Extension of Schwarzschild light cones white hole/black hole

1. Mar 18, 2015

binbagsss

The coordinates $u$ and $v$ are defined as $u=t+r*$, $v=t-r*$, where $r*=r+2M In(\frac{r}{2M}-1)$.
In $u,r,\theta,\phi$ coordinates the radially null geodesics are given by:
$\frac{du}{dr}=0$ for infalling,
$\frac{du}{dr}=2(1-\frac{2M}{r})^{-1}$ for outgoing.

In the $v,r,\theta,\phi$ coordinate the radially null geodesics are given by:
$\frac{du}{dr}=0$ for outgoing,
$\frac{du}{dr}=-2(1-\frac{2M}{r})^{-1}$ for infalling.

For $r<2M$ it is clear that the $(1-\frac{2M}{r})^{-1}$ term becomes negative.

Question:

So this results in the light cones on $u$ vs $r$ , first attachment, tilting over.
This explains why $r \leq 2M$ is a black-hole, within this region as time increases $r$ can only decrease.

My question is in a similar way trying to explain why, attachment two, demonstrates a white-hole.
So here the light-cones do the opposite, for $r>2M$ as time increases $r$ decreases and within $r \leq 2M$ as time increases $r$ increases.
So the definition of a white hole is that no signal from infinity can enter - how should this be obvious from th light cones?

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2. Mar 18, 2015

bluechic92

So from the light cones you understand in attachment one that the light cones are tilted towards the black hole and that is the direction of light as it approaches the event horizon/black hole.

For the second attachment: One way to look at it is that since the light cones tilt away it is impossible for light coming from infinity to approach a black hole. The path is in the opposite direction.

Edit: I'm not sure if the above made sense. I have actually done a problem like this ( and got it correct) and I know I have it somewhere... I'll give you a better explanation once I find it/ redo it. =) However, I think I had draw a picture similar to your attachment two+calculations and I simply said what I stated above. I can't remember the metric I used...

Last edited: Mar 18, 2015