Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the relation between white hole and the Schwarzschild solution commonly found in textbooks of physics and interpreted usually as black hole?

share|cite|improve this question
In Kruskal-Szekeres coordinates, white hole is the "past interior" (region iV). These coordinates cover the maximally extended Schwarzschild solution manifold – Trimok Jul 12 '13 at 8:50
related: – Ben Crowell Jul 12 '13 at 18:22

OT, but are you related to Adam Ondra, the rock climber? That guy is an animal.

In most areas of physics, we're accustomed to having global symmetries such as parity and time-reversal. These are interpreted as arising from symmetries of the fixed spacetime background, which is Minkowski space. In GR, we aren't guaranteed to have any such symmetries. In fact, for a generic spacetime there is not even any way to define global operators such as time-reversal or parity. (E.g., you can have spacetimes that aren't even time-orientable.)

However, GR locally has the full symmetry of the Poincare group, and we typically expect to find this embodied in some identifiable way in the solutions of the field equations. This may seem a little mysterious, because in Finkelstein's 1958 unidirectional membrane interpretation (Finkelstein 1958, Townsend 1997), the event horizon only allows matter to pass in one direction. This can be thought of as a kind of spontaneous symmetry breaking. If you take the Penrose diagram of the Schwarzschild spacetime and flip it upside-down, it seems pretty clear that this can be interpreted as a time-reversal. The time-reversed version of the membrane is one that only allows matter to emerge.

Of course in a vacuum spacetime, the whole notion of a direction of time is just an arbitrary convention anyway. But you can imagine test particles inside the spacetime, e.g., clouds of gas, which would have their own thermodynamic arrow of time. We can then compare the direction of this arrow of time with the direction in which the unidirectional membrane is oriented, and we really do get physically distinguishable black and white holes.

White holes are not expected to exist in our universe as real objects. This is because they can't form by astrophysical collapse. You can't have collapse if stuff is prohibited from going in!

You can also construct a maximally extended Schwarzschild spacetime (something that Finkelstein didn't understand at first -- there's a note added in proof at the end of the paper). This spacetime contains two regions that look like a black hole and a white hole, and two copies of Minkowski space. This is of theoretical interest only, because this spacetime can't form by gravitational collapse.

One thing that I had a hard time gaining intuition for was why we could only have event horizons that acted like unidirectional membranes. One thing that helped me here was to look on a Penrose diagram at the region of the black hole inside the event horizon. The singularity is spacelike, not timelike (see What is the definition of a timelike and spacelike singularity? ). When the singularity is, say, a past singularity for all observers, clearly we can't have stuff falling into it.

A white hole has positive mass and attracts other objects gravitationally. It is not the same thing as a negative-mass Schwarzschild metric (Glaser 2006). Any stuff that flies out of its event horizon decelerates as it moves away from the white hole.

Finkelstein, Finkelstein, Phys. Rev. 110, 965–967 (1958), "Past-Future Asymmetry of the Gravitational Field of a Point Particle," downloadable from his web page at ; also at



share|cite|improve this answer
"OT, but are you related to Adam Ondra, the rock climber? That guy is an animal." No, I am not, to my best knowledge. – Leos Ondra Jul 12 '13 at 18:16

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.