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What would happen if, beneath the event horizon, a photon was emitted outwards along the radius of the black hole? It's speed can't change to any observer in any reference frame, but it surely cannot escape the black hole either! Is there a flaw in my second assumption?

Poo2uhaha
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Light cones within the event horizon are tilted such that they are directed towards the singularity at $r=0$. Any observer would see the photon moving away from them at $c$ locally, but it'd still be moving towards the singularity rather than away from it.

Here's an spacetime illustration which might help: Finkelstein diagram in ingoing coordinates. from http://www.damtp.cam.ac.uk/user/tong/gr.html, Figure 44. All geodesics are 'ingoing' within the horizon.

Eletie
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  • Do I need an understanding of general relativity to get to grips with spacetime geodesics in this particular context (which I do not have)? – Poo2uhaha Dec 24 '20 at 18:28
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    @OliverL I'm not really too sure if you can understand if without GR. If you accept the diagram on faith (i.e. where the lightcones face), then you should understand the concept. Obviously if you want a deeper understanding of why the diagram looks this way GR might be necessary. Hope this helps! – Eletie Dec 24 '20 at 19:05