I mainly don't understand this fact: far away from the source, ##t## and ##t'## have the same meaning: they represent both the proper time given by the clock of an observer stationary with respect to the black hole and infinitely distant from it. So the observer can use a single clock to monitor...
I'm studying Eddington-Finkelstein coordinates for Schwarzschild metric. Adopting the coordinate set ##(t,r,\theta,\phi)##, the line element assumes the form:
$$
ds^2 = \left(1 - \frac{R_S}{r}\right)dt^2 - \left(1 - \frac{R_S}{r}\right)^{-1}dr^2 - r^2 [d\theta^2 + (\sin{\theta})^2d\phi^2],
$$...
Okay I start to understand the logic... but I still don't understand why the fact that a time reversal is discrete may cause ##(ds)^2## not to be in general invariant under such a transformation. Why "discreteness" implies that we can't regard it as a valid coordinate transformation?
D'Inverno introduces in sect. 14.1 the concept of stationary spacetime and static spacetime, without (at the beginning) mentioning Killing vector fields. Then he proves, for example for the concept of stationary spacetime, that the initial definition is equivalent for the metric to admit a...
But that proof does not start from the assumption that the metric is static... I try to summarize what I think D'Inverno says in the chapter up to sec. 14.3 (it's not easy since the topic is presented in a chaotic way). He defines a static metric as a stationary metric which admits a coordinate...
Hello,
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...