Penrose spacetime diagram for the Schwarzchild solution

Consider the metric corresponding to the Schwarzchild solution. It represents a Non-rotating Black hole. When we want to understand the causal structure of the spacetime we find the null geodesic equation.

Outgoing radial null geodesic - The outgoing null geodesics are drawn in region r>2m and r<2m. For r>2m, the null geodesics go to infinity as we expect. For r<2m, it goes and hits the singularity. But these are outgoing null geodesics in the region r<2m seems to be coming from the horizon at time t=-(infinity). Are these outgoing null geodesics in both the regions somehow connected at t=-(infinity)?

Ingoing radial null geodesic- The ingoing null geodesics are drawn similarly in region r<2m and r>2m. In the region r<2m, the geodesics come from the horizon and hits the singularity. They start at t=+(infinity) and hits the singularity at some finite time which is less that infinity. Do these null geodesics travel back in time in this region?

You shouldn't give much physical meaning to the coordinates inside the event horizon. In particular, $r$ is the coordinate that "acts like" time, and $t$ acts like space.

Something that is important to understand is that since the Schwarzschild coordinates don't cover all of the spacetime and become singular at the horizon, rigorously speaking they don't define a manifold: they define two pieces of a manifold, which a priori are not connected. A proper mathematical definition of the black hole spacetime uses (for example) the Kruskal-Szekeres coordinates, which cover the whole manifold. These are the coordinates used to make a Penrose diagram.

To answer your questions: the two sets of outgoing geodesics are not connected. This makes sense, since they sort of go in opposite directions. The ones inside the horizon go towards the singularity, and those outside go off to infinity. The ingoing geodesics are connected, which again makes sense: you just have one geodesic, which goes through the horizon. But "traveling back in time" is not a meaningful phrase, since the $t$ coordinate doesn't necessarily mean "time". You're right that along the geodesic it goes to $+\infty$ and back, but it doesn't travel back in time.

You can see this in the Kruskal spacetime (ignore regions III and IV):

I've drawn outgoing geodesics in green, and ingoing in orange.

• What tool did you use to draw this diagram? – magma Oct 9 '17 at 8:19
• @magma I found it on Wikipedia, see the link in my answer. – Javier Oct 9 '17 at 12:33
• I have one more question. Are the region IV and II causally connected? Is there any way to go from one region to another? – Khushal Oct 9 '17 at 13:47
• @Khushal as seen in the Penrose diagram, you can go from IV to II but not the other way around. – Javier Oct 9 '17 at 13:58