dec 15, 2015

I would say the Lagrangian was simply guessed/backwards-engineered.

I was afraid of that, Void -- that would mean there is no simple way to go about trying to generalise the results to other cases (some simple perfect fluids or AdS backgrounds or something). It is just the uncanny resemblance with $\sqrt{-g} R$ that made me think they are related; I tried integrating by parts out the second order terms -- didn't seem to help.

It could be possible to derive the Ernst Lagrangian by subtracting a possibly non-tensorial four-divergence from $R\sqrt{-g}$. The first thing to try would be to find out whether the difference between them is really a four-divergence.

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