I've been trying to calculate proper time in EF coordinates: $$ds^2 = (1-\frac{2M}{r})d^2\upsilon -2d\upsilon dr - r^2(d^2\theta +sin^2\theta d^2\phi)$$ As it is on a timelike worldline: $$ds^2 = d\tau ^2$$ but that means that for r<2M we'll get a negative under a root??
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1$\begingroup$ With the sign convention you are using you will have negative $ds^2$ for spacelike lines, not for timelike worldlines. Why do you think otherwise? $\endgroup$– DaleCommented Sep 18 at 14:14
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$\begingroup$ cause $$d\tau^2 = (1-\frac{2M}{r})d^2\upsilon -2d\upsilon dr -r^2 (d^2 \theta + sin^2 \theta d^2 \phi)$$ which basically only adds negative factors for r<2M $\endgroup$– Agatha HarknessCommented Sep 18 at 14:31
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$\begingroup$ You are assuming that $dv$ and $dr$ have the same sign. Also it is $dv^2$, not $d^2v$. Same with $\theta$ and $\phi$ $\endgroup$– DaleCommented Sep 18 at 14:43
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$\begingroup$ i agree about dr but $$d\upsilon$$ needs to be positive if we're looking at the future, doesn't it? $\endgroup$– Agatha HarknessCommented Sep 18 at 14:45
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1$\begingroup$ $r<2M$ is within an event horizon, isn’t it? $\endgroup$– controlgroupCommented Sep 18 at 14:48
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With the metric in this form a timelike worldline will always have a positive $ds^2$. It is not correct that it becomes negative for $r<2M$.
In this metric future-directed worldlines all have positive $dv$. So at $r<2M$ that means that $dr<0$. A positive $dv$ and a negative $dr$ mean that $-2\ dv \ dr$ is positive so it can keep $ds^2$ positive even for $r<2M$. The restriction to $dr<0$ for $r<2M$ is expected since it is not possible for a timelike worldline to go outwards at or below the horizon.
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