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Black holes and time invariance.

  1. Oct 13, 2013 #1
    Einstein's field equations are time invariant.
    So is it conceivable that a reverse black hole can exist i.e a "white hole"?

    Or would the second law of thermodynamics prevent such a thing?
  2. jcsd
  3. Oct 13, 2013 #2
    The white hole and the black hole are in fact one and the same. The white hole is the time reversal of a black hole but the black hole solution is symmetric under time reversal (assuming it is isolated and conveniently forgetting how the black hole formed to begin with). That means it reverses into itself and that a white hole is a black hole.
  4. Oct 13, 2013 #3


    Staff: Mentor

    This is true if by "time invariant" you mean "time symmetric"; I assume from the rest of your post that that's what you mean. ("Time invariant", taken literally, actually means something like "static", i.e., "nothing ever changes", which is clearly false as it stands: plenty of solutions to the field equations are not static.)

    However, it's important to distinguish the (true) claim that the field equations are time symmetric from the (false) claim that individual *solutions* of the equations must be time symmetric. What the time symmetry of the equations means is that solutions which are not time symmetric occur in pairs, with each member of the pair being the time reverse of the other.

    Mathematically, yes, but you have to be clear about which solutions of the field equations you are talking about. See my forthcoming post in response to dauto.

    Physically, yes, this is the standard belief: a white hole is a valid mathematical solution of the field equations (actually, there are two solutions to which the term "white hole" can be applied--see my forthcoming post), but it is not physically reasonable because it violates the second law, so actual white holes do not exist.
    Last edited: Oct 13, 2013
  5. Oct 13, 2013 #4


    Staff: Mentor

    No, they aren't. They are either two separate solutions of the EFE, or two separate regions of a particular solution of the EFE. See below.

    There is a pair of solutions to the EFE that meets this description: the "black hole" solution, in which an object collapses to form a black hole, and the "white hole" solution, in which a white hole expands to form an object. But these are two separate solutions, neither of which is time symmetric individually; each one is the time reverse of the other.

    Of the two, the black hole solution is considered physically reasonable (though highly idealized), whereas the white hole solution is not. The reason, as bcrelling suggested in the OP, is the second law of thermodynamics: an object collapsing into a black hole represents a large increase in entropy, therefore the time reverse of that, a white hole expanding to form an object, represents a large *decrease* in entropy.

    This applies to a *different* "black hole" solution, the maximally extended Schwarzschild spacetime, which is time symmetric and includes both a black hole region *and* a white hole region, as *separate* regions. This solution is not considered physically reasonable because there is no way for it to come into being in the first place.

    Which amounts to considering the single maximally extended solution instead of the separate black hole solution I described above.

    This is incorrect for both types of solutions described above.
  6. Oct 14, 2013 #5

    Ok, so the second law of thermodynamics would make the natural creation of a white hole unlikely.

    1) Would in be possible to create a white hole using an artificial method which outsources a greater entropy increase elsewhere?

    2) Assuming a white hole is created, would it require a constant influx of hawking radiation to be maintained?
  7. Oct 14, 2013 #6


    Staff: Mentor

    I don't know that this question has ever really been considered as you state it, but some of the speculations about "baby universes" being created within black holes are somewhat similar to this idea. I don't think we'll really be able to answer questions like this until we have a tested theory of quantum gravity, since the initial singularity inside a white hole, just like the final singularity inside a black hole, involves spacetime curvature going to infinity, which really means curvature large enough that quantum gravity effects dominate.

    *If* a white hole could be created, the obvious way to model the process would be as the time reverse of an evaporating black hole, which would entail this. However, it's by no means clear that that's the only possible model for a white hole. (In fact, it's not clear that the obvious model of an evaporating black hole is correct; there is a running debate among physicists about how to correctly model this process.) Once again, I don't think we'll really be able to answer this question until we have a tested theory of quantum gravity.
  8. Oct 14, 2013 #7
    I agree with everything PeterDonis said. That's why I said one has to conveniently forget how the B-hole formed to begin with. If you just assume it has been there for ever - Usually called a eternal-hole, Then the presence of the W-hole is unavoidable. There is no static solution for a B-hole without a W-hole. In the real world though what that means is that there are no static solutions at all and eternal holes (B-hole + W-hole) don't exist.
  9. Oct 14, 2013 #8
    After reading PeterDonis post a second time I realized I don't really agree with everything he said. I don't agree with the quote above. The white hole and the black hole are not really two separate regions of the spacetime. They are actually two separate sets of geodesics within the same region of the space time. All the geodesics leading to the hole behave as one would expect a B-hole to behave while all the geodesics leaving the hole behave as one would expect a B-hole to behave. The B-hole and the W-hole are actually intertwined with each other hence my statement that they are one and the same (which I stand by). The point of my previous post is that in the real world if you extend the W-hole geodesics backward in time you may never reach the horizon because it may reach a point in time before the horizon formed. Important to note that Hawking radiation exits the hole following these W-geodesics I'm talking about exactly as one would expect from a W-hole. So yes indeed, the W-hole exists and we call it a B-hole
  10. Oct 14, 2013 #9


    Staff: Mentor

    No, this is not correct. They are two separate regions of spacetime. The easiest way to see that is to look at a Kruskal diagram, such as the one on the Wikipedia page on Kruskal coordinates:


    The black hole is the region marked "II", and the white hole is the region marked "IV". They are not the same.

    This doesn't even make sense. Geodesics are individual curves, not regions of spacetime. A region of spacetime can be covered by sets of geodesics, but for any given region, there will be many sets of geodesics that cover it, so you can't equate regions of spacetime with sets of geodesics either.

    Also, as I said above, "the hole" is not one region, it's two. Furthermore, there are geodesics that pass through both regions: they leave the white hole (region IV in the diagram), travel through regions I or III, and enter the black hole (region II). So even to the extent it makes sense to compare the "behavior" of geodesics to the "behavior" of a region of spacetime, you can't equate the behavior of geodesics with the behavior of either the black hole or the white hole, since the two holes "behave" differently but the same geodesics can pass through both.

    I'm not sure what you mean by this, but I don't see anything in the math or the physics that corresponds to it.

    Backward in time to where? If you mean taking a geodesic in region I and extending it backward until you reach a portion within region IV, yes, you can do that, but if that's what you mean, then this...

    ...is false, because there is no point in time before the white hole horizon formed (the WH horizon is different from the BH horizon--that's obvious from the Kruskal diagram).

    If, OTOH, you mean taking geodesics in region I and extending them backward in time into the black hole region (region II), that's impossible; there are no such geodesics in the static, maximally extended spacetime.

    If you mean taking geodesics in a *different* spacetime, one containing an object that collapses to form a black hole, then yes, you can find geodesics that, if extended backward in time, will reach a time before the black hole formed. But there is no white hole region in this spacetime, so there are no such things as geodesics exiting the white hole.

    No, it doesn't. A spacetime with an evaporating black hole is *different* from the one in the Kruskal diagram; it's not static, and it has no white hole region so there are no "white hole geodesics" in this spacetime, just as there aren't in the classical spacetime where an object collapses to form a static (non-evaporating) black hole.

    Also, even if the black hole evaporates, that doesn't mean Hawking radiation "exits" the hole. Hawking radiation is formed just outside the event horizon, not within it. The event horizon is a global surface; it's the boundary of the region of spacetime from which light can't escape. Hawking radiation escapes, therefore it does not come from within the event horizon. (Another way of putting this is that an evaporating black hole's event horizon works differently from the horizon of a non-evaporating black hole, because the latter is static and the former is not.)
  11. Oct 14, 2013 #10
    You're correct. I was wrong. I was thinking of something else.
  12. Oct 21, 2013 #11
    I will try to push one more argument in defense of this claim.

    Imagine a black hole in a dead universe, filled only with heat (and with the single black hole). The heat has some temperature. The black hole absorbs the heat. The black hole emits the Hawking radiation of the same temperature. This is a black hole in a thermal equilibrium with the rest of the universe.

    This picture is completely time-symmetric. We have the black hole sucking heat and emitting the Hawking radiation (also heat), but if we reverse the time, we get exactly the same situation.
  13. Oct 21, 2013 #12


    Staff: Mentor

    This is true, but the spacetime you are describing here is *not* the maximally extended Schwarzschild spacetime; it's a different spacetime with a different structure. You could, I suppose, say that the term "white hole sucking heat and emitting radiation" is just as apt for this spacetime as "black hole sucking heat and emitting radiation", but that's a matter of terminology, not physics. The physics is that you are talking about something different than the situation that was being discussed in this thread.

    Also, in your description of the situation you left out a key item: how did the black hole form in the first place? Even if there is a mathematical solution to the laws of physics (which are here not just the Einstein Field Equation but the appropriate equations of quantum field theory governing the Hawking radiation) corresponding to an "eternal" black/white hole in thermal equilibrium with the rest of the universe, this mathematical solution is not physically reasonable because there is no way for it to come into being. A black hole in the real universe, even if it someday reached a state of thermal equilibrium with the rest of the universe, will have formed by gravitational collapse, so the spacetime that describes it, taken as a whole, will not be time symmetric.
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