# How does a black hole look like independently of coordinates?

Observing a black hole from far away, far it looks as if infalling matter accumulates on the horizon. If one falls in though, there is no horizon at all (or it always stays in front of you).

Which raises the question, what is the true, uncoordinated, spacetime manìfold? Or are we implicitly always coordinating?

• @planetmaker Haha! Yes, you always get wet. Well, tensors are coordinate independent, but how you show that in a coordinate free picture? Without getting wet... Jun 15, 2022 at 8:55
• @planetmaker In a fixed coordinate system you can use different units. Or adapt the coordinates to units. But you stay in the same frame. Now is there a frame from which we see the true hole? No. What's the true hole then? Jun 15, 2022 at 9:19
• I fail to follow your argument. I can describe a circle as cartesian $P(x,y): \{x^2 + y^2 = R^2\}$. Or shifted: $P(x,y): \{(x-x_0)^2 + (y-y_0)^2 = R^2\}$. Or I can describe it as polar $P(r,\phi) = \{r=R,\phi\in(0°;360°]\}$. (Or any other frame) It's different coordinate systems. It's the same circle. And I can choose whatever units (or even no units at all, if I norm it by whatever default radius $R \longrightarrow R/R_0$) Jun 15, 2022 at 11:56
• @ProfRob Yes, okay, but isn't the description dependent on the coordinates? It looks different in the coordinates used faraway. The hole has no interior as seen from there. But if you fall in it has an interior. Now how does it look in spacetime, Independent of coordinates? Jun 15, 2022 at 16:38
• @planetmaker Yes, of course, but a circle is a static spatial structure to which you look, in your example, in it's restframe. But has a hole a restframe? If you fall in it has a centre, if you're outside of it only a horizon. Jun 15, 2022 at 16:41