- #1

andyl01

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Hello,

I'm studying General Relativity using Ray D'Inverno's book.

. I don't understand what the author writes in paragraph 14.3 ("Static solutions") where he demonstrates that for a static spacetime there are no cross-terms (dx)^0 (dx)^\alpha in (ds)^2 , where x^0 is a timelike coordinate e x^\alpha is a spacelike coordinate. He says that the "assumption that the solution is static means that (ds)^2 is invariant under a time reversal about any origin of time"... but a time reversal is a coordinate transformation and (ds)^2 is a scalar, so is automatically invariant under a time reversal (for it is a coordinate transformation), and not only specifically for a static metric.

Soon after the author demonstrates (Eq. 14.22) that a metric does not contain cross terms of the type indicated above and so he concludes that the metric is static. I don't understand how to prove this, is there someone who can give me a detailed demonstration of this? Thank you.

I'm studying General Relativity using Ray D'Inverno's book.

*[Moderator's note: link deleted due to possible copyright issues.]*. I don't understand what the author writes in paragraph 14.3 ("Static solutions") where he demonstrates that for a static spacetime there are no cross-terms (dx)^0 (dx)^\alpha in (ds)^2 , where x^0 is a timelike coordinate e x^\alpha is a spacelike coordinate. He says that the "assumption that the solution is static means that (ds)^2 is invariant under a time reversal about any origin of time"... but a time reversal is a coordinate transformation and (ds)^2 is a scalar, so is automatically invariant under a time reversal (for it is a coordinate transformation), and not only specifically for a static metric.

Soon after the author demonstrates (Eq. 14.22) that a metric does not contain cross terms of the type indicated above and so he concludes that the metric is static. I don't understand how to prove this, is there someone who can give me a detailed demonstration of this? Thank you.

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