For instance, how far would a photon travel once passing the event horizon until it meets the singularity.
What is the circumference of an orbit as a function of the distance to the singularity?
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1$\begingroup$ How do I ask a good question?: "Have you thoroughly searched for an answer before asking your question? Sharing your research helps everyone. Tell us what you found and why it didn’t meet your needs. This demonstrates that you’ve taken the time to try to help yourself, it saves us from reiterating obvious answers, and above all, it helps you get a more specific and relevant answer!" $\endgroup$– Alfred CentauriCommented Aug 17, 2019 at 13:34
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$\begingroup$ Your two questions seem contradictory. How can a photon have a orbit around the singularity, with a radius and a circumference, and also meet the singularity? $\endgroup$– G. SmithCommented Aug 17, 2019 at 16:35
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1$\begingroup$ All four black hole metrics (Schwarzschild, Reissner-Nordström, Kerr, and Kerr-Newman) describe the spacetime inside as well as outside the event horizon. $\endgroup$– G. SmithCommented Aug 18, 2019 at 22:02
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$\begingroup$ @G. Smith My apologies, I should have said “of an orbit” so as to not imply that Inwas referring to the same proton. Why is your second comment a comment and not an answer? $\endgroup$– DigcoalCommented Aug 20, 2019 at 18:16
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$\begingroup$ @G. Smith Got it. So, if the original question asked “What...” instead of asking about the existence of said equations, then you would have made that comment an answer? $\endgroup$– DigcoalCommented Aug 20, 2019 at 18:20
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2 Answers
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All four black hole metrics (Schwarzschild, Reissner-Nordström, Kerr, and Kerr-Newman) describe the spacetime inside as well as outside the event horizon.
There are no photon orbits inside the event horizon. All photon trajectories inside lead to the singularity, as do the trajectories of massive particles.
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$\begingroup$ So would circle be a better term than orbit? I just wanted to know how the circumferences of circles change relative to spacetime outside the Event Horizon. In other words, if you described the interior of black hole using standard Euclidean Space, like it was flat space, and compared it to the space described by one of the equations you mentioned, what would be the difference between the circumference a kilometer from the singularity in both cases? Flat space would be 3.14 km^2, but what would it actually be at what would have been a kilometer? $\endgroup$– DigcoalCommented Aug 20, 2019 at 18:28
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$\begingroup$ That is a totally different question, and asking different questions is not what comments are for. $\endgroup$– G. SmithCommented Aug 20, 2019 at 18:30
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$\begingroup$ I’m trying to figure out how to word my question sufficiently to express my what I’m interested in finding out. Is there a better forum to discuss how to word what I am trying to ask? $\endgroup$– DigcoalCommented Aug 20, 2019 at 18:35
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$\begingroup$ As soon as I get a good idea of what YOU are looking for in a question, I will delete all this, research the equations you mentioned some more, and ask again. Thank you for your patience. $\endgroup$– DigcoalCommented Aug 20, 2019 at 18:36
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$\begingroup$ The geometry inside is hard to describe since the “radial” coordinate becomes temporal and the “time” coordinate becomes spatial. The question will probably be considered ill-posed because the inside geometry is more complicated than you are envisioning. $\endgroup$– G. SmithCommented Aug 20, 2019 at 18:42
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Certainly there are equations that describe spacetimes inside event horizons. They are the same set of equations that describe spacetimes outside event horizons. The equations don't break down until you get to the singularity. For example, the (maximally extended) Schwarzschild solution describes a spherically symmetric black hole. It works both inside and outside the horizon. You may have to do a coordinate transformation to get it to work in all the regions though.
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$\begingroup$ That’s much closer to what I’m asking about. Thank you. $\endgroup$– DigcoalCommented Aug 20, 2019 at 18:29
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$\begingroup$ "You may have to do a coordinate transformation to get it to work at the horizon though." - This is a common misconception. Such a transformation would be necessarily singular and therefore mathematically forbidden at the horizon. $\endgroup$ Commented Aug 29, 2019 at 16:32
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$\begingroup$ "the Schwarzschild metric describes a spherically symmetric, static, black hole. It works both inside and outside the horizon" - Another common misconception. The Schwarzschild metric is not static inside the horizon and therefore cannot be a part of the static solution. The original Schwarzschild solution did not include the interior region, but had the radius starting at zero from the horizon. The interior region was added decades later by others and its validity remains controversial despite a popular fascination with singularities "there". $\endgroup$ Commented Aug 29, 2019 at 16:40
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1$\begingroup$ You're right on the static point. I forgot the interior solution is no longer static. So I should also replace the "Schwarzschild metric" with "the maximally extended Schwarzschild solution". :) $\endgroup$– enumarisCommented Aug 30, 2019 at 2:47
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$\begingroup$ It doesn't work. You cannot make a non-static metric static simply by changing coordinates. A coordinate transformation does not change the nature of the metric. Specifically, the Kruskal-Szekeres metric depends on $r$, which is timelike inside the horizon. So the maximally extended solution in the Kruskal-Szekeres coordinates is not static inside and thus is just as invalid there as the inside solution in the Schwarzschild coordinates. Physics doesn't depend of the choice of coordinates. $\endgroup$ Commented Sep 2, 2019 at 6:05