# To an observer inside a black hole, why is the rest of the universe outside the black hole not a white hole?

To an observer inside a black hole, why is the rest of the universe outside the black hole event horizon not a white hole?

It is a region the observer could never enter. It is a region from which things can only emerge. It is also in the past.

"Obviously" the rest of the universe in not a "hole", so what in GR distinguishes a white hole from the area outside a black hole? What is a hole?

Conversely, from inside a white hole, the white hole event horizon has similar features to a black hole. It is in the observer's possible future because white holes can emit. It is not in the observer's inevitable future, because white holes do not have to emit. But, should the observer cross this horizon, they can never return to the interior of the white hole.

I am sure there is a key difference, but what is that?

• FWIW, I personally think that the question is not at all trivial and should not be deleted (unless you wish so for other reasons) Mar 24 at 20:45

In the maximally extended Schwarzschild spacetime there are two singularities, one in the past and one in the future. These singularities are local features of the spacetime meaning that you can test if you are local to the singularity using only local information (local tidal forces go to infinity and other nastiness).

The horizons are not local features. You cannot tell from information available locally at the horizon if you are at the horizon. Specifically:

It is an area the observer could never enter. It is an area from which things can only emerge. It is also in the past.

Is local information that is true for any null surface that any observer passes anywhere.

To determine if a specific null surface is a horizon requires global information. The black hole horizon is the null surface inside of which all maximally extended timelike paths go to the future singularity. The white hole horizon is the null surface inside of which all maximally extended timelike paths come from the past singularity. The two horizons are unambiguously distinct, so there is in principle no possibility of confusing the two, but both require non-local information to identify.

• Thank you for the effort, I appreciate it and get it. And thanks for the edits. Mar 24 at 23:28
• I have no doubt you are correct! I am just thinking, and also waiting for someone other than me to vote up your answer. Mar 25 at 13:15
• @fundagain as I said in the answer the main indication is that tidal forces become infinite at the singularity.
– Dale
Mar 25 at 21:00
• It is still just the tidal forces. That is what makes any singularity a singularity. The difference is that at the WH singularity they are decreasing from infinity while at the BH singularity they are increasing to infinity
– Dale
Mar 25 at 21:07
• Got it. Thank you. Mar 25 at 21:08

Kruskal coordinates (in the context of extended black holes, or worm-holes) on a black hole geometry are defined, from the Schwarzschild coordinates $${\displaystyle (t,r,\theta ,\phi )}$$, by replacing $$t$$ and $$r$$ by a new timelike coordinate $${\displaystyle T}$$ and a new spacelike coordinate $${\displaystyle X}$$:

$$T=\left(\frac{r}{2 G M}-1\right)^{1 / 2} e^{r / 4 G M} \sinh \left(\frac{t}{4 G M}\right)$$ $$X=\left(\frac{r}{2 G M}-1\right)^{1 / 2} e^{r / 4 G M} \cosh \left(\frac{t}{4 G M}\right)$$

for the exterior region $${\displaystyle r>2GM}{\displaystyle r>2GM}$$ outside the event horizon and:

$$T=\left(1-\frac{r}{2 G M}\right)^{1 / 2} e^{r / 4 G M} \cosh \left(\frac{t}{4 G M}\right)$$

$$X=\left(1-\frac{r}{2 G M}\right)^{1 / 2} e^{r / 4 G M} \sinh \left(\frac{t}{4 G M}\right)$$

The picture above shows the Kruskal diagram, illustrated for $$2GM=1$$. The quadrants are the black hole interior (II), the white hole interior (IV) and the two exterior regions (I and III). The dotted 45° lines, which separate these four regions, are the event horizons. The darker hyperbolas which bound the top and bottom of the diagram are the physical singularities. The paler hyperbolas represent contours of the Schwarzschild $$r$$ coordinate, and the straight lines through the origin represent contours of the Schwarzschild $$t$$ coordinate.

The two diagrams below serve to understand how the two coördinate systems, $$(t,r,\theta ,\phi )$$, and $$(X,T)$$ are related. When, for example, a spherical lightwave (for which $$r$$ is a constant at any time $$t$$) is traveling towards the black hole, then you can see these waves as the concentric circles in the first picture (where $$(x,y)$$ represent $$(r,t)$$ and the angle refers to a Lorenz transformation, which is not important here).

In the picture above you can see how the concentric waves travel in the Kruskal diagram. Observe that there is also a wave traveling in a downward direction. They travel backward in time towards the other black hole. Or forward in time from the other black hole. I.e., the other black hole represents a white hole. It seems as if the black and white hole have an infinite extent, but this is due to the coördinates. The waves end up at a point, the center of the concentric circles.

Maybe this gives you an idea.

There are no known physical circumstances that can lead to the existence of white holes. They are purely hypothetical (due to the math). Though Smolin says that they actually exist on the other side of black holes. The other side gives rise to other universes Smolin thinks some kind of a Natural selection "procedure" is happening. Likewise, he thinks that our big bang is similar to a white hole spitting out our universe, and as such there was a white hole geometry before the big bang. I'm not sure I understand this. Can time run backward?

To answer your question, an observer inside a black hole will "see" things on the outside of the black hole (away from the black hole horizon) not as a white hole. unless you accept that the universe emerged from a white hole. If the black hole is extended with a white hole (to form a wormhole) then he would never be able to see the white hole as all inside the white hole will not be able to reach him. It's a bit confusing though.